Biology 2008 STPM

Time: 1h 45 min

PAPER 1 1

2

3

4

5

A

Which molecule is made up of or contains glucose molecules? A Fructose B Cellulose C Ribonucleic acid D Deoxyribonucleic acid

6

Which of the following contributes to the secondary structure of a protein? A Peptide bond B Hydrogen bond C van der Waals force D Hydrophobic interaction

Which enzyme is involved in protein synthesis? A Ligase B DNA polymerase C RNA polymerase D Reverse transcriptase

7

Which statement best explains the polarity of water? A The angle between hydrogen atoms is 104.3°. B Oxygen is more electronegative than hydrogen. C Hydrogen is covalently bonded to oxygen to form water. D Polar compounds with partial charges tend to dissolve in water.

Which statement is true of transcription? A It begins with ATG and ends with TAG. B The sense strand is used as a template. C The DNA polymerase is used to synthesise DNA. D It uses 70S ribosome in prokaryote and 80S ribosome in eukaryote.

8

Which statement is true of noncompetitive inhibitor? A Its mode of action is reversible. B It binds directly to the enzyme at the active site. C Its binding to enzyme lowers the activation energy. D Its inhibitory effect can be reduced by increasing the substrate concentration.

9

Rubisco binds with both carbon dioxide and oxygen in A C3 plants B C4 plants C CAM plants D C4 and CAM plants

Which statement is not true of doublestranded DNA molecules? A Adenine pairs with thymine and guanine pairs with cytosine. B The nitrogenous bases hold both strands by a single hydrogen bond. C During DNA replication, a new strand is synthesised from 5' to 3' direction. D The sugar phosphate backbone of the polynucleotide is on the outside of the helix. Which organelle, in an animal cell, is spherical in shape and bounded by a single membrane?

© Majlis Peperiksaan Malaysia 2008

B C D

Lysosome Ribosome Microbody Mitochondrion

Actual 2008 STPM Biology Examination Paper

1

10

The following table shows the reactions of photosynthesis and their facts. Reaction

Fact

I Calvin cycle II Light reaction

11

Occurs in the stroma Produces oxygen Reduces NADP+ Uses ATP

p q r s

Which of the following is correct for the reactions and their facts? p q r s A I I II II B I II II I C II I I II D II II I I

The diagram below shows the conversion of pyruvate to ethanol during anaerobic respiration. CH3 C

O

C

O

OH

NADH

CH3

CO2

C Pyruvate decarboxylase

H

O

H

CH3

NAD+

C

OH

H

Alcohol dehydrogenase

Ethanol

Acetaldehyde

Pyruvate Which of the following allows glycolysis to continue? + The release of carbon dioxide A The regeneration of NAD C The addition of yeast to ethanol B The regeneration of NADH D 12

Which enzyme does

not

correspond to the catalytic reaction involved in glycolysis?

Enzyme B

Enolase Phosphoglucoisomerase

C

Phosphoglycerokinase

D

Phosphofructokinase

A

Catalytic reaction

Conversion of phosphoenolpyruvate to 2-phosphoglycerate 2-phosphoglycerate Conversion of glucose-6-phosphate to fructose-6phosphate Transfer of a phosphate group from 1, 3-biphosphoglycerate to ADP Transfer of a phosphate group from ATP to fructose-6phosphate

13

An example of saprophytic organism is A Mucor sp. B Taenia sp. C Rafflesia sp. D Periplenata sp.

14

The oxygen dissociation curve for the haemoglobin of an individual is further left compared to the oxygen dissociation curve for the haemoglobin of a normal individual. Which statements are true of that individual?

2


I The individual migrates from a low altitude to a high altitude. II The individual migrates from a high altitude to a low altitude. III The total haemoglobin and red blood cell count of the individual increases. IV The total haemoglobin and red blood cell count of the individual decreases. A I and III C II and III B I and IV D II and IV © Majlis Peperiksaan Malaysia 2008

15

16

The events which lead to the opening of a stoma are as follows: I The stoma is open. II The guard cells become turgid. III The water potential of the the guard cells decreases. IV Water enters the guard cells osmotically. V Potassium ions are actively transported into the guard cells. Which sequence of events is correct? A III, IV, V, II, I B III, V, IV, II, I C V, III, IV, II, I D V, IV, III, II, I The graph below shows the oxygen dissociation curves for two values of pH.

Percentage 100 of oxygen saturation

CO2

HHb Tissue cells Capillary wall

17

The diagram below shows the diffusion of carbon dioxide from respiring cells into the blood involving steps P, Q, R and S.


