CHAPTER 1 INTRODUCTION TO CELLS 2009 Garland Science Publishing 3rd Edition
Unity and Diversity of Cells 1-1 The smallest unit of life is a(n) (a) DNA molecule. (b) cell. (c) organelle. (d) virus. (e) protein. 1-2 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Cells can be very diverse: superficially, they come in various sizes, ranging from bacterial cells such as Lactobacillus, which is a few __________________ in length, to larger cells such as a frog’s egg, which has a diameter of about one __________________. Despite the diversity, cells resemble each other to an astonishing degree in their chemistry. For example, the same twenty __________________ are used to make proteins. Similarly, the genetic information of all cells is stored in their __________________. Although __________________ contain the same type of molecules as cells, their inability to reproduce themselves by their own efforts means that they are not considered living matter. amino acids DNA fatty acids plasma membranes
micrometer(s) millimeter(s) plants
viruses yeast meter
1-3 Which of the following statements about the basic chemistry of cells is TRUE? (a) All cells contain exactly the same proteins. (b) All proteins are constructed from the same 22 amino acids. (c) The genetic instructions in all cells are stored in DNA. (d) All organisms contain the same genes. (e) All of the above
1
1-4
Which of the following statements is TRUE? (a) Mutations are always harmful to an organism. (b) Mutation is essential for evolution to occur. (c) Mutation is the only source of genetic differences between parents and offspring in plants and animals. (d) Mutation always leads to evolution. (e) Mutations always lead to evolutionary “dead ends.”
Cells Under the Microscope 1-5
What unit of length would you generally use to give the measurements of a typical human cell? (a) Centimeters (b) Nanometers (c) Millimeters (d) Micrometers
1-6
A. B.
1-7
State whether you would use a phase-contrast light microscope, a fluorescence microscope, an electron microscope, or none of the above to do the following things: A. look at unstained living animal cells. B. look at ribosomes. C. look at an electron. D. look at a living cell expressing green fluorescent protein. E. do confocal microscopy.
What sets the limit on the size of structure that can be seen in a light microscope? Why are tissues usually cut into thin sections and stained before examination under a light microscope?
The Procaryotic Cell 1-8
Which of the following statements concerning procaryotes are TRUE? (a) They have no nucleus and hence no DNA. (b) They have no Golgi apparatus. (c) They can form simple multicellular organisms. (d) They include bacteria, yeast, and protozoans. (e) They are all able to live on inorganic energy sources.
The Eucaryotic Cell 1-9
The most reliable feature distinguishing a eucaryotic cell from a procaryotic cell is the (a) presence of a plasma membrane. (b) presence of a nucleus. (c) eucaryotic cell’s larger size. (d) presence of DNA. 2
1-10
Correct each of the following so that it becomes a TRUE statement about mitochondria. A. Mitochondria take in carbon dioxide and release oxygen. B. ADP is synthesized from ATP in mitochondria. C. Mitochondria are enclosed by two membranes, the outer one of which is highly folded. D. Mitochondria are thought to be derived from photosynthetic bacteria. E. Mitochondria are found in aerobic procaryotes.
1-11
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Eucaryotic cells are bigger and more elaborate than procaryotic cells. By definition, all eucaryotic cells have a __________________, usually the most prominent organelle in the eucaryotic cell. Another organelle found in essentially all eucaryotic cells is the __________________, which generates the chemical energy for the cell. On the other hand, a(n) __________________ can be found only in the cells of plants and algae, and performs photosynthesis. If we were to strip away the plasma membrane from a eucaryotic cell and remove all of its membrane-enclosed organelles, we would be left with the __________________, which contains many long, fine filaments of protein that are responsible for cell shape and structure and thereby form the cell’s __________________. chloroplast chromosome cytoskeleton
1-12
cytosol endoplasmic reticulum mitochondrion
nucleus ribosomes
On the schematic drawing of an animal cell in Figure Q1-12 match the labels given in the list below to the numbered label lines. A. Plasma membrane B. Nuclear envelope C. Cytosol D. Golgi apparatus E. Endoplasmic reticulum
Figure Q1-12 3
1-13
In an aerobic bacterium, where do you think most of the proteins responsible for cellular respiration are located?
1-14
Which one of the following statements is TRUE for mitochondria only, and not for both mitochondria and chloroplasts? (a) They are enclosed by a double membrane. (b) They are thought to be derived from procaryotes. (c) They cannot grow and reproduce when isolated from the cell. (d) They reproduce by dividing in two. (e) They are found in all aerobic eucaryotic cells.
1-15
You fertilize egg cells from a healthy plant with pollen (which contains the male germ cells) that has been treated with DNA-damaging agents. You find that some of the offspring have defective chloroplasts, and that this characteristic can be passed on to future generations. This surprises you at first because you happen to know that the male germ cell in the pollen grain contributes no chloroplasts to the fertilized egg cell and thus to the offspring. What can you deduce from these results?
1-16
Which of the following organelles is surrounded by two layers of membrane? (a) Endoplasmic reticulum (b) Nucleus (c) Lysosome (d) Peroxisome (e) Vacuole
1-17
In a eucaryotic cell specialized for secretion, which internal organelles would you expect to be particularly abundant?
1-18
From the list below, select the THREE cellular structures or compartments that are found in all cells. (a) Nucleus (b) Ribosomes (c) Cytosol (d) Mitochondria (e) Chloroplasts (f) Plasma membrane (g) Endoplasmic reticulum (h) Lysosomes
4
1-19
You have grown a culture of human cells and discover that it is heavily contaminated with bacteria. Which of the following procedures will most likely eliminate the bacteria without killing the human cells? (a) Treating the culture with a drug that causes microtubules to fall apart. (b) Diluting a small portion of the contaminated culture with 1000 times as much fresh nutrient broth and regrowing the cells. (c) Treating the culture with a drug that damages DNA. (d) Treating the culture with a drug that dissolves cell walls. (e) Treating the culture with a detergent that destroys cell membranes.
1-20
Circle the appropriate cell type in which the listed structure or molecule can be found. Note that the structure or molecule can be found in more than one type of cell. Structure or Molecule A. DNA
Cell Type plant
animal
B.
nucleus
animal
plant
bacterial
C.
plasma membrane
animal
plant
bacterial
D.
chloroplast
animal
plant
bacterial
E.
cell wall
animal
plant
bacterial
F.
lysosome
animal
plant
bacterial
G.
mitochondrion
animal
plant
bacterial
H.
Golgi apparatus
animal
plant
bacterial
bacterial
1-21
The protozoan Didinium feeds on other organisms by engulfing them. Why are bacteria, in general, unable to feed on other cells in this way?
1-22
The specialized cell types in the body of a multicellular organism are different from each other chiefly because (a) each cell type contains different genes. (b) different genes are switched on in different cell types. (c) some cell types have lost some of the genes that were present in the fertilized egg. (d) the fertilized egg divides by cell divisions that do not give rise to genetically identical cells. (e) the different cell types contain fundamentally different organelles.
1-23
List the following items in order of size from the smallest to the largest. A. Protein molecule B. Human fat cell C. Carbon atom D. Ribosome E. Yeast cell F. Mitochondrion
5
Model Organisms 1-24
Given what you know about the differences between procaryotic cells and eucaryotic cells, rate the following things as “good” or “bad” processes to study in the model organism, E. coli. A. formation of the endoplasmic reticulum B. DNA replication C. how the actin cytoskeleton contributes to cell shape D. how cells decode their genetic instructions to make proteins E. how mitochondria get distributed to cells during cell division
1-25
A. B.
In what way does a fungal cell structurally resemble a plant cell, rather than an animal cell? Which principal organelle does a plant cell contain that a fungal cell does not?
1-26
You wish to explore how mutations in specific genes affecting sugar metabolism might alter tooth development. Which organism is likely to provide the best model system for your studies and why? (a) Humans (b) Mice (c) E. coli (d) Arabidopsis (e) Gorillas
1-27
Circle the simplest model organism best used to study the following processes: Process Model Organism A.
programmed cell death
E. coli
yeast
C. elegans
B.
chloroplast function
C. elegans
Arabidopsis
Drosophila
C.
immunology
mouse
yeast
Arabidopsis
D.
development of a multicellular tissue
Drosophila
E. coli
yeast
1-28
When the genomes of distantly related organisms, such as a fly and a mouse, are compared, they are found to contain some genes that encode proteins with almost identical amino acid sequences. Explain how this happens.
1-29
Genes that have homologues in a variety of species have been discovered through the analysis of genome sequences. In fact, it is not uncommon for a homologous gene to encode a protein that looks similar in amino acid sequence in organisms as diverse as budding yeast, archeons, plants, and humans. Even more remarkably, many of these proteins can functionally substitute for their homologues in other organisms. Explain what it is about the origins of cells that makes it possible for proteins expressed by homologous genes to be functionally interchangeable in different organisms.
6
1-30
Your friend has just returned from a deep sea mission and claims to have found a new single-celled life form. He believes this new life form may not have descended from the common ancestor that all types of life on Earth share. However, he’s never taken Cell Biology, so he asks you determine whether his claim is true. In order to verify or dispute your friend’s claim, you realize that you must first make a list of characteristics common to all procaryotes and eucaryotes, so that you can check whether this new life form is similar or different from all other types of life on Earth. Name two basic characteristics that you could check to distinguish all procaryotes and eucaryotes from newly derived life forms.
How We Know: Life’s Common Mechanisms 1-31
One effective strategy for investigating how a particular cellular process works is to identify the gene required for the process to be carried out normally. A common technique used to identify these genes is to isolate organisms that are defective in particular cellular functions by randomly inducing mutations in individual genes. When a gene is mutant, the protein encoded by this gene no longer functions, and the organism shows a defect in the cellular function of interest; these organisms are considered mutants. Thus, scientists look for mutant organisms whose cells cannot carry out the process of interest and then determine the identity of the gene whose function has been altered. Sometimes scientists are interested in studying processes that are essential to the cell (required for the cell to live). Explain what temperature-sensitive mutants are and why they are helpful for the study of essential processes, especially in single-celled organisms such as yeast and bacteria.
7
Answers 1-1
(b)
1-2
Cells can be very diverse: superficially, they come in various sizes, ranging from bacterial cells such as Lactobacillus, which is a few micrometers in length, to larger cells such as a frog’s egg, which has a diameter of about one millimeter. Despite the diversity, cells resemble each other to an astonishing degree in their chemistry. For example, the same twenty amino acids are used to make proteins. Similarly, the genetic information of all cells is stored in their DNA. Although viruses contain the same type of molecules as cells, their inability to reproduce themselves by their own efforts means that they are not considered living matter.
1-3
(c)
1-4
(b)
1-5
(d)
1-6
A. B.
The wavelength of visible light. Most tissues are not transparent enough to be examined directly in a light microscope. Transparency is increased by slicing them into thin sections, and staining shows the different cellular structures in contrasting colors.
1-7
A. B. C. D. E.
phase-contrast light microscope electron microscope none of the above fluorescence microscope fluorescence microscope
1-8
(b) and (c)
1-9
(b)
1-10
A. B. C. D. E.
Mitochondria take in oxygen and release carbon dioxide. ATP is synthesized from ADP in mitochondria. Mitochondria are enclosed by two membranes, the inner one of which is highly folded. Mitochondria are thought to be derived from aerobic bacteria. Mitochondria are found in aerobic eucaryotes.
8
1-11
Eucaryotic cells are bigger and more elaborate than procaryotic cells. By definition, all eucaryotic cells have a nucleus, usually the most prominent organelle in the eucaryotic cell. Another organelle found in essentially all eucaryotic cells is the mitochondrion, which generates the chemical energy for the cell. On the other hand, a(n) chloroplast can be found only in the cells of plants and algae, and performs photosynthesis. If we were to strip away the plasma membrane from a eucaryotic cell and remove all of its membrane-enclosed organelles, we would be left with the cytosol, which contains many long, fine filaments of protein that are responsible for cell shape and structure and thereby form the cell’s cytoskeleton.
1-12
A. B. C. D. E.
1-13
In the plasma membrane. According to the theory of mitochondrial origin outlined in this chapter, the plasma membrane of the engulfed bacterium would become the inner mitochondrial membrane, where most of the proteins involved in cellular respiration are located.
1-14
(e)
1-15
Your results show that not all of the information required for making a chloroplast is encoded in the chloroplast’s own DNA; some, at least, must be encoded in the DNA carried in the nucleus. The reasoning is as follows. Genetic information is only carried in DNA, thus the defect in the chloroplasts must be due to a mutation in DNA. But all of the chloroplasts in the offspring (and thus all of the chloroplast DNA) must derive from those in the female egg cell, since chloroplasts only arise from other chloroplasts. Hence, all of the chloroplasts contain undamaged DNA from the female parent’s chloroplasts. In all the cells of the offspring, however, half of the nuclear DNA will have come from the male germ cell nucleus, which combined with the female egg nucleus at fertilization. Since this DNA has been treated with DNA-damaging agents, it must be the source of the heritable chloroplast defect. Thus, some of the information required for making a chloroplast is encoded by the nuclear DNA.
1-16
(b)
1-17
the endoplasmic reticulum and the Golgi apparatus
1-18
(b), (c), and (f)
1-19
(d)
Plasma membrane—3 Nuclear envelope—5 Cytosol—1; Golgi apparatus—2 Endoplasmic reticulum—4
Bacteria have cell walls, whereas mammalian cells do not.
9
1-20
A. B. C. D. E. F. G. H.
animal animal animal
animal animal animal
plant plant plant plant plant plant plant plant
bacterial bacterial bacterial bacterial
1-21
Didinium engulfs prey by changing its shape, and for this it uses its cytoskeleton. Bacteria have no cytoskeleton, and cannot easily change their shape because they are generally surrounded by a tough cell wall.
1-22
(b)
1-23
1—C, 2—A, 3—D, 4—F, 5—E, 6—B
1-24
A. B. C. D. E.
bad good bad good bad
1-25
A. B.
Like plant cells, fungal cells have cell walls. Chloroplasts
1-26
(b)
Mice are likely to provide the best model system. Mice have teeth and have long been used as a model organism. Mice reproduce relatively rapidly and the extensive scientific community that works on mice have developed techniques to facilitate genetic manipulations. Humans are not a model system. E. coli (a bacterium) and Arabadopsis (a plant) do not have teeth. Gorillas, although they have teeth, would not be a good model organism for many reasons. First, there is not an extensive scientific community working on the molecular and biochemical mechanisms of cellular behaviors in gorillas. Second, gorillas are expensive to house and, thus, perform experiments on. Third, gorillas do not reproduce rapidly, a characteristic desirable in model organisms. Finally, techniques for facile genetic manipulations on gorillas have not been extensively developed.
1-27
A. B. C. D.
C. elegans Arabidopsis mouse Drosophila
10
1-28
All living organisms are descended from a common ancestor. This means that their individual genes are also descended from common ancestral genes. Genes in different species that trace their descent back to a common ancestral gene in this way (that is, homologous genes) become different from one another through mutation and natural selection. However, the protein products of many genes are highly optimized for specific functions, involving precisely adjusted interactions of the protein molecule with other molecules in the cell. Almost any mutation altering the amino acid sequence of such a protein will be harmful and will be eliminated by natural selection. As a result, the amino acid sequence of the protein may remain almost unchanged over long periods of evolutionary time.
1-29
All living beings on Earth (and thus, all cells) are thought to be derived from a common ancestor. Solutions to many of the essential challenges that face a cell (such as the synthesis of proteins, lipids, and DNA) appear to have been achieved in this ancient common ancestor. The ancestral cell therefore possessed sets of proteins to carry out these essential functions. Many of the essential challenges facing modern-day cells are the same as those facing the ancestral cell, and the ancient solutions are often still effective. Thus, it is not uncommon for organisms to use proteins and biochemical pathways inherited from their ancestors. While these proteins often show some speciesspecific diversification, they still retain the basic biochemical characteristics of the ancestral protein. For example, homologous proteins often retain their ability to interact with a specific protein target, even in diverse cell types. Because the basic biochemical characteristics are retained, homologous proteins are capable of functionally substituting for one another.
1-30
Any two reasonable answers are OK. For example: 1. 2. 3.
The genomic information is encoded in nucleic acids. A particular set of twenty amino acids is used to make protein. Phospholipids are used to create cell membranes.
Nucleic acids, proteins made of amino acids, and phospholipids are all complex molecules produced and utilized by all known living cells on Earth. These compounds are not easily created in the absence of life. If one were to discover a new life form that did not contain these compounds, which are central to life as we know it, it would be likely that this new life form comes from an ancestor that used very different strategies to survive.
11
1-31
Temperature-sensitive mutants are organisms that contain a genetic mutation to make them sensitive to temperature. A temperature-sensitive mutant usually has a mutation in a gene that results in the production of a protein that does not function properly at a certain temperature (the restrictive temperature). At the permissive temperature, the mutant cells can live and reproduce normally. However, at the restrictive temperature, the cells will display the fatal defect. If a scientist is interested in an essential process, the scientist may set out to isolate mutant organisms defective in that essential process. In order to study an organism, it is important to be able to propagate it. However, a single-celled organism whose cells are defective in an essential process will die (and be unable to be propagated). Temperaturesensitive mutations permit the organisms to be propagated at the permissive temperature (where the proteins function normally) and allow the scientist to study the consequences of a lack of the essential gene function at the restrictive temperature (where the protein is defective).
12
CHAPTER 2 CHEMICAL COMPONENTS OF CELLS 2009 Garland Science Publishing 3rd Edition
Chemical Bonds 2-1 If the isotope 32S has 16 protons and 16 neutrons, how many protons and how many neutrons will the isotope 35S has? 2-2 A. B. C.
If 0.5 mole of glucose weighs 90 g, what is the molecular weight of glucose? What is the concentration, in grams per liter (g/l), of a 0.25 M solution of glucose? How many molecules are there in 1 mole of glucose?
2-3 Which of the following elements is LEAST abundant in living organisms? (a) Sulfur (b) Carbon (c) Oxygen (d) Nitrogen (e) Hydrogen 2-4 Your friend learns about Avogadro’s number and thinks it is so huge that there may not even be a mole of living cells on Earth. You have recently heard that there are about 50 trillion (50 × 1012) human cells in each adult human body, so you bet your friend $5 that there is more than a mole of cells on Earth. Once you learn that each human contains more bacterial cells (in the digestive system) than human cells, you are sure that you have won the bet. The human population is now more than 6 billion (6 × 106). What calculation can you show your friend to convince him you are right? 2-5 Atoms form covalent bonds with each other by (a) sharing protons. (b) sharing electrons. (c) transferring electrons from one atom to the other. (d) sharing neutrons. (e) attraction of positive and negative charges. 2-6 A carbon atom contains six protons and six neutrons. A. What are its atomic number and atomic weight? B. How many electrons does it have? C. What is its valence? How does this affect carbon’s chemical behavior? D. Carbon with an atomic weight of 14 is radioactive. How does it differ in structure from nonradioactive carbon? How does this difference affect its chemical behavior?
13
2-7
An ionic bond between two atoms is formed as a result of the (a) sharing of electrons. (b) loss of a neutron from one atom. (c) loss of electrons from both atoms. (d) loss of a proton from one atom. (e) transfer of electrons from one atom to the other.
2-8
Which of the following pairs of elements are likely to form ionic bonds? Use Figure Q2-8 if necessary.
Figure Q2-8 (a) (b) (c) (d) (e)
Hydrogen and hydrogen Magnesium and chlorine Carbon and oxygen Sulfur and hydrogen Carbon and chlorine
14
2-9
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Whereas ionic bonds form a(n) __________________, covalent bonds between atoms form a(n) __________________. These covalent bonds have a characteristic bond __________________ and become stronger and more rigid when two electrons are shared in a(n) __________________. Equal sharing of electrons yields a(n) __________________ covalent bond. If one atom participating in the bond has a stronger affinity for the electron, this produces a partial negative charge on one atom and a partial positive charge on the other. These __________________ covalent bonds should not be confused with the weaker __________________ bonds that are critical for the three-dimensional structure of biological molecules and for interactions between these molecules. charge covalent double bond ionic
length molecule noncovalent nonpolar
polar salt single bond weight
2-10
A. B.
In what scientific units is the strength of a chemical bond usually expressed? If 0.5 kilocalories of energy is required to break 6 × 1023 bonds of a particular type, what is the strength of this bond?
2-11
Approximately how many hydrogen bonds does it take to give an aggregate binding strength nearly equal to a single covalent bond? (a) 1 (b) 3 (c) 50 (d) 500 (e) 5000
2-12
After looking at Figure Q2-8 above, which of the following pairs of atoms do you expect to be able to form double bonds with each other? (a) Mg and Ca (b) C and Cl (c) S and O (d) C and H (e) He and O
15
2-13
A. B. C.
Sketch three different ways three water molecules could be held together by hydrogen bonding. On a sketch of a single water molecule, indicate the distribution of positive and negative charge (using the symbols δ + and δ –). How many hydrogen bonds can a hydrogen atom in a water molecule form? How many hydrogen bonds can the oxygen atom in a water molecule form?
2-14
Which of the following statements about hydrogen bonds are TRUE? (a) They are weak covalent bonds that are easily disrupted by heat. (b) They are weak bonds formed between hydrocarbons in water. (c) They are weak bonds formed between nonpolar groups. (d) They are weak bonds only formed in the presence of water. (e) They are weak bonds involved in maintaining the conformation of macromolecules.
2-15
Based on what you know about the properties of water, which of the following statements about methanol (CH3OH) are TRUE? (a) Methanol molecules form more hydrogen bonds than water molecules do. (b) The boiling point of methanol is higher than that of water. (c) Salts such as NaCl are less soluble in methanol than in water. (d) Methanol is a more cohesive liquid than water. (e) Methanol has a higher surface tension than water.
2-16
A. B. C.
What is the pH of pure water? What concentration of hydronium ions does a solution of pH 8 contain? Complete the following reaction: CH3COOH + H2O ↔
D.
Will the reaction in C occur more readily (be driven to the right) if the pH of the solution is high?
16
2-17
The amino acid histidine is often found in enzymes. Depending on the pH of its environment, sometimes histidine is neutral and other times it acquires a proton to become positively charged. Consider an enzyme with a histidine side chain that is known to play an important role in the function of the enzyme. It is not clear whether this histidine is required in its protonated or unprotonated state. To answer this question you measure enzyme activity over a range of pH, with the results shown in Figure Q2-17. Which form of histidine is necessary for the active enzyme?
Figure Q2-17
Molecules in Cells 2-18
Match the chemical groups shown in the first list with their names selected from the second list. List 1 A. –OH B. –C = O C. –COOH D. –CH3 E. –NH2
List 2 1. Amino 2. Aldehyde 3. Phosphate 4. Carboxyl 5. Carbonyl (ketone) 6. Methyl 7. Amido 8. Ester 9. Hydroxyl
17
2-19
Match the macromolecules shown in the first list with their small molecule building blocks from the second list. List 1 A. Polysaccharides B. DNA C. RNA D. Proteins E. Lipids
List 2 1. Amino acids 2. Deoxyribonucleotides 3. Aldehydes 4. Pyrophosphates 5. Ribonucleotides 6. Fatty acids 7. Sugars 8. Steroids
2-20
Which of the following are examples of isomers? 14C and 12C (a) (b) Alanine and glycine (c) Adenine and guanine (d) Glycogen and cellulose (e) Glucose and galactose
2-21
Which of the following is FALSE of condensation reactions? (a) Produce many biological polymers from monomers. (b) Consume H2O molecules. (c) Aid in storage of energy reserves. (d) Are the opposite of hydrolysis reactions. (e) Are usually catalyzed in cells by enzymes.
2-22
A. B. C.
How many carbon atoms does the molecule represented in Figure Q2-22 have? How many hydrogen atoms? What type of molecule is it?
Figure Q2-22
18
2-23
On the phospholipid molecule in Figure Q2-23 label each numbered line with a correct term selected from the list below. A. B. C. D. E. F. G. H. I. J.
Phosphate Nonpolar head group Glycerol Polar head group Saturated fatty acid Acetic acid Sugar Hydrophobic region Hydrophilic region Nonsaturated fatty acid
Figure Q2-23 2-24
Phospholipids can form bilayer membranes because they are (a) hydrophobic. (b) lipids. (c) amphipathic. (d) hydrophilic. (e) amphoteric.
2-25
A.
Write out the sequence of amino acids in the following peptide using the full names of the amino acids. Pro–Val–Thr–Gly–Lys–Cys–Glu
B. C.
Write the same sequence using the single letter code for amino acids. According to the conventional way of writing the sequence of a peptide or a protein, which is the C-terminal amino acid and which is the N-terminal amino acid in the above peptide? 19
2-26
Which of the following statements about amino acids is TRUE? (a) Twenty-two amino acids are commonly found in proteins. (b) Most of the amino acids used in protein biosynthesis have charged side chains. (c) Amino acids are often linked together to form branched polymers. (d) D- and L-amino acids are found in proteins. (e) All amino acids contain an NH2 and a COOH group.
2-27
Proteins can be modified by a reaction with acetate that results in the addition of an acetyl group to lysine side chains as shown in Figure Q2-27. The bond shown in the box in the acetylated lysine side chain is most like a(n)
Figure Q2-27 (a) (b) (c) (d) (f)
ester. peptide bond. phosphoanhydride bond. hydrogen bond. phosphoester bond.
2-28
DNA differs from RNA in (a) the number of different bases used. (b) the number of phosphates between the sugars in the sugar-phosphate backbone. (c) the kind of sugar found in the sugar-phosphate backbone. (d) one of the purines used. (e) the chemical polarity of the polynucleotide chain.
2-29
Which of the following statements is FALSE about ATP? (a) Contains high energy phosphoanhydride bonds. (b) Is sometimes called the “universal currency” in the energy economy of cells. (c) Can be incorporated into DNA. (d) Can be hydrolyzed to release energy to power hundreds of reactions in cells. (e) Comprises a sugar, phosphate groups, and a nitrogenous base.
20
2-30
Which of the following are likely to be disrupted by high concentrations of salt? (a) A lipid bilayer (b) The peptide bonds in a protein (c) A complex of two proteins (d) The sugar-phosphate backbone of a nucleic acid (e) An oil droplet in water
Macromolecules in Cells 2-31
You are trying to make a synthetic copy of a particular protein but accidentally join the amino acids together in exactly the reverse order. One of your classmates says the two proteins must be identical, and bets you $20 that your synthetic protein will have exactly the same biological activity as the original. After having read this chapter, you have no hesitation in staking your $20 that it won’t. What particular feature of a polypeptide chain makes you sure your $20 is safe, but that your project will have to be redone.
2-32
A protein chain folds into its stable and unique 3-D structure, or conformation, by making many noncovalent bonds between different parts of the chain. Such noncovalent bonds are also critical for interactions with other proteins and cellular molecules. From the list provided, choose the class(es) of amino acids that are most important for the interactions detailed below. A. Forming ionic bonds with negatively charged DNA B. Forming hydrogen bonds to aid solubility in water C. Binding to another water-soluble protein D. Localizing an “integral membrane” protein that spans a lipid bilayer E. Packing tightly the hydrophobic interior core of a globular protein acidic basic
2-33
nonpolar uncharged polar
DNA is negatively charged at physiological pH. A protein Z binds to DNA through noncovalent ionic interactions involving lysines. What will be the effect of acetylation of the lysine side chains (see Figure Q2-27) in protein Z on the strength of this binding? (a) It should increase because the acetylated lysine will form a greater number of ionic interactions with DNA. (b) It should decrease because the acetylated lysine no longer has a positive charge. (c) It should have no effect because the unmodified lysine would not have formed an ionic interaction with the DNA. (d) It should have no effect because the bond formed between lysine and the acetyl group still has a positive charge. (e) It should decrease unless the DNA can become more negatively charged.
21
2-34
You are studying a microorganism in which a “male” turns pink in the presence of a “female.” The male becomes pink because a protein X secreted by the female binds to and activates a protein Y on the male that is responsible for the color change. You have isolated a strain of the microorganism that produces a mutant form of protein X. This strain behaves normally at temperatures lower than 37°C, but at higher temperatures it cannot turn pink. Could any of the following changes in mutant protein X explain your results? If so, which ones, and explain why. (a) It makes an extra hydrogen bond to protein Y. (b) It makes fewer hydrogen bonds to protein Y. (c) It makes a covalent bond to protein Y. (d) It is completely unfolded at temperatures lower than 37°C. (e) It is completely unfolded at temperatures higher than 37°C. (f) It is unable to bind to protein Y at any temperature.
22
Answers 2-1
16 proteins and 19 neutrons
2-2
A. B. C.
180 daltons. A mole of a substance has a mass equivalent to its molecular weight expressed in grams. 45 g/l 6 ×1023 molecules
2-3
(a)
Sulfur is the least abundant element among the choices given.
2-4
Avogadro’s number, or 6 × 1023, is the number of atoms (or units) in a mole. If you multiply the number of people on Earth by the number of cells in the human body, then double it to account for the bacteria, you will calculate: (6 × 109) × (50 × 1012) × 2 = 6 × 1023. Thus, there must be much more than a mole of living cells on Earth, and you win $5.
2-5
(b)
2-6
A. B. C.
D.
2-7
(e)
2-8
(b)
The atomic number of carbon, which is the number of protons, is six. The atomic weight, which is the number of protons plus neutrons, is 12. The number of electrons, which equals the number of protons, is six. The valence is the minimum number of electrons that must be lost or gained to fill the outer shell of electrons. The first shell can accommodate two electrons and the second shell, eight. Carbon therefore has a valence of four because it needs to gain four additional electrons (or would have to give up four electrons) to obtain a full outermost shell. Carbon is most stable when it shares four additional electrons with other atoms (including other carbon atoms) by forming four covalent bonds. Carbon14 has two additional neutrons in its nucleus. As its electrons determine the chemical properties of an atom, carbon14 is chemically identical to carbon12.
Magnesium has a valence of two and chlorine has a valence of one. Thus, two chlorine atoms can each accept an electron donated by magnesium to yield a salt, designated as MgCl2 that contains twice as many Cl– chlorine anions as Mg2+ magnesium cations. All ions in this salt will have full outermost electron shells.
23
2-9
Whereas ionic bonds form a salt, covalent bonds between atoms form a molecule. These covalent bonds have a characteristic bond length and become stronger and more rigid when two electrons are shared in a double bond. Equal sharing of electrons yields a nonpolar covalent bond. If one atom participating in the bond has a stronger affinity for the electron, this produces a partial negative charge on one atom and a partial positive charge on the other. These polar covalent bonds should not be confused with the weaker noncovalent bonds that are critical for the three-dimensional structure of biological molecules and for interactions between these molecules.
2-10
A. B.
2-11
(c)
2-12
(c)
Sulfur and oxygen both require two electrons to fill their outer shell and can do so by sharing four electrons and forming a double bond.
2-13
A. B.
See Figure A2-13A. See Figure A2-13B.
kilocalories per mole (or kilojoules per mole) 0.5 kcal/mole
Figure A2-13 C. 2-14
Hydrogen can form one; oxygen two.
Choice (e) is the answer. Hydrogen bonds are critical for maintaining the conformation, or 3-D structure, of biological macromolecules like proteins and nucleic acids. Choice (a) is false because hydrogen bonds are not covalent. Choice (b) is false because the nonpolar-CH groups on hydrocarbons cannot form good hydrogen bonds, in water or out of it. Choice (c) is essentially another way of stating choice (b) and thus is false. Choice (d) is false because many molecules besides water can form hydrogen bonds and do so regardless of whether or not water is present. 24
2-15
Choice (c) is the answer. In methanol one of the hydrogens of a water molecule has been replaced by a nonpolar methyl group. Methanol will form fewer hydrogen bonds (thus choice (a) is false) and make fewer ionic interactions than water does. The ability of water to dissolve salts is a direct consequence of its ability to make ionic interactions. Salts are therefore less soluble in methanol. Choices (b), (d), and (e) are all false because the high boiling point, high degree of cohesion, and high surface tension of water are all a result of the extensive hydrogen bonding between water molecules. As methanol makes fewer hydrogen bonds, its boiling point will be lower, it will be less cohesive, and it will have a lower surface tension than water.
2-16
A. B. C. D.
2-17
Assuming the change in enzyme activity is due to the change in the protonation state of histidine, the enzyme must require histidine in the protonated, charged state. The enzyme is active only at low, acidic pH, where the proton (or hydronium ion) concentration is high and thus loss of a proton from histidine will be disfavored so that histidine is likely to be protonated.
2-18
A—9; B—5; C—4; D—6; E—1
2-19
A—7; B—2; C—5; D—1; E—6
2-20
(e)
2-21
Choice (b) is the answer. A condensation reaction releases a water molecule when forming polymers (like polysaccharide energy reserves) from monomeric units (like simple sugars), whereas the reverse hydrolysis reaction consumes a water molecule (thus, choices (a), (c), and (d) are correct). Most synthetic reactions in cells are catalyzed by enzymes (thus choice (e) is correct).
pH 7 10–8 M CH3COO– + H3O+ Yes. If the pH is high, then the concentration of hydronium ions will be low. Therefore the rightward reaction, which produces hydronium ions, will be favored.
Glucose and galactose are both six-carbon sugars and thus both have the formula C6H12O6. They are thus isomers of each other. 14C and 12C are examples of isotopes. Adenine and guanine are bases containing different numbers of nitrogen and oxygen atoms. Glycogen and cellulose are different polymers of glucose. Alanine and glycine are amino acids with quite different side chains, a methyl group and a hydrogen atom, respectively.
25
2-22
A. B. C.
20 carbon atoms 31 hydrogen atoms A fatty acid (Figure A2-22 is an arachidonic acid).
Figure A2-22 2-23
1—D; 2—A; 3—C; 4—J; 5—I; 6—H; 7—E
2-24
(c)
2-25
A. B. C.
2-26
Choice (e) is true. As their name implies, all amino acids have at lease one amino (NH2) group and at least one acidic carboxylic (COOH) group. It is through these two groups that they form peptide bonds. There are 20 common amino acids (choice (a) is false), and four or five of these have charged side chains (choice (b) is false). Each amino acid forms only two covalent bonds with other amino acids, one bond at the amino group and another at the carboxyl group (choice (c) is false); an exception to this is cysteine, because the side chains of two cysteines can form a covalent sulfhydryl bond to crosslink different regions of a polypeptide chain. Choice (d) is false because only L-amino acids are found in proteins.
2-27
(b)
The indicated bond is an amide. Like a peptide bond, it is formed by reaction between a carboxyl group and an amino group.
2-28
(c)
RNA contains the ribose sugar whereas DNA contains the deoxyribose sugar. They also differ in one of the pyrimidine bases used; RNA contains the pyrimidine uracil, while DNA instead contains thymine. All the other features are the same.
2-29
(c)
ATP is used in energy conversions, contains ribose, and can be incorporated into RNA. But synthesis of DNA requires the deoxyribose form of the nucleotide, dATP. All the other statements about ATP are true.
2-30
Choice (c) is correct. Noncovalent ionic interactions such as those that hold two proteins together are most likely to be disrupted by salt. Lipid bilayers (choice (a)) and a lipid droplet (choice (e)) are held together by “hydrophobic interactions” on which salt will have no effect. Choices (b) and (d)are examples of covalent bonds, which are not disrupted by salt.
proline-valine-threonine-glycine-lysine-cysteine-glutamic acid (or glutamate) PVTGKCQ C-terminal is glutamic acid (or glutamate); N-terminal is proline.
26
2-31
As a peptide bond has a distinct chemical polarity, a polypeptide chain also has a distinct polarity. (See Figure A2-31.) The reversed protein chains cannot make the same noncovalent interactions during folding and thus will not adopt the same 3-D structure as the original protein. The activities of these two proteins will definitely be different, since the activity of a protein depends on its 3-D structure. It is unlikely that the reverse chain will fold into any well-defined, and hence, functionally-useful structure at all, because it has not passed the stringent selective pressures imposed during evolution.
Figure A2-31 2-32
A. B. C. D. E.
basic uncharged polar uncharged polar, basic, and acidic nonpolar nonpolar
2-33
Choice (b) is correct. Unmodified lysine side chains are positively charged and hence attractive to the negatively charged DNA (thus choice (c) is incorrect). Because acetylation neutralizes the positive charge (thus choice (d) is incorrect), the acetylated form of protein Z will form fewer ionic bonds with DNA (thus choice (a) is incorrect), and thus the strength of the interaction will decrease. Choice (e) is incorrect, since increasing the number of negative charges on DNA would have no effect once the positive charge on the lysine has been neutralized.
2-34
Choices (b) and (e) are possible explanations. If protein X makes fewer hydrogen bonds to protein Y, the two proteins will bind less tightly and may come apart at temperatures above 37°C. Thermal motion is one of the forces that can disrupt the weak noncovalent bonds responsible for holding X and Y together. The male will, therefore, not be able to turn pink above 37°C. Weak noncovalent bonds are also responsible for folding X into the proper 3-D structure. If protein X is completely unfolded at elevated temperatures it will not be able to bind to protein Y, so choice (e) could be the correct answer. In contrast, choice (d) would explain a protein X that is able to bind to protein Y only at high temperatures, and would result in a strain that would turn pink only at high temperatures. Choice (a) would produce a protein X that would bind to protein Y more tightly than the normal protein, and would therefore be likely to bind (and turn pink) at temperatures above 37°C. If a covalent bond was made (choice (c)), it is unlikely that such a bond would be disrupted by any temperature in which the microorganism could survive; the microorganism would therefore turn pink at any temperature. Choice (f) would result in a strain that could not turn pink at any temperature.
27
CHAPTER 3 ENERGY, CATALYSIS, AND BIOSYNTHESIS 2009 Garland Science Publishing 3rd Edition
Catalysis and the Use of Energy by Cells 3-1 If you weigh yourself on a scale one morning then eat four pounds of food during the day, will you weigh four pounds more the next morning? Why or why not? Hint: What happens to the atoms contained in the food and the energy stored in the chemical bonds of food molecules? 3-2 Living organisms require a continual supply of energy to exist because they (a) defy the laws of thermodynamics. (b) convert it into heat energy, which powers biosynthetic reactions. (c) create order out of disorder inside their cells. (d) cause the entropy in the universe to decrease. (e) are closed systems isolated from the rest of the universe.
3-3 Life is thermodynamically possible because living things (a) release heat to the environment. (b) increase the degree of order in the universe. (c) reproduce themselves. (d) carry out energetically favorable reactions only. (e) can carry out a chain of reactions that is energetically unfavorable. 3-4 The energy required by a human cell to grow and reproduce is provided by (a) the generation of order inside it. (b) its anabolic metabolism. (c) its catabolic metabolism. (d) generation of heat. (e) its biosynthetic reactions.
3-5 Which of the following statements about photosynthesis are TRUE? (a) Photosynthesis is irrelevant to the existence of animals. (b) Photosynthesis converts light energy into heat energy and chemical bond energy. (c) Photosynthesis consumes activated carrier molecules. (d) Photosynthesis is the opposite of carbon fixation in the Earth’s carbon cycle. (e) Photosynthesis increases global warming.
29
3-6
A.
Complete the equation for respiration: Sugars + _________ → _________ + H2O + heat energy + _________ energy
B. 3-7
What does this process have in common with a fire that burns the polysaccharides in wood? How does it differ?
For each of the pairs A–D in Figure Q3-7, pick the more reduced member of the pair.
Figure Q3-7 3-8
Are the following statements TRUE or FALSE? Explain. A. If oxidation occurs in a reaction, it must be accompanied by a reduction. B. The hydrogenation of an unsaturated fatty acid to a saturated fatty acid, as in the conversion of vegetable oil to margarine, is an example of an oxidation reaction. C. The oxidation state of an atom influences its diameter.
30
3-9
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. By definition, catalysis allows a reaction to occur more __________________. Chemical reactions occur only when there is a loss of __________________ energy. Enzymes act more __________________ than other catalysts. A catalyst reduces the __________________ energy of a reaction. activation chemical bond completely favorable
free kinetic rapidly selectively
slowly unfavorable
3-10
Which of the following reactions are energetically favorable? (a) base + sugar + phosphate → nucleotide (b) amino acid + amino acid → peptide (c) CO2 + H2O → sugar (d) sucrose → CO2 + H2O (e) N2 + H2 → ammonia
3-11
Which of the following statements about enzymes is correct? (a) Catalysis of an energetically unfavorable reaction by an enzyme will enable that reaction to occur. (b) An enzyme can direct a molecule along a particular reaction pathway. (c) An enzyme can catalyze many chemically different reactions. (d) An enzyme can bind to many structurally unrelated substrates. (e) Enzymes are permanently altered after catalyzing a reaction.
3-12
Which of the following CANNOT be a reason that cells use enzymes rather than heat to increase the rate of biochemical reactions? (a) The temperature increase required to speed up a reaction by an appreciable extent is often huge. (b) Reactions cannot be coupled without enzymes. (c) An enzyme catalyzes just one or a very small number of different reactions; heat would affect all the reactions in a cell. (d) Enzymes change the equilibrium of biochemical reactions. (e) Enzymes can accelerate reactions to a much greater extent than can heat.
31
3-13
The graph Q3-13 is an energy diagram for an energetically favorable reaction of substrate S converted to product P. Indicate the following on the graph. A. The activation energy for the reaction B. The free energy change for the reaction C. Draw a new curve on the graph to indicate how an enzyme that converts S to P will change the energetics of the reaction.
Figure Q3-13 3-14
Energetically favorable reactions are those that (a) decrease the entropy of a system. (b) increase the free energy of a system. (c) have a positive ∆G. (d) decrease the free energy of a system. (e) create order in a system.
3-15
A.
Which one or more of the following reactions will NOT occur spontaneously under the standard conditions that specify ∆G°? (a) (b) (c) (d)
B.
ADP + Pi → ATP glucose-1-phosphate → glucose-6-phosphate glucose + fructose → sucrose glucose → CO2 + H2O
∆G° = +7.3 kcal/mole ∆G° = –1.7 kcal/mole ∆G° = +5.5 kcal/mole ∆G° = –686 kcal/mole
Which of the reactions in A could be coupled to any of the energetically unfavorable reactions to enable them to occur?
32
3-16
Energy diagrams for the conversion of a substrate S into a product P are shown in Figure Q3-16. The top diagram shows the original energy profile for the reaction. For each description of an altered reaction below, choose a matching altered energy diagram, labeled 1-6, from the figure.
Figure Q3-16 A. B. C. D.
∆G is more positive ∆G is more negative Reaction is catalyzed No change in ∆G and no change in activation energy. 33
3-17
Any reaction A ⇔ B is at equilibrium when (a) ∆G = 0. (b) ∆G° = 0. (c) [A] = [B]. (d) ∆G = ∆G°. (e) both forward and backward rates reach zero.
3-18
Consider the reaction X → Y in a cell at 37°C. At equilibrium, the concentrations of X and Y are 50 µM and 5 µM, respectively. Using the equations below and your new knowledge, answer the following questions. ∆G° = –0.616 ln Keq ∆G = ∆G° + 0.616 ln [Y]/[X] Recall that the natural log of a number z will have a negative value when z < 1, positive when z > 1, and 0 when z = 1. A. What is the value of Keq for this reaction? B. Is the standard free energy change of this reaction positive or negative? Is the reaction X → Y an energetically favorable or unfavorable reaction under standard conditions? C. What is the value of the standard free energy? Refer to Table 3-1 in the textbook or use a calculator. D. Imagine circumstances in which the concentration of X is 1000 µM and Y is 1 µM. Is conversion of X to Y favorable? Will it happen quickly? E. Imagine starting conditions in which the reaction X → Y is unfavorable, yet the cell needs to produce more Y. Describe two ways that this may be accomplished.
3-19
The equilibrium constant for the binding of a protein to its ligand can depend on all of the following EXCEPT the (a) number of noncovalent bonds formed between the protein and the ligand. (b) concentration of the ligand. (c) exact fit of the binding site to the ligand. (d) temperature. (e) pH.
34
3-20
Protein E can bind to two different proteins, S and I. The binding reactions are described by the following equations and values: E + S → ES E + I → EI
Keq for ES = 10 Keq for EI = 2
Based on the equilibrium constant values, which one of the following statements is TRUE? (a) E binds I more tightly than S. (b) When S is present in excess, no I molecules will bind to E. (c) The binding energy of the ES interaction is greater than that of the EI interaction. (d) Changing an amino acid on the binding surface of I from a basic amino acid to an acidic one will likely make the free energy of association with E more negative. (e) Binding of S to E will decrease the entropy of these two proteins, and thus it is an unfavorable reaction. 3-21
When the polymer X-X-X... is broken down into monomers, it is “phosphorylyzed” rather than hydrolyzed, in the repeated reaction: X-X-X... + P → X-P + X-X... (reaction 1) Given the ∆G° values of the reactions shown in Table Q3-21, what is the expected ratio of X-phosphate (X-P) to free phosphate (P) at equilibrium for reaction 1? (a) (b) (c) (d) (e)
1:106 1:104 1:1 104:1 106:1
Table Q3-21 X-X-X... + H2O → X + X-X... X + ATP → X-P + ADP ATP + H2O → ADP + P
∆G° = –4.5 kcal/mole ∆G° = –2.8 kcal/mole ∆G° = –7.3 kcal/mole
35
3-22
The following reactions take place in a cell located next to a blood vessel. X→Y Y + O2 → Z + CO2
∆G° = –10 kcal/mole ∆G° = +0.5 kcal/mole
Normally, the blood vessel brings in oxygen and takes away carbon dioxide, but years of overindulgence have taken their toll, and it has become completely clogged with cholesterol, cutting off the blood supply. Which of the following molecules would be expected to accumulate in large amounts? (a) X (b) Y (c) Z (d) Y and Z (e) X and Z 3-23
Figure Q3-23 shows graphs of energy diagrams for the unfavorable reaction X → Y, the favorable reaction Y → Z, and the composite reaction X → Z. A. B.
Draw lines near the composite reaction curve to indicate the free energy change for each of the three reactions: ∆G(XY), ∆G(YZ), and ∆G(XZ). Is the composite reaction favorable or unfavorable?
Figure Q3-23
36
3-24
A. B.
You are measuring the effect of temperature on the rate of an enzyme-catalyzed reaction. If you plot reaction rate against temperature, which of the graphs in Figure Q3-24 would you expect your plot to resemble? Explain why temperature has this effect.
Figure Q3-24 3-25
Are the following statements about diffusion TRUE or FALSE? A. The diffusion rate for small molecules is much slower in the dense gel of the cytoplasm than in pure water. B. The diffusion rate is faster for large macromolecules than for small molecules. C. The rate of diffusion determines the association rate of all binding reactions. D. The number of reactions per second that an enzyme catalyzes is independent of diffusion. E. Diffusion causes a molecule to move along a fairly straight path.
3-26
Which of the following features generally tend to increase the number of catalytic events an enzyme accomplishes per second? (a) Increased substrate concentration (b) More positive free energy of binding between enzyme and substrate (c) Decreased strength of binding between enzyme and product (d) Increased diffusion rate of substrate (e) Increased salt concentration
37
How We Know: Using Kinetics to Model and Manipulate Metabolic Pathways 3-27
The product Y of an enzymatic reaction absorbs light at the wavelength 260 nm and the product Z of another reaction absorbs at 340 nm, unlike the common substrate of the reactions. A spectrophotometer was used to measure the initial rate of production of Y and Z by the reactions shown in Figure Q3-27A. The initial rates were measured for several independent reactions, all containing equal amounts of enzyme A or enzyme B and differing amounts of substrate X. A graph was made of the initial reaction rate (v) plotted against the concentration of X ([X]) (Figure Q3-27). Given the data shown, are the following statements TRUE or FALSE?
Figure Q3-27 A. B. C. D. E.
Vmax(A) < Vmax(B). At [X] = 10 µM, the amounts of enzymes A and B limit the rate of reaction. KM(A) > KM(B). (Hint: [S] = KM when v = Vmax/2.) The turnover number of enzyme A is greater than the turnover number of enzyme B. (Hint: Turnover number = Vmax/[enzyme].) If [X] = 1 µM and both enzymes are present, most of the substrate will be converted to Y rather than Z.
38
3-28
Consider a description of an enzymatic reaction pathway that begins with the binding of substrate S to enzyme E and ends with the release of product P from the enzyme. E + S → ES → EP → E + P Under many circumstances, KM = [E] [S] / [ES] A. B.
What proportion of enzyme molecules are bound to substrate when [S] = KM? Recall that when [S] = KM, the reaction rate is Vmax/2. Does your answer to Part A make sense in light of this rate information?
Activated Carrier Molecules and Biosynthesis 3-29
Which of the following is NOT a crucial benefit of using enzymes to catalyze biological reactions? (a) Enzymes are highly selective in which reactions they catalyze. (b) Enzymes can drive an unfavorable reaction by coupling it to a favorable reaction, either directly or via activated carrier molecules. (c) Enzymes make reactions occur faster than without catalysis. (d) Enzymes change the equilibrium of a reaction to make it more favorable. (e) The activity of enzymes can be modulated by inhibitors and other small molecules to respond to the needs of the cell at each moment.
3-30
Consider an analogy between reaction coupling and money. In a simple economy, barter provides a means of direct exchange of material goods. For example, the owner of a cow may have excess milk and need eggs, whereas a chicken owner has excess eggs and needs milk. Provided these two people are in close proximity and can communicate, they may exchange or barter eggs for milk. But in a more complex economy, money serves as a mediator for the exchanges of goods or services. For instance, the cow owner with excess milk may not need other goods until three months from now or may want goods from someone who does not need milk. In this case, the “energy” from providing milk to the economy can be temporarily “stored” as money, which is a form of “energy” used for many transactions in the economy. Using barter and money as analogies, describe two mechanisms that can serve to drive an unfavorable chemical reaction in the cell.
3-31
A common means of providing energy to an energetically unfavorable reaction in a cell is by (a) generation of a higher temperature by the cell. (b) transfer of a phosphate group from the substrate to ADP. (c) enzyme catalysis of the reaction. (d) coupling of ATP hydrolysis to the reaction. (e) coupling of the synthesis of ATP to the reaction.
39
3-32
An anhydride formed between a carboxylic acid and a phosphate (Figure Q3-32A) is formed as a high-energy intermediate in some reactions in which ATP is used as the energy source. Arsenate mimics phosphate and can also be incorporated into a similar high-energy intermediate (Figure Q3-32B). The reaction profiles for the hydrolysis of these two high-energy intermediates are given in Figure Q3-32C. What is the effect of substituting arsenate for phosphate in this reaction?
Figure Q3-32 (a) (b) (c) (d) (e)
It forms a high-energy intermediate of lower energy. It forms a high-energy intermediate of the same energy. It decreases the stability of the high-energy intermediate. It increases the stability of the high-energy intermediate. It has no effect on the stability of the high-energy intermediate.
40
3-33
You are studying a biochemical pathway that requires ATP as an energy source. To your dismay, the reactions soon stop, partly because the ATP is rapidly used up and partly because an excess of ADP builds up and inhibits the enzymes involved. You are about to give up when the following table from a biochemistry textbook catches your eye. ∆G °
Hydrolysis reaction creatine + ATP
enzyme A
enzyme B
ATP + H2O pyrophosphate + H2O glucose 6-phosphate + H2O
enzyme D
enzyme E
creatine phosphate + ADP
+ 3 kcal/mole
ADP + phosphate
–7.3 kcal/mole
2 phosphate
–7 kcal/mole
glucose + phosphate
–3.3 kcal/mole
Which of the following reagents are most likely to revitalize your reaction? (a) A vast excess of ATP (b) Glucose 6-phosphate and enzyme E (c) Creatine phosphate and enzyme A (d) Pyrophosphate (e) Pyrophosphate and enzyme D 3-34
Which of the following statements is TRUE? (a) The oxidation of food molecules generates NAD+. (b) NADH and NADPH are found in mutually exclusive parts of the cell. (c) The ratio of NADPH:NADP+ is higher than the ratio of NADH:NAD+ because each molecule of NADPH is a stronger reducing agent than a molecule of NADH. (d) Many enzymes can use NADPH and NADH interchangeably. (e) One molecule of NADPH can cause the transfer of two hydrogen atoms.
3-35
Match the activated carrier molecules in List 1 with the groups they transfer, selected from List 2. Write the appropriate number beside each item in List 1. List 1 A. ATP B. Acetyl CoA C. NADPH D. Carboxylated biotin E. S-adenosylmethionine
List 2 1. –COO– 2. e– and H+ 3. Glucose 4. –PO43– 5. –CH3 6. Nucleotide 7. –COCH3 8. Amino acid 41
3-36
Which of the following processes must be coupled to an energetically favorable reaction in order to occur? (a) Conversion of protein into amino acids (b) Polymerization of amino acids into polypeptides (c) Conversion of glucose to carbon dioxide and water (d) Formation of a bilayer from phospholipids in water (e) The hydrolysis of ATP
3-37
The enzymes that catalyze the synthesis of macromolecules do not also catalyze their breakdown by hydrolysis because (a) enzymes can catalyze reactions in only one direction. (b) hydrolysis is not an energetically favorable reaction. (c) the hydrolytic reaction is not the reverse of the reaction pathway that is used for biosynthesis. (d) enzymes are destroyed immediately after synthesis is completed. (e) biosynthesis proceeds more rapidly than hydrolysis.
3-38
The energy required for the addition of a C nucleotide subunit (CMP) to a growing polynucleotide chain is originally derived from the hydrolysis of ATP. Explain how this is achieved.
42
Answers 3-1
No, you will not weigh four pounds more the next morning because only a small portion of the mass of the food will form components of the body. Much of the mass of food is released as CO2 that is breathed out into the atmosphere or is released into the environment as waste products. Most of the energy contained in the chemical bonds of the food molecules is converted to energy to maintain order among molecules in the body, energy to move and think, and energy for anabolic or biosynthetic reactions to rearrange the atoms from food into useful chemical structures (biological small molecules and macromolecules). As part of the process, a great deal of the bond energy is also converted to heat.
3-2
Choice (c) is the answer. Choice (a) is incorrect as no system, living or otherwise, can defy the laws of thermodynamics. Choice (b) is incorrect as living organisms do not use heat to power biochemical reactions. Heat is produced in the course of biochemical reactions. Choice (d) is incorrect: although living organisms are causing a local decrease in entropy, they cannot cause a decrease in the entropy of the universe as a whole as that would be a thermodynamic impossibility. Choice (e) is incorrect as living organisms are not closed systems.
3-3
(a)
3-4
Choice (c) is the answer. Catabolic reactions are the reactions in which a cell breaks down food molecules, releasing the energy held within their chemical bonds. Choices (a), (b), and (e) are energy-requiring processes.
3-5
Choice (b) is true. Photosynthesis harvests light energy from the sun and converts it into chemical bond energy. Choice (a) is false because food molecules and oxygen produced by photosynthesis are the sole source of the energy that powers nearly all living nonphotosynthetic organisms. Photosynthesis activates carrier molecules as intermediates in the “fixation” of inorganic carbon dioxide into organic sugar molecules (so choices (c) and (d) are false). By consuming carbon dioxide and producing oxygen photosynthesis lessens global warming cause by the greenhouse effect (choice (e) is false).
3-6
A. B.
By releasing heat to their environment, living things increase the entropy of the environment, thus compensating for the decrease in entropy inside cells. Living things, therefore, satisfy the second law of thermodynamics. They use special pathways for all their reactions that allow them to be energetically favorable.
Sugars + O2 → CO2 + H2O + heat energy + chemical bond energy Both respiration and burning are reactions that use oxygen gas to oxidize complex organic carbon molecules into CO2 + H2O. Burning is an uncontrolled oxidation in which the energy is all dissipated as heat; respiration is a multi-step, controlled oxidation that harnesses the energy in high-energy chemical bonds that are useful for anabolic reactions of cells.
43
3-7
A—ii; B—ii; C—i; D—ii. “More reduced” means having more electrons; gain of electrons can result in an increased negative charge or a decreased positive charge and can be due to an increase in the number of hydrogen atoms in a molecule.
3-8
A. B. C.
True. A redox reaction involves the complete or partial transfer of electrons from one molecule or atom to another. The donor is oxidized and the recipient is reduced in the reaction. False. Hydrogenation is a special kind of reduction reaction, involving receipt of an electron from a donor molecule and acquisition of a proton, usually from water. Hydrogenation increases the number of C-H bonds in a molecule. True. The diameter of an atom is influenced by the amount of negative charge, or electron density, surrounding it. The more reduced an atom becomes, the larger will be its electron cloud.
3-9
By definition, catalysis allows a reaction to occur more rapidly. Chemical reactions occur only when there is a loss of free energy. Enzymes act more selectively than other catalysts. A catalyst reduces the activation energy of a reaction.
3-10
(d)
3-11
(b)
3-12
(d)
Enzymes change only the activation energy of a reaction, not the free energy difference between reactants and products and thus cannot change the equilibrium concentrations of reactants and products.
44
3-13
See Figure A3-13.
Figure A3-13 A. B. C.
Activation energy is (a minus b). Change in free energy for the reaction is (b minus c). An enzyme will make the value of (a) smaller and leave the values for (b) and (c) unchanged.
3-14
(d)
3-15
A. B.
3-16
A—2; B—3; C—1; D—4. Graph 4 is the same as the graph for the original reaction in terms of the relative energetic differences between substrates, transition states, and products; the reaction diagram curve is simply positioned higher on the y-axis.
3-17
Choice (a) is correct. The value of ∆G for the reaction A ⇔ B is zero when there is no net tendency for either A → B or B → A, which is the definition of equilibrium. ∆G° is a constant and is thus always the same regardless of whether the reaction has reached equilibrium or not. Thus choices (b) and (d) are incorrect. Choice (c) is an incorrect answer; although a particular reaction might be at equilibrium when the concentration of substrate equals that of product, this is not true for most reactions. Choice (e) is not a definition of equilibrium, but of a reaction that is not occurring at all.
(a) and (c). Only reactions with a negative ∆G can occur spontaneously. Coupling of reaction (d) to either of the reactions (a) or (c) would provide an overall negative ∆G for the coupled reactions, thus enabling them to occur.
45
3-18
A. B.
C. D.
E.
Keq = [Y]/[X] = 5 µM/50 µM = 0.1. The standard free energy change, ∆G°, is positive because Keq is less than 1. Under standard conditions (equal concentrations of X and Y), the reaction X → Y is unfavorable. ∆G° = –0.616 ln Keq = –0.616 ln 0.1 = (–0.616) (–2.3) = 1.4 kcal/mol. Yes, the conversion is favorable because the value of [Y]/[X] is less than the equilibrium value. However, the speed of the reaction cannot be determined from the free energy difference. For example, combustion of this piece of paper is a highly favorable reaction, yet it will not happen in our lifetime without a catalyst. The cell may directly couple the unfavorable reaction to a second, energetically favorable reaction whose negative ∆G has a value larger than the positive ∆G of the X → Y reaction; the coupled reaction will have a ∆G equal to the sum of the component reactions. Alternatively, more X will be converted to Y if the concentration of Y drops; this may happen if Y is converted to Z in a second reaction or if Y is exported from the cell or compartment where the X → Y reaction occurs.
3-19
(b)
The equilibrium constant measures the strength of the interaction between a protein and its ligand and is independent of the concentration of either the protein or the ligand. The strength of the protein-ligand interaction increases as the number of noncovalent bonds between the two increases. The shape of the binding site affects the ability of the protein side chains to interact with portions of the substrate molecule. Both temperature and pH can disrupt noncovalent bonds that not only affect the binding, but are also responsible for keeping the protein folded and thus functional.
3-20
Choice (c) is true. The binding energy is the standard free energy of the binding reaction, and thus is proportional to ln Keq. As the binding energy increases, the equilibrium constant for the association reaction becomes larger. Choices (a) and (b) are false, because although E binds S more tightly than it does I, some E molecules will still be bound to I molecules in most circumstances; indeed, if the number of I molecules far exceeds the number of S molecules, more E molecules will be present in an EI complex than in an ES complex. Choice (d) is false; although not enough information is given to be certain, it is likely that binding is normally strengthened by an ionic interaction between a basic amino acid in I and an acidic amino acid in E—thus, if anything, the binding energy will be reduced by the amino acid change and the free energy change will be less (not more) negative. Choice (e) is false, because although the association of two molecules often does decrease their own entropy, it can increase the entropy of other molecules in the system. For example, heat release by the binding reaction can increase the entropy of the system and its surroundings.
46
3-21
(c)
Reaction 1 can be written as the sum of the three reactions given, since the ATP used in Step 2 is restored in Step 3. X-X-X... + H2O → X + X-X... X + ATP → X-P + ADP ADP + P → ATP + H2O
∆G° = –4.5 kcal/mole ∆G° = –2.8 kcal/mole ∆G° = +7.3 kcal/mole
Since ∆G° values are additive, ∆G°total = 0, and if ∆G° = 0, Keq = 1, meaning that [products]/[reactants] = 1, and the ratio of X-P to P is 1:1. 3-22
(b)
The constant removal of CO2 and replenishment of O2 by the blood normally drives the reaction Y → Z. Therefore, when CO2 is allowed to accumulate and O2 drops, Y will accumulate. Because the ∆G° of the first reaction is very negative, Y can accumulate to a very high level without causing significant amounts of X to build up.
3-23
A.
See Figure A3-23 for correct labeling of figure.
Figure A3-23
3-24
B.
Favorable
A. B.
Graph 1 By increasing thermal motion, increasing the temperature increases the rate of diffusion of components and the number of collisions of sufficient energy to overcome the activation energy. An increase in temperature will thus increase the reaction rate initially. However, enzymes are proteins and are held together by noncovalent interactions, so at very high temperatures, the enzyme will begin to denature and the reaction rate will fall.
47
3-25
A. B. C. D. E.
False. The diffusion rate is almost as fast in the cytoplasm and in water. False. The diffusion coefficient of a molecule decreases with increasing mass (shape is also a factor). False. Although some binding reactions are diffusion-limited, others require an unusually energetic collision to overcome an energy barrier. False. The diffusion rate influences how often an enzyme will encounter (and thus bind) its substrate. False. Diffusing molecules move in a “random walk,” in which they change direction frequently after colliding with other molecules.
3-26
Choices (a), (c), and (d) are correct. The higher the concentration and diffusion rate of substrate, the more frequently a free enzyme will collide with and bind its substrate. The less tightly an enzyme binds its product, in general, the faster the product will dissociate from the enzyme, leaving it free to bind a fresh substrate. Choice (b) is incorrect because a more positive free energy of binding indicates it is less energetically favorable and thus occurs less often. Choice (e) is likely to be false because many enzyme-substrate binding interactions rely on ionic bonds that are weakened by high salt concentrations; in addition, high salt concentrations can distort protein conformations.
3-27
A. B. C. D. E.
True True False False True
3-28
A. B.
When [S] is substituted for KM in the equation, it becomes clear that [E] = [ES]. Thus, half of the enzyme molecules are free and half are bound to the substrate. Yes. If half of the enzyme molecules are bound to the substrate, it makes intuitive sense that the reaction rate is half of the maximum possible rate or half of the rate observed when all of the enzyme molecules are bound to the substrate.
3-29
(d)
Enzymes cannot change the equilibrium of a reaction.
3-30
Barter is analogous to the direct coupling of a favorable to an unfavorable reaction by a single enzyme. Money is analogous to the storage of energy from a favorable reaction in the form of high-energy bonds in an activated carrier molecule. Such activated carrier molecules can drive a huge variety of other unfavorable reactions in the cell, either by being hydrolyzed to provide the needed energy for a reaction or by transferring the activated chemical group to another molecule.
3-31
(d)
48
3-32
Choice (c) is correct. The activation energy of the arsenate compound is extremely low, as can be seen from the reaction profile, meaning that its high-energy intermediate is very unstable and will be spontaneously hydrolyzed more rapidly than the phosphate compound. In fact, this hydrolysis occurs rapidly without enzyme catalysis, even in cellular conditions. Thus choices (d) and (e) are false. Choices (a) and (b) are false as more energy is released by the hydrolysis of the arsenoanhydride bond (as inferred by the greater difference in energy level between reactants and products in Figure Q3-32) so, by definition, the arsenoanhydride bond is said to have more energy than the phosphoanhydride bond.
