CLS Aipmt 15 16 XIII Zoo Study Package 2 Set 1 Chapter 3

Chapter

3

Biomolecules Solutions SECTION - A Objective Type Questions 1.

Which of the following is not strictly a biomacromolecule? (1) Proteins

(2) Lipids

(3) Polysaccharides

(4) Nucleic acid

Sol. Answer (2) Because the molecular weight of lipids does not exceed 800 Da but they come under the acid insoluble fraction. 2.

A secondary metabolite that acts as a toxin is (1) Carotenoids

(2) Curcumin

(3) Abrin

(4) Monoterpenes

(3) Gum

(4) Abrin

(3) An oligosaccharide

(4) A homopolymer

Sol. Answer (3) Carotenoid – Pigment Curcumin – Drug Monoterpenes – Terpenoids 3.

A secondary metabolite that is alkaloid in nature is (1) Codeine

(2) Anthocyanin

Sol. Answer (1) Anthocyanin – Pigment Gum – Polymeric substance Abrin – Toxin 4.

Peptidoglycan present in bacterial cell envelope is (1) Made up of cellulose

(2) A heteropolymer

Sol. Answer (2) Peptidoglycan is a heteropolysaccharide made up of two alternate amino sugar molecules i.e., N-acetyl glucosamines and N-acetyl muramic acid. 5.

In glycine the R group is replaced by (1) A methyl group

(2) Hydroxy methyl

(3) A carboxylic group

(4)

A hydrogen

Sol. Answer (4) Glycine is a simplest amino acid in which R-group is replaced by hydrogen. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

2 6.

Biomolecules

Solution of Assignment

Zwitterions are ionized species of (1) Acidic amino acids

(2) Basic amino acids

(3) Neutral amino acids

(4) All of these

Sol. Answer (4) Since amino acid can carry both +ve and –ve charges simultaneously, hence they are termed as Zwitterions. 7.

Non-essential amino acids (1) Must be obtained from food

(2) Are synthesized in our body

(3) Are not needed in our diet

(4) Both (2) & (3)

Sol. Answer (4) Because non-essential amino acid are synthesised in our body, so there is no need to take it in our diet. 8.

The most abundant protein in animal world is (1) Chitin

(2) Collagen

(3) Peptidoglycan

(4) Hyaluronic acid

Sol. Answer (2) Because chitin, peptidoglycan and hyaluronic acid, are polysaccharide, not protein. 9.

Proteins which catalyse biochemical reactions in the living world are known as (1) Enzymes

(2) Hormones

(3) Antibodies

(4) Receptor

Sol. Answer (1) Because almost all enzymes are protein which catalyse the biochemical reaction. Hormones can be steroid and amino-acid derivative also. Antibodies are glycoprotein. Receptors are associated with sensory reception. 10. The amino acids in a protein are held together by (1) Glycosidic bond

(2) Phosphodiester bond

(3) Peptide bond

(4) Hydrogen bond

Sol. Answer (3) Glycosidic bond : Formed between the two monosaccharides (sugar molecule). Phosphodiester bond : This bond is formed in nucleic acid i.e. DNA or RNA, between the phosphate and hydroxyl group of sugar. 11. A protein which exhibits  pleated structure is (1) Fibroin

(2) Haemoglobin

(3) Enzyme

(4) -Keratin

(3) Keratin

(4) All of these

Sol. Answer (1) Haemoglobin  Quaternary structure Enzymes  Tertiary structure -Keratin  Secondary structure 12. An example of protein with quaternary structure is (1) Myoglobin

(2) Haemoglobin

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

3

Sol. Answer (2) Myoglobin : Tertiary structure Keratin : Secondary structure 13. Lecithin is a (1) Type of wax

(2) Phospholipid

(3) Oil

(4) Simple fatty acid

Sol. Answer (2) Because lecithin made up of a molecule of glycerol, a phosphate group, 2 fatty acid molecules, choline (N-containing alcohol molecule) O

CH2

O R2

C

O

O

CH CH2

C

R1

O O

P

CH3 O

CH2

CH2

N

OH Lecithin

CH3 CH3

14. Lipids that insulate the nerve fibre are (1) Lecithin

(2) Cholesterol

(3) Suberin

(4) Glycolipids

Sol. Answer (4) The glycolipids are present in myelin sheath of nerve fibres. 15. The pentose sugar present in RNA is (1) Galactose

(2) Sucrose

(3) Ribose

(4) Fructose

Sol. Answer (3) Ribose is a monosaccharide sugar present in RNA. Deoxyribose is a derived monosaccharide sugar present in DNA. 16. Nucleoside is (1) Sugar + Nitrogenous base

(2) Sugar + Phosphate

(3) Nitrogenous base + Phosphate

(4) Purine + Pyrimidine

Sol. Answer (1) Nucleoside = Sugar + Nitrogenous base Nucleotide = Sugar + Nitrogenous base + Phosphate 17. In B-DNA, one full turn of the helical strand contains (1) 11 base pairs

(2) 8 base pairs

(3) 10 base pairs

(4) 9 base pairs

Sol. Answer (3) A form  11 base pairs B form  10 base pairs C form  9 base pairs D form  8 base pairs 18. In a DNA molecule adenine of one strand base pair with ____ on the other strand (1) Guanine

(2) Thymine

(3) Cytosine

(4) Both (1) & (3)

Sol. Answer (2) In DNA, A

T, C

G

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

4

Biomolecules

Solution of Assignment

19. In B-DNA, the rise per base pair would be (1) 4.3 Å

(2) 2.4 Å

(3) 3.4 Å

(4) 4.2 Å

Sol. Answer (3) In B-DNA, the rise per base pair would be 3.4 Å. 20. The nitrogenous bases of the two strands of DNA are joined by (1) Phosphodiester bond

(2) Hydrogen bond

(3) Glycosidic bond

(4) Peptide bond

Sol. Answer (2) Hydrogen bond : The bond formed between two polynucleotide strands of DNA. Glycosidic bond : Formed between the two monosaccharides (sugar). Peptide bond : Formed between 2 amino acids. Phosphodiester bond : Bond formed in nucleic acid i.e. DNA or DNA, between the phosphate and hydroxyl group of sugar. 21. The genetic material of Tobacco mosaic virus is (1) RNA

(2) DNA

(3) Protein

(4) NADPH

(3) Griffith

(4) Robert Brown

Sol. Answer (1) The genetic material of TMV is ssRNA. 22. The double helix model of DNA was proposed by (1) Berzelius

(2) Watson and Crick

Sol. Answer (2) The double helix model of DNA was proposed by Watson and Crick. 23. In the 5 end of a DNA molecule (1) The fifth carbon of pyrimidine base is free