S

O2+ Hb

H+ + HCO3– R

HbO2–

Red blood cell

18

What happens when the ventricles of the heart contract? I The semi-lunar valves open. II The semi-lunar valves close. III The atrioventricular valves open. IV The atrioventricular atrioventricular valves close. A I and III B I and IV C II and III D II and IV

19

Which response occurs when a person loses a lot of blood? A A decrease in renin secretion B An increase in the secretion of sodium ions C An increase in the production of angiotensin D A decrease in the production of aldosterone

20

Which condition causes the closing of a stoma? A The influx of potassium ions into the guard cells B The increase in the concentration of glucose in the guard cells C The decrease in the concentration of carbon dioxide in the guard cells D The increase in the concentration of abscisic acid when plants are exposed to stress

pH 7.2

Which statement about the curve is true? A The increase in pH is due to vigorous activities. B The increase in pH causes the curve to shift to the right. C The percentage of oxygen-saturated haemoglobin decreases when pH increases. D The shifting of the curve to the right is due to an increase in the concentration of blood carbon dioxide.

H2CO3

Which step requires carbonic anhydrase to proceed to the next? A P C R B Q D S

pH 7.4

50 100 Partial pressure of oxygen (mm Hg)

P

Q

50

0

CO2 + H2O


3

21

The function of the sympathetic nervous system is to A hold memory B secrete hormones C speed up heartbeat D regulate the osmotic pressure of blood

22

The diagram below shows a structure of a neuron.

26

In the process of initiating humoralimmune response in humans, the role of the helper T cell is to secrete A perforin to lyse the infected cell B interleukin I to stimulate B lymphocytes to differentiate into plasma and memory cells C interleukin II to stimulate B lymphocytes to differentiate into plasma and memory cells D interleukin II to stimulate T lymphocytes to differentiate into cytotoxic T and memory cells

27

Which of the following is not a stage of the embryonic development in humans? A Cleavage B Fertilisation C Gastrulation D Organogenesis

28

Which of the following is true of blastula? A A group of cells at the stage of cleavage B A solid ball of blastomeres formed at the early stage of cleavage C A hollow ball of cells marking the end stage of cleavage D A cup-shaped embryonic stage consisting of two or three layers of cells

29

The diagram below shows the structure of a monocot seed.

R

P Q

S

muscle cell

Which of the following contains acetylcholine? A P C R B Q D S 23

24

25

4

Which statement is not true of auxin? A It stimulates the division of cell in a stem. B It stimulates the elongation of coleoptile. C It promotes the formation of lateral shoot. D It inhibits the elongation of root at high concentration. Oestrogen and progesterone are used in contraceptive pills to A maintain the endometrium of the uterus B inhibit the production of gonadotropic hormones C stimulate the release of luteinising hormone D stimulate the release of follicle stimulating hormone What is injected into the body of a person bitten by a dog? A Dead bacteria B Weakened viruses C Serum containing antigens containing specific D Serum antibodies Actual 2008 STPM Biology Examination Paper

S P

Q R

Which of the following is responsible for producing hydrolytic enzymes after imbibition? A P C R B Q D S © Majlis Peperiksaan Malaysia 2008

30

The table below shows the functions of gibberellin, aleurone and water in the seed germination of peas. Component

The genotype and number of progeny obtained are as shown below. Genotype

Function

I Gibberellin X Synthesises digestive II Aleurone enzymes Y Triggers the III Water synthesis of αamylase Z Releases gibberellin Which functions correspond gibberellin, aleurone and water? I II III A Y X Z B Y Z X C Z X Y D Z Y X 31

to

I Reduces water loss II Reduces energy consumption for body heat III Inhibits insect growth

Q Estivation

Diapause

R

Which types of dormancy correspond to their significance? I II III A P Q R B P R Q C Q P R D R Q P 32.

A test cross between individuals of genotypes below is carried out as follows: A

B

a

b

a

b

X

a

b


B

a

b

a

B

A

b

A

b

a

B

a

b

a

b

40

11

9

40

33

A cross between red-flowered and white-flowered snapdragons produces all pink-flowered progeny. This type of interaction is called A epistasis B codominance C complete dominance D incomplete dominance

34

A type of gene mutation is as follows:

Significance

Hibernation

P

A

What is the map distance, in map units, between the two genes? A 9 C 20 B 11 D 40

The table below shows types of dormancy and their significance. Types of dormancy

Number

mRNA Amino acid

–AAA lys

UGG trp

GUG val

UCU– ser

Mutation mRNA Amino acid

–AAA lys

UGG trp

GAG glu

UCU– ser

The mutation results in A haemophilia B cystic fibrosis C sickle cell anaemia D thalassaemia major 35

