JEE Chemistry Biomolecules MCQs Set 22

Answer: (b) sucrose and amylose have 1,2-glycosidic linkages

Explanation:
1. Amylose has α-1,4 links and cellulose has β-1,4 links; maltose and lactose also have 1,4 links.
2. Sucrose has an α-1,β-2 link, but amylose has no 1,2 links; it has 1,4 links.
3. So the last statement is incorrect.

Teacher's Note:
1. Amylose is linear α-1,4.
2. Sucrose is joined from C-1 of glucose to C-2 of fructose.

Answer: (d) all of these

Explanation:
1. Collagen and keratin are structural proteins, enzymes and antibodies are proteins, and many hormones such as insulin are proteins.
2. All three roles are correct.

Teacher's Note:
1. Proteins are the most versatile biomolecules.
2. Haemoglobin transports oxygen.

Answer: (c) glycerol

Explanation:
1. Glycerol has three hydroxyl groups, each esterified with a long-chain fatty acid to give a triglyceride.
2. Such esters are fats (solid) or oils (liquid).

Teacher's Note:
1. Soap is made by alkaline hydrolysis of fats.
2. Glycerol is propane-1,2,3-triol.

Answer: (a) Uracil

Explanation:
1. RNA contains uracil in place of thymine.
2. Cytosine, guanine and adenine occur in both nucleic acids.

Teacher's Note:
1. Thymine is 5-methyluracil.
2. This is a basic fact about nucleic acids.

Answer: (c) Peptide bond

Explanation:
1. Denaturation destroys the secondary and tertiary structure by breaking hydrogen bonds, disulfide bridges, ionic and hydrophobic interactions.
2. The primary structure, the peptide bonds, remains intact.

Teacher's Note:
1. Denaturation does not change the amino acid sequence.
2. Examples: boiling an egg, curdling of milk.

Answer: (a) D-glucose and D-fructose

Explanation:
1. Hydrolysis of sucrose gives equal amounts of D-glucose and D-fructose; this mixture is called invert sugar.

Teacher's Note:
1. Honey contains mostly invert sugar.
2. The mixture is laevorotatory.

Answer: (a) I, II and IV only

Explanation:
1. Aspartame is Asp-Phe with the C-terminal carboxyl group esterified as the methyl ester, so I and II are correct.
2. Only one peptide bond is present (two amino acids), so it is not a tripeptide.
3. Its functional groups are the amino group, the carboxylic acid, the amide (peptide) link and the ester: four, so IV is correct.

Teacher's Note:
1. Count amino acid residues to name the peptide type.
2. The ester is on the phenylalanine end.

Answer: (d) epimers

Explanation:
1. They differ only in the configuration at C-2, so they are C-2 epimers (diastereomers that are not mirror images).

Teacher's Note:
1. Mannose is the C-2 epimer of glucose and galactose the C-4 epimer.
2. Anomers differ at C-1.

Answer: (d) an acetal linkage

Explanation:
1. The anomeric carbon of a sugar bears two oxygen atoms (the ring oxygen and the glycosidic oxygen), which makes it an acetal (or ketal) link.

Teacher's Note:
1. Glycosides are hydrolysed by acids, as acetals are.
2. They do not reduce Fehling's solution.

Answer: (a) Vitamin C

Explanation:
1. Vitamins A, D, E and K are fat soluble.
2. Vitamin C and the B-complex vitamins are water soluble, so they are not stored in the body.

Teacher's Note:
1. Fat-soluble: A, D, E, K.
2. Water-soluble vitamins need regular intake.

Answer: (c) is a source of energy during starvation

Explanation:
1. Glycogen is the storage polysaccharide of animals.
2. When blood glucose is low, the liver breaks glycogen down to glucose, giving energy.

Teacher's Note:
1. Glycogen is 'animal starch'.
2. It is more branched than amylopectin.

Answer: (c) Glucose and fructose

Explanation:
1. Phenylhydrazine reacts at C-1 and C-2 of an aldose (or C-2 and C-1 of a ketose), destroying the configuration at C-2 and forming a bis-hydrazone.
2. Glucose, mannose and fructose have the same configuration at C-3, C-4 and C-5, so they give the same osazone (glucosazone).
3. Galactose differs at C-4, so its osazone is different.