3-33
(c)
An excess of ATP will initially restore the reactions, but as ATP is hydrolyzed, ADP will build up and inhibit the enzymes again. Pyrophosphate does not look like ATP and is therefore unlikely to be used by the enzymes as an alternative energy source. Pyrophosphate + enzyme D will just heat things up. What you need is a high-energy source of phosphate that can convert ADP back to ATP. Since the ∆G° of the reaction, ATP + creatine → ADP + creatine phosphate, catalyzed by enzyme A is greater than zero, the addition of creatine phosphate and enzyme A can be used to form ATP from ADP, regenerating the ATP while also forming creatine as a waste product.
3-34
Choice (e) is correct. NADPH has two electrons and one proton more than NADP+ (like NADH compared to NAD+) and donates both electrons. Protons are always present in solution. So the recipient molecule effectively acquires two hydrogen atoms. Choice (a) is false because oxidation of food molecules produces NADH, not NAD+. Chioce (b) is false because NADH and NADPH can be found in the same parts of the cell, but are used for different functions; this is possible because the enzymes that recognize one do not recognize the other; thus choice (d) is also false. Choice (c) is false because the parts of NADPH and NADH that participate in reduction are identical, and thus they both have essentially the same reducing power.
3-35
A—4 (phosphate group); B—7 (acetyl group); C—2; D—1 (carboxyl group); E—5 (methyl group).
3-36
(b)
Polymerization of amino acids into polypeptides lead to the formation of peptide bonds that have higher energy than the free amino acids and also represents an increase in order. Hence, it can only be brought about via an input of energy. The other processes are thermodynamically spontaneous.
49
3-37
(c)
Hydrolysis is not the reverse of the reactions catalyzed by biosynthetic enzymes. For instance, the reactions involved in RNA biosynthesis are: polynucleotide(n) + NTP → polynucleotide(n + 1) + PPi PPi + H2O → 2 Pi. The reverse reactions are: 2 Pi → PPi + H2O PPi + polynucleotide(n + 1) → polynucleotide(n) + NTP, Not: polynucleotide(n + 1) + H2O → polynucleotide(n) + NMP (nucleoside monophosphate), which is the reaction by which RNA is hydrolyzed. (a) and (b) are untrue: enzymes catalyze both forward and reverse reactions, and hydrolysis is an energetically favorable reaction. (d) is untrue, since enzymes are unchanged by participating in catalysis. Whether (e) is true or not for any particular reaction is irrelevant.
3-38
In order to add the C nucleotide to the polynucleotide chain, it must be in the form of a CTP (cytidine triphosphate). Conversion of CMP to CTP occurs by the sequential transfer of two terminal phosphate groups from two molecules of ATP. Thus, hydroylysis of ATP is coupled to phosphorylation of CMP and then to phosphorylation of CDP. Subsequently, the reaction that adds CMP to the polynucleotide chain releases pyrophosphate (PPi), which is hydrolyzed to inorganic phosphate; this favorable reaction provides the energetic drive for the overall condensation (or polymerization) reaction.
50
CHAPTER 4 PROTEIN STRUCTURE AND FUNCTION 2009 Garland Science Publishing 3rd Edition
The Shape and Structure of Proteins 4-1 For each of the following sentences, fill in the blanks with the best word or phrase selected from the chapter on Protein Structure and Function. A protein is similar to a charm bracelet that has a linear linked chain or backbone with charms hanging off at regular intervals. Similar to the amino acid sequence of a protein, it is the sequence and identity of the charms that dictate the character of the bracelet. The part of a protein that is analogous to the bracelet chain is called the __________________ backbone. The parts of a protein that are analogous to the charms are called the __________________.
4-2 Which of the following statements is TRUE? (a) (b) (c) (d) (e)
Peptide bonds are the only covalent bonds that can link together two amino acids in proteins. The polypeptide backbone is free to rotate about each peptide bond. Nonpolar amino acids tend to be found in the interior of proteins. The sequence of the atoms in the polypeptide backbone varies between different proteins. A protein chain ends in a free amino group at the C-terminus.
51
4-3
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. A newly synthesized protein generally folds up into a __________________ conformation. All the information required to determine a protein’s conformation is contained in its amino acid __________________. On heating, a protein molecule will become __________________ due to breakage of __________________ bonds. On removal of urea, an unfolded protein can become __________________. The final folded conformation adopted by a protein is that of __________________ energy. composition covalent denatured highest
4-4
irreversible lowest noncovalent renatured
reversible sequence stable unstable
Typical folded proteins have a stability ranging from 7 to 15 kcal/mol at 37°C. Stability is a measure of the equilibrium between the folded (F) and unfolded (U) forms of the protein, with the unfolded form having a greater free energy. See Figure Q4-4. For a protein with a stability of 7.1 kcal/mol, calculate the fraction of protein that would be unfolded at equilibrium at 37°C. The equilibrium constant (Keq) is related to the free energy (∆G°) by the equation: Keq = 10–∆G°/1.42
Figure Q4-4 4-5
You wish to produce a human enzyme, protein A, by introducing its gene into bacteria. The genetically engineered bacteria make large amounts of protein A, but it is in the form of an insoluble aggregate with no enzymatic activity. Which of the following procedures might help you to obtain soluble, enzymatically active protein? Explain your reasoning. A. Make the bacteria synthesize protein A at a slower rate and in smaller amounts. B. Dissolve the protein aggregate in urea, then dilute the solution and gradually remove the urea. C. Treat the insoluble aggregate with a protease. D. Make the bacteria overproduce chaperone proteins in addition to protein A. E. Heat the protein aggregate to denature all proteins, then cool the mixture.
52
4-6
Which of the following statements about proteins is TRUE? (a) The three-dimensional structure of a protein can usually be predicted from knowledge of its amino acid sequence. (b) Two proteins having similar amino acid sequences will often have similar shapes. (c) Proteins containing fewer than 100 amino acids cannot fold into stable structures. (d) Most proteins contain more than 2000 amino acids. (e) The detailed three-dimensional structure of a protein can usually be determined by electron microscopy.
4-7
The α helix and β sheet are found in many different proteins because they are formed by (a) hydrogen bonding between the amino acid side chains most commonly found in proteins. (b) noncovalent interactions between amino acid side chains and the polypeptide backbone. (c) ionic interactions between charged amino acid side chains. (d) hydrogen bonding between atoms of the polypeptide backbone. (e) hydrophobic interactions between the many nonpolar amino acids.
4-8
A.
In the schematic diagram of a protein given in Figure Q4-8, label the three protein strands that are linked together in a β sheet with a “b”.
Figure Q4-8 B.
Is this β sheet parallel or antiparallel?
53
4-9
For each polypeptide sequence listed, choose from the options given below to indicate which secondary structure the sequence is most likely to form upon folding. The nonpolar amino acids are italicized. A. B. C.
Leu-gly-val-leu-ser-leu-phe-ser-gly-leu-met-trp-phe-phe-trp-ile Leu-leu-gln-ser-ile-ala-ser-val-leu-gln-ser-leu-leu-cys-ala-ile Thr-leu-asn-ile-ser-phe-gln-met-glu-leu-asp-val-ser-ile-arg-trp amphipathic α helix amphipathic β sheet
hydrophilic α helix hydrophilic β sheet
hydrophobic α helix
4-10
A helical structure (a) will contain two, three, four, or some other exact number of subunits per each turn of the helix. (b) that is right-handed if viewed from one end will appear to be left-handed if viewed from its other end. (c) can form only by joining together a string of identical protein molecules. (d) can form either within a single large molecule or from an assembly of separate molecules. (e) is usually maintained entirely by covalent bonds.
4-11
Drawn below are segments of β sheets, which are rigid pleated structures held together by hydrogen bonds between the peptide backbones of adjacent strands (Figure Q4-11). The amino acid side chains attached to the Cα carbons are omitted for clarity.
Figure Q4-11 A. B.
Indicate whether each structure is parallel or antiparallel. Draw the hydrogen bonds as dashed lines (- - - - - -).
54
4-12
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. The α helices and β sheets are examples of protein __________________ structure. A protein such as hemoglobin, which is composed of more than one protein __________________, has __________________ structure. A protein’s amino acid sequence is known as its __________________ structure. A protein __________________ is the modular unit from which many larger single-chain proteins are constructed. The three-dimensional conformation of a protein is its __________________ structure. allosteric domain helix
4-13
ligand primary quaternary
secondary subunit tertiary
You are digesting a protein 625 amino acids long with the enzymes Factor Xa and thrombin, which are proteases that bind to and cut proteins at particular short sequences of amino acids. You know the amino acid sequence of the protein and so can draw a map of where factor Xa and thrombin should cut it (Figure Q4-13). You find, however, that treatment with each of these proteases for an hour results in only partial digestion of the protein, as summarized under the figure. List the segments (A–E) of the protein that are most likely to be folded into compact, stable domains.
Figure Q4-13 4-14
Calculate how many different amino acid sequences there are for a polypeptide chain 10 amino acids long.
4-15
All known proteins in cells adopt a single stable conformation because (a) any chain of amino acids can fold up into only one stable conformation. (b) protein chains that can adopt several different conformations have been weeded out by natural selection. (c) chaperone proteins prevent the protein from adopting a preferred unstable conformation. (d) they are complexed with other molecules that keep them in that one particular conformation. (e) one conformation always has the most positive free energy. 55
4-16
A friend tells you that she has just discovered that the protein responsible for causing dogs to chase cars is a member of the MAP protein kinase family. In response to your blank stare, she adds that the yeast protein Fus3p, which is involved in response to a yeast hormone is also a MAP kinase family member. Although you still have no idea of what either a MAP kinase or Fus3p is, which one or more of the following statement(s) can you safely predict to be TRUE? Explain your reasoning. (a) The dog protein and Fus3p have mostly similar amino acid sequences. (b) The dog protein and Fus3p catalyze the transfer of a phosphate group to another molecule. (c) The dog protein phosphorylates the same type of molecule that Fus3p phosphorylates. (d) The dog protein and Fus3p have identical three-dimensional structures. (e) The dog protein is involved in response to hormones.
4-17
A hemoglobin molecule (a) is composed of four protein domains. (b) is a dimer of polypeptide chains. (c) has two binding sites for nitrogen gas. (d) is composed of two different types of protein subunits. (e) is composed of four identical protein subunits.
4-18
When purified samples of protein Y and a mutant version of protein Y are both washed through the same gel-filtration column, mutant protein Y runs through the column much slower than the normal protein. Which one or more of the following changes in the mutant protein is/are most likely to explain this result? (a) The loss of a binding site on the mutant protein surface through which protein Y normally forms dimers (b) A change that results in the mutant protein acquiring an overall positive instead of a negative charge (c) A change that results in mutant protein Y being larger than the normal protein (d) A change that results in mutant protein Y having a slightly different shape from the normal protein (e) The loss of a binding crevice in the mutant protein Y for a small molecule ligand
4-19
Examine the three protein monomers in Figure Q4-19. From the arrangement of complementary binding surfaces, which are indicated by similarly shaped protrusions and invaginations, decide whether each monomer could assemble into a defined multimer, a filament, or a sheet.
Figure Q4-19
56
4-20
For each of the following indicate whether the individual folded polypeptide chain forms a globular (G) or fibrous (F) protein molecule. A. Keratin B. Lysozyme C. Elastin D. Collagen E. Hemoglobin F. Actin
4-21
S–S bonds (a) are formed by the cross-linking of methionine residues. (b) are formed mainly in proteins that are retained within the cytosol. (c) stabilize but do not change a protein’s final conformation. (d) can be broken by oxidation through agents such as mercaptoethanol. (e) rarely form in extracellular proteins.
How Proteins Work 4-22
Which of the following statements is FALSE? (a) The three dimensional structure of a protein dictates its function by determining its binding specificity for other molecules. (b) Many proteins have more than one binding site. (c) Binding between protein and ligand generally involves noncovalent bonds. (d) Proteins are designed to bind their ligands as tightly as possible. (e) Changes in the amino acid sequence of a protein can decrease binding to a ligand, even if the altered amino acid does not lie in the binding site for the ligand.
4-23
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. The human immune system produces __________________ of different immunoglobulins, also called __________________, which enable the immune system to recognize and fight germs by specifically binding one or a few related __________________. The hypervariable structural element that forms the ligand binding site is comprised of several __________________. Purified antibodies are useful for a variety of experimental purposes, including protein purification using __________________ chromatography. affinity antibodies antigens β strands
billions coiled coils hundreds ion exchange
57
ligands loops size exclusion
4-24
For each functional class of enzymes in List 1, choose the kind of reaction it catalyzes from List 2. Use each item in List 2 no more than once. List 1 A. Kinase B. Phosphatase C. Polymerase D. Hydrolase E. Dehydrogenase F. ATPase G. Synthase H. Protease
4-25
List 2 1. Oxidation of substrate 2. Reduction of substrate 3. Addition of phosphate group 4. Removal of phosphate group 5. Cleavage of substrate using water 6. Condensation reaction in anabolic pathway 7. Repeated condensation reactions with similar building blocks to form a growing chain 8. Hydrolysis of nucleotide triphosphate 9. Hydrolysis of peptide bonds
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Any substance that will bind to a protein is known as its __________________. Enzymes bind their __________________ at the __________________. The enzyme hexokinase is so specific that it reacts with only one of the two __________________ of glucose. Enzymes catalyze a chemical reaction by lowering the __________________, because they provide conditions favorable for the formation of a __________________ intermediate called the __________________. Once the reaction is completed, the enzyme releases the __________________ of the reaction. activation energy active site free energy high-energy
4-26
inhibitors isomers ligand low-energy
products substrates transition state
One way that an enzyme can lower the activation energy required for a reaction is to bind the substrate(s) and distort its structure so that the substrate more closely resembles the transition state of the reaction. This mechanism will be facilitated if the shape and chemical properties of the enzyme’s active site are more complementary to the transition state than to the undistorted substrate, in other words, if the enzyme were to have higher affinity for the transition state than for the substrate. Knowing this, your friend looked in an organic chemistry textbook to identify a stable chemical that closely resembles the transition state of a reaction that converts X into Y. She generated an antibody against this transition state analog and mixed the antibody with chemical X. What do you think might happen?
58
4-27
Which of the following statements is FALSE? (a) Polypeptide chains are never covalently linked to lipids or sugars. (b) Most vitamins bind noncovalently to the active sites of particular enzymes and provide essential catalytic assistance. (c) Some catalytic functions would be difficult or impossible using only the chemistry offered by the 20 amino acid side chains. (d) Small molecules tightly bound to enzymes can form transient bonds during catalysis. (e) Proteins that interact with photons and gas molecules often do so by virtue of tightly bound small molecules.
How Proteins Are Controlled 4-28
The biosynthetic pathway for the two amino acids E and H is shown schematically in Figure Q4-28. You are able to show that E inhibits enzyme V, and H inhibits enzyme X. Enzyme T is most likely to be subject to feedback inhibition by __________________ alone.
Figure Q4-28 (a) (b) (c) (d) (e) 4-29
A B C E H
In general, a ligand that binds to only one conformation of an allosteric protein will stabilize the bound conformation. Oxygen binds to the “oxy” conformation of hemoglobin, while carbon dioxide and the small organic molecule BPG bind to the “deoxy” conformation, each at different sites. Which of the following statements is FALSE? (a) A high concentration of carbon dioxide will stimulate binding of BPG to hemoglobin. (b) A high concentration of oxygen will stimulate dissociation of BPG and carbon dioxide from hemoglobin. (c) A high concentration of carbon dioxide will cause oxygen to dissociate. (d) In the presence of BPG, a lower concentration of carbon dioxide is required to cause dissociation of oxygen. (e) A variant of hemoglobin that has a lower affinity for BPG will bind to oxygen less tightly than normal hemoglobin in the presence of the same level of BPG. 59
4-30
The movement of a ciliated protozoan is controlled by a protein called RacerX. When this protein binds to another protein found at the base of the cilia, it stimulates the cila to beat faster and the protozoan to swim faster. This ciliar protein, Speed, can be phosphorylated and can only bind to RacerX in its phosphorylated form. You have identified the threonine residue at which Speed is phosphorylated and changed it to an alanine residue. How would you expect the mutant protozoan to behave? (a) Swims fast occasionally (b) Always swims fast (c) Never swims fast (d) Switches rapidly back and forth between fast and slow swimming (e) Cannot move at all
4-31
Which of the following mutant protozoa would swim fast all of the time? (a) One lacking the protein kinase that phosphorylates the Speed protein (b) One lacking the Speed protein (c) One lacking the RacerX protein (d) One in which the protein phosphatase that dephosphorylates the Speed protein is produced in much greater amounts than normal (e) One lacking the protein phosphatase
4-32
Some of the enzymes that oxidize sugars to yield useable cellular energy (for example, ATP) are regulated by phosphorylation. For these enzymes, would you expect the inactive form to be the phosphorylated form or the dephosphorylated form? Explain your answer.
4-33
GTP-binding proteins (a) form a transient covalent bond with guanine nucleotides. (b) are generally activated by factors that increase their rate of GTP hydrolysis. (c) immediately release the GDP produced by GTP hydrolysis. (d) “reset” themselves by phosphorylating bound GDP. (e) do not readily exchange bound GDP for GTP unless stimulated to do so.
4-34
The Ras protein is a GTPase that functions in many growth-factor signaling pathways. In its active form, with GTP bound, it transmits a downstream signal that leads to cell proliferation; in its inactive form, with GDP bound, the signal is not transmitted. Mutations in the gene for Ras are found in many cancers. Of the choices below, which alteration of Ras activity is most likely to contribute to the uncontrolled growth of cancer cells? (a) A change that prevents Ras from being made. (b) A change that increases the affinity of Ras for GDP. (c) A change that decreases the affinity of Ras for GTP. (d) A change that decreases the affinity of Ras for its downstream targets. (e) A change that decreases the rate of hydrolysis of GTP by Ras.
60
4-35
A molecule of the motor protein Winnebago, supplied with ATP, is moving along a microtubule in the direction shown in Figure Q4-35.
Figure Q4-35 What will happen if you suddenly remove all ATP from the system by adding an enzyme that hydrolyzes ATP? (a) No change: Winnebago will continue to move from point A to point B. (b) Winnebago will wander back and forth along the microtubule. (c) Winnebago will move backwards (towards point A instead of point B). (d) Winnebago will stall on the microtubule. (e) Winnebago will continue to move from point A to point B, but at a slower rate. 4-36
Assembling the individual enzymes required for a multistep process into a protein machine is likely to increase the efficiency with which the entire process is carried out in all of the following ways EXCEPT by (a) increasing the rate at which the individual enzymes encounter their substrates. (b) ordering the reactions sequentially. (c) increasing the Vmax of the individual enzymes. (d) coordinating the regulation of the individual enzymes. (e) coordinating the movement of the enzymes.
4-37
Explain how new proteomic approaches may advance the goal of being able to predict the three-dimensional structure of a protein from its amino acid sequence. Why is this an important goal?
61
Answers 4-1
A protein is similar to a charm bracelet that has a linear linked chain or backbone with charms hanging off at regular intervals. Similar to the amino acid sequence of a protein, it is the sequence and identity of the charms that dictate the character of the bracelet. The part of a protein that is analogous to the bracelet chain is called the polypeptide backbone. The parts of a protein that are analogous to the charms are called the side chains.
4-2
Choice (c) is the answer. Choice (a) is untrue, as some proteins also contain covalent disulfide bonds (–S–S– bonds) linking two amino acids. Choice (b) is untrue, as the peptide bond is rigid. Choice (d) is untrue, as the sequence of atoms in the polypeptide backbone itself is always the same from protein to protein; it is the order of the amino acid side chains that differs. Choice (e) is untrue, as a protein chain has a carboxyl group at its C-terminus.
4-3
A newly synthesized protein generally folds up into a stable conformation. All the information required to determine a protein’s conformation is contained in its amino acid sequence. On heating, a protein molecule will become denatured due to breakage of noncovalent bonds. On removal of urea, an unfolded protein can become renatured. The final folded conformation adopted by a protein is that of lowest energy.
4-4
The ∆G° of the unfolding reaction is equal to the stability of the protein, 7.1 kcal/mol. At equilibrium, the ratio of unfolded to folded protein is Keq = [U]/[F] = 10–∆G°/1.42 = 10–7.1/1.42 = 10–5. Thus, one molecule in 100,000 is unfolded.
62
4-5
Choices A, B, and Dare all worth trying. Some proteins require molecular chaperones in order to fold properly within the environment of the cell. In the absence of chaperones, a partly folded polypeptide chain has exposed amino acids that can form noncovalent bonds with other regions of the protein itself and with other proteins, thus causing nonspecific aggregation of proteins. A. Since the protein you are expressing in bacteria is being made in large quantities, it is possible that there are not enough chaperone molecules in the bacterium to fold the protein. Expressing the protein slowly and at lower levels might increase the amount of properly folded protein. B. Removing the urea slowly and gradually often allows the protein to refold. Presumably, under less crowded conditions, the protein should be able to refold into its proper conformation. C. Treating the aggregate with a protease, which cleaves peptide bonds, will probably solubilize the protein by trimming it into pieces that do not interact as strongly with one another; however, chopping up the protein will also destroy its enzymatic activity. D. Overexpressing chaperone proteins might increase the amount of properly folded protein. Urea should solubilize the protein and completely unfold it. E. Heating can lead to the partial denaturation and aggregation of proteins to form a solid gelatinous mass, as when cooking an egg white, and rarely helps solubilize proteins.
4-6
(b)
4-7
(d)
4-8
A.
Since a protein’s three-dimensional structure is determined by its amino acid sequence, proteins with similar amino acid sequences will often have very similar shapes.
See Figure A4-8.
Figure A4-8 B.
Antiparallel 63
4-9
A.
B.
C.
4-10
(d)
4-11
A. B.
Hydrophobic α helix. Nearly all of the amino acid side chains in this sequence are nonpolar or hydrophobic, which favors the only hydrophobic option given in the list. Amphipathic α helix. In an ideal α helix, there are 3.6 residues per complete turn. Thus, an amphipathic helix with one hydrophobic side and one hydrophilic side will have, minimally, nonpolar side chains (N) repeating every third then next fourth amino acid: NxxNxxxNxxNxxxN. Polar side chains (P) will exhibit the same pattern, but shifted relative to the nonpolar side chains: for example, xxPxxxPxxPxxxPxxP. Amphipathic β sheet. Due to the zig-zag like structure of a β sheet, a sequence with alternating nonpolar and polar side chains may form an amphipathic β sheet that is hydrophobic on one side and hydrophilic on the other.
(A) is parallel and (B) is antiparallel See Figure A4-11.
Figure A4-11 4-12
The α helices and β sheets are examples of protein secondary structure. A protein such as hemoglobin, which is composed of more than one protein subunit, has quaternary structure. A protein’s amino acid sequence is known as its primary structure. A protein domain is the modular unit from which many larger single-chain proteins are constructed. The three-dimensional conformation of a protein is its tertiary structure.
64
4-13
Segments B and D. To cut the protein chain, Factor Xa and thrombin must bind to their preferred cutting sites. If these sites are folded into the interior of a stable protein domain, it will be much more difficult for the proteases to gain access to them than if they are part of a relatively unstructured part of the chain. Hence, sites that are folded inside of a protein domain are protected from cleavage by a protease. From the sizes of the fragments produced by digestion of the protein with Factor Xa, we can conclude that the enzyme does not cut at the sites in regions B or D, although it does cut in region E. From the sizes of the fragments produced by thrombin, we can conclude that this enzyme cuts at the sites in A, C, and E. Therefore, the segments of the protein that are most likely to be folded into compact stable domains are B and D.
4-14
The quantity 2010 = approximately 1013.
4-15
(b)
4-16
Choices (a) and (b) are the answers. Members of the same protein family have similar protein sequences, similar three-dimensional structures, and roughly similar chemical activities (for example, all of the serine proteases catalyze the cleavage of a peptide bond). So if the dog protein is a MAP protein kinase, it is similar in sequence to other MAP kinases (choice (a)) and most likely has kinase activity (the transfer of a phosphate group from ATP to another molecule) (choice (b)). However, the actual substrates and the physiological function of proteins in the same family can differ quite markedly, so it is unlikely that the dog protein phosphorylates the same type of molecule that Fus3p does or is involved in the same type of response that Fus3p mediates.
4-17
(d)
4-18
Choice (a) is the answer. Dimers formed by a normal protein will run through the gelfiltration column faster than a mutant protein Y monomer. Choices (b) and (e) are unlikely, as gel-filtration columns separate proteins on the basis of size, not charge or affinity for small molecules. Choice (c) is unlikely, since if the mutant protein were larger than normal it would be less able to enter the porous beads and would run through the column at a faster speed than the normal protein. Choice (d) is unlikely, since a small change in shape without a change in size would be unlikely to have a major effect on the behavior of a protein in gel-filtration column.
4-19
A. B. C.
4-20
A—F; B—G; C—F; D—F; E—G; F—G
4-21
Choice (c) is the answer. Choice (a) is incorrect since S–S bonds are formed between cysteines. Choice (b) is incorrect, as they are formed mainly in extracellular proteins. Choice (d) is incorrect; they are broken by mercaptoethanol, but by reduction not oxidation. Choice (e) is incorrect for the reason stated in choice (b).
Defined multimer with four subunits, called a tetramer. Sheet Filament
65
4-22
(d)
False. Most proteins need to release their ligand at some point. For example, hemoglobin would be useless as a carrier of oxygen if it never released the oxygen to the tissues that need it. In addition, enzymes could not be catalysts if they did not release the product after its formation from substrate.
4-23
The human immune system produces billions of different immunoglobulins, also called antibodies, which enable the immune system to recognize and fight germs by specifically binding one or a few related antigens. The hypervariable structural element that forms the ligand binding site is comprised of several loops. Purified antibodies are useful for a variety of experimental purposes, including protein purification using affinity chromatography.
4-24
A—3; B—4; C—7; D—5; E—1; F—8; G—6; H—9
4-25
Any substance that will bind to a protein is known as its ligand. Enzymes bind their substrates (or inhibitors) at the active site. The enzyme hexokinase is so specific that it reacts with only one of the two isomers of glucose. Enzymes catalyze a chemical reaction by lowering the activation energy, because they provide conditions favorable for the formation of a high-energy intermediate called the transition state. Once the reaction is completed, the enzyme releases the products of the reaction.
4-26
If your friend was lucky, she made a “catalytic antibody” that catalyzed the conversion of X into Y. Such catalytic antibodies have been isolated and shown to catalyze a variety of reactions, but with lower efficiency than genuine enzymes.
4-27
(a)
4-28
(c)
If E alone inhibited T, then it would be possible to shut down the pathway even if H were in low abundance. Likewise, if H alone inhibited T, it would be possible to shut down the pathway even if E were in low abundance. An enzyme is generally not inhibited by its substrate. Levels of C will build up only if both E and H are abundant and have inhibited V and X. It is more likely that C alone rather than B alone will inhibit T, since B will accumulate only after C has done so.
4-29
(e)
Since carbon dioxide and BPG stabilize the form of hemoglobin that oxygen cannot bind to, either BPG or carbon dioxide will stimulate the dissociation of oxygen; likewise, a high concentration of oxygen will stimulate the dissociation of carbon dioxide and BPG. Since BPG and carbon dioxide both bind to and stabilize the same form of hemoglobin, these two ligands should help each other bind. Since BPG binding stimulates the dissociation of oxygen, a variant of hemoglobin that does not bind BPG well should bind more tightly to oxygen.
66
4-30
(c)
The alanine side chain has no hydroxyl (OH) group and therefore cannot be phosphorylated. Because the altered Speed protein cannot be phosphorylated, it can never bind RacerX. When RacerX is not bound to Speed, the cilia beat more slowly and thus the mutant protozoan would never swim fast.
4-31
Choice (e) is the answer. The lack of the protein phosphatase would mean that the Speed protein could remain phosphorylated all the time, causing the organism to swim fast. A mutant missing RacerX (choice (c)) would not be able to swim fast at all and neither would one missing the protein kinase that phosphorylates Speed (choice (a)) or one lacking the Speed protein (choice (b)). One that overproduced the protein phosphatase (choice (d)) would keep the Speed protein permanently dephosphorylated and thus would also be unable to swim fast.
4-32
In general the inactive form is the phosphorylated form. The main purpose of glycolysis and the citric acid cycle is to generate ATP; thus the enzymes are inactive when ATP is high and active when ATP is low. It makes sense that cells would not want to have to phosphorylate their enzymes to turn them on when ATP levels are already low, since phosphorylation requires ATP.
4-33
Choice (e) is the answer. GTP-binding proteins generally hydrolyze GTP and then retain the bound GDP until stimulated to exchange GDP for GTP by some other protein. The conformational change driven by hydrolysis, therefore, is due to the loss of a single phosphate group, not the whole guanine nucleotide (choice (c)). G proteins do not form covalent intermediates with either GTP or GDP (choice (a)). Since the GTP-bound form is usually the active form, a factor that stimulates hydrolysis will inhibit the protein (choice (b)). G proteins are reset by nucleotide exchange, not by rephosphorylation of bound GDP (choice (d)).
4-34
(e)
This choice will increase the amount of Ras that is in the GTP-bound state and thus increase the strength of the proliferative signal. All of the other changes will decrease the strength of the proliferative signal sent through the pathway by Ras.
4-35
(d)
Motor proteins that are capable of unidirectional movement require ATP (or GTP) hydrolysis to drive them in one direction. This is because the conformational change is coupled to ATP hydrolysis, such that movement in the reverse direction requires ATP synthesis. Hence, if ATP is depleted, the protein will completely stop moving, being unable to move forward for lack of ATP and unable to move backward because ATP synthesis is thermodynamically unfavorable.
67
4-36
(c)
If the product of one enzyme is the substrate for another, assembling the enzymes into a machine will bring the enzymes closer to their substrates because the product will have to diffuse only a short distance to the next enzyme. If the enzymes are properly arranged spatially, one can easily imagine how the machine could also facilitate the ordering of reactions. Gathering the enzymes into a complex also makes it easier to regulate and move all of the enzymes together. A machine is only as good as each of its parts, however, and if an enzyme has a low turnover number, complexing it with other proteins is unlikely to increase the enzyme’s ability to convert substrate into product.
4-37
It is hoped that large-scale analyses of many protein structures simultaneously will quickly complete the list of all protein folding patterns and all protein domains that evolved in living organisms. The identification of the thousands of different protein folds—the structural units that form the basis of all proteins—may allow elucidation of the rules that determine the conformation adopted by each amino acid sequence. Then, when a scientist identifies a protein responsible for a particular disease or discovers a protein in a new organism, it will be immediately possible to hypothesize its structure and thus its function.
68
CHAPTER 5 DNA AND CHROMOSOMES 2009 Garland Science Publishing 3rd Edition
How We Know: Genes are Made of DNA 5-1 Using terms from the list below, fill in the blanks in the following brief description of the experiment with Streptococcus pneumoniae that identified which biological molecule carries heritable genetic information. Some terms may be used more than once. Cell-free extracts from S strain cells of S. pneumoniae were fractionated to __________________ DNA, RNA, protein, and other cell components. Each fraction was then mixed with __________________ cells of S. pneumoniae. Its ability to change these into cells with __________________ properties resembling the __________________ cells was tested by injecting the mixture into mice. Only the fraction containing __________________ was able to __________________ the __________________ cells to __________________ (or __________________ ) cells that could kill mice. carbohydrate DNA identify label
lipid nonpathogenic pathogenic purify
The Structure and Function of DNA 5-2 In a DNA double helix (a) (b) (c) (d) (e)
the two DNA strands are identical. purines pair with purines. thymine pairs with cytosine. the two DNA strands run antiparallel. the nucleotides are ribonucleotides.
69
R strain RNA S strain transform
5-3
On the diagram of a small portion of a DNA molecule in Figure Q5-3, match the labels below to the numbered label lines.
Figure Q5-3 (a) (b) (c) (d) (e) (f) 5-4
Base Sugar Phosphate Hydrogen bond 5′ end 3′ end
The structures of the four bases in DNA are given in Figure Q5-4.
Figure Q5-4 A. B.
Which are purines and which are pyrimidines? Which bases pair with each other in double-stranded DNA? 70
5-5
Using the structures in Figure Q5-4 as a guide, sketch the hydrogen bonds that form when the appropriate bases form base pairs in DNA. Hint: The bases in the figure are all drawn with the –NH– that attaches to the sugar at the bottom of the structure.
5-6
A ribbon model of the DNA double helix showing the major and minor grooves is reproduced in Figure Q5-6.
Figure Q5-6 A. B. 5-7
Draw on the figure to indicate the length of a single full helical turn. How many base pairs per turn does a DNA helix have?
Given the sequence of one strand of a DNA helix: 5′-GCATTCGTGGGTAG-3′ give the sequence of the complementary strand and label the 5′ and 3′ ends.
5-8
Which of the following sequences can fully base-pair with itself? A. 5′-AAGCCGAA-3′ B. 5′-AAGCCGTT-3′ C. 5′-AAGCGCAA-3′ D. 5′-AAGCGCTT-3′ E. 5′-AATTGGCC-3′
5-9
The DNA from two different species can often be distinguished by a difference in the (a) ratio of A + T to G + C. (b) ratio of A + G to C + T. (c) ratio of sugar to phosphate. (d) presence of bases other than A, G, C, and T. (e) number of strands in the helix.
71
5-10
When double-stranded DNA is heated, the two strands separate into single strands in a process called melting or denaturation. The temperature at which half of the duplex DNA molecules are intact and half have melted is defined as the Tm. A. Do you think Tm is a constant or is dependent on other small molecules in the solution? Do you think high salt concentrations increase, decrease, or have no effect on Tm? B. Under standard conditions, the expected melting temperature in degrees Celsius can be calculated using the equation Tm = 59.9 + 0.41 [%(G + C)] – [675/length of duplex]. Does the Tm increase or decrease if there are more G + C (and thus fewer A + T) base pairs? Does the Tm increase or decrease as the length of DNA increases? Why? C. Calculate the predicted Tm for a stretch of double helix that is 100 nucleotides long and contains 50% G + C content.
5-11
Consider the structure of the DNA double helix. A. You and a friend want to split a double-stranded DNA molecule so you each have half. Is it better to cut the length of DNA in half so each person has a shorter length, or to separate the strands and each take one strand? Explain. B. In the original 1953 publication describing the discovery of the structure of DNA, Watson and Crick wrote, “It has not escaped our notice that the specific pairings we have postulated immediately suggest a possible copying mechanism for the genetic material.” What did they mean?
5-12
A.
B.
In principle what would be the minimum number of consecutive nucleotides necessary to correspond to a single amino acid to produce a workable genetic code? Assume that each amino acid is encoded by the same number of nucleotides. Explain your reasoning. On average how often would the nucleotide sequence CGATTG occur in a DNA strand 4000 bases long? Explain your reasoning.
72
The Structure of Eucaryotic Chromosomes 5-13
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. In eucaryotic __________________, DNA is complexed with proteins to form __________________. The paternal and maternal copies of human Chromosome 1 are __________________, whereas the paternal copy of Chromosome 1 and the maternal copy of Chromosome 3 are __________________. Cytogeneticists can determine large-scale chromosomal abnormalities by looking at a patient’s __________________. Fluorescent molecules can be used to paint a chromosome by virtue of DNA __________________, and thereby to identify each chromosome using microscopy. bands chromatin chromosomes condensation
5-14
A. B. C.
5-15
extended homologous hybridization karyotype
kinetochore nonhomologous
Define a gene. Consider two different species of yeast that have similar genome size. Is it likely that they contain a similar number of genes? A similar number of chromosomes? Figure 5-15 in the textbook shows the G + C content and genes found along a single chromosome. Is there any relationship between the G + C content and the locations of genes?
The human genome is comprised of 23 pairs of chromosomes found in nearly every cell in the body. Answer the quantitative questions below by choosing one of the numbers in the following list: 23 46 A. B. C.
69 92
>200 >109
How many centromeres are in each cell? What is the main function of the centromere? How many telomeres per cell? What is their main function? How many replication origins per cell? What is their main function?
73
5-16
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Each chromosome is a single molecule of __________________ whose extraordinarily long length can be compacted by as much as __________________-fold during __________________ and tenfold more during __________________. This is accomplished by binding to __________________ that help package the DNA in an orderly manner so it can fit in the small space delimited by the __________________. The structure of the DNA-protein complex, called __________________, is highly __________________ over time. 10,000 100 1000 cell cycle cell wall chromatin
5-17
chromosome different DNA dynamic interphase lipids
mitosis nuclear envelope nucleolus proteins similar static
For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement about nucleosomes. A. Nucleosomes are present in [procaryotic/eucaryotic] chromosomes, but not in [procaryotic/eucaryotic] chromosomes. B. A nucleosome contains two molecules each of histones [H1 and H2A/H2A and H2B] as well as histones H3 and H4. C. A nucleosome core particle contains a core of histone with DNA wrapped around it approximately [twice/three times/four times]. D. Nucleosomes are aided in their formation by the high proportion of [acidic/basic/polar] amino acids in histone proteins. E. Nucleosome formation compacts the DNA into approximately [one-third/onehundredth/one-thousandth] of its original length.
74
5-18
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Interphase chromosomes contain both darkly staining __________________ and more lightly staining __________________. Genes that are being transcribed are thought to be packaged in a __________________ condensed type of euchromatin. Nucleosome core particles are separated from each other by stretches of __________________ DNA. A string of nucleosomes coils up with the help of __________________ to form the more compact structure of the __________________. The __________________ model describes the structure of the 30 nm fiber. The 30 nm chromatin fiber is further compacted by the formation of __________________ that emanate from a central __________________. 30 nm fiber active chromatin axis beads-on-a-string euchromatin
heterochromatin histone H1 histone H3 histone H4 less
linker loops more synaptic complex zigzag
5-19
In which of the following instances can the state of chromatin packing differ? Explain your reasoning. A. Between different cells of the same organism. B. In different stages of the cell cycle. C. In different parts of the same chromosome. D. In different members of a pair of homologous chromosomes.
5-20
Evidence suggests that the replication of DNA packaged into heterochromatin occurs later than the replication of other chromosomal DNA. What is the simplest possible explanation for this phenomenon?
5-21
If the mottled coloring of calico cats is due to X-chromosome inactivation, which of the following statements will be TRUE? (a) Calico cats can be male or female. (b) Female calico cats will be the same color as their mother. (c) The mottled color is due to X chromosomes repeatedly switching back and forth between active and inactive states during development. (d) Calico cats with identical patterns will be rare. (e) Coat color is determined by a single gene.
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5-22
A.
B.
5-23
Chromatin remodeling complexes act to regulate gene expression and other DNA processes that involve access to DNA. A characteristic component of chromatin remodeling complexes is a powerful ATPase. Why is ATPase activity needed? Imagine that you have developed a method to isolate all of the histones bound to a single human chromosome. You then examine histones from the inactive X chromosome in an individual female and compare them to histones from her active X chromosome. Do you think these two sets of histones will be the same? Explain.
Your friend is working in a lab to study how cells adapt to growth on different carbon sources. He grew half of his cells in the presence of glucose and the other half in the presence of galactose. Then he harvested the cells and isolated their DNA using a gentle procedure that leaves nucleosomes and some higher order chromatin structures intact. He treated the DNA briefly with a low concentration of M-nuclease, a special enzyme that easily degrades protein-free stretches of DNA. After removing all the proteins, he separated the resulting DNA on the basis of length. Finally, he used a procedure to visualize only those DNA fragments from a region near a particular gene called Sweetie or another gene called Salty. The separated DNA fragments are shown in Figure Q5-23. Each vertical column, called a lane, is from a different sample. DNA spots near the top of the figure represent DNA molecules that are longer than those near the bottom. Darker spots contain more DNA than fainter spots. The lanes are as follows: 1. 2. 3. 4. 5.
“marker” containing known DNA fragments of indicated lengths. grew cells in glucose, visualized DNA near Sweetie gene. grew cells in galactose, visualized DNA near Sweetie gene. grew cells in glucose, visualized DNA near Salty gene. grew cells in galactose, visualized DNA near Salty gene.
Figure Q5-23 76
A. B. C.
D. E.
The lowest spot (as observed in lanes 2, 4, and 5) has a length of about 150 nucleotides. Can you propose what it is and how it arose? What are the spots with longer lengths? Why is there a ladder of spots? Notice the faint spots and extensive smearing in lane 3, suggesting the DNA could be cut almost anywhere near the Sweetie gene after growth of the cells in galactose. This was not observed in the other lanes. What probably happened to the DNA to change the pattern between lanes 2 and 3? What kinds of enzymes might have been involved in changing the chromatin structure from lane 2 into lane 3? Do you think that gene expression of Sweetie is higher, lower, or the same in galactose compared to glucose? What about Salty?
77
Answers 5-1
Cell-free extracts from S strain cells of S. pneumoniae were fractionated to purify DNA, RNA, protein, and other cell components. Each fraction was then mixed with R strain cells of S. pneumoniae. Its ability to change these into cells with pathogenic properties resembling the S strain cells was tested by injecting the mixture into mice. Only the fraction containing DNA was able to transform the R strain cells to pathogenic (or S strain) cells that could kill mice.
5-2
(d)
5-3
(a)—6; (b)—3; (c)—2; (d)—5; (e)—1; (f)—4
5-4
A. B.
5-5
See Figure A5-5
Adenine and guanine are purines; cytosine and thymine are pyrimidines. Cytosine pairs with guanine and adenine with thymine.
Figure A5-5
78
5-6
A.
See Figure A5-6.
Figure A5-6 B.
There are approximately 10 base pairs per turn when the DNA has the standard conformation.
5-7
5′ CTACCCACGAATGC 3′.
5-8
D.
5-9
Choice (a) is the answer. Since the sequence of nucleotides in the DNAs from different species varies considerably, the ratio A-T to G-C base pairs will vary as well. Yet since A must always pair with T and G with C, the ratio of A+G to C+T must always be 1 (choice (b) is therefore incorrect). Choice (c) is incorrect, since the DNA backbone always contains a 1:1 ratio of sugar to phosphate. Choice (d) is incorrect, as all living organisms contain only A, G, C, and T in their DNA (but modifications of these bases occur in some organisms and under some circumstances, after their incorporation into the nucleotide chain). Choice (e) is incorrect because all genomic DNA is double stranded (except in some viruses).
5′-AAGCGCTT-3′ 3′-TTCGCGAA-5′ None of the other sequences will base-pair with themselves. This double-stranded DNA molecule has the same sequence whether read forward or backward, which is why it is described as palindromic. Note that the reading direction is always defined as 5′ to 3′. An example of a palindromic sentence is “Step on no pets.”
79
5-10
A.
B.
C.
5-11
A.
B.
5-12
A.
B.
5-13
Tm depends on the identity and concentration of other molecules in the solution. High salt concentrations are more effective at shielding the two negatively charged phosphate-sugar backbones in the double helix from each other, so the two strands repel each other less strongly. Thus, a high salt concentration stabilizes the duplex and increases the melting temperature. The Tm increases as the proportion of G + C bases increases and as the length increases. The thermal energy required for melting depends on how many hydrogen bonds between the strands must be broken. Each G-C base pair contributes three hydrogen bonds, whereas an A-T base pair contributes only two. Inserting values into the equation in part (B) gives Tm = 59.9 + (0.41 × 50) – (675/100) = 73.65°C, which is about twice the normal temperature of the human body and nearly too hot to touch. It is better to separate the stands and each take a single strand, because all of the information found in the original molecule is preserved in a full-length single strand but not in a half-length double-stranded molecule. Watson and Crick meant that the complementary base pairing of the strands allows a single strand to contain all of the information necessary to direct the synthesis of a new complementary strand. As there are 20 amino acids used in proteins, each amino acid would have to be encoded by a minimum of three nucleotides. For example, a code of two consecutive nucleotides could specify a maximum of 16 (42) different amino acids, excluding stop and start signals. A code of three consecutive nucleotides has 64(43) different members and thus can easily accommodate the 20 amino acids plus a signal to stop protein synthesis. As 46 (= 4096) different sequences of six nucleotides can occur in DNA, any given sequence of six nucleotides would occur on average once in a DNA strand 4000 bases long.
In eucaryotic chromosomes, DNA is complexed with proteins to form chromatin. The paternal and maternal copies of human Chromosome 1 are homologous, whereas the paternal copy of Chromosome 1 and the maternal copy of Chromosome 3 are nonhomologous. Cytogeneticists can determine large-scale chromosomal abnormalities by looking at a patient’s karyotype. Fluorescent molecules can be used to paint a chromosome by virtue of DNA hybridization, and thereby to identify each chromosome using microscopy.
80
5-14
A.
B.
C.
5-15
A.
B.
C.
A gene is a segment of DNA that stores the information required to specify the particular sequence found in a protein (or, in some cases, the sequence of a structural or catalytic RNA). A similar genome size indicates relatively little about the number of genes and virtually nothing about the number of chromosomes. For example, the commonly studied yeasts Saccharomyces cerevisiae (Sc) and Schizosaccharomyces pombe (Sp) are separated by roughly 400 million years of evolution, and both have a genome of 14 million base pairs. Yet Sc has 6500 genes packaged into 16 chromosomes and Sp has 4800 genes in 3 chromosomes. Regions of the chromosome with a high density of genes tend to have about 50% G + C, whereas those with few genes tend to have a lower G + C content. This is generally true in most organisms. There are 46 centromeres per cell, one on each chromosome. The centromeres play a key role in the distribution of chromosomes to daughter cells during mitosis. There are 92 telomeres per cell, two on each chromosome. Telomeres serve to protect the ends of chromosomes and to enable complete replication of the DNA of each chromosome all the way to its tips. There are far more than 200 replication origins in a human cell, probably around 10,000. These DNA sequences direct the initiation of DNA synthesis needed to replicate chromosomes.
5-16
Each chromosome is a single molecule of DNA whose extraordinarily long length can be compacted by as much as 1000-fold during interphase and tenfold more during mitosis. This is accomplished by binding to proteins that help package the DNA in an orderly manner so it can fit in the small space delimited by the nuclear envelope. The structure of the DNA-protein complex, called chromatin, is highly dynamic over time.
5-17
A. B. C. D. E.
Nucleosomes are present in eucaryotic chromosomes, but not in procaryotic chromosomes. A nucleosome contains two molecules each of histones H2A and H2B, as well as histones H3 and H4. A nucleosome core particle contains a core of histone with DNA wrapped around it approximately twice. Nucleosomes are aided in their formation by the high proportion of basic amino acids in histone proteins. Nucleosome formation compacts DNA into approximately one-third of its original length.
81
5-18
Interphase chromosomes contain both darkly staining heterochromatin and more lightly staining euchromatin. Genes that are being transcribed are thought to be packaged in a less condensed type of euchromatin. Nucleosome core particles are separated from each other by stretches of linker DNA. A string of nucleosomes coils up with the help of histone H1 to form the more compact structure of the 30 nm fiber. The zigzag model describes the structure of the 30 nm fiber. The 30 nm chromatin fiber is further compacted by the formation of loops that emanate from a central axis.
5-19
The state of chromosome packing can differ in all of the instances given. Chromatin differs most dramatically during the different stages of the cell cycle. But even within the same interphase chromosome, chromatin can be less tightly packed into euchromatin (especially in regions where genes are actively being expressed) or more densely packed into heterochromatin (particularly near centromeres and telomeres). In the cells of female mammals, one of the two homologous X chromosomes is entirely heterochromatic and transcriptionally inactive. Because different cell types in the same organism are expressing different genes, not all cells will have the same state of chromatin packing.
5-20
The DNA double helix in heterochromatin may be so tightly packed and condensed that it is inaccessible to the proteins that bind replication origins, including the DNA replication machinery. It may take extra time to remodel the chromatin to make it more accessible to the proteins required to initiate and carry out DNA replication.
5-21
Choice (d) is the answer. Since the pattern of X-chromosome inactivation is established randomly over several days of embryonic development, at a stage where the embryo has quite a few cells, it is unlikely that two cats will inactivate the same X chromosome in exactly the same set of cells. The other statements are false. Once inactivated, an X is not normally reactivated during the life of the individual (choice (c) is false). If the sectored coloring is due to X-chromosome inactivation, then the coat-color gene responsible for the sectoring must be on the X chromosome (although there may be several additional, different coat color genes located on other chromosomes). The fact that sectoring results from X-chromosome inactivation cannot (by itself) distinguish between the possibilities that coat color is determined by a single gene or by multiple genes (choice (e) is false). Since males have only one X chromosome that they receive from their mother and that never becomes inactivated, males will not be mottled (choice (a) is false). Female calico cats receive an X chromosome from the father and another from the mother, who may have different forms of the coat-color gene, and so female cats are not necessarily the same color as their mother (choice (b) is false).
82
5-22
A.
B.
5-23
A.
B.
C.
D.
E.
For DNA to be accessible—for transcription or some other process—noncovalent bonds must be broken in order to change interactions between DNA and its packaging proteins. This energetically unfavorable feat can be accomplished by harnessing the energy released by hydrolysis of ATP. A nucleosome is held together by a large number of weak, noncovalent interactions. For example, many ionic bonds are made between the approximately 300 negatively charged phosphate groups in the double-stranded DNA backbone and the positively charged amino acid side chains of the eight proteins in the histone octamer. The standard free energy required to dissociate a single nucleosome into its components (histone octamer and DNA) is estimated to be around 14 kcal/mol, roughly twice the amount of standard free energy liberated by the hydrolysis of ATP to ADP and phosphate. Based on the information presented in Chapter 5, it seems reasonable to propose that the nucleosomes on the heterochromatic inactive X chromosome might contain histone H3 tails with distinctive modifications. Indeed, in line with the histone code hypothesis, at the time an X chromosome is first becoming inactive the associated histone H3 becomes uniquely modified. The lowest spot represents DNA of a length similar to that of the segment of DNA found in a nucleosome core particle. Partial digestion with an enzyme like M-nuclease causes breaks in the DNA backbone primarily within the linker DNA or other DNA segments not bound tightly to histones. Thus, this band is probably comprised of the DNA bound tightly to a single histone octamer and it arose by cutting the linker DNA outside a single nucleosome core particle. The ladder of bands with longer lengths probably corresponds to stretches of DNA associated with increasing numbers of nucleosomes (1, 2, 3, 4, 5, and so on). In support of this proposal, adjacent bands differ in size by roughly 200 nucleotides, which is the length of DNA found in a nucleosome core particle plus neighboring linker DNA. This interpretation must mean that the M-nuclease digestion did not go to completion; because if all non-nucleosomal DNA were digested completely, the samples would contain only the 150 base pair fragment. Based on the ability of M-nuclease to cut anywhere near Sweetie after growth in galactose, it appears that the DNA is no longer protected from digestion by binding to histones. Perhaps the wrapping of DNA within the nucleosomes has been loosened considerably as in Figure 5-29. This change in the nucleosomes must be specific to the Sweetie gene as it is not seen at the Salty gene or throughout the genome. The main candidates for enzymes that catalyzed the nucleosome alterations near Sweetie are chromatin remodeling complexes and enzymes that covalently modify histone tails with methyl, acetyl, or phosphate groups. As the chromatin appears to have been loosened near Sweetie, it seems likely that Sweetie gene expression is increased when cells are grown in galactose rather than glucose, whereas Salty gene expression is likely to be the same under the two conditions. Perhaps the Sweetie gene contains instructions for a protein that is required for cells to metabolize galactose but not glucose.
83
CHAPTER 6 DNA REPLICATION, REPAIR, AND RECOMBINATION 2009 Garland Science Publishing 3rd Edition
DNA Replication 6-1 DNA replication is considered semiconservative because (a) after many rounds of DNA replication, the original DNA double helix is still intact. (b) each daughter DNA molecule consists of two new strands copied from the parent DNA molecule. (c) each daughter DNA molecule consists of one strand from the parent DNA molecule and one new strand. (d) new DNA strands must be copied from a DNA template. (e) an RNA primer must be used to initiate synthesis of the DNA strand.
6-2 If the genome of the bacterium E. coli requires about 20 minutes to replicate itself, how can the genome of the fruit fly Drosophila be replicated in only three minutes? (a) The Drosophila genome is smaller than the E. coli genome. (b) Eucaryotic DNA polymerase synthesizes DNA at a much faster rate than procaryotic DNA polymerase. (c) The nuclear membrane keeps the Drosophila DNA concentrated in one place in the cell, which increases the rate of polymerization. (d) Drosophila DNA contains more origins of replication than E. coli DNA. (e) Eucaryotes have more than one kind of DNA polymerase. 6-3 Answer the following questions about DNA replication. A. On a DNA strand that is being synthesized, which end is growing—the 3’ end, the 5’ end, or both ends? Explain your answer. B. On a DNA strand that is being used as a template, where is the copying occurring relative to the replication origin—3’ of the origin, 5’ , or both? 6-4 If DNA strands were paired in a parallel rather than antiparallel fashion, how would the replication of the DNA differ from that of normal double-stranded DNA? (a) Replication would not be semiconservative. (b) Replication origins would not be required. (c) The replication fork would not be asymmetrical. (d) The polymerase used would not be self-correcting. (e) Both new strands would be synthesized discontinuously.
85
6-5
On Figure Q6-5 of a replication bubble:
Figure Q6-5 A. B. C. D. E. F.
Indicate where the origin of replication was located (use O). Label the leading-strand template and the lagging-strand template of the righthand fork [R] as X and Y, respectively. Indicate by arrows the direction in which the newly made DNA strands (indicated by dark lines) were synthesized. Number the Okazaki fragments on each strand 1, 2, and 3 in the order in which they were synthesized. Indicate where the most recent DNA synthesis has occurred (use S). Indicate the direction of movement of the replication forks with arrows.
6-6
The lagging strand is synthesized discontinuously at the replication fork because (a) the lagging strand template is discontinuous. (b) DNA polymerase always falls off the template DNA every ten nucleotides or so. (c) DNA polymerase can polymerize nucleotides only in the 5′-to-3′ direction. (d) DNA polymerase removes the last few nucleotides synthesized whenever it stops. (e) None of the above
6-7
Is the following statement TRUE or FALSE? When bidirectional replication forks from adjacent origins meet, a leading strand always runs into a lagging strand. Explain your answer with a diagram illustrating sequential snapshots of the meeting of two adjacent replication forks.
6-8
Which one of the following statements about the newly synthesized strand of a human chromosome is correct? (a) It was synthesized from a single origin solely by continuous DNA synthesis. (b) It was synthesized from a single origin by a mixture of continuous and discontinuous DNA synthesis. (c) It was synthesized from multiple origins solely by discontinuous DNA synthesis. (d) It was synthesized from multiple origins by a mixture of continuous and discontinuous DNA synthesis. (e) It was synthesized from multiple origins by either continuous or discontinuous DNA synthesis, depending on which specific daughter chromosome is being examined. 86
6-9
You have discovered an “Exo–” mutant form of DNA polymerase in which the 3′-to-5′ exonuclease function has been destroyed but the ability to join nucleotides together is unchanged. Which of the following properties do you expect the mutant polymerase to have? (a) It will polymerize in both the 5′-to-3′ direction and the 3′-to-5′ direction. (b) It will polymerize more slowly than the normal Exo+ polymerase. (c) To replicate the same amount of DNA, it will hydrolyze fewer deoxyribonucleotides than will the normal Exo+ polymerase. (d) It will fall off the template more frequently than the normal Exo+ polymerase. (e) It will introduce fewer mutations into new strands than the normal Exo+ polymerase.
6-10
Replication of DNA requires a primer to initiate DNA synthesis because (a) DNA polymerase can add its first nucleotide only to an RNA chain. (b) DNA polymerase can add a nucleotide only to a base-paired nucleotide with a free 3′ end. (c) DNA polymerase can polymerize nucleotides only in the 5′-to-3′ direction. (d) DNA polymerase can polymerize DNA only in short fragments. (e) DNA polymerase has a 3′-to-5′ exonuclease activity.
6-11
Indicate whether each of the following statements is correct or incorrect. Explain your answers. (a) Primase is less accurate than DNA polymerase at copying a DNA template. (b) The RNA primer remains as a permanent part of the new DNA molecule. (c) Replication of the leading strand does not require primase. (d) Longer primers are required to synthesize longer DNA fragments. (e) Primase can join ribonucleotides to create an RNA strand, using a single-stranded DNA as a template, without the need for its own primer.
6-12
A.
B.
You are studying a strain of bacteria that carries a temperature-sensitive mutation in one of the genes required for DNA replication. The bacteria grow normally at the lower temperature, but when the temperature is raised they die. When you analyze the remains of the bacterial cells grown at the higher temperature you find evidence of partly replicated DNA. When the strands of this DNA are separated by heating, numerous single-stranded DNA molecules around 1000 nucleotides long are found. Which of the proteins listed below are most likely to be impaired in these mutant bacteria? Explain your answer. Next to the proteins listed below, write the number (1, 2, 3, and so on) that corresponds to the order in which the proteins function during the synthesis of a new stretch of DNA. DNA ligase DNA polymerase Helicase Initiator proteins
Primase Repair polymerase RNA nuclease Single-stranded binding protein
87
6-13
Which of the following proteins or protein complexes are most abundant near the replication fork? Why? (a) Single-strand binding protein (b) Sliding clamp (c) DNA polymerase (d) Helicase (e) Primase
6-14
A molecule of bacterial DNA introduced into a yeast cell is imported into the nucleus but fails to replicate. Where do you think the block to replication arises? Choose the protein or protein complex below that is most likely responsible for the failure to replicate bacterial DNA. Give an explanation for your answer. (a) Primase (b) Helicase (c) DNA polymerase (d) Sliding clamp protein (e) Initiator proteins
6-15
Indicate whether the following statements about plasmids are TRUE or FALSE. (a) Replication of plasmid DNA is independent of a replication origin. (b) Maintenance of plasmid over the course of several cell divisions requires telomerase activity. (c) DNA replication in plasmids is bidirectional. (d) Plasmids are experimentally useful for introducing specific DNA sequences into yeasts and bacteria. (e) Plasmids were used to identify the replication origins from human DNA.
6-16
Most cells in the body of an adult human lack the telomerase enzyme because its gene is turned off and thus not expressed. An important step in the conversion of a normal cell into a cancer cell, which circumvents normal growth control, is the resumption of telomerase expression. Explain why telomerase might be necessary for the ability of cancer cells to divide over and over again.
DNA Repair 6-17
A pregnant mouse is exposed to high levels of a chemical. Many of the mice in her litter are deformed, but when they are interbred with each other, all their offspring are normal. Which TWO of the following statements could explain these results? (a) In the deformed mice, somatic cells but not germ cells were mutated. (b) The original mouse’s germ cells were mutated. (c) In the deformed mice, germ cells but not somatic cells were mutated. (d) The toxic chemical affects development but is not mutagenic. (e) The original mouse was defective in DNA repair.
88
6-18
During DNA replication in a bacterium, a C is accidentally incorporated instead of an A into one newly synthesized DNA strand. Imagine this error was not corrected and has no effect on the ability of the progeny to grow and reproduce. A. After this original bacterium divides once, what proportion of its progeny would you expect to contain the mutation? B. What proportion of its progeny would you expect to contain the mutation after three more rounds of DNA replication and cell division?
6-19
Mismatch repair of DNA (a) is carried out solely by the replicating DNA polymerase. (b) involves cleavage of the DNA backbone to excise a stretch of single-stranded DNA containing a mispaired base. (c) preferentially repairs the leading strand to match the lagging strand. (d) makes replication 100,000 times more accurate. (e) can occur only on newly replicated DNA.
6-20
Which of the following DNA repair processes accurately restores genetic information only immediately after the DNA has been replicated? Explain your answer. (a) Repair of deamination (b) Repair of depurination (c) Mismatch repair (d) Repair of pyrimidine dimers
6-21
Several members of the same family were diagnosed with the same kind of cancer when they were unusually young. Which one of the following is the most likely explanation for this phenomenon? Possibly, the individuals with the cancer have (a) inherited a cancer-causing gene that was mutated in an ancestor’s somatic cells. (b) inherited a mutation in a gene required for DNA synthesis. (c) inherited a mutation in a gene required for mismatch repair. (d) inherited a mutation in a gene required for the synthesis of purine nucleotides. (e) independently accumulated multiple random mutations over a period of years leading to cancer.
6-22
If uncorrected, deamination of cytosine in DNA is most likely to lead to (a) substitution of an AT base pair for a CG base pair. (b) deletion of the altered CG base pair from the DNA. (c) conversion of the DNA into RNA. (d) generation of a thymine dimer. (e) None of the above.
6-23
Which of the following compounds is likely to be the most mutagenic? (a) One that depurinates DNA. (b) One that replaces adenine with guanine during DNA replication. (c) One that nicks the sugar-phosphate backbone. (d) One that causes thymidine dimers. (e) One that crosslinks together the two strands of the double helix. 89
6-24
You have made a collection of mutant fruit flies that are defective in various aspects of DNA repair. You test each mutant for its hypersensitivity to three DNA-damaging agents: sunlight, nitrous acid (which causes deamination of cytosine), and formic acid (which causes depurination). The results are summarized in Figure Q6-24, where a “yes” indicates that the mutant is more sensitive than a normal fly and blanks indicate normal sensitivity.
Figure Q6-24 A. B. 6-25
Which mutant is most likely to be defective in the DNA repair polymerase? What aspect of repair is most likely to be affected in the other mutants?
You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons. You notice that the coding sequence that actually directs the sequence of amino acids in the enzyme is very similar in the two organisms but that the surrounding sequences vary quite a bit. What is the most likely explanation for this? (a) Coding sequences are repaired more efficiently. (b) Coding sequences are replicated more accurately. (c) Coding sequences are packaged more tightly in the chromosomes to protect them from DNA damage. (d) Mutations in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences. (e) DNA repair enzymes preferentially repair the newly replicated strand to match the original strand.
90
DNA Recombination 6-26
Homologous recombination is initiated by double-strand breaks (DSBs) in a chromosome. DSBs arise from DNA damage caused by harmful chemicals or by radiation (for example, x-rays). During meiosis, the specialized cell division that produces gametes (sperm and eggs) for sexual reproduction, the cells intentionally cause DSBs in order to stimulate crossover homologous recombination. If there is not at least one occurrence of crossing-over within each pair of homologous chromosomes during meiosis, those non-crossover chromosomes will segregate randomly during division.
Figure Q6-26 A.
B.
C.
6-27
Consider the copy of chromosome 3 that you received from your mother. Is it identical to the chromosome 3 that she received from her mother (her maternal chromosome) or identical to the chromosome 3 she received from her father (her paternal chromosome) or neither? Explain. Starting with the representation in Figure Q6-26 of the double stranded maternal and paternal chromosomes found in your mother, draw two possible chromosomes you may have received from your mother. What does this indicate about your resemblance to your grandfather and grandmother?
Identify each statement below as TRUE or FALSE. Explain your answers. A. Site-specific recombination can repair sites of damaged DNA. B. Mobile genetic elements comprise nearly half of the human genome. C. Viruses probably evolved from intracellular mobile genetic elements. D. Genes that contain instructions for making motor proteins are called mobile genetic elements. E. Homologous recombination results in the accumulation of mobile genetic elements.
91
6-28
Which of the following DNA sequences are commonly carried on mobile genetic elements? You may choose more than one option. (a) Transposase gene (b) Holliday junction (c) Recognition site for transposase (d) Antibiotic resistance gene (e) Replication origin
6-29
Retrotransposons (a) are found only in eucaryotes. (b) can move by either the cut-and-paste mechanism or by a mechanism requiring an RNA intermediate. (c) include the transposable elements LINE-1, Alu, and Tn10. (d) can move only if they encode a reverse transcriptase. (e) were multiplied to high copy numbers in a common ancestor of all mammals.
6-30
Describe the likely consequence of introducing high levels of reverse transcriptase into a human embryo.
6-31
Some retrotransposons and retroviruses integrate preferentially into regions of the chromosome that are (a) packaged in euchromatin and (b) located outside the coding regions of genes that contain information for making a protein. Why might these mobile genetic elements have evolved this strategy?
6-32
All viruses (a) have single-stranded genomes. (b) lyse the cells they infect. (c) encode all of the enzymes needed to replicate themselves. (d) contain both nucleic acid and protein. (e) have the same size genomes.
6-33
The enzymes reverse transcriptase and DNA polymerase both synthesize DNA. Which of the following statements about them are TRUE? (a) Reverse transcriptase uses only an RNA template. DNA polymerase uses only a DNA template. (b) Reverse transcriptase can use either an RNA or a DNA template. DNA polymerase uses only a DNA template. (c) DNA polymerase is used only by cells. Reverse transcriptase is used only by viruses. (d) DNA polymerase uses deoxynucleotides. Reverse transcriptase uses ribonucleotides. (e) Reverse transcriptase and DNA polymerase both contain a 3′-to-5′ exonuclease activity.