In a plant species, a trisomic plant has 29 chromosomes and a monosomic plant has 27 chromosomes. How many chromosomes does a triploid plant have? Actual 2008 STPM Biology Examination Paper

5

A B 36

37

38

39

40

6

28 42

C D

81 87

A gene pool is defined as A the total number of the genes of all the individuals in a population B the sharing of genes between two populations through interbreeding C the random changes in the allelic frequency in a small breeding population D a population in which the allelic and genotype frequencies do not change from one generation to the next What parameter should be calculated in order to know whether a population is in Hardy-Weinberg equilibrium? A Gene pool B Genetic drift C Allele frequency D Gene substitution In a population of 100 000 people, 10 of them are albinos. The frequency of the albinism carriers is A 0.01 C 0.20 B 0.02 D 0.99 An operon consists of A the structural genes only B the regulator and promoter genes only C the structural and promoter genes only D the structural, promoter and regulator genes Which statement is true of the lactose operon in E. coli? A It allows the inducer to bind to the promoter. B It is active when lactose binds to the promoter. C It is active when the repressor is bound to allolactose. D It allows the expression of the structural genes in the absence of lactose. Actual 2008 STPM Biology Examination Paper

41

The transgenic plant is different from the wild type plant because A it is a triploid B it contains antibiotic resistance plasmid C it contains foreign gene in their genomes D its DNA is transcribed from 3' to 5' direction

42

The taxa Amphibia, Reptilia, Aves and Mammalia are of the same A class but of different orders B order but of different families C kingdom but of different phyla D phylum but of different classes

43

The skeleton of vertebrates differs from that of arthropods in terms of A support B shedding C protection D muscle attachment

44

Which statement is true of allopatric speciation? produces hybrid A Speciation species. B Individuals occupy an overlapping area. C Speciation is not caused by geographical factor. D Geographical barrier prevents gene flow between populations.

45

The graph below shows the result of a type of selection for a population.

Number of individuals

Before selection

After selection

Phenotype

What conclusions can be made about the graph? © Majlis Peperiksaan Malaysia 2008

I II III IV A B C D 46

Disruptive selection Directional selection Selection against extreme groups Selection against intermediate groups I and III I and IV II and III II and IV

C D

49

All ecosystems are dependent on energy input because A producers have a greater biomass than consumers B carnivores have a greater biomass than producers C decomposers process the greatest amount of energy in an ecosystem D energy transformation results in partial loss of usable energy to the environment

50

In a study of a population of tiger prawns, the mean mass is 12.3 g and the standard deviation is 1.2 g. Which conclusions can be made about the population? I 68% of the population have masses of 12.3 g – 14.7 g. II 68% of the population have masses of 11.1 g – 13.5 g. III 95% of the population have masses of 12.3 g – 14.7 g. IV 95% of the population have masses of 9.9 g – 14.7 g. A I and III C II and III B I and IV D II and IV

In which situation would an algal bloom occur in a pond? A After a heavy rain B Plenty of sunlight C A run-off from a nearby fertilised field D The presence of a large population of zooplankton

47

Which organism does not contribute to the primary productivity of an ecosystem? A Cyanobacteria B Purple bacteria C Saprotrophic bacteria D Chemosynthetic bacteria

48

The carrying capacity is defined as A the maximum population growth rate under ideal conditions B the change in the population size per individual per unit time


the maximum population size that can be supported by an ecosystem the members of a species that live in an ecosystem and have the potential to reproduce


7

PAPER 2 Section A

[40 marks]

all questions in this section. Answer all 1

The diagram below shows a U-tube containing two mixtures of glucose and sucrose solutions separated by a membrane that is permeable to water and glucose but not to sucrose. A

B

2 mol dm –3 glucose + 1 mol dm –3 sucrose

1 mol dm–3 glucose + 2 mol dm –3 sucrose

Membrane

Arm A is half-filled with a mixture of 1 mol dm–3 glucose and 2 mol dm–3 sucrose. Arm B is half-filled with a mixture of 2 mol dm –3 glucose and 1 mol dm–3 sucrose. The levels of the mixtures in both arms are initially the same. (a) Explain what is meant by diffusion. [2 marks]

(b) State the initial state, in terms of tonicity, of the mixture in arm A with respect to that in arm B. [1 mark ]

(c) Explain how the system achieves equilibrium. [2 marks]

(d) Complete the following diagram and labels for the system when the equilibrium is achieved.

Initial level ... mol dm–3 glucose + ... mol dm–3 sucrose

... mol dm–3 glucose + ... mol dm–3 sucrose

Membrane

[5 marks] 8



2

Ectotherms and endotherms have homeostasis mechanisms to regulate body temperature physiologically, structurally or via behavioural means. (a) Define homeostasis. [1 mark ]

(b) Sketch, on the same axes, and label two graphs to show the relationships between outside temperature and body temperature for ectotherms and endotherms.

