NEET Biology Biomolecules MCQs Set 23

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Chapter-wise Objective Questions: Biomolecules

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Multiple Choice Questions

Question 1: The number of chiral carbon atoms in α-D-(+)-glucose (cyclic form) is:
(a) Five
(b) Six
(c) Three
(d) Four
Show Answer & Explanation

Answer: (a) Five

Explanation:
1. In the cyclic (pyranose) form, carbons 1, 2, 3, 4 and 5 are all asymmetric.

Teacher's Note:
1. C-6 is a CH2OH group and is not chiral.
2. So glucose has five chiral carbons in the ring form.

Question 2: Arachidonic acid has a carboxyl group attached to an R group. This R group contains:
(a) 16 carbons
(b) 15 carbons
(c) 20 carbons
(d) 19 carbons
Show Answer & Explanation

Answer: (d) 19 carbons

Explanation:
1. Arachidonic acid has 20 carbons in all.

Teacher's Note:
1. One carbon is in the carboxyl group, so the R group has 19 carbons.
2. It is an unsaturated fatty acid with four double bonds.

Question 3: In the primary structure of a protein, imagined as a line, the left end represents:
(a) The first amino acid and its carboxyl terminal
(b) The last amino acid and its carboxyl terminal
(c) The first amino acid and its amino terminal
(d) The last amino acid and its amino terminal
Show Answer & Explanation

Answer: (c) The first amino acid and its amino terminal

Explanation:
1. The left end of the polypeptide chain is the first amino acid, which has the free amino (N) terminal.

Teacher's Note:
1. The right end has the last amino acid with the carboxyl (C) terminal.
2. Protein synthesis also starts at the N-terminal.

Question 4: Find the incorrect statement:
(a) Oils have a lower melting point (e.g. gingelly oil) and hence remain as oil in winters
(b) Adenosine, guanosine, cytosine, uridine and thymidine are nucleosides
(c) Adenylic acid, thymidylic acid, guanylic acid, uridylic acid and cytidylic acid are nucleotides
(d) Nucleic acids like DNA and RNA consist of nucleotides only
Show Answer & Explanation

Answer: (b) Adenosine, guanosine, cytosine, uridine and thymidine are nucleosides

Explanation:
1. Cytosine is a nitrogenous base, and the nucleoside is cytidine.

Teacher's Note:
1. The other statements are correct.
2. A nucleoside is a base plus a sugar.

Question 5: A polymeric polysaccharide consisting of only one type of monosaccharide is:
(a) Glycogen
(b) Cellulose
(c) Inulin
(d) All of the above
Show Answer & Explanation

Answer: (d) All of the above

Explanation:
1. Glycogen and cellulose are polymers of glucose, and inulin is a polymer of fructose.

Teacher's Note:
1. Each is a homopolymer.
2. Starch is also a homopolymer of glucose.

Question 6: The exoskeleton of arthropods has a complex polysaccharide called chitin, which consists of only one type of amino sugar. That is:
(a) N-acetyl-galactosamine
(b) N-acetyl-glucosamine
(c) β-fructose
(d) β-glucose
Show Answer & Explanation

Answer: (b) N-acetyl-glucosamine

Explanation:
1. Chitin is a polymer of N-acetyl-glucosamine.

Teacher's Note:
1. Fructose and glucose are not amino sugars.

Question 7: Which of the following organic compounds are present in the acid insoluble fraction but are not strictly biomacromolecules?
(a) Nucleic acids
(b) Polysaccharides
(c) Proteins
(d) Lipids
Show Answer & Explanation

Answer: (d) Lipids

Explanation:
1. Lipids come in the acid insoluble pool because they form vesicles from membrane breakup, but their molecular weight is below 800 Da.

Teacher's Note:
1. So lipids are not true macromolecules.
2. Nucleic acids, polysaccharides and proteins are macromolecules.