92
6-34
Once a retrovirus has integrated into the genome of a host cell, why would it not be possible to eradicate the virus by treating the infected cell with reverse transcriptase inhibitors?
6-35
Why do retroviruses need to package reverse transcriptase molecules into their virus particles even though they carry the gene for reverse transcriptase in their genomes?
93
Answers 6-1
Choice (c) is the answer. Choices (a) and (b) are false. Although choices (d) and (e) are correct statements, they are not the reasons that DNA replication is called semiconservative.
6-2
Choice (d) is the answer. Bacteria have one origin of replication and Drosophila has many. Choice (a) is incorrect because the Drosophila genome is bigger than the E. coli genome. Choice (b) is incorrect, as eucaryotic polymerases are not faster than procaryotic polymerases. Choices (c) and (e) are technically correct statements, but are not relevant to the question.
6-3
A. B.
The 3′ end. DNA polymerase can add nucleotides only to the 3′-OH end of a nucleic acid chain. Both, due to the bidirectional nature of chromosomal replication.
6-4
(c)
The antiparallel nature of the strands of normal DNA, combined with the 5′-to-3′ activity of the DNA polymerase, precludes the continuous synthesis of both strands and thus requires that the replication fork be asymmetrical. If the strands were parallel, both new strands could be synthesized continuously.
6-5
See Figure A6-5.
Figure A6-5 6-6
(a) (b) (c) (d) (e)
False False True False False
94
6-7
True. See Figure A6-7 for an illustration of the meeting of two adjacent replication forks.
Figure A6-7 6-8
(d)
Each newly synthesized strand in a daughter duplex was synthesized by a mixture of continuous and discontinuous DNA synthesis from multiple origins. Consider a single replication origin: The fork moving in one direction synthesizes a daughter strand continuously as part of leading-strand synthesis; the fork moving in the opposite direction synthesizes a portion of the same daughter strand discontinuously as part of lagging-strand synthesis.
6-9
Choice (c) is the answer. An Exo— polymerase will be unable to proofread and thus will hydrolyze fewer nucleotides than one that can proofread, because it cannot remove an incorrect nucleotide and try again to add the correct nucleotide (thus choice (e) is false). Choice (a) is unlikely because it postulates an entirely new function for the defective polymerase. Choice (b) is incorrect because if the rate of polymerization changes, it would become faster not slower, as a consequence of the Exo— polymerase lacking a proofreading function that might cause it to stop frequently to check its products and remedy errors. Choice (d) is unlikely, because the ability to stay on the template is due to the association of polymerase with additional proteins.
6-10
Choice (b) is the answer. Choices (a) and (d) are false statements. Choices (c) and (e) are true but are not the reason DNA polymerase requires a primer.
95
6-11
(a) (b)
(c)
(d) (e)
6-12
A.
B.
Correct. Primase lacks a 3′-to-5′ exonuclease activity and thus cannot proofread the nucleotide chain it makes. Incorrect. A nuclease removes the RNA primer after it provides its 3′-OH to prime the synthesis of DNA. Then enzymes involved in repair synthesis and ligation join the newly synthesized stretches of DNA to make a continuous strand containing only DNA. Incorrect. The leading strand requires primase to initiate DNA synthesis, although fewer RNA primers will be needed on the leading strand than the lagging strand. Incorrect. The length of the primer does not correlate with the length of the DNA strand that is synthesized. Correct. Like other RNA polymerases, primase does not require a primer to initiate synthesis of the nucleotide chain. The 1000-nucleotide fragments that accumulate in the mutant are likely to be Okazaki fragments. Thus, the mutant is likely to be defective in the function of RNA nuclease, repair polymerase, or DNA ligase. These proteins are required to remove the RNA primer, fill in the gap, and stitch together the Okazaki fragments, respectively. We could test which enzyme was defective by determining if the fragments contained a short stretch of RNA at the 5′ end (nuclease defective) or if the fragments annealed to the template leaving gaps (repair polymerase defective). DNA ligase—8 DNA polymerase—5 Helicase—2 Initiator proteins—1 Primase—4 Repair polymerase—7 RNA nuclease—6 Single-stranded binding protein—3
6-13
(a)
Single-strand binding protein is required to coat all single-stranded regions of DNA that form at a replication fork; this requires many molecules of single-strand binding protein at each fork. Only one or two molecules of each of the other proteins or protein complexes are required at each replication fork.
6-14
Choice (e) is the answer. DNA from all organisms is chemically identical except for the sequence of nucleotides. The proteins listed in choices (a) through (d) can act on any DNA regardless of its sequence. In contrast, the initiator proteins recognize specific DNA sequences at the origins of replication. These sequences differ between bacteria and yeast.
96
6-15
(a) (b) (c) (d) (e)
6-16
In the absence of telomerase, the lifespan of a cell and its progeny cells is limited. With each round of DNA replication, the length of telomeric DNA will shrink, until finally all the telomeric DNA will disappear. Without telomeres capping the chromosome ends, the ends might be treated like breaks arising from DNA damage or crucial genetic information might be lost. Cells whose DNA lacks telomeres will stop dividing or die. However, if telomerase is provided to cells, they may be able to divide indefinitely because their telomeres will remain a constant length despite repeated rounds of DNA replication.
6-17
Choice (a) or (d) is correct. Choice (b) cannot account for these results since a mutation in the original mouse’s germ cells would have no effect on the fetuses she was already carrying. Neither can choice (c), as mutations in the germ cells of the fetuses while in utero would have had no effect on their development, but might have led to mutant mice among their offspring. If the original mouse were defective in DNA repair (choice (e)), this would increase the number of mutations caused by the chemical but cannot explain the observed effects on her offspring and their offspring.
6-18
A.
B.
6-19
False False True True False
Half or 50%. DNA replication in the original bacterium will create two new DNA molecules, one of which will now carry a mismatched C-T base pair. So one daughter cell of that cell division will carry a completely normal DNA molecule; the other cell will have the molecule with the mutation mispaired to a correct nucleotide. A quarter or 25%. At the next round of DNA replication and cell division, the bacterium carrying the mismatched C-T will produce and pass on one normal DNA molecule from the undamaged strand containing the T and one mutant DNA molecule with a fully mutant C-G base pair. So at this stage, one out of the four progeny of the original bacterium is mutant. Subsequent cell divisions of these mutant bacteria will give rise only to mutant bacteria, while the other bacteria will give rise to normal bacteria. The proportion of progeny containing the mutation will, therefore, remain at 25%.
Choice (b) is the answer. Choice (a) is incorrect, as mismatch repair requires specialized repair proteins that act after the DNA is replicated or at other times during the life of a cell (thus choice (e) is incorrect). Choice (c) is incorrect because mismatch repair repairs the newly replicated leading strand or the lagging strand to match its template strand. Choice (d) is incorrect, since mismatch repair makes replication approximately 100 (not 100,000) times more accurate.
97
6-20
(c)
6-21
Choice (c) is the answer. In fact, affected individuals in some families with a history of early-onset colon cancer have been found to carry mutations in mismatch repair genes. Mutations arising in somatic cells are not inherited (choice (a) is incorrect). A defect in DNA synthesis or nucleotide biosynthesis would likely be lethal (choices (b) and (d) are incorrect). Choice (e) describes the way that most cancers arise in people late in life, but it is extremely unlikely that several individuals in the same family spontaneously acquired similar random mutations leading to the early onset of the same kind of cancer.
6-22
(a)
6-23
Choice (e) is the answer. Altogether, the DNA repair pathways described in Chapter 6 can easily repair the DNA damage described in options (a)–(d), these types of damage occur on one strand of the double helix, and thus can be repaired using the intact genetic information encoded on the complementary strand. When the two strands are crosslinked and thus both strands are damaged, entirely different and probably less accurate pathways of repair are required.
6-24
A.
B.
Mismatches occur most often as a result of replication errors, in which case the erroneous base is found only on the newly synthesized strand. In most eucaryotic cells, the preferential repair of the erroneous base instead of its pairing partner is thought to require recognition of nicks in the sugar-phosphate backbone of the newly synthesized DNA strand. These nicks are sealed soon after replication, so mismatch repair must occur in the short interval between passage of the replication fork and sealing of the sugar-phosphate backbone in order to accurately restore the original sequence. When mismatch repair occurs at other times, it has an equal probability of restoring the original information or setting the mutation. For the other kinds of DNA damage (U-G mismatch, abasic site, pyrimidine dimer) it is always evident to determine which strand is damaged and which strand contains reliable original information.
Mr. Self-Destruct is more likely than the other mutants to be defective in the DNA repair polymerase because Mr. Self-Destruct is defective in repair of all three kinds of DNA damage. The repair pathways for all three kinds of damage are similar in the later steps, including a requirement for the DNA repair polymerase. The other mutants are specific for a particular type of damage. Thus the mutations are likely to be in genes required for the first stage of repair, the recognition and excision of the damaged bases. Dracula and Mole are likely to be defective in the recognition or excision of thymidine dimers; Faust is likely to be defective in the recognition or excision of U-G mismatched base pairs; and Marguerite is likely to be defective in the recognition or excision of abasic sites.
98
6-25
Choice(d) is the answer. Mutations—whether they arise by mistakes in replication or by damage to the DNA that remains unrepaired—tend to hit the DNA fairly randomly (choices (a) and (b) are false). However, if a mutation occurs in a protein-coding sequence rather than in the surrounding DNA, it is more likely to cause a deleterious change that kills or impairs the organism and thereby decreases the likelihood that the mutation will be passed on to future generations. Since skinks and Komodo dragons share a common lizard ancestor, differences in their genomes have arisen during their divergence from this ancestor. Mutations in noncoding sequences are more likely to have no effect on the functioning of the organism and thus frequently get passed along to progeny. Choice (c) is incorrect, as genes that are being expressed tend to be more loosely packaged than the noncoding DNA. Choice (e) is true, but has no bearing on the phenomenon described.
6-26
A. B.
Neither. The copy of chromosome 3 you received from your mother is a hybrid of the ones she received from her mother and her father. See Figure A6-26. The right answers include any chromosome in which a portion matches the information from the paternal chromosome and the remainder matches the information from the maternal chromosome.
Figure A6-26
6-27
C.
Due to extensive crossing over, you resemble both your grandmother and your grandfather. If there were no crossing over, then you might have a much stronger resemblance to one than the other.
A.
False. Homologous recombination, not site-specific recombination, is sometimes used to repair sites of damaged DNA. Site-specific recombination is used mostly in the mobilization of mobile genetic elements. True. Forty-five percent of the human genome is comprised of mobile genetic elements. True. A good guess for how viruses evolved is that some mobile genetic elements acquired genes encoding coat proteins and other proteins required for packaging and cellular escape of the nucleic acids of mobile genetic elements. False. Cellular motor proteins are completely unrelated to mobile genetic elements. False. Site-specific recombination, not homologous recombination, is the primary mechanism for the accumulation of mobile genetic elements. Homologous recombination does sometimes aid in the repair of DNA damage caused by the excision of a mobile genetic element from the chromosome, but it does not aid in the insertion of the element into a new location.
B. C.
D. E.
99
6-28
Choices (a), (c), and (d) are correct. A Holliday junction is not a sequence, but a structural intermediate in homologous recombination (choice (b) is false). Mobile genetic elements often have no replication origin, since their movement and proliferation occurs by a type of repair DNA synthesis and their replication during cell division occurs by passage of a replication fork that originates elsewhere in the chromosome (choice (e) is false).
6-29
Choice (a) is the answer. Retrotransposons are found only in eucaryotes. Retrotransposons (by definition) move only through an RNA intermediate (choice (b) is false), and include LINE-1 and Alu sequences (but not the bacterial transposon Tn10, so choice (c) is false). Retrotransposons do not necessarily need to provide their own reverse transcriptase so long as there is an alternative source of reverse transcriptase in the cell (choice (d) is false). The LINE-1 and Alu retrotransposons that comprise about a third of the human genome are not identical in sequence or location to the retrotransposons in other mammals like mice (choice (e) is false).
6-30
The embryo would probably have severe developmental abnormalities or die. The huge numbers of retrotransposons littering the human genome are largely immobile due to the accumulation of disabling mutations. However, it is likely that at least a few of the millions of copies of the transposons still contain the sites necessary for retrotransposition, although they do not encode a functional reverse transcriptase. High levels of reverse transcriptase will probably cause many retrotransposition events. The resultant insertion of retrotransposons is likely to disable genes required for development or survival.
6-31
The most evolutionarily successful mobile genetic elements are those that are best at reproducing themselves. In order to increase the number of copies of a particular element, the element must meet two criteria: (1) it must not kill its host and (2) it must maximize its ability to continue reproducing. If an element inserts into the coding region of a gene, it might disable the gene and thereby confer a selective disadvantage in the reproduction or survival of its host. Thus, elements that devised a way to avoid insertion into coding regions probably were better able to increase their copy number throughout the human population. If an element inserts into a heterochromatic region of a chromosome, its genes may not be expressed and therefore it may become immobile. Elements that devised a way to direct insertion into euchromatin would be more likely to maintain mobility and thereby increase their copy number over time.
6-32
Choice (d) is the answer. All viruses contain both protein and nucleic acid. Viruses can have either double- or single-stranded genomes (choice (a)). Not all viruses lyse the cells they infect (choice (b)); for example, some bud out of the cell without killing it. Viruses can have as few as three genes or more than a hundred (choice (e)). No virus is able to replicate in the absence of a host cell (choice (c)).
100
6-33
Choice (b) is the answer. Reverse transcriptase can use an RNA or DNA template (thus choice (a) is false). Choice (c) is false because DNA polymerase is often used by DNA viruses. Choice (d) is false because both enzymes polymerize deoxynucleotides. Choice (e) is false because reverse transcriptase does not have proofreading exonuclease activity.
6-34
Once the virus has integrated into the genome, it has no further need for reverse transcriptase. Therefore, an inhibitor of reverse transcriptase may be able to block infection of other cells by viruses that bud off the infected cell, but it will not be able to eradicate the integrated virus. If the virus integrates into the genome of a cell with the potential to divide, it will be faithfully propagated along with all the genomic DNA to all progeny of that cell.
6-35
Retroviruses carry their own reverse transcriptase with them, as they must produce a double-stranded DNA copy of their genome before their genes can be transcribed and expressed.
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CHAPTER 7 FROM DNA TO PROTEIN: HOW CELLS READ THE GENOME 2009 Garland Science Publishing 3rd Edition
From DNA to RNA 7-1 RNA in cells differs from DNA in that (a) it contains the base uracil, which pairs with cytosine. (b) it is single-stranded and cannot form base pairs. (c) it is single-stranded and can fold up into a variety of structures. (d) the nucleotides are linked together in a different way. (e) the sugar ribose contains fewer oxygen atoms than does deoxyribose. 7-2 Transcription is similar to DNA replication in that (a) it requires a molecule of DNA helicase to unwind the DNA. (b) it uses the same enzyme as that used to synthesize RNA primers during DNA replication. (c) the newly synthesized RNA remains paired to the template DNA. (d) nucleotide polymerization occurs only in the 5’ -to-3’ direction. (e) an RNA transcript is synthesized discontinuously and the pieces then joined together. 7-3 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. In order for a cell’s genetic material to be utilized, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called __________________. Various kinds of RNAs are produced, each with different functions. __________________ molecules code for proteins, __________________ molecules act as adaptors for protein synthesis, __________________ molecules are integral components of the ribosome, while __________________ molecules are important for splicing of RNA transcripts. Incorporation mRNA pRNA translation
rRNA snRNA transcription
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transmembrane tRNA proteins
7-4
Match the following structures with their names:
Figure Q7-4 7-5
Imagine that an RNA polymerase is transcribing a segment of DNA that contains the sequence: 5′- AGTCTAGGCACTGA -3′ 3′- TCAGATCCGTGACT -5′ A. B.
If the polymerase is transcribing from this segment of DNA from left to right, which strand (top or bottom) is the template? What will be the sequence of that RNA (be sure to label the 5′ and 3′ ends of your RNA molecule)?
7-6
The sigma subunit of bacterial RNA polymerase (a) contains the catalytic activity of the polymerase. (b) remains part of the polymerase throughout transcription. (c) recognizes promoter sites in the DNA. (d) recognizes transcription termination sites in the DNA.
7-7
Which of the following might decrease the transcription of only one specific gene in a bacterial cell? (a) A decrease in the amount of sigma factor (b) A decrease in the amount of RNA polymerase (c) A mutation that introduced a stop codon into the DNA preceding the coding sequence of the gene (d) A mutation that introduced extensive sequence changes into the DNA preceding the transcription start site of the gene (e) A mutation that moved the transcription termination signal of the gene farther away from the transcription start site
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7-8
From the list below, pick THREE reasons why the primase that is used to make the RNA primer for DNA replication would not be suitable for gene transcription? (a) Primase initiates RNA synthesis on a single-stranded DNA template. (b) Primase can initiate RNA synthesis without the need for a base-paired primer. (c) Primase synthesizes only RNAs of around 5 to 20 nucleotides in length. (d) The RNA synthesized by primase remains base-paired to the DNA template. (e) Primase uses nucleotide triphosphates.
7-9
Indicate where the following processes take place by adding numbered labeling lines to the schematic diagram of a eucaryotic cell in Figure Q7-9.
Figure Q7-9 1. 2. 3. 4. 5.
Transcription Translation RNA splicing Polyadenylation RNA capping
7-10
Total nucleic acids are extracted from a culture of yeast cells and are then mixed with resin beads to which the polynucleotide 5′-TTTTTTTTTTTTTTTTTTTTTTTTT-3′ has been covalently attached. After a short incubation, the beads are then extracted from the mixture. When you analyze the cellular nucleic acids that have stuck to the beads, which of the following will be most abundant? (a) DNA (b) tRNA (c) rRNA (d) mRNA (e) Primary transcript RNA
7-11
Name three modifications that can be made to an RNA molecule in eucaryotic cells before the RNA molecule becomes a mature mRNA. 105
7-12
The length of a particular gene in human DNA, measured from the start site for transcription to the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the mRNA produced from this gene is 4000 nucleotides. What is the most likely reason for this discrepancy?
7-13
A fragment of human DNA containing the gene for a protein hormone with its regulatory regions removed is introduced into bacteria; although it is transcribed at a high level into RNA, no protein is made. When this RNA is extracted from the bacteria, mixed with human mRNA encoding the same hormone, and then examined in the electron microscope, you see the following structure (Figure Q7-13). Label each of the statements below as either “consistent” or “inconsistent” with your results and explain your reasoning.
Figure Q7-13 A. B. C. D.
The human DNA was inserted in the bacterial DNA next to a bacterial promoter and in its normal orientation. The human DNA was inserted in the bacterial DNA next to a bacterial promoter but in an orientation opposite to normal. The human DNA contained an intron. The human DNA acquired a deletion while in the bacterium.
7-14
Why is the old dogma “one gene—one protein” not always true for eucaryotic genes?
7-15
Is this statement TRUE or FALSE? Explain your answer. “Since introns do not contain protein coding information, they do not have to be removed precisely (meaning, a nucleotide here and there should not matter) from the primary transcript during RNA splicing.”
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7-16
You have discovered a gene (see Figure Q7-16 part A) that is alternatively spliced to produce several forms of mRNA in various cell types, three of which are shown in part B of Figure Q7-16. (Note that splicing is indicated by lines connecting the exons that are included in the mRNA). Your experiments have found that protein translation begins in exon 1. For all forms of the mRNA, the encoded protein sequence is the same in the regions of the mRNA that correspond to exons 1 and 10. Exons 2 and 3 are alternative exons used in different mRNA, as are exons 7 and 8. Which of the following statements about exons 2 and 3 is the most accurate? Explain your answer.
Figure Q7-16 (a) (b) (c)
Exons 2 and 3 must have the same number of nucleotides. Exons 2 and 3 must contain an integral number of codons (that is, the number of nucleotides divided by 3 must be an integer). Exons 2 and 3 must contain a number of nucleotides that when divided by 3, leaves the same remainder (that is, 0, 1, or 2).
From RNA to Protein 7-17
Which of the following statements about the genetic code are correct? (a) All codons specify more than one amino acid. (b) The genetic code is redundant. (c) All amino acids are specified by more than one codon. (d) The genetic code is different in procaryotes and eucaryotes. (e) All codons specify an amino acid.
107
NOTE: The following codon table is to be used for Problems Q7-18 – 7-23, Q7-29, and 7-33.
7-18
7-19
The following DNA sequence includes the beginning of a sequence coding for a protein. What would be the result of a mutation that changed the C marked by an asterisk to an A? 5′- AGGCTATGAATGGACACTGCGAGCCC.... * Which amino acid would you expect a tRNA with the anticodon 5′-CUU-3′ to carry? (Refer to Codon table provided above Q7-18.) (a) (b) (c) (d) (e)
7-20
Which of the following pairs of codons might you expect to be read by the same tRNA as a result of wobble? (Refer to Codon table provided above Q7-18.) (a) (b) (c) (d) (e)
7-21
Lysine Glutamate Glutamine Leucine Phenylalanine
CUU and UUU GAU and GAA CAC and CAU AAU and AGU CCU and GCU
Below is a segment of RNA from the middle of an mRNA. If you were told that this segment of RNA was part of the coding region of an mRNA for a large protein, give the amino acid sequence for the protein that is encoded by this segment of mRNA. (Refer to Codon table provided above Q7-18.) 5′- UAGUCUAGGCACUGA -3′
108
7-22
(Refer to Codon table provided above Q7-18.) One strand of a section of DNA isolated from the bacterium E. coli reads: 5′- GTAGCCTACCCATAGG -3′ A.
B. C. 7-23
Suppose that an mRNA is transcribed from this DNA using the complementary strand as a template. What will be the sequence of the mRNA in this region (make sure you label the 5′ and 3′ ends of the mRNA)? How many different peptides could potentially be made from this sequence of RNA, assuming translation initiates upstream of this sequence? What are these peptides? (Give your answer using the one letter amino acid code.)
A strain of yeast translates mRNA into protein with a high level of inaccuracy. Individual molecules of a particular protein isolated from this yeast have the following variations in the first 11 amino acids compared with the sequence of the same protein isolated from normal yeast cells (Figure Q7-23). What is the most likely cause of this variation in protein sequence? Explain your answer. (Refer to Codon table provided above Q7-18.)
Figure Q7-23 (a) (b) (c) (d) (e)
7-24
A mutation in the DNA coding for the protein A mutation in the anticodon of the isoleucine tRNA (tRNAIle) A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different amino acids A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different tRNA molecules A mutation in a component of the ribosome that allows binding of incorrect tRNA molecules to the A-site
Which of the following statements is TRUE? (a) Ribosomes are large RNA structures composed solely of rRNA. (b) Ribosomes are synthesized entirely in the cytoplasm. (c) rRNA contains the catalytic activity that joins amino acids together. (d) A ribosome consists of two equally sized subunits. (e) A ribosome binds one tRNA at a time.
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7-25
Figure Q7-25A shows the stage in translation when an incoming aminoacyl-tRNA has bound to the A-site on the ribosome. Using the components shown in Figure Q7-25A as a guide, show on Figure Q7-25B and Q7-25C what happens in the next two stages to complete the addition of the new amino acid to the growing polypeptide chain.
Figure Q7-25 7-26
A poison added to an in vitro translation mixture containing mRNA molecules with the sequence 5′-AUGAAAAAAAAAAAAUAA-3′ has the following effect: the only product made is a Met-Lys dipeptide that remains attached to the ribosome. What is the most likely way in which the poison acts to inhibit protein synthesis? (a) It inhibits binding of the small subunit of the ribosome to mRNA. (b) It inhibits peptidyl transferase activity. (c) It inhibits movement of the small subunit relative to the large subunit. (d) It inhibits release factor. (e) It mimics release factor.
110
7-27
In eucaryotes, but not procaryotes, ribosomes find the start site of translation by (a) binding directly to a ribosome-binding site preceding the initiation codon. (b) scanning along the mRNA from the 5′ end. (c) recognizing an AUG codon as the start of translation. (d) binding an initiator tRNA.
7-28
Figure Q7-28 shows an mRNA molecule.
Figure Q7-28 A.
B. 7-29
Match the labels given in the list below with the label lines in Figure Q7-28. (a) ribosome-binding site (b) initiator codon (c) stop codon (d) untranslated 3′ region (e) untranslated 5′ region (f) protein-coding region Is the mRNA shown procaryotic or eucaryotic? Explain your answer.
A tRNA for the amino acid lysine is mutated such that the sequence of the anticodon is 5′-UAU-3′ (instead of 5′-UUU-3′). Which of the following aberrations in protein synthesis might this tRNA cause? (Refer to Codon table provided above Q7-18.) (a) (b) (c) (d) (e)
Read through of stop codons Substitution of lysine for isoleucine Substitution of lysine for tyrosine Substitution of lysine for phenylalanine Substitution of lysine for the amino-terminal methionine
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7-30
You have discovered a protein that inhibits translation. When you add this inhibitor to a mixture capable of translating human mRNA and centrifuge the mixture to separate polyribosomes and single ribosomes, you obtain the results shown in Figure Q7-30. Which of the following interpretations are consistent with these observations?
Figure Q7-30 (a) (b) (c) (d)
7-31
The protein binds to the small ribosomal subunit and increases the rate of initiation of translation. The protein binds to sequences in the 5′ region of the mRNA and inhibits the rate of initiation of translation. The protein binds to the large ribosomal subunit and slows down elongation of the polypeptide chain. The protein binds to sequences in the 3′ region of the mRNA and prevents termination of translation.
The concentration of a particular protein X in a normal human cell rises gradually from a low point, immediately after cell division, to a high point, just before cell division, and then drops sharply. The level of its mRNA in the cell remains fairly constant throughout this time. Protein X is required for cell growth and survival, but the drop in its level just before cell division is essential for division to proceed. You have isolated a line of human cells that grow in size in culture but cannot divide, and on analyzing these mutants, you find that levels of X mRNA in the mutant cells are normal. Which of the following mutations in the gene for X could explain these results? (a) The introduction of a stop codon that truncates protein X at the fourth amino acid. (b) A change of the first ATG codon to CCA. (c) The deletion of a sequence that encodes sites at which ubiquitin can be attached to the protein. (d) A change at a splice site that prevents splicing of the RNA.
112
7-32
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed of two subunits: the __________________ subunit, which catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain, and the __________________ subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the __________________ -site of the ribosome. An incoming aminoacyltRNA carrying the next amino acid in the chain will bind to the __________________ -site by forming base pairs with the exposed codon in the mRNA. The __________________ enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the __________________ called release factor. Eventually, most proteins will be degraded by a large complex of proteolytic enzymes called the __________________. A central DNA E large
medium P peptidyl transferase polymerase protein
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proteosome RNA small T ubiquitin
7-33
After treating cells with a mutagen, you isolate two mutants. One carries alanine and the other carries methionine at a site in the protein that normally contains valine. After treating these two mutants again with mutagen, you isolate mutants from each that now carry threonine at the site of the original valine. Assuming that all mutations caused by the mutagen involve single nucleotide changes, deduce the codons that are used for valine, alanine, methionine, and threonine at the affected site. See Figure Q7-33. (Refer to Codon table provided above Q7-18.)
Figure Q7-33
RNA and the Origins of Life 7-34
According to current thinking, the minimum requirement for life to have originated on Earth was the formation of a (a) molecule that could provide a template for the production of a complementary molecule. (b) double-stranded DNA helix. (c) molecule that could direct protein synthesis. (d) molecule that could catalyze its own replication.
7-35
Which of the following reactions are known to be carried out by a ribozyme? (a) DNA synthesis (b) Transcription (c) RNA splicing (d) Protein hydrolysis (e) Polysaccharide hydrolysis
7-36
You are studying a disease that is caused by a virus, but when you purify the virus particles and analyze them you find they contain no trace of DNA. Which of the following molecules are likely to contain the genetic information of the virus? (a) Protein (b) RNA (c) Lipids (d) Carbohydrates
7-37
Give a reason why DNA makes a better material than RNA for storage of genetic information and explain your answer.
114
How We Know: Cracking the Genetic Code 7-38
An extraterrestrial organism (ET) is discovered whose basic cell biology seems pretty much the same as terrestrial organisms except that it uses a different genetic code to translate RNA into protein. You set out to break the code by translation experiments using RNAs of known sequence and cell-free extracts of ET cells to supply the necessary protein-synthesizing machinery. In experiments using the RNAs below, the following results were obtained when the 20 possible amino acids were added either singly or in different combinations of two or three: RNA 1: RNA 2:
5′–GCGCGCGCGCGCGCGCGCGCGCGCGCGC–3′ 5′–GCCGCCGCCGCCGCCGCCGCCGCCGCCGCC–3′
Using RNA 1, a polypeptide was produced only if alanine and valine were added to the reaction mixture. Using RNA 2, a polypeptide was produced only if leucine and serine and cysteine were added to the reaction mixture. Assuming that protein synthesis can start anywhere on the template, that the ET genetic code is nonoverlapping and linear, and that each codon is the same length (like the terrestrial triplet code), how many nucleotides does an ET codon contain? (a) 2 (b) 3 (c) 4 (d) 5 (e) 6 7-39
NASA has discovered an alien life form. You are called in to help them deduce the genetic code for this alien. Surprisingly, this alien life form shares many similarities with life on Earth: this alien uses DNA as its genetic material, makes RNA from DNA, and reads the information from RNA to make protein using ribosomes and tRNAs. Even more amazing, this alien uses the same 20 amino acids, like the organisms found on Earth, and also codes for each amino acid by a triplet codon. However, the scientists at NASA have found that the genetic code used by the alien life form is different than that used by life on Earth. The experiment that allowed the NASA scientists to draw this conclusion involved creating a cell-free protein synthesis system from alien cells and adding an mRNA made entirely of uracil (poly U). This led to the finding that poly U directs the synthesis of a peptide containing only glycine. NASA scientists have synthesized a poly AU mRNA and find that it codes for a polypeptide of alternating serine and proline residues. From these experiments, can you determine which codons code for serine and proline? Explain why or why not. Bonus question: Can you propose a mechanism for how the alien’s physiology is altered so that it uses a different genetic code from life on Earth, despite all the similarities?
115
Answers 7-1
Choice (c) is the correct answer. Choice (a) is untrue, since although RNA contains uracil, uracil pairs with adenine, not cytosine. Choice (b) is false because RNA can form base pairs with a complementary RNA or DNA sequence. Choice(d) is false. Choice (e) is false because ribose contains one more oxygen atom than deoxyribose.
7-2
Choice (d) is the correct answer. RNA polymerase unwinds only a few base pairs of the double helix at a time and does not need a helicase to do so, which is why choice (a) is incorrect. The enzyme used to make primers during DNA synthesis is indeed an RNA polymerase, but it is a special enzyme, primase, and not the enzyme that is used for transcription, which is why choice (b) is incorrect. Choice (c) is false. Choice (e) is incorrect because an RNA transcript is made by a single polymerase molecule that proceeds from the start site to the termination site without falling off.
7-3
In order for a cell’s genetic material to be utilized, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called transcription. Various kinds of RNAs are produced, each with different functions. mRNA molecules code for proteins, tRNA molecules act as adaptors for protein synthesis, rRNA molecules are integral components of the ribosome, while snRNA molecules are important for splicing of RNA transcripts.
7-4
A—4; B—1; C—2; D—3
7-5
A. B.
7-6
(c)
7-7
Choice (d) is the correct answer. Such changes would probably destroy the function of the promoter, making RNA polymerase unable to bind to it. Decreasing the amount of sigma factor or RNA polymerase (choices (a) or (b)) would affect the transcription of most of the genes in the cell, not just one specific gene. Introducing a stop codon before the coding sequence (choice (c)) would have no effect on transcription of the gene, since the transcription machinery does not recognize translational stops. Moving the termination signal farther away (choice (e)) would merely make the transcript longer.
7-8
Choices (a), (c), and (d) are the correct reasons. Choices (b) and (e) are true for both primase and RNA polymerase.
The bottom strand 5′- AGUCUAGGCACUGA -3′
116
7-9
See Figure A7-9.
Figure A7-9 7-10
(d)
mRNA is the only type of RNA that is polyadenylated, and this poly(A) tail would be able to base-pair with the strands of poly(T) on the beads and thus stick to them. DNA would not be found in the sample, as the poly(A) tail is not encoded in the DNA and long runs of T are rare in DNA.
7-11
1. 2. 3.
A poly(A) tail must be added. A 5′ cap must be added. Introns must be spliced out. (“Export from nucleus” is also an acceptable answer.)
7-12
The gene contains one or more introns.
7-13
B, C, and D are consistent with the results; A is inconsistent. B must be true for the RNA produced in the bacterium to be complementary to, and thus able to pair with, the mRNA from a human cell. If the human DNA had become inserted in its normal orientation next to the promoter (A), the corresponding portions of RNA would be identical (or at least very similar) in sequence and thus the two RNAs would not be complementary and would not pair. The loop formed in the hybrid tells us that one of the molecules contains sequences that the other is missing. This could come about either because the bacterial RNA was transcribed from human sequences that acquired a deletion (B) or because the human gene contains an intron (C).
7-14
The transcripts from some genes can be spliced in more than one way to give mRNAs containing different sequences, thus encoding different proteins. A single eucaryotic gene, therefore, may encode more than one protein.
117
7-15
False. Although it is true that the sequences within the introns are mostly dispensable, the introns must still be removed precisely because an error of one or two nucleotides would shift the reading frame of the resulting mRNA molecule and change the protein it encodes.
7-16
Choice (c) is the only answer that must be true for exons 2 and 3. Although choices (a) and (b) could be true, they don’t have to be. Because the protein sequence is the same in segments of the mRNA corresponding to exons 1 and 10, the choice of either exon 2 or exon 3 would not alter the reading frame. To maintain the normal reading frame, whatever it is, the alternative exons must have a number of nucleotides that when divided by 3 (the number of nucleotides in a codon) give the same remainder.
7-17
Choice (b) is the correct answer. The majority of the amino acids can be specified by more than one codon. Choice (a) is incorrect because each codon specifies only one amino acid. Choice (c) is incorrect because tryptophan and methionine are encoded by only one codon. Choice (d) is incorrect because, with a few minor exceptions, the genetic code is the same in all organisms. Choice (e) is incorrect because some codons specify translational stop signals.
7-18
The change creates a stop codon (TGA, or UGA in the mRNA) very near the beginning of the protein-coding sequence and in the correct reading frame (the beginning of the coding sequence is indicated by the ATG). Thus, translation would terminate after only four amino acids had been joined together, and the complete protein would not be made.
7-19
(a)
7-20
Choice (c) is the answer. These two codons differ only in the third position and also encode the same amino acid, which is the definition of wobble. Although the codons GAU and GAA (choice (b)) also differs only in the third position, they are unlikely in normal circumstances to be read by the same tRNA, as they encode different amino acids.
7-21
SLGT is the answer. (Reading frame two is the only reading frame that does not contain a stop codon.)
7-22
A. B.
C.
Lys (lysine). As is conventional for nucleotide sequences, the anticodon is given 5′ to 3′. The complementary base-pairing occurs between antiparallel nucleic acid sequences, and the codon recognized by this anticodon will therefore be 5′-AAG-3′.
5′-GUAGCCUACCCAUAGG -3′ Two. (There are three potential reading frames for each RNA. In this case, they are: GUA GCC UAC CCA UAG ... UAG CCU ACC CAU AGG..... AGC CUA CCC AUA GG?.... The center one cannot be used in this case, because UAG is a stop codon.) VAYP SLPIG Note: PTHR will not be a peptide because it is preceded by a stop codon. 118
7-23
Choice (c) is the correct answer. A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between amino acids would allow an assortment of amino acids to be attached to the tRNAIle. These assorted aminoacyl-tRNAs would then base-pair with the isoleucine codon and cause a variety of substitutions at positions normally occupied by isoleucine. Choice (a) is incorrect because a mutation in the gene encoding the protein would cause only a single variant protein to be made. Choice (e) is incorrect because a mutation in the ribosome that allows binding of any amino-acyltRNA to the A site would cause substitutions all over the protein, not only at isoleucine residues. Choices (b) and (d) are also incorrect. A mutation in the anticodon loop of tRNAIle (choice (b)) or a mutation in the isoleucine-tRNA synthetase that decreases its ability to distinguish between different tRNA molecules (choice (d)) would cause substitution of isoleucine for some other amino acid (which is the opposite of what is observed).
7-24
Choice (c) is the correct answer. Choice (a) is incorrect because ribosomes contain proteins as well as rRNA. Choice (b) is incorrect because rRNA is synthesized in the nucleus, and ribosomes are partly assembled in the nucleus. Choice (d) is incorrect because a ribosome consists of one small subunit and one large subunit. Choice (e) is incorrect because a ribosome must be able to bind two tRNAs at any one time.
7-25
See Figure A7-25.
Figure A7-25 119
7-26
Choice (c) is the correct answer. Either choice (a) or (b) would prevent all peptide bond formation. Choice (d) would have no affect on translation until the stop codon was reached. Choice (e) would be likely to result in a mixture of polypeptides of various lengths; a poison mimicking a release factor could conceivably cause only Met-Lys to be made, but this dipeptide would not remain bound to the ribosome.
7-27
Choice (b) is the correct answer. Choice (a) is true only for procaryotes. Choices (c) and (d) are true for both procaryotes and eucaryotes.
7-28
A. B.
(a)—3; (b)—2; (c)—4; (d)—6;(e)—1; (f)—5 The mRNA is procaryotic. It contains coding regions for more than one protein, as shown by the multiple initiation codons, each preceded by a ribosome-binding site. It contains an unmodified 5′ end, as shown by the three phosphate groups, and an unmodified 3′ end, as shown by the absence of a poly(A) tail.
7-29
(b)
The mutant tRNALys will be able to pair with the codon 5′-AUA-3′, which codes for isoleucine.
7-30
Choice (b) is the correct answer. The results in Figure Q7–30 show a marked decrease in the number of polyribosomes formed relative to normal. Polyribosomes form because the initiation of translation is fairly rapid: ribosomes can bind successively to the free 5′ end of an mRNA molecule and start translation before the first ribosome has had a chance to finish translating the message. Therefore, inhibition of the rate of initiation will tend to decrease the number of ribosomes in the polyribosome, and in the extreme case there will be only one ribosome per mRNA. Conversely, increasing the rate of initiation or slowing the rate of elongation would result in an increased number of ribosomes per polyribosome (up to a maximum point), making choices (a) and (c) false. Choice (d) is incorrect, as preventing termination would prevent release of the ribosomes at the end of the coding sequence and would be expected to “freeze” the assembled polyribosomes, so that the ratio of polyribosomes to ribosomes would be much as normal.
7-31
Choice (c) is the correct answer. The drop in level of protein X in the normal cell is most likely due to protein degradation, since levels of mRNA remain constant. The inability of the mutant cell to divide could be due to a mutation that inhibits protein degradation. This would be achieved by removal of sites for attachment of ubiquitin, which targets proteins for destruction. Choices (a), (b), and (d) would probably not produce the result described, as without the production of a functional protein X the mutant cells could not grow in size.
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7-32
Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed of two subunits: the large subunit, which catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain, and the small subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the P-site of the ribosome. An incoming aminoacyltRNA carrying the next amino acid in the chain will bind to the A-site by forming base pairs with the exposed codon in the mRNA. The peptidyl transferase enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the protein called release factor. Eventually, most proteins will be degraded by a large complex of proteolytic enzymes called the proteosome.
7-33
Given that only single nucleotide changes are involved, the only codons consistent with the changes are: GUG for valine, GCG for alanine, AUG for methionine, and ACG for threonine.
7-34
Choice (d) is the correct answer. Choice (a) is incorrect in that although this may have been a step in self-replication, it would not by itself be sufficient. Choices (b) and (c) are incorrect, as these stages in the evolution of the cell must have succeeded the formation of the first self-replicating molecules.
7-35
(c)
7-36
(b)
7-37
Three possible answers are: 1. The deoxyribose sugar of DNA makes the molecule much less susceptible to breakage compared to RNA, due to the lack of the hydroxyl group on carbon 2 of the ribose sugar. 2. DNA is double stranded and therefore the complementary strand provides a template from which damage can be repaired accurately. 3. The use of “T” in DNA instead of “U” (as in RNA) protects against the effect of deamination, a common form of damage. Deamination of T produces an aberrant base (methyl C), whereas deamination of U generates C, a normal base. The presence of an abnormal base eases the cell’s job of recognizing the damaged strand.
121
7-38
Choice (d) is the correct answer. An organism having codons with an even number of nucleotides (i.e., 2, 4, or 6) could read 5′-GCGCGCGCGC-3′ (RNA 1) in either of two ways, namely “GC GC GC GC...” or “CG CG CG CG...” Either one of the two amino acids alone could have supported protein synthesis, so you would not need them in combination (thus eliminating choices (a), (c), and (e)). An organism having three bases per codon could read the sequence 5′-GCCGCCGCCGCCGCC-3′ (RNA 2) in one of three ways, namely “GCC GCC GCC GCC...,” “CCG CCG CCG CGG...,” or “CGC CGC CGC CGC...,” and so again, any one of the three amino acids could have supported synthesis of a polypeptide, and you would not need to add all three amino acids to produce a polypeptide chain, thus eliminating choice (b). Only a five-nucleotide code gives you two different consecutive codons for RNA 1 and three different consecutive codons for RNA 2.
7-39
No, you cannot definitively determine what the codons that code for serine or valine are because it could be either UAU or AUA. Bonus: The alien aminoacyl-tRNA synthetases could adapt a different amino acid to each tRNA, thus matching an amino acid with a different codon compared to those codons used by life on Earth.
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CHAPTER 8 CONTROL OF GENE EXPRESSION 2009 Garland Science Publishing 3rd Edition
An Overview of Gene Expression 8-1 The distinct characteristics of different cell types in a multicellular organism are produced mainly by the differential regulation of the (a) replication of specific genes. (b) transcription of genes transcribed by RNA polymerase II. (c) transcription of housekeeping genes. (d) translation of mRNA. (e) packing of DNA into nucleosomes in some cells and not others. 8-2 In principle, a eucaryotic cell can regulate gene expression at any step in the pathway from DNA to the active protein. Place the types of control listed below at the appropriate places on the diagram in Figure Q8-2.
A. Translation control B. Transcriptional control C. RNA processing control D. Protein activity control
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How Transcriptional Switches Work 8-3
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. The genes of a bacterial __________________ are transcribed into a single mRNA. Many bacterial promoters contain a region known as a(n) __________________, to which a specific gene regulatory protein binds. Genes in which transcription is prevented are said to be __________________. Bacterial genes are regulated by small molecules such as tryptophan by the interaction of such molecules with __________________ DNA-binding proteins such as the tryptophan repressor. Genes that are being __________________ expressed are being transcribed all the time. allosteric constitutively induced
negatively operator operon
positively promoter repressed
8-4
A gene regulatory protein called HisP regulates the enzymes for histidine biosynthesis in the bacterium E. coli. HisP is an allosteric protein whose activity is modulated by histidine. Upon binding histidine, HisP alters its conformation, dramatically changing its affinity for the regulatory sequences in the promoters of the genes for the histidine biosynthetic enzymes. A. If HisP functions as a gene repressor, would you expect that HisP would bind more tightly or less tightly to the regulatory sequences when histidine is abundant? Explain your answer. B. If HisP functions as a gene activator, would you expect that HisP would bind more tightly or less tightly to the regulatory sequences when histidine levels are low? Explain your answer.
8-5
In class we talked about how bacterial cells can take up the amino acid tryptophan from their surroundings, or if the external supply is insufficient, they can synthesize trytophan by using enzymes in the cell. In some bacteria, the control of glutamine synthesis is similar to that of tryptophan synthesis, such that the glutamine repressor is used to inhibit the transcription of the glutamine operon, which contains the genes that code for the enzymes required for glutamine synthesis. Upon binding to cellular glutamine, the glutamine repressor binds to a site in the promoter of the operon. A. Why is glutamine-dependent binding to the operon a useful property for the glutamine repressor? B. What would you expect to happen to the regulation of the enzymes that synthesize glutamine in cells that express a mutant form of the glutamine repressor that cannot bind to DNA? C. What would you expect to happen to the regulation of the enzymes that synthesize glutamine in cells that express a mutant form of the glutamine repressor that binds to DNA even when no glutamine is bound to it? 124
8-6
In the absence of glucose, E. coli can proliferate using the pentose sugar arabinose. The ability of E. coli to utilize the sugar arabinose is regulated via the arabinose operon, depicted in Figure Q8-6. The araA, araB, and araD genes encode enzymes for the metabolism of arabinose. The araC gene encodes a gene regulatory protein that binds adjacent to the promoter of the arabinose operon. To understand the regulatory properties of the AraC protein, you engineer a mutant bacterium in which the araC gene has been deleted and look at the effect of the presence or absence of the AraC protein on the AraA enzyme.
Figure Q8-6 A.
B.
If the AraC protein works as a gene repressor, would you expect araA RNA levels to be high or low in the presence of arabinose in the araC– mutant cells? What about in the araC– mutant cells in the absence of arabinose? Explain your answer. Your findings from the experiment are summarized in Table 8.6. Table 8-6 Genotype araC+ (normal cells) araC– (mutant cells)
araA RNA Levels in the absence of arabinose in the presence of arabinose low high low low
Do the results in Table 8-6 indicate that the AraC protein regulates arabinose metabolism by acting as a gene repressor or a gene activator? Explain your answer.
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8-7
We have discussed how the lac operon (see Figure Q8-7) is controlled by both the CAP activator protein and the lac repressor. You create cells that are mutant in the gene coding for the lac repressor so that now these cells lack the lac repressor under all conditions. For these mutant cells, state whether the lac operon will be ON or OFF under the following situations and explain why.
Figure Q8-7 A. B. C. D. 8-8
In the presence of glucose and lactose In the presence of glucose and the absence of lactose In the absence of glucose and the absence of lactose In the absence of glucose and the presence of lactose
You have discovered an operon in a bacterium that is only turned on when sucrose is present and glucose is absent. You have also isolated three mutants that have changes in the upstream regulatory sequences of the operon and whose behavior is summarized in the Table 8-8. You hypothesize that there are two gene regulatory sites in the upstream regulatory sequence, A and B, which are affected by the mutations. For this question, a plus (+) indicates a normal site and a minus (–) indicates a mutant site that no longer binds its gene regulatory protein. Table 8-8
normal (A+ B+) mutant 1 mutant 2 mutant 3 A. B. C.
Transcription of the operon in different media Glucose Sucrose Glucose + sucrose only only OFF ON OFF OFF OFF OFF OFF ON ON OFF OFF OFF
If mutant 1 has sites A–B+, which of these sites is regulated by sucrose and which by glucose? Give the state (+ or –) of the A and B sites in mutants 2 and 3. Which site is bound by a repressor and which by an activator?
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8-9
Which one of the following is the main reason that a typical eucaryotic gene is able to respond to a far greater variety of regulatory signals than a typical procaryotic gene or operon? (a) Eucaryotes have three types of RNA polymerase. (b) Eucaryotic RNA polymerases require general transcription factors. (c) The transcription of a eucaryotic gene can be influenced by proteins that bind far from the promoter. (d) Procaryotic genes are packaged into nucleosomes. (e) The protein-coding regions of eucaryotic genes are longer than those of procaryotic genes.
8-10
Match the following types of RNAs with the main polymerase that transcribes them: Types of RNAs A. rRNA genes B. tRNA genes C. 5S rRNA genes D. protein coding genes
8-11
Polymerases 1. RNA polymerase I 2. RNA polymerase II 3. RNA polymerase III
List three ways that the process of eucaryotic transcription differs from the process of bacterial transcription.
127
8-12
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. In eucaryotic cells, general transcription factors are required for the activity of all promoters transcribed by RNA polymerase II. The assembly of the general transcription factors begins with the binding of the factor __________________ to DNA, causing a dramatic local distortion in the DNA. This factor binds at the DNA sequence called the __________________ box, which is typically located 25 nucleotides upstream from the transcription start site. Once RNA polymerase II has been brought to the promoter DNA, it must be released in order to begin making transcripts. This release process is facilitated by the addition of phosphate groups to the tail of RNA polymerase by the factor __________________. It must be remembered that the general transcription factors and RNA polymerase are not sufficient to initiate transcription in the cell and are affected by proteins bound thousands of nucleotides away from the promoter. Proteins that link the distantly bound gene regulatory proteins to RNA polymerase and the general transcription factors include the large complex of proteins called the__________________. The packing of DNA into chromatin also affects transcriptional initiation, and histone __________________ is an enzyme that can render the DNA less accessible to the general transcription factors. activator CAP deacetylase enhancer
lac ligase mediator TATA
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TFIIA TFIID TFIIH
8-13
Label the following structures on the figure below.
Figure Q8-13 A. B. C. D. 8-14
Activator protein RNA polymerase General transcription factors Mediator
How are most eucaryotic gene regulatory proteins able to affect transcription when their binding sites are far from the promoter? (a) By binding to their binding site and sliding to the site of RNA polymerase assembly (b) By looping out the intervening DNA between their binding site and the promoter (c) By unwinding the DNA between their binding site and the promoter (d) By attracting RNA polymerase and modifying it before it can bind to the promoter
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8-15
The expression of the BRF1 gene in mice is normally quite low, but mutations in a gene called BRF2 lead to increased expression of BRF1. You have a hunch that nucleosomes are involved in the regulation of BRF1 expression and so you investigate the position of nucleosomes over the TATA box of BRF1 in normal mice and in mice that either lack the BRF2 protein (BRF2–) or part of histone H4 (HHF –) (histone H4 is encoded by the HHF gene). Your results are summarized below. A normal functional gene is indicated by a plus (+). Mouse BRF2+ HHF+ BRF2– HHF+ BRF2+ HHF– BRF2– HHF–
Nucleosome positioning specific pattern random random random
Relative level of BRF1 mRNA 1 100 1 100
Which of the following conclusions CANNOT be drawn from your data? Explain your answer. (a) BRF2 is required for repression of BRF1. (b) BRF2 is required for the specific pattern of nucleosome positions over the BRF1 upstream region. (c) The specific pattern of nucleosome positioning over the BRF1 upstream region is required for BRF1 repression. (d) The part of histone H4 missing in HHF– mice is not required to form nucleosomes. 8-16
The yeast Gal4 gene encodes a transcriptional regulator that can bind DNA upstream of genes required for the metabolism of the sugar galactose and turns on these genes when necessary. The Gal4 protein contains two protein domains: a DNA-binding domain and an activation domain. The DNA-binding domain allows it to bind to the appropriate sites in the promoters of the galactose metabolism genes. The activation domain attracts histone-modifying enzymes and also binds to a component of the RNA polymerase II enzyme complex, attracting it to the promoter so the regulated genes can be turned on when Gal4 is also bound to the DNA. When Gal4 is expressed normally, the genes can be maximally activated. You decide to try to produce more of the galactose metabolism genes by overexpressing the Gal4 protein at levels fifty-fold greater than normal. You conduct experiments to show that you are overexpressing the Gal4 protein and that it is properly localized in the nucleus of the yeast cells. To your surprise, you find that too much Gal4 causes the galactose genes to be transcribed only at a low level. What is the most likely explanation for your findings?
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The Molecular Mechanisms That Create Specialized Cell Types 8-17
In principle, how many different cell types can an organism having four different types of gene regulatory proteins and thousands of genes create? (a) Up to 4 (b) Up to 8 (c) Up to 12 (d) Up to 16 (e) Thousands
8-18
A virus produces a protein X that activates only a few of the virus’s own genes (V1, V2, and V3) when it infects cells. The cellular proteins A (a zinc finger protein) and the cellular protein B (a homeodomain protein) are known to be repressors of the viral genes V1, V2, and V3. You examine the complete upstream gene regulatory sequences of these three viral genes and find that: 1. 2. 3.
V1 and V2 contain binding sites for the zinc finger protein A only. V3 contains a binding site for the homeodomain protein B only. The only sequence that all three genes have in common is the TATA box.
Label each one of the choices below as LIKELY or UNLIKELY to be an explanation for your findings. For each choice you label as UNLIKELY, explain why. A. B. C. D. E. 8-19
Protein X binds nonspecifically to the DNA upstream of V1, V2, and V3 and activates transcription. Protein X binds to a repressor and prevents the repressor from binding upstream of V1, V2, and V3. Protein X activates transcription by binding to the TATA box. Protein X activates transcription by binding to and sequestering protein A and protein B. Protein X represses transcription of the genes for proteins A and B.
In mammals, individuals with two X chromosomes are female while individuals with an X and a Y chromosome are male. It had long been known that a gene located on the Y chromosome was sufficient to induce the gonads to form testes, which is the main maledetermining factor in development, and people began to search for the product of this gene, the so-called testes-determining factor (TDF). For several years, the TDF was incorrectly thought to be a zinc-finger protein encoded by a gene called BoY. Which of the following is the best evidence that BoY could NOT be the TDF? Explain your answer. (a) Some XY individuals that develop into females have mutations in a different gene, SRY, but are normal at BoY. (b) BoY is not expressed in the adult male testes. (c) Expression of BoY in adult females does not masculinize them. (d) A few of the genes that are known to be expressed only in the testes have binding sites for the BoY protein in their upstream regulatory sequences, but most do not. (e) A gene encoding a protein whose amino acid sequence is very similar to that of the BoY protein is found on the X chromosome. 131
8-20
From the sequencing of the human genome, we believe that there are approximately 30,000 protein-coding genes in the genome, for which there are an estimated 1500 to 3000 transcription factors. If every gene has a tissue-specific and signal-dependent transcription pattern, how can such a small number of transcriptional regulatory proteins generate a much larger set of transcriptional patterns?
How We Know: Gene Regulation—The Story of Eve 8-21
The gene for a hormone necessary for insect development contains binding sites for three gene regulatory proteins called A, B, and C. Because the binding sites for A and B overlap, A and B cannot bind simultaneously. You make mutations in the binding sites for each of the proteins and measure hormone production in cells that contain equal amounts of the A, B, and C proteins. The results of your studies are summarized in Figure Q8-22. In each of the following sentences, choose one of the phrases within square brackets to make the statement consistent with the above results.
Figure Q8-22 A. B. C.
Protein A binds to its DNA binding site [more tightly/less tightly] than protein B binds to its DNA binding site. Protein A is a [stronger/weaker] activator of transcription than protein B. Protein C is able to prevent activation by [protein A only/protein B only/both protein A and protein B].
132
8-22
The Drosophila eve gene has a complex promoter containing multiple binding sites for four gene regulatory proteins: Bicoid, Hunchback, Giant, and Krüppel. Bicoid and Hunchback are activators of eve transcription while Giant and Krüppel repress eve transcription. The patterns of expression of these regulators are shown in Figure Q8-22A. As we discussed, the eve promoter contains modules that control expression in various stripes. You construct a reporter gene that contains the DNA 5 kb upstream of the eve gene, so that this reporter contains the stripe 3 module, the stripe 2 module, the stripe 7 module, and the TATA box, all fused to the lacZ reporter gene (which encodes the βgalactosidase enzyme), as shown in Figure Q8-22B. This construct results in expression of the β-galactosidase enzyme in three stripes, which correspond to the normal positions of stripes 3, 2, and 7.
Figure Q8-22
133
A. B.
By examining the overlap of sites on the strip 2 module, as depicted in Figure Q822B, what is the biological effect of having some of the gene regulatory protein binding sites overlap? You make two mutant versions in which several of the binding sites in the eve stripe 2 module have been deleted, as detailed below. Refer to Figure Q8-22B for the positions of the binding sites. (Note, however, that because many of the binding sites overlap, it is not possible to delete all of one kind of site without affecting some of the other sites.) Match the appropriate mutant condition with the most likely pattern of eve expression shown in Figure Q8-22C. Explain your choices. i. ii.
Deletion of the Krüppel binding sites in stripe 2. Deletion of the two bicoid binding sites in the stripe 2 module that are marked with an asterisk (*) in Figure Q8-22B.
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Answers 8-1
Choice (b) is the correct answer. The major cause of differences between different cell types is in the differential expression of protein-coding genes transcribed by polymerase II, since these genes encode not only the speciality proteins characteristic of different cell types, but also the gene regulatory proteins required to maintain and control this pattern of expression. Choice (a) is untrue, since all genes are replicated equally when cells divide. Choice (c) is untrue, since expression of housekeeping genes do not differ much from cell to cell, as they mainly encode the proteins that are necessary for all cells to live. Choice (d) is untrue, as the main stage at which gene expression is regulated is the initiation of transcription. Choice (e) is untrue, as DNA is packed into nucleosomes in all eucaryotic cells.
8-2
See Figure A8-2
Figure A8-2 8-3
The genes of a bacterial operon are transcribed into a single mRNA. Many bacterial promoters contain a region known as a(n) operator, to which a specific gene regulatory protein binds. Genes in which transcription is prevented are said to be repressed. Bacterial genes are regulated by small molecules such as tryptophan by the interaction of such molecules with allosteric DNA-binding proteins such as the tryptophan repressor. Genes that are being constitutively expressed are being transcribed all the time.
8-4
A. B.
If HisP functions as a gene repressor, it would bind more tightly to the regulatory sequences when histidine is abundant, because the histidine biosynthetic genes should be turned off when the cell has enough histidine. If HisP functions as a gene activator, it should bind more tightly to the regulatory sequences when histidine levels are low. When histidine levels are low, the cell needs to synthesize more histidine.
135
8-5
A.
B. C.
8-6
A.
B.
8-7
A. B.
C.
D.
If sufficient glutamine is present in cells, the glutamine repressor will block the synthesis of enzymes that would make more glutamine. Likewise, if cells are starved for glutamine, the unoccupied repressor would not bind to the DNA and the enzymes that synthesize glutamine would be induced. This allows for a direct connection between the levels of glutamine and the expression of glutamine synthesizing enzymes. The glutamine synthesis enzymes would be permanently ON, regardless of the level of glutamine in the cells. The glutamine synthesis enzymes would be always OFF, regardless of the level of glutamine in the cells, since the repressor is always bound to the DNA. These cells will not be able to grow unless glutamine is added to the medium. If the AraC protein acted as a gene repressor for the arabinose operon, araA RNA levels should be high in the presence or absence of arabinose when there is no AraC protein around. In fact, the araA RNA levels should be high all the time, regardless of the presence or absence of arabinose, since the AraA gene should be transcribed under all conditions in the absence of AraC. The results are consistent with AraC acting as a gene activator for the arabinose operon. A gene activator must bind to the promoter regions of the arabinose genes in order to stimulate their transcription. Thus, if the gene for the regulatory protein is deleted, the arabinose genes cannot be turned on. Operon OFF. CAP will not bind in the presence of glucose. Operon OFF. Although normally the lac repressor would bind in the absence of lactose, the lack of the lac repressor in this case does not matter because the presence of glucose means that the CAP protein will not bind and activate transcription. Operon ON. Normally in the absence of both glucose and lactose, the operon would be OFF. However, since the cells are lacking the lac repressor, the cells cannot sense the absence of lactose. Because the CAP protein will bind and activate transcription, the operon will be ON. Operon ON. The CAP protein will bind and activate transcription due to the presence of glucose. It does not matter whether the lac repressor gene is mutant, because there is lactose available.
8-8
A. B. C.
Site A is regulated by sucrose and site B by glucose. Mutant 2 (A+ B–); mutant 3 (A– B–) or (A– B+). Site A is bound by an activator and site B by a repressor.
8-9
(c)
The fact that, in eucaryotes, gene regulatory proteins can influence the initiation of transcription even when they are bound far away from the promoter means that there can be a very large number of gene regulatory sites affecting the same promoter. Thus, the initiation of transcription can be influenced by a great variety and number of different signals, each of which may induce the binding of different gene regulatory proteins to these regulatory regions.
136
8-10
A—1; B—3; C—3; D—2
8-11
Any three of these are acceptable. 1. Bacterial cells contain a single RNA polymerase whereas eucaryotic cells have three. 2. Bacterial RNA polymerase can initiate transcription without the help of additional proteins whereas eucaryotic RNA polymerases need general transcription factors. 3. In eucaryotic cells, gene regulatory proteins can influence transcriptional initiation thousands of nucleotides away from the promoter whereas bacterial regulatory sequences are very close to the promoter. 4. Eucaryotic transcription is affected by chromatin structure and nucleosomes whereas bacterial transcription is not.
8-12
In eucaryotic cells, general transcription factors are required for the activity of all promoters transcribed by RNA polymerase II. The assembly of the general transcription factors begins with the binding of the factor TFIID to DNA, causing a dramatic local distortion in the DNA. This factor binds at the DNA sequence called the TATA box, which is typically located 25 nucleotides upstream from the transcription start site. Once RNA polymerase II has been brought to the promoter DNA, it must be released in order to begin making transcripts. This release process is facilitated by the addition of phosphate groups to the tail of RNA polymerase by the factor TFIIH. It must be remembered that the general transcription factors and RNA polymerase are not sufficient to initiate transcription in the cell and are affected by proteins bound thousands of nucleotides away from the promoter. Proteins that link the distantly bound gene regulatory proteins to RNA polymerase and the general transcription factors include the large complex of proteins called the mediator. The packing of DNA into chromatin also affects transcriptional initiation, and histone deacetylase is an enzyme that can render the DNA less accessible to the general transcription factors.
8-13
See Figure A8-13.
Figure A8-13 137
8-14
(b)
Of the cases studied thus far, most eucaryotic gene regulatory proteins that act at a distance are thought to do so by looping out the intervening DNA while at the same time binding, via the mediator, to proteins that form the initiation complex at the promoter.
8-15
(c)
All the other conclusions can be drawn from the data. Since the BRF2+ HHF– mutant does not exhibit the specific pattern of nucleosome positioning yet still has a low level of BRF1 expression, and since the BRF2–HHF– mutant has high levels of BRF1 expression (indicating that HHF is not required for BRF1 expression), it would appear that repression of BRF1 could take place in the absence of nucleosome positioning. Since nucleosomes are formed in all cases, the missing portion of histone H4 is not required for their formation.
8-16
In order for Gal4 to work properly, the DNA-bound Gal4 must attract histone-modifying enzymes and recruit RNA polymerase to the promoter. If there is too much Gal4 in the cell, the non-DNA-bound Gal4 (or free Gal4) will compete with the DNA-bound Gal4 for binding to histone modifying enzymes and RNA polymerase. The excess amount of Gal4 forms non-productive complexes with histone modifying enzymes and RNA polymerase, preventing their recruitment to the promoter and lowering the level of transcription.
8-17
(d)
The type of cell is determined by the particular combination of gene regulatory proteins active within it. With four different proteins available, there is one possibility with no proteins at all and one with all four proteins. There are four possibilities with one protein each, six possible combinations of two different proteins, and four possible combinations of three different proteins.
8-18
A.
Unlikely. If protein X were to bind nonspecifically to DNA, it would not specifically regulate a particular subset of genes. Unlikely. If a single repressor were to bind upstream of, we would expect to have found a binding site common to all three genes, but there is none. Unlikely. If a protein X activated gene transcription were to bind to the TATA box, it would be likely to activate most genes transcribed by RNA polymerase II, since most genes contain TATA boxes. Likely Likely
B. C. D. E.
138
8-19
Choice (a) is the correct answer. XY individuals that develop as females presumably lack the testis-determining factor (TDF). If BoY is normal in these individuals, it would strongly suggest that BoY is not the TDF. Although expression of TDF is necessary for testes development, this does not mean that is must be expressed in adult males once the gonad has already formed. Similarly, even though TDF expression is sufficient to induce testis formation, once the structures are formed, TDF may not be able to exert any additional effect (thus choices (b) and (c) are not considered strong evidence against BoY being TDF). Choice (d) is not compelling evidence against BoY being the TDF, since the TDF will not necessarily bind upstream of all of the genes whose expression it influences; some of the genes it regulates directly probably encode other gene regulatory proteins that bind to regulatory sites different from the TDF site. The presence of a protein similar to BoY on the X chromosome (choice (e)) is not necessarily evidence for or against BoY being the TDF.
8-20
Gene regulatory proteins are generally used in combinations, thereby increasing the possible regulatory repertoire of gene expression with a limited number of proteins.
8-21
A. B. C.
8-22
A.
B.