[2 marks]

(c) State how ectotherms and endotherms obtain body heat. [1 mark ]

(d) Explain how ectotherms and endotherms are able to maintain body temperature. Ectotherms:

Endotherms: [6 marks] © Majlis Peperiksaan Malaysia 2008


9

3

The Michaelis-Menten plots for enzymatic reaction are given below. Reaction velocity/ v v 0 X Y

Z

Substrate

(a) Name the curves X , Y and Z . X : Y : Z : [3 marks]

(b) Explain why the shapes of the curves are as such. X : Y : Z : [6 marks]

(c) Which curve shows an inhibition that could not be recovered by the addition of more substrate? [1 mark ] 4

Since the time of Aristotle until the middle of 19th century, biologists divided organisms into two kingdoms: Plantae and Animalia. With the advancement in the field of microscopy, there were organisms that could neither be classified as plants nor animals. To overcome this problem, Margulis and Schwartz proposed the Five Kingdom System of Classification. (a) State the kingdoms, to which the organisms in the following table belong, according to Margulis and Schwartz. [5 marks] Kingdom Organism

10

Two Kingdom System

Mucor

Plantae

Amoeba

Animalia

Chlorophyta

Plantae

Euglena

Animalia/Plantae


Five Kingdom System


Bacteria

Plantae

Obelia

Animalia

Animalia

Marchantia

Plantae

Plantae

(b) State one characteristic used to classify each of the following organisms in the Five Kingdom System of Classification. [5 marks] Organism

Characteristic

Mucor Amoeba

Chlorophyta Euglena

Bacteria

Section B

[60 marks]

Answer any four questions from this section. 5

(a) Distinguish a bacteria chromosome from a eukaryotic chromosome. [4 marks] (b) With the aid of a labelled diagram, describe the structure and functions of the Golgi apparatus. [11 marks]

6

(a) The transportation of water molecules and mineral ions from the soil to the roots could occur via several pathways such as vacuole, apoplast and symplast. Describe these three pathways. [8 marks] (b) Among the mechanisms of the translocation of sugar through sieve tubes are the mass-flow hypothesis, the electro-osmosis and the cytoplasmic streaming. Describe these three mechanisms. [7 marks]

7

Explain the mechanism of action of adrenaline on the liver cell.

8

(a) State the principal target tissue and the action of each of the following female reproductive hormones. (i) Progesterone (ii) Oestrogen (iii) Oxytocin (iv) Prolactin [8 marks] (b) Explain the events involved in the development of a fruit after the process of double fertilisation. [7 marks]


[15 marks]


11

9

In a small population, 40 individuals are tasters of phenylthiocarbamide and 60 are nontasters. The dominant allele T controls the ability to taste phenylthiocarbamide. (a) (i) Determine the frequencies of allele t and allele T. [6 marks] (ii) Determine the number of individuals who are heterozygous in the population. [4 marks]

(b) If 20 non-tasters immigrate into the population, determine the new frequencies of allele t and allele T. [5 marks] 10

12

(a) State the first and the second laws of thermodynamics. [4 marks] (b) Using the laws of thermodynamics in (a), explain how energy could be transferred in a named ecosystem. [11 marks]



SUGGESTED ANSWERS

PAPER 1

1. B Cellulose Cellu lose is made up β-glucose which is a hexose. RNA and DNA contain pentose sugar. 2. B Hydrogen bonds hold the α-helix and the βpleated sheet protein secondary structures in place. van der Waals force and hydrophobic interaction hold the tertiary structure of the protein. Peptide bond binds the amino acids of the linear primary structure of the protein. 3. B The small small positive charge of the hydrogen hydrogen atom and the small negative charge of the oxygen atom of the water molecule (due the uneven distribution of the electrons) contribute to the polarity of water. 4. B The nitrogenous nitrogenous bases hold both the DNA strands by more than one hydrogen bond. 5. A Ribosome is not not bounded bounded by any any membrane; membrane; mitochondrion is bounded by double membrane. 6. C Ligase and DNA polymerase are involved in DNA replication; reverse transcriptase is involved in the formation of DNA from mRNA. 7. B Transcription uses the sense strand of the DNA as a template to form the complementary mRNA and does not involve DNA polymerase and ribosome. Transcription does not necessarily begin with ATG in the template strand. 8. A Non-competitive inhibitor binds to the allosteric site of an enzyme which does not lower the activation energy. Its mode of action can be reversible or non-reversible. Only the competitive inhibitory effect can be reduced by increasing the substrate concentration. 9. A C3 plants experience photorespiration the most, which is very much reduced or prevented in C4 and CAM plants. 10. B Light reaction produces oxygen and reduced NADP during the photolysis of water and non-cyclic photophosphorylation whereas ATP is used for phosphorylation during Calvin cycle which occurs in the stroma of the chloroplast. 11. A Only with the presence of NAD+, oxidation of glucose or the intermediate can occur during glycolysis. 12. A Enolase catalyses the formation of phosphoenolpyruvate from 2 phosphoglycerate. 13. A Taenia sp. and Rafflesia sp. are parasites and Periplenata sp. is holozoic. 14. A At high high altitude, altitude, the air is thin and and the partial pressure of oxygen is low. Therefore the affinity of the haemoglobin towards oxygen © Oxford Fajar Sdn. Bhd. (008974-7) 2008

