Viva Voce Orals in Biochemistry

Chapter–1

Chemistry of Carbohydrates 1. DEFINE CARBOHYDRATES. CA RBOHYDRATES. Ans.

Carbohydrates are defined as aldehyde aldehyde or keto derivatives of po polyhydric lyhydric alcohols. For example example : Glycerol on oxidation is converted to D-glyceraldehyde, which is a carbohydrate derived from the trihydric alcohol (glycerol). CH2 – OH CH – OH CH2 – OH

Oxidation

Glycerol (trihydric alcohol) 2. CLASSIFY CARBOHYDRATES. Ans.

H–C=O | H – C – OH | CH2 – OH OH D-glyceraldehyde (carbohydrate)

Carbohydrates are classified into (a) Monosa Monosaccha ccharid rides es (simple (simple sugars) sugars) exampl example: e: glucose. glucose. (b) Disacchari Disaccharides des (composed (composed of two two monosacchar monosaccharides) ides) example example:: Sucrose, Sucrose, Lactose. Lactose. (c) Oligosacc Oligosacchari harides des (consis (consisting ting of of 2–10 monosa monosacchar ccharides ides). ). (d ) Polysacchari Polysaccharides des (consisti (consisting ng of more more than 10 monosac monosaccharid charides), es), example: example: Starch, Starch, glycogen.

3. HOW MONOSACCHARIDES ARE CLASSIFIED? GIVE EXAMPLES. Ans. S . No .

1. 2. 3. 4. 5.

Monosaccharides are sub classified depending on the number of ‘C’ atoms present in them and the nature of sugar group present in them No. of ‘C’ atoms

C3 (Trioses) C4 (Tetroses) C5 ( Pe n to s es ) C6 (Hexoses) C7 (H ep to se s)

Na t u re of s ug a r g r o up (a ld ehyd e) al dos es

D-Glyceraldehyde D-E ryt hro se D- Ri bos e D- gl u co se , D -g al ac t os e ––

Nat u r e o f s u g ar gr o u p ( ke t o) ke to se s

Di-hydroxy acetone D-Erythrulose D - R i b o u l o se a n d D - x y l u l o s e D - Fr u c t o s e D - S e do h e p t u l o s e

Viva Voce/Orals in Biochemistry

2

4. WHAT ARE ‘D’ AND ‘L’ ‘L’ STEREO STERE O ISOMERS? ISOME RS? Ans.

Sugars which are related to ‘D’ glyceraldehyde in spatial configuration (struc (s tructure ture)) are called ‘D’ isomers and sugars which are related relate d to L-glyceraldehyde L-glyceraldehyde in spatial spatial configuration configuration (structure) are called L isomers. H–C=O | H – C – OH | CH2 – OH

H–C=O | OH – C – H | C H2 – OH OH

D-glyceraldehyde

L-glyceraldehyde

The orientation of H and OH groups of a sugar in carbon atom which is adjacent to last carbon atom are related to D-glyceraldehyde are called D-sugars. And the orientation of H and OH groups of a sugar in carbon atom which is adjacent to last carbon atom are related to L-glyceraldehyde are called L-sugars. 5. WHAT ARE OPTICAL ISOMERS? Ans.

The presence of asymmetric ‘C’ atom in the sugar confers optical activity on the compound. When a beam of polarised light is passed through optically active sugar the plane plane of polarised light is rotated to the right (Dextrorotatory) or to the left (Levorotatory). Dextrorotatory is designated as + sign and Levorotatory is designated as – sign. ‘D’ glucose is a dextrorotatory therefore it is designated as D (+) glucose and L glucose is levorotatory and therefore therefore it is designated as L(–) glucose.

6. WHAT ARE EPIMERS? GIVE EXAMPLES OF SUGARS AS EPIMERS? Ans.

When the stereo isomers (sugars) differing in the orientation of H+ and OH– groups in a single ‘C’ atom they are called epimers. Example:

Glucose and galactose are epimers at 4th ‘C’ atom. Because they differ in the spatial configuration of OH and H at 4th ‘C’ atom. Similarly glucose and mannose mannose are epimers of 2nd ‘C’ atom.

Chemistry of Carbohydrates

3

H – C1 = 0

H – C1 = 0

HO – C2 – H

H – C2 – H

H – C2 – H

HO – C 3 – H

HO – C3 – H

HO – C 3 – H

H – C 4 – OH

H – C4 – OH

HO – C4 – H

H – C 5 – OH

H – C5 – OH

H – C 5 – OH

CH2 – O H D-mannose (2nd ‘C’ atom) 7.

H – C1 = 0

CH2 – O H D-glucose

CH 2 – O H D-galactose (4th ‘C’ atom)

WHAT WHAT ARE ARE THE DERIVA DERIVATIV TIVES ES OF MONOSA MONOSACCHARI CCHARIDES DES AND WHAT WHAT ARE THEIR FUNCTIONS? Ans.

The following are the derivatives of monosaccharides and their important biological functions.

S . No .

Der i va t iv e

F or me d b y

Impo r ta nt f un ct io n s

1.

Glucuronic acid

Formed by oxidation of 6th ‘C’ atom of glucose

UDP glucuronic acid is involved in conjugation of bilirubin

2.

Glucosamime

2 n d OH g ro u p o f g lu c o s e i s replaced by NH2 group

These are present in the muco polysaccharides (Proteoglycans)

3.

Galactosamine

2nd OH group of galactose is replaced by NH2 group

4.

Deo xyri bose

2nd oxygen of D–ribose is removed

Present in DNA

5.

So rb ito l

R e du c ti o n o f al d eh yd e gr o up of glucose by aldol reductase

Accumulate in diabetes mellitus and forms cataract

6.

Ribitol

R e du c ti o n o f al d eh yd e gr o up of ribose

Present in riboflavin (B 2)

8. WHAT IS MUTA ROTATION ROTATION? ? Ans.

The gradual change in the specific rotation (optical activity) of a freshly prepared solution of monosaccharide until it remains constant on standing is called muta rotation.


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Example:

a–D-glucose + 112° (specific rotation) 9.

b–D-glucose + 19° (specific rotation)

Glucose at equilibrium + 52.5°

WHAT WHAT IS IS THE COMPOS COMPOSITIO ITION N OF THE THE FOLLOWI FOLLOWING NG DISACC DISACCHARID HARIDES ES AND HOW THE MONOSACCHARIDE UNITS ARE JOINED JOIN ED IN THEM AND WHAT IS THE BIOLOGICAL IMPORTANCE? (I) SUCROSE

(II) LACTOSE

(III) MALTOSE

Ans. S . No .