Protein A binds to its DNA binding site more tightly than protein B binds to its DNA binding site. Protein A is a weaker activator of transcription than protein B. Protein C is able to prevent activation by both protein A and protein B. Repressor (Krüppel and Giant) binding sites do not seem to overlap. Activator (Bicoid and Hunchback) binding sites also do not seem to overlap. Instead, the binding sites for repressor proteins seem to overlap with the binding sites for activator proteins. These overlapping binding sites allow for repressor and activator proteins to compete for binding to the DNA. It is thought that the binding of a repressor and an activator is mutually exclusive. The overlap between the repressor and activator binding sites allows eve expression to be exquisitely sensitive to the levels of repressors and activators in the cell and suggests that the repressors function by preventing activator binding. In fact, the repressors and activators can antagonize each other, allowing the creation of sharp stripes of transcription from smooth gradients of protein regulatory factors. i. Mutant embyo (B). When the Krüppel-binding sites are removed, the effects of the Krüppel repressor are eliminated. Stripe 2 expression now expands slightly in the posterior direction, which is to be expected since hunchback and bicoid expression extends slightly beyond the posterior end of stripe 2. ii. Mutant embryo (C). When two of the bicoid binding sites are removed, expression from the promoter is less sensitive to the effects of the bicoid activator. Thus, stripe 2 appears at its normal position, but the expression of β-galactosidase is decreased.
139
CHAPTER 9 HOW GENES AND GENOMES EVOLVE 2009 Garland Science Publishing 3rd Edition
Generating Genetic Variation 9-1 Which of the following statements are TRUE? (Note that more than one statement may be true.) (a) Asexual reproduction involves the formation of germ cells. (b) A mutation that arises in a mother’s somatic cell often causes a disease in her daughter. (c) All mutations in a typical asexually reproducing organism are passed onto progeny. (d) In an evolutionary sense, somatic cells exist only to help propagate germ line cells. (e) A mutation is passed on to offspring only if it is present in the germ line.
9-2 Transposable elements litter the genomes of primates and a few of them are still capable of moving to new regions of the genome. If a transposable element jumped into an important gene in one of your cells when you were a baby and caused a disease, is it likely that your child would also have the disease? Explain. 9-3 What is the most likely explanation for why the overall mutation rates in bacteria and humans are roughly similar? (a) The DNA replication and repair machinery cannot possibly be any more accurate. (b) Cell division needs to be fast. (c) Most mutations are silent. (d) There is a narrow range of mutation rates that offers an optimal balance between keeping the genome stable and generating sufficient diversity in a population. (e) It benefits a multicellular organism to have some variability among its cells.
9-4 For each statement below, indicate if it is TRUE or FALSE and explain why. (a) To address a challenge or develop a new function, evolution essentially builds from first principles, like an engineer, to find the best possible solution. (b) Nearly every instance of DNA duplication leads to a new functional gene. (c) A pseudogene is highly homologous to a functional gene but cannot be expressed due to mutation(s). (d) Most genes in vertebrates are unique, and only a few genes are members of multigene families. (e) Horizontal transfer is very rare and thus has had little influence on the genomes of bacteria.
141
9-5
Consider a gene with a particular function. Mutation X and mutation Y each cause defects in the function of the encoded protein. Yet a gene containing both mutations X and Y encodes a protein that works even better than the original protein. The odds that a single mutational event will generate both mutations X and Y are exceedingly small. Explain a simple way that an organism with a mutant gene containing both mutations X and Y could arise during evolution.
9-6
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Most variation between individual humans is in the form of __________________. __________________ may arise by recombination within introns and can create proteins with novel combinations of domains. Scientists and government regulators must be very careful when introducing herbicide-resistant transgenic corn plants into the environment, because if resistant weeds arise from __________________ then the herbicides could become useless. Families of related genes can arise from a single ancestral copy by __________________ and subsequent __________________. divergence exon shuffling gene duplication horizontal gene transfer
purifying selection single nucleotide polymorphisms synteny unequal crossing over
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9-7
Figure Q9-7 shows several possible substrates of exon shuffling. Horizontal lines and small filled circles represent chromosomes and centromeres, respectively. Exons are labeled A, B, C, and D. Homologous recombination or shuffling may take place at short, repeated homologous DNA sequences in introns; because DNA sequences have a polarity, the repeated sequences can be considered to have a head and a tail and thus are drawn as arrows. A recombinational crossover is indicated by a big X. Panel (A) shows that recombination between two direct repeats located on opposite sides of the centromere yields one circular product that contains a centromere and a second product that lacks a centromere and will therefore be lost when the cell divides. Panel (B) shows that recombination between inverted repeats flanking the centromere will keep the rearranged chromosome intact. Draw the products of recombination when the repeated sequences are located on different chromosomes, as shown in panels (C) and (D). Will these products be faithfully transmitted during cell division?
Figure Q9-7 9-8
Which of the following would contribute most to successful exon shuffling? (a) Shorter introns (b) A haploid genome (c) Exons that code for more than one protein domain (d) Introns that contain regions of similarity to one another (e) Inability of short stretches of amino acids to fold into discrete functional units
143
Reconstructing Life’s Family Tree 9-9
You are working in a human genetics laboratory that studies causes and treatments for eye cataracts in newborns. This disease is thought to be caused by a deficiency in an enzyme called galactokinase, but the human gene that encodes this enzyme has not yet been identified. At a talk by a visiting scientist, you learn about a strain of bakers yeast that contains a mutation called gal1– in its galactokinase gene. Because this gene is needed to metabolize galactose, the mutant strain cannot grow in galactose medium. Knowing that all living things evolved from a common ancestor and that distantly related organisms often have homologous genes that perform similar functions, you wonder if the human galactokinase gene can function in yeast. Since you have an optimistic temperament, you decide to pursue this line of experimentation. You isolate mRNA gene transcripts from human cells, use reverse transcriptase to make complementary DNA (cDNA) copies of the mRNA molecules, and ligate the cDNAs into circular plasmid DNA molecules that can be stably propagated in yeast cells. You then transform the pool of plasmids into gal1– yeast cells so that each cell receives a single plasmid. What do you think will happen when you spread the plasmid-containing cells on petri plates that contain galactose as a carbon source? How can this approach help you find the human gene encoding galactokinase?
9-10
A.
B.
9-11
When a mutation arises, it can have three possible consequences: beneficial to the individual, selectively neutral, or detrimental. Order these from most likely to least likely. The spread of a mutation in subsequent generations will, of course, depend on its consequences to individuals that inherit it. Order the three possibilities above from that which is most likely to spread and become over-represented in subsequent generations to that which is most likely to become under-represented or disappear from the population.
Some types of genes are more highly conserved than others. For each of the following pairs of gene functions, choose the one that is more likely to be highly conserved. (a) Genes involved in sexual reproduction vs. genes involved in sugar metabolism (b) DNA replication vs. developmental pathways (c) Hormone production vs. lipid synthesis
144
9-12
A hypothetical phylogenetic tree is shown in Figure Q9-12. Use this tree to answer the following questions.
Figure Q9-12 A. B. C. D.
9-13
How many years ago did species M and N diverge from their last common ancestor? How much nucleotide divergence is there on average between M and N? Are M and N more or less closely related to each other than P and S? In looking for functionally important nucleotide sequences, is it more informative to compare the genome sequences of M and N or those of M and Q?
For each statement below, indicate if it is TRUE or FALSE and explain why. (a) All highly conserved stretches of DNA in the genome are transcribed into RNA. (b) To find functionally important regions of the genome, it is more useful to compare species whose last common ancestor lived 100 million years ago rather than 5 million years ago. (c) Most mutations and genome alterations have neutral consequences. (d) Proteins required for growth, metabolism, and cell division are more highly conserved than those involved in development and in responding to the environment. (e) Introns and transposons tend to slow the evolution of new genes.
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9-14
A friend of yours has sequenced the genome of her favorite experimental organism, a kind of yeast. She wants to identify the locations of all the genes in this genome. To aid her search, she decides to collaborate with another researcher, one who has sequenced the genome of a distantly related yeast species. Luckily, the absence of introns simplifies the effort. She and her collaborator use a computer program to align similar stretches of DNA sequence from the two genomes. The program yields the graphical output shown in Figure Q9-14, where the horizontal lines represent a portion of the two genomic sequences and vertical lines indicate where the sequences differ. (No vertical line means the sequence is identical in the two yeasts.) Label both the functionally conserved regions and the divergent (nonconserved) sequences. Are all of the functionally conserved regions likely to be transcribed into RNA? If not, what might be the function of the nontranscribed conserved regions?
Figure Q9-14 9-15
The genomes of some vertebrates are much smaller than those of others. For example, the genome of the puffer fish Fugu is much smaller than the human genome, and even much smaller than those of other fish, primarily due to the small size of its introns. A. Describe a mechanism that might drive evolution toward small introns or loss of introns and could therefore account for the evolutionary loss of introns according to the “introns early” hypothesis. B. Describe a mechanism that might drive evolution toward more or larger introns and could thereby account for the evolutionary appearance of introns according to the “introns late” hypothesis.
Examining the Human Genome 9-16
For each statement below, indicate if it is TRUE or FALSE and explain why. (a). The increased complexity of humans compared to flies and worms is largely due to the vastly larger number of genes in humans. (b) Repeats of the CA dinucleotide are useful for crime investigations and other forensic applications. (c) The majority of single-nucleotide polymorphisms cause no observable functional differences between individual humans. (d) There is little conserved synteny between human and mouse genes. (e) The differences between multicellular organisms are largely explained by the different kinds of genes carried on their chromosomes.
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9-17
The number of proteins found in humans and other organisms can vastly exceed the number of genes. This is largely due to (a) protein degradation. (b) alternative splicing. (c) homologous genes. (d) synteny. (e) mutation.
HOW WE KNOW: COUNTING GENES 9-18
Explain how ESTs are identified and how they aid in finding the genes within an organism’s genome.
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Answers 9-1
C, D, E. A is false because it is sexual reproduction that requires the production of germline cells. B is false because mutations are carried in the genetic material and the only genetic material passed along to the offspring of a sexually-reproducing organism comes from a germ-line cell (not a somatic cell).
9-2
It is not likely that child would have the disease, because it is unlikely that the mutation is carried in the germ line. Probably the mutation occurred in a cell that gave rise to somatic cells and not germ cells. Only mutations in germ cells are passed onto progeny.
9-3
Choice (d) is the correct answer. Choices (a), (b), and (e) are probably false. Choice (c) is true but cannot explain the similar mutation rate.
9-4
(a)
(b) (c) (d) (e)
False. Evolution can work only by tinkering with the tools and materials on hand, not by starting from scratch to make completely new genes or pathways. New functions arise from the ancestral functions by a process of gradual mutational change, and thus may not represent the best possible solution to a problem. False. Many duplications are subsequently lost or become pseudogenes, and only a few evolve into new genes. True. Pseudogenes look very similar to normal genes but cannot produce a fulllength protein due to one or more disabling mutations. False. A large proportion of the genes in vertebrates (and many other species) are members of multigene families. False. By some estimates, 20% of the genomic DNA in some bacterial species arose by horizontal gene transfer.
9-5
The simplest way to evolve the new gene is by duplication and divergence. If the gene is duplicated, then the cell or lineage can maintain one functional, intact old copy of the original gene and can thus tolerate the disabling mutations in the other copy. The second copy can first be modified by the X or Y mutation that impairs its function; second, at some later time, the gene with the single mutation can acquire the additional mutation to yield the doubly mutant X+Y gene with the new or improved function.
9-6
Most variation between individual humans is in the form of single nucleotide polymorphisms. Exon shuffling may arise by recombination within introns and can create proteins with novel combinations of domains. Scientists and government regulators must be very careful when introducing herbicide-resistant transgenic corn plants into the environment, because if resistant weeds arise from horizontal gene transfer then the herbicides could become useless. Families of related genes can arise from a single ancestral copy by gene duplication and subsequent divergence.
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9-7
See Figure A9-7. The products of panel (C) will be segregated to progeny cells reliably. In contrast, one product in panel (D) will have two centromeres and the other will lack a centromere. The chromosome without a centromere will be rapidly lost as cells divide. The chromosome containing two centromeres probably will be broken during mitosis (see Chapter 19), and subsequently lost or severely damaged.
Figure A9-7 9-8
Choice (d) is the correct answer. Exon shuffling is facilitated by long introns (thus choice (a) is incorrect) and by short exons that each code for one protein domain (thus choice (c) is incorrect). Since exon shuffling can occur via recombination between introns, introns with regions of similarity to one another will facilitate shuffling. A haploid genome will probably be less prone to exon shuffling than a diploid genome (thus choice (b) is incorrect) because having two copies of each gene allows an organism to keep one copy of the gene as a backup while it shuffles the other copy. Exon shuffling is possible only because many proteins are modular, composed of short, folded domains that have discrete functional properties (thus choice (e) is incorrect).
9-9
On galactose medium, the original gal1– yeast cells cannot grow, nor can cells that received plasmids containing most human cDNA sequences. However, yeast cells that received a plasmid with the human galactokinase gene will probably be able to grow on galactose medium and produce many progeny. This kind of “selection” procedure is very powerful, because even if only 1 in 100,000 cells has the ability to grow under particular conditions it will be easy to find it. The other 99,999 cells will die in the petri plate and will therefore be invisible to the investigator. Indeed, scientists have found that the human galactokinase gene can function perfectly well in yeast and thus can “rescue” the defect of the gal1– mutant. It was initially astonishing that genes from humans can function properly in yeast, but this phenomenon has now been observed for many genes.
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9-10
A.
B.
Selectively neutral, detrimental, beneficial. Most nucleotide changes in the genome, or mutations, will have little to no effect on the fitness of the individual because many changes are not located in regions that encode a protein or regulate expression of a gene. Even changes within a coding region may not change the amino acid encoded or may cause a conservative amino acid change, for example from one small nonpolar amino acid to another. Most changes that have a functional consequence will interfere with the regulation of a gene or the behavior of the encoded protein, usually rendering it useless and occasionally making it harmful or yielding a new function. Only very rarely will a mutation improve the performance of the gene or its encoded protein. Beneficial, selectively neutral, detrimental. Individuals bearing beneficial mutations will be more likely to have more offspring than others in the population, and thus the beneficial mutations will become over-represented in the population in subsequent generations. Individuals bearing detrimental mutations will be likely to have fewer children and grandchildren, and thus these mutations will be culled from the population, though perhaps not eliminated.
9-11
(a) (b) (c)
Sugar metabolism DNA replication Lipid synthesis. These pathways or phenomena are fundamental to the growth and proliferation of all cells, including bacteria, and thus are likely to be highly conserved from species to species.
9-12
A. B.
M and N diverged 10 million years ago. There is an average of 2.0% nucleotide substitution in M compared to N (follow the path connecting the two species, which is twice the distance between each one and their common ancestor). Neither more nor less. They show roughly the same degree of relatedness. The sequence divergence between M and N is about 2.0%, the same as that between P and S. Both pairs of species diverged 10 million years ago. It is more informative to compare species that are separated by a greater evolutionary distance, thus comparing M and Q which diverged 20 million years ago will be better able to identify sequences likely to be important for function. Closely related species share many sequences by chance because insufficient time has been allowed for neutral mutations to accumulate.
C.
D.
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9-13
(a) (b)
(c)
(d)
(e)
9-14
False. Many highly conserved stretches of DNA are not transcribed but instead contain information critical for regulating where and when genes are expressed. True. Species that diverged recently have many identical stretches of DNA sequence by chance, whereas sequence similarity between species that diverged long ago is probably due to functional constraints. True. Most genomic changes do not alter the amino acid sequence of proteins or the regulatory properties of genes. Even some mutations that cause minor alterations have little functional consequence. True. All organisms need to perform a similar basic set of fundamental functions, such as those for metabolism, protein synthesis, and DNA replication. Proteins involved in these functions are shared by descent, and their evolution is constrained. Different species and cells are likely to require different developmental paths and to encounter different environmental challenges, so the proteins involved in these processes will tend to be more variable. For example, bacteria do not undergo elaborate developmental programs and so lack many of the regulators of development found in animals. False. Introns and transposons can act as sites where recombinational crossovers occur. Transposons can also catalyze genetic rearrangements. Rearrangements occurring within these sequences are less likely to be detrimental than those occurring elsewhere in the genome. In general only the short intron sequences required for splicing are important to intron function; alterations in sequences outside the splicing sites may have no consequences for intron function and thus will not be subject to purifying selection.
See Figure A9-14. Not all of the functionally conserved regions will be transcribed into RNA. Some of the functionally conserved regions are likely to encode RNA and others are likely to be critical for regulating when the gene is transcribed and when it is turned off. These nontranscribed regulatory regions may be conserved nearly as much as the coding regions.
Figure A9-14 9-15
A. B.
Loss of introns may be caused by spontaneous deletions or selection pressure to decrease the time or cost of DNA replication. Acquisition of intron sequences provides a selective advantage in the face of transposon insertions. According to this idea, introns became sinks for transposon and virus insertion to protect the rest of the genome. Alternatively, introns may provide another advantage to the host genome: by providing ample sites for crossing over, larger introns could facilitate exon shuffling and thus the generation of new genes with novel functions.
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9-16
(a)
(b)
(c) (d)
(e)
False. The number of genes differs only by about a factor of two. It is thought that the increased complexity of humans is due largely to differences in when and where the genes are expressed. Differential splicing may also be a major contributor to the relative complexity of humans. True. There are CA repeats in many locations throughout the genome. Because the number repeats at a given location varies greatly between individuals and families, it can be used as an identifying characteristic to match two samples (e.g., blood samples) from the same or related individuals. True. Nearly all single-nucleotide polymorphisms have no effect on the appearance or behavior of the individual, but a few cause important differences. False. Human and mouse chromosomes show extensive synteny, with blocks of chromosomal DNA exhibiting homologous genes arranged in the same order between the two species. False. Multicellular organisms are built from essentially the same toolbox of gene building blocks, but the parts are put together differently due to regulatory differences that dictate when and where and how much of each protein is made. Alternative splicing can also have an important role, as it can generate several proteins from a single gene in some species, yet the homologous gene in other species may produce only one protein.
9-17
Choice (b) is the correct answer. Alternative splicing can produce several different mRNA transcripts from a single gene, and these transcripts can be translated into several different but related proteins. Choices (c), (d), and (e) do not yield more protein species than genes. Protein degradation (choice (a)) can produce several proteins from a single gene, but this mechanism is used sparingly.
9-18
To identify expressed sequence tags, or ESTs, mRNA must first be isolated from cells. This mRNA is converted into complementary DNA (cDNA) using specialized nucleic acid polymerases. The nucleotide sequence of a short region of each cDNA is then determined. Each short sequence (or EST) corresponds to a portion of a gene that was expressed in the cells from which the mRNA was isolated; each sequence can be used as a tag to identify or manipulate the gene from which it came. A collection of ESTs can be fed into a computer to search for matches to the total genome sequence and thereby identify the sequences and chromosomal locations of many genes.
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CHAPTER 10 MANIPULATING GENES AND CELLS 2009 Garland Science Publishing 3rd Edition
Isolating Cells and Growing Them In Culture 10-1 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. The ability to purify uniform populations of cells from a tissue facilitates their study. In order to obtain a purified population of cells from a compact tissue, __________________ enzymes must be used to disrupt the adhesive bonds between cells. The __________________ cell sorter allows for isolation of a particular cell type by recognizing cell-surface differences. Once cells are isolated, their properties can be studied by, for example, biochemical experiments that aim to purify molecules from cells and reconstitute cellular reactions __________________ (that is, in a test tube). Although cells may be grown in culture, most vertebrate cells have a limited number of cell divisions they can undergo, often due to lack of expression of the __________________ enzyme. Cells that can divide indefinitely may result from a genetic change; such cells are considered to be __________________ and can be propagated in culture as a homogeneous cell line.
agitating de novo desiccated differentiated electrically excitable
fluorescence-activated immortalized in vitro in vivo nuclease
polymerase propagated proteolytic telomerase
10-2 Explain why the undifferentiated state of human embryonic stem cells is critical to their usefulness.
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How DNA Molecules Are Analyzed 10-3
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. A nuclease hydrolyzes the __________________ bonds in a nucleic acid. Nucleases that cut DNA only at specific short sequences are known as __________________. DNA composed of sequences from different sources is known as __________________. __________________ can be used to separate DNA fragments of different sizes. Millions of copies of a DNA sequence can be made entirely in vitro using the __________________ technique. DNA sequencing endonucleases exonucleases gel electrophoresis hydrogen nucleic acid hybridization
10-4
phosphodiester polymerase chain reaction recombinant DNA restriction nucleases ribonucleases
You have purified DNA from your recently deceased goldfish. Which of the following restriction nucleases would you use if you wanted to end up with DNA fragments of average size around 70 kb after complete digestion of the DNA? The recognition sequence for each enzyme is indicated in the right-hand column. “N” indicates that any nucleotide may be in this position and the enzyme will still cleave the site. (a) Sau3A I GATC (b) BamH I GGATCC (c) BsaJ I CCNNGG (d) Not I GCGGCCGC (e) Xza I GAAGGATCCTTC
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10-5
You have accidentally torn the labels off two tubes, each containing a different plasmid, and now do not know which plasmid is in which tube. Fortunately, you have restriction maps for both plasmids, shown in Figure Q10-5. You have the opportunity to test just one sample from one of your tubes. You have equipment for agarose gel electrophoresis, a standard set of DNA size markers, and the necessary restriction enzymes.
Figure Q10-5 A. B.
Outline briefly the experiment you would do to determine which plasmid is in which tube. Which restriction enzyme or combination of restriction enzymes would you use in this experiment?
155
Note: The following codon table is to be used for Problems Q10-6, Q10-24 and Q10-29.
10-6
You have sequenced a short piece of DNA and produced the gel shown below:
Figure Q10-6 A. B.
What is the sequence of the DNA, starting from the 5′ end? If you knew that this sequence is from the middle of a protein-coding cDNA clone, what amino acid sequence can be deduced from this sequence?
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10-7
You have sequenced a fragment of DNA and produced the gel shown in Figure Q10-7. Near the top of the gel, there is a section where there are bands in all four lanes (indicated by the arrow). Which of the following mishaps would account for this phenomenon? Explain your answer.
Figure Q10-7 (a) (b) (c) (d) (e)
You mistakenly added all four dideoxynucleotides to one of the reactions. You forgot to add deoxynucleotides to the reactions. You forgot to add dideoxynucleotides to each of the reactions. A fraction of the DNA you are sequencing was cut by a restriction nuclease. Your primer hybridizes to more than one area of the fragment of DNA you are sequencing.
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Nucleic Acid Hybridization 10-8
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. The technique of __________________ hybridization can be used to detect a specific RNA expression in a particular region of the brain. Northern blotting detects a specific sequence in __________________. Southern blotting detects a specific sequence in __________________. A short, single-stranded DNA is a(n) __________________. A piece of DNA used to detect a specific sequence in a nucleic acid by hybridization is known as a(n) __________________. DNA in situ in vitro oligonucleotide
polymerase chain reaction probe RNA vector
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10-9
Assume that defects in a hypothetical gene, X, have been linked to antisocial behavior. Two copies of a defective gene X predispose a child to bad behavior from childhood, while a single copy of the gene seems to produce no symptoms until adulthood. Since the effects of the gene can be counteracted if treatment is started early enough, a program of voluntary genetic testing is being carried out with delinquent prospective parents. Charles S. and Caril Ann F. have been arrested on charges of robbery and assault, and Caril Ann is pregnant with Charles’s child. You obtain DNA samples from Charles, Caril Ann, and the fetus. You digest these samples with Not I and use these samples to perform two Southern Blots, which you probe with two different oligonucleotide probes, A and B, that hybridize to different parts of the normal gene X, as shown in Figure Q10-9A. Your results are shown in Figure 10-9B.
Figure Q10-9 A. B. C.
Which of the three individuals have defects in gene X? Which individuals have a single defective gene and which have two defective copies of the gene? Indicate the nature (single base-pair mutation or deletion) and location of each individual’s defects on gene X.
10-10 Figure Q10-10 shows a restriction map of a piece of DNA containing your favorite gene. The arrow indicates the position and orientation of the gene in the DNA. In part B of the figure are enlargements showing the portions of the DNA whose sequences have been used to make oligonucleotide probes A, B, C, and D. Which of the oligonucleotides can be used to detect the gene in each of the following?
Figure 10-10 A. B.
A Southern blot of genomic DNA cut with Hind III. A Northern blot. 159
10-11 In situ hybridization can be used to determine the (a) sequence of a cloned gene. (b) distribution of proteins in tissues. (c) position of a cloned fragment of DNA on a plasmid. (d) size of a gene. (e) distribution of a given type of mRNA in different tissues. 10-12 You are interested in understanding how the brain works, and are using the fruit fly Drosophila as a model system to study brain development. You perform microarray analysis to try to determine genes expressed in the fly brain. For your microarray experiment, you first prepare cDNA from fly brains and label it with a red fluorochrome. Then, you isolate cDNA from whole flies and label it with a green fluorochrome. Next, you hybridize these cDNA populations to a microarray containing the Drosophila genes. From this, you obtain a list of genes that are specifically enriched in the brain (those that show up as a red spot on the microarray). You are disappointed because your favorite fly gene, tubby, does not appear on this list, even though you have repeated the microarray experiment 10 times and did not encounter any technical difficulties. The reason you thought tubby would appear on this list is that you believe tubby is important for brain development, since flies mutant in tubby have no brains. Not to be discouraged, you perform in situ analysis using the tubby DNA as a probe, and see that it is expressed at high levels in the fly brain of normal flies but not expressed in animals lacking the tubby gene. Why do you think tubby did not show up as a gene specifically enriched in the brain in your microarray experiment?
DNA Cloning 10-13 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Two fragments of DNA can be joined together by __________________. Restriction enzymes that cut DNA straight across the double helix produce fragments of DNA with __________________. A fragment of DNA is inserted into a __________________ in order to be cloned in bacteria. A __________________ library contains a collection of DNA clones derived from mRNAs. A __________________ library contains a collection of DNA clones derived from chromosomal DNA. blunt ends cDNA DNA ligase
DNA polymerase genomic probe
160
RNA staggered ends vector
10-14 Figure Q10-14 shows the recognition sequences and sites of cleavage for the restriction enzymes Sal I, Xho I, Pst I, and Sma I and a plasmid with the sites of cleavage for these enzymes marked.
Figure Q10-14 A.
After which of the following treatments described in choices 1 through 5 can the plasmid shown in Figure Q10-14 be recircularized simply by treating with DNA ligase? Assume that after treatment any small pieces of DNA are removed, and it is the larger portion of plasmid only that you are trying to recircularize. After digestion with 1. Sal I alone. 2. Sal I and Xho I. 3. Sal I and Pst I. 4. Sal I and Sma I. 5. Sma I and Pst I.
B.
In which of the cases 1–5 can the plasmid be recircularized by adding DNA ligase after the cut DNA has been treated with DNA polymerase in a mixture containing the four deoxynucleotides? Again assume that you are trying to recircularize the larger portion of plasmid.
10-15 Name three features that a cloning vector for use in bacteria must contain. Explain your answers. 10-16 Which of the following statements about genomic DNA libraries are FALSE? (a) The larger the size of the fragments used to make the library, the fewer colonies you will have to examine to find a clone that hybridizes to your probe. (b) The larger the size of the fragments used to make the library, the more difficult it will be to find your gene of interest once you have identified a clone that hybridizes to your probe. (c) The larger the genome of the organism from which a library is derived, the larger the fragments inserted into the vector will tend to be. (d) The smaller the gene you are seeking, the more likely it is that the gene will be found on a single clone. (e) The shorter the oligonucleotide used to probe the library, the greater the number of colonies to which the probe will hybridize. 161
10-17 A DNA library has been constructed by purifying chromosomal DNA from mice, cutting the DNA with the restriction enzyme Not I, and inserting the fragments into the Not I site of a plasmid vector. What information CANNOT be retrieved from this library? (a) Gene regulatory sequences (b) Intron sequences (c) Sequences of the telomeres (the ends of the chromosomes) (d) Amino acid sequences of proteins 10-18 You have the amino acid sequence of a protein and wish to search for the gene encoding for this protein in a DNA library. Using this protein sequence, you deduce a particular DNA sequence that can encode for this protein. Why is it unwise to use only this DNA sequence you have deduced as the probe for isolating the gene encoding your protein of interest from the DNA library? 10-19 What is the main reason for using a cDNA library rather than a genomic library to isolate a human gene from which you wish to make large quantities of the human protein in bacteria? 10-20 Some clones from cDNA libraries can have defects because of the way a cDNA library is constructed. For each of the dilemmas described in A–D, indicate which of the outcomes listed in 1–4 you might encounter. Outcomes may be used more than once. Dilemma A. The mRNA corresponding to the clone you are looking for was degraded at its 5′ end by a nuclease. B. The mRNA corresponding to the clone you are looking for was degraded at its 3′ end by a nuclease. C. The 5′ end of the reverse transcriptase product of the gene you are trying to clone hybridizes to sequences in the middle of the gene. D. The gene you are trying to clone has a long stretch of A’s in the middle of the coding sequence.
Outcomes 1. The 5′ part of the gene will be missing. 2. The 3′ part of the gene will be missing. 3. An internal fragment of the gene will be missing. 4. The gene will be missing from the library.
10-21 Why is a heat-stable DNA polymerase from a thermophilic bacterium (the Taq polymerase) used in the polymerase chain reaction rather than a DNA polymerase from E. coli or humans?
162
10-22 Which of the following is a limitation on the use of PCR to detect and isolate genes? (a) The sequence at the beginning and end of the DNA to be amplified must be known. (b) It also produces large numbers of copies of sequences beyond the 5′ or 3′ end of the desired sequence. (c) It cannot be used to amplify a particular sequence from a mixture of mRNAs. (d) It cannot be used to amplify cDNAs. (e) It will amplify only sequences present in multiple copies in the DNA sample. 10-23 You have an oligonucleotide probe that hybridizes to part of gene A from a eucaryotic cell. Indicate whether a cDNA library or a genomic DNA library will be more appropriate to use for the following applications. A. You want to study the promoter of a gene A. B. Gene A encodes a tRNA and you wish to isolate a piece of DNA containing the full-length sequence of the tRNA. C. You discover that Gene A is alternatively spliced and you want to see which predicted alternative splice products are actually produced in a cell. D. You want to find both gene A and the genes located near gene A on the chromosome. E. You want to express gene A in bacteria to produce lots of protein A. 10-24 Which of the restriction nucleases listed below can potentially cleave a segment of cDNA that encodes the peptide KIGDACF? (Refer to Codon table provided above Q10-6.) Restriction Nuclease Eco RI Hind III Nsi I
Recognition Sequence GAATTC AAGCTT ATGCAT
10-25 You want to amplify the DNA between the two stretches of sequence shown in Figure Q10-25. Of the listed primers, choose the pair that will allow you to amplify the DNA by PCR.
Figure Q10-25
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10-26 Your friend works at the Centers for Disease Control and has discovered a brand new virus that has been recently introduced into the human population. She has just developed a new assay that allows her to detect the virus using PCR products made from the blood of infected patients. The assay utilizes primers in the PCR reaction that hybridize to sequences in the viral genome. Your friend is distraught because of the result she obtained (see Figure Q10-26) when she looked at PCR products made from using the blood of three patients suffering from the viral disease, from her own blood, and from a leaf from her petunia plant. You advise your friend not to panic, as you believe she is missing an important control. Which one of these choices is the best control for clarifying the results of her assay. Explain your choice.
Figure Q10-26 (a) (b) (c) (d) (e)
A PCR reaction using blood from a patient that is newly infected but does not yet show symptoms. A PCR reaction using blood from a dog. A PCR reaction from an uninfected person. Repeating the experiment she’s already done with a higher concentration of primers. Repeating the experiments she’s already done with a new tube of polymerase.
DNA Engineering 10-27 Insulin is a small protein hormone that regulates blood sugar level, and is given by injection to people who suffer from the disease diabetes. Diabetics once used insulin purified from pig pancreas to control their diabetes. Give two reasons why the drug companies who produce insulin wanted to clone the human insulin gene to provide an alternative source of the hormone.
164
10-28 Scientists produce large quantities of RNA by transcription in vitro rather than by expression of cloned genes in vivo primarily because (a) in vitro transcription removes the need to clone the gene for the RNA into a DNA vector. (b) the viral RNA polymerases used for in vitro transcription are not active in vivo. (c) RNA molecules are present in only a few copies per cell and thus only small amounts of RNA can be prepared in vivo. (d) in vitro transcription eliminates the need to purify the RNA produced from the cell’s RNA molecules. 10-29 You have been hired to create a cat that will not cause allergic reactions in cat-lovers. Your co-workers have cloned the gene encoding a protein found in cat saliva, expressed the protein in bacteria, and shown that it causes violent allergic reactions in people. But you soon realize that even if you succeed in making a knockout cat lacking this gene, anyone who buys one will easily be able to make more hypoallergenic cats just by breeding them. Which of the following will ensure that people will always have to buy their hypoallergenic cats from you? (a) (b) (c) (d) (e)
Injecting the modified ES cells into embryos that have a genetic defect that will prevent the mature adult from reproducing. Implanting the injected embryos into a female cat that is sterile due to a genetic defect. Selling only the offspring from the first litter of the female cat implanted with the injected embryos. Surgically removing the sexual organs of all the knockouts before you sell them. Selling only male knockouts.
10-30 Your biochemist friend has isolated a protein he thinks is responsible for making you feel sleepy. Since he knows you’re taking Cell Biology, he wants you to help him isolate the gene encoding this protein. Unfortunately, since your friend could only isolate small amounts of protein, he was only able to obtain three short stretches of amino acid sequence from the protein: (a) (b) (c)
H-C-W-K-M R-S-L-L-S D-A-Q-W-Y
For each of the three peptides above, you design a set of DNA oligonucleotide probes that can be used to detect the gene in a library. Which of the three sets of oligonucleotide probes would be preferable to screen a library? Explain. (Refer to Codon table provided above Q10-6.)
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10-31 Your friend has isolated a protein present in the cheek cells of all straight-A seniors at your University. She says that this protein helps you remember everything you read and therefore will help cut down on the number of hours needed to study for exams, and calls the protein “geniuszyme”. She sequences the protein and designs a probe to isolate the gene that encodes geniuszyme. To make sure she designed the probe correctly, she consults with the company that cloned Factor VIII. They have 100% confidence that her probe will work. She also obtains a high quality liver cDNA library from the company and uses her probe to try to isolate the gene for geniuszyme. Unfortunately, she is unable to isolate any clones. A. What is the likely explanation for her failure? B. Not to be discouraged, your friend has obtained some genomic DNA isolated from the nuclei of liver cells and has made a genomic library from that DNA. Do you expect she will succeed in isolating the geniuszyme gene from this library? Why or why not?
How We Know: Sequencing the Human Genome 10-32 You have been asked to consult for a biotech company that is seeking to understand why some fungi can live in very extreme environments, such as the high temperatures inside naturally occurring hot springs. The company has isolated two different fungal species, F. cattoriae and W. gravinius, both of which can grow at temperatures exceeding 95°C. The company has determined the following things about these fungal species: Property
F. cattoriae
W. gravinius
Genome size Repetitive DNA?
1 MB 20% of genome contains large stretches of CG repeats
3 MB < 0.1% of genome
By sequencing and examining their genomes, the biotech company hopes to understand why these species can live in extreme environments. However, the company only has the resources to sequence one genome, and would like your input as to which species should be sequenced and whether you believe a shotgun strategy will work in this case. (Be sure to explain your answer).
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Answers 10-1
The ability to purify uniform populations of cells from a tissue facilitates their study. In order to obtain a purified population of cells from a compact tissue, proteolytic enzymes must be used to disrupt the adhesive bonds between cells. The fluorescence-activated cell sorter allows for isolation of a particular cell type by recognizing cell-surface differences. Once cells are isolated, their properties can be studied by, for example, biochemical experiments that aim to purify molecules from cells and reconstitute cellular reactions in vitro (that is, in a test tube). Although cells may be grown in culture, most vertebrate cells have a limited number of cell divisions they can undergo, often due to lack of expression of the telomerase enzyme. Cells that can divide indefinitely may result from a genetic change; such cells are considered to be immortalized and can be propagated in culture as a homogeneous cell line.
10-2
Because embryonic stem cells are undifferentiated, they can retain the ability to give rise to any tissue in the body. If the differentiation of an embryonic stem cell is guided appropriately in culture, they could potentially provide a source of cells capable of replacing or repairing tissues that have been damaged by injury or disease.
10-3
A nuclease hydrolyzes the phosphodiester bonds in a nucleic acid. Nucleases that cut DNA only at specific short sequences are known as restriction nucleases. DNA composed of sequences from different sources is known as recombinant DNA. Gel electrophoresis can be used to separate DNA fragments of different sizes. Millions of copies of a DNA sequence can be made entirely in vitro by the polymerase chain reaction technique.
10-4
(d)
A restriction enzyme that has a 4 base-pair recognition sequence cuts on average once every 44 or 256 base pairs; one that has a 6 base-pair recognition sequence cuts once every 46 or 4096 base pairs; one that has an 8 base-pair recognition sequence cuts once every 48 or 65,536 base pairs; one that has a 12 base-pair recognition sequence cuts once every 412 or 16 million base pairs. So to obtain fragments of around 70 kb in size, you would cut with an enzyme that recognizes an 8 base-pair site.
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10-5
A.
B.
You would first digest your sample with a combination of restriction enzymes selected so that they give a set of fragment sizes that could have come from only one of the plasmids. Then you would run the resulting mixture of DNA fragments on a gel alongside a set of size markers and determine the sizes of each fragment. By looking at the restriction maps, you should then be able to match your results to one of the plasmids. Digestion with any of the following combinations will enable you to distinguish which plasmid you have: Hind III + Bgl II; Eco RI + Bgl II; Eco RI + Bgl II + Hind III. The plasmids are the same size, so you cannot distinguish them simply by making a single cut (with Hind III) and determining the size of the complete DNA by gel electrophoresis. Nor can you distinguish them by cutting with all four restriction nucleases, since the set of fragment sizes produced from both plasmids will be the same. Cutting with Bam HI or Eco RI on their own is not sufficient because you will get bands of the same size from both plasmid A and plasmid B. The only difference between the two plasmids is the location of the Bgl II site relative to the two Bam HI sites, so if you cut with an enzyme that cuts outside the Bam HI fragment and with Bgl II, you will get different-sized fragments from the two plasmids.
10-6
A. B.
TAGACTGACCTG Arg-Leu-Thr (Only the second reading frame can be used, as reading frame 1 contains a stop codon (TAG), as does reading frame 3 (TGA).)
10-7
Choice (d) is the correct answer. If some of the DNA templates you are sequencing are cut at one specific site (as would be the case if the DNA were cut by a restriction enzyme), the polymerase will stop when it comes to the end of the DNA, giving rise to at least some product of one particular size in all the reaction mixtures. If this were the case, all four lanes will have a band of this particular size. In addition, you would get normal sequence from the full-length templates, and normal sequence from those templates in which the polymerase incorporated a dideoxynucleotide before encountering the end. The other options are incorrect: if you added all four dideoxynucleotides to one of the reactions (choice (a)), that lane would have a band at every position because the polymerase would stop at A’s, C’s, G’s, and T’s instead of at only one type of nucleotide. If you forgot to add deoxynucleotides to the reactions (choice (b)), you would not get any polymerization, and all of your lanes would be blank. If you forgot to add dideoxynucleotides (choice (c)), the reactions would not stop until the end of the DNA fragment, and all of the products would be full-length and would all be at the top of the gel. If your primer hybridized to more than one part of the fragment of DNA you were sequencing (choice (e)), your gel would look as though two different sequences had been superimposed onto each other.
10-8
The technique of in situ hybridization can be used to detect a specific RNA expression in a particular region of the brain. Northern blotting detects a specific sequence in RNA. Southern blotting detects a specific sequence in DNA. A short, single-stranded DNA is called a(n) oligonucleotide. A piece of DNA used to detect a specific sequence in a nucleic acid by hybridization is known as a(n) probe. 168
10-9
A. B. C.
10-10 A. B.
All three are affected. The two parents have a single defective copy of the gene; the fetus has two defective copies. As seen in the two blots in Figure Q10-9B, Caril Ann and Charles each have one full-length copy of gene X (the bands at the top of the gel), which hybridizes with both probe A and probe B. The fetus does not. The blot with probe A shows that Caril Ann and the fetus have a short fragment of gene X that hybridizes with probe A only, indicating that this copy of gene X has a deletion somewhere other than in the region recognized by probe A. The second blot (with probe B) shows that Charles and the fetus have a short fragment that hybridizes with probe B, indicating that this copy of gene X has a deletion somewhere other than in the region recognized by probe B. Since the shortened gene found in Charles does not show up on the probe A blot, this deletion must be in the region of A; similarly, since the shortened gene found in Caril Ann does not show up on the probe B blot, her deletion must be in the region of B. The fetus has inherited two defective copies of gene X, one from each parent. All four oligonucleotide probes. Oligonucleotide probe B and oligonucleotide probe D. Both the upper and lower strands of DNA are present in genomic Southern blots, so all four oligos will hybridize to either Southern blot. (Oligonucleotides A and B will still be able to hybridize to genomic DNA cut with Bgl II, since they can still base-pair to the individual fragments that result from the digest.) Northern blots contain only RNA, which has the sequence of the upper strand of the DNA. Hence, only oligonucleotide probe B and oligonucleotide probe D will hybridize to a Northern blot.
10-11 (e) 10-12 The tubby gene is expressed in all tissues, including the brain. A red spot on a microarray is indicative of a DIFFERENCE in gene expression between the two RNA samples being compared. You may expect in this experiment that the tubby gene would be a yellow spot (a gene expressed at equal levels in both samples). An in situ experiment looks at the RNA level directly in the tissue of interest, which is why in this case you see ample levels of tubby RNA. 10-13 Two fragments of DNA can be joined together by DNA ligase. Restriction enzymes that cut DNA straight across the double helix produce fragments of DNA with blunt ends. A fragment of DNA is inserted into a vector in order to be cloned in bacteria. A cDNA library contains a collection of DNA clones derived from mRNAs. A genomic library contains a collection of DNA clones derived from chromosomal DNA.
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10-14 A. B.
1 and 2. When Sal I and Xho I cut DNA, the staggered ends left behind will match up by base-pairing and thus can be joined by ligase alone. 1, 2, and 4. Sma I cuts and leaves a blunt end. Addition of DNA polymerase and the four deoxynucleotides will fill in the 5′ overhangs generated by digestion with Sal I and Xho I, leaving blunt ends. Blunt ends can be joined by DNA ligase. However, 3′ overhangs will not be filled in (i.e., those generated by Pst I), as DNA polymerase moves in a 5′ to 3′ direction. DNA ligase will not join 3′ overhangs to blunt ended DNA, which are the situations presented in choices 3 and 5.
10-15 A cloning vector for use in bacteria must contain: 1. a bacterial replication origin (to allow the plasmid to be replicated); 2. at least one unique restriction site (to allow easy insertion of foreign DNA); and 3. an antibiotic-resistance gene or some other selectable marker gene (to allow selection for bacteria that have taken up the recombinant plasmids). 10-16 Choice (c) is the correct answer. The size of the fragments that are left after a restriction digest does not depend on the total size of the genome; it depends on the sequence of the genome and the frequency with which the restriction enzyme recognition site is found in the genome. Choices (a) and (b) are true: as a limiting case, you can think of what would happen if a fragment the size of the entire genome is inserted into the bacterial vector. In this case, you would have to screen only one colony to find the clone that hybridizes to your probe, but it will be very difficult to find out where on the insert your gene of interest lies. Choice (d) is true: the larger the gene you are seeking, the more likely it is that there will be a restriction fragment in the gene (or that the gene will be broken if the DNA was fragmented by random shearing), and hence the less likely it is that the entire gene will be found in one clone. Choice (e) is true: as the size of your oligo probe decreases, the chance of finding that sequence randomly in the genome increases (just as the number of restriction sites increases when the size of the recognition site decreases). 10-17 (c)
The very ends of all of the chromosomes are unlikely to be Not I sites, meaning that the fragments containing the ends of the chromosomes will not be able to insert into the bacterial vector (since they have not been cut by Not I at both ends) and will be lost from the library. All sequences present in genomic DNA (which includes regulatory sequences and introns) should be present in a genomic library. The coding sequence of the gene (and hence the amino acid sequence of the encoded protein) is also present in a genomic clone, although it is interrupted by intron sequences and therefore somewhat difficult (but not impossible) to determine.
10-18 Because most amino acids can be encoded by more than one codon, a given sequence of amino acids could be encoded by a number of different nucleotide sequences. Probes corresponding to all these possible sequences have to be synthesized in order to be sure of including the one that corresponds to the actual nucleotide sequence of the gene and thus will hybridize with it.
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10-19 The gene isolated from a genomic library would still contain introns, and bacteria do not contain the biochemical machinery for removing introns by RNA splicing. The same gene isolated from the cDNA library will have already had its introns removed. 10-20 A. B.
C.
D.
Outcome 1 would occur. If the mRNA is degraded from the 5′ end, it would still be reverse transcribed and would end up in the library as a clone lacking its 5′ end. Outcome 4 would occur. If the mRNA was degraded from the 3′ end, it would be missing its 3′ poly(A) tail. In the construction of a cDNA library, only molecules that still have their poly(A) tail will be reverse transcribed, so mRNAs missing their 3′ end will not be represented in the library. Outcome 1 would occur. If the 5′ end hybridizes to sequences in the middle of the gene, the “hairpin” formed when the single-stranded DNA loops back on itself to form the primer for DNA polymerase will be very large. After this loop is digested, the remaining double-stranded DNA fragment will be missing the 5′ end of the gene. Outcome 2 would occur. If the gene has a long stretch of internal A’s, the poly(T) primer used in the reverse transcription step could hybridize to the internal poly(A) stretch rather than to the poly(A) tail, and the resulting cDNA will have lost its 3′ end.
10-21 The PCR technique involves heating the reaction at the beginning of each cycle to separate the newly synthesized DNA into single strands so that they can act as templates for the next round of DNA synthesis. Using a heat-stable polymerase avoids having to add it afresh for each round of DNA replication. 10-22 Choice (a) is the correct answer. In order to construct primers that will bracket the desired gene, you have to know the sequence at the beginning and end of the DNA to be copied. Although a few copies of sequences beyond the ends of the desired sequence are made in the early cycles, these soon become a negligible fraction of the total DNA synthesized, and thus choice (b) would not be an appropriate usage of PCR and certainly not a limitation. You can use PCR to amplify a particular RNA sequence (choice (c)) by using the appropriate primers to first guide the reverse transcriptase reaction that makes a DNA copy of the RNA and then to guide the synthesis of the complementary DNA strand. PCR can be used to amplify a sequence from any DNA, including cDNAs (choice (d)). PCR is extremely sensitive and can detect and amplify a particular sequence even if it is present only in a single copy in the DNA sample (choice (e)).
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10-23 A. B. C. D. E.
Genomic library. (cDNAs are produced from mRNAs; therefore, the promoters will not be included in a cDNA library.) Genomic library. (cDNAs are usually constructed using an oligo dT primer; tRNAs do not have poly(A) tails. If the cDNA library were made using random primers, it would be unlikely to contain the full-length version of the tRNA.) cDNA library. (Since cDNAs are produced from mRNAs, isolating cDNAs would tell you which splice variants are produced in a cell.) Genomic library. (A genomic DNA fragment can contain the genes next to your gene of interest; a cDNA will not.) cDNA library. (Bacteria do not have the ability to remove introns, which may exist in DNA isolated from a genomic library.)
10-24 The nucleotide sequence that can encode the peptide KIGDACF is below: K I G D A C F AAA ATT GGT GAT GCT TGT TTT G C C C C C C A A A G G The enzyme Nsi I cleaves at ATGCAT. 10-25 The appropriate PCR primers are primer 1 (5′-GACCTGTGGAAGC) and primer 8 (5′-TCAATCCCGTATG). The first primer will hybridize to the bottom strand and prime synthesis in the rightward direction. The second primer will hybridize to the top strand and prime synthesis in the leftward direction. (Remember that strands pair antiparallel.) The middle two primers in each list (primers 2, 3, 6, and 7) would not hybridize to either strand. The remaining pair of primers (4 and 5) would hybridize, but would prime synthesis in the wrong direction—that is, outward, away from the central segment of DNA. Each of these wrong choices has been made at one time or another in most laboratories that use PCR. In most cases the confusion arises because the conventions for writing nucleotide sequences have been ignored. By convention, nucleotide sequences are written 5′ to 3′, with the 5′ end on the left. For double-stranded DNA, the 5′ end of the top strand is on the left.
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10-26 (c)
Primers can sometimes hybridize to unintended sequences and produce unintended products. The appropriate control for your friend’s experiment would be DNA from an uninfected person; that way she would be able to determine whether the bands present in the PCR from her blood truly correspond to product generated from viral DNA rather than cross-hybridization to DNA sequences in the human genome, since the bands would be absent from a person uninfected by the virus in the former case only. Doing PCR from an infected but asymptomatic person would not be useful, as it would not allow your friend to distinguish whether she is infected. Although doing PCR from dog blood should not give any viral bands, any nonspecific products would likely be different from a dog compared to your friend. The absence of PCR fragments in the petunia lane suggests that there is no viral contaminant in any of your friend’s reagents, so using a new tube of polymerase is not the solution. Increasing the concentration of primers will exacerbate any nonspecific hybridization.
10-27 Any two of the following would be acceptable. 1. Cloning the gene allows human insulin to be produced in large quantities from bacteria or other cells carrying the cloned DNA sequence. 2. It is easier and less costly to extract the same amount of insulin from a bacterial culture than from pig pancreas. 3. Insulin made in a bacterial culture and then purified will be free of any possible contaminating viruses that pigs (and any other whole animal) tend to harbor. 4. The pig protein has slight differences from the human protein, which can lead to side effects on prolonged use. Whenever possible, a human protein would be preferred for clinical treatment of this sort. 10-28 (d)
In order to produce RNA by transcription in vitro, you must first clone the DNA of the gene you wish to transcribe in order to get a large amount of pure template. Many RNAs are produced at high levels in cells. Viral RNA polymerases are able to transcribe RNA in vivo to especially high levels, since their purpose is to make high levels of viral proteins.
10-29 Choice (d) is the correct answer. Neutering all the knockout animals that you sell is the only option of the five listed that will prevent happy pet owners from becoming happy pet breeders. The situation described in choice (a), will not allow you to make any knockout cats because the first litter (which will at best have a few mosaics in which one copy of the gene has been knocked out in the germ cells) will be sterile and you will not be able to mate them. The genotype of the female cat in which you implant the embryos has no effect on the genotype of the embryos which is why choice (b) is incorrect. Choice (c) is incorrect because the first litter will yield mosaic cats that still have one copy of the allergen-producing gene in their cells, and therefore will not be hypoallergenic. Selling only male knockouts does no good, because they can be mated to normal females and the heterozygous offspring can be backcrossed to yield homozygous hypoallergenic cats (choice (e)).
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10-30 (a)
H-C-W-K-M. There is the least amount of degeneracy in the nucleotides that could code for this peptide. (See below) H C W K M CAC TGC TGG AAA ATG T T G R S L L S AGA AGC TTA TTA AGC G T G G T CGA TCA CTA CTA TCA G G G G G C C C C C T T T T T D A Q W Y GAC GCA CAA TGG TAC T G G T C T
10-31 A. B.
Geniuszyme is not expressed in the liver. Since cDNA is made from mRNA, a cDNA library is a reflection of the genes expressed in a particular tissue. Yes, she should be able to isolate the gene, because genomic DNA is essentially the same in all tissues.
10-32 Even though the genome of F. cattoriae is smaller, the fact that the W. gravinius genome contains less repetitive DNA makes it more attractive to sequence. Repetitive DNA makes the assembly of sequenced fragments difficult. Shotgun sequencing would not be an unrealistic approach for W. gravinius, because the genome of W. gravinius contains little repetitive DNA and is relatively small. The genome of H. influenzae is 1.83 MB and was successfully sequenced using the shotgun approach. (For comparison, the genome of S. cerevisiae is 14 MB.)
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CHAPTER 11 MEMBRANE STRUCTURE 2009 Garland Science Publishing 3rd Edition
The Lipid Bilayer 11-1 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. The specialized functions of different membranes are largely determined by the __________________ they contain. Membrane lipids are __________________ molecules, composed of a hydrophilic portion and a hydrophobic portion. All cell membranes have the same __________________ structure, with the __________________ of the phospholipids facing into the interior of the membrane and the __________________ on the outside. The most common lipids in most cell membranes are the __________________. The head group of a glycolipid is composed of __________________.
amphipathic cholesterol fatty acid tails glycolipids hydrophilic head groups
hydrophobic lipid bilayer lipid monolayer lipids phosphatidylcholine
phosphatidylserine phospholipids proteins sterols sugars
11-2 Which of the following membrane lipids does not contain a fatty acid tail? (a) Phosphatidylcholine (b) A glycolipid (c) Phosphatidylserine (d) Sphingomyelin (e) Cholesterol
11-3 Which of the following statements regarding lipid membranes is TRUE? (a) Phospholipids will spontaneously form liposomes in nonpolar solvents. (b) In eucaryotes, all membrane-enclosed organelles are surrounded by one lipid bilayer. (c) Membrane lipids diffuse within the plane of the membrane. (d) Membrane lipids frequently flip-flop between one monolayer and the other. (e) The preferred form of a lipid bilayer in water is a flat sheet with exposed edges.
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11-4
A bacterium is suddenly expelled from a warm human intestine into the cold world outside. Which of the following adjustments might the bacterium make to maintain the same level of membrane fluidity? (a) Increase the length of the hydrocarbon tails in its membrane phospholipids. (b) Increase the proportion of unsaturated hydrocarbon tails in its membrane phospholipids. (c) Increase the proportion of hydrocarbon tails with no double bonds in its membrane phospholipids. (d) Decrease the amount of cholesterol in the membrane. (e) Decrease the amount of glycolipids in the membrane.
11-5
Which of the following statements regarding the fatty acid tails of phospholipids is FALSE? (a) Phospholipids with unsaturated tails make the bilayer more fluid because the tails contain fewer hydrogen atoms and thus form fewer hydrogen bonds with each other. (b) Saturated phospholipid tails pack more tightly against each other than do unsaturated tails. (c) Most membrane phospholipids have one fully saturated tail. (d) Phospholipid tails in a membrane can interact with each other via van der Waals interactions. (e) Fatty acid tails vary in length.
11-6
New membrane synthesis occurs by (a) the spontaneous aggregation of free phospholipids into a new bilayer in the aqueous environment of the cell. (b) incorporation of phospholipids into both faces of a preexisting membrane by enzymes attached to each face. (c) incorporation of phospholipids into one face of a preexisting membrane and their random redistribution to both faces by flippases. (d) incorporation of phospholipids into one face of a preexisting membrane and their specific redistribution by flippases.
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11-7
Three phospholipids X, Y, and Z are distributed in the plasma membrane as indicated in Figure Q11-7. For which of these phospholipids does a flippase probably exist?
Figure Q11-7 (a) (b) (c) (d) (e)
X only Z only X and Y Y and Z X and Z
11-8
Where does most new membrane synthesis take place in a eucaryotic cell? (a) In the Golgi apparatus (b) In the endoplasmic reticulum (c) In the plasma membrane (d) In the mitochondria (e) On ribosomes
11-9
A small membrane vesicle containing a transmembrane protein is shown below. Assume this membrane vesicle is in the cytoplasm of a cell.
Figure Q11-9 A. B.
C.
Label the cytosolic and non-cytosolic faces of the membrane vesicle. This membrane vesicle will undergo fusion with the plasma membrane. Sketch the plasma membrane after vesicle fusion, indicating the new location of the vesicle membrane and the transmembrane protein carried by the membrane vesicle. On your drawing for B, label the original cytosolic and noncytosolic faces of the vesicle membranes as it resides in the plasma membrane. Also label the extracellular space and the cytosol. Indicate the amino and carboxyl terminus of the inserted transmembrane protein. 177
11-10 Why are glycolipids found on the extracellular, but not the cytoplasmic, surface of the plasma membrane? (a) Flippases transport them from the cytosolic face. (b) The enzymes that produce them are present only on the extracellular surface of the plasma membrane. (c) The enzymes that add the sugar groups are confined to the inside of the endoplasmic reticulum and the Golgi apparatus. (d) The oligosaccharides on glycolipids are cleaved off by enzymes found only in the cytosol. (e) They flip spontaneously, after incorporation, due to the hydrophilic sugar head groups.
Membrane Proteins 11-11 A group of membrane proteins, which can be extracted only from membranes using detergents, are all found to have a similar amino acid sequence at their carboxyl terminus: -KKKKKXXC (where K stands for lysine, X stands for any amino acid, and C stands for cysteine). This sequence is essential for their attachment to the membrane. What is the most likely way in which the carboxyl-terminal sequence attaches these proteins to the membrane? (a) The cysteine residue is covalently attached to a membrane lipid. (b) The peptide spans the membrane as an α helix. (c) The peptide spans the membrane as part of a β sheet. (d) The positively charged lysine residues interact with an acidic integral membrane protein. 11-12 A strain of bacteria secretes a toxin that can lyse human red blood cells. You are able to partially purify the toxin and find that it is a small protein. Furthermore, the toxin is capable of rendering liposomes made of pure phospholipids permeable to many different ions. What type of protein is the bacterial toxin likely to be? (a) A flippase (b) A β-barrel protein (c) A protease (d) A protein containing a single hydrophobic α helix. (e) An enzyme that adds carbohydrate groups to lipids.
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11-13 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. There are several ways that membrane proteins can associate with the cell membrane. Membrane proteins that extend through the lipid bilayer are called __________________ proteins and have __________________ regions that are exposed to the interior of the bilayer. On the other hand, membrane-associated proteins do not span the bilayer and instead associate with the membrane through an α-helix that is __________________. Other proteins are __________________ attached to lipid molecules that are inserted in the membrane. __________________ membrane proteins are linked to the membrane through noncovalent interactions with other membrane-bound proteins. amphipathic cortical covalently detergent
hydrophilic hydrophobic integral micelle
noncovalently peripheral transmembrane unfolded
11-14 Which of the following statements regarding membrane proteins is FALSE? (a) Integral membrane proteins often precipitate (form insoluble aggregates) in aqueous solutions lacking detergents. (b) Some hydrophobic amino acids in membrane proteins are not in contact with the lipid bilayer. (c) Peripheral membrane proteins can be dissociated from membranes using a gentle detergent. (d) Strong detergents can completely unfold both membrane and nonmembrane proteins. (e) In transmembrane proteins that form an aqueous pore through the membrane, the pore is lined with hydrophobic amino acid side chains. 11-15 Proteins often span the plasma membrane as an α- helix. If it takes 20 amino acids to span a lipid bilayer, which of the three 20-amino acid sequences listed below is the most likely candidate for such a transmembrane segment? Explain your choice. (a) IALIVFGVFAGVIGGILLIS (b) KITPVVKHKHDKIDTPLLIR (c) DTYYYRRREADDDDDLLISD
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11-16 Proteins that form a ß-barrel pore in the membrane have several ß-strands that span the membrane. The amino acid side chains facing the inside of the pore would be hydrophilic whereas the amino acid side chains facing the lipid bilayer would be hydrophobic. Which of the three 10-amino acid sequences listed below is the most likely candidate for a transmembrane ß-strand in a ß-barrel protein? Explain your choice. (a) ADFKLSVELT (b) AFLVLDKSET (c) AFDKLVSELT 11-17 Which of the following functions does the cell cortex perform? (a) It influences the passage of small molecules into and out of the cell. (b) It allows cells to change shape and to move. (c) It lubricates the cell. (d) It restricts the movement of certain proteins in the lipid bilayer. (e) It supports and strengthens the membrane. 11-18 You have isolated two mutants of a normally pear-shaped microorganism that have lost their distinctive shape and are now round. One of the mutants has a defect in a protein you call A and the other has a defect in a protein you call B. You grind up each type of mutant cell and normal cells separately and separate the plasma membranes from the cytoplasm. You then wash the membrane fraction with a low concentration of urea (which will unfold proteins and disrupt their ability to interact with other proteins) and centrifuge the mixture. The membranes and their constituent proteins form a pellet while the proteins liberated from the membranes by the urea wash remain in the supernatant. When you check each of the factions for the presence of A or B, you obtain the results given below.
Normal cells Mutant A Mutant B
First cell extract Membrane Cytosol A and B no A or B A B A B
After urea wash and centrifugation Membrane Supernatant A B A no A or B A no A or B
Which of the following statements are consistent with your results? (a) Protein A is an integral membrane protein that interacts with B, a peripheral membrane protein that is part of the cell cortex. (b) Protein B is an integral membrane protein that interacts with A, a peripheral membrane protein that is part of the cell cortex. (c) Proteins A and B are both integral membrane proteins. (d) The mutation in A affects its ability to interact with B. (e) The mutation in A affects its ability to interact with the membrane.
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11-19 Detergents and phospholipids are both amphipathic molecules. However, when exposed to water, detergents aggregate into small clusters called micelles while phospholipids form closed spherical liposomes. Explain what the difference is between the structure of a detergent compared to the structure of a phospholipid that causes detergents to form micelles instead of liposomes. 11-20 Diversity among the oligosaccharide chains found in the carbohydrate coating of the cell surface can be achieved in which of the following ways? (a) Varying the types of sugar monomers used (b) Varying the types of linkages between sugars (c) Varying the number of monomers in the chain (d) Varying the number of branches in the chain (e) All of the above 11-21 Which of the following statements about the carbohydrate coating of the cell surface is FALSE? (a) It is not usually found on the cytosolic side of the membrane. (b) It can play a role in cell-cell adhesion. (c) The arrangement of the oligosaccharide side chains are highly ordered, much like the peptide bonds of a polypeptide chain. (d) Specific oligosaccharides can be involved in cell-cell recognition. (e) It can protect the cell surface from mechanical and chemical damage. 11-22 Cell membranes are fluid and thus proteins can diffuse laterally within the lipid bilayer. However, sometimes the cell needs to localize proteins to a particular membrane domain. Name three mechanisms a cell can use to restrict a protein to a particular place in the cell membrane.
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How We Know: Measuring Membrane Flow 11-23 Fluorescence recovery after photobleaching (FRAP) is a technique that allows visualization of diffusion within the membrane. You set out to perform FRAP using four different samples of cells. In sample 1, you have fluorescently labeled a phospholipid. In samples 2, 3, and 4 you have fluorescently labeled membrane proteins X, Y, and Z, respectively. You photobleach an area of the membrane in each sample and record the rate of recovery of fluorescence. The data you obtain are shown in the graphs below:
Figure Q11-23 A. B.
Using the data from these graphs, list the proteins in order of their ability to diffuse in the membrane, from fastest to slowest. Given the data on the graphs above, is the following statement a good hypothesis: protein X and protein Z are always present in the cell as part of the same protein complex? Explain your answer.
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11-24 You perform single-particle tracking (SPT) experiments on three proteins, A, B, and C, and obtain the following pathways:
Figure Q11-24 A. B.
Which protein displays a pattern of motion similar to a protein that is anchored to the cytoskeleton? If you were to find a cell containing a membrane where 10% of its membrane proteins were anchored to the cytoskeleton, how do you think this would affect the fluidity of the lipids in the membrane?
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Answers 11-1
The specialized functions of different membranes are largely determined by the proteins they contain. Membrane lipids are amphipathic molecules, composed of a hydrophilic portion and a hydrophobic portion. All cell membranes have the same lipid bilayer structure, with the fatty acid tails of the phospholipids facing into the interior of the membrane and the hydrophilic head groups on the outside. The most common lipids in most cell membranes are the phospholipids. The head group of a glycolipid is composed of sugars.
11-2
(e)
11-3
(c)
11-4
Choice (b) is the correct answer. At colder temperatures, the membrane will be less fluid. Hence, in order to maintain the status quo, the bacterium will have to take measures to increase membrane fluidity. Increasing the length of the hydrocarbon tails (choice (a)) would decrease membrane fluidity, while decreasing the number of glycolipids (choice (e)) would have little or no effect. Decreasing the proportion of fatty acid tails with no double bonds (fully saturated) (choice (c)) would decrease membrane fluidity. Bacteria do not have cholesterol (choice (d)).
