JEE Chemistry Biomolecules MCQs Set 23

Answer: (d) D-glucose

Explanation:
1. Cellulose is a linear polymer of β-D-glucose units joined by β-1,4 glycosidic bonds.
2. Hydrolysis gives D-glucose.

Teacher's Note:
1. Both starch and cellulose give glucose.
2. The difference is the configuration of the glycosidic bond.

Answer: (a) heat

Explanation:
1. Secondary structure is held by hydrogen bonds between peptide groups.
2. Heating disrupts hydrogen bonds and unfolds helices and sheets.
3. Beta-mercaptoethanol breaks disulfide bridges (tertiary structure), and dilute KCl has little effect.

Teacher's Note:
1. Heat denatures proteins by breaking weak interactions.
2. Disulfide bridges belong to tertiary structure.

Answer: (d) Ala-Gly

Explanation:
1. The N-terminal unit (with the free amino group) is alanine, since its R group is CH3.
2. The C-terminal unit (with the free carboxyl group) is glycine.
3. By convention the sequence is written from the N-terminus, giving Ala-Gly.

Teacher's Note:
1. Write peptides from N-terminus to C-terminus.
2. Ala-Gly and Gly-Ala are different compounds.

Answer: (b) in the basic medium fructose undergoes enolisation and isomerises to an aldose (glucose and mannose)

Explanation:
1. Tollens' reagent is alkaline.
2. Under basic conditions fructose tautomerises through an enediol to glucose and mannose, which have an aldehyde group, and these reduce silver ions.
3. Secondary alcohol groups are not oxidised by Tollens' reagent.

Teacher's Note:
1. This conversion is the Lobry de Bruyn-van Ekenstein rearrangement.
2. α-hydroxy ketones reduce Tollens' reagent.

Answer: (a) alkaline hydrolysis

Explanation:
1. Heating a fat with aqueous NaOH (saponification) gives glycerol and the sodium salts of the fatty acids, which are soaps.

Teacher's Note:
1. Acid hydrolysis gives free fatty acids.
2. Trans-esterification with methanol gives biodiesel.

Answer: (c) mutarotation

Explanation:
1. A freshly dissolved α- or β-glucose slowly equilibrates with the other anomer through the open-chain form.
2. The rotation changes until it reaches the equilibrium value, a phenomenon called mutarotation.

Teacher's Note:
1. α-D-glucose starts at +112° and β at +19°; both go to +52.7°.
2. Needs a free anomeric OH.

Answer: (a) 19

Explanation:
1. A-T pairs form two hydrogen bonds and G-C pairs form three.
2. The strand A, T, G, C, C, T, A, A has five A/T bases (2 bonds each = 10) and three G/C bases (3 bonds each = 9), so 10 + 9 = 19.

Teacher's Note:
1. Count only the bases of one strand; each is paired with its complement.
2. A=T has two bonds; G=C has three.

Answer: (c) cysteine

Explanation:
1. Cysteine has the side chain CH₂SH, which contains the thiol group.
2. Cystine is the disulfide dimer of cysteine, methionine has a thioether and cytosine is a base.

Teacher's Note:
1. Two cysteines oxidise to cystine through an S-S bridge.
2. Methionine has S-CH₃.

Answer: (a) lipid

Explanation:
1. Proteins (helices and sheets), DNA (double helix) and RNA (hairpins) have secondary structures.
2. Lipids are not polymers and have no secondary structure.

Teacher's Note:
1. Secondary structure applies to polymers.
2. Lipids form aggregates such as bilayers.

Answer: (d) the genetic code

Explanation:
1. Each triplet of bases (codon) on mRNA specifies one amino acid.
2. The correspondence between codons and amino acids is the genetic code; translation is the process of protein synthesis.

Teacher's Note:
1. The genetic code is a table; translation is a process.
2. There are 64 codons and 20 common amino acids.

Answer: (c) n-hexane

Explanation:
1. HI is a strong reducing agent; prolonged heating reduces all the oxygen-containing groups of glucose.
2. The final product is n-hexane, which showed that glucose has an unbranched six-carbon chain.

Teacher's Note:
1. This reaction proves the straight chain of glucose.
2. Red phosphorus is often added to regenerate HI.

Answer: (c) hydrogen bonding

Explanation:
1. Codon-anticodon pairing is complementary base pairing, held together by hydrogen bonds.

Teacher's Note:
1. The pairing is weak and reversible, as needed for translation.
2. Same type of force as in DNA base pairs.

Answer: (a) 2nd

Explanation:
1. Ribose in RNA has a hydroxyl group at C-2'.
2. In 2-deoxyribose of DNA that carbon has only hydrogen atoms.