Su gar

Comp o s it i o n a n d l i nk a g e

Bi o lo gi c al importance

1.

Sucr Sucros osee (no (nonn red reduc ucin ingg dis disac acch char arid ide) e) a D–glucose and b D–fructose linked Table sugar by a 1 b 2 glycosidic linkage.

2.

Lact Lactos osee (red (reduc ucin ingg disa disacc ccha hari ride de))

b D–galactose and b D–glucose. Present in milk b 1, 4 glycosidic linkage

3.

Maltose (r (reducing di disaccharide)

2 a D–glucose units a 1, 4 glycosidic linkage.

Obtained by hydrolysis of starch

10. WHAT WHAT ARE THE SOLUBLE SOLUBLE AND AND INSOLUBLE INSOLUBLE PORTIONS PORTIONS OF STARCH AND WHAT ARE THE DIFFERENCES BETWEEN BE TWEEN THEM? Ans.

S. No .

Starch is physically separated into two components. They are amylose and amylopectin. Differences between amylose and amylopectin. Amylose

A my l o pe c t i n

1.

Solu Solubl blee port portio ionn of of sta starc rchh (10– (10–20 20 %)

Insoluble portion of starch (80–90 %)

2.

Gluc Glucos osee uni units ts are are joi joine nedd by by a 1, 4 glycosidic linkages and give straight chain structure

Glucose units are joined by a 1, 4 and a 1, 6 glycosidic linkage and give branching.

3.

Mole Molecu cula larr weig weight ht is less less

Molecular weight is more

Chemistry of Carbohydrates

5

11. WHAT WHAT ARE THE THE MAJOR DIFFERENCES DIFFERENCES BETWEEN AMYLOSE AMYLOSE AND CELLULOSE? Ans. S . No .

Amylose

Ce l l u l o s e

1.

A com compo pone nent nt of star starch ch whic whichh is is pre prese sent nt in rice, wheat and pulses

Present in fibrous portion of plant (wood) and widely occurs in nature.

2.

It ha has nu nutritive va value

It is not important from nutrition point of view. However it prevents the constipation by expulsion of feces.

3.

Gluc Glucos osee uni units are joine oinedd bya bya 1,4 glycosidic linkages and is hydrolysed by amylase

Glucose units are joined b 1,4 glycosidic linkages and not hydrolysed by amylase but hydrolysed by bacterial cellulose.

12. WHAT WHAT ARE THE MAJOR MAJOR DIFFERENCES DIFFERENCES BETWEEN GLYCOGEN GLYCOGEN AND AMYLOPECTIN? Ans. S . No .

Gl y c o g e n

A my l o p e c t i n

1.

It is is an an anim animal al pol polys ysac acch char arid idee stor stored ed in in liver and muscle

It is a component of starch, which is a plant polysaccharide.

2.

Gluc Glucos osee uni units ts are are joi joine nedd by by bot bothh a–1,4 and a–1,6 glycosidic linkages. The degree of branching is more i.e., about 0.09 (more) i.e., one end glucose unit for each 11 glucose units

Glucose units are joined by a 1,4 and a 1,6 glycosidic linkages. The degree of branching is 0.04 (less) i.e. one end group for each 25 glucose units.

13. WHAT ARE MUCOPOLYSAC MUCOPOLYSACCHARIDES CHARIDES (GLYCOSAMINOGLYCANS) (GLYCOSAMINOGLYCANS) AND WHAT IS THEIR BIOLOGICAL BIOL OGICAL IMPORTANCE? IM PORTANCE? Ans.

S . No .

1.

Mucopolysaccharides are complex carbohydrates consisting of amino sugars and uronic acids. These are hyaluronic acid, heparin, and various chondriotin sulfates. Na me of t h e mucopolysaccharides

Hyaluronic acid

Re pe at in g un it

N–acetyl glucosamine and D– glucuronic acid joined by bb- 1,3 glycosidic 1,4 and b-1,3 linkages

B i o l o g i c a l i mp o r t a n c e

Present in synovial synovial fluid. It acts as a lubricant and shock absorbent in joints. (contd...)


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2.

Heparin

Repeating uni t consists of sulfated glucosamine and sulfated iduronic acid joined by a1, 4 glycosidic linkages.

It is an anti coagulant present in liver, lung and blood.

3.

Cho Chondroitin 4 sulf ulfate

Repe Repeat atin ingg unit unit cons consis ists ts of b– glucuronic acid and N–acetyl galactosamine joined by b–1, 3 and b–1, 4 linkages. (Sulfated at 4th position).

Present in cartilage, bone and cornea.

4.

C hondroitin 6 su lfate

S tr tr uc uc tu tu re re i s s im im il il ar ar t o chondroitin 4 sulfate except N– acetyl galactosamine is sulfated at 6th position.

Present in cartilage, bone and cornea.

5.

Dermatan sulfate

It is simi lar to c hondr ioti n sulfate except the uronic acid is ‘L’ iduronic acid.

Present in skin.

Chapter–2

Chemistry of Lipids 1. DEFINE LIPIDS. Ans.

The lipids are heterogeneous group of compounds which are insoluble in water but soluble in non-polar solvents such as ether, chloroform and benzene. All lipids invariably contain fatty acids.

2. CLASSIFY LIPIDS.

Lipids are classified into three groups: 1. Simple lipids (Alcohol (Alcohol + Fatty acids i.e. i.e. glycerol glycerol + 3 FA’s) 2. Complex lipids (Compound (Compound lipids) lipids) (Alcohol (Alcohol + FA’s + groups) groups) (a) Phospolipid Phospolipidss (Alcohol (Alcohol + FA + Phospho Phosphoric ric acid acid + ‘N’ ‘N’ base/other base/other group. group. (b) Glycolip Glycolipids ids [Sphin [Sphingosi gosine ne (Alcoho (Alcohol) l) + FA + Carbo Carbohydr hydrate]. ate]. 3. Derived lipids: These are obtained by the the hydrolysis hydrolysis of simple simple and complex lipids. Examples: Fatty acids, Glycerol, Steroids etc.

Ans.