11-5
(a)
11-6
(d)
11-7
(c)
11-8
(b)
The remaining answers are false. Phospholipids form bilayers only in polar solvents. Nuclei and mitochondria are enclosed by two membranes. Membrane lipids rarely flip-flop between one monolayer and the other. The preferred form of a lipid bilayer in water is a closed sphere, so that the hydrophobic groups at the edges of the bilayer can avoid contact with water.
Unsaturated fatty acid tails do have fewer hydrogen atoms and do interact less well with one another but not for the reason stated. The decrease in interaction is due to a decrease in van der Waals interactions between the hydrocarbon tails because they can pack less closely. Hydrocarbon chains cannot form hydrogen bonds with each other.
As phospholipids are initially inserted into the cytosolic face of the lipid bilayer, flippases are required to move X and Y to the opposite face, as they cannot spontaneously flip across the bilayer.
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11-9
A.
See Figure A11-9A.
Figure A11-9A B., C. See Figure A11-9B.
Figure A11-9B 11-10 (c) 11-11 (a)
The peptide is extremely hydrophilic and is, therefore, unlikely to be inserted into the lipid bilayer. It is also too short to span the membrane as an α-helix. Although it is possible that the lysines interact with an acidic membrane protein, if such an interaction were solely responsible for attaching the protein to the membrane, it would not require a detergent to remove the protein, since ionic interactions are disrupted by milder treatments such as salt washes and pH changes.
11-12 Choice (b) is the correct answer. Insertion of a pore-forming protein into the lipid bilayer will have the effects noted by making the red blood cell unable to regulate its internal ion composition. The type of protein most likely to form the sort of nonspecific pore described is a β-barrel protein. Enabling membrane lipids to flip from one layer of the bilayer to the other should not affect permeability (choice (a)). A protease would have no effect on liposomes made of pure phospholipids (choice (c)). Proteins containing a single hydrophobic α-helix would not form a channel through the membrane (choice (d)). Addition of carbohydrate groups to lipids should not make the bilayer any more permeable to ions (choice (e)).
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11-13 There are several ways that membrane proteins can associate with the cell membrane. Membrane proteins that extend through the lipid bilayer are called transmembrane proteins and have hydrophobic regions that are exposed to the interior of the bilayer. On the other hand, membrane-associated proteins do not span the bilayer and instead associate with the membrane through an α-helix that is amphipathic. Other proteins are covalently attached to lipid molecules that are inserted in the membrane. Peripheral membrane proteins are linked to the membrane through noncovalent interactions with other membrane-bound proteins. 11-14 Choice (e) is the correct answer. In transmembrane proteins that form an aqueous pore through the membrane, the pore is lined with hydrophilic amino acid side chains. The other statements are all true. Integral membrane proteins often precipitate in aqueous solutions because of their stretches of hydrophobic amino acids (choice (a)). Proteins also contain hydrophobic amino acids in parts of the protein other than the membranespanning region, for example, in the cores of their extracellular or cytoplasmic domains (choice (b)). Peripheral membrane proteins are attached to the membrane by noncovalent interactions with other membrane proteins, making their membrane association relatively weak and thus disruptable by gentle detergents (choice (c)). 11-15 Choice (a) is the correct answer. When an α-helix traverses a lipid bilayer, the amino acid side chains are exposed to the hydrophobic portion of the lipid bilayer and thus should consist of nonpolar side chains. The sequence given in choice (a) contains nonpolar amino acids. Choices (b) and (c) are inappropriate, as they both contain several negatively and positively charged amino acids. (Note that in addition, choice (b) contains two prolines, which would render this sequence unable to form an α-helical structure and therefore lead to exposure of the hydrogen-bonding moieties in the polypeptide background to the nonpolar environment of the lipid bilayer.) 11-16 (a)
In a β-sheet, the amino acid side chains project alternately above and below the plane of the sheet. Therefore, every other amino acid side chain will face the same side of the strand. If a β-sheet were part of a β-barrel pore, that would mean that one side of the sheet would face the lipid bilayer while the other side would face the hydrophilic pore. This would necessitate an alternation between hydrophilic and hydrophobic amino acid side chains so that one side of the sheet would be hydrophobic while the other side was hydrophilic. The amino acids in choice (a) alternate between amino acids with nonpolar (hydrophobic) side chains and amino acids with polar (hydrophilic) side chains.
11-17 (b), (d), and (e)
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11-18 (a), (d) The results from the extracts of normal cells show that protein A is an integral protein that remains in the membrane through all the treatments, while protein B is a peripheral protein that can be removed from the membrane by urea. In the cell extracts from the mutants in A, the protein A still remains in the membrane, but the B protein does not. This is consistent with the mutation in A affecting its interaction with B. The same results are obtained when the B protein is mutant, which is consistent with the idea that A and B interact. The loss of an interaction between an integral membrane protein and a protein in the cortex would be more likely to result in a change in cell shape than the loss of an interaction between an integral membrane protein and a protein on the exterior of the cell. 11-19 A detergent has one hydrophobic tail while a phospholipid has two hydrophobic tails. This difference is reflected in the geometry of the molecules, with detergent molecules being shaped more like cones whereas phospholipids are more cylindrical. The cone-like detergent molecules will aggregate into micelles, with the hydrophilic head group on the outside and the hydrophobic tail group on the inside. On the other hand, the cylindrical shape of a phospholipid means that when phospholipids aggregate, the formation of a bilayer is most energetically favorable, with the hydrophobic tails on the inside of the two-layered sheet and the hydrophilic heads facing the water molecules. However, it is not energetically favorable for a phospholipid bilayer to exist as a sheet, because the exposed free edges would lead to exposure of the hydrophobic tails to water. Therefore, phospholipid bilayers spontaneously close to form spherical liposomes. 11-20 (e) 11-21 (c)
The sugars in an oligosaccharide side chain attached to the cell surface can be joined together in many different ways and in varied sequences.
11-22 Any combination of these four answers is acceptable: 1. The protein can be attached to the cell cortex inside the cell. 2. The protein can be attached to the extracellular matrix outside the cell. 3. The protein can be attached to other proteins on the surface of a different cell. 4. The protein can be restricted in its ability to diffuse by a diffusion barrier, such as that set up by specialized junctional proteins at a tight junction. 11-23 A.
B.
X diffuses the fastest, followed by Z, with Y barely diffusing at all. (The faster the recovery of fluorescence in the bleached area, the greater the diffusion coefficient of the protein and the faster the protein diffuses.) It is unlikely that X and Z are part of the same protein complex because then the rate of diffusion of X and Z should be the same.
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11-24 A. B.
Protein C displays the least amount of motion and therefore is the one most likely to be tethered to the cytoskeleton. The fluidity of the lipid bilayer should not be affected by the anchoring of membrane proteins. The lipid molecules should still be able to flow around anchored proteins much like water flows around rocks in a stream. (Note that the fluidity of the lipid bilayer is affected by the degree of saturation found in the hydrocarbon tails, the length of the hydrocarbon tails, and in animal cells, the amount of cholesterol in the membrane.)
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CHAPTER 12 MEMBRANE TRANSPORT 2009 Garland Science Publishing 3rd Edition
Principles of Membrane Transport 12-1 The most abundant intracellular cation is (a) Na+ (b) Ca2+ (c) H+ (d) K+ (e) positively charged macromolecules. 12-2 Circle the molecule in each pair that is more likely to diffuse through the lipid bilayer.
12-3 A hungry yeast cell lands in a vat of grape juice and begins to feast on the sugars there, producing carbon dioxide and ethanol in the process:
Unfortunately, the grape juice is contaminated with proteases (enzymes that attack some of the transport proteins in the yeast cell membrane), and the yeast cell dies. Which of the following could account for the yeast cell’s demise? (a) (b) (c) (d) (e)
Toxic buildup of carbon dioxide inside the cell Toxic buildup of ethanol inside the cell Diffusion of ATP out of the cell Inability to import sugar into the cell Inability to take water into the cell
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12-4
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. A molecule moves down its concentration gradient by __________________ transport, but requires __________________ transport to move up its concentration gradient. Carrier proteins and ion channels function in membrane transport by providing a __________________ pathway through the membrane for specific polar solutes or inorganic ions. __________________ are highly selective in the solutes they transport, binding the solute at a specific site and changing its conformation in order to transport the solute across the membrane. On the other hand, __________________ discriminate among solutes mainly on the basis of size and electrical charge. active amino acid amphipathic
carrier proteins hydrophilic hydrophobic
ion channels noncovalent passive
Carrier Proteins and Their Functions 12-5
Your friend isolates a new species of yeast that can grow using either ethanol (CH3CH2OH) or acetate (CH3COO–) as the yeast’s sole carbon source. He measures the rate of uptake of these carbon sources by the yeast as a function of concentration of carbon source and obtains the following graphs:
Figure Q12-5 A. B. 12-6
Is molecule A or B more likely to utilize a carrier protein to mediate its transport? Why? Given what you know about membrane transport, is molecule A more likely to be ethanol or acetate? Why?
Name the three main types of active transport.
190
12-7
Which of the following descriptions are TRUE of the bacterial protein bacteriorhodopsin? It is a (a) light-driven pump. (b) proton pump. (c) passive transporter. (d) coupled transporter. (e) transmembrane protein.
12-8
The Aeroschmidt weed contains an ATP-driven ion pump in its vacuolar membrane that pumps potentially toxic heavy metal ions such as Zn2+ and Pb2+ into the vacuole. The pump protein exists in a phosphorylated and an unphosphorylated form and works in a similar way to the Na+-K+ pump of animal cells. To study its action, you incorporate the unphosphorylated form of the protein into phospholipid vesicles containing K+ in their interiors. (You ensure that all of the protein molecules are in the same orientation in the lipid bilayer.) When you add Zn2+ and ATP to the solution outside such vesicles, you find that Zn2+ is pumped into the vesicle lumen. You then expose vesicles containing the pump protein to the solutes as shown in Table 12-8A. Table 12-8A Outside Inside
A B C 2+ 2+ Zn + ATP Zn Zn + ATP K+ 2+
D Zn2+ K+
E ATP
F ATP K+
You then determine the amount of phosphorylated and unphosphorylated ATP-driven ion pump protein in each sample. Your results are summarized in Table 12-8B, where “–” indicates an absence of a type of protein and “+” indicates the presence of a type of protein. Table 12-8B phosphorylated protein present unphosphorylated protein present
A +
B –
C –
D –
E –
F ++
++
++
++
++
++
–
What would you expect to happen if you treat vesicles as in lane F, but before determining the phosphorylation state of the protein, you wash away the outside buffer and replace it with a buffer containing only Zn2+? (a) Nothing will happen. (No Zn2+ will move into the vesicle; no K+ will move out of the vesicle; the phosphorylation state of the protein will not change.) (b) No Zn2+ will move into the vesicle; no K+ will move out of the vesicle; the protein will become unphosphorylated. (c) A small amount of Zn2+ will move into the vesicle; no K+ will move out of the vesicle; the phosphorylation state of the protein will not change. (d) A small amount of Zn2+ will move into the vesicle; no K+ will move out of the vesicle; the protein will become unphosphorylated. (e) A small amount of Zn2+ will move into the vesicle; a small amount of K+ will move out of the vesicle; the phosphorylation state of the protein will not change. 191
12-9
Explain why Na+ is commonly used to drive the coupled inward transport of nutrients in animal cells.
12-10 Ouabain inhibits the active uptake of glucose into epithelial cells by (a) binding to the glucose-Na+ symport. (b) opening K+ channels. (c) changing the pH of the cell. (d) increasing the intracellular concentration of Na+. (e) depleting the cell of ATP. 12-11 Which of the following statements is TRUE? (a) Amoebae have carrier proteins that actively pump water molecules from the cytoplasm to the cell exterior. (b) Bacteria and animal cells rely on the Na+–K+ pump in the plasma membrane to prevent lysis due to osmotic imbalances. (c) The Na+–K+ pump allows animal cells to thrive under conditions of very low ionic strength. (d) The cell wall surrounding plant cells prevents osmosis. (e) The Na+–K+ pump helps to keep both Na+ and Cl– ions out of the cell. 12-12 The Ca2+ pumps in the plasma membrane and endoplasmic reticulum are examples of (a) ATP-driven pumps. (b) coupled transporters. (c) passive carrier proteins. (d) ion channels. (e) symports. 12-13 Ca2+ pumps in the plasma membrane and endoplasmic reticulum are important for (a) maintaining osmotic balance. (b) preventing Ca2+ from altering the behavior of molecules in the cytosol. (c) providing enzymes in the endoplasmic reticulum with Ca2+ ions that are necessary for their catalytic activity. (d) maintaining a negative membrane potential. (e) helping cells import K+. 12-14 Do you agree of disagree with the following statement? Explain your answer. A symporter would function as an antiporter if its orientation in the membrane were reversed (that is, if the portion of the protein normally exposed to the cytosol faced the outside of the cell instead).
192
12-15 You have prepared lipid vesicles (spherical lipid bilayers) that contain Na+–K+ pumps as the sole membrane protein. All of the Na+–K+ pumps are oriented in such a way that the portion of the molecule that normally faces the cytosol is in the inside of the vesicle and the portion of the molecule that normally faces the extracellular space is on the outside of the vesicle. Assume that each pump transports one Na+ ion in one direction and one K+ ion in the other direction during each pumping cycle (see Figure Q12-15 for how the Na+–K+ pump normally functions in the plasma membrane).
Figure 12-15 Predict what would happen in each of the following conditions: A. The solutions inside and outside the vesicles contain both Na+ and K+ ions but no ATP. B. The solution outside the vesicles contains both Na+ and K+ ions; the solution inside contains both Na+ and K+ ions and ATP. C. The solution outside contains Na+; the solution inside contains Na+ and ATP.
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12-16 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. For an uncharged molecule, the direction of passive transport across a membrane is determined solely by its __________________ gradient. On the other hand, for a charged molecule, an additional force called the __________________ must also be considered. The net driving force for a charged molecule across a membrane therefore has two components and is referred to as the __________________ gradient. Active transport allows the movement of solutes against this gradient. The carrier proteins called __________________ transporters utilize the movement of one solute down its gradient to provide the energy to drive the uphill transport of a second gradient. When this transporter moves both ions in the same direction across the membrane, it is considered a(n) __________________; if the ions move in opposite directions, the transporter is considered a(n) __________________. antiport ATP hydrolysis concentration
coupled electrochemical light-driven
membrane potential symport uniport
Ion Channels and the Membrane Potential 12-17 Ion channels (a) only open in response to a signal of some kind. (b) require input of energy in order to function. (c) have no limit to the rate at which they can transport ions. (d) can transport both negative and positive ions through the same channel. (e) allow passage of ions in both directions. 12-18 A gated ion channel (a) stays continuously open when stimulated. (b) opens more frequently in response to a given stimulus. (c) opens more widely the stronger the stimulus. (d) remains closed if unstimulated.
194
12-19 You have patch-clamped a single voltage-gated ion channel in a membrane and have obtained the following recording (see Figure Q12-19A).
[Figure Q12-19A] For this channel, match the recordings depicted in Figure Q12-19B with the appropriate choices 1 through 4. 1. A channel that is closed all the time. 2. A channel that is open all the time. 3. A channel where the ion flow has been reversed. 4. A channel spending more time in the open configuration.
[Figure Q12-19B]
195
12-20 For each of the following sentences, fill in the blank with the appropriate type of gating for the ion channel described. You can use the same type of gating mechanism more than once. A. The acetylcholine receptor in skeletal muscle cells is a __________________ ion channel. B. __________________ ion channels are found in the hair cells of the mammalian cochlea. C. __________________ ion channels in the mimosa plant propagate the leafclosing response. D. __________________ ion channels respond to changes in membrane potential. E. Many receptors for neurotransmitters are __________________ ion channels. 12-21 The high intracellular concentration of K+ in a resting animal cell is partly due to (a) the K+ leak channels in the plasma membrane. (b) the Na+–K+ pump in the plasma membrane. (c) voltage-gated K+ channels in the plasma membrane. (d) intracellular stores of K+ in the endoplasmic reticulum. (e) the membrane potential. 12-22 State whether you agree or disagree with the following statement. Explain your answer. The membrane potential arises from movements of charge that leave ion concentrations practically unaffected and result in only a very slight discrepancy in the number of positive and negative ions on the two sides of the membrane.
Ion Channels and Signaling in Nerve Cells 12-23 Match the numbered lines with the following structures: A. Nerve terminal B. Cell body C. Axon D. Dendrite
Figure Q12-23 196
12-24 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. The action potential is a wave of __________________ that rapidly spreads along the neuronal plasma membrane. This wave is triggered by a local change in the membrane potential to a value that is __________________ negative than the resting membrane potential. The action potential is propagated by the opening of __________________gated channels. During an action potential, the membrane potential changes from __________________ to __________________. The action potential travels from the neuron’s __________________ along the __________________ to the nerve terminals. Neurons chiefly receive signals at their highly branched __________________. anions axon cell body cytoskeleton dendrites
depolarization hyperpolarization less ligand more
negative neutral positive pressure voltage
12-25 On the diagram of an action potential shown below, draw a dashed line showing what would have happened to the membrane potential after the initial depolarizing stimulus if there had been no voltage-gated Na+ channels in the membrane.
Figure Q12-25
197
12-26 When a neuron receives a signal, an action potential can be triggered. A graph of the traveling membrane depolarization assumes a characteristic shape as the depolarization moves down the axon (i.e., the peak height and length of depolarization is constant). What property of the voltage-gated Na+ channel ensures that the action potential assumes this characteristic form? Explain your answer. 12-27 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Neurons communicate with each other through specialized sites called __________________. Many neurotransmitter receptors are ligand-gated ion channels that open transiently in the __________________ cell membrane in response to neurotransmitters released by the __________________ cell. Ligand-gated ion channels in nerve cell membranes convert __________________ signals into __________________ ones. Neurotransmitter release is stimulated by the opening of voltage-gated __________________ in the nerve terminal membrane. acetylcholine receptor Ca2+ channels chemical electrical
GABA receptor K+ channels Na+ channels postsynaptic
presynaptic synapses
12-28 Inhibitory neurotransmitters such as GABA often open Cl– channels. Explain how the opening of a Cl– channel will affect the firing of an action potential. 12-29 K+ leak channels are important for setting the resting membrane potential. If you were to add a drug that inhibits K+ leak channels in neurons, would this make it easier or harder to trigger an action potential in the treated neurons? Why?
198
How We Know: Squid Reveal Secrets of Membrane Excitability 12-30 Studies on the squid giant axon were instrumental in our current understanding of how action potentials are generated. You decide to do some experiments on the squid giant axon yourself. A. You remove the cytoplasm in an axon and replace it with an artificial cytoplasm that contains twice the normal concentration of K+ by adding KOAc, where OAc– is an anion that is impermeable to the membrane. In this way you double the internal concentration of K+ while maintaining bulk electrical balance of the cytoplasmic solution. Will this make the resting potential of the membrane more or less negative? B. You add NaCl to the extracellular fluid and effectively double the amount of extracellular Na+ cation. How does this affect the action potential? C. You replace half of the NaCl in the extracellular fluid with choline chloride. (Choline is a monovalent cation much larger than Na+. Note that the presence of choline will not impede the flow of Na+ through its channels.) How will this affect the action potential?
199
Answers 12-1
(d)
12-2
The two basic properties governing the likelihood of whether a molecule will diffuse through a lipid bilayer are the size of the molecule and the charge of the molecule. A smaller molecule will be more likely to diffuse through the lipid bilayer than a larger molecule. A nonpolar (hydrophobic) molecule will be more likely to diffuse through the lipid bilayer than a polar molecule, which is more likely to diffuse through the lipid bilayer than a charged molecule. A. benzene (small nonpolar vs. larger uncharged) B. ethanol (polar vs. charged) C. glucose (large polar vs. very large highly charged) D. O2 (nonpolar vs. polar) E. adenosine (polar vs. highly charged)
12-3
Choice (d) is the correct answer. Because the lipid bilayer is permeable to carbon dioxide and ethanol, destroying membrane proteins is unlikely to affect their exit (choices (a) and (b)). The lipid bilayer is also permeable to water (choice (e)). On the other hand glucose requires a membrane transport protein to be imported into the cell. ATP, which is a highly charged molecule, also requires a transport protein to cross a membrane and thus could not be lost from the cell by simple diffusion (choice (c)).
12-4
A molecule moves down its concentration gradient by passive transport, but requires active transport to move up its concentration gradient. Carrier proteins and ion channels function in membrane transport by providing a hydrophilic pathway through the membrane for specific polar solutes or inorganic ions. Carrier proteins are highly selective in the solutes they transport, binding the solute at a specific site and changing its conformation in order to transport the solute across the membrane. On the other hand, ion channels discriminate among solutes mainly on the basis of size and electrical charge.
200
12-5
A.
B.
Molecule B is more likely to utilize a carrier protein. From the graph, the yeast cell’s uptake of molecule B levels off when the concentration of carbon source is high; this indicates a saturation point for the uptake of molecule B. Carrier proteins work by specifically binding their solutes and transferring the solute molecules across the membrane one at a time, down the concentration gradient of the solute. Thus, the rate at which a solute can be transported by a carrier protein is limited by the number of carrier proteins in the membrane and will reach a maximum when the solute concentration is high enough for solute molecules to saturate the carrier proteins. Since the graph indicates that the uptake of molecule B reaches a saturation point at high concentrations, molecule B is more likely to utilize a carrier protein. Molecule A is more likely to be ethanol. Uptake of molecule A is proportional to its concentration, while the uptake of molecule B reaches a saturation point at high concentration. The linear dependence of molecule A upon its concentration suggests that molecule A can diffuse into the cell. Since ethanol (a small polar molecule) is more likely to diffuse into the cell compared to acetate (a charged molecule), ethanol is likely to be molecule A. Since acetate is charged, it is more likely to require a carrier protein in order to be transported across the membrane.
12-6
ATP-driven transport, coupled transport, light-driven transport
12-7
(a), (b), and (e)
12-8
(d)
12-9
Na+ is commonly used to drive coupled transport in animal cells because a steep concentration gradient of Na+ (high outside and low inside) is maintained by the Na+–K+ pump. Na+ readily flows back into the cell down this gradient because of the negative membrane potential. The energy provided by the flow of Na+ down this steep electrochemical gradient can be harnessed by coupled transporters.
12-10 (d)
If the pump is mechanistically similar to the Na+–K+ pump, then the transport of ions is driven by ATP hydrolysis, the pump is transiently phosphorylated; phosphorylation is stimulated by one ion and dephosphorylation is stimulated by the other ion. Since all of the protein is in the phosphorylated form in the absence of Zn2+ (lane F), Zn2+ is probably required for dephosphorylation. K+, then, probably binds to the dephosphorylated form and stimulates the ATPase/autophosphorylation. So, if Zn2+ is added to the phosphorylated pump, Zn2+ will stimulate dephosphorylation, trigger a conformational change, and be injected into the vesicle. K+ will stimulate the kinase activity of the pump, but since there is no ATP to be hydrolyzed in the interior of the vesicle, no phosphorylation and hence no movement of K+ will occur.
Ouabain inhibits the Na+–K+ pump and, therefore, leads to an increase in the intracellular concentration of Na+, which is continually leaking into the cell. Uptake of glucose into epithelial cells occurs via a Na+-glucose symport, which uses the Na+ gradient to drive movement of glucose into the cell.
201
12-11 Choice (e) is the correct answer. The Na+–K+ pump keeps Na+ out directly by pumping it out and keeps Cl– out indirectly by helping to maintain the negative membrane potential. Cells do not have pumps for moving water molecules across the membrane (choice (a)), since the lipid bilayer is permeable to water. Bacteria do not have Na+–K+ pumps in their plasma membranes (choice (b)). The Na+–K+ pump cannot directly remove water molecules from the cell; it helps maintain osmotic balance by pumping out the Na+ that leaks in, which would not help if the cell is in a solution of very low ionic strength (choice (c)). The plant cell wall is permeable to water and therefore cannot prevent osmosis (choice (d)). 12-12 (a) 12-13 Choice (b) is the correct answer. The major purpose of the Ca2+ pumps is to keep the cytosolic concentration of Ca2+ low. When Ca2+ does move into the cytosol, it alters the behavior of many proteins; hence Ca2+ is a powerful signaling molecule. It is not involved in the catalytic activity of ER enzymes (choice (c)). Since the levels of Ca2+ are very low relative to the levels of K+ and Na+, the Ca2+ gradient does not have a significant effect on the osmotic balance of the cell (choice (a)) or the membrane potential (choice (d)). They are not involved in K+ import (choice (e)). 12-14 Disagree. A symporter functions by transporting two solutes in the same direction while an antiporter transports two different solutes to opposite sides of the membrane. Therefore, flipping the symporter around would not change its characteristics into that of an antiporter, as it should still transport both solutes in the same direction. 12-15 A. B.
C.
Without any ATP to provide energy for the Na+–K+ pumps, no ions will be pumped. The pumps will utilize the energy from ATP hydrolysis to transport Na+ out of the vesicles and K+ into the vesicles. (The pumps will stop working when either the amount of ATP inside the vesicle is depleted or, when the K+ outside of the vesicles is depleted.) The pump will bind a molecule of Na+, causing the ATPase activity to hydrolyze ATP and transfer the phosphate group onto the pump. A conformational change will occur, leading to release of Na+ from the vesicle. However, since there is no K+ outside of the vesicle, the pump will get stuck at that step and subsequent steps of the cycle will not occur.
202
12-16 For an uncharged molecule, the direction of passive transport across a membrane is determined solely by its concentration gradient. On the other hand, for a charged molecule, an additional force called the membrane potential must also be considered. The net driving force for a charged molecule across a membrane therefore has two components and is referred to as the electrochemical gradient. Active transport allows the movement of solutes against this gradient. The carrier proteins called coupled transporters utilize the movement of one solute down its gradient to provide the energy to drive the uphill transport of a second gradient. When this transporter moves both ions in the same direction across the membrane, it is considered a(n) symport; if the ions move in opposite directions, the transporter is considered a(n) antiport. 12-17 Choice (e) is the correct answer. Ions can pass either way through a channel; the direction in which flow takes place depends on the concentration gradient and the membrane potential. Some ion channels require a specific stimulus to open them; others, such as K+ leak channels, do not (choice (a)). Ion channels are passive transporters and therefore require no energy source in order to function (choice (b)). Ion channels are very fast relative to carrier proteins but are limited by the rate at which ions can move through the channel (choice (c)). An ion channel allows specific positive or negative ions to pass, but not both (choice (d)). 12-18 (b) 12-19 1—c; 2—d; 3—b; 4—a 12-20 A. B. C. D. E.
The acetylcholine receptor in skeletal muscle cells is a ligand-gated ion channel. Stress-activated ion channels are found in the hair cells of the mammalian cochlea. Voltage-gated ion channels in the mimosa plant propagate the leaf-closing response. Voltage-gated ion channels respond to changes in membrane potential. Many receptors for neurotransmitters are ligand-gated ion channels.
12-21 (b), (e) The Na+–K+ pump continually transports K+ into the cell. The negative membrane potential also helps to retain K+ in the cell. The K+ leak channels allow K+ to move both into and out of the cell and so do not contribute to the accumulation of K+ in the cell. 12-22 Agree. The membrane potential arises from a thin layer of ions that are close to the membrane. Only a small number of ions (compared to the number of ions present) must move across the membrane to set up the membrane potential. 12-23 A—4; B—1; C—3; D—2
203
12-24 The action potential is a wave of depolarization that rapidly spreads along the neuronal plasma membrane. This wave is triggered by a local change in the membrane potential to a value that is less negative than the resting membrane potential. The action potential is propagated by the opening of voltage-gated channels. During an action potential, the membrane potential changes from negative to positive. The action potential travels from the neuron’s cell body along the axon to the nerve terminals. Neurons chiefly receive signals at their highly branched dendrites. 12-25 See figure A12-25.
Figure A12-25 12-26 The ability of the voltage-gated Na+ channel to adopt an inactivated conformation allows for the action potential to move along the membrane in a wave-like fashion. During the firing of an action potential, the voltage-gated Na+ channel will open and then, after a delay, adopt the inactive form. When the voltage-gated Na+ channel adopts an inactive form, the membrane potential begins to return to its resting potential and cannot support the production of another action potential (i.e., further membrane depolarization) until the channel changes from an inactive to a closed state. 12-27 Neurons communicate with each other through specialized sites called synapses. Many neurotransmitter receptors are ligand-gated ion channels that open transiently in the postsynaptic cell membrane in response to neurotransmitters released by the presynaptic cell. Ligand-gated ion channels in nerve cell membranes convert chemical signals into electrical ones. Neurotransmitter release is stimulated by the opening of voltage-gated Ca2+ channels in the nerve terminal membrane.
204
12-28 To trigger an action potential, a threshold level of membrane depolarization must be achieved to allow for the opening of voltage-gated Na+ channels. As an action potential is initiated, the initial opening of voltage-gated Na+ channels causes Na+ to enter the cell. This makes the membrane potential less negative and opens even more voltage-gated Na+ channels, creating a positive feedback loop that triggers the action potential. The opening of Cl– channels by binding of inhibitory neurotransmitters permits an influx of Cl–, which makes the membrane potential more negative. The entry of Cl– into the cell counteracts the effect of the incoming Na+ on the membrane potential, making it more difficult to depolarize the membrane and therefore, more difficult to reach the threshold potential needed to generate an action potential. 12-29 The K+ leak channel inhibitor would make it easier to trigger an action potential in the neuron. Inhibition of the K+ leak channel changes the permeability of the membrane with respect to K+. The membrane potential is affected by the relative permeability of Na+ and K+. In order for an action potential to fire, a threshold membrane potential must be achieved. By changing the permeability of the membrane to K+ with the addition of the inhibitor, the membrane is now closer to the threshold needed to fire an action potential because the cell is now partially depolarized (i.e., addition of the inhibitor leads to a less negative membrane potential). Therefore, less Na+ needs to flow into the neuron for the threshold potential to be achieved, making it easier to trigger an action potential. 12-30 A.
B.
C.
Increasing the concentration of K+ in the squid axon cytoplasm will make the membrane potential more negative. Doubling the amount of K+ increases the driving force for K+ to move out of the cell, leaving the inside of the cell more negative and thus decreasing the membrane potential. (Remember, from the Nernst equation, the driving force for an ion across a membrane is proportional to the ratio of the concentration of the ion on the outside to the concentration of the ion on the inside.) Doubling the amount of Na+ in the extracellular fluid will increase the height of the peak of the action potential. Again, this is because now the driving force for Na+ to enter the cell is greater than it was before. Thus, when Na+ channels open, the flux of Na+ ions is now greater. (Remember that flux is the number of ions entering per second.) The action potential in this case will reach a height that is less than that normally achieved. (Choline is added in this case to maintain bulk electrical neutrality. Since Na+ channels are not permeable to choline, they do not contribute to the electrochemical gradient.) You have now reduced the concentration of Na+ by one half and thus decreased the driving force for Na+ to enter the cell.
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CHAPTER 13 HOW CELLS OBTAIN ENERGY FROM FOOD 2009 Garland Science Publishing 3rd Edition
The Breakdown of Sugars and Fats 13-1 Which of the following stages in the breakdown of the piece of toast you had for breakfast generates the most ATP? (a) Digestion of starch to glucose (b) Glycolysis (c) The citric acid cycle (d) Oxidative phosphorylation (e) Conversion of pyruvate to acetyl CoA
13-2 The advantage to the cell of the gradual oxidation of glucose during cellular respiration compared with its combustion to CO2 and H2O in a single step is (a) more free energy is released for a given amount of glucose oxidized. (b) no energy is lost as heat. (c) energy can be extracted in usable amounts. (d) more CO2 is produced for a given amount of glucose oxidized. (e) less O2 is required for a given amount of glucose oxidized 13-3 The final metabolite produced by glycolysis is (a) acetyl CoA. (b) pyruvate. (c) 3-phosphoglycerate. (d) glyceraldehyde 3-phosphate. (e) fatty acids. 13-4 Which of the following steps or processes in aerobic respiration include the production of carbon dioxide? (a) Breakdown of glycogen (b) Glycolysis (c) Conversion of pyruvate to acetyl CoA (d) Oxidative phosphorylation (e) The citric acid cycle
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13-5
On a diet consisting of nothing but protein, which of the following is the most likely outcome? (a) Loss of weight because amino acids cannot be used for the synthesis of fat. (b) Muscle gain because the amino acids will go directly into building muscle. (c) Tiredness because amino acids cannot be used to generate energy. (d) Excretion of more nitrogenous (ammonia-derived) wastes than with a more balanced diet. (e) Production of more carbon dioxide than with a more balanced diet.
13-6
Figure Q13-6 represents a cell lining the gut. Draw numbered labeled lines to indicate exactly where inside a cell the following processes take place.
Figure Q13-6 1. 2. 3. 4. 5. 6. 7.
Glycolysis Citric acid cycle Conversion of pyruvate to activated acetyl groups Oxidation of fatty acids to acetyl CoA Glycogen breakdown Release of fatty acids from triacylglycerols Oxidative phosphorylation
13-7
Each of the ten steps of glycolysis is catalyzed by a different enzyme. Steps 1 and 3, catalyzed by hexokinase and phosphofructokinase, involve the hydrolysis of one ATP molecule. Steps 7 and 10, catalyzed by phosphoglycerate kinase and pyruvate kinase, each generates a single ATP molecule. At first glance, it seems that the final ATP yield is zero because there are two ATP-hydrolysis steps and two ATP-formation steps. How can the net yield of glycolysis be two ATP molecules per glucose molecule?
13-8
The oxidation of sugars by glycolysis (a) occurs only in aerobic organisms. (b) generates carbon dioxide. (c) produces a net gain of ATP. (d) occurs in mitochondria. (e) uses NADH as a source of energy.
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13-9
Phosphorylation of glucose (GLC) to produce glucose 6-phosphate (G6P) is the first step in glucose metabolism after entry into cells. Thermodynamically, it is perfectly valid to consider the cellular phosphorylation of glucose as the sum of two reactions. (1) (2) NET:
GLC + Pi → ATP + H2O → GLC + ATP →
G6P + H2O ADP + Pi G6P + ADP
∆G° = 3.3 kcal/mole ∆G° = –7.3 kcal/mole
But biologically it makes no sense at all. Hydrolysis of ATP (reaction 2) in one part of the cell can have no effect on phosphorylation of glucose (reaction 1) elsewhere in the cell. A. How does the cell manage to link these two reactions? B. What is the ∆G° for the net reaction? 13-10 What purpose is served by the phosphorylation of glucose to glucose 6-phosphate by the enzyme hexokinase as the first step in glycolysis? (a) It helps drive the uptake of glucose from outside the cell. (b) It generates a high-energy phosphate bond. (c) It converts ATP to a more useful form. (d) It enables the glucose 6-phosphate to be recognized by phosphofructokinase, the next enzyme in the glycolytic pathway. (e) It oxidizes one of the carbon atoms to yield usable energy. 13-11 Which reaction does the enzyme phosphoglucose isomerase catalyze? (a) glucose → glucose 6-phosphate (b) fructose 6-phosphate → fructose 1,6-bisphosphate (c) glucose 6-phosphate → fructose 6-phosphate (d) glucose → glucose 1-phosphate (e) glucose → fructose 13-12 Give the full names of the reactants indicated by question marks in Figure Q13-12.
Figure Q13-12
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13-13 For each statement below, indicate whether it is TRUE or FALSE. Explain why. A. Only aerobic organisms can do glycolysis, suggesting that glycolysis evolved rather recently. B. Oxidation of a molecule requires the removal of electrons and can occur even if there is no oxygen involved in the reaction. C. For a cell to capture energy from oxidation of food molecules, it is better to release the energy in small packets so it can be stored in activated carrier molecules. D. Fermentation produces more ATP than glycolysis. E. One turn of the citric acid cycle generates two molecules of CO2. F. The breakdown of one molecule of glucose during glycolysis results in one molecule of pyruvate. G. NADH is more reduced than NAD+. H. The reactions of the citric acid cycle do not directly require the presence of oxygen. 13-14 Which of the following cells rely exclusively on glycolysis to supply them with ATP? (a) Anaerobically growing yeast. (b) Aerobic bacteria. (c) Skeletal muscle cells. (d) Plant cells. (e) Protozoa. 13-15 In anaerobic conditions, skeletal muscle produces (a) lactate and CO2. (b) ethanol and CO2. (c) lactate only. (d) ethanol only. (e) lactate, ethanol, and CO2. 13-16 In mammals, liver cells are able to convert lactate to pyruvate. What purpose does this serve for the organism? (a) It is an important way of generating more NADH for the organism. (b) It is an important way of generating NAD+. (c) It allows the organism to grow in anaerobic conditions. (d) It allows the lactate to be productively utilized. (e) It is an important way for the body to generate heat. 13-17 Anaerobically growing yeast further metabolizes the pyruvate produced by glycolysis to CO2 and ethanol as part of a series of fermentation reactions. A. What other important reaction occurs during this fermentation step? B. Why is this reaction (i.e., the answer to part A) essential for the anaerobically growing cell?
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13-18 In the absence of oxygen, cells consume glucose at a high, steady rate. When oxygen is added, glucose consumption drops precipitously and is then maintained at the lower rate. Why is glucose consumed at a high rate in the absence of oxygen and at a low rate in its presence? 13-19 Glycolysis and the citric acid cycle are comprised of several types of reactions that occur sequentially and serve to harvest energy from the oxidation of carbon atoms. From the two lists below, match the general class of enzyme from list 1 with the type of reaction catalyzed from list 2. List 1 A. Kinase
B. Isomerase C. Dehydrogenase D. Synthase or synthetase
List 2 1. Generation of product with the same chemical formula as the substrate but different connections between atoms. 2. Transfer of phosphate group from one molecule to another. 3. Formation of additional carbon-carbon bonds. 4. Oxidation of a substrate.
13-20 The first energy-generating steps in glycolysis begin when glyceraldehyde 3-phosphate undergoes an energetically favorable reaction in which it is simultaneously oxidized and phosphorylated by the enzyme glyceraldehyde 3-phosphate dehydrogenase to form 1,3bisphosphoglycerate, with the accompanying conversion of NAD+ to NADH. In a second energetically favorable reaction catalyzed by a second enzyme, the 1,3bisphosphoglycerate is then converted to 3-phosphoglycerate, with the accompanying conversion of ADP to ATP. Which of the following statements is TRUE? (a) The reaction glyceraldehyde 3-phosphate → 1,3-bisphosphoglycerate should be inhibited when levels of NADH fall. (b) The ∆Gº for the oxidation of the aldehyde group on glyceraldehyde 3-phosphate to form a carboxylic acid is more negative than the ∆Gº for ATP hydrolysis. (c) The high-energy bond to the phosphate group in glyceraldehyde 3-phosphate contributes to driving the reaction forward. (d) The cysteine side chain on the enzyme is oxidized by NAD+. (e) The overall reaction glyceraldehyde 3-phosphate → 3-phosphoglycerate has a positive ∆G.
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13-21 The simultaneous oxidation and phosphorylation of glyceraldehyde 3-phosphate described in Question 13-20 involves the formation of a highly reactive covalent thioester bond between a cysteine side chain (reactive group –SH) on the enzyme (glyceraldehyde 3-phosphate dehydrogenase) and the oxidized intermediate (see arrow in Figure Q1321A). If the enzyme had a serine (reactive group –OH) instead of a cysteine at this position, which could form only a much lower-energy bond to the oxidized substrate (see arrow in Figure Q13-21B), how might this new enzyme act?
Figure Q13-21 (a) (b) (c) (d) (e)
It would oxidize the substrate and phosphorylate it without releasing it. It would oxidize the substrate but not release it. It would phosphorylate the substrate on the 2 position instead of the 1 position. It would behave just like the normal enzyme. It would use ATP instead of Pi to phosphorylate the substrate.
13-22 Acetyl CoA is (a) synthesized from pyruvate in the mitochondrial intermembrane space. (b) the intermediate through which food molecules are completely metabolized to carbon dioxide in animal cells. (c) synthesized from pyruvate and CoA in a reaction that also generates NADH, CO2, and ATP. (d) synthesized by the breakdown of fatty acids in the cytosol. (e) an intermediate in the oxidation of glucose in anaerobic skeletal muscle.
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13-23 Assuming complete oxidation, which of the fatty acids shown in Figure Q13-23 will generate the most ATP through cellular respiration? Why?
Figure Q13-23 13-24 During a single turn of the citric acid cycle (a) the two carbon atoms from acetyl CoA that enter the cycle are completely oxidized to CO2. (b) three molecules of ATP are generated. (c) three molecules of NADH are generated. (d) an acetyl group is added to citric acid. (e) three molecules of CO2 are generated. 13-25 Explain why the following statement is FALSE. “One mole of oxaloacetate is required for every mole of acetyl CoA that is metabolized via the citric acid cycle.” 13-26 Cells oxidizing acetyl groups via the citric acid cycle require molecular oxygen in order to (a) oxidize the acetyl groups to CO2. (b) regenerate NAD+. (c) regenerate FADH2. (d) regenerate CoA. (e) oxidize fatty acids to acetyl groups. 13-27 Given a mixture of all the enzymes of the citric acid cycle plus acetyl CoA, which of the following sets of additions could support conversion of acetyl CoA to carbon dioxide? Explain why. (a) Water, NAD+, GDP, phosphate, FAD+ (b) Water, NAD+, GDP, phosphate, FAD+, oxaloacetate (c) Water, NAD+, GDP, phosphate, FAD+, citrate (d) Water, NAD+, GDP, phosphate, FAD+, citrate, coenzyme A 13-28 The last reaction of the citric acid cycle, which regenerates oxaloacetate (OAA) from malate (MAL), has a very positive ∆G° = 7.1 kcal/mole. Despite its unfavorable equilibrium position, material must flow through this reaction quite readily in mitochondria—otherwise the cycle could not turn. How is flow through the cycle accomplished in the face of such an overwhelmingly positive ∆G°? 213
13-29 Consider the schematic of the citric acid cycle shown in Figure Q13-29. The cycle begins with the formation of a 6-carbon (6C) molecule from joining the 4-carbon (4C) oxaloacetate and the 2-carbon (2C) acetyl group. By about halfway through the cycle, two carbons have been lost as carbon dioxide and a 4C molecule labeled D is regenerated. List two reasons why the cycle does not stop after production of D and instead must continue through a series of different 4C intermediates.
Figure Q13-29 13-30 Which of the following statements regarding electron transport is TRUE? (a) Only high-energy electrons from NADH can be used to drive the electron transport chain. (b) The proteins involved in electron transport couple oxidation to phosphorylation in much the same way that glyceraldehyde 3-phosphate dehydrogenase couples oxidation to phosphorylation in glycolysis. (c) Electron transport occurs only in eucaryotes. (d) Molecular oxygen is required as a donor of electrons to the electron transport chain. (e) Electrons passing along the electron transport chain move to successively lower energy states. 13-31 In the final stage of the oxidation of food molecules, a gradient of protons is formed across the inner mitochondrial membrane, which is normally impermeable to protons. If cells were exposed to an agent that causes the membrane to become freely permeable to protons, which of the following effects would you expect to observe? (a) Cells would be completely unable to synthesize ATP. (b) NADH would build up. (c) Carbon dioxide production would cease. (d) Consumption of oxygen would fall. (e) The ratio of ATP to ADP in the cytoplasm would fall.
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13-32 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase may be used more than once. Oxidative phosphorylation is a process that occurs in the __________________ of mitochondria. It requires an electron-transport chain that operates on the high-energy electrons taken from the activated carrier molecules __________________ and __________________ that are produced by glycolysis and the citric acid cycle. These electrons are transferred through a series of molecules, and the energy released during these transfers is used to generate a gradient of __________________, or __________________. Since their concentration is much __________________ outside than inside the mitochondria, the flow of __________________, or __________________, down the concentration gradient is energetically very __________________ and can thus be coupled to the production of ATP from ADP. Thus, oxidative phosphorylation refers to the oxidation of __________________ and __________________ molecules and the phosphorylation of __________________. Without this process, the yield of ATP from each glucose molecule would be __________________ decreased. ADP ATP cytosol electrons FADH2 favorable glucose
GTP H+ higher inner membrane lower matrix moderately
NAD+ NADH Pi protons severely slightly unfavorable
How We Know: Unraveling the Citric Acid Cycle 13-33 Consider the following statement: Oxaloacetate acts catalytically to aid in the oxidation of pyruvate found in suspensions of minced pigeon muscles. A. What was the original evidence for this statement? B. How did these sorts of experiments aid in the identification of intermediates in the citric acid cycle?
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Storing and Utilizing Food 13-34 Which of the following statements is TRUE? (a) Plant cells store all their food reserves as starch, whereas animals store all their food reserves as glycogen. (b) Glycogen stores more energy than starch because glycogen molecules have many more branch points that can be hydrolyzed. (c) Animal cells can convert fatty acids to sugars. (d) Plants synthesize starch for the same reason that animals synthesize glycogen. (e) Protein is an important form of energy storage in animal cells under normal conditions. 13-35 In humans, glycogen is a more useful food storage molecule than fat because (a) a gram of glycogen produces more energy than a gram of fat. (b) it can be utilized to produce ATP under anaerobic conditions whereas fat cannot. (c) it binds water and therefore is useful in keeping the body hydrated. (d) for the same amount of energy storage, glycogen occupies less space in a cell than does fat. (e) glycogen can be carried to cells via the bloodstream whereas fats cannot. 13-36 The intermediates of the citric acid cycle are constantly being depleted because they are used to produce many of the amino acids needed to make proteins. These intermediates must therefore be replenished by the conversion of pyruvate to oxaloacetate by the enzyme pyruvate carboxylase. Bacteria, but not animal cells, have additional enzymes that can carry out the reaction acetyl CoA + isocitrate → oxaloacetate + succinate. Which of the following compounds will not support the growth of animal cells when used as the major source of carbon in food, but will support the growth of nonphotosynthetic bacteria? (a) Pyruvate (b) Glucose (c) Fatty acids (d) Carbon dioxide (e) Fructose 13-37 Pyruvate can be converted into many other molecules by various biosynthetic and metabolic pathways, which makes it a central hub in the regulation of cellular metabolism. Which of the following molecules is not made from pyruvate? (a) Oxaloacetate (b) Ethanol (c) NADH (d) Lactate (e) Acetyl CoA
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13-38 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. A carbon atom in a CO2 molecule in the atmosphere eventually becomes a part of one of the enzymes that catalyzes glycolysis in one of your cells. The CO2 first enters a cell in a corn leaf where photosynthesis fixes the carbon to make it part of a sugar molecule, which travels from the leaf to an ear of corn where it is stored as part of a polysaccharide __________________ molecule in the corn seed. You then eat a corn chip made from the corn seed. You digest the corn seed, and the free __________________ travels in your bloodstream, eventually being taken up by a liver cell and stored as __________________. When required, this storage molecule breaks down into glucose-1-phosphate, which enters the glycolytic pathway. Glycolysis produces __________________, which is converted into acetyl CoA, which enters the __________________. Several intermediates in this process can provide the carbon skeleton for production of __________________, which are then incorporated into the enzymes that catalyze steps in glycolysis. amino acids carbon fixation citric acid cycle fatty acid fermentation galactose glucose glycogen
insulin lactate nucleotides oxidative phosphorylation pyruvate starch triacylglycerol
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Answers 13-1
Choice (d) is the correct answer.Oxidative phosphorylation produces about 28 ATP molecules. Choice (a) produces no ATP; choice (b) nets 2 ATP; choice (c) produces 1 GTP; and choice (e) produces no ATP.
13-2
Choice(c) is the correct answer. Choice (a) is untrue as the same overall amount of free energy is released by glucose oxidation, whatever the route. Choice (b) is untrue as a proportion of the energy released is still lost as heat. Choices (d) and (e) are untrue as the same amount of CO2 will be released and O2 consumed by the oxidation of glucose to CO2 and H2O, whatever the route.
13-3
(b)
13-4
Choices (c) and (e) are the correct answers. To obtain the maximal energy from pyruvate, all three carbons can be fully oxidized to carbon dioxide. First, one carbon is oxidized and released as carbon dioxide when pyruvate is converted to acetyl CoA. Second, the two remaining carbons are fully oxidized and released as carbon dioxide by virtue of the citric acid cycle.
13-5
Choice (d) is the correct answer. Because ammonia is given off when amino acids are metabolized to yield energy but is not given off when sugars and fats are metabolized, you would expect more nitrogenous waste to be excreted. Choice (c) is incorrect since amino acids can be converted into pyruvate and acetyl CoA and used to generate energy. If more amino acids are consumed than are used, the body will not store them as protein in muscle tissue but rather will store them as fat, so choices (a) and (b) are incorrect. Amino acid metabolism does not produce more carbon dioxide than carbohydrate or fat metabolism, so choice (e) is incorrect.
13-6
See Figure A13-6.
Insert Figure A13-16
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13-7
The net yield of glycolysis is two ATP molecules per glucose molecule because the steps that require ATP occur once per glucose molecule whereas those that generate ATP occur twice per glucose molecule. This is because Step 4, catalyzed by aldolase, breaks down a 6-carbon intermediate into two 3-carbon intermediates. Each 3-carbon intermediate is processed by the remaining steps and thus Steps 7 and 10 occur twice per original 6carbon glucose molecule.
13-8
Choice (c) is the correct answer. Glycolysis, the step-wise oxidation of glucose to pyruvate, produces a net gain of two ATP molecules per starting glucose molecule. Glycolysis occurs in both anaerobic and aerobic organisms; the citric acid cycle, but not glycolysis, generates carbon dioxide and occurs in mitochondria (thus choices (a), (b), and (d) are incorrect). Glycolysis does not use NADH and instead produces it (thus choice (e) is incorrect).
13-9
A.
B.
The cell links these two reactions by using the same enzyme to catalyze both of them. In essence, the enzyme hexokinase binds both GLC and ATP substrates and catalyzes the direct transfer of a phosphate group from ATP onto the GLC. Reactions 1 and 2 do not actually occur because free Pi is never produced. The ∆G° for the net reaction is –4.0 kcal/mole because the free energy is the sum of component reactions and is independent of the pathway used to convert substrates into products. ∆G° = 3.3 kcal/mole – 7.3 kcal/mole = –4.0 kcal/mole.
13-10 Choice (a) is the correct answer. It helps drive the uptake of glucose from outside the cell. Choice (b) is incorrect since the phosphate transferred to the glucose is not held by a high-energy covalent bond. Choice (c) is incorrect since the reaction converts ATP to ADP, which is not useful as an energy source for most cellular reactions, even though it still has one high-energy bond. Choice (d) is incorrect since the next enzyme in the pathway is phosphoglucose isomerase, not phosphofructokinase. Choice (e) is incorrect since the reaction does not involve oxidation of carbon nor does it yield usable energy. 13-11 Choice (c) is the correct answer. The isomerase part of the enzyme name indicates that it catalyzes an isomerization reaction, and the phosphoglucose part of the name indicates the type of substrate used. The enzyme that catalyzed reaction (e) would be called glucose isomerase. The enzymes that catalyzed reactions (a), (b), and (d) would be called kinases, because they transfer phosphate groups from one molecule to another. 13-12 phosphoenolpyruvate, adenosine diphosphate, pyruvate, adenosine triphosphate
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13-13 A. B. C. D. E. F. G. H.
13-14 (a)
False. All organisms can do glycolysis, suggesting that it evolved rather early. True. Oxidation is the loss of electrons. The mechanism of oxidation often involves reaction with oxygen atoms, but instead can occur in other ways. True. If all the energy of oxidation were released at once, as in combustion, most of the energy would be lost as heat and not captured in usable forms. False. Fermentation by itself produces no ATP, but only regenerates NAD+ so that glycolysis can continue to operate and to produce ATP. True. Each turn of the citric acid cycle essentially exchanges the two carbons from the input acetyl group into two carbons completely oxidized as CO2. False. Glycolysis converts each molecule of glucose into two molecules of pyruvate. True. NADH is more reduced than NAD+, because it contains more electrons. True. Oxygen is not one of the substrates in the reactions of the citric acid cycle. Instead oxygen is needed for the citric acid cycle only indirectly, to regenerate the oxidized forms of NAD+ and FAD, which are substrates of the citric acid cycle. All the other cells can carry out oxidative phosphorylation to generate additional ATP.
13-15 (c) 13-16 Choice (d) is the correct answer. Lactate is a metabolic dead end and it cannot be utilized any further by cell. Pyruvate, on the other hand, can either be converted to acetyl CoA and metabolized in the citric acid cycle or converted to glucose by a process called gluconeogenesis. Conversion of lactate to pyruvate uses up NAD+ (making choice (b) incorrect) and generates NADH, but the reason animal cells make lactate in the first place is that they are suffering from an excess of NADH (and thus choice (a) is incorrect). Choice (c) is incorrect since no mammal can survive without oxygen. The amount of heat generated by converting lactate to pyruvate is negligible, so choice (e) is incorrect. 13-17 A. B.
NADH → NAD+. Under anaerobic conditions, it is the only means of regenerating the NAD+ required for glycolysis, the main energy-generating pathway of an anaerobically growing yeast cell.
13-18 Glucose is consumed at a much higher rate in the absence of oxygen because less usable energy can be harvested from glucose in the absence of oxygen. Regardless of the presence of oxygen in the environment, cells need about the same amount of energy in the form of ATP. In the absence of oxygen, a glucose molecule yields only 2 ATP molecules and thus many glucose molecules must be consumed to satisfy the energetic needs of the cell. In the presence of oxygen, a glucose molecule yields about 30 ATP molecules. 13-19 A—2; B—1; C—4; D—3
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13-20 Choice (b) is the correct answer. This is another way of stating that the energetically favorable oxidation of glyceraldehyde 3-phosphate provides sufficient energy to ultimately drive the energy-requiring step of ATP synthesis from ADP. Choice (a) is untrue: NADH is an end product of the reaction G-3-P to 1,3-PG and therefore high (not low) levels of it would inhibit the reaction. Choice (c) is untrue: the reactions do not involve the 3-phosphate group on glyceraldehyde 3-phosphate at all. Choice (d) is untrue, since the cysteine on the enzyme is important in making a covalent intermediate with the substrate and is not oxidized by NAD+. Choice (e) is untrue, since if the reaction had an overall positive ∆G, it could not be used to power the energetically unfavorable reactions of ATP and NADH synthesis. 13-21 Choice (b) is the correct answer. The phosphorylation and release of the product from the normal enzyme is possible because the high-energy thioester bond formed between the oxidized substrate and enzyme can be attacked by a phosphate molecule. If the bond between oxidized substrate and enzyme is a much lower-energy bond, the enzyme will not be able to transfer the oxidized substrate to a phosphate group, and substrate and enzyme will remain covalently bound. Choices (a), (c), and (d) could not happen, as none of the bonds in the substrate molecule is reactive enough to be broken by a phosphate group. Choice (e) would not happen because the enzyme does not bind ATP. 13-22 Choice (b) is the correct answer. Some food molecules can enter the citric acid cycle at points other than acetyl CoA, but only molecules that enter as acetyl CoA are completely oxidized to carbon dioxide. Acetyl CoA is made in the mitochondrial matrix, not the intermembrane space, so choice (a) is incorrect. Choice (c) is incorrect since no ATP is directly generated by the conversion of pyruvate to acetyl CoA; carbon dioxide and NADH are the only other products of the reaction. Choice (d) is incorrect as the breakdown of fatty acids to produce acetyl CoA occurs in the mitochondrial matrix. Choice (e) is incorrect since acetyl CoA is not an intermediate in the anaerobic fermentation reaction that coverts pyruvate to lactate. 13-23 (B)
This will produce 2 NADH, 2 FADH2, and 3 acetyl CoA on complete oxidation. Because of the double bond in A, this fatty acid will produce 2 NADH and 3 acetyl CoA, but only 1 FADH2, since the initial oxidation step using FAD+ that reduces the two-carbon unit –CH2CH2– to –CH=CH– is not needed for twocarbon units already containing a double bond. So although the amount of acetyl CoA entering the citric acid cycle will be the same for A and B, fewer reducing equivalents will eventually enter the electron transport chain from the oxidation of A, thus less ATP will be produced.
13-24 (c) 13-25 Only small amounts of oxaloacetate are required relative to the amount of acetyl CoA metabolized because oxaloacetate is regenerated after every round of the citric acid cycle.
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13-26 Choice (b) is the correct answer. The citric acid cycle generates high-energy electrons that are passed to NAD+ to form NADH. NADH then donates these electrons to the electron transport chain that drives oxidative phosphorylation, regenerating the NAD+ needed to keep the citric acid cycle going. The electrons from NADH are passed via the electron transport chain to oxygen. 13-27
Choices (b) and (d) are correct. To get the cycle turning you need water, NAD+, GDP, phosphate, FAD, coenzyme A, and at least one intermediate of the citric acid cycle, which are all provided in choice (d). In addition, choice (b) will be sufficient because oxaloacetate reacts with acetyl CoA to release a molecule of coenzyme A, which can then be reused. Choice (c) would produce a small amount of CO2 initially from the added citrate, but the cycle could not continue since citrate has to go through a step requiring coenzyme A to complete the cycle. Choice (a) will not work, as this set does not include any citric acid cycle intermediate.
13-28 The mitochondrion obtains a flow through this reaction by maintaining high concentrations of substrates and low concentrations of products, so that the ∆G is negative despite a positive ∆G°. The concentration ratio of products to substrates, [OAA][NADH]/[MAL][NAD+], must be substantially smaller than the equilibrium constant K to overcome a positive ∆G° value. 13-29 The two reasons are to keep the cycle going and to harvest more energy. (1) The cycle must regenerate oxaloacetate, which acts catalytically in the citric acid cycle to aid in the oxidation of many acetyl groups. (2) Additional reactions are required to more fully oxidize the carbons and harvest energy by producing several activated carrier molecules, including GTP, FADH2, and NADH. 13-30 Choice (e) is the correct answer. Electrons passing along the electron transport chain move to successively lower energy states. Choice (a) is untrue as electrons from FADH2 can be used as well. Choice (b) is untrue as the two mechanisms of coupling oxidation to phosphorylation are quite different. Oxidative phosphorylation involves the oxidation of NADH to NAD+ by proteins of the electron transport chain. Electron transport then causes the formation of a proton gradient across a membrane, which drives ATP synthesis. In contrast, glyceraldehyde 3-phosphate dehydrogenase action involves the reduction of NAD+ to NADH and uses the ∆G° of glyceraldehyde oxidation to form a high-energy bond that can be attacked directly by a phosphate group. Choice (c) is untrue, as electron transport occurs in the plasma membrane of procaryotes. Choice (d) is untrue, as molecular oxygen acts as an acceptor, not a donor, for electrons.
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13-31 Choice (e) is the correct answer. If the inner mitochondrial membrane became permeable to protons, the electron transport chain would continue to oxidize NADH to NAD+, transport electrons and pump protons, so the consumption of oxygen would not fall (choice (d)). However, the energy stored by the protons would be immediately dissipated as heat when they flowed back across the membrane and thus could not drive the synthesis of ATP. But, NADH would not build up (choice (b)), and the citric acid cycle and glycolysis would continue (and thus CO2 would still be produced, contrary to choice (c)). Since glycolysis and the citric acid cycle produce 2 molecules of ATP and one molecule of GTP (which can be converted to ATP), respectively, ATP production would not completely cease (choice (a)), but it would be very much less than normal. 13-32 Oxidative phosphorylation is a process that occurs in the inner membrane of mitochondria. It requires an electron-transport chain that operates on the high-energy electrons taken from the activated carrier molecules NADH and FADH2 that are produced by glycolysis and the citric acid cycle. These electrons are transferred through a series of molecules, and the energy released during these transfers is used to generate a gradient of protons, or H+. Since their concentration is much higher outside than inside the mitochondria, the flow of protons, or H+, down the concentration gradient is energetically very favorable and can thus be coupled to the production of ATP from ADP. Thus, oxidative phosphorylation refers to the oxidation of NADH and FADH2 molecules and the phosphorylation of ADP. Without this process, the yield of ATP from each glucose molecule would be severely decreased. 13-33 A.
B.
Experiments showed that minced pigeon muscles contained large amounts of pyruvate, but oxidized this compound rather slowly, so that little oxygen was consumed and little carbon dioxide was produced. When a tiny amount of oxaloacetate was added to such muscle preparations, large amounts of oxygen and carbon dioxide were consumed and produced, respectively. If the added oxaloacetate was simply being oxidized fully, the oxygen consumption and carbon dioxide production would be expected to increase only slightly, but instead the large amount of oxygen consumed suggested that each molecule of added oxaloacetate aided in the oxidation of many molecules of some other substance. Analogous experiments showed that the addition of several other compounds, like succinate and fumarate, had the same consequences as adding oxaloacetate. This was interpreted as evidence that these compounds are intermediates in the same pathway and can be converted into, or are derived from, oxaloacetate, which was later demonstrated more directly.
13-34 Choice (d) is the correct answer. Both starch and glycogen are storage polymers of glucose. Choice (a) is false, since both plants and animals can also store food as fats and oils. Choice (b) is false, as although glycogen synthesis requires ATP, no ATP is generated by its hydrolysis to monomers (so the number of branch points is irrelevant to energy storage). Choice (c) is false, as animal cells cannot do this. Choice (e) is false, as the use of protein for energy occurs only under starvation conditions.
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13-35 Choice (b) is the answer. The breakdown of glycogen to glucose does not require oxygen; the glucose can then enter glycolysis and generate ATP by a fermentation process that produces lactic acid. In contrast, fats are broken down to acetyl CoA that must enter the citric acid cycle, which requires oxygen to keep turning. Choice (a) is incorrect, as a gram of glycogen (wet or dry) produces less energy than a gram of fat. Choice (c) is incorrect, as the water bound by glycogen is not useful in keeping the body hydrated and merely contributes to making the glycogen weigh a lot. Choice (d) is incorrect, as the actual mass of glycogen required to store the same amount of energy is six-fold greater than the amount of fat. Choice (e) is incorrect, as fats can be carried in the bloodstream. If the energy stored in glycogen is required by other cells, glycogen is broken down to glucose, and the glucose is then released into the bloodstream. 13-36 Choice (c) is the correct answer. In oxidative metabolism, fatty acids can only be converted to acetyl CoA, which is completely oxidized to carbon dioxide through the citric acid cycle. In addition, bacteria can use some of this acetyl CoA as a source of carbon atoms to replenish the citric acid cycle, whereas animals cannot. Choices (a), (b), and (e) are incorrect, since glucose and fructose can be converted to pyruvate, and hence to citric acid cycle intermediates, in both animal and bacterial cells, while choice (d) is incorrect, since carbon dioxide cannot be used as a main carbon source by either nonphotosynthetic bacteria or animal cells. 13-37 (c)
Pyruvate cannot be converted into NADH, but it can be converted into the other metabolites in one or two steps.
13-38 A carbon atom in a CO2 molecule in the atmosphere eventually becomes a part of one of the enzymes that catalyzes glycolysis in one of your cells. The CO2 first enters a cell in a corn leaf where photosynthesis fixes the carbon to make it part of a sugar molecule, which travels from the leaf to an ear of corn where it is stored as part of a polysaccharide starch molecule in the corn seed. You then eat a corn chip made from the corn seed. You digest the corn seed, and the free glucose travels in your bloodstream, eventually being taken up by a liver cell and stored as glycogen. When required, this storage molecule breaks down into glucose-1-phosphate, which enters the glycolytic pathway. Glycolysis produces pyruvate, which is converted into acetyl CoA, which enters the citric acid cycle. Several intermediates in this process can provide the carbon skeleton for production of amino acids, which are then incorporated into the enzymes that catalyze steps in glycolysis.