Question 8: Match: (i) Drugs (ii) Polymeric substances (iii) Alkaloids (iv) Pigments with (a) gums, cellulose (b) morphine, codeine (c) carotenoids, anthocyanins (d) vinblastin, curcumin.
(a) i-c, ii-a, iii-b, iv-d
(b) i-d, ii-b, iii-a, iv-c
(c) i-d, ii-a, iii-b, iv-c
(d) i-a, ii-d, iii-b, iv-c
Show Answer & Explanation

Answer: (c) i-d, ii-a, iii-b, iv-c

Explanation:
1. Vinblastin and curcumin are drugs, gums and cellulose are polymeric substances, morphine and codeine are alkaloids, and carotenoids and anthocyanins are pigments.

Teacher's Note:
1. So the match is i-d, ii-a, iii-b, iv-c.
2. All these are secondary metabolites.

Question 9: Which of the following statements is not correct?
(a) There is no uncatalysed metabolic conversion in living systems
(b) The living state is an equilibrium steady state to be able to perform work
(c) The living state and metabolism are synonymous
(d) The catalysts which hasten the rate of a given metabolic conversion are also proteins, named enzymes
Show Answer & Explanation

Answer: (b) The living state is an equilibrium steady state to be able to perform work

Explanation:
1. The living state is a non-equilibrium steady state, and a system at equilibrium cannot perform work.

Teacher's Note:
1. The other statements are correct.
2. Metabolism is the sum of the chemical reactions in a living cell.

Question 10: Match the component with the percentage of the total cellular mass: (i) Proteins (ii) Carbohydrates (iii) Nucleic acids (iv) Lipids with (a) 5-7% (b) 2% (c) 3% (d) 10-15%.
(a) i-d, ii-c, iii-a, iv-b
(b) i-c, ii-d, iii-a, iv-b
(c) i-b, ii-c, iii-a, iv-d
(d) i-d, ii-a, iii-c, iv-b
Show Answer & Explanation

Answer: (a) i-d, ii-c, iii-a, iv-b

Explanation:
1. Proteins make 10-15%, carbohydrates 3%, nucleic acids 5-7% and lipids 2% of the cellular mass.

Teacher's Note:
1. Water is the most abundant at 70-90%.
2. Proteins are the most abundant organic compounds.

Question 11: Which of the following statements is not correct?
(a) The power of enzymes is incredible indeed
(b) Inorganic catalysts do not work efficiently at high temperatures and high pressure
(c) Almost all enzymes are proteins
(d) Zinc is a cofactor for the proteolytic enzyme carboxypeptidase
Show Answer & Explanation

Answer: (b) Inorganic catalysts do not work efficiently at high temperatures and high pressure

Explanation:
1. Inorganic catalysts work efficiently at high temperature and pressure, while enzymes are denatured at high temperature.

Teacher's Note:
1. The other statements are correct.
2. Enzymes work at ordinary temperatures.

Question 12: A competitive inhibitor of an enzyme:
(a) Increases Km without affecting Vmax
(b) Decreases Km without affecting Vmax
(c) Increases Vmax without affecting Km
(d) Decreases both Vmax and Km
Show Answer & Explanation

Answer: (a) Increases Km without affecting Vmax

Explanation:
1. A competitive inhibitor competes for the active site, so more substrate is needed to reach half of Vmax and Km rises.

Teacher's Note:
1. At very high substrate concentration Vmax is still reached.

Question 13: Prosthetic groups are organic compounds tightly bound to the apoenzyme. Haem is the prosthetic group of:
(a) Peroxidase
(b) Catalase
(c) Carboxypeptidase
(d) Both 1 and 2
Show Answer & Explanation

Answer: (d) Both 1 and 2

Explanation:
1. Haem is the prosthetic group of peroxidase and catalase.

Teacher's Note:
1. Carboxypeptidase has zinc as a cofactor, which is a metal ion.
2. Haem is an iron-containing group.