3. CLASSIFY THE FATTY ACIDS. GIVE EXAMPLES. Ans.

Fatty acids are classified into (a) Satu Satura rate tedd fat fatty ty acid acids. s. (b) Unsat Unsatur urat ated ed fatt fattyy aci acids ds.. (c) Cycl Cyclic ic fatt fattyy acid acids. s.

Saturated fatty acids:

All the naturally occuring fatty acids have even number of ‘c’ atoms. Most predominant fatty acids which occur in nature are: (i) Palmitic acid H3C – (CH2)14– COOH (C16 fatty acid) (ii) Stearic acid H3C – (CH2)16– COOH (C18 fatty acid) Unsaturated fatty acids:

These contain one or more than one double bonds in them. They are: (i) Oleic Oleic acid (most (most common common unsatur unsaturated ated fatty fatty acid) acid) 18:1;9 18:1;9


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(ii) Poly unsaturated unsaturated fatty acids (PUFA) Linoleic Linoleic acid two (2) double double bonds, bonds, Linoleni Linolenicc acid three (3) double bonds and Arachidonic acids four (4) double bonds. 4. WHAT ARE ESSENTIAL FATTY ACIDS? ACIDS ?

The following polyunsaturated fatty acids are called essential fatty acids. 1. Linoleic 18:2;9,12 (not synthesised in the body) 2. Linolenic 18:3;9,12,15 (not synthesised in the body) 3. Arachidonic 20:4;5,8,11,14 (semi essential. It can be synthesised from linoleic acid)

Ans.

5. WHAT IS RANCIDITY? RANCIDITY ? Ans.

The unpleasant odour and taste developed by natural fats upon ageing is called rancidity.

6. DEFINE SAPONIFI SAPONIFICATIO CATION N NUM NUMBER, BER, IODINE NUMBER AND ACID ACID NUMBER. NUMBER. Ans.

(a)

Saponification number:

It is defined as milligrams of KOH required to saponify one

gram of fat. (b) (c)

Iodine number: It is the grams of Iodine absorbed by 100 gms. of fat. Acid number: It is defined as milligrams of KOH required to neutralise the fatty acids

present in one gram of fat. Acid number indicates degree of rancidity due to free acid. 7. WHAT WHAT ARE ARE PHOSPHO PHOSPHOLIP LIPIDS? IDS? BRIE BRIEFLY FLY OUTL OUTLINE INE THE THE STRUCTURE STRUCTURE AND FUNCTIONS OF PHOSPHOLIPIDS. Ans.

Phospholipids are made up of alcohol, fatty acid, phosphoric acid and a nitrogenous base or other group. These are classified into two major types: (a) Glycerophos Glycerophosphol pholipid ipidss (Glycerol (Glycerol containin containing). g). (b) Sphingoph Sphingophospho ospholipi lipids ds (Sphingosi (Sphingosine ne containing) containing)..

Glycerophospholipids

These phospholipids contain a common substance phosphatidic acid + a ‘N’ base or other group. Diagramatic Representation of Phospholipid

FA (Fatty acid) FA PA (Phosphoric acid) – N-base Glycerol Phosphatidic acid is composed of glycerol + 2 FAs + Phosphoric acid. ‘N’ base is attached to Phosphoric acid residue in the phospolipids.

Chemistry of Lipids

9

The following are the different types of glycerophospholipids which differ in the ‘N’ base. S .No

Name of th e Phospholipid

‘N’ b a se / ot h er gr o u p

Functions

1.

Lecithin

Choline

(a) Presen Presentt in cell cell membr membrane aness (b) Involv Involved ed in the the format formation ion of of cholesterol esters and lipo proteins (c) Involv Involved ed in the the forma formatio tionn of lung surfactant (dipalmitoyl lecithin) and the defect in its synthesis results in the development of RESPIRATORY DISTRESS SYNDROME (RDS) (d ) Posses Possesses ses amphip amphipath athic ic properties and forms micells in the digestion and absorption of lipids.

2.

Cephalin

Ethanolamine

Pres ent in the bi omem bran es.

3.

Ph os ph at id yl s er in e

Serine

Pres ent in the bi omem bran es.

4.

P ho s p h a t i d y l i n o s i t ol

Inosi tol

By h or mo ne a go ni st it i s cl e av ed t o 1. D A G an an d IP IP 3 wh wh i c h ac ac t as as second messengers in signal transduction

5.

Plasmalogen s (Resembles cephalin)

Ist ‘C’ ether linkage (unsaturated alcohol)

10 % ph osp ho lip id s of br ai n and muscle

6.

Cardiolipin (Di –phosphatidyl glycerol)

PA–glycerol –PA

Present in mitochondrial membranes

Sphingophospholipids (Sphingosine containing)

Spingomyelin contains (i) Sphin Sphingos gosin inee (compl (complex ex amino amino alcoh alcohol ol)) (ii) Fatty acid (iii) Phos Phosph phor oric ic aci acid (iv ) Choline Sphingosine + FA is called ceramide. Sphingomyelins are present in the nervous system. 8. WHAT ARE ARE GLYCOLIPIDS AND WHAT ARE THEIR IMPORTANT IMPORTANT FUNCTIONS? FUNCTION S?

Glycolipids contain contain ceramide (Sphingosine (Sphingosine + FA) and sugar. These are: (a) Cerebros Cerebrosides ides (Galacto (Galacto or glucocer glucoceramid amides). es). (b) Sulfatides Sulfatides (Sulpho (Sulpho galacto galacto ceramide) ceramide) present present in myelin myelin..

Ans.


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(c) Gangliosi Gangliosides des (Glucosyl (Glucosyl ceramide ceramide + sialic sialic acid). Glycolipi Glycolipids ds are present in the the nervous tissues tissues and cell membranes. 9. WHAT IS IS THE BASIC STRUCTURE OF THE STEROID? Ans.

Steroid possesses cyclopentano, perhydro phenanthrene nucleus. Phenantherene has three A, B, C six membered rings to which a five membered cyclopentane ring is attached. The whole system is called CPP nucleus with 19 positions. 12 18 17 1 2

19 10

3 4

5

11 9

6

13 8

14

16 15

7

10. 10. NAME THE BIOLOGICALLY BIOLOGICALLY IMPORTANT SUBSTANCES POSSESSING CPP NUCLEUS. Ans.