(2) The fifth carbon of purine base is free

(3) The fifth carbon of pentose sugar is free

(4) Both (1) & (3)

Sol. Answer (3) One end of the strand is called 5 end where the fifth carbon of the pentose sugar is free and the other is called 3'end where the third carbon of pentose sugar is free. 24. The pitch of the B-DNA is (1) 36 Å

(2) 3.4 Å

(3) 34 Å

(4) 3.6 Å

Sol. Answer (3) In B-DNA, one turn of helical strand has 10 nucleotides. The base pairs of DNA are stacked 3.4 Å part.  Pitch of the DNA = 3.4 Å × 10 = 34 Å 25. The primary precursor for the production of cholesterol in our body is (1) Acetic acid

(2) Citric acid

(3) Ethyl alcohol

(4) Methanol

Sol. Answer (1) In biosynthetic pathway or anabolic pathway, the acetic acid is converted into cholesterol in liver. 26. Adenosine triphosphate is a (1) Protein

(2) Derived monosaccharide

(3) Simple lipid

(4) Nucleotide

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

5

Sol. Answer (4) Because ATP is made up of adenine, ribose sugar and three phosphate. 27. Metabolic flux is regulated by (1) Enzymes

(2) Sugars

(3) Phospholipids

(4) Sterols

Sol. Answer (1) Metabolic flux is the rate of turn over of molecules through a metabolic pathway. 28. In DNA, uracil is replaced by (1) Thymine

(2) Thiamine

(3) Cytosine

(4) Adenine

(3) Oligosaccharide

(4) Simple lipids

Sol. Answer (1) In DNA, thymine is present. In RNA, uracil is present. 29. Ribozymes are _______ that behave like enzymes (1) Proteins

(2) Ribonucleic acids

Sol. Answer (2) Because ribozymes are the RNA which act as an enzyme. 30. The most abundant enzyme in the biosphere is (1) Collagen

(2) RuBisCO

(3) Trypsin

(4) Insulin

Sol. Answer (2) Ribulose bisphosphate carboxylase-oxygenase (RuBisCO) is the most abundant protein in the whole of the biosphere. 31. What is the fate of pyruvic acid under anaerobic conditions in our body? (1) It gets converted into methyl alcohol (2) It gets converted into acetyl CoA (3) It gets converted into lactic acid (4) It gets converted into glycogen Sol. Answer (3) Under anaerobic condition, the glucose is converted into lactic acid in muscle. 32. Organic compounds that are tightly bound to apoenzyme is (1) Prosthetic group

(2) Apoenzyme

(3) Metal ions

(4) Co-enzymes

Sol. Answer (1)

Enzyme (Holoenzyme) Protein part (Apozyme)

Non-protein part (Cofactor) Organic compound

Tightly bound (Prosthetic group)

Inorganic

Loosely bound (Co-enzyme)

It includes metal ions 2+ 2+ i.e., Zn , Mg , etc.

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

6

Biomolecules

Solution of Assignment

33. A competitive inhibitor of succinate dehydrogenase is (1) p-amino benzoic acid

(2) Malonate

(3) Succinate

(4) Sulphanilamide

Sol. Answer (2) Competitive inhibitor of succinate dehydrogenase are are malonate, while succinate is substrate. 34. Bacterial pathogen can be controlled by (1) p-amino benzoic acid (2) Malonate

(3) Sulphanilamide

(4) All of these

Sol. Answer (3) Sulpha drugs are derivatives of sulphanilamide, which inhibit the synthesis of folic acid in bacteria by competing with p-amino benzoic acid (PABA) required for the synthesis of folic acid by folic acid synthetase. 35. Non-protein part of enzyme is known as (1) Apoenzyme

(2) Cofactor

(3) Inorganic catalyst

(4) Active site

Sol. Answer (2)

Enzyme (Holoenzyme) Protein part (Apozyme)

Non-protein part (Cofactor)

Organic compound

Tightly bound (Prosthetic group)

Inorganic

Loosely bound (Co-enzyme)

It includes metal ions 2+ 2+ i.e., Zn , Mg etc.

SECTION - B Objective Type Questions 1.

In maltose glycosidic bond is formed between (1) Carbon 1 of one glucose molecule and carbon 4 of second glucose molecule (2) Carbon 2 of one glucose molecule and carbon 3 of second glucose molecule (3) Carbon 3 of one glucose molecule and carbon 4 of second glucose molecule (4) Carbon 1 of one glucose molecule and carbon 6 of second glucose molecule

Sol. Answer (1)

CH2OH H OH

CH2OH O

H

H OH

H

H

OH

O

H

1

4

O

H OH

H

H

OH

H OH

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

2.

Biomolecules

7

The given structure represents a monosaccharide known as

O

HOCH2 OH

OH OH

(1) Ribose

OH

(2) Glucose

(3) Fructose

(4) Raffinose

Sol. Answer (1) Ribose is a pentose sugar. 3.

Chitin present in the exoskeletons of arthropods is (1) Protein

(2) Polysaccharide

(3) Lipid

(4) Derived monosaccharide

Sol. Answer (2) Because chitin is a polymer of nitrogen containing glucose derivative known as N-acetyl glucosamine. 4.

The given amino acid is _______ in nature.

COOH H2N–C–H CH2.COOH (1) Acidic

(2) Basic

(3) Neutral

(4) Aromatic

Sol. Answer (1) Because given structure has an extra carboxylic group. 5.

The structure of protein which gives a three dimensional view is (1) Primary structure

(2) -helix

(3) -pleated sheet

(4) Tertiary structure

Sol. Answer (4) Tertiary structure of protein gives a three dimensional view. 6.

The product of the given reaction would be a

CH2OH CHOH

(1) Monoglyceride

HOOC–R1 +

HOOC–R2

CH2OH

HOOC–R3

Glycerol

Fatty acid

(2) Diglyceride

(3) Triglyceride

(4) Both (1) & (3)

Sol. Answer (3) Because given structure has three fatty acids. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

8 7.

Biomolecules

Solution of Assignment

The given fatty acid is known as CH3(CH2)7 CH = CH (CH2)7 COOH (1) Palmitic acid

(2) Oleic acid

(3) Stearic acid

(4) Arachidonic acid

Sol. Answer (2) (CH3)2(CH2)7CH = CH(CH2)7COOH – Oleic acid CH3(CH2)14COOH – Palmitic acid CH3(CH2)16COOH – Stearic acid Arachidonic acid : CH3(CH2)4CH = CHCH2CH = CHCH2CH = CHCH2CH = CH(CH2)3COOH (20 C) 8.