15. C

16. D

17. A

18. B

19. C

20. D

21. C

22. A 23. C 24. B

25. D

26. C

27. B

28. C

must be high as displayed by the shifting of the oxygen dissociation curve to the left. Having more haemoglobin and red blood cells would certainly increase the efficiency of oxygen uptake. This is the potassium accumulation hypothesis used to explain the opening and closing of the stomata. Decrease in pH is due to the increase in the partial pressure of carbon dioxide in blood during vigorous exercise and therefore shifting the oxygen dissociation to the right so that the percentage of oxygen-saturated haemoglobin is decreased to release oxygen. Carbonic anhydrase catalyses the formation of carbonic acid when carbon dioxide is dissolved in water. During ventricular systole, the higher pressure in the ventricle forces the semi-lunar valves to open and shut the atrioventricular valves to ensure blood is flowed one way in the heart. Loss of blood results in loss of sodium which causes an increase in angiotensin and aldosterone so that more sodium is reabsorbed from the collecting ducts to the vasa recta. High concentration of abscisic acid acid during water stress causes the stoma to close. Influx of potassium, increase in glucose concentration in guard cells (decreased water potential) and lower carbon dioxide concentration (that is, higher pH) causes the stoma to open. The systematic systematic nervous system consists consists of adrenergic motor nerves which can send impulses to the pacemaker of the heart to speed up heartbeat. P is a synaptic knob with vesicles containing acetylcholine (a neurotransmitter). Auxin promotes promot es the formation of adventitious roots at low concentration. When gonadotrophic hormones are not produced, follicles will not develop in the ovary and no eggs are released. The specific antibodies will help to destroy the antigens introduced into one’s body through the dog’s bite quickly. Humoral-immune response involves B lymphocytes which will proliferate with the help of interleukin II. Fertil isation of the egg by a sperm will produce a zygote which then develops into the embryo. Blastula is a hollow ball of of cells cells containing containing a fluid-filled cavity (blastocoel). Actual 2008 STPM Biology Examination Paper

13

29. A P is the aleurone (protein) layer which can be activated to produce hydrolytic enzymes. enzymes. 30. A Imbibition of water causes the release of gibberellin from the embryo which then stimulates the aleurone to synthesise hydrolytic enzymes. 31. C During hibernation, the animal reduces its metabolic rate to a minimum to keep body heat and body cells alive. Estivation is a dormant stage of an animal’s life due to prolonged drought. Diapause is a dormant period in insects due to unsuitable photoperiod and other adverse conditions. 32. C Calculate the COV as the map distance, that is (11 + 9) ————————– × 100% (40 + 11 + 9 + 40) 33. D Incomplete dominance results in an intermediate phenotype. 34. C This type of mutation is caused by a base substitution which results in a slightly different protein formed but is still functional. 35. B The diploid plant (2n) has 28 chromosomes. Therefore a triploid plant (3n) has 42 chromosomes. 36. A A gene pool refers to the sum total of all genes and their different alleles of all the individuals in a sexually reproducing population. 37. C Hardy-Weinberg equilibrium states that the allele frequency does not change from generation to generation. 10 38. B q 2 = ———– = 0.0001; q = 0.01; p = 1 – 0.01 = 100000 0.99; 2 pq = 2(0.99)(0.01) = 0.02 39. C An operon consists of a promoter, an operator and a cluster of structural genes that are regulated and expressed as a single unit. 40. C Allolactose will bind bind to the repressor repressor which will then detach from the operator to activate the lactose operon. 41. 41. C This can be done through genetic engineering. 42. D These different classes classes come from the same phylum, chordata. 43. B Arthropods shed their exoskeleton (ecdysis) to grow which does not occur in vertebrates with endoskeleton. 44. D Allopatric speciation is the the formation of new new species due to subpopulations being isolated by a geographical barrier. 45. B It is a disruptive selection where environment does not favour the intermediates. 46. C Too much organic material in the pond causes algal bloom or eutrophication. 47. 47. C Saprotrophic bacteria are consumers (decomposers). 48. C The maximum maximum population size that can be supported by an ecosystem with no environmental changes. 14