The following substances have CPP nucleus in their structures: (a) Choles olestterol erol.. (b) Pro Provit vitami amin –D3 (7 dehydro cholesterol). (c) Bile acids (cholic (cholic acid acid and and chenodeo chenodeoxy xy cholic cholic acid). (d ) Steroid hormones (glucocorticoids, (glucocorticoids, mineralocorticoids mineralocorticoids and sex hormones).

11. 11. WHAT WHAT ARE THE GENERAL GENERAL FEATURES OF STRUCTURE STRUCTURE OF CHOLESTEROL. Ans.

Cholesterol is a 27 carbon containing compound. It has (a) CPP nu nucleus. (b) OH gro group up at 3rd 3rd posi positi tion on.. 5 (c) D i.e. a double bond is present between 5th and 6th carbon atoms. (d ) A side side chain chain at 17th 17th posit positio ion. n.

12. 12. NAME THE IMPORTANT IMPORTANT COLOR REACTION OF CHOLESTEROL CH OLESTEROL AND WHAT IS ITS IMPORT I MPORTANCE? ANCE? Ans.

The important color reaction of cholesterol is LIEBERMANN BURCHARD REACTION. The chloroform solution of cholesterol is treated with acetic anhydride and sulphuric acid which gives a red color and this color quickly changes from blue to green. This reaction is the basis of quantitative estimation of cholesterol.

Chemistry of Lipids

11

13. 13. WHAT ARE EICOSANOIDS AND WHAT ARE THEIR FUNCTIONS? Ans.

The compounds derived from arachidonic acid are called Eicosanoids. These are 6 primary PGs

(a) Pros Prosta tagl glan andi dins ns (PGs (PGs)) 8 secondary PGs Major functions : They cause contraction of pregnant uterus and therefore they can be used in the induction of labor or medical termination of pregnancy (MTP). (b) Prostacyclins (PGI 2): These prevent platelet platelet aggregation and act as vasodialators. (c) Thromboxanes (TX 2): These cause platelet aggregation and forms thrombus. (d ) Leucotrienes : These are formed as the products of mast cell degradation. SRS S RS – A is formed in the anaphylaxis. This substance consists of leukotrienes C 4, D4 and E4. 14. WHAT ARE w 3 FATTY FATTY ACIDS? WHAT IS THE CLINICAL SIGNIFICANCE OF THESE FATTY ACIDS? Ans.

The w3 fatty acids have double bond between 3rd carbon (3rd from w carbon atom left side) and 4th carbon. Examples:

a – Linolenic acid (present in plant oils). 1 2 3 4 H3C — CH2 – C = C -- CH2 – C = C -- CH 2 – C = C – (CH 2)7 – COOH H H H H H H Omega Carbon

16 15 14 13 12 Double bond is present between 3rd cabron (from ω carbon) and 4th carbon

11

10

9

18 : 3; 9, 12, 15 The other w3 fatty acids are (a) Eicosap Eicosapent entanoi anoicc acid 20:5; 20:5; 5, 8, 11, 11, 14, 17 (b) Docosa Docosa pentan pentanoic oic acid 22:5; 22:5; 7, 10, 13, 13, 16, 16, 19 19 (c) Docosahexa Docosahexanoic noic acid 22:6; 22:6; 4, 7, 10, 10, 13, 13, 16, 16, 19 19 a, b, c, are present in fish oils. The w3 fatty acids prevent thrombosis and hence these are used for the prevention of coronary artery heart disease (CAHD).

Chapter–3

Chemistry of Proteins 1.

WHAT WHAT ARE ARE THE THE DIFFEREN DIFFERENT T LEVELS LEVELS OF ORGANIZ ORGANIZATIO ATION N OF STRUCTURE STRUCTURE OF A PROTEIN? Ans . (a) (b) (c) (d )

2.

Prim Primar aryy stru struct ctur ure. e. Seco Second ndar aryy stru struct ctur ure. e. Tert Tertia iary ry stru struct ctur ure. e. Quat Quater erna nary ry stru struct ctur ure. e.

WHAT WHAT ARE THE THE MAIN MAIN FEATU FEATURES RES OF OF PRIMAR PRIMARY Y STRUCTURE STRUCTURE OF A PROTEIN? Ans.

Primary structure comprises the sequence or order of amino acids in the polypeptide chains and location of disulphide bonds in them. Peptide bond is the main force which maintains primary structure: H R | | H2N — C — CO — N — C — COOH COOH | | | R H H Peptide bond The polypeptide chain has: (i) One ‘N’ termin terminal al amino amino acid (Ist (Ist amino acid acid on left termin terminal al of polypep polypeptide tide chain chain having free amino group). (ii) One ‘C’ termin terminal al amino acid acid (last (last amino acid acid having having free carboxy carboxyll group). group). In between the amino acids are joined by peptide bonds. Polypeptide chain CT A.A.

NT A.A.

3. WHAT IS SECONDARY SECONDARY STRUCTURE OF A PROTEIN AND GIVE EXAMPLES? Ans.

The regular folding and twisting of the polypeptide chain brought about by hydrogen bonds is called secondary structure of protein.

Chemistry of Proteins

13

Examples:

(i) a– helix structure. (A) (A) Fibr Fibrou ouss prote protein ins: s: (a) Keratin of hair, hair, nails nails and skin. (b) Myosin Myosin and tropomy tropomyosin osin of muscles muscles.. (B) (B) Glob Globul ular ar prot protei eins ns Hemoglobin (ii) b–pleated sheet Silk fibroin. Salient features of a –helix

(a) (b) (c) (d )

The distance distance travelled travelled per per turn of helix helix is 0.54 0.54 nm and 3.6 3.6 amino acid acid residues residues take part. The dista distance nce betwe between en adjace adjacent nt resid residues ues 0.15 0.15 nm. nm. Proline Proline and hydroxy hydroxy proline proline disrupt disrupt helix. helix. Hydroge Hydrogenn bonds bonds and Van der der Waal’s Waal’s forc forces es stabil stabilize ize a– helix structure. N CC

C C N C

N C Distance per turn of helix C 3.6 residues

N

C C C

C

N

C C

C

N

C

N

N

N N

C

C C

CN

C C

0.15 nm

Figure 1 a –helix structure

Salient features of b – pleated sheet

(a) (b) (c) (d ) (e)

It is is a shee sheett rath rather er than than rod. rod. The dist distance ance betwee betweenn adjacen adjacentt A.A is 3.5 3.5 Å. Sheets Sheets are composed composed of of two (2) (2) or more than than two two (2) polyp polypepti eptide de chains. chains. And also also anti anti-pa -paral rallel lel sheets sheets.. Ans also also in in oppo opposit sitee direc directio tion. n.