Nucleic acids exhibit (1) Secondary structure

(2) Tertiary structure

(3) Quaternary structure (4) Both (2) & (3)

Sol. Answer (1) Because nucleic acid (DNA) has double helical structure and it comes under secondary structure. 9.

The backbone of a DNA molecule is made up of (1) Adenine and guanine

(2) Sugar-phosphate-sugar chain

(3) Cytosine and thymine

(4) All of these

Sol. Answer (2) Backbone of a DNA molecule is made up of Sugar – Phosphate – Sugar chain

P 5'

3' S

A

T

S

P

P S

G

C

S

P

P = Phosphate S = Sugar A = Adenine T = Thymine G = Guanine C = Cytosine

P S

T

A

5' P

S 3'

10. If the sequence of bases in one of the DNA strand is A G G A G A A, then the sequence of bases in the other complementary strand of DNA would be (1) C C T T C T T

(2) T C T C T C C

(3) T C C T C T T

(4) C C T C T C T

Sol. Answer (3) Because A = T, G  C 11. Which of the following statements is correct? (1) Biocatalysts accelerate the rate of a given metabolic reaction (2) Biocatalysts are generally protein (3) Enzyme catalyst differ from inorganic catalyst (4) All of these Sol. Answer (4) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

9

12. In which of the following energy is released? (1) Conversion of glucose into pyruvate

(2) Formation of proteins from amino acids

(3) Conversion of glucose into lactic acid

(4) Both (1) & (3)

Sol. Answer (4) Anaerobic combustion of glucose releases energy in the form of ATP. 13. Adenosine Triphosphate (ATP) liberates high energy by the breakdown of (1) Glycosidic bond

(2) Hydrogen bond

(3) Phosphate bond

(4) Both (1) & (3)

Sol. Answer (3) ATP  ADP + iP (inorganic phosphate) 14. Which of the following statements is incorrect w.r.t. inorganic catalysts? (1) They do not occur in living cells

(2) They are not specific for any reaction

(3) They get damaged at high temperature

(4) They work efficiently at high pressure

Sol. Answer (3) Inorganic catalyst work efficiently at high temperature and pressure, but enzyme get damaged at high temperature (i.e. above 40ºC) except thermophillic enzyme. 15. Enzymes catalyse biochemical reactions by (1) Lowering the activation energy (2) Increasing the activation energy (3) Establishing stable bonds with substrate (4) Increasing temperature Sol. Answer (1) Enzyme lowers the activation energy and thus increases the rate of reaction. 16. Read the following : (a) Low temperature preserves the enzyme (b) Enzyme activity increases above optimum temperature (c) Enzyme gets denatured at high temperature (d) Competitive inhibitor competes with the product formed Which of the following statement are true? (1) (a) & (c)

(2) (b) & (d)

(3) (c) & (d)

(4) (a) & (b)

Sol. Answer (1) (b)

Statement is wrong because enzyme activity decreases above optimum temperature.

(d)

Statement is wrong because competitive inhibitor completes with substrate for active site, not with product formed.

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

10

Biomolecules

Solution of Assignment

17. Enzymes catalysing the breakdown of larger molecules into smaller molecules are (1) Hydrolases

(2) Isomerases

(3) Ligase

(4) Both (1) & (3)

Sol. Answer (1) Because, hydrolases are the enzyme which catalyse the breakdown of larger molecules into smaller molecules with the addition of water. Isomerases : They are the enzymes which catalyse the rearrangement of molecular structure to form isomers. Ligases : They help in joining C – O, C – S, C – N etc. bonds by using energy of ATP. 18. Michaelis constant (Km) value of enzyme is substrate concentration at which velocity of reaction is (1) Vmax

(2) One third Vmax

(3) Half Vmax

(4) One fifth Vmax

Sol. Answer (3)

Vmax Velocity of reaction

Vmax 2

Km

[s]

19. _______ catalyses covalent bonding of two substrates. (1) Invertase

(2) Amylase

(3) Glutamate pyruvate transaminase

(4) PEP carboxylase

Sol. Answer (4) Because this is a ligase enzyme. 20. The enzyme that catalyses the conversion of glucose-6-phosphate into fructose-6-phosphate is (1) A ligase

(2) An isomerase

(3) A lyase

(4) A hydrolase

Sol. Answer (2) isomerase   Glucose-6-phosphate   fructose-6-phosphate

21. Study the following statements : (a) The substrate binds to the active site of the enzyme (b) Enzymes isolated from thermophilic organisms get denatured at 50°C (c) The active site of enzyme breaks the chemical bonds of the product (d) Prosthetic groups are tightly bound to the apoenzyme Select the option which includes all correct statements : (1) (a) & (c)

(2) (c) & (d)

(3) (b) & (c)

(4) (a) & (d)

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

11

Sol. Answer (4) (b) Statement is wrong, because thermophillic enzymes don't get denatured at high temperature, they work efficiently at high temperature (750ºC). (c) Statement is wrong, because enzyme don't break chemical bonds of product but of substrate. 22. One of the following is correct sequence of carbohydrates in the order of increasing complexity of chemical structure (1) Sucrose, starch, oligosaccharide, maltose, triose

(2) Triose, maltose, sucrose, oligosaccharide, starch

(3) Triose, glucose, maltose, oligosaccharide, starch

(4) Oligosaccharide, triose, starch, sucrose, maltose

Sol. Answer (3) Triose – 3 C containing molecule Glucose – Monosaccharide (6 C) Maltose – Disaccharide (2 Monosaccharide units) Oligosaccharide – (2–10) Monosaccharide units Starch – Polysaccharides 23. Glucose is stored as glycogen in (1) Pancreas

(2) Bone

(3) Kidney

(4) Liver

Sol. Answer (4) Because liver is a gland where glucose get converted into glycogen. 24. A cellulose molecule is formed by the polymerisation of glucose. The number of glucose molecules present in a cellulose is (1) 600

(2) 6000

(3) 60,000

(4) 60

Sol. Answer (2) A cellulose molecule is formed by polymerisation of 6000 molecules of  glucose. 25. Which of the following are basic amino acids? (1) Glycine and Alanine

(2) Lysine and Arginine

(3) Glutamic acid and Aspartic acid

(4) Histidine and Proline

Sol. Answer (2) Lysine and Arginine  Basic amino acid  +ve charged Glutamic acid and Aspartic acid  Acidic amino acid  –ve charge 26. Which of the following aminoacids is involved in the formation of Heme? (1) Tryptophan

(2) Tyrosine

(3) Glycine

(4) Histidine

Sol. Answer (3) Because amino-acid glycine provides nitrogen and carbon atoms for the synthesis of heme. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