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CHAPTER 14 ENERGY GENERATION IN MITOCHONDRIA AND CHLOROPLASTS 2009 Garland Science Publishing 3rd Edition
Mitochondria and Oxidative Phosphorylation 14-1 Which of the following statements are likely to be TRUE? Note that more than one statement may be selected. (a) Organisms that could carry out fermentation evolved before those that could carry out aerobic respiration. (b) Organisms that could carry out oxygen-producing photosynthesis evolved before those that could carry out fermentation reactions. (c) Eucaryotic organisms were the first to evolve mechanisms of chemiosmotic coupling. (d) Both photosynthesis and aerobic respiration first evolved in procaryotes. (e) Aerobic respiration arose as an adaptation to increasing levels of oxygen in the atmosphere that had been produced by photosynthesis. 14-2 Is the following statement TRUE or FALSE? Explain your answer. The most important contribution of the citric acid cycle to energy metabolism is the extraction of high-energy electrons during the oxidation of acetyl CoA to CO2
14-3 Is the following statement TRUE or FALSE? Explain your answer. Each respiratory enzyme complex in the electron-transport chain has a greater affinity for electrons than its predecessors, so that electrons pass sequentially from one complex to another until they are finally transferred to oxygen, which has the greatest electron affinity of all. 14-4 The citric acid cycle generates NADH and FADH2, which are then used in the process of oxidative phosphorylation to make ATP. If the citric acid cycle (which does not use oxygen) and oxidative phosphorylation are separate processes, as they are, then why is it that the citric acid cycle stops almost immediately upon removal of O2? 14-5 What three essential items are missing from the following list of cellular components required to make ATP by chemiosmotic coupling? ADP; ATP synthase; protons; electron-transport chain; proton pump; membrane.
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14-6
Electron transport is coupled to ATP synthesis in mitochondria, chloroplasts, and the thermophilic bacterium Methanococcus. Which of the following are likely to affect the coupling of electron transport to ATP synthesis in ALL of these systems? Note that more than one statement may be selected. (a) A potent inhibitor of cytochrome oxidase. (b) The removal of oxygen. (c) The absence of light. (d) An ADP analogue that inhibits ATP synthase. (e) Dinitrophenol (permeabilizes membranes to protons).
14-7
In which of the four compartments of a mitochondrion are each of the following located? A. Porin B. The mitochondrial genome C. Citric acid cycle enzymes D. Proteins of the electron transport chain E. ATP synthase F. Membrane transport protein for pyruvate
14-8
Which of the following statements about mitochondria is FALSE? (a) Protons are pumped from the intermembrane space into the matrix. (b) ATP is synthesized in the matrix. (c) Mitochondria can change shape. (d) The outer membrane is permeable to protons. (e) The inner membrane is folded into cristae.
14-9
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Mitochondria can use both __________________ and __________________ as fuel. __________________ produced in the citric acid cycle donates electrons to the electron-transport chain. The citric acid cycle oxidizes __________________ and produces __________________ as a waste product. __________________ acts as the final electron acceptor in the electron-transport chain. The synthesis of ATP in mitochondria is also known as __________________. acetyl groups carbon dioxide chemiosmosis fatty acids glucose NAD+
NADH NADP+ NADPH oxidative phosphorylation oxygen pyruvate
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14-10 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. NADH donates electrons to the __________________ of the three respiratory enzyme complexes in the mitochondrial electron-transport chain. __________________ is a small protein that acts as a mobile electron carrier in the respiratory chain. __________________ transfers electrons to oxygen. Electron transfer in the chain occurs in a series of __________________ reactions. The first mobile electron carrier in the respiratory chain is __________________. cytochrome c cytochrome oxidase first NADH dehydrogenase oxidation oxidation-reduction phosphorylation
plastoquinone reduction second the cytochrome b-c1 complex third ubiquinone
14-11 Which of the following statements is TRUE? (a) Because the electrons in NADH are at a higher energy than the electrons in reduced ubiquinone, the NADH dehydrogenase complex can pump more protons than can the cytochrome b-c1 complex. (b) The pH in the mitochondrial matrix is higher than the pH in the intermembrane space. (c) The proton concentration gradient and the membrane potential across the inner mitochondrial membrane tend to work against each other in driving protons from the intermembrane space into the matrix. (d) The difference in proton concentration across the inner mitochondrial membrane has a much larger effect on the total proton-motive force than does the membrane potential. (e) All of the free energy released by the net transfer of electrons from NADH to O2 is captured in the form of the proton gradient. 14-12 Some bacteria can live both aerobically and anaerobically. How does the ATP synthase in the plasma membrane of the bacterium enable such bacteria to keep functioning in the absence of oxygen?
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14-13 Bongkrekic acid is an antibiotic that inhibits the ATP/ADP transport protein in the inner mitochondrial membrane. Which of the following will allow electron transport to occur in mitochondria treated with bongkrekic acid? (a) Placing the mitochondria in anaerobic conditions (b) Adding FADH2 (c) Permeabilizing the inner membrane to protons (d) Inhibiting the ATP synthase (e) Increasing the concentration of ATP in the matrix 14-14 Which of the following types of ion movement might be expected to require cotransport of protons from the intermembrane space to the matrix, inasmuch as it could not be driven by the membrane potential across the inner membrane? (Assume that each ion being moved is moving against its concentration gradient.) (a) Import of Ca2+ into the matrix from the intermembrane space (b) Import of acetate ion into the matrix from the intermembrane space (c) Exchange of Fe2+ in the matrix for Fe3+ in the intermembrane space (d) Exchange of ATP from matrix for ADP in the intermembrane space (e) Exchange of Ca2+ in the matrix for Na+ in the intermembrane space 14-15 The relationship of free-energy change (∆G) to the concentrations of reactants and products is important because it predicts the direction of spontaneous chemical reactions. Consider, for example, the hydrolysis of ATP to ADP and inorganic phosphate (Pi). The standard free-energy change (∆G°) for this reaction is –7.3 kcal/mole. The free-energy change depends on concentrations according to the following equation: ∆G = ∆G° + 1.42 log10 ([ADP] [Pi]/[ATP]) A.
B.
In a resting muscle, the concentrations of ATP, ADP, and Pi are approximately 0.005 M, 0.001 M, and 0.010 M, respectively. What is ∆G for ATP hydrolysis in resting muscle? What is the ∆G for ATP synthesis in resting muscle? What will ∆G equal when the hydrolysis reaction reaches equilibrium? At [Pi] = 0.010 M, what will be the ratio of [ATP] to [ADP] at equilibrium?
14-16 How many molecules of ATP are produced by the electron-transport chain and the ATP synthase from the oxidation of the 18-carbon fatty acid derivative stearyl CoA? The first steps in the oxidation of stearyl CoA occur in the mitochondrion and generate 9 molecules of acetyl CoA, 8 molecules of NADH, and 8 molecules of FADH2.
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How We Know: How Chemiosmotic Coupling Drives ATP Synthesis 14-17 The respiratory chain is relatively inaccessible to experimental manipulation of intact mitochondria. After disrupting mitochondria with ultrasound, however, it is possible to isolate functional submitochondrial particles, which consist of broken cristae that have resealed inside-out into small closed vesicles. In these vesicles the components that originally faced the matrix are now exposed to the surrounding medium. A. How might such an arrangement aid in the study of electron transport and ATP synthesis? B. Consider an anaerobic preparation of such submitochondrial particles. If the flow of protons through ATP synthase is blocked by an inhibitor and a small amount of oxygen is added, do you predict that the preparation will consume oxygen in respiration reactions? Will the medium outside the particles become more acidic or basic? Explain. 14-18 In classic experiments that helped convince investigators of the veracity of the chemiosmotic hypothesis, artificial membrane vesicles were reconstituted with purified bacteriorhodopsin (which is a light-driven H+ pump from a photosynthetic bacterium) and purified ATP synthase from ox heart mitochondria. In these vesicles the membrane complexes are oriented so that protons are pumped into the vesicle and ATP synthesis occurs on the outer surface. When ADP and Pi were added to the medium surrounding the vesicles and the vesicles were illuminated with appropriate light, the interior of the vesicles became acidic and ATP was produced. A. If, instead, the ATP synthase molecules were randomly oriented so that about half faced the outside of the vesicle and half faced the inside, would you expect ATP to be synthesized? Explain your answer. B. If the bacteriorhodopsin molecules were randomly oriented, would you expect ATP to be synthesized? Explain your answer.
Electron-Transport Chains and Proton Pumping 14-19 Distinguish between a proton, a hydrogen atom, a hydride ion, and a hydrogen molecule. 14-20 Which of the following statements is TRUE? (a) Only compounds with negative redox potentials can donate electrons to other compounds under standard conditions. (b) Compounds that donate one electron have higher redox potentials than those of compounds that donate two electrons. (c) The ∆E′0 of a redox pair does not depend on the concentration of each member of the pair. (d) The free energy change, ∆G, for an electron transfer reaction does not depend on the concentration of each member of a redox pair. (e) If the E′0 of the reaction AH2 → A + 2 H+ + 2e– is 600 mV and the E′0 of the reaction H2O → 1/2 O2 + 2 H+ + 2e– is 820 mV, then the transfer of electrons from water to AH2 must be favorable under standard conditions. 229
14-21 Consider a redox reaction between molecules A and B. Molecule A has a redox potential of –100 mV and molecule B has a redox potential of +100 mV. For the transfer of electrons from A to B, is the ∆G° positive or negative or zero? Under what conditions will the reverse reaction, transfer of electrons from B to A, occur? 14-22 For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement. “An electron bound to a molecule with low affinity for electrons is a [high/low] energy electron. Transfer of an electron from a molecule with low affinity to one with higher affinity has a [positive/negative] ∆G° and is thus [favorable/unfavorable] under standard conditions. If the reduced form of a redox pair is a strong electron donor with a [high/low] affinity for electrons, it is easily oxidized; the oxidized member of such a redox pair is a [weak/strong] electron acceptor.” 14-23 Which of the following reactions have a large enough free energy change to enable it to be used, in principle, to provide the energy needed to synthesize one molecule of ATP from ADP and Pi under standard conditions? See Table 14-23. Recall that ∆G° = –n (0.023) ∆E′0 and ∆E′0 = E′0 (acceptor) – E′0 (donor). (a) The reduction of a molecule of pyruvate by NADH (b) The reduction of a molecule of cytochrome b by NADH (c) The reduction of a molecule of cytochrome b by reduced ubiquinone (d) The oxidation of a molecule of reduced ubiquinone by cytochrome c (e) The oxidation of cytochrome c by oxygen Table 14-23 Reaction NADH → NAD+ + H+ + 2e– Lactate → pyruvate + 2H+ + 2e– Reduced ubiquinone → ubiquinone + 2H+ + 2e– Cytochrome b (Fe2+) → cytochrome b (Fe3+) + e– Cytochrome c (Fe2+) → cytochrome c (Fe3+) + e– H2O → 1/2O2 + 2H+ + 2e–
E′0 –320 mV –190 mV 30 mV 70 mV 230 mV 820 mV
14-24 Is the following statement TRUE or FALSE? Explain your answer. Most cytochromes have a higher redox potential (higher affinity for electrons) than iron-sulfur centers, which is why the cytochromes tend to serve as electron carriers near the O2 end of the respiratory chain.
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14-25 Which of the following statements is TRUE? (a) Ubiquinone is a small hydrophobic protein containing a metal group that acts as an electron carrier. (b) A 2Fe2S iron-sulfur center carries one electron while a 4Fe4S center carries two. (c) Iron-sulfur centers generally have a higher redox potential than do cytochromes. (d) Mutation of hydrophobic amino acids near the heme group of cytochrome c to acidic amino acids is likely to increase the redox potential of cytochrome c. (e) Mitochondrial electron carriers with the highest redox potential generally contain copper ions and/or heme groups. 14-26 Which of the following is not an electron carrier that participates in the electron transport chain? (a) Cytochrome (b) Quinone (c) Rhodopsin (d) Copper ion (e) Iron-sulfur center
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14-27 In 1925, David Keilin used a simple spectroscope to observe the characteristic absorption bands of the cytochromes that participate in the electron-transport chain in mitochondria. A spectroscope passes a very bright light through the sample of interest and then through a prism to display the spectrum from red to blue. If molecules in the sample absorb light of particular wavelengths, dark bands will interrupt the colors of the rainbow. His key discovery was that the absorption bands disappeared when oxygen was introduced and then reappeared when the samples became anoxic. Subsequent findings demonstrated that different cytochromes absorb different frequencies of light. When light of a characteristic wavelength shines on a mitochondrial sample, the amount of light absorbed is proportional to the amount of a particular cytochrome present in its reduced form. Thus, spectrophotometric methods can be used to measure how the amounts of reduced cytochromes change over time in response to various treatments. If isolated mitochondria are incubated with a source of electrons such as succinate, but without oxygen, electrons enter the respiratory chain, reducing each of the electron carriers almost completely. When oxygen is then introduced, the carriers oxidize at different rates, as can be seen from the decline in the amount of reduced cytochrome (see Figure Q14-27). Note that cytochromes a and a3 cannot be distinguished and thus are listed as cytochrome (a + a3). How does this result allow you to order the electron carriers in the respiratory chain? What is their order?
Figure Q14-17 14-28 When molecular oxygen (O2) picks up one electron it becomes converted to the superoxide radical O2–. This radical is potentially damaging to cells as it will avidly pick up another three electrons from a wide variety of cellular molecules. How do cells avoid this happening during cellular respiration? 14-29 The antibiotic antimycin A blocks the electron-transport chain in bacterial cells. Specifically, antimycin A inhibits the flow of electrons between cytochrome b and cytochrome c1. A. After treatment with this antibiotic, which electron carriers in the chain will accumulate in the reduced form? B. Which compounds will be likely to be oxidized more than before treatment? C. Will the ATP:ADP ratio in the cell rise or fall?
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Chloroplasts and Photosynthesis 14-30 Both mitochondria and chloroplasts use electron transport to pump protons, creating an electrochemical proton gradient, which drives ATP synthesis. Are protons pumped across the same (analogous) membranes in the two organelles? Is ATP synthesized in the analogous compartments? Explain your answers. 14-31 Ions are able to diffuse freely from (a) one thylakoid space to another thylakoid space in the same chloroplast. (b) the cytosol to the intermembrane space. (c) the intermembrane space to the stroma. (d) the stroma to the thylakoid space. (e) the interior of one granum to another in the same chloroplast. 14-32 Indicate if each of the following features is found in chloroplasts (C), mitochondria (M), or both (B). A. Inner membrane B. Thylakoid C. Ribosomes D. Stroma E. ATP synthase F. Grana G. Citric acid cycle H. Matrix I. Electron-transfer reactions J. Carbon fixation K. Cristae 14-33 Write out the list of substrates and products for the overall net reaction that occurs in the “light” stage of photosynthesis. Make the same list for the net overall reaction in the “dark” stage of photosynthesis.
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14-34 If you shine light on chloroplasts and measure the rate of photosynthesis as a function of light intensity, you get a curve that plateaus at a fixed rate of photosynthesis, x, as shown in Figure Q14-34.
Figure Q14-34 Which of the following conditions will increase the value of x? (a) Increasing the number of chlorophyll molecules in the antennae complexes (b) Increasing the number of reaction centers (c) Adding a powerful oxidizing agent (d) Decreasing the wavelength of light used (e) Increasing the rate at which chlorophyll molecules are able to transfer electrons to one another 14-35 If you add a compound to illuminated chloroplasts that inhibits the NADP+ reductase, NADPH generation ceases, as expected. Ferredoxin, however, does not accumulate in the reduced form because it is able to donate its electrons not only to NADP+ (via NADP+ reductase), but also back to the cytochrome b6-f complex. Thus, in the presence of the above compound, a “cyclic” form of photosynthesis occurs in which electrons flow in a circle from ferredoxin, to the cytochrome b6-f complex, to plastocyanin, to photosystem I, to ferredoxin. What will happen if you now also inhibit photosystem II? (a) Less ATP will be generated per photon absorbed. (b) ATP synthesis will cease. (c) Plastoquinone will accumulate in the oxidized form. (d) Plastocyanin will accumulate in the oxidized form. (e) Ferredoxin will accumulate in the reduced form.
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14-36 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. In the carbon fixation process in chloroplasts, carbon dioxide is initially added to the sugar __________________. The final product of carbon fixation in chloroplasts is the three-carbon compound __________________. This is converted into __________________ (which can be directly used by the mitochondria), into __________________ (which is exported to other cells), and into __________________ (which is stored in the stroma). The carbon fixation cycle requires energy in the form of __________________ and reducing power in the form of __________________. 3-phosphoglycerate ATP glyceraldehyde 3-phosphate NADH NADPH
pyruvate ribose 1,5-bisphosphate ribulose 1,5-bisphosphate starch sucrose
14-37 The enzyme ribulose bisphosphate carboxylase (rubisco) normally adds carbon dioxide to ribulose 1,5-bisphosphate. However, it will also catalyze a competing reaction in which O2 is added to ribulose 1,5-bisphosphate to form 3-phosphoglycerate and phosphoglycolate. Assume that phosphoglycolate is a compound that cannot be used in any further reactions. If O2 and CO2 have the same affinity for rubisco, which of the following is the lowest ratio of CO2 to O2 at which a net synthesis of sugar can occur? (a) 1:3 (b) 1:2 (c) 1:1 (d) 2:1 (e) 3:1 14-38 A suspension of the cyanobacterium Chlamydomonas is actively carrying out photosynthesis in the presence of light and CO2. If you turned off the light, how would you expect the amounts of ribulose 1,5-bisphosphate and 3-phosphoglycerate to change over the next minute? How about if you left the light on but removed the CO2?
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The Origins of Chloroplasts and Mitochondria 14-39 Which of the phylogenetic trees in Figure Q14-39 is the most accurate? (The mitochondria and chloroplasts are from maize, but are treated as independent “organisms” for the purposes of this question.)
Figure Q14-39
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14-40 The biology of Methanococcus jannaschii is of interest to evolutionary biologists because (more than one answer may be correct) (a) phylogenetically it is thought to be more closely related to the proposed ancestor cell than are organisms such as cyanobacteria and yeast. (b) its small genome is approximately the same size as the genome of the proposed ancestor cell. (c) it lives under conditions that resemble the environment in which the first living cells may have dwelt. (d) its method of carbon fixation is thought to have given rise to the dark reactions used in modern cyanobacteria and chloroplasts. (e) the proteins in the electron-transport chain that Methanococcus uses to generate energy are thought to have given rise to those in the electron-transport chain used in modern mitochondria and aerobic bacteria. 14-41 Describe what is meant by “nitrogen fixation.”
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Answers 14-1
Choices (a), (d) and (e) are correct. Fermentation probably evolved before aerobic respiration because the early earth had a very low concentration of atmospheric oxygen (choice (a)). Fermentation probably evolved before photosynthesis because it is far simpler and more fundamental than photosynthesis or aerobic respiration (thus b is unlikely). Once photosynthetic reactions caused an increase in the concentration of free oxygen, organisms that were able to catalyze reactions of aerobic respiration came to have an enormous selective advantage (choice (e)). Chemiosmotic coupling, which forms the basis of both aerobic respiration and photosynthesis, first evolved in prokaryotes (thus choice (c) is incorrect). It is thought that the ancestor of eucaryotic cells enveloped a bacterial cell capable of aerobic respiration, which then became the mitochondria; a similar mechanism of envelopment and symbiosis that occurred later probably originated the chloroplasts (choice (d)).
14-2
True. When the citric acid cycle is operating as a cycle of reactions, the primary products from a single molecule of pyruvate are 3 CO2, 1 GTP, 3 reduced NADH, and 1 reduced FADH2. The electrons in NADH and FADH2 are passed into the electron transport chain to generate ATP via oxidative phosphorylation. Although a single GTP is made directly by the citric acid cycle, many molecules of ATP are made by oxidative phosphorylation of the 3 NADH molecules and 1 FADH2.
14-3
True. An electron carrier with low electron affinity carries the electron in a high-energy state. Transfer of an electron from a low-affinity carrier to a high-affinity carrier is energetically favorable. Thus, transfer of an electron to compounds with sequentially higher electron affinities is thermodynamically favorable and releases energy that can be coupled to the energetically unfavorable pumping of protons against an electrochemical gradient. Oxygen has very high affinity for electrons, and thus serves as an efficient electron sink for the entire chain.
14-4
The citric acid cycle stops almost immediately upon removal of oxygen because several steps in the cycle require the oxidized forms of NAD+ and FAD. In the absence of oxygen, these electron carriers can be reduced by the citric acid cycle reactions but cannot be reoxidized by the electron-transport chain that participates in oxidative phosphorylation.
14-5
(1) inorganic phosphate (Pi ); (2) an electron donor like NADH or FADH2 that provides high-energy electrons (or high-energy electrons); (3) an electron acceptor.
14-6
Choices (d) and (e) are the correct answers. All chemiosmotic coupling systems involve a proton gradient that is utilized by an ATP synthase that binds ADP and phosphorylates it. Hence all chemiosmotic systems will be affected by agents that prevent ADP from binding the synthase or that dissipate the proton gradient. Cytochrome oxidase and oxygen are required only for mitochondria and aerobic bacteria (not Methanococcus); light is required only for chloroplasts and photosynthetic bacteria (not Methanococcus).
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14-7
A. B. C. D. E. F.
Porin = outer membrane mitochondrial genome = matrix citric acid cycle enzymes = matrix proteins in the electron-transport chain = inner membrane ATP synthase = inner membrane transport protein for pyruvate = inner membrane.
14-8
Choice (a) is the correct answer. Protons are NOT pumped from the intermembrane space into the matrix, but are instead pumped in the reverse direction. All the other statements are true.
14-9
Mitochondria can use both pyruvate and fatty acids directly as fuel. NADH produced in the citric acid cycle donates electrons to the electron-transport chain. The citric acid cycle oxidizes acetyl groups and produces carbon dioxide as a waste product. Oxygen acts as the final electron acceptor in the electron-transport chain. The synthesis of ATP in mitochondria is also known as oxidative phosphorylation.
14-10 NADH donates electrons to the first of the three respiratory enzyme complexes in the mitochondrial electron-transport chain. Cytochrome c is a small protein that acts as a mobile electron carrier in the respiratory chain. Cytochrome oxidase transfers electrons to oxygen. Electron transfer in the chain occurs in a series of oxidation-reduction reactions. The first mobile electron carrier in the respiratory chain is ubiquinone. 14-11 Choice (b) is the correct answer. The pumping of protons out of the matrix into the intermembrane space creates a difference in proton concentration between the two sides of the membrane, with the matrix at a higher pH (i.e., more alkaline) than the intermembrane space, which tends to equilibrate with the cytosol that has a neutral pH. The electrons in NADH are at a higher energy than the electrons in reduced ubiquinone but proton pumping is not determined simply by the energy of the electron donors (choice (a)). Instead, the number of protons that can be pumped by each complex is determined by the difference in energy between the electrons in each substrate/product pair (i.e., the difference between the electrons in NADH and reduced ubiquinone, compared to that between reduced ubiquinone and reduced cytochrome c). The proton concentration gradient and the membrane potential generated by the electron-transport chain work in the same direction (choice (c)), creating a steep electrochemical gradient for protons across the membrane. Choice (d): The difference in proton concentration has a smaller effect on the total proton-motive force than does the membrane potential. Choice (e): All of the free energy released by the net transfer of electrons from NADH to O2 cannot be captured in the form of the proton gradient; some of the free energy released by the reactions is lost as heat. 14-12 In the absence of oxygen, the respiratory chain no longer pumps protons and thus no proton electrochemical gradient is generated across the bacterial membrane. In these conditions the ATP synthase uses some of the ATP generated by glycolysis in the cytosol to pump protons out of the bacterium, thus forming the proton gradient across the membrane that the bacterium requires to import vital nutrients by coupled transport. 239
14-13 (c)
Inhibition of the ATP/ADP translocase prevents export of ATP generated by oxidative phosphorylation in exchange for an import of ADP into the matrix. The ensuing buildup of ATP at the expense of ADP inhibits the ATP synthase. Since protons are no longer being used to power the ATP synthase, the proton gradient is not dissipated; the increasingly steep proton gradient makes it increasingly difficult for the electron-transport proteins to pump protons out of the matrix and electron transport quickly stops. Hence, permeabilizing the inner membrane to protons will allow electron transport to resume (although no ATP will be synthesized).
14-14 Choices (b) and (e) are the correct answers. Since the inside of the membrane (the mitochondrial matrix) is more negative than the outside, in principle, any traffic resulting in an increase in the positive charge in the matrix can be driven by the membrane potential. Hence, import of Ca2+ into the matrix, and exchange of Fe2+ (or ATP) in the matrix for Fe3+ (or ADP) in the intermembrane space can be driven by the membrane potential and need not require the cotransport of protons down the pH gradient. Import of acetate ion into the matrix and exchange of Ca2+ in the matrix for Na+ in the intermembrane space, in contrast, result in an increase in the amount of negative charge in the matrix and cannot be driven by the charge difference between the two mitochondrial compartments. 14-15 A.
B.
The ∆G for hydrolysis is –11.1 kcal/mole. This result is calculated by plugging values into equation given: ∆G = –7.3 kcal/mole + 1.42 log10 ([0.001 M] [0.010 M]/[0.005 M]) = –7.3 kcal/mole + 1.42 log10 (0.002) = –11.1 kcal/mole. The ∆G for synthesis is +11.1 kcal/mole because the forward and reverse reactions always have the same numerical value for ∆G, but with the sign reversed. At equilibrium, the ∆G is equal to zero by definition. The ratio of [ATP] to [ADP] at equilibrium is less than one to 10 million. This result is calculated by setting ∆G = 0, so that 1.42 log10 ([ADP] [Pi]/[ATP]) = –∆G° = 7.3 kcal/mole. Solving for [ADP]/[ATP], the equation becomes log10 ([ADP] [0.010]/[ATP]) = 7.3/1.42 = 5.14; then [ADP]/[ATP] = (105.14)/(0.010) = 13.8 × 106. Thus, the reciprocal [ATP]/[ADP] is 7.2 × 10–8.
14-16 113 molecules. Since each molecule of acetyl CoA produces 3 molecules of NADH and 1 molecule of FADH2, the oxidation of stearyl CoA generates a total of 35 molecules of NADH and 17 molecules of FADH2. From each molecule of NADH and FADH2, 2.5 and 1.5 molecules of ATP can be formed, respectively. Not counted in this calculation are the 9 molecules of GTP produced by the citric acid cycle, prior to electron transport.
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14-17 A.
B.
14-18 A.
B.
This arrangement of components within the vesicles allows experimental manipulation of the medium surrounding the vesicles in order to examine the consequences of different conditions in the mitochondrial matrix. The medium can be altered by changing pH, adding electron carriers and oxygen, and providing ADP and Pi, for example. Oxidation of electron carriers, consumption of oxygen, and production of ATP can be measured in the medium. By changing the composition of the medium, it should be possible, for example, to identify the electron carriers that can donate electrons from the matrix to the transport chain, to assess the redox potentials of various components of the transport chain, and to determine the dependence of ATP synthesis on the pH gradient across the membrane and on the ATP:ADP ratio. Respiration reactions will rapidly consume at least some of the added oxygen. During the anaerobic conditions, the electron carriers in the electron transport chain were reduced; upon addition of oxygen, electrons will be transferred to oxygen, thereby reducing the oxygen and oxidizing the carriers. Concomitant with the electron flow, protons will be pumped from the medium into the vesicles, thereby making the medium slightly more basic and the inside of the vesicles acidic. Inhibition of the ATP synthase will not have an immediate effect on oxygen consumption or proton pumping. However, the proton concentration inside the vesicles will quickly become too high to continue the activity of the electron transport-coupled proton pumping and thus electron transport and oxygen consumption will cease. If the ATP synthase molecules were randomly oriented, you would still expect ATP to be synthesized, although at about half the rate. The molecules that were oriented correctly would make ATP; the oppositely oriented ATP synthase molecules would be inert. If bacteriorhodopsin was randomly oriented, you would expect much less ATP to be synthesized. In vesicles with equal numbers of oppositely oriented bacteriorhodopsin molecules, no pH difference would be generated upon exposure to light because the proton pumping in both directions would be equal. In vesicles with an excess of outwardly directed proton pumps, the pH difference would be in the wrong direction to be utilized by ATP synthase and thus, no ATP would be made. In vesicles with an excess of inwardly directed proton pumps, a pH difference in the right orientation would be generated; thus, those vesicles would be capable of synthesizing some ATP.
14-19 A proton is a hydrogen atom that has lost its single electron and, thus, is positively charged. A hydride ion is a hydrogen atom that has gained an extra electron and thus is negatively charged. A hydrogen atom is a proton plus one electron; it is neutral. A hydrogen molecule is a pair of hydrogen atoms that share their two electrons in a covalent bond; it is neutral.
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14-20 Choice (c) is the correct answer. By definition, E′0 refers to the standard state of equal concentrations of each member of the redox pair. Therefore ∆E′0 does not vary with the actual concentrations. Compounds with positive redox potentials can donate electrons to other compounds under standard conditions, so long as the electron acceptor has a higher (more positive) redox potential, thus option (a) is incorrect. Compounds that are able to donate only one electron do not necessarily have higher redox potentials than those of compounds that are able to donate two electrons, thus option (b) is incorrect. (Water, for example, has a very high redox potential.) Although the ∆E′0 of a reaction is directly proportional to the ∆G°′ of a reaction and both are independent of the concentrations of substrates and products, the ∆G depends on these concentrations, thus option (d) is incorrect. In option (e), the transfer of electrons from water to AH2 is unfavorable. This redox reaction gives a ∆E′0 = ∆E′0 (acceptor) –E′0 (donor) = 600 mV – 820 mV = –220 mV. Since a negative ∆E′0 corresponds to a positive standard free-energy, this electron transfer will not occur under standard conditions. 14-21 The ∆G° is negative. The sign of ∆G° is opposite of that of ∆E′0 = E′0 (acceptor) – E′0 (donor). The acceptance of electrons by B from A has a ∆E′0 = 100 + 100 = 200 . The reverse reaction, the donation of electrons from B to A, has a positive ∆G° and thus is unfavorable under standard conditions. Remember that, by definition, the concentrations of A and its redox pair A′ are equal under standard conditions; likewise, the concentration of B is equal to the concentration of its redox pair B′. B will be able to donate electrons to A only when [B]>[B′] and/or [A]<[A′] to such an extent that the ∆G for electron transfer becomes negative. 14-22 An electron bound to a molecule with low affinity for electrons is a high energy electron. Transfer of an electron from a molecule with low affinity to one with higher affinity has a negative ∆G° and is thus favorable under standard conditions. If the reduced form of a redox pair is a strong electron donor with a low affinity for electrons, it is easily oxidized; the oxidized member of such a redox pair is a weak electron acceptor. 14-23 (b), (e) In order for a reaction to drive ATP synthesis under standard conditions, the ∆G°′ of the reaction must be less than –7.3 kcal/mol. Since ∆G°′ = –n (0.023) ∆E′0, the value of ∆E′0 must be greater than 317 mV/n, where n is the number of electrons transferred. ∆E′0 is 130 mV for the reduction of a molecule of pyruvate by NADH, 390 mV for the reduction of a molecule of cytochrome b by NADH, 40 mV for the reduction of a molecule of cytochrome b by ubiquinone, 200 mV for the oxidation of a molecule of ubiquinone by cytochrome c, and 590 mV for the oxidation of cytochrome c by oxygen. The number of electrons transferred in each of the above reactions is 2, 1, 1, 1, and 1, respectively. Thus only reactions (b) and (e) are sufficient to drive ATP synthesis. 14-24 True. The flow of electrons down the respiratory chain moves from carriers with lower affinity for electrons to ones with higher affinity and finally to oxygen, which has the highest affinity of all. Thus, the relative positions of cytochromes and iron-sulfur centers make sense in terms of the natural flow of electrons to oxygen. 242
14-25 Choice (e) is the correct answer. Cytochrome oxidase, which is the last carrier in the mitochondrial electron-transport chain and thus has the highest redox potential, contains copper ions and a heme group. Ubiquinone is not a protein and does not contain a metal group (choice (a)). Both 2Fe2S and 4Fe4S centers carry one electron (choice (b)). Ironsulfur centers generally have a lower redox potential than do cytochromes (choice (c)). The heme group in cytochrome c contains a charged iron ion. The interiors of proteins are often hydrophobic, favoring a relatively high redox potential, since reduction of the iron ion decreases its charge, and charges are energetically unfavorable in a hydrophobic environment. Mutation of hydrophobic residues near the heme group of cytochrome c to acidic (negatively charged) amino acids will neutralize the charge on the iron atom and most likely decrease the redox potential, by making the uptake of an electron less energetically favorable (choice (d)). 14-26 (c) 14-27 This result allows you to order the electron carriers in the respiratory chain because when oxygen is added, the last carrier in the chain will be oxidized first. This is because oxygen is the final sink for the electrons that flow through the chain, and it participates directly in a redox reaction with the last electron carrier. The wave of oxidation will then proceed backwards through the chain toward the first electron carrier in the chain; this is because the oxidation of each carrier will convert it to a form that can accept electrons from the “upstream” carrier in the chain, thereby oxidizing sequentially each upstream carrier. The order of cytochromes in the respiratory chain is the reverse of the order in which they are oxidized (i.e., the order in which the reduced form is lost). Listed from first to last, the cytochromes in the chain are b, c1, c, (a + a3). 14-28 Oxygen is held tightly by the cytochrome oxidase of the respiratory chain until it has accepted the four electrons (plus the 4 H+) required to convert it to 2 molecules of water. Superoxide is therefore never released into the cellular milieu. 14-29 A. B.
C.
Cytochrome b, quinone, the NADH dehydrogenase complex, and NADH will accumulate in the reduced form. Cytochrome c1, cytochrome c, and the cytochrome oxidase complex are more likely to be oxidized because they will still be able to donate electrons to O2, but will not receive any more electrons from the early part of the pathway. The ATP:ADP ratio will fall because electron transport, the source of most cellular ATP, will be stopped.
14-30 No, protons are pumped across different membranes in the two organelles. Protons are pumped across the inner membrane into the intermembrane space in mitochondria. By contrast, they are pumped across the thylakoid membrane into the thylakoid space in chloroplasts. Yes, ATP is synthesized in the corresponding compartments in the two organelles: in the matrix in mitochondria and in the stroma in chloroplasts.
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14-31 (a), (b), and (e) The thylakoid spaces are continuous with one another, and grana are merely stacks of thylakoids. Ions can flow from one thylakoid space to another and therefore from the interior of one granum to another. Since the outer membrane is porous, ions are also able to freely diffuse from the cytosol to the intermembrane space. 14-32 A—B; B—C; C—B; D—C; E—B; F—C; G—M; H—M; I—B; J—C; K—M 14-33 Light reaction: H2O + NADP+ + H+ + ADP + Pi → O2 + NADPH + ATP. Dark reaction: CO2 + NADPH + ATP → (CH2O) + NADP+ + H+ + ADP + Pi. (CH2O) here indicates a one-carbon unit of a carbohydrate; other ways of representing the sugar produced are also valid. 14-34 (b)
The rate of photosynthesis will increase with increasing light intensity until all of the reaction centers are being hit directly by photons. At saturating levels of light, the rate of photosynthesis is limited by the number of reaction centers that are still capable of being excited and hence can be increased only by increasing the number of reaction centers or by increasing the rate at which the reaction centers are restored to their low energy state. Increasing the number of chlorophyll molecules in the antennae complexes, the energy per photon of light, or the rate at which chlorophyll molecules are able to transfer energy electrons to one another will have no effect on either of these parameters. Adding a powerful oxidizing agent might, if anything, interfere with the reduction of the reaction center back to its resting state.
14-35 (c)
If you now inhibit photosystem II, you will deprive plastoquinone, which can still donate its electrons to the b6-f complex, of an electron source. Hence, plastoquinone will accumulate in its oxidized form. In contrast, all of the other components downstream of plastoquinone will be able to cycle between their oxidized and reduced states. ATP synthesis will continue, since electrons are still being fed through the b6-f complex, and in fact the same amount of ATP will be generated.
14-36 In the carbon fixation process in chloroplasts, carbon dioxide is initially added to the sugar ribulose 1,5-bisphosphate. The final product of carbon fixation in chloroplasts is the three-carbon compound glyceraldehyde 3-phosphate. This is converted into pyruvate (which can be directly used by the mitochondria), into sucrose (which is exported to other cells), and into starch (which is stored in the stroma). The carbon fixation cycle requires energy in the form of ATP and reducing power in the form of NADPH.
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14-37 (e)
Three molecules of O2 are required to form 3 molecules of 3-phosphoglycerate and 3 molecules of phosphoglycolate. In order to break even (i.e., simply to keep the Calvin cycle going with no net sugar produced), you need to have enough 3phosphoglycerate to synthesize ribulose 1,5-bisphosphate again. Therefore, for every 3 molecules of O2 that react with ribulose 1,5-bisphosphate, you need to generate 2 additional molecules of 3-phosphoglycerate. For every 3 molecules of CO2 that go into the Calvin cycle, one molecule of 3-phosphoglycerate is formed. So if you have at least 6 molecules of CO2 per 3 molecules of O2 going through the Calvin cycle, you will break even. Only if you have a ratio of CO2 to O2 higher than 6:3 (2:1) can you have a net synthesis of carbohydrate.
14-38 In the absence of light and presence of CO2, ribulose 1,5-bisphosphate will decrease and 3-phosphoglycerate will increase. The presence of CO2 allows ribulose 1,5-bisphosphate to be converted to 3-phosphoglycerate, but in the absence of light (and therefore in the presence of reduced amounts of NADPH and ATP), 3-phosphoglycerate will accumulate because subsequent reactions require NADPH and ATP. In the absence of CO2, ribulose 1,5-bisphosphate will accumulate (and 3-phosphoglycerate will decrease) because its conversion to 3-phosphoglycerate is dependent on CO2. 14-39 C
Mitochondria are most closely related to Bacillus, and chloroplasts to cyanobacteria. Maize (a eucaryote) is more closely related to Giardia (a simple eucaryote) than it is to bacteria (procaryotes).
14-40 (a), (c) Methanococcus jannaschii is an archaebacterium that lives in very hot, anaerobic thermal vents that are thought to resemble the conditions under which the first cells lived. Like many organisms that live under these unpleasant conditions, Methanococcus jannaschii seems to be more closely phylogenetically related to the proposed ancestor cell than it is to most other organisms. However, the genome of Methanococcus is perhaps 18 times larger than the genome of the suspected ancestor cell, which may have had only 100 or so genes at most. The reactions that Methanococcus uses in carbon fixation and energy generation use many proteins that are different from those used by chloroplasts, cyanobacteria, mitochondria, and aerobic bacteria. 14-41 Nitrogen fixation is the conversion of gaseous nitrogen (N2) from the environment into a form (like ammonia, NH3) that cells can use for biosynthesis of nitrogenous biomolecules, for example, the production of the bases in nucleotides. It is analogous to carbon fixation, which is the conversion of carbon atoms in gaseous CO2 into a form (like sugar) that can likewise be used for biosynthesis.
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CHAPTER 15 INTRACELLULAR COMPARTMENTS AND TRANSPORT 2009 Garland Science Publishing 3rd Edition
Membrane-Enclosed Organelles 15-1 Name the membrane-bounded compartments in a eucaryotic cell where each of the functions listed below takes place. A. Photosynthesis B. Transcription C. Oxidative phosphorylation D. Modification of secreted proteins E. Steroid hormone synthesis F. Degradation of worn-out organelles G. New membrane synthesis H. Breakdown of lipids and toxic molecules
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15-2
Label the structures of the cell indicated by the lines on the figure below:
Figure Q15-2 A. B. C. D. E. F. G. H. I. J. 15-3
nucleus free ribosomes rough endoplasmic reticulum Golgi apparatus cytosol endosome plasma membrane lysosome mitochondrion peroxisome
You discover a fungus that contains a strange star-shaped organelle not found in any other eucaryotic cell you have seen. On further investigation you find the following 1. the organelle possesses a small genome in its interior. 2. the organelle is surrounded by two membranes. 3. vesicles do not pinch off the organelle membrane. 4. the interior of the organelle contains proteins similar to those of many bacteria. 5. the interior of the organelle contains ribosomes. How might this organelle have arisen?
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Protein Sorting 15-4
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Plasma membrane proteins are inserted into the membrane in the __________________. The address information for protein sorting in a eucaryotic cell is contained in the __________________ of the proteins. Proteins enter the nucleus in their __________________ form. Proteins that remain in the cytosol do not contain a __________________. Proteins are transported into the Golgi apparatus via __________________. The proteins transported into the endoplasmic reticulum by __________________ are in their __________________ form. amino acid sequence endoplasmic reticulum folded
15-5
Golgi apparatus plasma membrane protein translocators
sorting signal transport vesicles unfolded
What would happen in each of the following cases? Assume in each case that the protein involved is a soluble protein, not a membrane protein. A. You add a signal sequence (for the ER) to the amino-terminal end of a normally cytosolic protein. B. You change the hydrophobic amino acids in an ER signal sequence into charged amino acids. C. You change the hydrophobic amino acids in an ER signal sequence into other, hydrophobic, amino acids. D. You move the amino-terminal ER signal sequence to the carboxyl-terminal end of the protein.
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15-6
You are trying to identify the peroxisome-targeting sequence in the thiolase enzyme from yeast. The thiolase enzyme normally resides in the peroxisome and therefore must contain amino acid sequences that are used to target the enzyme for import into the peroxisome. To identify the targeting sequences, you create a set of hybrid genes that encode fusion proteins containing part of the thiolase protein fused to another protein, histidinol dehydrogenase (HDH). HDH is a cytosolic enzyme required for the synthesis of the amino acid histidine and cannot function if it is localized in the peroxisome. You genetically engineer a series of yeast cells to express these fusion proteins instead of their own versions of these enzymes. If the fusion proteins are imported into the peroxisome, the HDH portion of the protein cannot function and the yeast cells cannot grow on media lacking histidine. You obtain the following results:
Figure Q15-6 What region of the thiolase protein contains the peroxisomal targeting sequence? Explain your answer.
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15-7
What is the role of the nuclear localization sequence in a nuclear protein? (a) It is bound by cytoplasmic proteins that direct the nuclear protein to the nuclear pore. (b) It is a hydrophobic sequence that enables the protein to enter the nuclear membranes. (c) It aids protein unfolding in order for the protein to thread through nuclear pores. (d) It prevents the protein diffusing out of the nucleus via nuclear pores.
15-8
A gene regulatory protein, A, contains a typical nuclear localization signal but surprisingly is usually found in the cytosol of cells. When the cell is exposed to hormones, protein A moves from the cytosol into the nucleus where it turns on genes involved in cell division. When you purify protein A from cells that have not been treated with hormones, you find that protein B is always complexed with it. To determine the function of protein B, you engineer cells lacking the gene for protein B. You compare normal and defective cells by using differential centrifugation to separate the nuclear fraction from the cytoplasmic fraction and then separate the proteins in these fractions by gel electrophoresis. You identify the presence of protein A and protein B by looking for their characteristic bands on the gel. The gel you run is shown below:
Figure Q15-8 On the basis of these results, what is the function of protein B? Explain your conclusion and propose a mechanism for how protein B works. 15-9
Which of the following statements about import of proteins into mitochondria are TRUE? (a) The signal sequences on mitochondrial proteins are usually carboxyl terminal. (b) The first stage of import of a mitochondrial protein is across the outer membrane into the intermembrane space. (c) Most mitochondrial proteins are not imported from the cytosol but are synthesized inside the mitochondria. (d) Mitochondrial proteins are translocated across the inner and outer membranes simultaneously. (e) Mitochondrial proteins cross the membrane in their native, folded state.
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15-10 Proteins destined to enter the endoplasmic reticulum (a) are transported across the membrane after their synthesis is complete. (b) are synthesized on free ribosomes in the cytosol. (c) begin to cross the membrane while still being synthesized. (d) cross the membrane in a folded state. (e) all remain within the endoplasmic reticulum. 15-11 After isolating the rough endoplasmic reticulum from the rest of the cytoplasm, you purify the RNAs attached to it. Which of the following proteins do you expect the RNA from the rough endoplasmic reticulum to encode? (a) Soluble secreted proteins (b) ER membrane proteins (c) Mitochondrial membrane proteins (d) Plasma membrane proteins (e) Ribosomal proteins 15-12 Briefly describe the mechanism by which the presence of an internal stop-transfer sequence in a protein causes the protein to become embedded in the lipid bilayer as a transmembrane protein with a single membrane-spanning region. Assume that the protein has an amino terminal signal sequence and just one internal hydrophobic stop-transfer sequence. 15-13 Using genetic engineering techniques, you have created a set of proteins that contain two (and only two) conflicting signal sequences that specify different compartments. Predict which signal would win out for the following combinations. Explain your answers. A. Signals for import into the nucleus and import into the ER. B. Signals for export from the nucleus and import into the mitochondria. C. Signals for import into mitochondria and retention in the ER. 15-14 A protein traverses the plasma membrane three times in the orientation shown below (N = amino terminus, C = carboxyl terminus; the hydrophobic membrane-spanning regions are shown as open boxes). This protein is known to have a signal sequence that is cleaved by signal peptidase in the ER.
Figure Q15-14 Sketch the ER membrane and the arrangement of the newly synthesized protein chain after it has completed its entry into the ER membrane but before any action of signal peptidase. Be sure to label the cytosol, the ER lumen, the signal sequence, and the amino and carboxyl termini of the protein in your diagram. 252
15-15 The figure below shows the orientation of a multipass transmembrane protein after it has completed its entry into the ER membrane (part A) and after it gets delivered to the plasma membrane (part B). This protein has an amino-terminal signal sequence (depicted as the dark grey membrane spanning box), which is cleaved off in the endoplasmic reticulum by signal peptidase. The other membrane-spanning domains in the protein are depicted as open boxes. Given that any hydrophobic membrane-spanning domain can act as either a start-transfer or a stop-transfer region, draw the final consequences of the actions described below on the orientation of the protein in the plasma membrane. Be sure to indicate on your drawing the extracellular space, the cytosolic face, and the plasma membrane, as well as the amino- and carboxyl-termini of the protein.
Figure Q15-15 A. B.
Deleting the first signal sequence. Changing the hydrophobic amino acids in the first, cleaved, sequence to charged amino acids. C. Changing the hydrophobic residues in every other transmembrane sequence to charged residues, starting with the first, cleaved, signal sequence. 15-16 Examine the multipass transmembrane protein shown in Figure Q15-16. What would you predict would be the effect of converting the first hydrophobic transmembrane segment to a hydrophilic segment? Sketch the arrangement of the modified protein in the ER membrane.
Figure Q15-16
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Vesicular Transport 15-17 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Proteins are transported out of a cell via the __________________ or __________________ pathway. Fluids and macromolecules are transported into the cell via the __________________ pathway. All proteins being transported out of the cell pass through the __________________ and the __________________. Transport vesicles link organelles of the __________________ system. The formation of __________________ in the endoplasmic reticulum stabilizes protein structure. carbohydrate disulfide bonds endocytic endomembrane endoplasmic reticulum endosome exocytic
Golgi apparatus hydrogen bonds ionic bonds lysosome protein secretory
15-18 Name two functions of the protein coat of vesicles that bud from membranous organelles used in vesicular transport. 15-19 An individual transport vesicle (a) contains only one type of protein in its lumen. (b) will fuse with only one type of membrane. (c) is endocytic if it is traveling toward the plasma membrane. (d) is enclosed by a membrane with the same lipid and protein composition as the membrane of the donor organelle.
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15-20 In class we have discussed how v-SNAREs and t-SNARES mediate the recognition of a vesicle with its target membrane so that a vesicle displaying a particular type of vSNARE will only fuse with a target membrane containing a complementary type of tSNARE. It is also known that in some cases, v-SNAREs and t-SNAREs may also mediate fusion of identical membranes. In yeast cells, right before the formation of a new cell, vesicles derived from the vacuole will come together and fuse to form a new vacuole destined for the new cell. Unlike the situation we’ve discussed in class, the vacuolar vesicles contain both v-SNAREs and t-SNAREs. Your friend is trying to understand the role of these SNAREs in the formation of the new vacuole and wants to consult with you regarding the interpretation of his data. Your friend has designed an ingenious assay for the fusion of vacuolar vesicles utilizing alkaline phosphatase. The protein alkaline phosphatase is made in a “pro” form that must be cleaved in order for the protein to be active. Your friend has designed two different strains of yeast: strain A produces the “pro” form of alkaline phosphatase (proPase), while strain B produces the protease that can cleave pro-Pase into the active form (Pase). Neither strain has the active form of the alkaline phosphatase, but when vacuolar vesicles from the strains A and B are mixed, fusion of vesicles generates active alkaline phosphates, whose activity can be measured and quantified.
Figure Q15-20A Your friend has taken each of these yeast strains and further engineered them so that they express only the v-SNAREs, the t-SNAREs, both (the normal situation), or neither SNARE. He then isolates vacuolar vesicles from all strains and tests the ability of each variant form of strain A to fuse with each variant form of strain B, using the alkaline phosphatase assay. The data are shown in the graph depicted in Figure Q15-20B. On this graph, the SNARE present on the vesicle of the particular yeast strain is indicated as “v” (for the presence of the v-SNARE) and “t” (for the presence of the t-SNARE).
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Figure Q15-20 B What does his data say about the requirements for v-SNAREs and t-SNAREs in the vacuolar vesicles? Be sure to comment on whether it is important to have a specific type of SNARE (that is, v- or t-SNARE) on each vesicle.
Secretory Pathway 15-21 N-linked oligosaccharides on secreted glycoproteins are attached to (a) nitrogen atoms in the polypeptide backbone. (b) the serine or threonine in the sequence Asn-X-Ser/Thr. (c) the amino terminus of the protein. (d) the asparagine in the sequence Asn-X-Ser/Thr. (e) the aspartic acid in the sequence Asp-X-Ser/Thr. 15-22 Name two types of protein modification that can occur in the ER but not in the cytosol. 15-23 If you were to remove the ER-retention signal from a protein that normally resides in the ER lumen, where do you expect the protein will ultimately end up? Be sure to explain your reasoning, for full credit.
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15-24 Match the set of labels below with the numbered label lines on Figure 15-24.
Figure Q15-24 A. B. C. D. E.
Cisterna Golgi stack Secretory vesicle trans Golgi network cis Golgi network
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15-25 A plasma membrane protein carries an oligosaccharide containing mannose (Man), galactose (Gal), sialic acid (SA), and N-acetylglucosamine (GlcNAc). These sugars are added to the protein as it proceeds through the secretory pathway. First, a core oligosaccharide containing Man and GlcNAc is added, followed by Gal, Man, SA, and GlcNAc in a particular order. Each addition is catalyzed by a different transferase acting at a different stage as the protein proceeds through the secretory pathway. You have isolated mutants defective for each of the transferases, purified the membrane protein from each of the mutants, and identified which sugars are present in each mutant protein. The results are summarized in Table Q15-25. Table Q15-25 Cell lacking: A. Oligosaccharide protein transferase B. Galactose transferase C. SA transferase D. GlcNAc transferase
Sugars present in the purified protein Man Gal SA GlcNAc – – – – +
–
–
+
+ +
+ –
– –
+ less than in normal cells
From these results, match each of the transferases (A, B, C, D) to its subcellular location selected from the list below. (Assume that each location contains only one enzyme.) 1. 2. 3. 4.
Central Golgi cisternae cis Golgi network ER trans Golgi network
15-26 For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement. New plasma membrane reaches the plasma membrane by the [regulated/constitutive] exocytosis pathway. New plasma membrane proteins reach the plasma membrane by the [regulated/constitutive] exocytosis pathway. Insulin is secreted from pancreatic cells by the [regulated/constitutive] exocytosis pathway. The interior of the trans Golgi network is [acidic/alkaline]. Proteins that are constitutively secreted [aggregate/do not aggregate] in the trans Golgi network. 15-27 In a cell capable of regulated secretion, what are the three main classes of proteins that must be separated before they leave the trans Golgi network?
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Endocytic Pathways 15-28 Name three possible fates for an endocytosed molecule that has reached the endosome. 15-29 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Eucaryotic cells are continually taking up materials from the extracellular space by the process of endocytosis. One type of endocytosis is __________________, which involves utilizing __________________ proteins to form small vesicles containing fluids and molecules. After these vesicles pinch off from the plasma membrane, they will fuse with the __________________, where the uptaken materials are sorted. A second type of endocytosis is __________________, which is used to take up large vesicles that can contain microorganisms and cellular debris. Macrophages are especially suited for this process, as they extend __________________ (sheetlike projections of their plasma membrane) to surround the invading microorganisms. chaperone cholesterol clathrin endosome
Golgi apparatus mycobacterium phagocytosis pinocytosis
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pseudopods rough ER SNARE transcytosis
15-30 Fibroblast cells from patients W, X, Y, and Z, who each have a different inherited defect, all contain “inclusion bodies,” which are lysosomes filled with undigested material. You wish to identify the cellular basis of these defects. The possibilities are: 1. 2. 3.
a defect in one of the lysosomal hydrolases. a defect in the phosphotransferase that is required for mannose-6-phosphate tagging of the lysosomal hydrolases. a defect in the mannose-6-phosphate receptor, which binds mannose-6-phosphate tagged lysosomal proteins in the trans Golgi network and delivers them to lysosomes.
You find that when some of these mutant fibroblasts are incubated in media in which normal cells have been grown, the inclusion bodies disappear. This leads you to suspect that lysosomal hydrolases are being secreted by the constitutive exocytic pathway in normal cells and are being taken up by the mutant cells. (It is known that some mannose6-phosphate receptor molecules are found in the plasma membrane and can take up and deliver lysosomal proteins via the endocytic pathway.) You incubate cells from each patient with media from normal cells and media from each of the other mutant cell cultures, and get the following results.
Cell Line Normal W X Y Z
From normal cells
From cultures of W cells
Media From cultures of X cells
+ – + + +
+ – + + +
+ – – – –
From cultures of Y cells
From cultures of Z cells
+ – – – +
+ – – + –
+ indicates that the cells appear normal; – indicates that the cells still have inclusion bodies. For each patient (W, X, Y, Z) indicate which of the defects (1, 2, 3) they are most likely to have. 15-31 How is it that the low pH of lysosomes protects the rest of the cell from lysosomal enzymes in case the lysosome breaks?
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How We Know: Tracking Protein and Vesicle Transport 15-32 You have created a GFP fusion to a protein that is normally secreted from yeast cells. Since you have learned about the use of temperature-sensitive mutations in yeast to study protein and vesicle transport, you obtain a collection of three mutant yeast strains, each one defective in some aspect of the protein secretory process. Being a good scientist, you of course, also obtain a wild-type control strain. You decide to examine the fate of your GFP fusion protein in these various yeast strains and engineer the mutant strains to express your GFP fusion protein. However, in your excitement to do the experiment, you realize that you did not label any of the mutant yeast strains and no longer know which strain is defective in what process. You end up numbering your strains with the numbers 1 through 4, and then you carry out the experiment anyway, obtaining the following results (note that the black dots represent your GFP fusion protein):
Figure Q15-32 Name the process defective in each of these strains. Remember that one of these strains is your wild-type control.
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Answers 15-1
A B. C. D. E. F. G. H.
15-2
Photosynthesis = chloroplast Transcription = nucleus Oxidative phosphorylation = mitochondrion Modification of secreted proteins = Golgi apparatus and rough endoplasmic reticulum (ER) Steroid hormone synthesis = smooth ER Degradation of worn-out organelles = lysosome New membrane synthesis = ER Breakdown of lipids and toxic molecules = peroxisome
See Figure A15-2.
Figure A15-2 15-3
A genome, a double membrane, ribosomes, and proteins similar to those found in bacteria are evidence for an organelle having evolved from an engulfed bacterium.
15-4
Plasma membrane proteins are inserted into the membrane in the endoplasmic reticulum. The address information for protein sorting in a eucaryotic cell is contained in the amino acid sequence of the proteins. Proteins enter the nucleus in their folded form. Proteins that remain in the cytosol do not contain a sorting signal. Proteins are transported into the Golgi apparatus via transport vesicles. The proteins transported into the endoplasmic reticulum by protein translocators are in their unfolded form.
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15-5
A. B. C. D.
15-6
The protein will now be transported into the ER lumen. The altered signal sequence will not be recognized and the protein will remain in the cytosol. The protein will still be delivered into the ER. It is the distribution of hydrophobic amino acids that is important, not the actual sequence. The protein will not enter the ER. Because the carboxyl terminus of the protein is the last part to be made, the ribosomes synthesizing this protein will not be recognized by the SRP and carried to the ER.
The peroxisomal targeting sequence lies between amino acids number 100 and number 125. Any fusion protein containing this sequence can be targeted for import into the peroxisome (because the yeast cannot grow on media lacking histidine) while the fusion proteins lacking this region do not target the fusion protein for import into the peroxisome (because the yeast do grow on media lacking histidine). The most important pieces of data are the fusion proteins containing amino acids 100–200 of the thiolase protein fused to HDH and the fusion protein containing amino acids 1–125 of the thiolase protein fused to HDH. Both of these fusion proteins do not allow growth on media lacking histidine and can be used to define the minimal region necessary for targeting thiolase for import into the peroxisome. (Note that although these experiments show that amino acids 100–125 are necessary, these experiments do not show that this region is sufficient for peroxisomal targeting. It is possible that amino acids 100–125 is sufficient, or, it could be that this region collaborates with redundant signals between amino acids 1–100 or 125–200.)
15-7
(a)
15-8
The data on the gel shows that protein A is always found in the nucleus in the absence of protein B. Therefore, any mechanism that is proposed must explain this result. On possible answer is that protein B binds protein A and masks the nuclear localization signal. In the presence of hormone, protein B interacts with the hormone, which changes its conformation so that it can no longer bind protein A. When protein B no longer binds to protein A, the nuclear localization signal on protein A is now exposed and protein A can enter the nucleus. Therefore, in the absence of protein B, the nuclear localization signal on protein A is always exposed and protein A resides in the nucleus. Another possible answer is that protein B binds protein A and sequesters it by keeping protein A in some subcellular compartment, away from the nucleus. In the presence of hormone, protein B interacts with the hormone, changing its conformation so that it can no longer bind to protein A. When protein B is not present, protein A can enter the nucleus in the presence or absence of hormone.
15-9
(d)
15-10 (c) 263
15-11 (a), (b), and (d) The rough ER consists of ER membranes and polyribosomes that are in the process of translating and translocating proteins into the ER membrane and lumen. Thus all proteins that end up in the lysosome, Golgi apparatus, or plasma membrane, or are secreted, will be encoded by the RNAs associated with the rough ER. Mitochondrial and ribosomal proteins are translated on free cytosolic ribosomes. 15-12 The amino-terminal signal sequence initiates translocation and the protein chain starts to thread through the translocation channel. When the stop-transfer sequence enters the translocation channel, the channel discharges both the signal sequence and the stoptransfer sequence sideways into the lipid bilayer. The signal sequence is then cleaved, so that the protein remains held in the membrane by the hydrophobic stop-transfer sequence. 15-13 A.
B.
C.
The protein would enter the ER. The signal for a protein to enter the ER is recognized as the protein is being synthesized and will end up either in the ER or on the ER membrane. Proteins destined for the nucleus get recognized by cytosolic nuclear transport proteins once they are fully synthesized and fully folded. The protein would enter in the mitochondria. In order for a nuclear export signal to work, the protein would have to end up in the nucleus first and thus would need a nuclear import signal for the nuclear export signal to get utilized. The protein would enter the mitochondria. In order to be retained in the ER, the protein needs to enter the ER. Since there is no signal for ER import, the ER retention signal would not function.
15-14 The N-terminal signal sequence initiates translocation of the protein across the ER membrane. The signal sequence will be cleaved off by signal peptidase, leaving the amino-terminus of the protein in the luminal side of the ER membrane. Upon fusion to the plasma membrane, the amino terminus of the protein will reside in the extracellular space.
Figure A15-14
264
15-15 A.
Deleting the first signal sequence completely would convert the next membranespanning domain into an internal start-transfer signal and would invert the orientation of the protein (see Figure A15-15A).
B.
Changing the hydrophobic amino acids to charged amino acids destroys the ability of the sequence both to act as a signal sequence and to become a membrane-spanning sequence. Therefore, the adjacent membrane spanning domain will now become an internal start-transfer sequence and the protein will be inverted, as seen above in part A. The mutated signal sequence would not get cleaved off, since it would remain on the cytoplasmic side of the membrane and signal peptidase is found only inside the ER (see Figure A15-15B).
C.
Mutating every other membrane spanning region so that they are now charged (and thus cannot span the membrane) would decrease the number of transmembrane regions and increase the size of the internal loops between membrane-spanning regions (see Figure 15-15C).
Figure A15-15 15-16 As shown in Figure A15-16, elimination of the first transmembrane segment (by making it hydrophilic) would be expected to reverse the orientation of the protein in the membrane. What originally was the second transmembrane segment (a stop-transfer signal), would now be read as a start-transfer signal and would have the opposite orientation in the membrane—as would all the remaining transmembrane segments. Although the N-terminus would still be in the ER lumen, all the rest of the external parts of the protein would swap positions so that what was in the cytosol would now be in the ER lumen, and vice versa.
Figure A15-16 265
15-17 Proteins are transported out of a cell via the secretory or exocytic pathway. Fluid and macromolecules are transported into the cell via the endocytic pathway. All proteins being transported out of the cell pass through the endoplasmic reticulum and the Golgi apparatus. Transport vesicles link organelles of the endomembrane system. The formation of disulfide bonds in the endoplasmic reticulum stabilizes protein structure. 15-18 1. 2.
The proteins in the coat help shape the membrane into a bud. The proteins in the coat can also select cargoes for transport.
15-19 Choice (b) is the correct answer. An individual vesicle may contain more than one type of protein in its lumen (choice (a)), all of which will contain the same sorting signal (or will lack specific sorting signals). Endocytic vesicles (choice (c)) generally move away from the plasma membrane. The vesicle membrane will not necessarily contain the same lipid and protein composition as the donor organelle, since the vesicle is formed from a selected subset of the organelle membrane from which it budded (choice (d)). 15-20 In order to get maximal levels of vacuolar vesicle fusion, vesicles from each strain must carry both v-SNAREs and t-SNARES. Experiment 1, which represents the normal scenario, is the only experiment where 100% alkaline phosphatase activity is measured. However, as long as complementary SNAREs are present on the vesicles, some vesicle fusion does occur (see experiments 3, 4, 6, 7, 8, 9). If both vesicles are missing either vSNAREs (experiment 2) or t-SNAREs (experiment 5) or both SNAREs (experiment 10 and 11), the level of fusion is very low. It does not matter whether a t- or v-SNARE is on the vesicle of a particular strain, as long as the vesicle from the other strain contains a complementary SNARE (compare experiments 3 and 4, 6 and 7, and 8 and 9). 15-21 (d) 15-22 1. 2.
Proteins in the ER can undergo disulfide bond formation. (This does not occur in the cytosol because of its reducing environment.) Proteins in the ER can undergo glycosylation. (Glycosylating enzymes are not found in the cytosol.) (Signal-sequence cleavage is also an acceptable answer, although not really what this question is referring to.)
15-23 The protein would end up in the extracellular space. Normally, the protein would go from the ER to the Golgi apparatus, get captured because of its ER-retrieval signal, and return to the ER. However, without the ER-retrieval signal, the protein would evade capture, ultimately leave the Golgi via the default pathway, and become secreted into the extracellular space. The protein would not be retained anywhere else along the secretory pathway, as it presumably has no signals to promote such localization since it normally resides in the ER lumen. 15-24 A—3; B—1; C—5; D—4; E—2
266
15-25 A—3 (oligosaccharide protein transferase = ER) B—1 (galactose transferase = central Golgi cisternae) C—4 (SA transferase = trans Golgi network) D—2 (GlcNAc transferase = cis Golgi network Proteins are modified in a stepwise fashion in the Golgi apparatus, with early steps taking place in the cis Golgi, intermediate steps taking place in the central Golgi cisternae, and late steps occurring in the trans Golgi network. If each enzyme produces the substrate for the next step, then a mutant lacking the enzyme that catalyzes the addition of the first sugar will be missing all of the sugars, a mutant lacking the enzyme that catalyzes the addition of the second sugar will contain the first sugar but will lack the other three, and so on. By this logic, mannose and GlcNAc must be the first sugars added, additional GlcNAc the second, galactose the third, and SA the last. Hence, the oligosaccharide protein transferase must be in the ER, the GlcNAc transferase in the cis Golgi, the galactose transferase in the central Golgi, and the SA transferase in the trans Golgi. 15-26 New plasma membrane reaches the plasma membrane by the constitutive exocytosis pathway. New plasma membrane proteins reach the plasma membrane by the constitutive exocytosis pathway. Insulin is secreted from pancreatic cells by the regulated exocytosis pathway. The interior of the trans Golgi network is acidic. Proteins that are constitutively secreted do not aggregate in the trans Golgi network. 15-27 The three main classes of proteins that must be sorted before they leave the trans Golgi network in a cell capable of regulated secretion are (1) those destined for lysosomes, (2) those destined for secretory vesicles, and (3) those destined for immediate delivery to the cell surface. 15-28 1. 2. 3.
recycled to the original membrane destroyed in the lysosome transcytosed across the cell to a different membrane.