Question 14: Enzymes that catalyse removal of groups from substrates by mechanisms other than hydrolysis, leaving double bonds, are:
(a) Hydrolases
(b) Ligases
(c) Lyases
(d) Isomerases
Show Answer & Explanation

Answer: (c) Lyases

Explanation:
1. Lyases split C-C, C-O and C-N bonds by means other than hydrolysis and leave double bonds.

Teacher's Note:
1. Hydrolases use water, ligases join molecules, and isomerases rearrange them.

Question 15: Which testifies that cofactors play a crucial role in the catalytic activity of the enzyme?
(a) Catalytic activity is lost when the cofactor is removed from the enzyme
(b) Co-enzymes are organic compounds and their association with the apoenzyme is permanent
(c) Catalytic activity is increased when the cofactor is removed
(d) Co-enzymes serve as cofactors in a number of different enzyme catalysed reactions
Show Answer & Explanation

Answer: (a) Catalytic activity is lost when the cofactor is removed from the enzyme

Explanation:
1. When a cofactor is removed from the holoenzyme, the apoenzyme loses its catalytic activity.

Teacher's Note:
1. Coenzymes are loosely and transiently bound.

Question 16: One molecule of an enzyme can convert two molecules of substrate into product in 5 minutes. Ten molecules of enzyme and 30 molecules of substrate are mixed in a test tube. At the end of 10 minutes the test tube will have:
(a) Products, enzyme molecules and 5 molecules of substrate
(b) Products only
(c) Products and enzyme molecules
(d) Products and 5 molecules of substrate
Show Answer & Explanation

Answer: (c) Products and enzyme molecules

Explanation:
1. One enzyme converts 4 substrate molecules in 10 minutes, so 10 enzymes can convert 40 molecules.
2. Only 30 are present, so all the substrate is used.

Teacher's Note:
1. Enzymes are not used up, so the tube holds products and enzyme molecules.

Question 17: Which of the following is an example of non-competitive and reversible type of inhibition?
(a) CN poisoning of cytochrome oxidase
(b) Inhibition of hexokinase by glucose-6-phosphate
(c) Inhibition of succinic dehydrogenase by malonate
(d) Both 1 and 3
Show Answer & Explanation

Answer: (b) Inhibition of hexokinase by glucose-6-phosphate

Explanation:
1. Glucose-6-phosphate binds hexokinase at a site other than the active site, and the inhibition is reversible.

Teacher's Note:
1. Malonate is a competitive inhibitor, and cyanide binds cytochrome oxidase irreversibly.

Question 18: Arrange the steps of catalytic action of an enzyme in order: (A) the enzyme releases the products and the free enzyme is ready to bind another substrate molecule (B) the active site, now in close proximity to the substrate, breaks the chemical bonds and forms the enzyme-product complex (C) the binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate (D) the substrate binds to the active site of the enzyme.
(a) A, B, C, D
(b) D, B, C, A
(c) D, C, B, A
(d) D, C, A, B
Show Answer & Explanation

Answer: (c) D, C, B, A

Explanation:
1. The substrate first binds the active site (D), the enzyme changes shape to grip it (C), bonds are broken to form the enzyme-product complex (B) and the products are released (A).

Teacher's Note:
1. So the order is D, C, B, A.
2. This is the induced fit model.

Question 19: The active site of an enzyme is a crevice or pocket into which the substrate fits. These active sites are present on:
(a) Primary structure of protein
(b) Secondary structure of protein
(c) Tertiary structure of protein
(d) All of the above
Show Answer & Explanation

Answer: (c) Tertiary structure of protein

Explanation:
1. The active site is formed by the folding of the polypeptide chain into a three dimensional (tertiary) shape.

Teacher's Note:
1. It is not present in the primary or secondary structure alone.
2. Loss of tertiary structure destroys enzyme activity.

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Biomolecules Objective Questions & Solutions for NEET Biology

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