12

Biomolecules

Solution of Assignment

27. Which one of the following is alcoholic amino acid pair? (1) Tyrosine and serine

(2) Threonine and serine

(3) Phenylalanine and tyrosine

(4) Tryptophan and phenylalanine

Sol. Answer (2) Because these two amino acids contain alcohol group. 28. Lysine is an essential amino acid because (1) It is very rare (2) It has a high nutritive value (3) It is an important constituent of all proteins (4) It is not formed in the body and has to be provided through diet Sol. Answer (4) Essential amino acids are not synthesised in body. 29. A nucleotide is made up of (1) (Base-Sugar-Phosphate)n (2) Nitrogenous base and sugar (3) Nitrogenous base, sugar and phosphate (4) Phosphate and N-base Sol. Answer (3) Nucleotide is made up of nitrogenous base, sugar and phosphate Nucleic acid(DNA, RNA) – Made up of (Base – Sugar – Phosphate)n 30. RNA is a polymer of (1) Ribonucleotides

(2) Deoxyribonucleotides

(3) Deoxyribonucleosides

(4) Ribonucleosides

Sol. Answer (1) It is made up of ribose sugar, phosphate and nitrogenous bases (A, U, G, C). 31. t-RNA constitutes about (1) 70-80% of the total RNA

(2) 15% of the total RNA

(3) 5% of the total RNA

(4) 1-3% of the total RNA

Sol. Answer (2) r-RNA  70 – 80% of total RNA t-RNA  15% of total RNA m-RNA  2–5% of total RNA 32. In DNA, cytosine pairs with (1) Guanine

(2) Thymine

(3) Adenine

(4) Uracil

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

13

Sol. Answer (1) A = T (2 hydrogen bonds between adenine and thymine) G  C (3 hydrogen bonds between cytosine and guanine) 33. All enzymes are proteins, except (1) Trypsin

(2) Pepsin

(3) Steapsin

(4) Ribozyme and Ribonuclease-P

Sol. Answer (4) Because, all enzymes are proteins, except, ribozyme and ribonuclease-P, which are RNA act as enzyme. 34. Cyanide kills an animal by (1) Killing the brain cells (2) Competitive inhibitor of enzyme cytochrome oxidise (3) Inhibiting cytochrome oxidase, a mitochondrial enzyme essential for cellular respiration by Non-competitive inhibition (4) Killing the cells of cardiac muscles Sol. Answer (3) Cyanide poisoining is an example of non competitive inhibition of enzyme. 35. Electron transferring enzymes belong to (1) Transferases

(2) Oxidoreductases

(3) Lyases

(4) Isomerases

Sol. Answer (2) The main enzymes of electron transport chain are dehydrogenase, reductase, cytochrome oxidase, which belongs to the oxidoreductase class of an enzyme.

SECTION - C Previous Years Questions 1.

In sea urchin DNA, which is double stranded 17% of the bases were shown to be cytosine. The percentages of the other three bases expected to be present in this DNA are [AIPMT-2015] (1) G 8.5%, A 50%, T 24.5%

(2) G 34%, A 24.5%, T 24.5%

(3) G 17%, A 16.5%, T 32.5%

(4) G 17%, A 33%, T 33%

Sol. Answer (4) 2.

Which one of the following statements is incorrect?

[AIPMT-2015]

(1) The presence of the competitive inhibitor decreases the Km of the enzyme for the substrate (2) A competitive inhibitor reacts reversibly with the enzyme to form an enzyme-inhibitor complex (3) In competitive inhibition, the inhibitor molecule is not chemically changed by the enzyme (4) The competitive inhibitor does not affect the rate of breakdown of the enzyme-substrate complex Sol. Answer (1) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

14 3.

Biomolecules

Solution of Assignment

Select the option which is not correct with respect to enzyme action:

[AIPMT-2014]

(1) Substrate binds with enzyme at its active site (2) Addition of lot of succinate does not reverse the inhibition of succinic dehydrogenase by malonate (3) A non - competitive inhibitor binds the enzyme at a site distinct from that which binds the substrate (4) Malonate is a competitive inhibitor of succinic dehydrogenase Sol. Answer (2) Inhibition of succinic dehydrogenase by malonate is an example of competitive inhibition. This is reversible reaction. On increasing the substrate (succinate) concentration the effect of inhibitor is removed and Vmax remain same. 4.

Which one of the following is a non - reducing carbohydrate? (1) Maltose

(2) Sucrose

(3) Lactose

[AIPMT-2014] (4) Ribose 5-phosphate

Sol. Answer (2) Lactose, Maltose, Ribose 5-phosphate all are reducing sugars. 5.

A phosphoglyceride is always made up of:

[NEET-2013]

(1) Only an unsaturated fatty acid esterified to a glycerol molecule to which a phosphate group is also attached (2) A saturated or unsaturated fatty acid esterified to a glycerol molecule to which a phosphate group is also attached (3) A saturated or unsaturated fatty acid esterified to a phosphate group which is also attached to a glycerol molecule (4) Only a saturated fatty acid esterified to a glycerol molecule to which a phosphate group is also attached Sol. Answer (2) 6.

The essential chemical components of many coenzymes are: (1) Nucleic acids

(2) Carbohydrates

(3) Vitamins

[NEET-2013] (4) Proteins

Sol. Answer (3) 7.

Transition state structure of the substrate formed during an enzymatic reaction is : (1) Permanent but unstable

(2) Transient and unstable

(3) Permanent and stable

(4) Transient but stable

[NEET-2013]

Sol. Answer (2) 8.

Macromolecule chitin is

[NEET-2013]

(1) Phosphorus containing polysaccharide

(2) Sulphur containing polysaccharide

(3) Simple polysaccharide

(4) Nitrogen containing polysaccharide

Sol. Answer (4) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

9.

Biomolecules

15

Which one out of A – D given below correctly represents the structural formula of the basic amino acid? A NH2 | H C COOH H | CH2 | CH2 | C OH O

B

C

NH2 | C COOH | CH2 | OH

CH2OH | CH2 H | CH2 | NH2

D NH2 C

COOH

CH2 | CH2 | CH2 | CH2 | NH2

[AIPMT (Prelims)-2012] (1) A

(2) B

(3) C

(4) D

Sol. Answer (4) Because in 'D', it has an extra amino group, because of which it carry +ve charge which comes under basic amino acid. 10. Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds in the living tissues. Identify the category shown and the one blank component “X” in it

HOCH2 O "X"

OH OH [AIPMT (Prelims)-2012] Category

Component

(1) Nucleotide

Adenine

(2) Nucleoside

Uracil

(3) Cholesterol

Guanin

(4) Amino acid

NH2

Sol. Answer (2) Because the given structure doesn't have phosphate group so it is nucleoside and "X" is uracil because it consists ribose sugar. 11. For its activity, carboxypeptidase requires (1) Copper

(2) Zinc

[AIPMT (Mains)-2012] (3) Iron

(4) Niacin

Sol. Answer (2) Carboxypeptidase require Zn2+ metal ion. 12. Which one of the following biomolecules is correctly characterised?