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( f arrangem gement ent of of polype polypepti ptide de chain chainss in b–pleated sheet is parallel pleated sheet. f ) The arran (g) Peptide Peptide chains chains are side by by side side and lie lie in the same same directio direction. n. 4.

WHAT WHAT ARE ARE THE THE SALIEN SALIENT T FEATURES FEATURES OF OF TRIPLE TRIPLE HELIX HELIX STRUCTU STRUCTURE RE OF COLLAGEN? Ans.

Collagen consists of approximately 1000 A.As. The repeating unit of collagen structure is represented by (Gly–X–YN)n. About 100 of X positions are prolines and 100 of Y positions are hydroxy prolines. Due to presence of proline, hydroxy proline and glycine in the structure of collagen, these A.As prevent a– helix and b– pleated structure and forms triple helix structure. Collagen is composed of three (3) polypeptide chains. Each a chain is twisted into left–handed helix of three (3) residues per turn. Three (3) a chains then form a rod like structure.

Forces of triple helix

(a) Hydrogen bonds: Left handed helices are held together together by interchain hydrogen hydrogen bonds. (b) Lysi Lysine ne or leuc leucin inee bon bond. d. Covalent cross links

Collagen fibers are further stabilized by the formation of covalent cross-links both within and between triple helical units. This occurs by lysyl oxidase which causes oxidative deamination of e– NH2 group of lysine and hydroxy lysine residues to form aldehydes.

α

–chains

Triple Helix Figure 2. Triple Helix structure of collagen

5. WHAT WHAT ARE ARE THE SALIENT SALIENT FEATURES FEATURES OF TERTIAR TERTIARY Y STRUCTURE OF A PROTEIN? Ans.

The looping and winding of the secondary structure of a protein by other associative forces between the amino acid residues which give three dimensional conformation is called tertiary structure.


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In the tertiary structure, proteins fold into compact structure. The tertiary structure reflects overall shape of the molecule. Example: Myoglobin. The polypeptide chains are folded in such a way that hydrophobic side chains are burried inside and its polar side chains are on the surface (exterior). Forces involved in the tertiary structure

(a) (b) (c) (d )

Hydr Hydrog ogen en bond bonds. s. Hydr Hydrop opho hobi bicc intera interact ctio ions ns.. Van Van der der Waal Waal’s ’s forc forces es.. Disu Disulp lphi hide de bond bonds. s.

6. What are the the Salient Salient Features Features of Quatern Quaternary ary Structur Structure e of a Protein? Protein? Ans.

The association of catalytically or functionally active small subunits of protein is called quaternary structure. The quaternary structure of protein is due to proteins having more than one polypeptide chain. Examples: (i) Hemoglo Hemoglobin bin

(Hb– (Hb–A): A): 4 poly polypep peptid tidee chains chains (a2b2). (ii) Lactate Lactate dehydrog dehydrogenase enase (LDH) (LDH) – 4 polypep polypeptide tide chains chains (H or M). (iii) Creatin Creatinee kinase kinase (CK) (CK) – 2 polypep polypeptide tide chain chainss (B or M). M). Forces involved for the stabilization of quaternary structure (a) Hydr Hydrop opho hobi bicc intera interact ctio ions ns.. (b) Hydr Hydrog ogen en bond bonds. s. (c) Ionic bo bonds.

7.

WHAT WHAT ARE THE THE MAIN MAIN FEATUR FEATURES ES OF STRUCT STRUCTURE URE OF INSULI INSULIN? N? Ans.

Insulin is a protein hormone secreted by the b– cells of islets of Langerhans. Insulin consists of two (2) polypeptide chains (‘A’ chain and ‘B’ chain) and composed of 51 A.As. The molecular weight is 5734. ‘N’ terminal A.A.

‘C’ terminal A.A.

Chai n

Total No. of A.As.

A

21

Glycine

As par azin e

B

30

Phenylalanine

Th re on in e

Two chains are joined by (two) 2 inter disulphide bonds. Ist bond is formed by 7th cystein of ‘A’chain and 7th cystein of B chain. 2nd bond is formed by 20th cystein cystein of ‘A’ chain and 19th cystein of ‘B’ ‘B’ chain. One intra disulphide bond is present in the ‘A’ chain formed by 6th cystein to 11th cystein.


16

8. What What is end Grou Group p Analy Analysis sis? ?

End group analysis is the method devised for the identification of ‘N’ terminal and ‘C’ terminal A.As. in the polypeptide chains of a protein.

Ans.

9. GIVE ONE EXAMPLE EXAMPLE OF IDENTIFI IDENTIFICATION CATION OF ‘N’ ‘N’ TERMINAL TERMINAL AMINO AMINO ACID AND ‘C’ TERMINAL AMINO AMI NO ACID?

(A) Method for the identification identification of ‘N’ ‘N’ terminal amino acid. The ‘N’ terminal amino acid reacts with Sanger’s reagent (1 fluoro 2, 4 dinitrobenzene) and forms corresponding DNP – amino acid.

Ans.

H | R – C – NH2 + F | COOH

NO 2 NO2 FDNB

H | R–C–N– | COOH

NO2 NO2 + DNP – A.A. + HF

(B) Method for the identification of ‘C’ terminal amino acid. ‘C’ terminal A.A. is identified by treating with carboxy peptidase which splits terminal peptide bond and releases ‘C’ terminal A.A. which can be identified by chromatography. Carboxy peptidase

‘C’ terminal A.A. + Peptide minus one A.A.

10. WHAT ARE THE BASIC, ACIDIC, ACIDIC, ‘S’ CONTAINING, CONTAINING, HYDROXY, BRANCHED CHAIN AND AROMA AR OMATIC TIC AMINO ACIDS AND WHAT ARE THE IMINO ACIDS? Ans. S . No .

Category of A.As.

Ex amp le s

(i)

Basic

Ly si ne , ar gi ni ne , o rn it hi ne , hi st id in e (slightly basic)

(ii )

Acidic

Glutamic acid, aspartic acid

(iii )

‘S’ containing

M et h i on i n e , c ys t e i ne an d c y s ti n e

(iv)

Branched chain

Valine, isoleucine and leucine

(v)

Hydroxy

Serin e an d th reonine

(vi)

Aromatic

P he n yl a la n in e , t y ro s in e a n d t ry p to p ha n

(vii )

Imino acids

Proline and hydroxy proline.


