Solved Exercise For Class 12 Chemistry Chapter 14 Biomolecules | Maharashtra Board

Solved Exercise For Class 12 Chemistry Chapter 14 Biomolecules | Maharashtra Board

1. Select the most correct choice.

1. CH2OH-CO-(CHOH)4 -CH2OH is an example of

a. Aldohexose

b. Aldoheptose

c. Ketotetrose

d. Ketoheptose

answer - d. Ketoheptose

2. Open chain formula of glucose does not contain

a. formyl group

b. anomeric hydroxyl group

c. primary hydroxyl group

d. secondary hydroxyl group

Answer : anomeric hydroxyl group

3. Which of the following does not apply to CH2NH2 - COOH?

a. Neutral amino acid

b. L - Amino acid

c. Exists as zwitter ion

d. Natural amino acid

Answer : L - Amino acid

4. Tryptophan is called essential amino acid because

a. it contains aromatic nucleus.

b. it is present in all the human proteins.

c. it cannot be synthesised by human body

d. it is essential constituent of enzymes

Answer : it cannot be synthesised by human body

5. A disulphide link gives rise to the following structure of a protein.

a. Primary

b. Secondary

c. Tertiary

d. Quaternary

Answer : Tertiary

6. RNA .

a. A - U base pairing 

b. P-S-P-S backbone

c. double helix

d. G - C base pairing

Answer : RNA has A - U base pairing

2. Give scientific reasons :

1. The disaccharide sucrose gives negative Tollens test while the disaccharide maltose gives a positive Tollens test. 

Answer :

1. The structure of sucrose contains glycosidic linkage between C-1 of a-glucose and C-2 of fructose.

Since the potential aldehyde and ketone groups of both the monosaccharide units are involved in the formation of the glycosidic bond (i.e ., a, B-1,2- glycosidic bond), sucrose is a non-reducing sugar and gives negative Tollen's test.

2. The glycosidic bond in maltose is in between C-1 of one glucose ring and C-4 of the other (i.e ., a -1,4-glycosidic linkage).

3. The hemiacetal group at C-1 of the second ring is not involved in the glycosidic linkage. Hence, maltose is a reducing sugar and gives positive Tollen's test.

2. On complete hydrolysis DNA gives equimolar quantities of adenine and thymine.

Answer :

1. Both the strands of DNA double helix are complementary to each other.

2. That is a number of bases on each strand are equal and complementary to each other.

3. As adenine pairs with thymine; the number of adenine bases on one strand and thymine on another are equal in number. Thus, on complete hydrolysis DNA gives equimolar quantities of adenine and thymine.

3. a-Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass.

Answer : This is due to the peculiar structure called zwitter ion structure of a -amino acids.

1. a-Amino acid molecule contains both acidic carboxyl (-COOH) group as well as basic amino (-NH2) group.

2. Proton transfer from the acidic group to the basic group of amino acid forms a salt, which is a dipolar ion called zwitter ion.

3. Thus, a-amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass.

4. Hydrolysis of sucrose is called inversion.

Answer :

1. Sucrose (C12H22011) is dextrorotatory (+66.5). 

On hydrolysis with dilute acid or an enzyme called invertase, sucrose gives equimolar mixture of D-(+)-glucose and D-(-)fructose.

2. Since the laevorotation of fructose (-92.4) is larger than the dextrorotation of glucose (+52.7), the hydrolysis product has net laevorotation. Hence, hydrolysis of sucrose is also called inversion of sucrose.

5. On boiling egg albumin becomes opaque white.

Answer :

1. Proteins when subjected to high temperature undergo disruption of noncovalent interactions which are responsible for the specific shape of protein. That is, it undergoes denaturation.

2. Denaturation disturbs the specific structure of egg albumin which causes a change in the physical properties. Thus, on boiling egg albumin becomes opaque white.

6. The following statement apply to DNA only, some to RNA only, and some to both. Label them accordingly. 

 

a. The polynucleotide is double stranded (..........).

Answer : The polynucleotideis double stranded.(DNA)

b. The polynucleotide contains uracil (...........).

Answer : The polynucleotide contains uracil. (RNA).

c. The polynucleotide contains

D-ribose (............).

answer : The polynucleotide contains D-ribose (DNA).

d. The polynucleotide contains Guanine (..........).

Answer : The polynucleotide contains Guanine (both DNA & RNA).

2. Write the sequence of the complementary strand for the following segment of a DNA molecule.

a. 5' - CGTTTAAG - 3'

answer : 3' - GCAAATTC - 5'

b. 5' - CCGGTTAATACGGC - 3'

answer : 3' - GGCCAATTATGCCG - 5'

3.  Writes the names and schematic represention of all the possible dipeptides formed from alanine, glycine and tyrosine. 

answer : 

1. Glycylglycine: Gly-Gly

2. Alanylalanine: Ala-Ala

3. Tyrosyltyrosine: Tyr-Tyr

4. Glycylalanine: Gly-Ala

5. Alanylglycine: Ala-Gly

6. Glycyltyrosine: Gly-Tyr

7. Tyrosylglycine: Tyr-Gly

8. Tyrosylalanine: Tyr-Ala

9. Alanyltyrosine: Ala-Tyr

4. Give two evidences for presence of formyl group in glucose.

answer : . Glucose gets oxidized to a six-carbon monocarboxylic acid called gluconic acid on reaction with bromine water which is a mild oxidizing agent. Thus, the carbonyl group in glucose is in the form of formyl (–CHO).

2. Hemiacetal group of glucopyranose structure is a potential aldehyde (formyl) group. It imparts reducing properties to glucose. Thus, glucose gives positive Tollen’s test or Fehling test.