JEE Chemistry Biomolecules MCQs Set 20

Answer: (b) amylopectin

Explanation:
1. Amylopectin is the branched component of starch, with C1-C4 chain links and C1-C6 branch points.
2. Amylose is linear, and cellulose is a linear polymer of beta-glucose.
3. Glycogen is therefore like amylopectin (but more highly branched).

Teacher's Note:
1. Glycogen is called animal starch.

Answer: (d) Glycogen

Explanation:
1. Excess glucose in animals is stored as glycogen in the liver and muscles.
2. Amylose and amylopectin are the storage polysaccharides of plants, and cellulose is structural.

Teacher's Note:
1. Glycogen is the animal reserve carbohydrate.

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

Explanation:
1. Sucrose is formed by a glycosidic link between C1 of alpha-D-glucose and C2 of beta-D-fructose.
2. Hydrolysis gives an equimolar mixture of D-glucose and D-fructose (invert sugar).

Teacher's Note:
1. The mixture is called invert sugar because the optical rotation changes from positive to negative.

Answer: (c) hydrogen bonds

Explanation:
1. In an alpha-helix the C=O of each amino acid hydrogen bonds with the N-H of the amino acid four residues ahead.
2. These intramolecular hydrogen bonds hold the helix; peptide bonds define the primary structure.

Teacher's Note:
1. Both alpha-helix and beta-pleated sheets are secondary structures held by hydrogen bonds.

Answer: (c) Sucrose

Explanation:
1. A disaccharide is non-reducing when the anomeric carbons of both units are involved in the glycosidic bond.
2. In sucrose the C1 of glucose is joined to the C2 of fructose, so no free aldehyde or ketone group remains.
3. Maltose, lactose and cellobiose have a free hemiacetal group.

Teacher's Note:
1. Sucrose does not show mutarotation or reduce Tollens' reagent.

Answer: (b) Ascorbic acid

Explanation:
1. Ascorbic acid is vitamin C.
2. Aspartic acid is an amino acid, adipic acid is a dicarboxylic acid used in nylon-6,6 and saccharic acid is the dicarboxylic acid obtained by oxidising glucose with nitric acid.

Teacher's Note:
1. Vitamin C is water soluble and its deficiency causes scurvy.

Answer: (a) 5' and 3'

Explanation:
1. In nucleic acids the phosphate group joins the 5'-OH of one nucleotide to the 3'-OH of the next.
2. This gives the sugar-phosphate backbone with 5' to 3' direction.

Teacher's Note:
1. The base is at C1' and the phosphate ester at C5' of a nucleotide.

Answer: (b) nucleotides

Explanation:
1. A nucleotide = base + pentose sugar + phosphate.
2. Nucleic acids (DNA and RNA) are long chains of nucleotides joined by phosphodiester links.
3. Nucleosides lack the phosphate.

Teacher's Note:
1. Nucleoside + phosphate = nucleotide.

Answer: (c) It exists mainly in the furanose form

Explanation:
1. Glucose has an aldehyde group and six carbon atoms (aldohexose), is reduced by HI/red P to n-hexane, and reduces Tollens' and Fehling's reagents.
2. It exists mainly in the six-membered pyranose ring form, not the furanose form.

Teacher's Note:
1. Glucose = pyranose; fructose = furanose (in combined form, e.g. in sucrose).

Answer: (a) primary structure of proteins

Explanation:
1. The linear sequence of amino acids joined by peptide bonds is the primary structure.
2. Secondary refers to local folding (helix, sheet), tertiary to overall 3D folding and quaternary to the arrangement of several chains.

Teacher's Note:
1. Primary structure is determined by the gene sequence.

Answer: (c) Thymine

Explanation:
1. RNA contains adenine, guanine, cytosine and uracil.
2. Thymine (5-methyluracil) is found only in DNA.

Teacher's Note:
1. Remember: thymine in DNA, uracil in RNA.