15-29 Eucaryotic cells are continually taking up materials from the extracellular space by the process of endocytosis. One type of endocytosis is pinocytosis, which involves utilizing clathrin proteins to form small vesicles containing fluids and molecules. After these vesicles pinch off from the plasma membrane, they will fuse with the endosome, where the uptaken materials are sorted. A second type of endocytosis is phagocytosis, which is used to take up large vesicles that can contain microorganisms and cellular debris. Macrophages are especially suited for this process, as they extend pseudopods (sheetlike projections of their plasma membrane) to surround the invading microorganisms.
267
15-30 W—3 (defect in mannose-6-phosphate receptor) X—2 (defect in phosphotransferase) Y—1; Z—1 (defect in lysosomal hydrolases); these will be defects in two different lysosomal acid hydrolases. A cell that has no mannose-6-phosphate receptor will be able to make all the lysosomal hydrolases properly, but will not be able to send them to the lysosome and will also not be able to scavenge hydrolases from the external media. Hence, this cell line cannot be rescued by culture media that has had lysosomal hydrolases secreted into it and thus will not be rescued by any of the media tested here. A cell line that has no phosphotransferase will be able to scavenge hydrolases from the external medium, but since all of the cell’s own hydrolases will lack the mannose-6-phosphate tag, it will be rescued only by media from a cell line that is able to make all of the hydrolases. Cell lines missing one hydrolase will be rescued by media from any cell line that is able to secrete that hydrolase in a mannose-6-phosphate tagged form; in addition, media from cultures of cells missing a hydrolase will rescue any cell line with another type of defect. 15-31 The lysosomal enzymes are all acid hydrolases, which have optimal activity at the low pH (about 5.0) found in the interior of lysosomes. If a lysosome were to break, the acid hydrolases would find themselves at pH 7.2, the pH of the cytosol, and would therefore do little damage to cellular constituents. 15-32 Strain A has protein accumulating in the ER, which means that this cell has a mutation that blocks transport from the ER to the Golgi apparatus. Strain B has secreted protein, and therefore is your wild-type control. Strain C has protein accumulating in the Golgi apparatus, and thus has a mutation that blocks exit of proteins from the Golgi apparatus. Strain D has protein accumulating in the cis-Golgi network, and thus has a mutation that blocks the travel of proteins through the Golgi cisternae.
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CHAPTER 16 CELL COMMUNICATION 2009 Garland Science Publishing 3rd Edition
General Principles of Cell Signaling 16-1 Cell lines A and B both survive in tissue culture containing serum but do not proliferate. Addition of Factor F stimulates proliferation of cell line A but not cell line B. Cell line A produces a normal receptor protein (R), and, in order to see what the role of R might be, you introduce this receptor protein into cell line B, using recombinant DNA techniques. You then test all of your various cell lines in the presence of serum for their response to factor F, with the results summarized in Table Q16-1.
Which of the following can be concluded from your results? Please explain your choices. (a) Binding of Factor F to its receptor is required for proliferation of cell line A but not cell line B. (b) Binding of receptor R to its ligand is required for proliferation of cell line B but not cell line A. (c) Receptor R probably does not bind to Factor F. (d) Cell line B does not normally have a receptor for Factor F.
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16-2
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Cells can signal to each other in various ways. A signal that must be relayed to the entire body is most efficiently sent by __________________ cells, which produce hormones that are carried throughout the body through the bloodstream. On the other hand, __________________ methods of cell signaling do not require the release of a secreted molecule and are used for very localized signaling events. During __________________ signaling, the signal remains in the neighborhood of the secreting cell and thus acts as a local mediator on nearby cells. Finally, __________________ signaling involves the conversion of electrical impulses into a chemical signal. Cells receive signals through a __________________, which can be an integral membrane protein or can reside inside the cell. amplification contact-dependent endocrine epithelial
G-protein K+ channel neuronal paracrine
phosphorylation receptor target
16-3
All members of the steroid hormone receptor family (a) are cell-surface receptors. (b) do not undergo conformational change. (c) are found only in the cytoplasm. (d) interact with signal molecules that diffuse through the plasma membrane. (e) regulate sexual development.
16-4
Can signaling through the steroid hormone/steroid hormone receptor pathway lead to amplification of the original signal? If so, how?
16-5
When the neurotransmitter acetylcholine is applied to skeletal muscle cells, it binds the acetylcholine receptor and causes the muscle cells to contract. Succinylcholine, which has been demonstrated to bind the acetylcholine receptor on skeletal muscle cells, is a muscle relaxant. Succinylcholine is a chemical analog of acetylcholine and is sometimes used by surgeons as a muscle relaxant. Propose a model for why succinylcholine causes muscle relaxation instead of contraction, as acetylcholine does.
16-6
The local mediator nitric oxide stimulates the intracellular enzyme guanylyl cyclase by (a) way of a G-protein-mediated mechanism. (b) way of a receptor tyrosine kinase. (c) diffusing into cells and stimulating the cyclase directly. (d) way of a cell-surface receptor linked to an intracellular signaling pathway. (e) activation of intracellular protein kinases.
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16-7
Name the three main classes of cell-surface receptors.
16-8
Which of the following statements are TRUE? (a) All hydrophilic signaling molecules bind to cell-surface receptors that contain at least one membrane-spanning domain. (b) All extracellular signaling molecules are transported across the plasma membrane into the cytosol by their receptor. (c) A cell-surface receptor capable of binding only one natural ligand can mediate only one kind of response. (d) Cells having the same set of receptors can respond in different ways to the same ligand molecules. (e) Foreign substances that occupy receptor sites normally bound by signal molecules always induce the same response that is produced by the natural ligand.
16-9
Intracellular steroid hormone receptors have binding sites for a signal molecule and a DNA sequence. Give one reason to explain how different cells that have identical intracellular steroid hormone receptors can activate different genes when the receptors bind the same signal molecule.
16-10 Your friend is studying a segment of a newly discovered virus that carries an enhancer of gene expression that confers glucocorticoid (a hormone) responsiveness on genes that are linked to it. He constructs two versions of a reporter gene, one which has only a minimal promoter linked to it (which contains sites for RNA polymerase binding), while the other reporter gene has both a minimal promoter plus the viral enhancer attached to it. The reporter gene allows him to measure the amount of transcription that occurs from each construct. Your friend puts each of these constructs into two different cell lines and examines the activity of the reporter gene in each cell line, as shown in the figure below. He is puzzled by his data and asks for your help in interpreting the findings.
Figure Q16-10 A. B.
From this data, can you tell if both cell lines contain glucocorticoid receptors? Why? Propose a possible explanation for the difference in the transcription of the reporter gene in cell lines 1 and 2 after introduction of the construct containing the viral enhancer in the presence of glucocorticoid.
271
G-Protein-Linked Receptors 16-11 Indicate by writing “yes” or “no” whether amplification of a signal could occur at the particular steps described below. Explain your answers. A. An extracellular signal molecule binds and activates a cell surface receptor. B. The activated G-protein-linked receptors cause Gα to separate from Gβ and Gγ. C. Adenylyl cyclase produces cyclic AMP. D. cAMP activates protein kinase A. E. Protein kinase A phosphorylates target proteins. 16-12 When a trimeric G protein is activated by a cell-surface receptor (a) the β subunit exchanges its bound GDP for GTP. (b) the GDP bound to the α subunit is phosphorylated to form bound GTP. (c) it dissociates into a free β subunit and an αγ subunit. (d) the α subunit exchanges its bound GDP for GTP. (e) it dissociates into an active α subunit and an inactive βγ subunit.
272
16-13 Acetylcholine acts at a G-protein-linked receptor on heart muscle to make the heart beat + more slowly by the effect of the G protein on a K channel, as shown in Figure Q16-13. Which one or more of the following would enhance this effect of acetylcholine?
Figure Q16-13 (a) (b) (c) (d) (e)
A high concentration of a non-hydrolyzable analog of GTP. Modification of the acetylcholine-receptor-linked G protein α subunit by cholera toxin. Mutations in the acetylcholine receptor that weaken the interaction between the receptor and acetylcholine. Mutations in the acetylcholine receptor that weaken the interaction between the receptor and the G protein. Mutations in the G protein β subunit that weaken the interaction between βγ and + the K channel.
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16-14 During the mating process, yeast cells respond to pheromones that are secreted from other cells. These pheromones bind to G-protein-linked receptors on the surface of the responding yeast cell and lead to activation of G proteins inside the cell. When a wildtype yeast cell senses the pheromone, its cellular physiology changes in preparation for mating: the cell stops growing until a mating partner is found. If yeast cells do not undergo the appropriate cellular responses upon sensation of pheromone, they are considered sterile. Yeast cells that are defective in one or more components of the G protein have characteristic phenotypes in the absence and presence of the pheromone, which are listed in Table 16-14. Table 16-14 Mating phenotypes of various mutant and nonmutant strains of yeast Phenotype MUTATION MINUS PHEROMONE PLUS PHEROMONE None (wild type) α subunit deleted β subunit deleted γ subunit deleted α and β deleted α and γ deleted β and γ deleted
normal growth arrested growth normal growth normal growth normal growth normal growth normal growth
arrested growth, mating response arrested growth, sterile normal growth, sterile normal growth, sterile normal growth, sterile normal growth, sterile normal growth, sterile
Which of the following models is consistent with the data from the genetic analysis of these mutants? Explain your answer. (a) Gα activates the mating response but is inhibited when bound by Gβγ. (b) Gβγ activates the mating response but is inhibited when bound by Gα. (c) The Gαβγ trimer is inactive; either free Gα or free Gβγ is capable of activating the mating response. (d) The Gαβγ trimer is active; both free Gα and free Gβγ are required to inhibit the mating response (e) The Gαβγ trimer inhibits the mating response; free Gα and free Gβγ have no activity. 16-15 Receipt of extracellular signals can cause changes in cellular physiology. These signals cause changes by altering the cell’s cytoplasmic machinery. Some changes occur on a very fast time scale (i.e., in less than a second) whereas other changes occur long after the receipt of a signal (i.e., hours). A. Name an intracellular molecular change that could result in a quick change in cell physiology. B. Name an intracellular molecular change that could lead to a slow change in cellular physiology. C. Explain why the response you named in A results in a quick change while the response you named in B results in a slow change.
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16-16 A calmodulin-regulated kinase (CaM kinase) is involved in learning and memory. This kinase is able to phosphorylate itself, and the kinase activity of the phosphorylated form 2+ is independent of the intracellular concentration of Ca . Thus the kinase stays active 2+ after Ca levels have dropped. Mice completely lacking this kinase have severe learning defects but are otherwise normal. A. Each of the following mutations also leads to defective memory. For each case explain the reason why. (1) A mutation that prevents the kinase from binding to ATP. (2) A deletion of the calmodulin-binding part of the kinase. (3) A mutation that destroys the site of autophosphorylation. B. What would be the effect on the kinase activity of CaM kinase if there were a mutation that reduced the strength of binding of the kinase to the phosphatase responsible for inactivating the kinase? 16-17 Activated G-protein-linked receptors activate G proteins by reducing the strength of GDP binding to the α subunit of the G protein, allowing GDP to dissociate and GTP, which is present at much higher concentrations, to bind. How do you suppose the activity of a G protein would be affected by a mutation that causes the affinity of the α subunit for GDP to be reduced without significantly changing its affinity for GTP? 16-18 When adrenaline binds to adrenergic receptors on the surface of a muscle cell, it activates a G-protein, initiating a signaling pathway where the activated Gα subunit activates adenylyl cyclase, increasing cAMP levels in the cell; the cAMP molecules then activate a cAMP-dependent kinase that, in turn, activates enzymes that result in the breakdown of muscle glycogen, thus lowering glycogen levels. You obtain muscle cells that are defective in various components of the G-protein signaling pathway. Indicate how glycogen levels would be affected in the presence of adrenaline in the following cells. Would they be higher, lower, or the same as normal cells that are treated with adrenaline? A. In cells that lack adenylyl cyclase? B. In cells that lack the receptor? C. In cells that lack cAMP phosphodiesterase? D. In cells that have a Gα subunit that cannot hydrolyze GTP but can interact properly with Gβ and Gγ. 16-19 The rod photoreceptors in the eye are extremely sensitive to light. Rod cells sense light through a signal transduction cascade involving light-activation of a G-protein coupled receptor that activates a G-protein, which activates cyclic GMP phosphodiesterase. Describe how you would expect the addition of the following drugs to affect the lightsensing ability of the rod cells. Explain your answers. A. A drug that is an inhibitor of cyclic GMP phosphodiesterase? B. A drug that is a nonhydrolyzable analog of GTP?
275
Enzyme-Linked Receptors 16-20 The growth factor Superchick stimulates proliferation of cultured chicken cells. The receptor that binds Superchick is a receptor tyrosine kinase, and many tumor cell lines have mutations in the gene that encodes this receptor. Which of the following types of mutations would be expected to induce uncontrolled cell proliferation? (a) A mutation that prevents localization of the receptor to the plasma membrane. (b) A mutation that prevents dimerization of the receptor. (c) A mutation that destroys the kinase activity of the receptor. (d) A mutation that prevents recognition of the receptor by phosphatases. (e) A mutation that prevents ligand binding. 16-21 Male cockroaches with mutations that strongly reduce the function of a receptor tyrosine kinase X, are oblivious to the charms of their female comrades. This particular receptor tyrosine kinase binds to a small molecule secreted by sexually mature females. Most males carrying loss-of-function mutations in the gene for Ras protein are also unable to respond to females. You have just read a paper in which the authors describe how they have screened cockroaches that are mutant in receptor tyrosine kinase X for additional mutations that partially restore the ability of males to respond to females. These mutations decrease the function of a protein that the authors call Z. Which of the following types of proteins could Z be? Explain your answer. (a) A protein that stimulates exchange of GDP for GTP by the Ras protein. (b) A protein that stimulates hydrolysis of GTP by the Ras protein. (c) A protein kinase activated by the Ras protein. (d) An adaptor protein that mediates the binding of the receptor tyrosine kinase X to the Ras protein. (e) A transcription factor required for the expression of the Ras gene. 16-22 A kinase will act as an integrating device in signaling if it (a) phosphorylates more than one substrate. (b) catalyzes its own phosphorylation. (c) is activated or inhibited to more than one protein. (d) has at least two sites at which the kinase itself can be phosphorylated by different kinases. (e) initiates phosphorylation cascades. 16-23 Antibodies are Y-shaped molecules that have two identical binding sites. Imagine that you have obtained an antibody that is specific for the extracellular domain of a receptor tyrosine kinase. When the antibody binds the receptor, it brings together two receptor tyrosine kinase molecules. If cells containing the receptor tyrosine kinase were exposed to the antibody, would you expect the kinase to be activated, inactivated, or unaffected? Explain your reasoning.
276
16-24 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. There are many different signaling pathways that cells can use for signal transduction. Some of them may involve signals that are gases, such as __________________, which can diffuse easily into cells. Other signaling pathways use cytokines as the signal, which bind to cytokine receptors. Cytokine receptors have no intrinsic enzyme activity but are associated with cytoplasmic tyrosine kinases called __________________s, which become activated upon cytokine binding to its receptor and go on to phosphorylate and activate cytoplasmic gene regulatory proteins called __________________s. Some signal transduction pathways involve chains of proteins that phosphorylate each other, as seen in the __________________ cascade. Lipids can also partake in relaying signals in the cell, as seen when phospholipase C cleaves the sugar-phosphate head off a lipid molecule to generate the two small messenger molecules __________________ (which remains imbedded in the plasma membrane) and __________________ (which diffuses into the cytosol). cyclic GMP DAG IP3 JAK
MAP-kinase NO Ras SMAD
277
STAT TGF-β
16-25 When activated by the signal, the platelet-derived growth factor (PDGF) receptor phosphorylates itself on multiple tyrosines (as indicated below by the circled Ps; the numbers next to these Ps indicate the amino acid number of the tyrosine). These phosphorylated tyrosines serve as docking sites for proteins that interact with the activated PDGF-receptor. These proteins are indicated in the figure below, and include the proteins A, B, C, and D. The PDGF-receptor is activated when it binds to the signal, PDGF. One of the cellular responses to PDGF is an increase in DNA synthesis, which can be measured by incorporation of radioactive thymidine into the DNA. To determine whether protein A, B, C, and/or D are responsible for activation of DNA synthesis, you construct mutant versions of the PDGF-receptor that retain one or more tyrosine phosphorylation sites. You express these mutant versions in cells that do not make a PDGF-receptor. In these cells, the various versions of the PDGF-receptor are expressed normally and in response to PDGF binding, become phosphorylated on whichever tyrosines remain. You measure the level of DNA synthesis in cells that express the various mutant receptors and obtain the data shown below.
Figure Q16-25 A. B. C.
From these data, which, if any, of these proteins A, B, C, and D are involved in the stimulation of DNA synthesis by PDGF? Why? Which, if any, of these proteins inhibit DNA synthesis? Why? Which, if any, of these proteins appear to play no detectable role in DNA synthesis? Why?
278
16-26 Two protein kinases, PK1 and PK2, work sequentially in a signal transduction pathway. You create cells that contain inactivating mutations in either PK1 or PK2 and find that these cells no longer respond to an extracellular signal. You also create cells containing a version of PK1 that is permanently active and find that the cells behave as if they are receiving the signal even when the signal is not present. When you introduce the permanently active version of PK1 into cells that have an inactivating mutation in PK2, you find that these cells also behave as though they are receiving the signal even when no signal is present. A. From these results, does PK1 activate PK2 or does PK2 activate PK1? Explain your answer. B. You now create a permanently active version of PK2 and find that cells containing this version behave as if they are receiving the signal even when the signal is not present. What do you predict will happen if you introduce the permanently active version of PK2 into cells that have an inactivating mutation in PK1? 16-27 The last common ancestor to plants and animals was a unicellular eucaryote. Thus, it is thought that multicellularity and the attendant demands for cell communication arose independently in these two lineages. This evolutionary viewpoint accounts nicely for the vastly different mechanisms that plants and animals use for cell communication. Fungi use signaling mechanisms and components that are very similar to those used in animals. Which of the phylogenetic trees shown in Figure 16-27 does this observation support?
Figure Q16-27
279
How We Know: Untangling Cell Signaling Pathways 16-28 Your friend is studying mouse fur color and has isolated the G-protein-linked receptor responsible for determining fur color as well as the signal that activates the receptor. She finds that, upon addition of the signal to pigment cells (cells that produce the pigment determining fur color), cAMP levels rise in the cell. She starts a biotech company and the company isolates more components of the signaling pathway responsible for fur color. Using transgenic mouse technology, the company genetically engineers mice that are defective in various proteins involved in determining fur color. The company obtains the following results: normal mice have beige (very light brown) fur color mice lacking the signal have white fur mice lacking the G-protein coupled receptor have white fur mice lacking cAMP phosphodiesterase have dark brown fur Your friend has also made mice that are defective in the Gα subunit of the G protein in this signaling pathway. This defective Gα works normally except that, once it binds GTP, it cannot hydrolyze GTP to GDP. What color do you predict the fur of these mice will be? Why? 16-29 Bacteria undergo chemotaxis toward amino acids, which usually indicates the presence of a food source. Chemotaxis receptors bind a particular amino acid and cause changes in the bacterial cell that induce the cell to move toward the source of the amino acid. Four types of chemotaxis receptors that mediate responses to different amino acids have been identified in a bacterium. The receptors are called ChrA, ChrB, ChrC, and ChrD. Each one of these receptors specifically senses serine, aspartate, glutamate, or glycine, although you do not know which receptor senses which amino acid. You have been given a wild-type bacterial strain that contains all four receptors, as well as various mutant bacterial strains that are lacking one or more of the receptors. To figure out which receptor senses which amino acid, you conduct experiments where you fill a capillary tube with an amino acid to attract the bacteria, dip the capillary tube into a solution containing bacteria, remove the capillary tube after 5 minutes, and count the number of bacteria in the capillary tube. Your results are shown in Table 16-29. Table 16-29 Chemotaxis in wild-type and mutant strains of bacteria NUMBER OF CELLS IN CAPILLARY TUBE Strain Intact Receptors Serine Aspartate Glutamate Glycine None 1 ChrA, ChrB, ChrC, ChrD 590 1050 950 66 5 2 ChrA, ChrB, ChrC 7 840 770 130 8 3 ChrC, ChrD 340 7 590 6 6 4 ChrA, ChrC, ChrD 550 6 650 41 5 5 ChrB, ChrC, ChrD 700 590 850 9 8 From these results, indicate which receptor is used for which amino acid. 280
Answers 16-1
(a), (c) Since cell line A proliferates only in the presence of F and since cell line B can proliferate in the absence of F once R is expressed, binding of factor F must be required for proliferation of cell line A but not cell line B. Because expression of R is sufficient to induce proliferation of cell line B in the absence of F, there is some other factor, X, presumably in the serum, that is capable of binding to R and stimulating B. Since R binds to X, it probably does not bind to F, since most receptors do not bind more than one ligand. We do not know the effect of eliminating receptor R from cell line A; therefore, we cannot say if binding of receptor R to its ligand is required for proliferation cell line A. Since R is not the receptor for F, we cannot say that cell line B does not contain a receptor for F. Cell line B could have a receptor for F and respond to F in ways that have nothing to do with cell proliferation.
16-2
Cells can signal to each other in various ways. A signal that must be relayed to the entire body is most efficiently sent by endocrine cells, which produce hormones that are carried throughout the body through the bloodstream. On the other hand, contact-dependent methods of cell signaling do not require the release of a secreted molecule and are used for very localized signaling events. During paracrine signaling, the signal remains in the neighborhood of the secreting cell and thus acts as a local mediator on nearby cells. Finally, neuronal signaling involves the conversion of electrical impulses into a chemical signal. Cells receive signals through a receptor, which can be an integral membrane protein or can reside inside the cell.
16-3
Choice (d) is the correct answer. All members of the steroid hormone receptor family are intracellular proteins (thus choice (a) is not correct) that interact with molecules that can diffuse through the plasma membrane unaided. Steroid hormone receptors shuttle between the cytoplasm and the nucleus, which is why choice (c) is incorrect. They are activated by either steroids or other hydrophobic signaling molecules and typically undergo a large conformational change upon binding to the signaling molecule (choice (b)). This conformational change activates the steroid hormone receptors, allowing them to promote or inhibit the transcription of the appropriate genes. Steroid hormone receptors are important for many aspects of human physiology, including, but not limited to sexual development (choice (e)).
16-4
Since the interactions of the signal molecule with its receptor and of the activated receptor with its gene are both one-to-one, there is no amplification in this part of the signaling pathway. The signal can, however, be amplified when the target gene is transcribed, since multiple copies of mRNA are usually produced from a gene once it has been switched on, and multiple copies of protein can be made from each mRNA molecule.
281
16-5
Although succinylcholine can bind the acetylcholine receptor, this association does not cause the acetylcholine receptor to alter its conformation and stimulate the intracellular signaling necessary for muscle contraction (which involves, in this case, the opening of ion channels). Succinylcholine prevents normal muscle contraction by competing with acetylcholine for binding to the acetylcholine receptor, thus blocking its action.
16-6
(c)
16-7
Ion-channel-linked receptors; G-protein-linked receptors; enzyme-linked receptors.
16-8
Choices (a) and (d) are correct answers. A hydrophilic molecule cannot diffuse across the membrane and can therefore only affect a cell if it binds to a cell-surface receptor that has at least one component that spans the bilayer (choice (a)). Cells having the same set of receptors can respond in different ways to the same ligand molecules, since the response to a receptor also depends on the types of intracellular relay pathways present in the cell (choice (d)). Most signal molecules remain bound to the extracellular domain of the receptor, while the intracellular domain mediates signal transduction. Although many signal molecules are endocytosed with their receptor, they remain inside membranebounded compartments and are therefore not transported into the cytosol (choice (b)). A cell-surface receptor capable of binding only one natural ligand can mediate more than one kind of response, depending on a variety of conditions, such as the types of intracellular signaling molecules present in the cell type on which the receptor is being expressed (choice (c)). Foreign substances that occupy receptor sites normally bound by extracellular signal molecules can sometimes induce the same response that is produced by the natural ligand, but they often block the binding of the natural ligand without stimulating an intracellular signal (choice (e)).
16-9
Cells can respond differently to activated intracellular steroid hormone receptors for several reasons. 1. The cells may differ in the gene regulatory proteins they express for modulating the activity of the activated receptor. 2. Different cells could have their DNA modified in different ways, either through differences in chromatin configuration or differences in the methylation-state of the DNA, which may affect the interaction of the activated receptor with its DNA.
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16-10 A.
B.
16-11 A. B. C. D.
E.
16-12 (d)
Both cell lines appear to contain the glucocorticoid receptor, because both cell lines demonstrate increases in reporter gene activity in response to addition of glucocorticoid. With the introduction of the construct containing the glucocorticoid-responsive viral enhancer, cell line 1 showed a 1000-fold increase and cell line 2 showed an 80-fold increase. Four reasonable explanations for this difference are (any one would be correct): 1. Cell line 1 contains a protein that does not exist in cell line 2. This protein cooperates with the glucocorticoid receptor to activate transcription of the reporter gene, leading to higher levels of the reporter gene in cell line 1 than cell line 2. 2. Cell line 2 contains a slightly defective glucocorticoid receptor that does not bind DNA as well as the normal glucocorticoid receptor in cell line 1, thus leading to lower levels of reporter gene activity in cell line 2 than cell line 1. 3. Cell line 1 contains a mutant glucocorticoid receptor that binds DNA much more strongly than the normal glucocorticoid receptor found in cell line 2. Therefore, higher levels of reporter gene activity are seen in cell line 1 than cell line 2. 4. Cell line 2 contains a protein that does not exist in cell line 1. This protein acts as an antagonist to the glucocorticoid receptor to lower the level of transcription from the reporter gene, leading to lower levels of reporter gene activity in cell line 2 than cell line 1. No. Each signal molecule activates only one receptor. No. Each G-protein-linked receptor only activates one heterotrimeric G protein. Yes. Each activated adenylyl cyclase molecule can generate many molecules of cAMP. No. In unstimulated cells, protein kinase A is held inactive in a protein complex. cAMP binding to the complex induces a conformational change, releasing the active protein kinase A. Therefore, 1 cAMP cannot activate more than 1 molecule of active protein kinase A. Yes. Each activated protein kinase A molecule can phosphorylate many molecules of target proteins. The α subunit exchanges its bound GDP for GTP. (e) is incorrect since the βγ subunit may have an effect instead of or as well as the α subunit. The other statements are simply untrue.
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16-13 (a), (b) The heart is induced to beat more slowly by binding of acetylcholine to a G+ protein-linked receptor, whose activated βγ subunit binds to and opens K channels. High concentrations of a non-hydrolyzable analogue of GTP or modification of the acetylcholine-receptor-linked G protein α subunit by cholera toxin (which also prevents GTP hydrolysis) will increase the length of time that + the G protein βγ subunit remains free of α and able to activate the K channel; they will therefore enhance the effect of acetylcholine. Receptor mutations that weaken its interaction with either acetylcholine or the G protein, or, β mutations + that weaken the interaction between βγ and the K channel, all make it more + difficult for the signal to proceed from the receptor to the K channel. 16-14 (b)
Single mutations in the Gβ and Gγ subunits of the G-protein allow for growth in the absence of pheromone and display a sterile phenotype in the presence of pheromone, while loss of the Gα subunit causes growth arrest in either the presence or absence of pheromone. Since arrested growth is a normal response that occurs when cells sense and bind pheromone, it must be the Gβγ subunits that normally activate the cellular mating response when released from the Gα subunit. In cells lacking Gα, Gβγ inappropriately causes growth arrest because Gα normally inhibits the action of Gβγ until Gβγ is released from Gα upon receptor binding to pheromone. This interpretation is consistent with the analysis of the double mutants where Gα is deleted together with either Gβ or Gγ. In these double mutants, normal growth is seen in the absence of pheromone because Gβ and Gγ act together as a complex to activate targets; lack of either Gβ or Gγ leads to no response to pheromone, even when Gα is not around to inhibit the response.
16-15 A
Any answer that involves modification of existing cellular components is correct. For example, protein phosphorylation, protein dephosphorylation, protein ubiquitination, lipid phosphorylation, and lipid cleavage are all examples of correct answers. Alterations in gene expression. Modification of an existing cellular component can happen quickly, as it involves components (both the modifying protein as well as the substance being modified) that already exist in the cell. Changes in cellular physiology due to changes in gene expression can take much longer, as the gene will need to be transcribed, the mRNA translated, and proteins accumulated to high enough levels to instigate change.
B. C.
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16-16 A.
B.
Since a complete lack of the CaM kinase causes a loss of memory, we can assume that mutations that lead to inactivation of the kinase would also have a deleterious effect on memory. (1) Protein kinases have a binding site for ATP, which is the source of the phosphate used for phosphorylation; if the kinase cannot bind ATP it will be inactive. (2) Since CaM kinases are activated by binding to calmodulin in the presence of Ca2+, deletion of the calmodulin-binding portion would inactivate the kinase. (3) A mutation that destroys the site of autophosphorylation will also impair the normal function of the kinase, since the kinase will become inactive as soon as Ca2+ levels drop. Mutations that make the kinase less likely to bind the phosphatase will increase the time that the kinase remains active following a transient increase in Ca2+ levels.
16-17 The mutant G protein would be constantly active. Each time the α subunit hydrolyzed GTP to GDP, the GDP would spontaneously dissociate, allowing GTP to bind and reactivate the α subunit, especially because the intracellular concentration of GTP is higher than the intracellular concentration of GDP. Normally, GDP is tightly bound by the α subunit, which keeps the G protein in its inactive state until release of GDP is stimulated by interaction with an appropriate activated G-protein-linked receptor. 16-18 A. B. C. D.
higher higher lower lower
16-19 A.
A drug that inhibits cyclic GMP phosphodiesterase would decrease any visual responses. Normally, cyclic GMP is continually being produced in the eye. The perception of light by a rod cell normally leads to activation of cyclic GMP phosphodiesterase, which then hydrolyzes cyclic GMP molecules; this causes Na+ channels to close, which changes the membrane potential and alters the signal sent to the brain. If cyclic GMP phosphodiesterase is blocked, levels of cyclic GMP would remain high and there would be no cellular response to light. A drug that is a nonhydrolyzable analog of GTP would lead to an extended visual response. This is because a nonhydrolyzable analog of GTP would cause prolonged activation of the G protein in response to light. Continued activation of the G protein would keep cyclic GMP phosphodiesterase levels higher than normal, leading to a prolonged period of lowered levels of cyclic GMP; this in turn would cause Na+ channels to be closed longer than normal, leading to a prolonged change in the membrane potential and an extended visual response.
B.
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16-20 (d)
Tyrosine receptor kinases are usually activated by ligand-induced dimerization, which allows the receptors to phosphorylate themselves and activate intracellular signaling proteins that are stimulated by the phosphorylated receptor. After it is activated, the receptor is dephosphorylated, and thereby inactivated, by a phosphatase. Therefore, a mutation in the phosphatase will inappropriately increase the activity of the receptor and lead to uncontrolled cell proliferation. Mutations that prevent localization of the receptor to the plasma membrane, dimerization of the receptor (including mutations that prevent ligand binding), or autophosphorylation will inactivate the receptor.
16-21 (b)
Mutations that increase the activity of Ras should mimic the effect of stimulating the receptor tyrosine kinase X in a receptor-independent fashion. Since the intracellular concentration of GTP is higher than that of GDP, some proportion of the Ras molecules is expected to be GTP-bound and active; ridding the cells of a protein, which stimulates GTP hydrolysis will increase this pool of active Ras. Mutants that cannot stimulate exchange of GDP for GTP by Ras will have the same phenotype as mutants lacking Ras, as will mutants lacking a transcription factor required for expression of the Ras gene. Mutants lacking the kinase activated by Ras will be unable to transmit any signal from Ras onward. Defects in an adaptor protein that mediates the binding of receptor X to Ras will have no further effect on a mutant already lacking the receptor.
16-22 Choices (c) and (d) are the correct answers. Integrating devices are able to relay signals from more than one signaling pathway. Being activated or inhibited by more than one protein or having sites that can be phosphorylated by different kinases allows a kinase (or any other signaling molecule) to be affected by more than one upstream signal. Choices (a), (b), and (e) affect the output signal that a kinase is able to produce, not its ability to integrate incoming signals. 16-23 The receptor tyrosine kinase will probably become activated upon binding of the antibody molecule. This is because receptor tyrosine kinases are usually activated by signal-induced dimerization. When receptor tyrosine kinase molecules are brought together, their cytoplasmic kinase domains become activated and each receptor phosphorylates the other.
286
16-24 There are many different signaling pathways that cells can use for signal transduction. Some of them may involve signals that are gases, such as NO, which can diffuse easily into cells. Other signaling pathways use cytokines as the signal, which bind to cytokine receptors. Cytokine receptors have no intrinsic enzyme activity but are associated with cytoplasmic tyrosine kinases called JAKs, which become activated upon cytokine binding to its receptor and go on to phosphorylate and activate cytoplasmic gene regulatory proteins called STATs. Some signal transduction pathways involve chains of proteins that phosphorylate each other, as seen in the MAP-kinase cascade. Lipids can also partake in relaying signals in the cell, as seen when phospholipase C cleaves the sugar-phosphate head off a lipid molecule to generate the two small messenger molecules DAG (which remains imbedded in the plasma membrane) and IP3 (which diffuses into the cytosol). 16-25 A.
B.
C.
16-26 A.
B.
Proteins A and D are involved in stimulating DNA synthesis. PDGF-receptors that can bind to only A or D (see experiments 2 and 5) can stimulate DNA synthesis to about 50% of normal amounts (which is represented by experiment 1). A and D are both needed and are used in an additive fashion; this is evident from experiment 6: when a PDGF-receptor can bind both A and D, DNA synthesis levels are close to that obtained with the normal receptor. Protein B is an inhibitor of DNA synthesis. Consequently, receptors with binding sites for B and D (see experiment 7) stimulate a lower DNA synthesis rate than do receptors that only bind D (experiment 5). Protein C plays no detectable role in DNA synthesis. Receptors that can only bind C (experiment 4) activate DNA synthesis about as much as receptors that don’t bind any proteins (experiment 9; our negative control). Furthermore, the binding of protein C does not affect the response mediated by protein D when the receptor can bind both C and D (experiment 8). Normally, PK2 activates PK1. We are told that PK1 and PK2 normally work sequentially in a signal transduction pathway. If PK1 is permanently activated, a response is seen independent of whether PK2 is present. If PK1 activated PK2, no response should be seen if PK1 were activated in the absence of PK2. You would predict that no response to signal would be observed. This is because PK2 normally needs to activate PK1 in order for the cells to respond to the signal. When PK2 is permanently activated in the absence of PK1, PK1 is not there to relay the signal to affect cellular changes.
16-27 Choice (b) is the correct answer (see Figure Q16-27). The similarities in signaling mechanisms between animals and fungi support the phylogenetic tree in which fungi branched from the animal lineage after plants and animals separated. This branching order is supported by a wide variety of other data, including genomic sequence comparisons.
287
16-28 These mice will have dark brown fur. The inability to hydrolyze GTP to GDP will lead to inappropriate activation of the signaling pathway that makes pigment. Too much pigment will be produced, as seen in the mice lacking cAMP phosphodiesterase (which lack the ability to dampen the signal), and the mice will end up with dark brown fur. 16-29 ChrA senses glycine, ChrB senses aspartate, ChrC senses glutamate, and ChrD senses serine. To figure this out, you must match the pattern of intact receptors with the pattern of responses to the various amino acids. For example, ChrD is missing in strain #2, which does not sense serine. Therefore, ChrD is the receptor used to sense serine. Since we know that one of the receptors senses glutamate, but all strains respond to glutamate, ChrC must be the sensor for glutamate as it is present in all strains.
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CHAPTER 17 CYTOSKELETON 2009 Garland Science Publishing 3rd Edition 17-1 Indicate which of the three major classes of cytoskeletal elements each statement below refers to. A. monomer that binds ATP B. includes keratin and neurofilaments C. important for formation of the contractile ring during cytokinesis D. supports and strengthens the nuclear envelope E. their stability involves a GTP cap F. used in the eucaryotic flagellum G. a component of the mitotic spindle H. can be connected through desmosomes I. directly involved in muscle contraction J. abundant in filopodia 17-2 Identify the cytoskeletal structures in the epithelial cells shown in Figure Q17-2.
Intermediate Filaments
17-3 Phosphorylation of nuclear lamins regulates their assembly and disassembly during mitosis. You add a drug to cells that are undergoing mitosis that inhibits the activity of an enzyme that dephosphorylates nuclear lamins. What do you predict will happen to these cells? Why?
289
17-4
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Intermediate filaments are mainly found in cells that are subject to mechanical stress. Some rare human genetic diseases are caused by mutations in genes that disrupt intermediate filaments. For example, the skin of people with epidermolysis bullosa simplex is very susceptible to mechanical injury; people with this disorder have mutations in their __________________ genes, the intermediate filament found in epithelial cells. These filaments are usually connected from cell to cell through junctions called __________________s. The main filaments found in muscle cells belong to the __________________ family; people with disruptions in these intermediate filaments can have muscular dystrophy. In the nervous system, __________________s help strengthen the extremely long extensions often present in nerve cell axons; disruptions in these intermediate filaments can lead to neurodegeneration. People who carry mutations in the gene for __________________, an important protein for crosslinking intermediate filaments, have a disease that combines symptoms of epidermolysis bullosa simplex, muscular dystrophy, and neurodegeneration. desmosome keratin kinase
lamin neurofilament plectin
synapse vimentin
Microtubules 17-5
Place the following in order of size, from the smallest to the largest. A. Protofilament B. Microtubule C. α-tubulin D. Tubulin dimer E. Mitotic spindle
290
17-6
In the three cell outlines in Figure Q17-6 indicate how the microtubules will be arranged, showing clearly their free and attached ends. On each figure indicate the plus end and the minus end for one of the microtubules.
Figure Q17-6 17-7
The graph in Figure Q17-7 shows the time-course of the polymerization of pure tubulin in vitro. You can assume that the starting concentration of free tubulin is much higher than it is in cells.
Figure Q17-7 A. B. 17-8
Explain the reason for the initial lag in the rate of microtubule formation. Why does the curve level out after point C?
The hydrolysis of GTP to GDP carried out by tubulin molecules (a) provides the energy needed for tubulin to polymerize. (b) occurs because the pool of free GDP has run out. (c) tips the balance in favor of microtubule assembly. (d) allows the behavior of microtubules called dynamic instability. 291
17-9
The microtubules in a cell form a structural framework that can have all the following functions EXCEPT (a) holding internal organelles such as the Golgi apparatus in particular positions in the cell. (b) creating long thin cytoplasmic extensions that protrude from one side of the cell. (c) strengthening the plasma membrane. (d) moving materials from one place to another inside a cell.
17-10 Do you agree or disagree with this statement? Explain your answer. Minus end-directed microtubule motors (like dyneins) deliver their cargo to the periphery of the cell, whereas plus end-directed microtubule motors (like kinesins) deliver their cargo to the interior of the cell. 17-11 Kinesins and dyneins (a) are inhibited by colchicine and taxol. (b) move along both microtubules and actin filaments. (c) often move in opposite directions to each other. (d) derive their energy from GTP hydrolysis. (e) have tails that bind to the filaments. 17-12 Match the following labels to the numbered lines on Figure Q17-12.
Figure Q17-12 A. B. C. D.
Minus end of microtubule. Tail of motor protein. Cargo of motor protein. Head of motor protein.
Which one of the two motors shown above is most likely a kinesin? Explain your answer.
292
17-13 Some lower vertebrates such as fish and amphibians can control their color by regulating specialized pigment cells called melanophores. These cells contain small, pigmented organelles, termed melanosomes, that can be dispersed throughout the cell, making the cell darker, or aggregated in the center of the cell to make the cell lighter. You purify the melanosomes from melanophores that have either aggregated or dispersed melanosomes and find that: 1. 2.
aggregated melanosomes co-purify with dynein dispersed melanosomes co-purify with kinesin
Given this set of data, propose a mechanism for how aggregation and dispersal of melanosomes occur. 17-14 Consider Figure Q17-14, which shows two isolated outer doublet microtubules from a eucaryotic flagellum with their associated dynein molecules.
Figure Q17-14 A. B. C.
Sketch what will happen to this structure if it is supplied with ATP. Sketch what will happen to this structure if the linking proteins are removed and it is supplied with ATP. In a complete flagellum, what would happen if all the dynein molecules were active at the same time?
293
17-15 You are curious about the dynamic instability of microtubules and decide to join a lab that works on microtubule polymerization. The people in the lab help you grow some microtubules in culture using conditions that allow you to watch individual microtubules under a microscope. You can see the microtubules growing and shrinking, as you expect. The professor who runs the lab gets in a new piece of equipment, a laser beam that is very fine and can be used to sever microtubules. She is very excited and wants to sever growing microtubules at their middle, using the laser beam. A. Do you predict that the newly exposed microtubule plus ends will grow or shrink? Explain your answer. B. What do you expect would happen to the newly exposed plus ends if you were to grow the microtubules in the presence of an analog of GTP that cannot be hydrolyzed and then severed the microtubules in the middle using a laser beam?
Actin Filaments 17-16 Do you agree or disagree with the following statement? Explain your answer. Nucleotide hydrolysis plays a similar role in actin polymerization and tubulin polymerization. 17-17 Actin binding proteins bind to actin and can modify its properties. You purify a protein, Cap1, that appears to bind and cap one end of an actin filament, although you do not know whether it binds the plus or the minus end. To determine which end of the actin filament your protein binds to, you decide to examine the effect Cap1 has on actin polymerization by measuring the kinetics of actin filament formation in the presence and the absence of Cap1 protein. You obtain the following results (see Figure Q17-17). Do you think Cap1 binds the plus or the minus ends of actin? Explain your reasoning.
Figure 17-17 17-18 In the cell, the concentration of actin monomer is higher than the concentration required for purified actin monomers to polymerize in vitro. Thymosin is a protein that can bind actin monomers. If you were to add a drug that inhibits the ability of thymosin to bind actin monomers, what effect would this have on actin polymerization? Explain your answer.
294
17-19 In the budding yeast, activation of the GTP-binding protein Cdc42 occurs upon binding of an external signal (pheromone) to a G-protein linked receptor. Activation of Cdc42 promotes actin polymerization. What would you predict would happen to actin polymerization, compared to pheromone treated cells, if you were to add: A. pheromone + an inhibitor of G-protein linked receptors? B. pheromone + a nonhydrolyzable analogue of GTP? 17-20 Which of the following statement(s) are FALSE? (a) Some myosin tails can bind the plasma membrane. (b) Actin filaments are usually excluded from the cell cortex. (c) Integrins are transmembrane proteins that can bind to the extracellular matrix. (d) ARPs can promote the formation of branched actin filaments. (e) Cytochalasins prevent actin polymerization.
Muscle Contraction 17-21 Do you agree or disagree with the following statement? Explain your answer. When skeletal muscle receives a signal from the nervous system to contract, the signal from the motor neuron triggers the opening of a voltage-sensitive Ca++ channel in the muscle cell plasma membrane, allowing Ca++ to flow into the cell. 17-22 Which of the following structures shortens during muscle contraction? (a) Myosin filaments (b) Flagella (c) Sarcomeres (d) Actin filaments
295
17-23 You isolate some muscle fibers to examine what regulates muscle contraction. When you bathe the muscle fibers in a solution containing ATP and Ca++, you see muscle contraction (experiment 3 in Table Q17-23). Ca++ is necessary, as solutions containing ATP alone or nothing do not stimulate contraction and thus the muscle remains in a relaxed state (experiments 1 and 2 in Table Q17-23). From what you know about the mechanism of muscle contraction, fill in your predictions of whether the muscle will be contracted or relaxed for experiments 4, 5, and 6. Explain your answers in the space below the table. Table Q17-23. Experiment number 1 2 3 4 5 6
added to muscle fibers
muscle state
nothing ATP ATP and Ca++ ATP, Ca++, and a drug that inhibits troponin from binding Ca++ ATP and a drug that inhibits binding of tropomyosin to actin a nonhydrolyzable analogue of ATP
relaxed relaxed contracted
Extra credit: In what state would the muscle be if you added Ca++ but no ATP?
How We Know: Pursuing Motor Proteins 17-24 You are interested in studying kinesin movements. You therefore prepare silica beads and coat them with kinesin molecules so that each bead, on average, contains only one kinesin molecule attached to it. You add these kinesin-coated beads to a preparation of microtubules you have polymerized. Using video microscopy, you watch the kinesin move down the microtubules. A. Kinesin-GFP has been measured to move along microtubules at a rate of 0.3 µm/second, and single-molecule studies have revealed that kinesin moves along microtubules progressively, with each step being 8 nm. How many steps can the kinesin molecule take in 4 seconds, assuming that the kinesin stays attached to the microtubule the entire 4 seconds? B. Since each kinesin molecule is thought to take approximately 100 steps before falling off the microtubule, will you see your silica beads detach from the microtubule during your 4 seconds of observation? C. What would you predict would happen to the kinesin-coated silica beads if you were to add AMP-PNP (a nonhydrolyzable ATP analogue)?
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Answers 17-1
A. B. C. D. E. F. G. H. I. J.
actin intermediate filaments actin intermediate filaments microtubules microtubules microtubules intermediate filaments actin actin
17-2
A—microtubules; B—intermediate filaments; C—actin filaments (microfilaments).
17-3
Cells should become arrested in mitosis. Normally, the lamins are phosphorylated during mitosis, causing disassembly of the nuclear envelope. At the end of mitosis, the nuclear lamins are dephosphorylated, causing the lamins to reassemble. Inhibition of this last step should therefore prevent the nuclear lamins from reassembling after mitosis.
17-4
Intermediate filaments are mainly found in cells that are subject to mechanical stress. Some rare human genetic diseases are caused by mutations in genes that disrupt intermediate filaments. For example, the skin of people with epidermolysis bullosa simplex is very susceptible to mechanical injury; people with this disorder have mutations in their keratin genes, the intermediate filament found in epithelial cells. These filaments are usually connected from cell to cell through junctions called desmosomes. The main filaments found in muscle cells belong to the vimentin family; people with disruptions in these intermediate filaments can have muscular dystrophy. In the nervous system, neurofilaments help strengthen the extremely long extensions often present in nerve cell axons; disruptions in these intermediate filaments can lead to neurodegeneration. People who carry mutations in the gene for plectin, an important protein for crosslinking intermediate filaments, have a disease that combines symptoms of epidermolysis bullosa simplex, muscular dystrophy, and neurodegeneration.
17-5
C, D, A, B, E
297
17-6
See Figure A17-6.
Figure A17-6 17-7
A.
B.
Before they can polymerize to form microtubules, tubulin molecules have to form small aggregates that act as nucleation centers. This aggregation step is slow, as the molecules have to come together in the right configuration. This is why there is a lag phase before microtubules start to be formed. After point C an equilibrium point has been reached where the rates of polymerization and depolymerization are exactly balanced.
17-8
(d)
The hydrolysis of GTP to GDP occurs after a GTP-bound tubulin molecule is incorporated into a microtubule, and it makes the microtubule more susceptible to disassembly. It is the resulting switch in microtubule stability that gives rise to the phenomenon known as dynamic instability.
17-9
(c)
One function of actin filaments, but not microtubules, is to provide a meshwork beneath the plasma membrane that helps to form and strengthen this membrane. Microtubules have all of the other functions that are described.
17-10 Disagree. The plus ends of microtubules usually point toward the cell periphery while the minus ends point toward the cell center. This is because the γ-tubulin in the centrosome serves to nucleate microtubule growth. Since the centrosomes are near the center of the cell, the minus ends of microtubules are located there. Therefore, a minusend directed microtubule motor would direct its cargo toward the center of the cell while a plus-end directed microtubule motor would direct its cargo toward the cell periphery. 17-11 (c) 17-12 A—4; B—2; C—1; D—3 The top motor is more likely to be kinesin, as kinesins usually move toward the plus end of the microtubules. 298
17-13 The melanosomes are transported in the cell on microtubules. When it is advantageous for the animal to become lighter, a signal is sent to the pigment cell that causes the melanosomes to associate with dynein. Since dynein is a minus-end directed motor, it will transport the melanosomes toward the center of the cell, causing the melanosomes to aggregate in the center and the cell to take on a lighter appearance. When the animal wants to become darker, a signal is sent to the pigment cell that causes the melanosomes to associate with kinesin. Kinesin is usually a plus-end directed motor and will move the melanosomes away from the center of the cell so that they are more dispersed, making the cell look darker. 17-14 A. B.
See Figure A17-14A. See Figure A17-14B. (Note to instructor: this question should be marked as correct only if the microtubule is shown bending in the correct direction (in A) and the correct microtubule is shown pushed forward (in B)).
Figure A17-14 C.
The flagellum will not bend because there is no significant relative motion of one microtubule doublet to another, since each is trying to push its neighbor forward at the same time. For the flagellum to bend, sets of dynein molecules on one side of the flagellum must be selectively activated.
299
17-15 A.
B.
The newly exposed microtubule plus ends will most likely shrink if the microtubules are severed in the middle. This is because a microtubule grows by adding GTP-carrying subunits to the plus end. The GTP is hydrolyzed over time, leaving only a cap of GTP-carrying subunits at the plus end with the remainder of the tubulin protofilament containing GDP-carrying subunits. Therefore, if you sever a growing microtubule at its middle, you will most likely create a plus end that contains GDP-carrying subunits. The GDP-carrying subunits are less tightly bound than the GTP-carrying subunits and will peel away from each other, causing depolymerization of the microtubule and shrinkage. If you were to polymerize the microtubules in the presence of a nonhydrolyzable analogue of GTP and then severed the microtubules with a laser, the newly exposed plus end would contain a GTP cap and so would likely continue to grow.
17-16 Agree. ATP-bound actin monomers are added onto actin filaments; GTP-bound tubulin subunits are added onto the growing end of a microtubule. Nucleotide hydrolysis occurs after addition of the subunit onto the filament in both actin and microtubules. ATP hydrolysis in actin polymerization reduces the strength of binding between monomers in the actin filaments; GTP hydrolysis during tubulin polymerization reduces the strength of binding between the tubulin subunits in the microtubule. Because the nucleotide hydrolysis decreases the strength of binding between subunits, depolymerization is enhanced in both actin filaments and microtubules. 17-17 You would predict that Cap1 binds the plus ends of actin, as it appears to inhibit actin polymerization. Actin filaments grow through the addition of monomers to the plus end of the actin filament. A capping protein that binds the plus ends of actin can block monomer addition to the actin filament. Thus, less actin polymerization will be seen in the presence of the Cap1 protein. 17-18 Addition of a drug that keeps thymosin from binding actin monomer will increase the rate of actin polymerization in the cell. Thymosin normally binds actin monomers in the cell to sequester monomers until they are needed for actin polymerization. The reason actin monomers do not spontaneously form filaments in the cell, despite the high concentration of actin monomer, is because the monomers are normally bound by actin monomer binding proteins (such as thymosin) and thereby prevented from adding onto the end of an actin filament. 17-19 A. B.
17-20 (b)
Less actin polymerization. Cdc42 will not be able to be activated by the Gprotein linked receptor. More actin polymerization. Cdc42 will be more active, as it will bind the nonhydrolyzable form of GTP and not be able to be turned off. Much of the actin in the cell is concentrated in the cell cortex, the region of the cell just beneath the plasma membrane.
300
17-21 Disagree. The rise in intracellular Ca++ during muscle contraction comes from an intracellular source. The Ca++ is released from the lumen of the sarcoplasmic reticulum, which is a specialized region of endoplasmic reticulum inside a muscle cell. The signal from the nerve terminal triggers an action potential in the muscle cell plasma membrane, which causes a voltage sensitive transmembrane protein in the membranous transverse tubules to open a Ca++ release channel in the membrane of the sarcoplasmic reticulum. 17-22 (c)
Sarcomeres contain actin filaments and myosin filaments that slide past each other during muscle contraction, leading to shortening of the sarcomere; the actin filaments and myosin filaments do not change in length. Flagella are microtubule-based structures that are not present on muscle cells.
17-23 See table below. Table A17-23 Experiment number 1 2 3 4 5 6
added to muscle fibers
muscle state
nothing ATP ATP and Ca++ ATP, Ca++, and a drug that inhibits troponin from binding Ca++ ATP and a drug that inhibits binding of tropomyosin to actin a nonhydrolyzable analogue of ATP
relaxed relaxed contracted relaxed contracted relaxed
In experiment 4, the muscle will be relaxed because troponin will not be able to bind Ca++. By preventing troponin from binding to Ca++, troponin will not be able to undergo the conformational change that causes tropomyosin to alter its association with actin. This altered association is normally required for myosin to bind actin. In the absence of troponin regulation by Ca++, myosin cannot bind actin and the muscle cannot contract. In experiment 5, the muscle will contract because tropomyosin cannot bind to actin. If tropomyosin cannot bind actin, myosin can. The presence of ATP means that it will be available for myosin to hydrolyze, causing muscle contraction. In experiment 6, the muscle will remain relaxed. The presence of Ca++ will induce a conformational change in troponin that causes tropomyosin to shift, exposing actin for myosin to bind. However, when myosin binds, the myosin molecule will attach and then release because the myosin will bind the nonhydrolyzable analogue of ATP. Because no ATP hydrolysis can occur, the muscle will remain in the relaxed state. Extra credit: The muscle will be in a rigor state. If Ca++ is added without ATP, troponin can bind Ca++, undergo a conformational change, which causes tropomyosin to shift and exposes the actin for the myosin to bind. However, myosin will bind actin and remain in the attached state, because a myosin head lacking a bound nucleotide is locked onto the actin until nucleotide binding occurs. 301
17-24 A.
B.
C.
You would expect the kinesin molecule to travel 150 steps. The calculation is as follows: 0.3 µm = 300 nm. Therefore, in 4 seconds, the kinesin molecule could travel 1200 nm if it were to move at a rate of 0.3 µm/second. Since the step size is 8 nm, 1200 nm/8 nm/step = 150 steps. Yes, you should see silica beads detach sometime during the four seconds of observation, as each bead will take more than 100 steps in that 4-second time frame (see A above). If you were to add AMP-PNP, you would no longer see the silica beads moving down the microtubule. It is thought that one molecule of ATP is hydrolyzed per step that kinesin takes; without ATP hydrolysis, translocation of the beads will be inhibited. However, you may still see the beads associated with the microtubules, as AMP-PNP does not inhibit the association of kinesin with the microtubule.
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CHAPTER 18 CELL DIVISION 2009 Garland Science Publishing 3rd Edition
An Overview of M Phase 18-1 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. The cell division cycle encompasses an alternation between __________________, which appears as a period of dramatic activity under the microscope, and a preparative period called __________________, which consists of three phases called __________________, __________________, and __________________. During M phase, the nucleus is divided in a process called __________________, and the cytoplasm is split in two in a process called __________________. The cell-cycle control system relies on sharp increases in the activities of regulatory proteins called __________________ or __________________ to trigger S and M phases. Inactivation of __________________ is required to terminate M phase after chromosome segregation. APC Cdks condensation cyclin-dependent kinases cytokinesis G1 phase G1-Cdk
G2 phase interphas intraphase kinesins M phase M-Cdk meiosis
metaphase microtubules mitosis myosins S phase S-Cdk
18-2 Which of the following does not occur during M phase in animal cells? (a) Condensation of chromosomes (b) Break down of nuclear envelope (c) Loosening of attachments to other cells (d) Growth of the cell (e) Fragmentation of the endoplasmic reticulum
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19-3
Before chromosome segregation in M phase, the chromosomes and the segregation machinery must be appropriately prepared. Indicate whether the following statements are TRUE or FALSE. If FALSE, change a single noun to make the statement TRUE. A. Sister chromatids are held together by condensins from the time they arise by DNA replication until the time they separate near the end of mitosis. B. Cohesins are required to make the chromosomes more compact and thus to prevent tangling between different chromosomes. C. The mitotic spindle is comprised of actin filaments and myosin filaments. D. Microtubule-dependent motor proteins and microtubule polymerization and depolymerization are responsible for the organized movements of chromosomes during mitosis. E. The centromere nucleates a radial array of microtubules called an aster, and its duplication is triggered by S-Cdk. F. Each microtubule-organizing center contains a pair of centrioles and hundreds of γ-tubulin rings that nucleate the growth of microtubules.
19-4
Describe the names, primary molecular components, and basic mechanisms of the two distinct cytoskeletal machines that are assembled to carry out the mechanical processes of mitosis and cytokinesis.
19-5
Examine the schematic representation of centrosome duplication in Figure Q19-5. By analogy with DNA replication, would you classify centrosome duplication as conservative or semi-conservative? Explain your answer.
Figure Q19-5
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Mitosis 19-6
Which stage of mitosis in an animal cell does each part of Figure Q19-6 represent?
Figure Q19-6 19-7
A living cell from the lung epithilium of a newt is shown at different stages in M phase in Figure Q19-7. Order these light micrographs into the correct sequence and identify the stage in M phase that each represents.
Figure Q19-7
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19-8
Name the stage of M phase in which the following events occur. Place the numbers 1-8 next to the letter headings to indicate the normal order of events. A. Alignment of the chromosomes at the spindle equator B. Attachment of microtubules to chromosomes C. Breakdown of nuclear envelope D. Pinching of cell in two E. Separation of two centrosomes to form mitotic spindle F. Reformation of the nuclear envelope G. Condensation of the chromosomes H. Separation of sister chromatids
19-9
The cytoskeleton of an animal cell changes dramatically between G1 and early M phase (prophase) of the cell cycle. For each of the following sentences, choose one of the options enclosed in square brackets that best describes the changes to the cytoskeleton and its components. Before mitosis, the number of centrosomes must [increase/decrease]. At the beginning of [anaphase/prophase] in animal cells, the centrosomes separate in a process driven partly by interactions between the [plus/minus] ends of microtubules arising from different centrosomes. This separation forms the bipolar mitotic spindle and is associated with a sudden [increase/decrease] in the dynamic instability of microtubules. Compared to an interphase microtubule array, a mitotic aster contains a [smaller/larger] number of [longer/shorter] microtubules. Extracts from M phase cells exhibit [increased/decreased/unchanged] rates of microtubule polymerization and increased frequencies of [catastrophes/rescues]. The changes in microtubule dynamics are largely due to [enhanced/reduced] activity of MAPs and [increased/decreased] activity of catastrophins. The new balance between polymerization and depolymerization of tubulin is necessary for the microtubules to bind the [sister chromatids/daughter chromosomes] and align them at the metaphase plate.
19-10 Match each of the main classes of spindle microtubules from List 1 with their functions and features from List 2. List 1 A. Interpolar microtubules B. Aster microtubules C. Kinetochore microtubules
List 2 1. Stabilized by interactions with each other via motor proteins 2. Interact with the cell cortex 3. Link chromosomes to a spindle pole 4. Required for separation of duplicated centrosomes 5. Depolymerize to promote anaphase A 6. Depolymerize to promote anaphase B
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19-11 Disassembly of the nuclear envelope (a) occurs during prophase. (b) involves the separation of the inner nuclear membrane from the outer nuclear membrane. (c) results in the conversion of the nuclear envelope into protein-free membrane vesicles. (d) is triggered by the phosphorylation of integrins. (e) must occur in order for kinetochore microtubules to form in animal cells. 19-12 Consider an animal cell that has eight chromosomes (four pairs of homologous chromosomes) in G1 phase. Indicate how many of each of the following features the cell will have at mitotic prophase. A. Sister chromatids B. Centromeres C. Kinetochores D. Centrosomes E. Centrioles 19-13 Which of the following statements about kinetochores is TRUE? (a) A kinetochore often binds to more than one microtubule. (b) They contain DNA-binding proteins that recognize sequences at the telomere of the chromosome. (c) Kinetochore proteins bind to the tubulin molecules at the very tip of the plus end of microtubules. (d) Kinetochores assemble on chromosomes that lack centromeres. (e) Kinetochores on sister chromatids face in the same direction. 19-14 Is the following statement TRUE or FALSE? Explain your answer. After the nuclear envelope breaks down, microtubules gain access to the chromosomes and, every so often, a randomly probing microtubule connects with a kinetochore and captures the chromosome. 19-15 A friend declares that chromosomes are held at the metaphase plate by microtubules that push on each chromosome from opposite sides. Which of the following observations best supports your belief that microtubules are pulling on the chromosomes, not pushing them? (a) The jiggling movement of chromosomes at the metaphase plate. (b) The way that chromosomes behave when the attachment between sister chromatids is severed. (c) The way that chromosomes behave when the attachment to one kinetochore is severed. (d) The shape of chromosomes as they move toward the spindle poles at anaphase. (e) The behavior of the spindle when colchicine is added to cells.
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19-16 It is estimated that as many as 25% of kinetochore microtubules and 75% of interpolar microtubules are not anchored to the centrosome. In spite of that, all the microtubules are focused tightly at the spindle pole. Why do you suppose the microtubule ends that are not attached to centrosomes do not splay out away from the poles? Hint: Think about how minus- and plus-end directed motor proteins can move and what the motors might bind. 19-17 Many anticancer drugs act on microtubules. It was, perhaps, initially surprising that high concentrations of some successful anticancer drugs (like colchicines) inhibit microtubule polymerization and others (like taxol) stabilize microtubules. At lower concentrations, both classes of drugs appear to block mitosis and induce apoptotic death of proliferating cells. Why might these drugs with opposite effects on microtubules both prevent mitosis? At what stage of mitosis do you expect treated cells will be blocked? What will happen if cells treated with these drugs do not arrest at a particular stage of mitosis? 19-18 Which of the following statements is TRUE? (a) Anaphase is triggered by the phosphorylation of proteins that hold the sister chromatids together. (b) Anaphase A must be completed before anaphase B can take place. (c) In cells where anaphase B predominates, the spindle will elongate much less than in cells where anaphase A dominates. (d) In anaphase A, both kinetochore and polar microtubules shorten. (e) In anaphase B, microtubules associated with the cell cortex shorten. 19-19 Which of the following is expected to impair anaphase B and not anaphase A? Explain your answer. (a) An antibody against myosin (b) ATPγS, a nonhydrolyzable ATP analog that binds to and inhibits ATPases (c) An antibody against motor proteins that move from the plus end of microtubules to the minus end (d) An antibody against motor proteins that move from the minus end of microtubules towards the plus end (e) Colchicine 19-20 Which of the following precede re-formation of the nuclear envelope during M phase in animal cells? (a) Formation of the contractile ring (b) Decondensation of chromosomes (c) Reassembly of the nuclear lamina (d) Activation of the anaphase promoting comples (APC) (e) Transcription of genes on chromosomes
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19-21 A serine to alanine mutation in the phosphorylation site of a lamin protein will (a) cause cells to arrest at telophase. (b) stabilize the nuclear lamina and thereby prevent its disassembly during mitosis. (c) prevent nuclear pore assembly. (d) prevent mitotic spindle formation. (e) prevent chromosome condensation. 19-22 Which of the following structures break into fragments prior to cell division? (a) Mitochondria (b) Nuclear membrane (c) Centrosomes (d) Lysosomes (e) Ribosomes (f) Chloroplasts (g) Endoplasmic reticulum
How We Know: Building the Mitotic Spindle 19-23 The lengths of microtubules in various stages of mitosis depend on the balance between the activities of catastrophins, which destabilize microtubules, and microtubuleassociated proteins (MAPs), which stabilize them. If you created cells with an increased number of catastrophin molecules, do you predict the length of the mitotic spindle will be longer, shorter, or unchanged relative to the corresponding stage of mitosis in wild-type cells? What do you predict for a cell with increased numbers of MAPs? Explain your reasoning.