[AIPMT (Mains)-2012]

(1) Alanine amino acid – Contains an amino group and an acidic group anywhere in the molecule (2) Lecithin – a phosphorylated glyceride found in cell membrane (3) Palmitic acid – an unsaturated fatty acid with 18 carbon atoms (4) Adenylic acid – adenosine with a glucose phosphate molecule Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

16

Biomolecules

Solution of Assignment

Sol. Answer (2) Option (1) is wrong, because alanine contains an amino group and an acidic group at -carbon in the molecule. Option (3) is wrong, because palmitic acid is a saturated fatty acid with 16C atoms. Option (4) is wrong, because adenylic acid is form by adding phosphate group to adenosine.

y-axis

13. The curve given below show enzymatic activity with relation to three conditions (pH, temperature and substrate concentration)

x-axis

What do the two axes (x and y) represent? x-axis

[AIPMT (Prelims)-2011]

y-axis

(1) Enzymatic activity

Temperature

(2) Enzymatic activity

pH

(3) Temperature

Enzyme activity

(4) Substrate concentration

Enzymatic activity

Sol. Answer (3) y Enzymatic activity

pH or Temperature

x

14. Which one of the following structural formulae of two organic compound is correctly identified along with its related function O CH2—O—C—R

O A

R2—C—O—CH

O

CH2—O—P—O—CH2—CH2 N+

OH NH2 B

N N

CH3

CH3 CH3

N NH

[AIPMT (Prelims)-2011]

(1) A : Lecithin – a component of cell membrane (2) B : Adenine – a nucleotide that makes up nucleic acids (3) A : Triglyceride – major source of energy (4) B : Uracil – a component of DNA Sol. Answer (1) Given structure 'A' is Leathin and 'B' is Adenine. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

17

15. The figure given below shows the conversion of a substrate into product by an enzyme. In which one of the four options (1–4) the components of reaction labelled as A, B, C and D are identified correctly?

A C D B Substrate Product [AIPMT (Mains)-2010]

Progress of Reaction (1)

A Potential energy

B Transition state

(2)

Transition state

Potential energy

(3)

Activation energy without enzyme

Transition state

(4)

Activation energy with enzyme

Transition state

C Activation energy with enzyme Activation energy without enzyme Activation energy with enzyme

D Activation energy without enzyme Activation energy with enzyme Potential energy

Activation energy without enzyme

Potential energy

Sol. Answer (2) 16. Three of the following statements about enzymes are correct and one is wrong. Which one is wrong? [AIPMT (Mains)-2010] (1) Enzymes require optimum pH for maximal activity (2) Enzymes are denatured at high temperature but in certain exceptional organisms they are effective even at temperatures 80°-90°C (3) Enzymes are highly specific (4) Most enzymes are proteins but some are lipids Sol. Answer (4) Because most enzymes are protein but some are RNA i.e. Ribonuclease-P and ribozyme. 17. Which one of the following pairs is wrongly matched? (1) Alcohol

– Nitrogenase

(2) Fruit juice

– Pectinase

(3) Textile

– Amylase

(4) Detergents

– Lipase

[AIPMT (Prelims)-2009]

Sol. Answer (1) 18. Carbohydrates are commonly found as starch in plant storage organs. Which of the following five properties of starch (a - e) make it useful as a storage material? [AIPMT (Prelims)-2008] a. Easily translocated

b. Chemical non-reactive

c. Easily digested by animals

d. Osmotically inactive

e. Synthesized during photosynthesis The useful properties are (1) Both a & e

(2) Both b & c

(3) Both b & d

(4) a, c & e

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

18

Biomolecules

Solution of Assignment

Sol. Answer (3) Carbohydrates like glycogen and starch are relatively easy to store because of the following advantages. 

They are stored in bulk.



They are chemically non-reactive.



They are osmotically inactive.

19. An organic substance bound to an enzyme and essential for its activity is called (1) Coenzyme

(2) Holoenzyme

(3) Apoenzyme

[AIPMT (Prelims)-2006] (4) Isoenzyme

Sol. Answer (1) Organic compound which bound to an enzyme is either coenzyme or prosthetic group. 20. Telomerase is an enzyme which is a : (1) Repetitive DNA

(2) RNA

[AIPMT (Prelims)-2005] (3) Simple protein

(4) Ribonucleoprotein

Sol. Answer (4) 21. Which of the following is the simplest amino acid? (1) Tyrosine

(2) Asparagine

[AIPMT (Prelims)-2005] (3) Glycine

(4) Alanine

Sol. Answer (3) Because in glycine, the R-group is replaced by hydrogen.

H HOOC

C

NH2

H = R group 22. Enzymes, vitamins and hormones can be classified into a single category of biological chemicals, because all of these [AIPMT (Prelims)-2005] (1) Enhance oxidative metabolism (2) Are conjugated proteins (3) Are exclusively synthesized in the body of a living organism as at present (4) Help in regulating metabolism Sol. Answer (4) (1) is incorrect, because enzymes can both enhance and inhibit the oxidative metabolism. (3) is incorrect, different enzymes, vitamins and hormones are synthesized in the body at different situations. (2) is incorrect, because all hormones and enzymes are not conjugated protein. 23. Carbohydrates, the most abundant biomolecules on earth, are produced by

[AIPMT (Prelims)-2005]

(1) All bacteria, fungi and algae

(2) Fungi, algae and green plant cells

(3) Some bacteria, algae and green plant cells

(4) Viruses, fungi and bacteria

Sol. Answer (3) Autotrophic organism produce glucose by photosynthesis which is a carbohydrate. Heterotrophic are dependent on autotrophes and don't produce carbohydrate and fungi are saprobic organisms. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

24. Which of the following statements regarding enzyme inhibition is correct ?

19

[AIPMT (Prelims)-2005]

(1) Non-competitive inhibition of an enzyme can be overcome by adding large amount of substrate (2) Competitive inhibition is seen when a substrate competes with an enzyme for binding to an inhibitor protein (3) Competitive inhibition is seen when the substrate and the inhibitor compete (4) Non-competitive inhibitors often bind to the enzyme irreversibly Sol. Answer (3) (1) is wrong, because competitive inhibition of an enzyme can be overcome by adding large amount of substrate. (2) is wrong, because competitive inhibtion is seen when a substrate competes with an inhibitor for binding to the active site of enzyme. (4) is wrong, because non-competitive inhibitor often bind to enzyme irreversibly. 25. The catalytic efficiency of two different enzymes can be compared by the

[AIPMT (Prelims)-2005]

(1) The Km value

(2) The pH optimum value

(3) Formation of the product

(4) Molecular size of the enzyme

Sol. Answer (1) 26. The four elements that make up 96% of all the elements found in a living system are (1) C, H, O and P

(2) C, N, O and P

(3) H, O, C and N

(4) C, H, O and S

Sol. Answer (3) % weight of human body : C = 18.5; O = 65.0; N = 3.3; H = 0.5 27. High cholesterol patients are advised to use (1) Ghee, butter and oils

(2) Groundnut oil, margarine and vegetable oils

(3) Fatty oil and butter

(4) Cheese, dalda and ghee

Sol. Answer (2) Because these oils contain polyunsaturated fatty acids (i.e. more than one double bond) and they are usually recommended by doctors to person having hypertension, high blood cholesterol and other cardiovascular diseases. 28.