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11. 11. WHICH IS THE SIMPLEST AMINO ACID AND WHICH AMINO ACID LACKS THE ASYMMETRIC ‘C’ ATOM? Ans.

Glycine.

12. 12. WHAT ARE THE BIOLOGICALLY BIOLOGICALLY IMPORTANT IMPORTANT PEPTIDES PEPTID ES AND WHAT ARE THEIR IMPORTANT IMPORTANT FUNCTIONS? FUNCTION S? Ans. S . No .

Name o f t he pe p t i d e

(i) Glutathione (GSH)

No. o f A.A s. pr es ent a nd is nature

3 amino acids

Impo r ta nt fu fun ct io n s

Takes part in the reduction reactions and in the detoxification o f H2O 2 which maintains the integrity of RBC membrane.

(ii ) Thyrotrophin 3 amino acids. It is pr produced in Releases TSH from the anterior pituitary gland releasing hormone (TRH) the hypothalamus (iii ) Angiotensin – II

8 amino acids formed from the Stimulates aldosterone secretion angiotensin I by the action of from the zona gloumerulosa cells converting enzyme. of adrenal cortex. Powerful vaso constrictor ( -- B.P)

(iv) Vasopressin (ADH)

9 am amino ac acids se secreted by by th the Causes absorption of water from the renal tubules. posterior pituitary gland Decapeptide (10 A.As.) formed It is converted to angiotensin – II by the action of renin on by the CE which - - BP angiotensinogen (a (a2 globulin) (hypertension).

(v) Angiotensin – I

13. WHAT ARE SIMPLE PROTEINS? Ans.

(a) (b) (c) (d ) (e) ( f f )

Simple proteins consists of only amino acids. These are : Albumins: (Egg white) white) soluble soluble in in water. Precipitated by full full saturation saturation with (NH4)2SO4. Globulins: Globulins: Soluble Soluble in dilute dilute neutral neutral salt solution solution (egg yolk and myosin of muscle). It is precipitated by half saturation with (NH4)2 SO4. Glutenins: Glutenins: Soluble Soluble in dilute acids and and alkalis. alkalis. Examples: Examples: Gluteli Glutelinn of wheat and oryzenin oryzenin of rice. Prolamines are alcohol alcohol soluble soluble proteins. proteins. Examples: Zein of corn and gliadin of wheat. wheat. Histones Histones are basic basic protein proteins. s. Globin Globin of hemoglo hemoglobin bin and nucleo nucleohist histone. one. Protamines Protamines are basic basic proteins proteins and present present in in nucleoprotein nucleoproteinss of sperm sperm (nucleoprotami (nucleoprotamines). nes).


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14. WHAT WHAT ARE CONJUGATED PROTEINS? PROTEINS? GIVE GIVE EXAMPLES. EXAMPLES. Ans.

Conjugated proteins contain amino acids and a non–protein prosthetic group: (a) Chromoproteins – Hemoglobin (Heme + globin). (b) Nucleoproteins – Nucleic acid + proteins. (c) Glycoprotein – Carbohydrate + protein. (d ) Phosphoproteins – Phosphoric acid + protein (casein of milk). (e) Lipoproteins – Lipid + proteins (chylomicrons, VLDL, LDL etc.).

15. 15. WHAT ARE DERIVED PROTEINS? GIVE EXAMPLES. Ans.

Derived proteins are formed by treatment with heat, acids and by the hydrolysis of native proteins. Example :

Primary derived proteins – Denatured proteins and metaproteins. Secondary derived proteins are formed by progressive hydrolysis of proteins. Proteoses, peptones and peptides. 16. WHAT WHAT IS ELECTROPHORESIS ELECTROPHORESIS AND WHAT WHAT ARE THE IMPORTANT IMPORTANT APPLICATIONS OF ELECTROPHORESIS? Ans.

Movement of charged particle in the electric field either towards cathode or anode when subjected to an electric current is called electrophoresis. The following factors influence the movement of particles during the electrophoresis. (a) Elec Electr tric ic cur curre rent nt.. (b) Net Net char charge ge of of the the part partic icle le.. (c) Size Size and and sha shape pe of of the the part partic icle le.. (d ) Type Type of of supp suppor orti ting ng med media ia.. (e) Buff Buffer er solu soluti tion on..

Important Applications of Electrophoresis

The technique of electrophoresis is used to separate and identify the (i) Seru Serum m prot protei einns (ii) Seru Serum m lipo lipopr prot otei eins ns (iii) Bloo Bloodd hem hemog oglo lobi bins ns 17. 17. WHAT ARE THE DIFFERENT TYPES OF ELECTROPHORESIS? Ans.

( a)

(a) Moving boundary electrophoresis electrophoresis : This technique was first introduced by TISELIUS in 1937 (b) Zone electrophoresis: In this type of electrophoresis different types of of supporting media are used. These are; Pape Paperr elec electr trop opho hore resi siss

(i) Wh What atma mann filt filter er pape paperr


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(ii) Cell Cellul ulos osee acet acetat atee ( b) Gel Gel elect electro roph phor ores esis is (i) Agarose. (ii) Polyacrylamid Polyacrylamidee

gel (used for the the separatio separationn of isoenzymes). isoenzymes).

