CBSE Class 12 Chemistry Biomolecules MCQs Set 11

Answer: (a) I and II

Explanation:
1. Structures I and II differ only in the configuration of the –OH group at C1 (the anomeric carbon); the rest of the chain is the same (D-glucose).
2. Hence, I (α-form) and II (β-form) are anomers.

Teacher's Note:
1. Anomers differ ONLY at the hemiacetal carbon C1; if any other carbon differs, the pair is not anomeric.
2. In a Fischer ring form, –OH of C1 on the right is the α-anomer of a D-sugar and on the left is the β-anomer.

Answer: (a) I, II, III

Explanation:
1. D or L configuration is decided by the position of the –OH group on the lowest asymmetric carbon (the one farthest from the carbonyl group).
2. In all three structures this –OH is on the right, so all three have D configuration, whatever the sign of rotation.

Teacher's Note:
1. D/L tells the configuration, not the sign of rotation – D-fructose is laevorotatory (−) but still D.
2. Always look only at the chiral carbon just above the bottom CH₂OH group.

Answer: (c) It is present in furanose form.

Explanation:
1. Glucose exists in the six-membered pyranose form (α and β), not in the five-membered furanose form.
2. The other three statements are true.

Teacher's Note:
1. Glucose → pyranose (6-membered ring); fructose → furanose (5-membered ring) – a frequently asked contrast.

Answer: (c) Pentaacetate of glucose does not react with hydroxylamine.

Explanation:
1. In glucose pentaacetate the anomeric –OH is also acetylated, so the ring cannot open to give a free –CHO group.
2. Hence it does not react with hydroxylamine; this can be explained only by the cyclic structure.

Teacher's Note:
1. The open-chain structure fails to explain three facts: no Schiff's test, no reaction with NaHSO₃, and pentaacetate not reacting with NH₂OH.

Answer: (d) Both (b) and (c)

Explanation:
1. Glucose forms an oxime with hydroxylamine and a cyanohydrin with HCN; both are reactions of a carbonyl group.
2. Reaction with HI only shows the straight six-carbon chain.

Teacher's Note:
1. Carbonyl presence is shown by NH₂OH and HCN; that it is an aldehyde is shown separately by oxidation with Br₂ water.

Answer: (c)

Explanation:
1. In pair (c), the two cyclic structures differ only in the configuration at C1 (the –OH is on the right in one and on the left in the other).
2. Such cyclic isomers differing only at C1 are anomers (α- and β-D-glucose).

Teacher's Note:
1. Open-chain structures can never be anomers – anomers exist only for the cyclic (hemiacetal) forms.
2. Pair (b) is a mirror-image pair (D- and L-glucose), i.e. enantiomers, not anomers.

Answer: (d) non-reducing sugar

Explanation:
1. In sucrose, the glycosidic bond is between C1 of glucose and C2 of fructose, i.e. between the two reducing groups.
2. No free aldehyde or ketone group is left, so sucrose is a non-reducing sugar (it is a disaccharide).

Teacher's Note:
1. A disaccharide is reducing only if one anomeric carbon is free; in sucrose both are tied up in the glycosidic bond.

Answer: (d) 80–85% of starch

Explanation:
1. Starch consists of amylose (15–20%, water-soluble) and amylopectin (80–85%, water-insoluble).

Teacher's Note:
1. Remember the pair: amylose 15–20% (soluble, unbranched), amylopectin 80–85% (insoluble, branched).

Answer: (a) cellulase

Explanation:
1. Cellulose has β-glycosidic linkages, which are hydrolysed by the enzyme cellulase.
2. Human beings lack cellulase, so cellulose cannot be digested.

Teacher's Note:
1. Invertase hydrolyses sucrose, zymase ferments glucose and urease hydrolyses urea – match each enzyme to its substrate.

Answer: (b) Amylopectin

Explanation:
1. Amylopectin is also a branched polymer of α-D-glucose with C1–C4 linkages in the chain and C1–C6 linkages at the branches.
2. Glycogen is more highly branched, but its structure is similar to amylopectin.

Teacher's Note:
1. Glycogen is called ‘animal starch’ because its structure resembles amylopectin.

Answer: (d) Glycogen

Explanation:
1. Carbohydrate is stored in the animal body as glycogen, mainly in the liver, muscles and brain.

Teacher's Note:
1. Plants store starch; animals store glycogen.

Answer: (a) Glycine

Explanation:
1. In glycine (H₂N–CH₂–COOH), R = H, so the α-carbon has two H atoms and is not chiral.