49. D Energy is always lost to the environment during transformation and need to be replaced to sustain the ecosystem. 50. D Statistically, 68% of the population have masses 12.3 (+/–) one standard deviation and 95% of the population have masses 12.3 (+/–) two standard deviations. PAPER 2 Section A

1. (a) Diffusion is the net movement movement of ions or molecules from a region where they are at higher concentration to a region of lower concentration, that is, to move down a concentration gradient until an equilibrium is reached. (b) Take Note: Tonicity is the abi lity of a solution to cause a region to gain or lose water. Tonicity of a solution depends in part on the concentration of solutes that cannot cross the membrane (nonpenetrating solutes), relative to that in the region itself. The mixture in arm A has more nonpenetrating solutes than the mixture in arm B, hence water will tend to leave arm B and move into arm A. (c) Water diffuses diffuses across across the membrane from arm B to arm A until the concentration of the sucrose solutions on both sides of the membrane are equal. (d) A B

1.5 mol dm –3 glucose + 1.5 mol dm –3 sucrose

1.5 mol dm–3 glucose + 1.5 mol dm–3 sucrose

2. (a) Homeostasis is the maintenance of the internal environment of a living organism at a constant level or within a narrow range of limits. (b) ut

er

40 p

e

ar

endotherm

30 l

et

m

20 nI

et

nr

a

ectotherm

10 0

10

20

30 40 50 External temperature/°C

© Oxford Fajar Sdn. Bhd. (008974-7) 2008

(c) Ectotherms obtain heat from the environment while endotherms rely on metabolic heat to maintain their body temperature. (d) Ectotherms Ectotherms:: These re ly on behavioura l and structural modifications to adapt to environmental conditions to maintain body temperature. When cold, ectotherms may bask in the sun, press themselves to warm surfaces or orientate body position relative to sun’s position. When hot, they move to shady places, wallow in water or open the mouths to increase heat loss. Endotherms Endotherms:: These mainta in the body temperature by balancing the heat produced from metabolism in the body against heat loss by the body. When hot, there is increased sweating, increased vasodilation of arterioles and rate of metabolism decreases. When cold, decreased sweating, increased vasoconstriction of arterioles, shivering and increased metabolic rate take place. 3. (a) X : Rate of enzyme activity without presence of inhibitor Y : Rate of enzyme activity with competitive inhibitor Z : Rate of enzyme activity with noncompetitive inhibitor (b) X : At low substrate co ncentration, the reaction velocity rises almost linearly with increasing substrate concentration. Since an enzyme molecule can only catalyse a certain number of reactions in a given time, the velocity of the reaction reaches maximum rate as the substrate concentration increases. Y : Competitive inhibitor and substrate compete for the same active site of the enzyme. In order to achieve the same velocity reached in the absence of the inhibitor, a higher substrate concentration is needed to overcome the competitive competition. Maximum reaction velocity can be reached if there is sufficient substrate. Z : The non-competitive inhibitor and substrate bind to different sites on the enzyme. Thus, the maximum reaction velocity decreases. (c) Curve Z . © Oxford Fajar Sdn. Bhd. (008974-7) 2008

4. (a)

Organism

(b)

Five kingdom system

Mucor

Fungi

Amoeba

Protoctista

Chlorophyta

Protoctista

Euglena

Protoctista

Bacteria

Prokaryotae

Organism

Characteristic

Mucor

– No chlorophyll; do not photosynthesise – Heterotrophic – Cell walls contain chitin rather than cellulose

Amoeba

– Unicellular organisms – Heterotrophic – Lack cell walls

Chloro Chlorophy phyta ta – Unicel Unicellul lular, ar, motile motile – Cells are surrounded by cell walls – Photosynthetic Euglena

– Unicellular organisms – Lacks a cellulose cell wall – Photosynthetic

Bacteria

– Lack true nuclei, whereby their DNA are not enclosed by a nuclear membrane and lie free in the cytoplasm. – No organelles bound by a double membrane. – Has a cell wall that consists of peptidoglycan.