(iii) SDS–PAGE. (iv ) Iso–electr Iso–electric ic focussing focussing (protein (proteinss seperated seperated in a medium possessi possessing ng a stable pH gradien gradient). t). (v ) Immuno electrophores electrophoresis is (for (for the separation separation of of immunoglo immunoglobulin bulins). s). 18. 18. WHAT ARE THE MAIN STEPS INVOLVED INVOLVED IN THE SEPARATION SEPARATION OF SERUM PROTEINS BY AGAROSE GEL ELECTROPHORESIS? Ans. (a)

(b) (c) (d ) (e) ( f f ) (g) (h) ( i) ( j j)

Preparation Preparation of barbitone barbitone buffer (vernal (vernal buffer (pH 8.6). At pH 8.6 all the serum protein proteinss carry the negative negative charge and they migrate from cathode to anode. anode. Preparat Preparation ion of 1% 1% agaros agarosee solut solution ion.. Preparati Preparation on of agarose agarose gel gel slide slide (layer (layer the molten molten agarose agarose on the the slide. slide. Application Application of the the serum serum sample sample with with the help of cover cover slip. slip. Filling Filling of electrophor electrophoretic etic tank with with the barbiton barbitonee buffer and slide slide to be kept on the the tank. Power supply supply to be switched switched on and and 7 mA current to be passed passed per each each slide for about about 30 minutes. Slide to be kept in in fixative fixative solution for 30 minutes and then in the dehydrating dehydrating solution solution for 4 hours. Slide Slide to be kept in the the incubat incubator or overni overnight ght at at 37° C. Slide is stained with with amido amido black stain for for 30 minutes and and then is destained destained with 7% acetic acid and should be washed with water. Quantita Quantitatio tionn is is done by densitom densitometer eter.. The following fractions of serum proteins are separated depending upon the rate of migration in order of rate of migration. (i) Albumin

Fastest moving fraction

(ii) a1 globulin

Next fast moving fraction

(iii) a2 globulin

Just ahead of b globulins

(iv ) b globulin

Just ahead of g globulins

(v ) ¡ globulin

Least mobile fraction and it is almost near the point of application

19. 19. WHAT IS CHROMA CHR OMAT TOGRAPHY AND A ND WHAT ARE THE IMPORTANT APPLICATIONS OF CHROMATOGRAPHY? CHROMATOGRAPHY? Ans.

Chromatography is a special technique by means of which a group of similar substances are separated by a continuous redistribution between two phases.


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(i) Stat Statio iona nary ry phas phasee (ii) Movin ving phase ase Many variety of attractive forces act between the stationary phase and the substances to be separated resulting in the separation of molecules. Important Applications of Chromatography

This special technique is used for the separation and identification of (i) Sugars Sugars (for the diagno diagnosis sis of inborn inborn errors errors of carbohy carbohydrat dratee metabolism metabolism). ). (ii) Amino acids acids (for (for the diagnosis diagnosis inborn inborn errors errors of amino acid acid metabolism, metabolism, example: example: phenylketonuria (PKU)). (iii) Lipid fractio fractions ns mainly mainly for the the separation separation of of phospholi phospholipids pids to to find out out Lecithin: Spingomyelin ratio in the Hyaline membrane disease (RDS). 20. 20. WHAT WHAT ARE THE DIFFERENT TYPES OF CHROMAT C HROMATOGRAPHY? OGRAPHY? Ans. (a)

(b) (c) (d ) (e)

There are various types of chromatography. Paper chromatogra chromatography phy (stationary (stationary phase is whatman filter paper and mobile phase is a solvent system. Example : Butanol, acetic acid and water) used for the the separation of amino acids. Thin layer layer chromatograp chromatography hy (glass (glass plates coated coated with with silica silica gel acts as station stationary ary phase and and solvent system as mobile phase). TLC is used for the separation of phospholipids and amino acids. ): A separation technique in which the mobile phase is a Gas liquid chromatography (GLC ): gas. GLC is used for the separation of volatile substances like fatty acids. technique separation is based in differences in the ion Ion exchange chromatography: In this technique exchange affinities of the sample components. ): A type of of liquid chromatography that High performance liquid chromatography chromatography ( HPLC HPLC ): uses an efficient column containing small particles of stationary phase. HPLC is carried out under liquid pressure in the chromatographic matrix. This improves speed and resolution for the efficient separation of substances.

21. 21. WHAT ARE THE MAIN STEPS OF PAPER CHROMAT C HROMATOGRAPH OGRAPHY Y IN THE SEPARATION OF AMINOACIDS? Ans.

(i) The chromatographic chromatographic chamber is is saturated saturated with with solvent solvent system system (butanol, acetic acid and and water). The solvent system is kept on the upper portion of the chamber in the glass trough (descending (descending paper chromatography). (ii) A drop of of the sample sample contain containing ing unknown unknown amino acids acids is appli applied ed about about 5 cms from the end of whatman filter paper. Known standard amino acids a cids are also applied applied on the line line of point point of application. (iii) The filter filter paper paper slip slip is dipped dipped in the solvent solvent system system and is is hung above downwards. downwa rds. (iv ) As the solvent solvent moves from from above above downwards downwards by the capillary capillary action action the the amino acids also move depending upon their R f values.


21

(v ) Once the solvent solvent reaches reaches the other other end of of the paper paper the paper is removed from the the chamber and is dried and sprayed with ninhydrin and acetone and heated. heate d. Purple spots of amino acids will be developed. (vi ) Amino Amino acids acids will will be identif identified ied by by calculat calculating ing R f values. (vii ) Amino acids with large large nonpolar groups groups move with a fastest rate. Pheny Phenylal lalani anine, ne, isoleu isoleucin cinee and leucine. The amino acids with polar groups are least mobile amino acids. 22. 22. WHAT WHAT IS Rf VALUE? WHAT IS THE IMPORTANCE IMPORTANCE OF Ans. R f value

R F VALUE?

is the relative fraction value. This is calculated by the formula. R f

valu valuee =

Distance travelled by the substances(A.A) Distance travelled by the solvent front

By R f value the unknown amino acids present in a sample can be identified. For instance the R f value of unknown spot is x and the R f value of leucine is also same x. The unknown spot is identified as leucine because its R f value corresponds to the R f value of leucine standard.

Chapter–4

Enzymes 1. WHAT WHAT ARE THE THE FACT FACTORS ORS INFLUENCING INFLUENCING ENZYME ENZYME CATA CATALYZED REACTION? REACTION? Ans.

Enzyme concentration, substrate concentration, temperature and pH.

2. WHAT WHAT IS THE EFFEC EFFECT T OF ENZYME ENZYME CONC CONCENTR ENTRATI ATION? ON? Ans.

When the other factors are kept constant the increase in enzyme concentration increases velocity. It maintains a linear relationship obeying Ist order kinetics.

3. WHAT WHAT IS THE EFFEC EFFECT T OF SUBSTR SUBSTRATE ATE CONCEN CONCENTRAT TRATION ION? ? Ans.

4.

WHAT IS KM VALUE AND WHAT IS ITS IMPORTANCE? IMPORTANCE? Ans.

5.

At low concentration of the substrate, the increase in substrate, increases the velocity until the enzyme is said to be saturated with the substrate molecules (Ist order kinetics). Beyond this point on further increase of substrate there is no increase of velocity (zero order kinetics).