Essential amino acid is (1) Phenylalanine

(2) Glycine

(3) Aspartic acid

(4) Serine

Sol. Answer (1) Essential amino acids are leucine, isoleucine, valine, tryptophane, phenylalanine, lysine and methionine. 29. Lipids are insoluble in water because lipid molecules are (1) Hydrophilic

(2) Hydrophobic

(3) Neutral

(4) Zwitter ions

Sol. Answer (2) Hydro-means water, phobic - means repeling or hating Lipids are hydrophobic that's why they are not soluble in water. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

20

Biomolecules

Solution of Assignment

30. The major role of minor elements inside living organisms is to act as (1) Co-factors of enzymes

(2) Building blocks of important amino acids

(3) Constituents of hormones

(4) Binders of cell structure

Sol. Answer (1) Minor elements basically includes Zn, Mg, K, Ni, Co, NAD+, NADP+, they all come under the category of cofactor. 31. Nucleotides are building blocks of nucleic acids. Each nucleotide is a composite molecule formed by (1) Base-sugar-phosphate

(2) Base-sugar-OH

(3) (Base-sugar-phosphate)n

(4) Sugar-phosphate

Sol. Answer (1) Nucleotide : Base + Sugar + Phosphate Nucleic acid : (Base + Sugar + Phosphate)n 32. About 98 percent of the mass of every living organism is composed of just six elements including carbon, hydrogen, nitrogen, oxygen and (1) Sulphur and magnesium

(2) Magnesium and sodium

(3) Calcium and phosphorus

(4) Phosphorus and sulphur

Sol. Answer (4) 33. Which of the following is a neutral amino acid? (1) Glutamine

(2) Arginine

(3) Valine

(4) Asparagine

Sol. Answer (3) Glutamine, Asparagine  Amide group containing amino-acid. Arginine  Basic amino acid 34. The most unsaturated fatty acid is (1) Linoleic acid

(2) Oleic acid

(3) Linolenic acid

(4) Arachidonic acid

Sol. Answer (4) Arachidonic acid : It is 20 C containing unsaturated fattty acid (C20H32O2) with four double bond. Oleic acid – 1 double bond Linoleic acid – 2 double bonds Linolenic acid – 3 double bonds 35. Which of the following is a nucleotide? (1) Thymidine

(2) Cytosine

(3) Thiamine

(4) Uridylic acid

Sol. Answer (4) Thymidine – Nucleoside; Thiamine – Vitamin; Cytosine – Nitrogen base; Uridylic acid – Nucleotide Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

21

36. Which of the following is incorrect regarding the amino acids and their functions? (1) Tyrosine

: Converted into epinephrine hormone and used in the synthesis of melanin pigment

(2) Glycine

: Involved in the formation of heme

(3) Tryptophan

: Helps in the synthesis of auxin hormone

(4) Histidine

: Can be converted into histamine by the removal of amino group

Sol. Answer (4) Histamine is derived from amino acid histidine through decarboxylation. 37. Which of the following is the diagrammatic representation of phospholipid lecithin?

O

O CH2 – O – C – R1 O

CH2 – O – C – R1 O

(1) CH – O – C – R2 O

(2) CH – O – C – R2 O +

CH2 – O – P – O – CH2 – CH2 – NH3

CH2 – O – C – R3

OH

O

O CH2 – O – C – R1 O

O

(3) CH – O – C – R2 O

CH3

(4) R2 – C – O – CH

OH

O

CH2 – O – P – OH

CH2 – O – P – O – CH2 – CH2 –+N – CH3 Sol. Answer (3)

CH2 – O – C – R1

OH

CH3

Lecithin is a phospholipid present in cell membrane 38. Which of the following is not a secondary metabolite of plant cell? (1) Rubber

(2) Chlorophyll

(3) Essential oil

(4) Tannins

(3) Lipid

(4) Protein

Sol. Answer (2) Rubber  Polymeric substance Essential oil Tannins  Pigments 39. The most abundant molecule in cell is (1) Water

(2) Carbohydrate

Sol. Answer (1) Water = 70 – 90% of the total cellular mass Carbohydrate = 3% Lipid = 2% Protein = 10–15% Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

22

Biomolecules

Solution of Assignment

40. Which are the most diverse molecules in the cell? (1) Lipids

(2) Mineral salts

(3) Proteins

(4) Carbohydrates

Sol. Answer (3) Because protein are made up of 20 different amino acids and are heteropolymers of amino acids. 41. The most abundant organic compound on earth is (1) Protein

(2) Cellulose

(3) Lipids

(4) Steroids

(3) Vitamin

(4) Skin pigment

Sol. Answer (2) Plant cell wall is made up of cellulose. 42. Haemoglobin is a type of (1) Carbohydrate

(2) Respiratory pigment

Sol. Answer (2) Because haemoglobin get binds to oxygen and help in the transportation of O2. 43. Collagen is (1) Fibrous protein

(2) Globular protein

(3) Lipid

(4) Carbohydrate

Sol. Answer (1) Because collagen are thread like proteins, which are insoluble in water. 44. Maltose is formed of two molecules of (1) Fructose

(2) Lactose

(3) Glucose

(4) Sucrose

Sol. Answer (3) Fructose = Monosaccharides Lactose = Galactose + Glucose Sucrose = Glucose + Fructose Maltose = Glucose + Glucose 45. A polysaccharide which is synthesized and stored in liver cells is (1) Arabinose

(2) Glycogen

(3) Lactose

(4) Galactose

(3) Brown algae

(4) Green algae

Sol. Answer (2) Glycogen is a polymer of glucose. 46. Agar is commercially obtained from (1) Red algae

(2) Blue-green algae

Sol. Answer (1) Agar is a type of mucopolysaccharide and is obtained from red algae. It is used as culture medium in laboratory. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

