JEE Chemistry Biomolecules MCQs Set 21

Answer: (d) Enzymes are specific biological catalysts that possess well defined active sites

Explanation:
1. Enzymes are protein catalysts with a specific three-dimensional active site (lock-and-key).
2. They denature at high temperature, are usually homogeneous catalysts and can be inhibited or poisoned.

Teacher's Note:
1. Enzymes work best near body temperature and a specific pH.

Answer: (b) A hormone

Explanation:
1. Insulin is a peptide hormone secreted by the pancreas.
2. It is not a catalyst (enzyme or coenzyme) and not an antibiotic.

Teacher's Note:
1. Hormones are chemical messengers; diabetes is linked to lack of insulin action.

Answer: (c) hydrogen bonding

Explanation:
1. The two strands are held together by hydrogen bonds between complementary bases (A=T with two, G-C with three).
2. This base pairing gives the double helix.

Teacher's Note:
1. Hydrogen bonding explains base pairing and the stability of the helix.

Answer: (c) D-glucose

Explanation:
1. Cellulose is a linear polymer of beta-D-glucose units joined by C1-C4 beta-glycosidic links.
2. Complete hydrolysis gives only D-glucose.

Teacher's Note:
1. Humans cannot digest cellulose because they lack the enzyme to break the beta links.

Answer: (c) Both of the above

Explanation:
1. A zwitterion needs both a basic group and an acidic group in the same molecule: the acid donates a proton to the amine, giving dipolar ions.
2. Amino acids (-NH2, -COOH) and sulphanilic acid (-NH2, -SO3H) both form zwitterions.

Teacher's Note:
1. Sulphanilic acid exists as a zwitterion with high melting point and low solubility in organic solvents.

Answer: (d) right, left, right, right

Explanation:
1. Oxime formation changes only the CHO group and leaves the configuration of C-2 to C-5 unchanged.
2. In D-(+)-glucose the OH groups are right, left, right, right.

Teacher's Note:
1. Mnemonic for D-glucose in the Fischer projection: right, left, right, right.

Answer: (b) hydrogen bonding

Explanation:
1. The two strands are connected through hydrogen bonds between base pairs.
2. Phosphodiester links hold the sugars of one strand together, and glycosidic bonds link base and sugar.

Teacher's Note:
1. Do not confuse links between bases (hydrogen bonds) with links along the strand (phosphodiester).

Answer: (b) beri-beri

Explanation:
1. Thiamine (B1) deficiency causes beri-beri.
2. Convulsions are linked to B6 deficiency, cheilosis to B2 and sterility to vitamin E.

Teacher's Note:
1. B1, thiamine, beri-beri.

Answer: (b) alpha-D-glucopyranose and beta-D-fructofuranose

Explanation:
1. Sucrose is alpha-D-glucopyranosyl-(1,2)-beta-D-fructofuranoside.
2. Glucose is in the pyranose form and fructose in the furanose form.

Teacher's Note:
1. Galactose + glucose gives lactose, not sucrose.

Answer: (c) (+)-Lactose is a reducing sugar and does not exhibit mutarotation

Explanation:
1. Lactose has a free anomeric hydroxyl on its glucose unit, so it reduces Tollens' reagent and shows mutarotation.
2. Statement (c) is therefore false.
3. The other statements are correct (beta-1,4 link; 8 OH groups).

Teacher's Note:
1. Reducing sugars with a free hemiacetal group always show mutarotation.

Answer: (c) (1) and (2)

Explanation:
1. Denaturation breaks hydrogen bonds and other weak interactions, so secondary and tertiary structures are lost and DNA strands separate.
2. The primary structure (peptide bonds) is not affected.

Teacher's Note:
1. Boiling an egg denatures albumin without breaking its peptide bonds.

Answer: (a) (+)-Sucrose

Explanation:
1. Mutarotation requires a free anomeric hydroxyl (hemiacetal) group.
2. Sucrose has both anomeric carbons tied up in the glycosidic bond, so it does not mutarotate.
3. Lactose, maltose and fructose have a free hemiacetal group.