Teacher's Note:
1. The name 2-deoxyribose gives the answer.
2. The 2'-OH makes RNA more susceptible to hydrolysis.

Answer: (b) 40

Explanation:
1. Each round of replication doubles the number of molecules: 10 × 2 × 2 = 40.

Teacher's Note:
1. After n rounds N molecules become N x 2^n.
2. Replication is semi-conservative.

Answer: (d) glucose only

Explanation:
1. Starch is a polymer of α-D-glucose (amylose and amylopectin).
2. Complete hydrolysis therefore gives only glucose.

Teacher's Note:
1. Maltose is the intermediate disaccharide.
2. Sucrose gives glucose and fructose.

Answer: (a) Both are made of similar monomers

Explanation:
1. Cellulose is made of glucose units and DNA of nucleotides (a base, a deoxyribose sugar and a phosphate), so their monomers are different.
2. The other statements are true.

Teacher's Note:
1. Cellulose has glycosidic C-O-C links; DNA has N-glycosidic links.
2. Monomers differ.

Answer: (b) an oligosaccharide

Explanation:
1. Glucose has six carbons (hexose) and an aldehyde group (aldose), and it is a carbohydrate.
2. It cannot be hydrolysed into smaller sugars, so it is a monosaccharide, not an oligosaccharide.

Teacher's Note:
1. Oligosaccharides give 2 to 10 monosaccharide units on hydrolysis.
2. Monosaccharides are not hydrolysed further.

Answer: (a) laevorotatory

Explanation:
1. Fructose rotates plane polarised light to the left by 92° and glucose to the right by 52°, so the equimolar mixture rotates to the left by about 20°.

Teacher's Note:
1. Sucrose itself is dextrorotatory (+66°).
2. A racemic mixture would have equal and opposite rotations.

Answer: (d) L-lactic acid

Explanation:
1. During strenuous exercise oxygen supply is insufficient, so pyruvate is reduced to L-lactic acid by anaerobic glycolysis.
2. Lactic acid builds up in muscle and causes fatigue.

Teacher's Note:
1. Anaerobic respiration in muscle gives lactic acid.
2. Yeast give ethanol instead.

Answer: (d) an amide (peptide) bond

Explanation:
1. The biuret test gives a violet colour with copper(II) ions in alkali for compounds containing two or more peptide (amide) bonds, so it detects proteins and peptides.

Teacher's Note:
1. Dipeptides do not give the test.
2. Ninhydrin detects free α-amino groups.

Answer: (b) It is optically active

Explanation:
1. Glycine, H₂N-CH₂-COOH, is a protein amino acid with amino and carboxyl groups.
2. The α carbon has two hydrogens, so it is achiral and optically inactive.

Teacher's Note:
1. Glycine is the simplest amino acid.
2. All other protein amino acids are chiral.

Answer: (d) anomers

Explanation:
1. The α and β forms differ only in configuration at C-1, the new stereocentre formed on ring closure (the anomeric carbon).
2. Such isomers are called anomers.

Teacher's Note:
1. Epimers differ at any one stereocentre other than the anomeric one.
2. Anomers are a special type of epimer.

Answer: (b) C₁₂H₂₂O₁₁

Explanation:
1. Two hexose units (2C₆H₁₂O₆ = C₁₂H₂₄O₁₂) combine with loss of one water molecule, giving C₁₂H₂₂O₁₁.
2. Sucrose, maltose and lactose all have this formula.

Teacher's Note:
1. Disaccharide = 2 monosaccharides - H2O.
2. The same applies to any pair of hexoses.

Answer: (b) Gly-Ala-Phe-Glu-His-Gly-Ala-Leu-Val

Explanation:
1. Overlap the fragments: Gly-Ala-Phe and Phe-Glu-His give Gly-Ala-Phe-Glu-His.
2. His-Gly-Ala continues it to Gly-Ala-Phe-Glu-His-Gly-Ala, and Gly-Ala-Leu and Ala-Leu-Val extend it to Gly-Ala-Phe-Glu-His-Gly-Ala-Leu-Val.
3. This sequence has nine residues and contains all five tripeptides.

Teacher's Note:
1. Look for the overlapping residues between fragments.
2. Check that every given tripeptide appears in the final sequence.

Answer: (c) an aldotriose

Explanation:
1. The simplest aldose is glyceraldehyde, which has three carbon atoms, so it is an aldotriose.

Teacher's Note:
1. Glyceraldehyde is the reference compound for the D/L system.
2. Dihydroxyacetone is the simplest ketose.