JEE Chemistry Biomolecules MCQs Set 24

Answer: (c) 4

Explanation:
1. A free dipeptide has one N-terminal NH3+ and one C-terminal COO-, a net charge of zero when the side chains are neutral.
2. A lysine side chain ((CH2)4NH2) adds a positive charge at pH 7, and an acidic side chain (CH2COOH) adds a negative charge.
3. Peptides IV, VI, VIII and IX contain lysine and are positively charged: 4 in all (III and VII are negative; I, II and V are neutral).

Teacher's Note:
1. Count basic and acidic side chains.
2. Peptide VI has two lysines and is more positive.

Answer: (a) Starch and cellulose

Explanation:
1. Starch and cellulose are polymers made only of glucose units.
2. Sucrose (glucose + fructose) and lactose (glucose + galactose) are disaccharides.

Teacher's Note:
1. Starch: α-glucose units; cellulose: β-glucose units.
2. Glycogen is also a glucose polysaccharide.

Answer: (a) The antiparallel sheet is more stable than the parallel sheet

Explanation:
1. In the antiparallel sheet the N-H and C=O groups of adjacent strands lie directly opposite each other, so the hydrogen bonds are linear and strong.
2. In the parallel sheet the hydrogen bonds are angled and weaker, so the antiparallel arrangement is more stable.

Teacher's Note:
1. Linear hydrogen bonds are the strongest.
2. Silk fibroin is an antiparallel sheet.

Answer: (a) Sucrose

Explanation:
1. Glycogen (glucose units), haemoglobin (amino acids) and DNA (nucleotides) are polymers.
2. Sucrose is a disaccharide made of two units only.

Teacher's Note:
1. A polymer has many repeating units.
2. Sucrose is a dimer of different sugars.

Answer: (c) hemiacetal

Explanation:
1. The ring carbon C-1 carries an OH group and a ring oxygen (-O-), both attached to the same carbon that came from an aldehyde.
2. A carbon with OH and OR groups, derived from an aldehyde, is a hemiacetal.

Teacher's Note:
1. Its methyl derivative (OCH3 in place of OH) is an acetal.
2. Fructose rings are hemiketals.

Answer: (d) Guanine

Explanation:
1. Uracil, cytosine and thymine are pyrimidines with a single six-membered ring.
2. Guanine (like adenine) is a purine with fused five- and six-membered rings.

Teacher's Note:
1. Pyrimidines: C, T, U. Purines: A, G.
2. Remember 'CUT the pie'.

Answer: (a) α-D-glucose and β-D-glucose are enantiomers

Explanation:
1. The two glucose forms differ only at C-1, so they are anomers, which are diastereomers.
2. Enantiomers are non-superimposable mirror images, and they would differ at all stereocentres (D versus L).
3. The penta-acetate has no free CHO group in the cyclic form, so it does not react with hydroxylamine.

Teacher's Note:
1. Anomers are not mirror images.
2. The cyclic penta-acetate cannot open to give a free aldehyde.

Answer: (b) Two molecules react in the same manner (forming the hydrazone groups) while the third reacts differently (oxidising the C-2 hydroxyl group and being cleaved to aniline and ammonia)

Explanation:
1. The first molecule forms the phenylhydrazone with the carbonyl group.
2. The second molecule oxidises the adjacent CHOH group to a carbonyl group and is itself reduced to aniline and ammonia.
3. The third molecule then condenses with the new carbonyl group, giving the osazone.

Teacher's Note:
1. Osazone formation consumes three moles of phenylhydrazine, producing aniline and ammonia as by-products.
2. Only C-1 and C-2 of the sugar are involved.

Answer: (a) 16

Explanation:
1. This aldohexose has four chiral carbons (C-2 to C-5).
2. Stereoisomers = 2^4 = 16, and all are optically active because the molecule has no internal symmetry.

Teacher's Note:
1. Apply 2^n with n chiral centres.
2. Aldohexoses have 16 stereoisomers (8 D and 8 L).

Answer: (d) a carbohydrate

Explanation:
1. A glycoprotein is a protein conjugated with a carbohydrate, which is its non-protein (prosthetic) part.

Teacher's Note:
1. Lipoproteins have lipid; nucleoproteins have nucleic acid.
2. Prosthetic groups are non-amino acid parts.

Answer: (b) the laevorotation of fructose is greater than the dextrorotation of glucose

Explanation:
1. Hydrolysis gives equal amounts of D-glucose ([α] = +52°) and D-fructose ([α] = -92°).
2. The magnitude of the fructose rotation is larger, so the net rotation is negative.

Teacher's Note:
1. Equimolar mixture: (+52 - 92)/2 = -20°.
2. The sign change is called inversion.