Section B

5. (a) – Bacteria DNA is naked, i.e. it is not incorporated in chromosomes but is a single, circular strand lying free in the cytoplasm. There is no nucleus. – Eukaryotic DNA is linear and incorporated with proteins (including histones) and RNA in chromosomes. There is a nucleus bound by a double membrane nuclear envelope. Actual 2008 STPM Biology Examination Paper

15

(b)

• newly formed cis-cisterna (receiving side)

smooth cisternal membrane cisternal space

dictyosome (a stack of cisternae) secretory vesicles peripheral tubule

•

•

mature trans-cisterna (shipping side) to cell surface

Structure • Golgi apparatus apparatus consists consists of a stack of flattened, membrane bound sacs called cisternae. • The concave-shaped cisternae nearer the cell surface membrane is known as the trans Golgi network. • The convex-shaped convex-shaped cis Golgi network near the ER is formed from the fusion of transport vesicles from endoplasmic reticulum. Functions • Golgi apparatus receives vesicles from ER, stores and modifies the proteins by adding sugar molecules to form glycoprotein. • The secretory vesicles produced by the Golgi apparatus contain zymogen (e.g. pepsinogen, trypsinogen), mucin, hormones and neurotransmitters and they release their contents to the cell’s exterior by exocytosis • The fusion of secretory vesicles with the plasma membrane maintains the membrane which is used to form phagocytic vacuoles and pinocytic vesicles. • Golgi apparatus is involved in the formation of lysosomes containing hydrolytic enzymes. • In plant cells, the Golgi apparatus secretes polysaccharides for the formation of cell plates and cell walls. • It is also involved involved in the formation formation of peroxisomes.

6. (a) •

• • •

16

Water is absorbed mainly by root hairs which are cellular extensions of root hair cells. The root root hairs increase the surface area of the roots enormously. Dissolved substances build up in root hair cells by diffusion and active transport. This accumulation accumulation of solutes solutes gives the root hair cells a lower water potential than that of the water in the soil.


(b) •

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As a result, water water enters enters a root hair cell by osmosis, increasing its water potential above that of its neighbours. Water is then then drawn drawn in in by osmosis from root hair cells into the adjacent cortical cells. Water moves from cell to cell in the root along a water potential gradient. There are three possible routes. (i) Apoplast pathway. – Water passes freely through the cell walls from one cell to another. – As water is pulled pulled up the xylem due to transpirational pull, the cohesive forces between water molecules ensure that water is drawn across adjacent cell walls. (ii) Symplast pathway. – Water diffuses through the cytoplasm of adjacent cells. – The cytoplasm of adjacent cells is interconnected by cytoplasmic strands called plasmodesmata which pass through pores in the cell walls. (iii) Vacuolar pathway. – Water moves along through the vacuoles as well as the cytoplasm. – Water has to move across the the endodermis by the symplast pathway because the cell walls of the endodermis are impregnated with a waterproof waxy material called suberin called Casparian strip. The Casparian strip blocks the water passing along the cell walls (apoplast pathway). – The endodermal endodermal cells cells actively actively secrete mineral salts into the xylem vessels of the root. This lowers the water potential in the xylem vessels. Water from the root cells is drawn into the xylem by osmosis. In the mass-flow hypothesis, the sucrose and other organic solutes synthesised in the leaf are actively loaded by transfer cells and companion cells into the sieve tube. This lowers lowers the water potential potential of of the solution in the sieve tubes. As a result, water is drawn from the xylem in the leaf into the sieve tubes by osmosis. The entry of water produces a high hydrostatic pressure in the sieve tube. © Oxford Fajar Sdn. Bhd. (008974-7) 2008

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In the roots, the sucrose is actively transported into the tissues for respiration, the synthesis of cell walls or changed into starch for storage. The water water potential in the root cells cells decreases. Water is drawn from the sieve tube into the root cells by osmosis. The hydrostatic hydrostatic pressure pressure in the leaf is higher in the source compared to the pressure in the roots. This causes causes passive mass flow of water and solutes such as sugars, amino acids from the leaf to the roots. In the the electro-osmosis electro-osmosis mechanism, the companion cells contain numerous mitochondria which generate ATP required to remove potassium ions (K+) from one side of the sieve plate into the companion cell. The potassium ions (K+) are then secreted on the other side of the sieve plate, creating a potential gradient across the sieve plate. This causes an electro-osmotic flow of water molecules and dissolved solutes through the sieve pores to the adjacent sieve tube element. The cytoplasmic streaming mechanism involves a circular movement of cytoplasm from one end of the sieve tube element to the other end. The solutes pass through the sieve pores by active transport. This helps to account for the bidirectional movements along the sieve tubes.

Adrenaline is a non-steroid hormone that is insoluble in lipid and cannot diffuse through the plasma membrane. Adrenaline acts as a first first messenger messenger and binds to a specific receptor protein in the plasma membrane. This binding causes a change in conformation in the receptor. This increases increases the affinity affinity of the receptor receptor to bind with the G-protein in the membrane. Once activated, the G-protein moves to stimulate the enzyme adenyl cyclase. The enzyme, adenyl cyclase catalyses the conversion of ATP to cAMP (cyclic adenine monophosphate) within the liver cell. cAMP acts as a second messenger. It diffuses diffuses into the liver liver cell cell where where it initiates a complex chain reaction. cAMP activates an enzyme enzyme that catalyses a reaction which activates another enzyme and so on.


