JEE Chemistry Biomolecules MCQs Set 06

Answer: (d) (i) and (ii)

Explanation:
1. Denaturation destroys secondary and tertiary structure, so (i) is true.
2. DNA denaturation separates the two strands, so (ii) is true.
3. The primary structure remains intact, so (iii) is false.

Teacher's Note:
1. Only weak forces break during denaturation.
2. Peptide bonds stay unchanged.

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

Explanation:
1. Maltose is made of two alpha-D-glucose units linked through C1 to C4 alpha-glycosidic bond.
2. Hydrolysis with maltase gives only glucose.

Teacher's Note:
1. Maltose is malt sugar, a reducing sugar.
2. It is produced by the hydrolysis of starch.

Answer: (d) right, left, right, right

Explanation:
1. The aldehyde group of D-glucose becomes CH=NOH. The stereocentres are untouched.
2. D-Glucose has OH groups at C-2 right, C-3 left, C-4 right and C-5 right.
3. So the oxime shows right, left, right, right.

Teacher's Note:
1. Remember the order for D-glucose as R, L, R, R from top to bottom.
2. Epimers differ at one of these positions.

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

Explanation:
1. Maltose has two alpha-D-glucopyranose units joined by an alpha-1,4 glycosidic link.

Teacher's Note:
1. Starch gives maltose on partial hydrolysis.
2. Maltose is a reducing sugar.

Answer: (b) sugar, phosphate and bases

Explanation:
1. Each nucleotide has a sugar, a phosphate and a nitrogenous base.
2. These units form the polynucleotide chain.

Teacher's Note:
1. Nucleoside is sugar plus base.
2. Nucleotide is nucleoside plus phosphate.

Answer: (b) the sugar in RNA is ribose and in DNA is 2-deoxyribose

Explanation:
1. RNA contains ribose and DNA contains 2-deoxyribose.
2. Arabinose is not found in either.

Teacher's Note:
1. Deoxy means lacking the OH at C-2.
2. RNA is single stranded, DNA is double stranded.

Answer: (d) Sucrose

Explanation:
1. Sucrose has no free hemiacetal group because both anomeric carbons are in the glycosidic bond.
2. The others are reducing.

Teacher's Note:
1. Non-reducing sugars do not reduce Tollens' or Fehling's.
2. They also do not show mutarotation.

Answer: (a) DNA to RNA to proteins

Explanation:
1. DNA is transcribed into m-RNA, and m-RNA is translated into protein.

Teacher's Note:
1. Reverse transcription in retroviruses is an exception.
2. The dogma was proposed by Francis Crick.

Answer: (a) 2-amino-6-oxopurine (2-amino-6-hydroxypurine)

Explanation:
1. Guanine is a purine base with an amino group at C-2 and a keto (oxo) group at C-6.
2. 6-Aminopurine is adenine, and 5-methyl-2,4-dioxopyrimidine is thymine.

Teacher's Note:
1. Purines (adenine, guanine) have a fused two-ring skeleton.
2. Pyrimidines (cytosine, thymine, uracil) have a single six-membered ring.

Answer: (b) 20

Explanation:
1. Proteins in the human body are built from 20 different alpha-amino acids.
2. Four bases form 64 triplet codons, which code for these 20 amino acids and the stop signals.

Teacher's Note:
1. The number of amino acids is not four.
2. The code is degenerate, so several codons code for one amino acid.

Answer: (a) amylopectin has 1,4 alpha-linkages and 1,6 alpha-linkages

Explanation:
1. Amylose is a linear chain with only 1,4 alpha links.
2. Amylopectin is branched with 1,4 alpha links in the chains and 1,6 alpha links at the branch points.
3. So the statement in option (a) is correct.

Teacher's Note:
1. Both are polymers of alpha-D-glucose.
2. The 1,6 links make amylopectin branched.

Answer: (d) Glycine

Explanation:
1. A zwitterion needs both an acidic group and a basic group in the same molecule.
2. Glycine has COOH and NH₂, so it exists as ⁺NH₃CH₂COO^-.

Teacher's Note:
1. Aniline has only a basic group and benzoic acid only an acidic group.
2. Zwitterions are why amino acids have high melting points.

Answer: (c) DNA is a polymer of D-ribose

Explanation:
1. DNA is a polymer of nucleotides containing 2-deoxy-D-ribose, not D-ribose.
2. The other statements are correct.

Teacher's Note:
1. D-ribose is the sugar in RNA.
2. Cellulose is made of beta-1,4 linked glucose units.

Answer: (a) Tryptophan, lysine, histidine

Explanation:
1. Tryptophan, lysine and histidine cannot be made in the body in sufficient amounts and must be taken in the diet.
2. Serine is non-essential.

Teacher's Note:
1. Serine appears in options b, c and d, so those are all wrong.
2. Learn the list of essential amino acids once.

Answer: (d) All of these

Explanation:
1. All three observations suggest that the aldehyde group in glucose is not free.
2. They are explained by the cyclic hemiacetal structure.

Teacher's Note:
1. Cyclic structure also explains mutarotation and the existence of anomers.
2. Glucose has a six-membered pyranose ring.

Answer: (a) 5 and 3

Explanation:
1. The phosphate joins the 5 position of one sugar to the 3 position of the next sugar.
2. This repeats along the chain.

Teacher's Note:
1. Hence DNA chains have a 5 to 3 direction.
2. C-1 carries the base and is not used.

Answer: (c) It is present in furanose form

Explanation:
1. Glucose is an aldohexose (a), and prolonged heating with HI gives n-hexane, showing a straight chain of six carbons (b).
2. It exists mainly in the pyranose (six-membered) form, not furanose.
3. It also fails Schiff's test (d).

Teacher's Note:
1. Fructose is the one that mainly exists in the furanose form in disaccharides.
2. Glucose forms a pyranose ring.

Answer: (c) Sucrose

Explanation:
1. Sucrose is made by linking the anomeric carbons of glucose and fructose.
2. No free hemiacetal group is left, so it does not reduce Tollens' or Fehling's reagent.
3. Maltose, lactose and cellobiose are reducing.

Teacher's Note:
1. A glycosidic link at both anomeric carbons makes a sugar non-reducing.
2. Trehalose is another non-reducing disaccharide.

Answer: (c) C-1 of one unit and C-4 of the other

Explanation:
1. Maltose has an alpha-1,4-glycosidic bond.
2. The anomeric C-1 of one glucose joins the C-4 hydroxyl of the second glucose.

Teacher's Note:
1. The second glucose keeps a free C-1, so maltose is reducing.
2. Cellobiose has a beta-1,4 bond.

Answer: (d) Starch is not hydrolysed by the enzyme amylase present in saliva

Explanation:
1. Salivary amylase starts the hydrolysis of starch to maltose, so (d) is incorrect.
2. Statements a, b and c are the standard facts about starch.

Teacher's Note:
1. Amylose gives deep blue with iodine, and amylopectin gives a purple-red colour.
2. Digestion of starch begins in the mouth.

Answer: (d) All of these

Explanation:
1. Proteins are polymers of amino acids and occur in all cells.
2. Phosphoproteins (casein) contain phosphorus, thyroglobulin contains iodine, and metalloproteins contain iron, copper, zinc or manganese.
3. So all statements are true.

Teacher's Note:
1. Conjugated proteins have non-protein parts.
2. Haemoglobin, haemocyanin and carbonic anhydrase are examples.