Answer: (d) Vitamin B12

Explanation:
1. Water-soluble vitamins are usually excreted and must be supplied regularly.
2. Vitamin B12 is an exception and is stored in the liver.

Teacher's Note:
1. B12 is stored in the liver for years; the other B vitamins are not stored.

Answer: (d) Uracil

Explanation:
1. DNA contains adenine, guanine, cytosine and thymine.
2. Uracil replaces thymine in RNA.

Teacher's Note:
1. Uracil occurs only in RNA.

Answer: (c) The pentaacetate of glucose does not react with hydroxylamine

Explanation:
1. The pentaacetate and oxime formation can be explained by the open-chain form with five OH groups and a CHO group.
2. If the open-chain aldehyde were present in the pentaacetate, it would react with hydroxylamine; its failure to do so shows that the CHO group is absent, i.e. glucose exists in a cyclic hemiacetal form.

Teacher's Note:
1. Cyclic structure also explains why glucose does not give Schiff's test or the sodium hydrogensulphite test.

Answer: (b) ribose sugar and uracil

Explanation:
1. RNA has ribose (with a 2'-OH group) and the bases A, G, C and U.
2. DNA has deoxyribose and the bases A, G, C and T.

Teacher's Note:
1. DNA: deoxyribose and thymine; RNA: ribose and uracil.

Answer: (a) 18 molecules of ATP

Explanation:
1. Fixing six \(\ce{CO2}\) molecules into one glucose molecule in the Calvin cycle needs 18 ATP and 12 NADPH.

Teacher's Note:
1. Each CO2 fixed needs 3 ATP and 2 NADPH, so 6 x 3 = 18 ATP.

Answer: (b) Sugars

Explanation:
1. Molisch's test uses alpha-naphthol and conc. \(\ce{H2SO4}\); carbohydrates are dehydrated to furfural derivatives that give a violet ring.

Teacher's Note:
1. Molisch's test is the general test for carbohydrates.

Answer: (b) 2nd

Explanation:
1. Ribose has an OH on C-2', whereas 2-deoxyribose has only H at C-2'.
2. This is the only difference between the sugars of RNA and DNA.

Teacher's Note:
1. Deoxy means missing the 2'-OH.

Answer: (d) mutarotation

Explanation:
1. Alpha and beta anomers interconvert through the open-chain form until an equilibrium mixture is reached, so the rotation drifts from +111 or +19 degrees to +52.5 degrees.
2. This phenomenon is mutarotation.

Teacher's Note:
1. Mutarotation is shown by reducing sugars (free anomeric OH).

Answer: (c) C=O and -OH

Explanation:
1. Carbohydrates are polyhydroxy aldehydes or ketones.
2. They contain carbonyl (C=O) and hydroxyl (-OH) groups.

Teacher's Note:
1. Aldoses carry CHO, ketoses carry C=O in the chain.

Answer: (c) anomers

Explanation:
1. The two forms differ only in configuration at C-1, the anomeric carbon.
2. Isomers that differ only at the anomeric carbon are anomers (epimers differ at any other single carbon).

Teacher's Note:
1. Anomers = epimers at C-1.

Answer: (d) isomers of glucose that differ in configuration at carbon one (C-1)

Explanation:
1. Anomers are the alpha and beta cyclic forms that differ only at C-1, the new stereocentre created on ring closure.

Teacher's Note:
1. A racemic mixture is not the same as an anomeric mixture.

Answer: (c) cytosine and thymine

Explanation:
1. Pyrimidines have a single six-membered ring: cytosine, thymine (DNA) and uracil (RNA).
2. Adenine and guanine are purines.

Teacher's Note:
1. Remember: CUT the pyrimidines (cytosine, uracil, thymine).

Answer: (b) C1' and C5'

Explanation:
1. In a nucleoside the base is joined to C-1' of the sugar by an N-glycosidic bond.
2. In a nucleotide the phosphate is esterified at C-5'.

Teacher's Note:
1. Base at 1', phosphate at 5'.