Teacher's Note:
1. Glycine is the only common α-amino acid without a chiral carbon.

Answer: (a) primary structure of proteins.

Explanation:
1. The specific sequence in which amino acids are linked in a polypeptide chain is the primary structure of the protein.

Teacher's Note:
1. Primary = sequence; secondary = α-helix/β-sheet (H-bonding); tertiary = overall folding; quaternary = arrangement of subunits.

Answer: (a) Insulin

Explanation:
1. Insulin is a globular protein; its polypeptide chains coil to give a spherical shape and it is soluble in water.
2. Keratin, collagen and myosin are fibrous proteins.

Teacher's Note:
1. Globular proteins (insulin, albumins, haemoglobin) are water-soluble; fibrous proteins (keratin, myosin, collagen) are insoluble.

Answer: (c) Hydrogen bonds

Explanation:
1. In the α-helix, the –NH group of each amino acid residue is hydrogen bonded to the >C=O group of an adjacent turn of the helix.

Teacher's Note:
1. Peptide bonds make the primary structure; the secondary structure is held by hydrogen bonds.

Answer: (d) Vitamin B₁₂

Explanation:
1. Vitamin B₁₂ is the only water-soluble vitamin that can be stored in the body (in the liver).
2. Other B vitamins are excreted in urine and must be supplied regularly.

Teacher's Note:
1. Water-soluble vitamins are generally not stored – B₁₂ is the well-known exception.

Answer: (b) Ascorbic acid

Explanation:
1. Ascorbic acid is vitamin C.

Teacher's Note:
1. Aspartic acid is an amino acid, adipic acid is used for nylon-6,6 and saccharic acid comes from oxidation of glucose by HNO₃.

Answer: (b) nucleotides

Explanation:
1. Nucleic acids are long chains of nucleotides (base + sugar + phosphate) joined by phosphodiester linkages.

Teacher's Note:
1. Nucleoside = base + sugar; nucleotide = base + sugar + phosphate. The polymer unit is the nucleotide.

Answer: (a) 5′ and 3′

Explanation:
1. The phosphodiester linkage joins C5′ of the sugar of one nucleotide to C3′ of the sugar of the next nucleotide.

Teacher's Note:
1. C1′ of the sugar carries the base; C3′ and C5′ are used for the phosphate links.

Answer: (c) Thymine

Explanation:
1. RNA contains A, G, C and U; thymine (T) is present only in DNA.

Teacher's Note:
1. Thymine is in DNA, uracil is in RNA; A, G and C are common to both.

Answer: (a) 2nd

Explanation:
1. RNA contains ribose, which has an –OH at C2′; DNA contains 2-deoxyribose, which has no –OH at C2′.

Teacher's Note:
1. The ‘2-deoxy’ in 2-deoxyribose itself tells you which carbon has lost its oxygen.

Answer: (b) Guanine

Explanation:
1. In DNA, cytosine always pairs with guanine (through three hydrogen bonds) and adenine pairs with thymine.

Teacher's Note:
1. Pairs to remember: A–T (two H-bonds) and G–C (three H-bonds).

Answer: (c) 1 molecule of glucose + 1 molecule of fructose

Explanation:
1. Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives 1 molecule of glucose + 1 molecule of fructose.
2. C₁₂H₂₂O₁₁ (Cane sugar) + H₂O \( \xrightarrow{H^{+}} \) C₆H₁₂O₆ (D-Glucose) + C₆H₁₂O₆ (D-Fructose)

Teacher's Note:
1. A disaccharide gives exactly two monosaccharide units on hydrolysis, so option (b) with three units is ruled out at once.

Answer: (c) ‘a’ carbon of glucose and ‘b’ carbon of fructose.

Explanation:
1. C-adjacent to oxygen atom in the cyclic structure of glucose or fructose is known as anomeric carbon. In above structure, ‘a’ of glucose and ‘b’ of fructose are present adjacent to the ring oxygen atom and carry the glycosidic link. Both carbons differ in configurations of the hydroxyl group.

Teacher's Note:
1. The anomeric carbon is the old carbonyl carbon: C1 in an aldose like glucose, C2 in a ketose like fructose.
2. In sucrose both anomeric carbons are tied up in the glycosidic bond, which is why sucrose is a non-reducing sugar.

Answer: (d) Uracil

Explanation:
1. Uracil is not present in DNA, instead thymine is present.

Teacher's Note:
1. Thymine is 5-methyluracil; DNA uses it instead of uracil.