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19-24 Last summer, while working in the laboratory of Professor Mike Rotubulis, you discovered a new protein that regulates microtubule dynamics. First, you isolated proteins from a cellular extract that bound to a tubulin affinity column. You then separated the proteins from each other by loading the mixture of proteins on an ionexchange column, eluting the column with increasing salt concentration and collecting small “fractions” of protein as they dripped from the column. To test if each fraction contained microtubule regulators, you mixed it with fluorescent tubulin and purified centrosomes and analyzed the reaction microscopically to measure the size of the aster formed. You found that fractions 8, 9, and 10 promoted formation of an unusually large aster, consisting of long microtubules. Because electrophoretic separation of the fractions on a gel revealed a plentiful protein with an apparent molecular weight of 98 kD, you named the protein p98. A. Does p98 behave like a MAP or a catastrophin? Hint: Does p98 promote or inhibit microtubule growth? B. Propose two ways that p98 might change the dynamic behavior of microtubules to account for the observed change in microtubule length. Hint: There are four simple possible mechanisms. C. Video microscopy of fluorescent tubulin in reactions with purified centrosomes allowed you to follow the behavior of individual microtubules over time. You graphed the changes in microtubule length in the absence (Figure Q19-24A) and presence (Figure Q19-24B) of recombinant p98. Five representative microtubules are shown for each condition. Does p98 alter the rate of microtubule growth or shrinkage? Does p98 alter the frequency of catastrophe or rescue? Explain your answers. D. After demonstrating the consequences of p98 on microtubule dynamics in vitro using purified components, you want to determine if the protein has the same effects in a complex cellular extract that naturally contains p98. Briefly describe an experiment that will allow you to determine what p98 normally does in mitotic extracts.
Figure Q19-24
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Cytokinesis 19-25 If a cell just entering mitosis is treated with nocodazole, which destabilizes microtubules, the nuclear envelope breaks down and chromosomes condense, but no spindle forms and mitosis arrests. Why do cells arrest in mitosis? 19-26 The cleavage furrow (a) is a puckering of the plasma membrane caused by constriction of a ring of filaments attached to the membrane. (b) begins to form at the end of telophase. (c) will not begin to form in the absence of a mitotic spindle. (d) always forms perpendicular to the interpolar microtubules. (e) always forms in the middle of the cell. 19-27 Cytokinesis in animal cells (a) requires ATP. (b) leaves a small circular scar of actin filaments on the inner surface of the plasma membrane. (c) is often followed by phosphorylation of integrins in the plasma membrane. (d) is generally accompanied by rearrangement of filaments making up the cell cortex. (e) is assisted by motor proteins that pull on microtubules attached to the cell cortex.
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Answers 19-1
The cell division cycle encompasses an alternation between M phase, which appears as a period of dramatic activity under the microscope, and a preparative period called interphase, which consists of three phases called G1 phase, S phase, and G2 phase. During M phase, the nucleus is divided in a process called mitosis, and the rest of the cell is split in two in a process called cytokinesis. The cell-cycle control system relies on sharp increases in the activities of regulatory proteins called cyclin-dependent kinases or Cdks to trigger S and M phases. Inactivation of M-Cdk is required to terminate M phase after chromosome segregation.
19-2
(d)
Cell size increases throughout interphase and not during M phase. All of the other phenomena are observed in M phase.
19-3
A.
False. Sister chromatids are held together by cohesins from the time they arise by DNA replication until the time they separate near the end of mitosis. False. Condensins are required to make the chromosomes more compact and thus to prevent tangling between different chromosomes. False. The contractile ring is comprised of actin filaments and myosin filaments. True. False. The centrosome nucleates a radial array of microtubules called an aster, and its duplication is triggered by S-Cdk. True.
B. C. D. E. F. 19-4
Mitosis, or the separation of chromosomes and their distribution to daughter cells, is accomplished by the bipolar mitotic spindle. The spindle is composed of microtubules and a variety of microtubule-dependent motor proteins. It is bipolar because in most animal cells two centrosomes, or microtubule organizing centers, are located at opposite poles of the cell and nucleate asters of microtubules that can interact with each other. Microtubules and microtubule motors also interact with the chromosomes and control their movement. Cytokinesis, or the division of an animal cell into two daughter cells, is accomplished by the contractile ring, which is composed of actin and myosin filaments and is located just under the plasma membrane. As the ring constricts, it pulls the membrane inward, ultimately dividing the cell in two.
19-5
Centrosome duplication is semi-conservative. The paired centrioles in the centrosome separate, and each serves to nucleate synthesis of a new centriole. As a consequence, each new centrosome consists of one old and one new centriole. Thus, centrosome duplication is analogous to DNA replication in which a new duplex consists of one old strand and one newly replicated strand.
19-6
A—telophase; B—prophase; C—anaphase; D—prometaphase
19-7
The correct order (and stage) are listed as follows: E (prophase), D (prometaphase), C (metaphase), A (anaphase), F (telophase), and B (cytokinesis).
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19-8
A. B. C. D. E. F. G. H.
5, metaphase 4, prometaphase 3, prometaphase 8, cytokinesis 2, prophase 7, telophase 1, prophase 6, anaphase
19-9
Before mitosis, the number of centrosomes must increase. At the beginning of prophase in animal cells, the centrosomes separate in a process driven partly by interactions between the plus ends of microtubules arising from different centrosomes. This separation forms the bipolar mitotic spindle and is associated with a sudden increase in the dynamic instability of microtubules. Compared to an interphase microtubule array, a mitotic aster contains a larger number of shorter microtubules. Extracts from M phase cells exhibit unchanged rates of microtubule polymerization and increased frequencies of catastrophes. The changes in microtubule dynamics are largely due to reduced activity of MAPs and increased activity of catastrophins. The new balance between polymerization and depolymerization of tubulin is necessary for the microtubules to bind the sister chromatids and align them at the metaphase plate.
19-10 A—1, 4; B—2, 6; C—3, 5 19-11 Choice (e) is the correct answer. In animal cells, kinetochore microtubules cannot form if the chromosomes in the nucleus are separated from the microtubules in the cytoplasm by the nuclear envelope. (But in some other cells, like the yeast S. cerevisiae, the nuclear envelope never breaks down and yet chromosomes can attach to microtubules emanating from the spindle poles within the nucleus.) The nuclear envelope disassembles during prometaphase (so choice (a) is false) by breaking up into vesicles containing lipids from both the outer and inner envelope (so choice (b) is false). Integral membrane proteins of the nuclear envelope and some of the nuclear lamins remain associated with the vesicles (so choice (c) is false). Phosphorylation of lamins (not integrins) triggers breakdown of the nuclear lamina (so choice (d) is false). 19-12 A—16; B—16; C—16; D—2; E—4 19-13 Choice (a) is the correct answer. The DNA-binding proteins of the kinetochore recognize the centromere, not the telomere (choice (b)), and thus cannot bind to chromosomes lacking centromeres (choice (d)). If kinetochore proteins bound to the very tips of microtubules, they would soon fall off, since the microtubules are dynamic and are constantly losing the monomers at the ends of the filament (choice (c)). Kinetochores on sister chromatids must face in opposite directions to facilitate their attachment to microtubules from opposite spindle poles (choice (e)).
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19-14 True. Microtubules nucleated by the centrosomes grow outward toward the chromosomes in a highly dynamic process, alternately growing and shrinking. When they eventually attach to the kinetochore of a chromosome, they become stabilized and are referred to as kinetochore microtubules. 19-15 Choice (c) is the correct answer. When the attachment to one kinetochore is severed, the whole chromosome moves to the opposite pole, showing that the kinetochore microtubules are pulling on their attached chromatid, not pushing it. The jiggling movement (choice (a)) is simply a sign that the chromosomes are subject to forces from both sides. When the attachment between sister chromatids is severed (choice (b)), both daughter chromosomes move toward their respective poles, but they could either be pushed there or pulled there. Similarly, the shape of the chromosomes as they move toward the pole (choice (d)) indicates that the chromosomes are being moved via an attachment at the centromere, but does not indicate whether they are being pushed or pulled. The disappearance of the spindle when colchicine is added to the cells simply shows that the spindle is composed of polymerizing and depolymerizing microtubules (choice (e)). 19-16 Microtubules that are not directly connected to the centrosomes are held in place by minus-end directed motors, which link them to the other microtubules and tend to “walk” them toward the spindle pole. 19-17 Mitosis requires that the microtubules be dynamically growing and shrinking to probe the cell cortex and seek attachments to kinetochores. Progression through mitosis requires that sister kinetochores be attached to opposite spindle poles and that dynamic microtubules exert opposing tension on the sister kinetochores. Static microtubules will be unable to execute these requirements, thus cells are likely to arrest prior to anaphase by virtue of the spindle assembly checkpoint. If cells do not arrest, cytokinesis will occur despite the absence of the microtubule-dependent movements of anaphase A and anaphase B that serve to segregate chromosomes to opposite poles of the cell. The failure of chromosome segregation will produce multinucleate cells or cells with an aberrant number of chromosomes. 19-18 Choice (e) is the correct answer. Anaphase is triggered by proteases that cleave the proteins that hold the sister chromatids together, thus (a) is false. Anaphase A and anaphase B generally occur at the same time, thus (b) is false. In cells where anaphase B predominates, the spindle will elongate more than in cells where anaphase A predominates, thus (c) is false. In anaphase A, only the kinetochore microtubules shorten, thus (d) is false.
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19-19 Choice (d) is the correct answer. The motor proteins used in anaphase B to push the interpolar microtubules apart move toward the plus end of microtubules, whereas the motor proteins used in anaphase A move toward the minus end. Myosin (choice (a)) is not involved in either anaphase A or anaphase B. Motor proteins require ATP hydrolysis (they are ATPases) and are used in both anaphase A and anaphase B, so both types of anaphase will be affected by ATPγS (choice (b)). The motor proteins used in anaphase A move toward the minus end of microtubules as do the motor proteins attached to the cell cortex used in anaphase B (choice (c)). Microtubules are also used in both anaphase A and anaphase B, so colchicine will affect both (choice (e)). 19-20 The correct answer is choices (a) and (d). Activation of the APC triggers anaphase, which must occur prior to the re-formation of the nuclear envelope in telophase. The contractile ring in animals begins to assemble in anaphase. The chromosomes do not decondense and the lamina cannot be assembled until the formation of the nuclear envelope is complete and nuclear proteins (including most of the lamins) have been imported through the pores. Transcription does not begin until the chromosomes decondense. 19-21 Choice (b) is the correct answer. A Ser to Ala mutation in the phosphorylation site of a lamin will prevent the lamin from being phosphorylated. Since the nonphosphorylated form is the form that polymerizes into a lattice, such a mutation will stabilize the nuclear lamina and prevent its dissasembly and nuclear envelope breakdown at prometaphase. Telophase involves repolymerization of lamins and therefore will not be blocked by the mutation (choice (a)). Lamins are not part of the nuclear pore complex (choice (c)). The mitotic spindle begins to form before the nucleus has broken down, forming a sort of cage around the nucleus (choice (d)). Lamins are not involved in chromosome condensation (choice (e)). 19-22 The correct answer is choices (b) and (g). The ER and the nuclear membrane (which is continuous with the ER) both fragment at mitosis. Mitochondria, lysosomes, ribosomes, and chloroplasts are all small and numerous organelles and do not need to be broken into small pieces to ensure roughly equal segregation to both daughter cells. Centrosomes do not fragment: they duplicate in an orderly and elaborate way that is poorly understood.
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19-23 In a cell with excessive catastrophin molecules, the balance between catastrophins and MAPs will be disrupted. Increased catastrophin activity relative to the MAP activity will lead to an increased frequency of microtubule catastrophes, the sudden shift from growth to shrinkage. Thus microtubules will be shorter on average because they spend less time growing slowly and more time shrinking rapidly. Shorter microtubules may be assembled into a shorter mitotic spindle. Conversely, increased MAP activity relative to catastrophin activity will stabilize microtubules by enhancing polymerization or inhibiting depolymerization, thereby promoting formation of longer microtubules and perhaps a longer mitotic spindle. However, in some cases the normal balance between catastrophins and MAPs has been shown to be necessary for formation of a bipolar spindle. It is possible that increasing the amount of MAP or catastrophin regulatory proteins will cause formation of microtubules that are too long or too short to be assembled into a spindle at all. 19-24 A. B.
C.
D.
p98 behaves like a MAP, or microtubule associated protein, because it promotes microtubule growth. The four possible mechanisms by which a protein can promote microtubule growth are (1) increasing the rate of polymerization, (2) decreasing the rate of depolymerization, (3) inhibiting the occurrence of catastrophes (when a microtubule abruptly shifts from growing to shrinking), and (4) stimulating the occurrence of rescues (when a microtubule switches from shrinking to growing). p98 increases the rate of microtubule growth, as seen by the steeper slopes of the lines when p98 was added. The data do not allow a conclusion about the rate of microtubule shrinkage. The protein decreases the rate of catastrophes: two of the five microtubules in the absence of p98 underwent catastrophe (a sudden and rapid decline in microtubule length), whereas none did so in the presence of p98. No rescues were observed, so it is uncertain if p98 alters the rescue rate. The p98 protein can be removed from a Xenopus egg mitotic extract using antibodies that specifically recognize p98. The depleted extract with little to no p98 protein can then be mixed with sperm nuclei, centrosomes, and fluorescent tubulin. Based on the previous experiments, the p98-depleted extract would be expected to have shorter and more dynamic microtubules. Ideally, this kind of experiment would be supplemented with the production and examination of intact cells in which p98 function has been abolished by mutation or another means.
19-25 Cells arrest in mitosis due to a checkpoint. Nocodozole treatment disassembles microtubules, preventing formation of a spindle. Because the nucleus breaks down and chromosomes condense, these events must be independent of aster formation by centrosomes, for example, and of any other microtubule-dependent process. Mitosis finally arrests because the unattached chromosomes trigger a signal that engages the spindle-attachment checkpoint, which halts the cycle.
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19-26 The correct answer is choices (a), (c), and (d). The cleavage furrow begins to form before the end of telophase, thus (b) is incorrect. Although the furrow eventually becomes independent of the mitotic spindle, the furrow requires the mitotic spindle for its establishment. The furrow always forms perpendicular to the interpolar microtubules about midway between the poles, but since the spindle is normally displaced in some cells, cell division does not always occur in the middle of the cell, thus (e) is incorrect. 19-27 The correct answer is choices (a) and (d). All cell movement requires ATP, and in cytokinesis actin and myosin molecules are moving relative to one another to cause contraction of the contractile ring. The myosin is an ATPase that hydrolyzes ATP to power this movement. The assembly of the contractile ring requires a general rearrangement of the filaments in the cell cortex. The contractile ring completely disassembles after mitosis (choice (b)). Phosphorylation of integrins, which weakens their hold on the extracellular matrix and allows cells to round up, generally precedes cytokinesis and is part of the general rearrangement of cell structure (including the cell cortex) that accompanies cell division (choice (c)). Microtubules do not play an important role in animal cytokinesis (choice (e)).
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CHAPTER 19 GENETICS, MEIOSIS, AND THE MOLECULAR BASIS OF HEREDITY 2009 Garland Science Publishing 3rd Edition
The Benefits of Sex 19-1 Why is sexual reproduction more beneficial to a species living in an unpredictable environment than to one living in a constant environment? 19-2 Which of the following statements is TRUE? (a) Diploid organisms reproduce only sexually. (b) All sexually reproducing organisms have at least two copies of each gene at some stage in their life cycle. (c) Gametes have only one chromosome. (d) Another name for the unfertilized egg cell is the zygote. (e) Mutations in somatic cells are passed on to individuals of the next generation. 19-3 Imagine a diploid sexually reproducing organism, Diploidus sexualis, that contains three pairs of chromosomes. This organism is unusual in that no recombination between homologous chromosomes occurs during meiosis. A. Assuming that the chromosomes are distributed independently during meiosis, how many different types of sperm or egg cells can a single individual of this species produce? B. What is the likelihood that two siblings of this species will be genetically identical?
19-4 A certain type of worm has two genders: males that produce sperm and hermaphrodites that produce both sperm and eggs. The diploid adult has four homologous pairs of chromosomes that undergo very little recombination. Given a choice, the hermaphrodites prefer to mate with males, but just to annoy the worm, you pluck a hermaphrodite out of the wild and fertilize its eggs with its own sperm. Assuming all the resulting offspring are viable, what fraction do you expect to be genetically identical to the parent worm? Assume that each chromosome in the original hermaphrodite is genetically distinct from its homologue. (a) (b) (c) (d) (e)
All None Half 1/16 1/256 333
20-5
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each words or phrase may be used more than once. To reproduce sexually, an organism must create haploid __________________ cells, or __________________, from diploid cells via a specialized cell division called __________________. During mating, the father’s haploid cells, called __________________ in animals, fuse with the mother’s haploid cells, called __________________. Cell fusion produces a diploid cell called a __________________, which undergoes many rounds of cell division to create the entire body of the new individual. The cells produced from the initial fusion event include __________________ cells that form most of the tissues of the body as well as the __________________-line cells that give rise to the next generation of progeny. allele bivalent eggs gametes
germ meiosis mitosis pedigree
pollen somatic sperm zygote
20-6
Which of the following statements is FALSE? (a) Asexual reproduction typically gives rise to offspring that are genetically identical. (b) Germ-line cells have half the number of chromosomes as somatic cells. (c) Sexual reproduction allows for a wide variety of gene combinations. (d) A zygote can only be produced through asexual reproduction. (e) Gametes are specialized sex cells.
20-7
Is the following statement TRUE or FALSE? Explain. Somatic cells leave no progeny and thus, in an evolutionary sense, exist only to help create, sustain, and propagate the germ cells.
20-8
Sexual reproduction is a large drain on the limited resources of an individual. Nonetheless, sexual reproduction is common. In fact, to allow sexual reproduction, many elaborate anatomical structures, cellular processes, and chemical signals have evolved. For example, flowers exist entirely to further the goal of sexual reproduction and many plants have enlisted the help of bees and birds to aid in the dissemination of their germ cells. Describe one reason why the majority of multicellular organisms have evolved to reproduce sexually instead of relying solely on asexual reproduction.
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Meiosis 20-9
Indicate whether the following is TRUE for meiosis, mitosis, both, or neither. A. Formation of a bivalent B. Products are genetically identical C. Chromosomes condense D. All paternal chromosomes segregate to one cell E. Involves DNA replication
20-10 Meiosis is a highly specialized cell division in which several events occur in a precisely defined order. Please order the meiotic events listed below. 1. Loss of cohesins near centromeres 2. Chromatid pairing 3. Chromosome condensation 4. Chromosome replication 5. Degradation of cohesins bound to chromosome arms 6. Formation of chiasmata (chiasmata = plural of chiasma) 7. Homolog pairing 8. Alignment of chromosomes at the metaphase plate 20-11 You have received exactly half of your genetic material from your mother, who received exactly half of her genetic material from her mother (your grandmother). A. Explain why it is unlikely that you share exactly one quarter of your genetic material with your grandmother, and instead it is more accurate to say that in general people receive an average of one quarter of their genetic endowment from each grandparent. B. Consider a gene on chromosome 3 that you received from your grandmother. Is it likely you received an entire chromosome 3 from your grandmother? Why or why not? C. What portion of your genetic material do you share with your sibling? Your aunt? Your cousin?
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20-12 Figure Q20-12 is a diagram of chromosomes during meiosis.
Figure Q20-12 A. B.
On the diagram, indicate which label lines correspond to the following items: (1) sister chromatids, (2) homologous chromosomes, (3) bivalent, (4) chiasma. On the figure, draw as small circles (oooo) the cohesin “glue” that is released in meiotic division I, and draw as small crosses (xxxx) the cohesin glue that is released in meiotic division II.
20-13 For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement. Starting with a single diploid cell, mitosis produces [two/four] [identical/different] [haploid/diploid] cells, whereas meiosis yields [two/four] [identical/different] [haploid/diploid] cells. This is accomplished in meiosis because a single round of chromosome [replication/segregation] is followed by two sequential rounds of [replication/segregation]. Mitosis is more like meiosis [I/II] than meiosis [I/II]. In meiosis I, the kinetochores on sister chromatids behave [independently/coordinately] and thus attach to microtubules from the [same/opposite] spindle. The cohesin-mediated glue between [chromatids/homologs] is regulated differently near the centromeres than along the chromosome arms. Cohesion is lost first at the [centromeres/arms] to allow segregation of [chromatids/homologs] and is lost later at the [centromeres/arms] to trigger segregation of [chromatids/homologs].
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20-14 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. An organism that has five homologous pairs of chromosomes can produce, in the absence of recombination, __________________ genetically different types of gametes. Recombination will __________________ the number of genetically different gametes possible. 5 8 10
15 25 32
decrease increase
20-15 Meiosis includes a recombination checkpoint that is analogous to the checkpoints in cell cycle progression. Recombination in meiosis is initiated by double-stranded breaks in the DNA. The broken end of a DNA molecule finds the corresponding sequence on a homologous chromosome and exchanges a chromosomal segment with its homolog, thereby repairing the break. Ongoing recombination sends a negative regulatory signal that prevents cells from entering meiotic division I. A. Mutations in several genes inactivate the recombination checkpoint. What do you predict will happen if a cell proceeds through meiotic division I before completing recombination? B. What will happen if a cell fails to initiate recombination and proceeds through meiotic division I? Meiotic division II?
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20-16 In some fungi, cell division during meiosis gives rise to an ordered spore sac containing a row of four haploid spores, as shown in Figure Q20-16A. The position of each spore within the sac reflects its relation to its neighbors, i.e., spores that result from the same meiosis II division are positioned next to each other. You notice that a strain of the fungus produced by crossing a dark-colored strain with a light-colored strain gives rise mostly to spore sacs as shown in Figure Q20-16B, with a few spore sacs like those in Figure Q20-16C. Which of the following can be concluded from the above observations? Explain your reasoning.
Figure Q20-16 (a) (b) (c) (d) (e)
Meiosis I and meiosis II in your fungus occur in the reverse order from that which occurs in humans. Recombination in the fungus can occur during prophase I. Recombination in the fungus cannot occur during prophase II. The spore sacs in panel C result from recombination between the centromere and the gene responsible for spore color. Recombination has occurred between the gene responsible for spore color and the end of the chromosome arm.
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20-17 In mammals, there are two sex chromosomes, X and Y, which behave like homologous chromosomes during meiosis. Normal males have one X chromosome and one Y chromosome, and normal females have two X chromosomes. Males with an extra Y chromosome (XYY) occasionally are found. Which of the following could give rise to such an XYY male? Explain your answer. (a) Nondisjunction in the first meiotic division of spermatogenesis; normal meiosis in the mother. (b) Nondisjunction in the second meiotic division of spermatogenesis; normal meiosis in the mother. (c) Nondisjunction in the first meiotic division of oogenesis; normal meiosis in the father. (d) Nondisjunction in the second meiotic division of oogenesis; normal meiosis in the father. (e) Nondisjunction in the first meiotic division of both the mother and father’s gametes.
Mendel and the Laws of Inheritance 20-18 Is the following statement TRUE or FALSE? Explain. The phenotype of an organism reflects all of the alleles carried by that individual. 20-19 Do you agree or disagree with the following statement? Explain your answer. If a diploid organism has 16 chromosomes (and thus 8 pairs of homologous chromosomes), that organism can produce only 28 genetically different gametes. 20-20 With respect to gene E on the chromosome drawn in Figure Q20-20, which gene is least likely to behave according to Mendel’s law of independent assortment? Explain your answer.
Figure 20.20
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20-21 Cystic fibrosis is due to mutations in a single gene that lies on chromosome 7. Only homozygous mutant (ff) individuals are sick; homozygous wild-type (FF) and heterozygous(Ff) individuals are healthy. A healthy married couple has one child with cystic fibrosis and the wife is pregnant with a second child. A. What is the genotype of the mother? The father? B. What is the chance that the second child will have cystic fibrosis? 20-22 Sickle cell anemia is caused by a mutation in the hemoglobin allele. Individuals with two mutant alleles have sickle cell anemia. Individuals homozygous and heterozygous for the mutant gene are more resistant to malaria than those with two wild-type alleles. Is this mutation dominant, recessive, or co-dominant?
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20-23 Your friend has obtained some pea seeds from the Abbey of St. Thomas in Brno, where Gregor Mendel worked. He is very excited because not only did he obtain some yellow and green pea seeds from true-breeding plants (like the ones used in the famous experiment described in your book), he was also able to obtain some purple pea seeds from a true-breeding plant. First, your friend takes the true-breeding yellow and green pea seeds, repeats the cross that Gregor Mendel did, and obtains the same results where he sees 100% yellow-seeded pea plants in the F1 generation and 75% yellow-seeded pea plants and 25% green-seeded pea plants in the F2 generation. His results are illustrated in Figure Q20-23A below. Your friend decides to set up two more crosses. First, for cross #2, he crosses the true-breeding purple-seeded pea plants to the true-breeding yellowseeded pea plants. The results from this cross are shown in Figure Q20-23B below. Next, for cross #3, he crosses the true-breeding purple-seeded pea plants to the truebreeding green-seeded pea plants. These results are shown in Figure Q20-23C.
Figure Q20-23 341
Given these results, if you were to take the purple-seeded pea plants produced in the F1 generation in cross #2 and cross them to the purple-seeded pea plants produced in the F1 generation of cross #3, what do you expect the phenotype of the progeny would look like? Explain your answer. 20-24 You are given two true-breeding strains of hamster. One strain has white fur color while the other strain has a dark brown fur color. When you cross the white fur strain to the dark brown fur strain, you obtain F1 progeny that have a light brown fur color. When you cross the F1 progeny with each other, you obtain 25% of the hamsters having white fur, 25% of the hamsters having dark brown fur, and 50% of the hamsters with light brown fur. How many genes crucial for fur coloration differ between the two starting strains? Explain your answer. Extra credit: Propose a molecular mechanism for how fur color is determined in this species of hamster.
Genetics as an Experimental Tool 20-25 Conditional alleles are mutant gene versions that encode proteins that can function normally at the permissive condition but are defective at the restrictive condition; one commonly used condition is temperature. Conditional alleles are especially useful to geneticists because they permit the study of essential genes. At the permissive temperature, the organism lives normally. When the organism is shifted to the nonpermissive temperature, the effect of inactivating the gene can be studied. Which of the three types of mutations shown in Figure Q20-25 is most likely to lead to a conditional allele? Explain your answer.
Figure Q20-25
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20-26 You performed a genetic screen in C. elegans and found two mutations (#1 and #2) that cause antisocial behavior. You want to learn if these are mutations in the same gene or different genes, so you cross mutant #1 and mutant #2 together to perform a complementation test. Imagine that the mutations are in different genes: mutation #1 is in gene M and mutation #2 in N.
Figure Q20-26 A.
B.
C. D.
What is the outcome of a complementation test if #1 and #2 are both recessive mutations (m and n, respectively)? Fill in the empty oval in Figure Q20-28A and indicate if the offspring exhibits social or antisocial behavior. What is the outcome if #1 is dominant and #2 is recessive (M* and n)? Fill in the empty oval in Figure Q20-28B and indicate if the offspring exhibits social or antisocial behavior. For the case in A., would the phenotype of the offspring be different if the two mutations were in the same gene? For the case in B., would the phenotype of the offspring be different if the two mutations were in the same gene? Are complementation tests useful for analyzing dominant mutations?
20-27 Do you agree or disagree with the following statement? Explain your answer. A trait that is found at a low frequency in the population has to be a recessive trait.
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20-28 You are studying a diploid yeast strain that normally utilizes glucose as an energy source but can use maltose when no glucose is present. You are interested in understanding how this yeast strain utilizes maltose as an alternative energy source. To begin to understand maltose metabolism, you undertake a genetic screen to isolate genes involved in maltose metabolism by screening for yeast that cannot grow when maltose is the sole energy source. You isolate 6 different mutants, all of which are recessive, and name these alleles mal1, mal2, mal3, mal4, mal5, and mal6. Next, you isolate gametes from the homozygous diploid mutant yeast strains and perform crosses between the different strains to do complementation analysis, because you wish to determine whether the mutations are likely to affect the same or different genes. Your results are shown in the table below: Table Q20-28. Complementation analysis of mal genes. mutation mal1 mal2 mal3 mal4 mal5 mal6 mal1 – – + + + – mal2 – – + + + – mal3 + + – + – + mal4 + + + – + + mal5 + + – + – + mal6 – – + + + – “+” = growth on maltose-containing media; “–” = no growth on maltose-containing media. In how many genes are you likely to have isolated mutations? Which alleles appear to affect the same genes? Explain your answer. 20-29 Gene A is located near gene B on chromosome 13 in humans. A mutation in the germ line of an individual with the haplotype AB generates gametes with the genotype Ab. Many descendents of this founder individual carry the b mutation, which predisposes carriers to high blood pressure. Initially, all descendents who inherit the b mutation also inherit the neighboring A allele. Through the generations, fewer and fewer descendents with the b mutation carry the A allele, and instead they have the a allele. (Individuals with A and a are equally healthy and fit.) Explain how the b and A alleles are separated.
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20-30 Shown in Figure Q20-30 is a genetic pedigree of a family with several members affected by a heritable disease. Affected individuals are shown in black and healthy individuals are shown in white. Males are shown as boxes and females as circles. Can a single mutation explain the pattern of inheritance? Is the mutation responsible for the disease dominant or recessive? Is the mutation carried on the X chromosome, the Y chromosome, or an autosome?
Figure Q20-30
How We Know: Reading Genetic Linkage Maps 20-31 You are trying to map a human gene thought to be involved in cat allergies. Because you know this gene is on chromosome 20, you decide to examine the linkage of several SNPs located on chromosome 20 with respect to the gene involved in cat allergies. You have obtained DNA from 10 individuals and know whether they are allergic to cats. Your SNP results are shown in Table Q20-31. Table Q20-31. SNP tests for chromosome 20. (+ = presence of SNP) allergic to cats? yes no no yes yes yes no yes yes no A. B.
SNP1
SNP2
SNP3
SNP4
SNP5
– – – + + – + + – +
+ + – + – – + + – –
+ – – + + + – + + –
+ – – – + + – + + –
– + + – + – + – + –
Which SNP is most likely to be tightly linked to the gene involved in cat allergies? Explain your answer. Of the SNPs tested above, which is likely to be the next closest to the gene responsible for the allergic state? Why?
20-32 Any two human beings typically have an estimated 0.1% difference in their nucleotide sequences, which is equivalent to about 3 million nucleotide differences. These differences are the basis of the SNPs used to construct genetic linkage maps. Some of these SNPs actually lie in the region of the DNA that codes for the protein yet have no effect on the phenotype of individuals carrying the SNP on both homologous chromosomes. Explain how some SNPs can lie within the portion of the DNA that codes for the protein and yet have no discernable effect on the protein’s activity. 345
Answers 20-1
The real benefit in sexual reproduction seems to be that parents produce children that are genetically unlike either parent and that are not genetically identical to each other. This provides more variation in the population than asexual reproduction could provide, and is an advantage if the environment is variable, since any one combination of the parents’ characteristics, however well adapted to the prevailing conditions, may or may not be the best in a new situation.
20-2
Choice (b) is the correct answer. The sexual life cycle involves an alternation of diploid and haploid phases; and these haploid cells combine by the process of mating or fertilization to produce a new diploid. Some diploid organisms (e.g., many plants) are capable of asexual reproduction (choice (a)). Gametes have only one member of each pair of homologous chromosomes, but since most organisms have more than one pair of homologous chromosomes, most gametes have more than one chromosome (choice (c)). Choice (d) is untrue because the zygote is formed by fusion of the egg and sperm. Choice (e) is untrue because only mutations in germ cells are passed on to progeny individuals.
20-3
A.
B.
Eight different types. With respect to each of the three chromosomes, an individual can produce two kinds of gametes. The gamete can receive the copy that the individual received from “mom” or the copy from “dad.” Thus, with three chromosomes there are 2 × 2 × 2 = 8 possible gametes. 1/64. The mother and father together can produce 8 × 8 = 64 different genetic combinations.
20-4
(d)
The parent is most likely heterozygous for each chromosome, and so can produce 24 = 16 types of eggs and 24 = 16 types of sperm. Any of the eggs produced will be able to give rise to an adult that is identical to the parent, but in order to do so, it must be fertilized by the right type of sperm. For each type of egg, only one of the 16 possible sperm will produce a diploid that is identical to the parent. Therefore, 1 out of 16 of the offspring should be identical to the parent. In other words, a sexually reproducing organism with several chromosomes has a relatively high probability of producing genetically distinct offspring even when the parent mates to itself.
20-5
To reproduce sexually, an organism must create haploid germ cells, or gametes, from diploid cells via a specialized cell division called meiosis. During mating, the father’s haploid cells, called sperm in animals, fuse with the mother’s haploid cells, called eggs. Cell fusion produces a diploid cell called a zygote, which undergoes many rounds of cell division to create the entire body of the new individual. The cells produced from the initial fusion event include somatic cells that form most of the tissues of the body as well as the germ-line cells that give rise to the next generation of progeny.
20-6
(d)
The fusion of two haploid gametes produces a zygote during sexual reproduction.
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20-7
True. Only germ cells contribute genetic material to the next generation of organisms, and thus only germ cells leave an evolutionary legacy in the gene pool of the species. The only contribution of somatic cells to subsequent generations arises through the assistance they provide to dissemination of the genetic material in the germ cells. In accordance with this, mutations that arise in somatic cells are not passed along to offspring.
20-8
A definitive explanation of the evolutionary advantages of sexual reproduction over asexual reproduction is elusive, but several benefits seem clear. (1) The reshuffling of genes that occurs during sexual reproduction generates most of the diversity among organisms within a species. Having a large variety of different genetic combinations in a population may help guarantee that at least a few individuals will survive after a sudden and unpredictable change in the environment. This is why some ecologists are concerned about the global trend toward agricultural monoculture, i.e., limiting planting variety to only a small number of genetically identical subspecies of plants. (2) The competition among individuals in a population for mating partners may increase the frequency of advantageous or more highly functional genes in the gene pool. This is because individuals will choose mating partners who are stronger, healthier, and more successful than their peers. (3) Sexual reproduction may help to eliminate deleterious recessive genes from a population. A recessive deleterious mutation in a gene may be “unmasked” in a haploid germ cell because, unlike the situation in a heterozygous diploid cell, a normal allele would not be present to provide the gene function. Because there is competition among germ cells for fertilizing partner cells, gametes with recessive deleterious mutations are less likely to form zygotes.
20-9
A. B. C. D. E.
meiosis mitosis both neither both
20-10 4, 2, 7, 6, 3, 8, 5, 1. Chromosome replication (4) occurs during meiotic S phase, and immediately after replication the resultant sister chromatids are paired tightly by cohesins (2). During meiotic G2, or prophase, the two homologous chromosomes pair (7) and undergo recombination to produce reciprocal exchanges of genetic material that are visible as chiasmata (6). Meiotic division I begins with chromosome condensation (3) and proceeds to metaphase, when chromosomes align at a central plane (8). Anaphase of meiotic division I is triggered by loss of the cohesin glue on the chromosome arms (5), which allows the homologs to be segregated. Anaphase of meiotic division II is triggered by degradation of cohesins near the centromere (1), which allows sister chromatids to be segregated.
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20-11 A.
B.
C.
20-12 A. B.
It is unlikely that you share exactly one quarter of your genetic material with your grandmother, but it is true that organisms receive an average of one quarter of their genes from each grandparent. This is because when cells in your mother’s germ line were undergoing meiosis, the chromosomes that she received from your grandmother and your grandfather were shuffled by recombination and then randomly assorted into the gametes. Each of these gametes probably received slightly less or slightly more than half of its genetic material from your grandmother. Thus, fusion of your mother’s gamete with your father’s created a zygote that shared approximately, but not exactly, one quarter of its genes with your grandmother; this zygote divided repeatedly to form all the cells in your body. It is unlikely that you received the entire chromosome 3 from your grandmother, because each chromosome undergoes at least one recombinational crossover with its homolog to ensure proper chromosome segregation in meiotic division I. You share an average of one half of your genetic material with your sibling, one quarter with your aunt, and one-eighth with your cousin. Since you and your sibling each get 1/2 of each parent’s DNA, you have approximately 1/2 of your genetic material in common with your sibling. As your mother and her sister share approximately 1/2 of their genetic material, you and your aunt share 1/2 × 1/2 = 1/4. As your cousin has 1/2 of your aunt’s genetic material, you share 1/2 × 1/4 = 1/8. See Figure A20-12. (1) a, (2) c, (3) b, (4) d. See Figure A20-12.
Figure A20-12
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20-13 Starting with a single diploid cell, mitosis produces two identical diploid cells, whereas meiosis yields four different haploid cells. This is accomplished in meiosis because a single round of chromosome replication is followed by two sequential rounds of segregation. Mitosis is more like meiosis II than meiosis I. In meiosis I, the kinetochores on sister chromatids behave coordinately and thus attach to microtubules from the same spindle. The cohesin-mediated glue between chromatids is regulated differently near the centromeres than along the chromosome arms. Cohesion is lost first at the arms to allow segregation of homologs and is lost later at the centromeres to trigger segregation of chromatids. 20-14 32; increase. Since homologous chromosomes assort randomly at meiosis and a gamete has two choices for each chromosome (because sexual organisms are diploid), there are 5 2 , or 32, possible genetically different gametes. Recombination effectively increases the number of types of chromosomes and therefore increases the number of genetically different gametes. 20-15 A.
B.
If a cell proceeds through meiotic division I even though its chromosomes are broken and incompletely repaired, segregation will be disastrous. Some recombination intermediates will be pulled to opposite spindle poles, thus breaking the DNA. Other chromosome fragments lacking centromeres will not be attached to microtubules and thus will segregate randomly, causing some meiotic products to have too little DNA and others to have too much. In the absence of recombination, the homologs will not segregate from each other properly in meiotic division I but the sister chromatids will segregate normally in meiotic division II. In meiotic division I, the unrecombined homologous chromosomes will not be held together by chiasmata. Thus, the homologous chromosomes will line up independently on the metaphase plate and segregate randomly, causing chromosome nondisjunction. Some products will have both homologs of a given chromosome and others will have none. In meiotic division II, the events will proceed normally and sister chromatids will be properly segregated to opposite poles. Nonetheless, because the chromosome sets at the start of meiotic division II were unevenly distributed, the gametes produced after meiotic division II will be aneuploid (i.e., will have an incorrect number of chromosomes).
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20-16 (b) and (d) In normal meiosis, the chromosomes are distributed as shown in Figure A2016A (the dark-colored chromosome carries the allele for dark pigment). The homologs are separated from each other during meiotic division I and the sister chromatids are separated during meiotic division II. This pattern of chromosome segregation, in the absence of recombination, always gives the pattern of spores that was shown in the question (Figure A20-16B). To yield the rare oddball spore sacs, there must be recombination between the two homologous chromosomes. Such recombination always occurs when both homologs are in the same cell (i.e., in prophase I of meiosis).
Figure A20-16 (a)
(b)
(c)
Incorrect. If meiosis I and meiosis II were reversed, we would expect a completely random pattern of spores in spore sacs, since the two homologs could go to either cell in the second division, as shown in Figure Q20-16C. This cannot explain the observation that most of the spore sacs appear like Figure Q20-16B. Correct. The oddball spore sacs must result from recombination between the homologs carrying different alleles of the color gene. Recombination between the homologs occurs during prophase I. Incorrect. The results shown cannot distinguish whether recombination has occurred during prophase II. During prophase II, there is only one homolog in each cell; if recombination took place between the two sister chromatids of this homolog we would not be able to detect it, as the sister chromatids are identical (except for recombinational exchanges).
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(d)
(e)
Correct. Segregation of chromosomes is mediated by microtubules that pull on the centromeres of chromosomes. Thus, oddball segregation patterns arise from recombination (and thus loss of a covalent connection) between the centromere and the gene of interest. Incorrect. If recombination had taken place only between the gene for spore color and the end of that chromosome arm, it would look as if no recombination had taken place (Figure Q20-16B) because the centromere would still be covalently connected to the gene of interest. See also the answer to option (d).
20-17 Choice (b) is the correct answer. Nondisjunction in one or the other meiotic division will give rise to gametes with two sex chromosomes instead of one. Since the only source of Y chromosomes is the father, the gametes that produce an XYY male must be X (a normal egg) and YY (an abnormal sperm). Nondisjunction in the second meiotic division of spermatogenesis could give rise to a YY gamete that fertilizes a normal egg and thereby creates an individual with the karyotype XYY (see Figure A20-17). Choice (a) is incorrect because nondisjunction in the first meiotic division of spermatogenesis would result in two XY sperm and two sperm with no sex chromosome. Choices (c), (d), and (e) are incorrect because nondisjunction during oogenesis cannot yield an individual with two Y chromosomes.
Figure A20-17 20-18 False. The phenotype, or the observable traits, of an organism often does not fully reflect its genotype, or the catalog of all alleles in the chromosomes. This is because some alleles are dominant (call these A) and other alleles are recessive (a). An individual who is heterozygous (Aa) for a dominant allele will have the same phenotype as one who is homozygous (AA). 20-19 Disagree. There are two distinct mechanisms to get genetic variation in gametes. One method involves the reassortment of chromosomes during meiosis. This method would result in a diploid organism with 16 chromosomes producing 28 genetically different gametes. However, a much greater number of genetically different gametes can be produced due to the recombination between chromosomes that occur during every meiosis.
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20-20 Gene F is least likely to behave according to Mendel’s law of independent assortment because it lies the closest to gene E and is least likely to be separated from gene E by recombination. 20-21 A. B.
The genotypes of the mother and father are the same: Ff. The only way that a child can have the disease is if both parents are carriers of the cystic fibrosis gene. The chance that the second child will have cystic fibrosis is one quarter. The chance that the mother will transmit her mutant f allele to the offspring is one half, multiplied by an equal chance that the father will transmit his mutant f allele to the offspring: 1/2 × 1/2 = 1/4. (The existence of a first child with cystic fibrosis does not change the probability of another child having the disease.)
20-22 The classification of all mutations depends on the phenotype under consideration. With regard to the sickle cell anemia phenotype, the mutant allele is recessive because a heterozygous individual has the same healthy phenotype as a homozygous wild-type individual. With regard to the malaria phenotype, the mutant allele is dominant. This is because a heterozygous individual has the same phenotype as a homozygous mutant individual, namely resistance to malaria. 20-23 You would expect the progeny to be 75% purple-seeded plants and 25% yellow seeded plants. The crosses are diagramed below, where we designate the purple-seeded causing allele as P, the yellow-seeded causing allele as Y, and the green-seeded causing allele as G. Note that P, Y, and G are all alleles of the same gene. Table A20-23A: Cross #1 F2 results (F1 = YG × YG) allele Y G
Y YY (yellow) YG (yellow)
G YG (yellow) GG (green)
Table A20-23B: Cross #2 F2 results (F1 = YP × YP) allele Y P
Y YY (yellow) YP (purple)
P YP (purple) PP (purple)
Table A20-23C: Cross #3 F2 results (F1 = GP × GP) allele G P
G GG (green) GP (purple)
P GP (purple) PP (purple)
Table A20-23D: Your cross (or Cross #2 F1 × Cross #3 F1 = YP × GP) allele G P
Y GY (yellow) YP (purple) 352
P GP (purple) PP (purple)
20-24 One gene involved in determining fur color differs between these two strains. The white allele is not fully recessive to the dark brown allele. Therefore, any animal that is heterozygous (and has one white allele and one dark brown allele) will be light brown in fur color. Thus the white allele and the dark brown allele are co-dominant. When two heterozygous animals are crossed together, 25% will be homozygous for one allele, 25% will be homozygous for the other allele, and 50% will heterozygous, exactly what is observed in the cross you performed. Extra credit answer: Any reasonable answer is fine. There are many possible answers, including the two below: 1. The dark brown and white alleles are alleles of a gene involved in synthesizing pigment. Hamsters with the white allele do not produce any pigment; hamsters with the dark brown allele produce pigment. When hamsters are heterozygous, they have one chromosome that carries the white allele and does not produce any pigment. Therefore, they produce half the amount of pigment needed normally and thus are light brown instead of dark brown. 2. The dark brown and white alleles are alleles of a gene involved in a microtubule motor important for transporting the pigment to the correct place in the cell. Hamsters with the white allele produce a defective version of the microtubule motor and thus no pigment is properly transported, leading to white fur. The hamsters with the dark brown allele produce a properly functioning microtubule motor. When hamsters are heterozygous, they have one chromosome that carries the white allele and one chromosome that carries the dark brown allele; therefore, only half the amount of functioning microtubule motor is being produced in the heterozygote. Because the amount of microtubule motors is limiting in the cell, half the amount of properly functioning microtubule motor can only transport half the amount of pigment, leading to a light brown fur. 20-25 A single nucleotide substitution is most likely to cause a conditional allele. Nucleotide addition and nucleotide deletion usually will cause a shift in the reading frame of the protein, leading to a string of many amino acid substitutions (and an altogether different protein) and possibly even the formation of a stop codon, which would result in the premature truncation of the protein. In contrast, a single nucleotide substitution would be more likely to result in an amino acid substitution that would disrupt protein function in a conditional manner. For example, this amino acid substitution may affect protein folding at high temperature due to a greater charge; at the permissive temperature, this protein may still fold properly and function normally.
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20-26 A. B.
Social. See Figure A20-26A. Antisocial. See Figure A20-26B.
Figure A20-26 C.
D.
Yes, different. If the two mutations were in the same gene, it is probable that the offspring would have an antisocial phenotype, unlike the case in panel (A). Such a cross is overwhelmingly likely to produce an antisocial offspring, because two mutations in the same gene are unlikely to complement one another. However, there are rare cases in which two different mutations in the same gene can complement one another. This can be observed if, for example, a gene encodes a protein with two functions and each mutation impairs a different function. No, not different. If the dominant and recessive mutations were in the same gene, the phenotype of the offspring would still be antisocial. Thus, complementation tests do not work for dominant mutations: the phenotype of the offspring when the two mutations are in the same gene is the same as the phenotype of the offspring when the mutations are in different genes. The phenotype of the offspring will always be mutant, because a single M* allele is sufficient for the mutant phenotype.
20-27 Disagree. The frequency of a trait in a population has nothing to do with whether this trait is dominant or recessive. In order to test dominance or recessiveness, the segregation of this trait must be observed. For example, the defective version of the gene involved in Huntington’s disease occurs at a relatively low frequency in the population, but behaves in a dominant fashion to cause disease.
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20-28 Three genes: mal1, mal2, and mal6 are mutations in one gene; mal3 and mal5 are mutations in a second gene; mal4 is the only mutation you identified in a third gene. These results can be deduced from the complementation analysis. For example, the diploid yeast containing the mal1 and mal6 alleles cannot grow on maltose-containing media, and therefore these alleles are likely to be alleles of the same gene. However, the diploid yeast containing the mal1 and the mal4 alleles do grow on maltose-containing media, and these alleles are, therefore, likely to be alleles of different genes. 20-29 Eventually, the b mutation will be separated from the A allele by meiotic recombination. Meiotic recombination exchanges portions of homologous chromosomes, and thereby generates great diversity among the gametes of each individual. The locations of the 1–5 exchanges per chromosome during meiosis in humans are more or less random. Thus, each passing generation increases the cumulative likelihood of a recombinational crossover between any two neighboring genes (A and b). Such a recombinational crossover in an individual heterozygous for both genes (Ab and aB) will separate two alleles that were originally linked (yielding AB and ab). 20-30 Yes, a single mutation can explain the pattern. The mutation responsible for the disease is recessive, since two unaffected parents can produce affected offspring. The mutation is likely carried on the X chromosome, since males (having only one X) are more likely to have the disease than their sisters (having two X chromosomes). If the mutation were on an autosome, male and female offspring would have equal probability of having the disease. If the mutation were located on the Y chromosome, all males in the family would be affected. 20-31 A.
B.
SNP3 is most likely to be tightly linked to the gene involved in cat allergies because it is found in all people who are allergic to cats and none of the people who aren’t allergic to cats. It is possible that this SNP is responsible for the allergic response, but the proof of this would require additional studies. SNP4 is probably the next closest to the gene involved in cat allergies because it is found in all but one person who is allergic to cats and none of the people who aren’t allergic to cats. SNP1, SNP2, and SNP5 seem to segregate randomly in both people allergic and not allergic to cats and therefore are probably more distant from the gene involved in cat allergies.
20-32 Because of the redundancy of the genetic code, in which more than one codon can code for the same amino acid, some single nucleotide changes may not cause changes in the amino acid coded for by that codon. Furthermore, some amino acid substitutions are neutral (e.g., substitution of a small uncharged amino acid for another amino acid with the same properties) and thus do not change the function of the protein even though the amino acid has been changed.
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CHAPTER 20 TISSUES AND CANCER 2009 Garland Science Publishing 3rd Edition
Extracellular Matrix and Connective Tissues 20-1 Both multicellular plants and animals have (a) cells capable of locomotion. (b) cells with cell walls. (c) vascular systems containing circulating cells. (d) a cytoskeleton composed of actin filaments, microtubules and intermediate filaments. (e) tissues composed of multiple different cell types.
20-2 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Plants are sedentary and thus their cells have different needs than cells found in motile animals. For example, in plant cells, __________________ generates the turgor pressure that drives cell growth. Plants have cell walls, but cell growth is possible in the developing tissue because the __________________ cell walls are expandable. The plant vascular system is composed of two types of complex tissues. The __________________ is important for the transport of organic solutes in the plant. The __________________ can include cell walls that have been strengthened by the deposition of a __________________ network, which serves to mechanically reinforce the matrix and makes the tissue waterproof. The outer protective covering of the plant, also known as the dermal tissue, contains __________________, which regulate gas exchange. cellulose lignin secondary collagen membranous stomata epidermis osmosis translocation glycine phloem xylem ground tissue primary
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20-3 A. Indicate the direction in which the plant cell shown in Figure Q20-3 is most likely to grow.
Figure Q 20-3 B. Explain your answer.
20-4 Indicate whether the following molecules are found in plants, animals, or both. A. intermediate filaments B. cell walls C. microtubules D. cellulose E. collagen 20-5 A major distinction between the connective tissues in an animal and other main tissue types such as epithelia, nervous tissue, or muscle is (a) the ability of connective tissue cells such as fibroblasts to change shape. (b) the abundant extracellular matrix in connective tissues. (c) that connective tissues can withstand mechanical stresses. (d) the exceptional amounts of polysaccharides in the extracellular matrix of connective tissues. (e) the numerous connections connective tissue cells make with each other.
20-6 What are the main structures providing tensile strength in A. animal connective tissue? B. animal epithelium? C. plant cell walls?
20-7 Do you agree or disagree with the following statement? Explain your answer. Like many other extracellular proteins, newly synthesized collagen molecules undergo post-translational processing inside the cell to convert them into their mature form, and are then secreted and self-assemble into fibrils in the extracellular space.
20-8 Fibroblasts organize the collagen of the extracellular matrix by (a) cutting and rejoining the fibrils. (b) crawling along existing fibrils and adjusting the packing of collagen molecules within them. (c) twisting fibrils together to make ropelike fibers. (d) pulling the collagen into sheets or cables after it has been secreted. (e) controlling the local Ca2+ concentration so as to regulate collagen assembly.
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20-9 Proteoglycans in the extracellular matrix of animal tissues (a) provide tensile strength. (b) form a dense compact matrix. (c) are linked to microtubules through the plasma membrane. (d) are polysaccharides composed of glucose subunits. (e) provide cartilage with the ability to resist compression.
20-10 Which of the following statement(s) are FALSE? (a) Proteoglycans can act as filters to regulate which molecules pass through the extracellular medium. (b) The negative charge associated with proteoglycans attracts cations, which cause water to be sucked into the extracellular matrix. (c) Proteoglycans are a major component of compact connective tissues, but relatively unimportant in watery tissues such as the jellylike substance in the interior of the eye. (d) Proteoglycans help tissues resist mechanical compression. (e) Glycosaminoglycans are components of proteoglycan.
Epithelial Sheets and Cell-Cell Junctions
20-11 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. __________________ join the intermediate filaments in one cell to those in the neighboring cell. __________________ anchor intermediate filaments in a cell to the extracellular matrix. __________________ involve cadherin connections between neighboring cells and are anchorage sites for actin filaments. __________________ allow for the exchange of small molecules from one cell to its adjacent cell. __________________ prevent the leakage of molecules between adjacent cells. adherens junctions gap junctions highway junctions desmosomes hemidesmosomes tight junctions
20-12 A basal lamina (a) is a thin layer of connective tissue cells and matrix underlying an epithelium. (b) is a thin layer of extracellular matrix underlying an epithelium. (c) is attached to the apical surface of an epithelium. (d) is impermeable to small organic molecules. (e) separates epithelial cells from each other.
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20-13 Label the five different types of cell-cell junctions shown in Figure Q20-13, and identify the apical and basal surfaces of the epithelium.
Figure Q20-13
20-14 Match the molecules (List 1) with the cell structures in which they play a part (List 2). A cell structure may be listed more or less than once. List 1 List 2 List 1 A. Cadherin B. Cellulose C. Collagen D. Connexons E.Occludins F. Claudins G. Keratins
Tissue Maintenance and Renewal
List 2 1. Desmosomes 2. Cell wall 3. Gap junctions 4. Nucleus 5. Basal lamina 6. Plasmodesmata 7. Tight junctions
20-15 Name the three key mechanisms important for maintaining the organization of cells into tissues. 20-16 Place the following in order of their replacement times from shortest to longest. A. Epidermal cell B. Nerve cell C. Bone matrix D. Red blood cell E. Cell lining the gut
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A stem cell divides into two daughter cells and one of the daughter cells goes on to 20-17 become a terminally differentiated cell. What is the typical fate of the other daughter cell?
20-18 Your friend is a pioneer in ES cell research. In her research, she uses an ES cell line that originated from an inbred strain of laboratory mice called FG426. She has just figured out methods that allow her to grow an entire liver from an ES cell and has successfully grown 10 livers. She demonstrates that the newly grown livers are functional by successfully transplanting one of the new livers into a FG426 laboratory mouse. You are particularly excited about this, because you have a sick pet mouse, Squeaky. You are very attached to Squeaky, as you found him when you were out camping in New Hampshire. Unfortunately, Squeaky has developed liver disease and will not live much longer without a liver transplant. After you see your friend on TV talking about her new method for growing mouse livers, you immediately grab your cell phone to ask her whether Squeaky could have one of the newly grown livers. Just as you are about to dial your friend, you remember something you learned in cell biology and realize that instead, you should ask your friend about possibly using therapeutic cloning for Squeaky’s benefit. A. Why do you think that one of the newly grown livers may not work in Squeaky? B. Explain why therapeutic cloning would solve this problem.
Cancer 20-19 A malignant tumor is more dangerous than a benign tumor because (a) its cells are proliferating faster. (b) it metastasizes by traveling through the digestive tract. (c) it causes neighboring cells to mutate. (d) its cells attack and phagocytose neighboring normal tissue cells. (e) its cells invade other tissues.
20-20 A certain mutation in the receptor for epidermal growth factor (EGF) causes the mutated receptor protein to send a positive signal along associated intracellular signaling pathway even when EGF is not bound to it. This leads to abnormal cell proliferation in the absence of growth factor. On the basis of this information, would you class the gene for the EGF receptor as a tumor-suppressor gene or a potential oncogene? Explain your answer.
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20-21 Ras is a GTP-binding protein that is often defective in cancer cells. Normal cells are often stimulated to divide by receiving a signal from a growth factor through a receptor tyrosine kinase. When the receptor tyrosine kinase binds the growth factor, Ras is stimulated to bind GTP. Ras in turn activates proteins that promote cell proliferation. A common mutation in cancerous cells causes Ras to behave as though it were bound to GTP all the time.
A. Why is this mutation advantageous to cancerous cells? B. Your friend decides that the signaling pathway involving the Ras protein is a good target for drug design, since the Ras protein is often defective in cancer cells. Your friend designs a drug that will turn off the receptor tyrosine kinase by preventing its ability to dimerize. Do you think that this will affect cells that have a defective Ras protein that acts as if it were always bound to GTP? Why or why not?
20-22 People who inherit one copy of the Rb (retinoblastoma) gene that is normal and one copy that is mutated—that is, people who are heterozygous for Rb—have a greatly increased risk of cancer. Given this information, do you agree or disagree with the following statement? Explain your answer. The Rb mutation must have a dominant effect, which means that it must result in an increase in Rb function. Thus, Rb in its mutant form must be an oncogene.
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20-23 Figure 20-23 shows a sequence of mutations that might underlie the development of colorectal cancer. Given what you have learned about the p53 protein in Chapter 18, explain why the loss of p53 is advantageous to cancerous cells.
Figure Q20-23
20-24 Drugs that block the function of oncogenic proteins hold great promise in the fight against cancer. Should cancer researchers also be attempting to design drugs that will interfere with the products of tumor suppressor genes like p53? Why or why not?
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How We Know: Making Sense of the Genes that Are Critical for Cancer 20-25 Your studies in cell biology have revealed how APC, a protein encoded by a tumor suppressor gene frequently inactivated in people with colorectal cancer, functions in the Wnt signaling pathway (see Figure Q20-25), and this has inspired you to do further research on Wnt signaling. You would like to design a drug to treat people with colorectal cancer. Given the pathway above and the knowledge that a majority of human colorectal tumors harbor mutations in the APC gene, name a protein from the pathway above that would be a good target for an activity-blocking drug. Explain your answer.
Figure Q20-25
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Answers
20-1 Choice (e) is the correct answer. Choice (d) is incorrect as plant cells do not have intermediate filaments. Plant cells are not capable of locomotion (choice (a)) and also do not have a vascular system with circulating cells (choice (c)). Only plant cells have cell walls (choice (b)).
20-2 Plants are sedentary and thus their cells have different needs than cells found in motile animals. For example, in plant cells, osmosis generates the turgor pressure that drives cell growth. Plants have cell walls, but cell growth is possible in the developing tissue because the primary cell walls are expandable. The plant vascular system is composed of two types of complex tissues. The phloem is important for the transport of organic solutes in the plant. The xylem can include cell walls that have been strengthened by the deposition of a lignin network, which serves to mechanically reinforce the matrix and makes the tissue waterproof. The outer protective covering of the plant, also known as the dermal tissue, contains stomata, which regulate gas exchange. 20-3 A. Vertically—see Figure A20-3. B. Cellulose fibers are highly resistant to stretching so a plant cell tends to grow, under the stimulus of turgor pressure, in a direction perpendicular to the orientation of the fibers in the cell wall.
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20-4 A. animals B. plants C. both D. plants E. animal
20-5 (b) Cells in many other types of tissue can change shape (e.g., muscle cells when they contract), and can withstand mechanical stress (e.g., cells of the epidermis). Connective-tissue cells tend to be scattered throughout extracellular matrix, and unlike, for example, nerve cells or epithelial cells, make few or no contacts with each other. 20-6 A. Collagen fibers B. Intermediate filaments C. Cellulose fibers
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20-7 Disagree. Newly synthesized molecules of collagen are secreted from the cell in an immature form as procollagen, and the peptides at the ends of the procollagen molecules then have to be cleaved off in the extracellular space before assembly of fibrils can occur. This ensures that collagen fibrils will not assemble prematurely inside of the cell.
20-8 (d) 20-9 (e) 20-10 (c) Proteoglycans are MORE important in the interior of the eye, where collagen is very scarce, than in compact connective tissues such as bone and tendon, whose structure is dominated by collagen. 20-11 Desmosomes join the intermediate filaments in one cell to those in the neighboring cell. Hemidesmosomes anchor intermediate filaments in a cell to the extracellular matrix. Adherens junctions involve cadherin connections between neighboring cells and are anchorage sites for actin filaments. Gap junctions allow for the exchange of small molecules from one cell to its adjacent cell. Tight junctions prevent the leakage of molecules between adjacent cells. 20-12 (b) 20-13 1, apical surface; 2, tight junction; 3, adherens junction; 4, desmosome junction; 5, gap junction; 6, hemidesmosome junction; 7, basal surface. 20-14 A—1; B—2; C—5; D—3; E—7; F—7; G—1 20-15 1. Cell communication 2. Selective cell-cell adhesion 3. Cell memory
20-16 Cell lining the gut (few days) < epidermal cell (1 or 2 months) < red blood cell (4 months) < bone matrix (10 years) < nerve cell (lifetime). 20-17 The other daughter cell typically remains a stem cell. 20-18 A. For organ transplantation to be successful, the donor and the recipient should be as close a genetic match as possible to minimize the risk of immunological rejection. Because you found Squeaky in the fields of New Hampshire, Squeaky is likely to have genetic differences with the FG426 inbred laboratory mice. B. Therapeutic cloning may help save Squeaky because it is a method for generating ES cells from Squeaky himself—cells that will be genetically identical to Squeaky. Hopefully, your friend will be successful when applying her methods to the ES cell line derived from Squeaky and can grow a functional liver for him.
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20-20 The mutation described leads toward cancerous cell behavior (excessive proliferation) by making the gene product hyperactive. The mutant gene is therefore, by definition, an oncogene. This effect is seen even if only one copy of the gene is affected; in other words, the mutation is dominant, as is typical for an oncogene. Mutations that delete an EGF receptor gene would be expected to have either no effect or an inhibiting effect on cell division. Thus the normal EGF receptor is classed as a potential oncogene (a proto- oncogene).
20-21 A. A Ras mutation that causes Ras to behave as though it is bound to GTP all the time is advantageous to cancer cells because Ras is then activated and turns on the activities of proteins required for cell proliferation. If the cell proliferation proteins are always turned on, the cancer cell will be able to proliferate at an unregulated rate, outgrowing its normal neighbors. The ability to proliferate in a signal-independent fashion is one of the hallmarks of a cancer cell. B. Unfortunately, a drug that blocks activation of the receptor that activates Ras will be unlikely to have a useful effect on a cell containing mutant Ras protein that acts as though it is always bound to GTP. A mutant Ras protein of this type behaves as though it is constantly activated. Because Ras acts downstream of the receptor, the activating mutation makes its effect felt, regardless of the state of the receptor on which Ras activation would normally depend: mutant Ras that is always active is no longer dependent on the receptor for activation. Therefore, blocking the ability of the receptor to dimerize and activate Ras will probably not have an effect on cells containing the mutant Ras protein.
20-22 Disagree. It is true that the Rb mutation is dominant, in the sense that a person who is heterozygous (inherits one normal copy and one mutant copy of the gene) is likely to show the mutant trait (that is, will be cancer-prone). However, this does not mean that the mutation in Rb has to be causing an increase in Rb gene function; in fact, the opposite is true—the propensity for cancer arises from a loss of Rb gene function. Therefore, Rb should be classified as a tumor suppressor gene and not as an oncogene. Most people have two functional Rb genes in each of their cells. Thus, for one of their cells to turn cancerous by losing Rb function, both copies of the gene in that cell must be inactivated or lost, a two-step process. However, in a person born lacking one copy of the Rb gene, each cell is only one step away from complete loss of Rb function. Consequently, such a person has a high risk that at least one of the cells in the body will undergo a mutation that precipitates cancer. In this way, at the level of the whole person, the Rb loss-offunction mutation is dominant, even though at the level of the individual cell it is recessive.
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p53 is a protein that is activated and stabilized when DNA is damaged. When p53 is 20-23 active, it stops the cell cycle in order to give the cell time to repair its damaged DNA, or, if that is not feasible, it causes the cell to commit suicide by apoptosis. Cells lacking p53 will continue to replicate their DNA, will avoid suicide, and will go through cell division even when the DNA has been damaged, producing mutant daughter cells. Repeated rounds of cell division under these circumstances perpetuate the genetic damage and allow still more mutations to occur. Some of these newly accumulated mutations will give the cell an increased ability to survive, proliferate, and metastasize, resulting in invasive cancer.
20-24 Oncogenic proteins lead toward cancer, because they have an excessive or unregulated activity as compared with the corresponding normal proteins. By blocking this activity with a drug molecule that simply clogs the active site of the oncogenic protein, one can remove the danger. For a tumor suppressor gene, the danger lies in a loss of function, and there is generally no simple way for a drug molecule to restore a protein function that has been lost. It is hard to see, therefore, how one could achieve any useful effect on cancer by means of drugs that interfere with tumor suppressor gene products. A drug that simply inhibited their function would be expected to promote, not cure, cancer. 20-25 There are two possible answers; β -catenin and TCF. Since most colorectal cancers consist of cells that lack APC, drugs that interfere directly with the Wnt signal protein, or with its receptor in the cell membrane, or with the intracellular signaling protein that acts upstream of APC will not be very useful: if APC is missing, large quantities of active β - catenin will be produced even in the presence of such drugs. However, drugs that β -catenin or TCF activity would be good candidates, provided that they do not interfere with vital functions of these proteins elsewhere in the body.
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