It is the substrate concentration at which it gives half the maximum velocity. Low Km indicates high affinity towards substrate. High Km indicates low affinity towards substrate. Example : Hexokinase – Low Km Glucokinase – High Km

WHAT WHAT IS IS THE OPTI OPTIMUM MUM TEM TEMPE PERA RATURE TURE OF ENZYMES ENZYMES PRESENT PRESENT IN HUMAN HUMAN BODY? Ans.

37 – 45°C.

6. GIVE GIVE EXAMPL EXAMPLES ES OF OF OPTIM OPTIMUM UM pH pH OF SOME SOME OF ENZYMES? ENZYMES? Ans.

Pepsin 1. 1 .5 – 2.0 ACP 4.9 – 5.0

T r yps in 8. 0 ALP 9.5 – 10 Amylase 6.8–6.9

7. WHAT WHAT IS COMP COMPETIT ETITIVE IVE INHIB INHIBITI ITION ON AND AND GIVE EXAM EXAMPLES PLES? ? Ans.

In competitive inhibition, the inhibitor inhibitor closely resemble the structure of substrate. substrat e. It combines with enzyme and forms E.I. complex thus preventing products formation.

Enzymes

23

Examples:

8.

Succinate dehydrogenase (SDH) inhibited by Malonate Xanthine oxidase (X.O.) inhibited by allopurinol allopurinol Folate reductase reductas e inhibited by methotrexate. methotre xate.

WHAT WHAT IS ALLOST ALLOSTERI ERIC C INHIBITIO INHIBITION? N? GIVE EXAMPL EXAMPLE E OF ALLOSTER ALLOSTERIC IC INHIBITOR INHIBI TOR AND A ND ALLOSTERIC ALLOSTERI C ACTIVAT ACTIVATOR? OR? Ans.

The inhibitor inhibitor binds with enzyme at allosteric site and prevents prevents the binding of substrate substr ate at substrate binding site. Enzyme Example :

Inhibitor

Aspartate transcarbamoylase HMG–CoA reductase Activator

Enzyme

1. N– Acetyl glutamate 2. Acetyl CoA 9.

CT P Cholesterol Carbamoyl phosphate synthase Pyruvate carboxylase

NAME NAM E THE THE CARDIAC CARDIAC E ENZ NZYM YMES. ES. HOW HOW DO YOU YOU INTER INTERPRET PRET THEM THEM IN AMI? AMI? Ans.

LDH, C.K. AST Ist C.K. is raised (MBCK clinches the diagnosis). Enzyme

Start–rising

Peak

Persists upto

C.K. 4 hrs 24 hrs 48–72 hours AST 12 hrs 24 hrs 4th day LDH 24 hrs . 72 hrs 8–10 days Flipped LDH (Ratio of LDH1:LDH2 >1) clinches the diagnosis of AMI. 10. NAME THE ENZYMES WHICH WHICH ARE RAISED IN HEPAT HEPATOCELLULAR OCELLULAR JAUNDICE. Ans.

ALT and AST.

11. NAME THE ENZYMES ENZYMES WHICH ARE RAISED IN OBSTRUCTIVE OBSTRUCTIVE JAUNDICE. JAUNDICE. Ans.

ALP, 5-NT, GGT.

12. 12. NAME THE ENZYME WHICH IS RAISED IN ACUTE PANCREATITIS? Ans.

Serum amylase.

13. 13. WHAT WHAT ARE ISO-EN ISO-ENZYME ZYMES? S? Ans.

Iso-enzymes are physically distinct forms of same catalytic activity. They differ in physical characters (like electrophoretic mobility) and chemical composition and are obtained from different sources.


24

14. GIVE EXAMPLES EXAMPLES OF ISO–ENZYMES? ISO–ENZYMES? Ans.

LDH (1 to 5) CK (1 to 3) LDH1 – H4 LDH4 – HM HM3

LDH2 – H3M LDH5 – M4

LDH3 – H2M2

15. CLASSIFY THE ENZYMES ACCORDIN ACCORDING G TO I.U.B. SYSTEM SYSTEM OF CLASSIFICATION? Ans. S l . No .

According to I.U.B. system enzymes are classified into 6 major classes. These are: Major class

A c ti o n

Examp le

1.

O x i d o r e d u ct a s e s

C at a l yz e th e ox i d at i on r ed u c ti o n reactions

Alcohol dehydrogenase

2.

Tr an sf era se s

Transfer the groups from one substrate to other substrate

Aspartate amino transferase

3.

Hyd rol ase s

Breaks the bonds by the introduction of water molecule

Amylase

4.

Lyases

Cleaves the substrate other than hydrolysis

Aldolase

5.

Isome rases

In te rco nv ert s var io us is ome rs .

Phosphotrio isomerase

6.

Ligases

Ca taly se t he sy nthe tic reac tion s by break down of phosphate bonds

Glutamine synthetase

16. 16. WHAT IS THE MECHANISM MECHANISM OF ACTION OF ENZYME? Ans.

According to MICHAELIS-MENTEN theory, enzyme combines with the substrate and forms enzyme – substrate complex, which immediately breaks down into products and liberates the enzyme. E+S ( E nz ym e )

E s c o m p le x (Substrate)

P+E (Products)

17. 17. WHAT IS CO-ENZYME? GIVE EXAMPLES OF CO-ENZYMES? Ans.

The enzymes become active only when they combine with the prosthetic groups. The prosthetic group is called co-enzyme Apo enzyme + Co-enzyme – Holo enzyme (Inactive) (Prosthetic group) – (Active)

Enzymes

25

The co-enzymes are non-protein moieties, dialisable, heat stable and low molecular weight substances. Most of the co-enzymes consist of one of the B-group B-group vitamins. Co-enzyme TDP FMN ü ý FAD þ

B-group vitamins B1

–

B2

NAD NAD ü ý NADP þ

–

Niacin

–

18. 18. WHAT ARE THE METHODS OF REGULATION OF ENZYME ACTIVITY? ACTIVITY? Ans.

Activation (a) Allo Allost ster eric ic regu regula lati tion on Inhibition Phosphorylation (b) Cova Covale lent nt modi modifi fica cati tion on Dephosphorylation (c) Induct Inductio ionn and and repres repressio sion. n.

Viva Voce Orals in Biochemistry

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