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CBSE Class 12 Chapter Test: BIOMOLECULES
General Instructions:
- This question paper consists of 16 questions divided into 5 sections: A, B, C, D, and E.
- Section A comprises 6 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
- Section B comprises 2 Very Short Answer questions of 2 marks each.
- Section C comprises 3 Short Answer questions of 3 marks each.
- Section D comprises 1 Case-Based question of 4 marks.
- Section E comprises 2 Long Answer questions of 5 marks each.
Directions for Q7 & Q8: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Reason (R): Denaturation involves the breaking of peptide bonds between amino acids.
Reason (R): The letter 'D' before glucose represents its dextrorotatory nature.
(i) All six carbon atoms are linked in a straight chain.
(ii) The carbonyl group present is an aldehyde group.
(iii) The presence of five hydroxyl (-OH) groups.
(i) Glycosidic linkage
(ii) Invert sugar
(iii) Nucleotide
Vitamins are organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism. Vitamins are broadly classified into two groups depending upon their solubility in water or fat. Vitamins which are soluble in fat and oils but insoluble in water are kept in one group. These are stored in liver and adipose (fat storing) tissues. Vitamins which are soluble in water must be supplied regularly in diet because they are readily excreted in urine and cannot be stored in our body.
(a) Name any two fat-soluble vitamins. [1 Mark]
(b) Why must Vitamin C be supplied regularly in our diet? [1 Mark]
(c) Identify the diseases caused by the deficiency of Vitamin A and Vitamin D. [2 Marks]
(a) What is meant by the denaturation of proteins? Which levels of protein structure are altered during denaturation? [2 Marks]
(b) What happens to the water present in the egg white when an egg is boiled? [1 Mark]
(c) Write two key differences between globular and fibrous proteins. [2 Marks]
(b) Write the products obtained when D-glucose reacts with:
(i) Hydroxylamine (NH2OH)
(ii) Nitric acid (HNO3) [2 Marks]
(c) Explain the difference between the α-helix and β-pleated sheet structures of proteins. [2 Marks]
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SOLUTIONS & MARKING SCHEME
1. Answer: (d) Sucrose
Reasoning: In sucrose, the glycosidic linkage is formed between C-1 of α-glucose and C-2 of β-fructose. Since the reducing groups (aldehyde and ketone) of both monosaccharides are involved in the bond formation, it has no free reducing group and acts as a non-reducing sugar.
2. Answer: (b) Gluconic acid
Reasoning: Bromine water is a mild oxidizing agent. It oxidizes only the aldehyde (-CHO) group of glucose into a carboxyl (-COOH) group, yielding a six-carbon acid called gluconic acid.
3. Answer: (d) Uracil
Reasoning: DNA contains Adenine, Guanine, Cytosine, and Thymine. RNA contains Adenine, Guanine, Cytosine, and Uracil. Thus, Uracil replaces Thymine in RNA.
4. Answer: (b) Vitamin B12
Reasoning: Pernicious anemia, a condition where red blood cells are deficient in hemoglobin, is specifically caused by a lack of Vitamin B12 (Cyanocobalamin).
5. Answer: (c) Peptide linkage
Reasoning: Amino acids are linked together by peptide bonds (or peptide linkages), which are amide bonds (-CO-NH-) formed by the condensation of the carboxyl group of one amino acid with the amino group of another, eliminating a water molecule.
6. Answer: (c) Enzymes increase the activation energy of reactions.
Reasoning: This statement is incorrect. Like all catalysts, enzymes decrease (lower) the activation energy of the biochemical reactions they catalyze, thereby increasing the rate of the reaction.
7. Answer: (c) A is true but R is false.
Reasoning: Assertion is true: Denaturation alters the secondary and tertiary structures, but the primary structure (amino acid sequence) remains intact. Reason is false: Denaturation breaks hydrogen bonds and uncoils helices, but it does not break the strong covalent peptide bonds.
8. Answer: (c) A is true but R is false.
Reasoning: D-Glucose is indeed dextrorotatory (rotates plane-polarized light to the right), which is indicated by the "(+)" sign. However, the letter 'D' represents the relative configuration (the -OH group on the lowest chiral carbon is on the right, relative to D-glyceraldehyde). It does NOT denote optical rotation.