Teacher's Note:
1. Epimers at C-2 and the corresponding ketose give the same osazone.
2. Only the first two carbons are affected.

Answer: (d) Glucose

Explanation:
1. Glucose is an aldehyde sugar that reduces ammoniacal silver nitrate (Tollens' reagent) to metallic silver, which coats the glass.

Teacher's Note:
1. Tollens' test gives a silver mirror.
2. Sucrose is non-reducing.

Answer: (d) lysine

Explanation:
1. Lysine has a second primary amino group on a four-carbon chain, which is strongly basic (pKa of the side chain about 10.5).
2. Histidine's imidazole is only weakly basic, while asparagine and serine are neutral.

Teacher's Note:
1. Basic amino acids: lysine, arginine, histidine.
2. Arginine's guanidine group is the most basic of all.

Answer: (a) glyceraldehyde

Explanation:
1. Glyceraldehyde (HOCH₂-CHOH-CHO) has one chiral carbon and is the smallest chiral aldose.
2. The others have more carbons and several chiral centres.

Teacher's Note:
1. Dihydroxyacetone has no chiral carbon.
2. Glyceraldehyde is the reference for D and L.

Answer: (b) dehydration

Explanation:
1. Concentrated sulfuric acid removes water from the carbohydrate molecules, leaving carbon, which appears as a black mass.

Teacher's Note:
1. Carbohydrates are 'hydrates of carbon'.
2. The acid acts as a dehydrating agent.

Answer: (d) 4

Explanation:
1. R1 = H gives glycine (achiral). R2 = CH3 is alanine (chiral), and R3 = phenyl on the α carbon is phenylglycine (chiral).
2. Two stereocentres give 2^2 = 4 stereoisomers, all optically active.

Teacher's Note:
1. Count stereocentres.
2. A glycine unit adds no stereocentre.

Answer: (d) Vitamin B12

Explanation:
1. Vitamin B12 (cyanocobalamin) has a corrin ring with a cobalt atom at the centre.
2. The other vitamins are organic compounds without metals.

Teacher's Note:
1. Cyanocobalamin: cobalt.
2. Deficiency causes pernicious anaemia.

Answer: (c) 8

Explanation:
1. Fructose has three chiral carbons (C-3, C-4 and C-5) in its open-chain form.
2. Number of stereoisomers = 2^3 = 8.

Teacher's Note:
1. Count carbons with four different groups.
2. Ketohexoses have one less chiral carbon than aldohexoses.

Answer: (d) the iodine test

Explanation:
1. Starch gives a deep blue colour with iodine because iodine fits in the amylose helix; cellulose gives no colour.
2. Both give glucose on hydrolysis and both give a positive Molisch test.

Teacher's Note:
1. Iodine-starch complex is blue-black.
2. Neither polysaccharide is a reducing sugar.

Answer: (d) 9

Explanation:
1. Each cleavage of a linear chain produces one additional chain.
2. 4000 units split into chains of 400 units gives 4000/400 = 10 chains, which needs 10 - 1 = 9 cleavages.

Teacher's Note:
1. n pieces need n - 1 cuts on a linear chain.
2. Number of chains = total units / average length.

Answer: (b) m-RNA

Explanation:
1. Messenger RNA carries the genetic message as codons (triplets of bases).
2. t-RNA carries the complementary anticodon.

Teacher's Note:
1. Codon on mRNA, anticodon on tRNA.
2. r-RNA is a component of ribosomes.

Answer: (a) steroids

Explanation:
1. Proteins, nucleic acids and polysaccharides such as glycogen are polymers of repeating units.
2. Steroids are lipids made of fused rings and are not polymeric.

Teacher's Note:
1. Biopolymers have many repeating monomers.
2. Steroids include cholesterol and sex hormones.

Answer: (a) optical activity

Explanation:
1. Carbohydrates are classified as sugars and non-sugars, as reducing or non-reducing, and as mono-, oligo- and polysaccharides according to their hydrolysis.
2. Optical activity is not used for classification.

Teacher's Note:
1. Classification uses structure, reducing power and taste.
2. Optical activity applies to individual compounds.

Answer: (b) lipid and protein

Explanation:
1. The membrane is a phospholipid bilayer in which proteins act as channels and carriers, giving selective permeability.

Teacher's Note:
1. Lipids give the barrier, proteins the transport.
2. Phospholipids are the chief membrane lipids.