JEE Chemistry Biomolecules MCQs Set 25

Answer: (b) glucose

Explanation:
1. Starch is a polymer of α-D-glucose; amylase and maltase hydrolyse it to maltose and then glucose.

Teacher's Note:
1. Cellulose also gives glucose.
2. Sucrose gives glucose and fructose.

Answer: (b) planar and usually found in a trans conformation

Explanation:
1. Resonance gives the C-N bond partial double bond character, so the six atoms of the peptide group lie in one plane.
2. The trans arrangement is normally preferred because it avoids steric clash between side chains.
3. The group is polar, since both the C=O and N-H groups can hydrogen bond.

Teacher's Note:
1. Rotation about C-N is restricted.
2. Proline can form cis peptide bonds more often.

Answer: (c) Arabinose

Explanation:
1. Glucose, fructose and galactose all contain six carbon atoms and are hexoses.
2. Arabinose has five carbon atoms (C₅H₁₀O₅), so it is an aldopentose.

Teacher's Note:
1. Ribose, xylose and arabinose are the common pentoses.
2. The name ending -ose marks sugars; the prefix gives the number of carbons.

Answer: (a) 4

Explanation:
1. Alanine is fixed at the C-terminal; the other three amino acids fill the first three positions in 3! = 6 ways.
2. The N-terminal amino acid is chiral unless it is glycine, so remove the arrangements that start with Gly: 2 arrangements.
3. 6 - 2 = 4 sequences.

Teacher's Note:
1. Glycine's α carbon is not chiral.
2. Count arrangements, then exclude those with Gly first.

Answer: (c) lysine - ninhydrin; furfural - 1-naphthol; benzyl alcohol - ceric ammonium nitrate; styrene - KMnO₄

Explanation:
1. Amino acids such as lysine give a purple colour with ninhydrin.
2. Furfural gives a coloured product with 1-naphthol (Molisch-type test).
3. Alcohols such as benzyl alcohol give a colour change with ceric ammonium nitrate, and styrene (alkene) decolourises KMnO₄.

Teacher's Note:
1. Match each functional group with its characteristic test.
2. Ninhydrin is the standard test for α-amino acids.

Answer: (a) 11

Explanation:
1. L-Glucose differs from D-glucose at C-2, C-3, C-4 and C-5.
2. D-Allose differs from D-glucose at C-3 only.
3. So D-allose and L-glucose differ at C-2, C-4 and C-5, and the sum is 2 + 4 + 5 = 11.

Teacher's Note:
1. Mirror image inverts all stereocentres.
2. Then undo the one inversion at C-3 that allose shares with L-glucose.

Answer: (d) glycogen

Explanation:
1. Animals store glucose as glycogen in the liver and muscles; plants store it as starch.

Teacher's Note:
1. Glycogen is a branched polymer of glucose.
2. It is also called animal starch.

Answer: (c) α-1,4 and β-1,4

Explanation:
1. Amylose is a linear chain of α-D-glucose joined by α-1,4 links.
2. Cellulose is a linear chain of β-D-glucose joined by β-1,4 links.

Teacher's Note:
1. α links give a coiled chain; β links a straight chain.
2. Amylopectin adds α-1,6 branches.

Answer: (b) All of these

Explanation:
1. The cyclisation of the open chain creates a new stereocentre at C-1, giving α and β forms.
2. Each form can be crystallised from a suitable solvent, and in water the rotation of either one slowly changes to an equilibrium value (mutarotation).

Teacher's Note:
1. The α form is obtained from hot ethanol; β from hot pyridine or acetic acid.
2. The statements are standard facts.

Answer: (d) Glucose : mutarotation

Explanation:
1. Sucrose is a disaccharide and a non-reducing sugar, because both anomeric carbons are involved in the glycosidic link.
2. Fructose is a ketose.
3. Glucose shows mutarotation, the change in optical rotation of a fresh solution as the α and β anomers interconvert.

Teacher's Note:
1. Check each term against its definition.
2. Mutarotation needs a free anomeric hydroxyl group.

Answer: (d) nucleic acids

Explanation:
1. Genetic information is stored in DNA (or RNA in some viruses), both nucleic acids.

Teacher's Note:
1. DNA stores genes.
2. Proteins are the products of gene expression.

Answer: (b) configuration at the glycosidic linkage

Explanation:
1. Both disaccharides are glucose dimers with 1,4 links and pyranose rings.
2. Maltose has an α-1,4 link whereas cellobiose has a β-1,4 link, so they differ in the configuration at the glycosidic link.

Teacher's Note:
1. Both have the formula C₁₂H₂₂O₁₁.
2. The α/β difference gives different shapes and enzyme specificity.

Answer: (c) sorbitol

Explanation:
1. The carbonyl group of glucose is reduced to a CH2OH group by sodium amalgam, giving the hexahydric alcohol sorbitol (glucitol).

Teacher's Note:
1. Oxidation with nitric acid gives saccharic acid, with bromine water gluconic acid.
2. Reduction gives a polyol.

Answer: (b) (2), (3) and (4) only

Explanation:
1. Starch is a mixture of amylose and amylopectin, so (1) is false and (2) is true.
2. Both have α-1,4 links between α-D-glucose units, and amylopectin is branched through α-1,6 links, so (3) and (4) are true.