9. Solution:
Zwitterion: In an aqueous solution, the carboxyl group (-COOH) of an amino acid can lose a proton, and the amino group (-NH2) can accept that proton. This forms a neutral dipolar ion containing both a positive and a negative charge, known as a Zwitterion.
Structure: +H3N - CH(R) - COO-
10. Solution:
Anomers: These are cyclic stereoisomers of a monosaccharide (like glucose) that differ in configuration only at the hemiacetal carbon (C-1).
Two anomers of D-glucose: α-D-Glucose and β-D-Glucose.
Anomeric Carbon: Carbon-1 (C-1) is the anomeric carbon in glucose.
11. Solution:
(i) Straight chain: Prolonged heating of glucose with Hydrogen Iodide (HI) yields n-hexane.
(ii) Aldehyde group: Reaction of glucose with mild oxidizing agent like Bromine water oxidizes it to Gluconic acid.
(iii) Five -OH groups: Reaction of glucose with Acetic anhydride yields glucose pentaacetate, indicating the presence of five hydroxyl groups.
12. Solution:
Difference between DNA and RNA:
1. Sugar: DNA contains β-D-2-deoxyribose sugar; RNA contains β-D-ribose sugar.
2. Bases: DNA contains Thymine (T) as a pyrimidine base; RNA contains Uracil (U) instead of Thymine.
3. Structure: DNA has a double-stranded helical structure; RNA is generally single-stranded.
(Optional 4th point: DNA is responsible for heredity; RNA is responsible for protein synthesis).
13. Solution:
(i) Glycosidic linkage: An oxide linkage formed by the loss of a water molecule that joins two monosaccharide units together.
(ii) Invert sugar: The equimolar mixture of D-glucose and D-fructose obtained by the hydrolysis of sucrose. The process reverses the optical rotation from dextrorotatory (sucrose) to laevorotatory (the mixture), hence the name.
(iii) Nucleotide: The basic building block of nucleic acids. It is composed of three components: a nitrogenous base, a pentose sugar, and a phosphoric acid group.
14. Solution (Case-Based):
(a) Fat-soluble vitamins: Vitamin A, Vitamin D, Vitamin E, or Vitamin K. (Any two).
(b) Vitamin C is a water-soluble vitamin. Because it dissolves in water, it is readily excreted in urine and cannot be stored in the body's fatty tissues. Therefore, it must be supplied regularly in the diet to prevent deficiency.
(c)
- Deficiency of Vitamin A causes: Night blindness (or Xerophthalmia).
- Deficiency of Vitamin D causes: Rickets (in children) or Osteomalacia (in adults).
15. Solution:
(a) Denaturation: When a protein in its native form is subjected to physical changes (like heat) or chemical changes (like pH variation), its hydrogen bonds are disrupted. Globules unfold and helices uncoil, causing the protein to lose its biological activity.
Altered structures: Secondary (2°) and Tertiary (3°) structures are destroyed. Primary (1°) structure remains intact.
(b) Boiling egg: The water present in the egg white is absorbed or gets trapped inside the coagulated protein network (due to denaturation) via hydrogen bonding, turning it into a solid.
(c) Globular vs Fibrous Proteins:
1. Shape: Globular proteins are highly folded into a spherical shape; Fibrous proteins have polypeptide chains running parallel to form fiber-like structures.
2. Solubility: Globular proteins are generally soluble in water (e.g., Insulin); Fibrous proteins are insoluble in water (e.g., Keratin).
16. Solution:
(a) The failure to give the 2,4-DNP test indicates that D-glucose does not possess a free aldehyde (-CHO) group in its normal state. This proves that D-glucose exists predominantly in a cyclic hemiacetal structure.
(b) Products of D-glucose:
(i) With Hydroxylamine (NH2OH): Forms Glucose oxime.
(ii) With Nitric acid (HNO3): Both the terminal -CHO and -CH2OH groups are oxidized to form a dicarboxylic acid called Saccharic acid.
(c) α-helix vs β-pleated sheet:
- α-helix: The polypeptide chain coils into a right-handed screw structure, stabilized by intramolecular hydrogen bonds between the -NH group of one amino acid and the >C=O group of an adjacent turn.
- β-pleated sheet: Polypeptide chains are stretched out to maximum extension and laid side by side, stabilized by intermolecular hydrogen bonds, resembling the pleated folds of drapery.
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