23

47. Which of the following groups consists of polysaccharides only? (1) Sucrose, glucose and fructose

(2) Maltose, lactose and fructose

(3) Glycogen, sucrose and maltose

(4) Glycogen, cellulose and starch

Sol. Answer (4) Sucrose  Disaccharide;

Fructose  Monosaccharide

Glucose  Monosaccharide;

Maltose  Disaccharide

Lactose  Disaccharide;

Glycogen  Polysaccharide

Cellulose  Polysaccharide;

Starch  Polysaccharide

48. Lactose is composed of (1) Glucose + galactose

(2) Fructose + galactose

(3) Glucose + fructose

(4) Glucose + glucose

Sol. Answer (1) Lactose is a disaccharide. 49. Cellulose, the most important constituent of plant cell wall is made up of (1) Branched chain of glucose molecules linked by 1, 4 glycosidic bond in straight chain and 1, 6 glycosidic bond at the site of branching (2) Unbranched chain of glucose molecules linked by 1, 4 glycosidic bond (3) Branched chain of glucose molecules linked by 1, 6 glycosidic bond at the site of branching (4) Unbranched chain of glucose molecules linked by 1, 4 glycosidic bond Sol. Answer (2) Glycogen and starch are branched polymer of glucose while cellulose is unbranched polymer of  glucose. 50. A person is eating boiled potato. His food contains (1) Cellulose, which can be digested by cellulase (2) Starch, which cannot be digested (3) Lactose, which cannot be digested (4) DNA, which can be digested by pancreatic DNAase Sol. Answer (4) (1) is wrong, because our gut can't produce cellulose. (2) is wrong because potato contain starch and it can be digested. (3) is wrong, because lactose is absent in potato. 51. Which of the following is a reducing sugar? (1) Galactose

(2) Gluconic acid

(3) -methyl galactoside

(4)

Sucrose

Sol. Answer (1) Because it can reduce Cu2+ ions into Cu+ ion. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

24

Biomolecules

Solution of Assignment

52. An important step in the manufacture of pulp for paper industry is the (1) Preparation of pure cellulose (2) Treatment of wood with chemicals that break down cellulose (3) Removal of oils present in the wood by treatment with suitable chemicals (4) Removal of water from the wood by prolonged heating at approximately 50°C Sol. Answer (1) Because the raw material for paper is cellulose. 53. 1-4 linkages are present in (1) Cellulose

(2) Chitin

(3) Starch

(4) Both (1) & (2)

Sol. Answer (4) Cellulose –

(1, 4) linkage (-glucose)

Chitin

–

(1, 4) linkage (N-acetyl glucosamine)

Starch

–

(1, 4) (in straight chain) of amylose and amylopectin (1, 6) (at branching)

54. Which of the following statements is false? (1) Cellulose is the most abundant organic compound in the biosphere (2) Cellulose is an unbranched polymer with -1,4 glycosidic bonds (3) Rayon and cellophane are chemically similar to cellulose xanthate (4) Cellulose can be digested by the herbivores by -amylase, produced by the glandular cells of their alimentary canal Sol. Answer (4) Cellulose can be digested by the herbivores because they have special type cellulose digesting micro-organism for the digestion of cellulose. 55. If the total amount of adenine and thymine in a double-stranded DNA is 60%, the amount of guanine in this DNA will be (1) 15%

(2) 20%

(3) 30%

(4) 40%

Sol. Answer (2) A + T = 60% Then, C + G = 40%, when 20%  C and 20% = G 56. DNA has equal number of adenine and thymine residues (A = T) and equal number of guanine and cytosine (G = C). These relationships are known as (1) Chargaff's rule

(2) Coulomb's law

(3) Le Chatelier's principle

(4) Van't Hoff plot

Sol. Answer (1) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

25

57. Which of the following pairs of nitrogenous bases of nucleic acids is mismatched with the category mentioned against it? (1) Adenine, Thymine – Purines

(2) Thymine, Uracil – Pyrimidines

(3) Uracil, Cytosine – Pyrimidines

(4) Guanine, Adenine – Purines

Sol. Answer (1) Adenine and Guanine  Purines Thymine, uracil, cytosine  Pyrimidines 58. In a DNA molecule (1) There are two strands which run antiparallel-one in 5  3 direction and other in 3  5 (2) The total amount of purine nucleotides and pyrimidine nucleotides is not always equal (3) There are two strands which run parallel in the 5  3 direction (4) The proportion of adenine in relation to thymine varies with the organism Sol. Answer (1) In DNA, two antiparallel strands are coiled around a common axis. 59. Which purine base is found in RNA? (1) Thymine

(2) Uracil

(3) Cytosine

(4) Guanine

Sol. Answer (4) Because Thymine  Present in DNA, not in RNA Cytosine, uracil  Pyrimidine 60. Similarity in DNA and RNA is that (1) Both are polymers of nucleotides

(2) Both have similar pyrimidines

(3) Both have similar sugar

(4) Both are genetic material in man

Sol. Answer (1) Nucleotide is a monomeric unit of nucleic acid. 61. Length of one loop of B-DNA is (1) 3.4 nm

(2) 0.34 nm

(3) 20 nm

(4) 10 nm

Sol. Answer (1) Pitch of B-DNA = 34 Å  3.4 nm (∵ 1 Å = 0.1 nm) 62. Which of the following enzymes is used to join bits of DNA? (1) Ligase

(2) Primase

(3) DNA polymerase

(4) Endonuclease

Sol. Answer (1) Class VI

 Ligase enzyme is used to join the bits of DNA.

Primase

 Is an enzyme use to attach primer.

DNA polymerase  is an enzyme use to add nucleotide in template strand of DNA. Endonuclease

 It is an restriction enzyme which causes internal cleavage of DNA.