Teacher's Note:
1. Amylose is linear; amylopectin is branched.
2. Test each statement separately.

Answer: (c) 14

Explanation:
1. Aspartic acid is acidic, so x = (2 + 4)/2 = 3.
2. Arginine is basic, so y = (9 + 13)/2 = 11.
3. x + y = 14.

Teacher's Note:
1. Acidic amino acid: average the two lowest values.
2. Basic amino acid: average the two highest.

Answer: (c) arginine

Explanation:
1. A pI near 11 indicates a strongly basic amino acid.
2. Arginine has a guanidinium side chain with pKa about 12.5; pI = (9.04 + 12.48)/2, about 10.8.
3. Glycine (5.97), histidine (7.59) and proline (6.3) have much lower pI values.

Teacher's Note:
1. Basic amino acids have pI above 7.
2. Arginine is the most basic.

Answer: (b) S1, S2 and S4 are true, S3 is false

Explanation:
1. All monosaccharides reduce Tollens' and Fehling's reagents (S1 true).
2. Bromine water oxidises aldoses to aldonic acids but does not react with ketoses (S2 true).
3. Aldoses differing only at C-2 are epimers, not anomers, since anomers differ at the anomeric carbon (S3 false); osazone formation affects only C-1 and C-2 (S4 true).

Teacher's Note:
1. Anomers differ at C-1 of the cyclic form; epimers differ at one other carbon.
2. Bromine water oxidises the CHO group only.

Answer: (c) (2) and (3) only

Explanation:
1. Invert sugar comes from hydrolysis of sucrose (not maltose, which gives only glucose), so (1) is wrong.
2. It is an equimolar mixture of D-glucose and D-fructose, so (2) is correct, and its rotation is (52 - 92)/2 = -20°, so (3) is correct.
3. Bromine water oxidises glucose to gluconic acid, not saccharic acid, which needs nitric acid, so (4) is wrong.

Teacher's Note:
1. Average the specific rotations for an equimolar mixture.
2. Saccharic acid needs a strong oxidant such as HNO3.

Answer: (c) carbon atoms 3, 4 and 5 have the same configuration in both

Explanation:
1. Osazone formation involves only C-1 and C-2.
2. The osazone is identical when the remaining carbons C-3, C-4 and C-5 have the same configuration, which is the case for glucose and fructose.

Teacher's Note:
1. Being hexoses or reducing sugars is not enough: galactose gives a different osazone.
2. Only the first two carbons change.

Answer: (b) 4

Explanation:
1. Essential amino acids cannot be made by the body and must come from the diet.
2. Of the amino acids listed, valine, leucine, isoleucine and threonine are essential.
3. Glycine, alanine, cysteine and serine are non-essential.

Teacher's Note:
1. Mnemonic: PVT TIM HALL covers all ten essential amino acids.
2. Non-essential ones can be synthesised in the body.

Answer: (a) It joins protein molecules to each other by covalent bonds between chains

Explanation:
1. The peptide bond is an amide with partial double bond character because the nitrogen lone pair is delocalised onto the carbonyl group with the carbonyl group.
2. The C=O and N-H groups form hydrogen bonds, which hold the secondary structure together.
3. Peptide bonds join amino acids within a chain; neighbouring chains are held by hydrogen bonds and other interactions, not by peptide links.

Teacher's Note:
1. Resonance makes the peptide group planar.
2. Hydrogen bonds between the groups stabilise helices and sheets.

Answer: (c) prosthetic group

Explanation:
1. A conjugated protein consists of a protein part and a non-protein part, which is called the prosthetic group (for example, haem in haemoglobin).

Teacher's Note:
1. Glycoproteins have a carbohydrate prosthetic group.
2. Simple proteins give only amino acids on hydrolysis.

Answer: (c) inversion

Explanation:
1. Sucrose is dextrorotatory ([α] = +66.5°) but the glucose-fructose mixture is laevorotatory, because fructose (-92°) outweighs glucose (+52.5°).
2. The change of sign of rotation is called inversion, and the mixture is called invert sugar.

Teacher's Note:
1. Invertase catalyses this reaction.
2. Equimolar mixture rotation = (52.5 - 92)/2, about -20°.

Answer: (a) Carbon atoms 3, 4 and 5 have the same configuration in both

Explanation:
1. The osazone retains the structure of C-3 to C-6 of the sugar and C-1 and C-2 are converted into the same hydrazone groups.
2. Hence identical osazones need identical configuration at C-3, C-4 and C-5; one hexose may be an aldose and the other a ketose.

Teacher's Note:
1. Glucose, mannose and fructose give the same osazone.
2. The configuration at C-1 and C-2 is lost.

Answer: (c) has a cyclic structure (hemiacetal form)

Explanation:
1. A free aldehyde would not form two products with only one molecule of alcohol; it would form a dimethyl acetal.
2. The cyclic hemiacetal OH at C-1 is replaced by OCH3, and the anomeric carbon can have two configurations, giving two glycosides.

Teacher's Note:
1. Two anomeric glycosides are evidence for the ring structure.
2. Glycosides are acetals and do not reduce Fehling's solution.