Answer: (b) diastase

Explanation:
1. Diastase (amylase) hydrolyses starch to maltose, a disaccharide.
2. Maltase then hydrolyses maltose to glucose.

Teacher's Note:
1. Starch to maltose to glucose.
2. Lactase acts on lactose.

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

Explanation:
1. The α and β forms of D-glucose differ at C-1 (they are anomers), and glucose and fructose are functional isomers rather than epimers, so (1) and (2) are wrong.
2. Glucose shows mutarotation and sucrose hydrolyses to glucose and fructose, so (3) and (4) are correct.

Teacher's Note:
1. Epimers are both aldoses (or both ketoses).
2. Anomers differ at C-1.

Answer: (b) hydrogen bonds

Explanation:
1. Base pairs are held by hydrogen bonds, giving DNA the double helix and RNA hairpins, just as hydrogen bonds give protein helices and sheets.

Teacher's Note:
1. Secondary structure is stabilised by H-bonds.
2. Covalent bonds make the primary structure.

Answer: (d) an N-glycosidic linkage

Explanation:
1. The ring nitrogen of the base (N-1 of pyrimidines or N-9 of purines) is joined to C-1' of the sugar by an N-glycosidic bond.

Teacher's Note:
1. Phosphodiester links connect nucleotides.
2. Hydrogen bonds join base pairs.

Answer: (c) show no net movement

Explanation:
1. At the isoelectric point the amino acid exists as a zwitterion with zero net charge, so it does not migrate towards either electrode.

Teacher's Note:
1. At pH below pI it moves to the cathode; above pI, to the anode.
2. Proteins precipitate most easily at the pI.

Answer: (a) two different sugars (deoxyribose and ribose)

Explanation:
1. DNA gives 2-deoxy-D-ribose and RNA gives D-ribose.
2. They have different formulas (C₅H₁₀O₄ and C₅H₁₀O₅), so they are not isomers.

Teacher's Note:
1. They differ in an oxygen atom at C-2.
2. Neither epimers nor isomers.

Answer: (a) C-1' and C-5'

Explanation:
1. The nitrogen base is attached through an N-glycosidic bond to C-1' of the sugar.
2. The phosphate group is esterified to the C-5' hydroxyl group of the sugar (and joins C-3' of the next unit).

Teacher's Note:
1. Nucleoside = base + sugar; nucleotide = base + sugar + phosphate.
2. Primed numbers refer to the sugar.

Answer: (a) glycine, alanine and phenylalanine

Explanation:
1. R = H is glycine, R = CH3 is alanine and R = CH2C6H5 is phenylalanine.

Teacher's Note:
1. Identify each amino acid from its side chain.
2. Glycine is the only one with R = H.

Answer: (a) Quinoline

Explanation:
1. DNA contains adenine, guanine, cytosine and thymine; quinoline is not a nucleic acid base.

Teacher's Note:
1. Only four bases occur in DNA.
2. Quinoline is a benzopyridine.

Answer: (a) Formation of a penta-acetate

Explanation:
1. Acetylation with acetic anhydride replaces each OH by an acetate group.
2. Glucose gives a penta-acetate, which shows that five OH groups are present; cyanohydrin formation, Fehling's and Tollens' tests show the carbonyl group.

Teacher's Note:
1. Counting acetyl groups is a standard way to count hydroxyl groups.
2. Fehling's and Tollens' tests detect an aldehyde group.

Answer: (a) water

Explanation:
1. A peptide bond forms by condensation: the OH of the carboxyl group and an H of the amino group are lost as H₂O.

Teacher's Note:
1. Hydrolysis is the reverse reaction.
2. A tripeptide releases two water molecules.

Answer: (d) Cysteine

Explanation:
1. Two cysteine residues can be oxidised to form a disulfide (S-S) bridge, a covalent cross-link that stabilises the tertiary structure.

Teacher's Note:
1. Insulin has disulfide bridges.
2. Cystine is the oxidised dimer.

Answer: (d) Only II

Explanation:
1. The Van Slyke method measures the nitrogen gas evolved from a free primary amino group on reaction with HNO₂.
2. Proline is a secondary amine (the nitrogen is part of a ring), so it does not release nitrogen.
3. Cysteine, histidine and valine each have a primary α-amino group.

Teacher's Note:
1. Primary aliphatic amines release nitrogen with nitrous acid.
2. Secondary amines give nitrosoamines instead.

Answer: (a) 2.77

Explanation:
1. Aspartic acid is an acidic amino acid, so the zwitterion of net charge zero lies between the first two ionisations.
2. pI = (pKa1 + pKa2)/2 = (1.89 + 3.65)/2 = 2.77.

Teacher's Note:
1. For acidic amino acids average the two lowest pKa values.
2. For basic amino acids average the two highest.