Teacher's Note:
1. Non-reducing sugars do not show mutarotation.

Answer: (d) enolisation of fructose followed by conversion to an aldehyde by base

Explanation:
1. Tollens' reagent is alkaline.
2. In basic medium fructose undergoes keto-enol tautomerism to glucose and mannose (aldoses via enediol), which are oxidised by Tollens' reagent.

Teacher's Note:
1. Alpha-hydroxy ketones also reduce Tollens' reagent after isomerisation in base.

Answer: (c) adenine and thymine; guanine and cytosine

Explanation:
1. Base pairing in DNA is A with T (two hydrogen bonds) and G with C (three hydrogen bonds).
2. Uracil is not present in DNA.

Teacher's Note:
1. Chargaff's rule: A = T and G = C.

Answer: (c) D-sugar component

Explanation:
1. The bases and the phosphate are achiral.
2. The chirality comes from the D-ribose or D-2-deoxyribose sugar, which has several stereocentres.

Teacher's Note:
1. Natural nucleic acids contain only D-sugars.

Answer: (c) Thiol

Explanation:
1. Two cysteine residues, each with an -SH (thiol) group, are oxidised to a disulphide bond (-S-S-).
2. This cross-link contributes to tertiary structure.

Teacher's Note:
1. Disulphide bridges are formed by oxidation of two thiol groups.

Answer: (a) five

Explanation:
1. In the pyranose ring, C-1 (anomeric), C-2, C-3, C-4 and C-5 are chiral.
2. C-6 is a CH2OH group and the ring oxygen is not a carbon.
3. Therefore there are 5 chiral carbons.

Teacher's Note:
1. The open-chain form has four chiral carbons; ring closure creates the fifth at C-1.

Answer: (a) The C-N bond length in proteins is longer than the usual C-N single bond length

Explanation:
1. The nitrogen lone pair is delocalised onto the carbonyl oxygen, so the C-N bond has partial double-bond character.
2. This makes the bond shorter than an ordinary C-N bond and keeps the group planar, so (a) is incorrect.

Teacher's Note:
1. Restricted rotation about the peptide C-N bond controls protein folding.

Answer: (d) three

Explanation:
1. The first molecule forms the hydrazone at C-1.
2. The second oxidises the neighbouring C-2 OH to C=O (and is itself reduced to aniline and ammonia), and the third forms the second hydrazone at C-2.
3. So three molecules are used.

Teacher's Note:
1. Glucose, fructose and mannose give the same osazone since they differ only at C-1 and C-2.

Answer: (d) a globular protein

Explanation:
1. Haemoglobin is an oxygen-carrying protein with four polypeptide chains folded into a compact globular shape.

Teacher's Note:
1. Globular proteins are water soluble; fibrous proteins like keratin are not.

Answer: (d) configuration

Explanation:
1. The two anomers have the same ring size and the same number of OH groups.
2. They differ in the spatial arrangement (configuration) of the OH at C-1.

Teacher's Note:
1. Anomers are diastereomers differing at the anomeric centre.

Answer: (a) Starch is a polymer of alpha-glucose

Explanation:
1. Starch (amylose and amylopectin) consists of alpha-D-glucose units.
2. Amylose is a component of starch, not cellulose; proteins contain about 20 different amino acids; the furanose ring of fructose has four carbons and one oxygen.

Teacher's Note:
1. Cellulose has beta-glucose; starch has alpha-glucose.

Answer: (b) 3

Explanation:
1. The alpha-amino acid with this formula is serine, \(\ce{HOCH2-CH(NH2)-COOH}\).
2. Its functional groups are amino, carboxyl and hydroxyl, so there are three.

Teacher's Note:
1. Count each distinct kind of functional group once per occurrence: here one each of -NH2, -COOH and -OH.

Answer: (b) 4

Explanation:
1. Maltose is the repeat disaccharide of starch, with an alpha-1,4 glycosidic link between C-1 of one glucose and C-4 of the other.
2. So x = 4.

Teacher's Note:
1. Cellobiose has a beta-1,4 link; maltose an alpha-1,4 link; lactose a beta-1,4 link.