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This cascade effect where the action of one enzyme in turn activates another enzymatic reaction results in many product molecules. It brings brings about a rapid and amplified response to the hormone. This reaction reaction ends with the activation of glycogen phosphorylase that catalyses the breakdown of glycogen into glucose phosphate.

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8. (a)

(i) – The target tissue is the endometrium layer in the uterus. – The hormone hormone helps to maintain maintain the thickness and vascularisation of the endometrium layer in the uterus as preparation for implantation of blastocyst. (ii) – The target tissue is the endometrium layer in the uterus. – The hormone stimulates the thickening and vascularisation of the endometrium of the uterus and the development of glands in the uterine wall after menstruation. (iii) – The target tissue is the uterine muscle or mammary glands. – The hormone causes uterine contractions at birth and stimulates milk flow from the mammary glands. (iv) The target tissue is the mammary glands.It stimulates mammary glands to secrete milk. (b) • After fertilisation, the fertilised ovule develops into a seed. • In the embryo sac of a fertilised ovule, there are triploid/endosperm nucleus and a diploid nucleus/zygote. • The triploid nucleus divides rapidly by mitosis to form a food store. • In monocotyledonous seeds, the endosperm continues to grow and is kept as the seed’s food store. • In dicotyledonous seeds, the endosperm becomes absorbed by the developing cotyledons which provide the food store for the germinating seed. • The zygote in the ovule divides producing two daughter cells of different sizes: terminal cell and basal cell. • The large basal cell divides repeatedly to form the suspensor which connects the embryo to the integuments. • The smaller terminal cell undergoes repeated division and forms the embryo. The embryo cells then begin to form either one or two cotyledons. Actual 2008 STPM Biology Examination Paper

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9. (a)

Embryo cells between the cotyledons differentiate to form the apical meristem of the shoot, while embryo cells near the suspensor becomes the root apical meristem. The outer integument of the ovule becomes the testa (seed coat). Nutrients for the growing embryo and food store are supplied by the disintegration of surrounding nucellus cells in the embryo sac. The mature ovary becomes the fruit, the the outer wall is known as pericarp. (i) Frequency of recessive homozygous 60 individuals (= q2), tt = —— 100 60 100

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of allele, t (= q) = —— ∴ Frequency of = 0.77 Since p + q = 1. Frequency of allele, T (= p) =1–q = 1 – 0.77 = 0.23 (ii) The number of individuals who are heterozygous in the population = 2pq × 100 = 2 × 0.77 x 0.23 × 100 = 35 (b) Frequency of recessive homozygous

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80 individuals (=q2), tt = —— 120

∴ Frequency of allele, t (= q) =

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= 0.82 Since p + q = 1. Frequency of allele, T (= p) = 1 – p = 1 – 0.82 = 0.18 10. (a) •

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The first law of the thermodynamics states that energy cannot be created or destroyed but can be transformed from one form to another. The second law of thermodynamics states that when energy is transformed from one form to another, a small amount of energy is lost as heat.


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The sun is the source of energy, in the form of solar energy, for a food chain in a forest ecosystem. The green plants and trees in the forest, forest, which are the primary producers trap approximately 1 to 5% of light energy that reaches the earth’s surface for photosynthesis and is changed into chemical energy in the form of organic food molecules. Herbivores, such as the deer, deer, which which are are the primary consumers in the forest, feed on the plants. Energy is transferred from the primary producers (first trophic level) to the herbivores (second trophic level). Usually, only a small proportion (10%) of the energy stored in plant tissues is transferred to the herbivores. This is because most plants die and decompose without being used by herbivores. Energy is also lost to the surrounding as heat released during respiration. In the herbivores, herbivores, some energy is lost as heat released from respiration, excretory wastes (urine) and faeces. Some will be used for growth of new tissues and reproduction. Secondary consumers, usually carnivores such as tigers, feed on the primary consumers (herbivores). Energy is transferred from the second trophic level to the third trophic level. Only about about 10% of the energy in one trophic level ends up in the next trophic level. As energy energy is transferred along a food chain, large losses of energy occur at each transfer. Therefore, each trophic level receives less energy than the trophic level below it. After the fourth or fifth trophic level, only a small amount of energy is left which may be insufficient to support more trophic levels. This is why the number of trophic levels in an ecosystem is usually limited to four or five.


Biology 2008 STPM

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