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

26

Biomolecules

Solution of Assignment

63. The 3-5 phosphodiester linkages inside a polynucleotide chain serve to join (1) One DNA strand with the other DNA strand

(2) One nucleoside with another nucleoside

(3) One nucleotide with another nucleotide

(4) One nitrogenous base with pentose sugar

Sol. Answer (3) Because nucleotide is a monomeric unit of nucleic acid which are joined together by 3' – 5' phosphodiester bond. 64. ATP is (1) Nucleotide

(2) Nucleoside

(3) Nucleic acid

(4) Vitamin

Sol. Answer (1) Because ATP (Adenosine triphosphate) is made up of adenine, ribose sugar and 3 phosphate groups. 65. The role of an enzyme in a reaction is to/as (1) Decrease activation energy

(2) Increase activation energy

(3) Inorganic catalyst

(4) None of these

Sol. Answer (1)

Potential energy

Transition state Activation energy without enzyme Activation energy with enzyme

Substrate energy

Product Progress of reaction 66. Which of the following factor(s) do(es) not affect enzyme activity? A. Temperature

B. pH

C. Enzyme concentration

D. Product concentration

E. Substrate concentration

F. Activation energy

(1) C only

(2) C & D

(3) D only

(4) F only

Sol. Answer (4) Because enzyme activity is affected by temperature, pH, enzyme concentration, product concentration and substrate concentration. Enzyme are not affected by activation energy but it lowers down the activation energy. 67. A competitive inhibitor of succinic dehydrogenase is (1) Malate

(2) Malonate

(3) Oxaloacetate

(4) Both (2) & (3)

Sol. Answer (4) Competitive inhibitors of succinic dehydrogenase are malate, malonate, oxaloacetate. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

27

68. Which of the following is a typical example of ‘feedback inhibition’? (1) Cyanide and cytochrome reaction (2) Sulpha drugs and folic acid synthesizer bacteria (3) Allosteric inhibition of hexokinase by glucose 6-phosphate (4) Reaction between succinic dehydrogenase and succinic acid Sol. Answer (3) (1) Cyanide and cytochrome reaction  Example of non-competitive inhibition. (2) Sulpha drugs and folic acid synthesizer bacteria are example of non-competitive inhibition. (4) Reaction between succinic dehydrogenase and succinic acid are example of competitive inhibition. 69. Which factor is responsible for inhibition of enzymatic process during feedback? (1) Substrate

(2) Enzyme

(3) End product

(4) Temperature

Sol. Answer (3) Feedback inhibition is also konwn as End product inhibition or allosteric modulation. 70. Which of the following is true for competitive enzyme inhibition? (1) Decrease in Vmax and Km

(2) Unchanged Vmax and decrease in Km

(3) Unchanged Vmax and increase in Km

(4) Increase in Vmax and Km

Sol. Answer (3) Vmax = constant; Km = increased Vmax

Reaction velocity

No Inhibitor

With inhibitor

Vmax 2

Km

K'm

[S]

71. The Michaelis constant Km is (1) Numerically equal to ½ Vmax (2) Dependent on the enzyme concentration (3) Numerically equal to the substrate concentration that gives half maximal velocity (4) Increased in the presence of non-competitive inhibitor Sol. Answer (3)

vmax Velocity of reaction

vmax 2

Km

[s]

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

28

Biomolecules

Solution of Assignment

72. If an enzyme has been given the EC code 5.2.1.7, it is likely to be involved in (1) Digestion

(2) Redox reaction

(3) Isomerization

(4) Molecular breakdown

Sol. Answer (3) EC code 5.2.1.7, in this the first digit represents class of enzyme. And the class V of enzyme is isomerases, which catalyse the isomerisation reaction. 73. Prosthetic group is a part of holoenzyme. It is (1) Loosely attached organic part (2) Loosely attached inorganic part (3) Non-protein organic part firmly attached with apoenzyme (4) None of these Sol. Answer (3)

Enzyme (Holoenzyme) Protein (Apoenzyme)

Non-protein (Cofactor)

Organic compound

Tightly bound (Prosthetic group)

Inorganic compound

Loosely bound (Coenzyme)

It includes metal ions 2+ 2+ 2+ i.e., Zn , Mg , Ca etc.

74. Which of the following has carbohydrate as prosthetic group? (1) Glycoprotein

(2) Chromoprotein

(3) Lipoprotein

(4) Nucleoprotein

Sol. Answer (1) It is a conjugated protein in which protein is conjugated with carbohydrate. 75. Mark the mismatched pair. (1) Cellulose

:

Unbranched polymer with ,1-4 glycosidic linkage

(2) Cellophane

:

Cellulose xanthate

(3) Carboxypeptidase

:

Exopeptidase, Mg2+ acts as a co-factor

(4) Aminopeptidase

:

Exopeptidase, cleaves the peptide bond at N-terminal end

Sol. Answer (3) Carboxypeptidase requires Zn2+ as cofactor. 76. Apoenzyme is (1) Always a protein

(2) Often a metal

(3) Always an inorganic compound

(4) Often a vitamin

Sol. Answer (1) Protein part of enzyme is known as apoenzyme and non-protein part of enzyme is known as cofactor. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

Biomolecules

29

SECTION - D Assertion-Reason Type Questions 1.

A : Cofactor of an enzyme may be a prosthetic group. R : NAD derived from niacin is a co-enzyme.

Sol. Answer (2) Organic compound firmly attach to protein part of enzyme is prosthetic group. 2.

A : Linolenic acid is an essential fatty acid. R : Linolenic acid cannot be synthesised in human beings.

Sol. Answer (1) PUFA (Polyunsaturated Fatty Acids) are essential fatty acid. 3.

A : Hormones are not enzymes, but they can stimulate the release of enzymes. R : Hormones are used up in metabolism, but enzymes can act over and over again.

Sol. Answer (2) Enzymes are not used up in metabolism. 4.

A : Tertiary structure of protein molecules makes them biologically active. R : It is native configuration of protein molecules maintained by multiple covalent bonds only.

Sol. Answer (3) Because enzyme exists in tertiary structure in which sides group comes close and form an active site. 5.

A : Dextrins are intermediate polysaccharides formed during hydrolysis of starch into sugar. R : Ascorbic acid is a sugar derivative.

Sol. Answer (2) Ascorbic acid is sugar acid. 6.

A : Non-competitive inhibitors have no effect on Vmax. R : In non-competitive inhibition, inhibitor and substrate bind at same sites on the enzyme.

Sol. Answer (4) Non-competitive inhibitors decrease Vmax. 7.

A : The polypeptide coil of collagen helix is strengthened by the estabilishment of hydrogen bond between > NH - group of glycine residue of each strand with –CO group of other two strand. R : In collagen helix locking effect also occurs with the help of proline and hydroxyproline amino acid.

Sol. Answer (2) In collagen, there are generally three polypeptides coil around one another. 8.

A : Allosteric enzymes do not show a typical Michaelis Menten constant or behaviour. R : All enzymes work at same pH.

Sol. Answer (3) Allosteric enzymes don't obey Michaelis Menten constant. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

30 9.

Biomolecules

Solution of Assignment

A : amylase of wheat endosperm has 16 isoenzymes. R : In competitive inhibition Vmax decreases.

Sol. Answer (3) Isoenzymes are the isomeric form of same enzyme. 10. A : Tertiary structure of protein is absolutely necessary for many biological activities of proteins. R : In protein, only right handed helices are observed. Sol. Answer (2) Enzymes have tertiary structure of protein.







Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456