Answer: (b) sucrose

Explanation:
1. In sucrose, glucose and fructose are linked through their reducing groups (C1 of glucose and C2 of fructose), so no free carbonyl group is available.

Teacher's Note:
1. Glucose, maltose and lactose all have a free hemiacetal carbon and reduce Tollens' and Fehling's reagents; sucrose does not.

Answer: (a) peptide bond

Explanation:
1. Amino acids join through –CO–NH– (peptide) bonds formed between the –COOH of one and the –NH₂ of the next, with loss of water.

Teacher's Note:
1. Match the linkage to the biomolecule: peptide – proteins, glycosidic – carbohydrates, phosphodiester – nucleic acids.

Answer: (c) A is true, but R is false.

Explanation:
1. Deoxyribose is a carbohydrate (a pentose sugar present in DNA), so A is correct.
2. R is incorrect: carbohydrates are not hydrates of carbon. Deoxyribose (C₅H₁₀O₄) does not fit C_x(H₂O)_y, while acetic acid (C₂H₄O₂) fits it but is not a carbohydrate.

Teacher's Note:
1. Define carbohydrates as optically active polyhydroxy aldehydes or ketones (or compounds giving them on hydrolysis), not by the formula.

Answer: (c) A is true, but R is false.

Explanation:
1. The (+) sign shows that glucose is dextrorotatory, so A is correct.
2. ‘D’ represents the configuration (–OH on the right at the lowest chiral carbon, related to D-glyceraldehyde), not the direction of rotation; so R is incorrect.

Teacher's Note:
1. D/L = configuration; (+)/(−) or d/l = direction of rotation. They are independent.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Reducing sugars have a free hemiacetal (anomeric) carbon, so the ring can open and close again.
2. Hence one pure anomer changes into an equilibrium mixture of α- and β-anomers, and the specific rotation changes to a constant value. R explains A.

Teacher's Note:
1. Glucose: α (+111°) and β (+19.2°) both change to the equilibrium value +52.5°.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In sucrose, C1 of α-D-glucose is linked to C2 of β-D-fructose, so neither reducing group is free.
2. Hence sucrose does not reduce Tollens’ or Fehling’s reagent; R explains A.

Teacher's Note:
1. Maltose and lactose are reducing because one anomeric carbon is free; sucrose is not.

Answer: (c) A is true, but R is false.

Explanation:
1. A is correct: sucrose (+66.5°) on hydrolysis gives invert sugar, which is laevorotatory.
2. R is incorrect: hydrolysis gives EQUAL amounts of glucose (+52.5°) and fructose (−92.4°); the change of sign is because the laevorotation of fructose is greater than the dextrorotation of glucose.

Teacher's Note:
1. Invert sugar is an equimolar mixture; its net rotation is negative because |−92.4°| > +52.5°.

Answer: (d) A is false, but R is true.

Explanation:
1. Maltose has an α-glycosidic linkage: C1 of one α-D-glucose is linked to C4 of another glucose unit. So A is incorrect and R is correct.

Teacher's Note:
1. Maltose and starch have α-linkages; lactose and cellulose have β-linkages.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. All α-amino acids except glycine have a chiral α-carbon, so they are optically active (A correct).
2. Natural amino acids are mostly L, which is true, but configuration does not explain optical activity; chirality does.

Teacher's Note:
1. Optical activity comes from a chiral carbon; L or D only tells which enantiomer it is.

Answer: (d) A is false, but R is true.

Explanation:
1. Glycine is a non-essential amino acid; it is synthesised in the body, so it need not be taken through diet. A is incorrect, R is correct.

Teacher's Note:
1. Essential amino acids (valine, leucine, lysine, etc.) must come from diet; non-essential ones (glycine, alanine) are made in the body.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are correct: proteins are polymers of α-amino acids, and denaturation destroys 2° and 3° structures.
2. But R is about denaturation, which does not explain the composition of proteins.

Teacher's Note:
1. During denaturation the primary structure (peptide bonds) stays intact; only the H-bond-dependent 2° and 3° structures are lost.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The active site binds the substrate and holds it in a position suitable for attack by the reagent, so the reaction occurs effectively. R explains A.

Teacher's Note:
1. Enzyme action is often described by the lock-and-key model: the substrate fits into the active site.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Fat-soluble vitamins (A, D, E, K) are stored in the liver and adipose (fat storing) tissues.
2. Hence vitamin D can be stored, and R explains A.

Teacher's Note:
1. Fat-soluble vitamins are stored; water-soluble ones (B group, C) are excreted in urine (except B₁₂).