CBSE Class 12 Chemistry Biomolecules Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. The number of chiral carbon atoms present in β-D-(+)-glucose (pyranose form) is:
Answer:
5
In the cyclic (pyranose) form C1, C2, C3, C4 and C5 are all chiral.

Teacher's Note:
1. In the open-chain form glucose has only 4 chiral carbons (C2–C5); ring closure makes C1 a new chiral centre.

Question 2. Name polysaccharides which is stored in the liver of animals.
Answer:
Glycogen.

Teacher's Note:
1. Glycogen is called 'animal starch' – its structure is similar to amylopectin but more highly branched.

Question 3. What structural feature is required for a carbohydrate to behave as reducing sugar?
Answer:
The carbonyl group of any one monosaccharide present in carbohydrate should be free.

Teacher's Note:
1. A free hemiacetal (or hemiketal) carbon can open up to give the free –CHO or >C=O group that reduces Tollens' or Fehling's reagent.

Question 4. Give the significance of (+) sign in the name D-(+)-glucose.
Answer:
(+) sign indicates dextro-rotatory nature of glucose.

Teacher's Note:
1. D and (+) are independent: D tells the configuration at C5, (+) is the measured direction of optical rotation.

Question 5. Glucose is an aldose sugar but it does not react with sodium hydrogen sulphite. Give reason.
Answer:
The –CHO group reacts with –OH group at C-5 to form a cyclic hemiacetal.

Teacher's Note:
1. This is one of the standard pieces of evidence for the cyclic structure of glucose.

Question 6. Why is sucrose called invert sugar?
Answer:
When sucrose is hydrolysed by water, the optical rotation of solution changes from positive to negative.

Teacher's Note:
1. Mention the reason in exams: fructose (−92.4°) rotates more strongly than glucose (+52.5°), so the mixture becomes laevorotatory.

Question 7. Name the amino acid which is not optically active.
Answer:
Glycine.

Teacher's Note:
1. Glycine (H₂N–CH₂–COOH) has R = H, so its α-carbon has two identical H atoms and is not chiral.

Question 8. Give reason: Amylase present in the saliva becomes inactive in the stomach.
Answer:
HCl present in stomach decreases the pH.

Teacher's Note:
1. Enzymes work in a narrow pH range; salivary amylase works near pH 7, while the stomach is strongly acidic.

Question 9. Which forces are responsible for the stability of α-helical structure of proteins?
Answer:
Hydrogen bonding.

Teacher's Note:
1. In the α-helix, H-bonds form between the –NH of one peptide unit and the C=O of the fourth unit along the same chain.

Question 10. Which nucleic acid is responsible for carrying out protein synthesis in the cell?
Answer:
RNA (Ribonucleic acid)

Teacher's Note:
1. DNA stores the message; the RNAs (mRNA, tRNA, rRNA) carry it out to build proteins.

Question 11. When RNA is hydrolysed, there is no relationship among quantities of different bases obtained. What does this fact suggest about structures of RNA?
Answer:
RNA is single stranded.

Teacher's Note:
1. In double-stranded DNA, A = T and G = C because of base pairing; no such equality in RNA means no complementary second strand.

Question 12. What type of linkage holds together the monomers of DNA and RNA?
Answer:
Phosphodiester linkage.

Teacher's Note:
1. The phosphate links C5′ of one nucleotide's sugar to C3′ of the next.

Question 13. Give the Haworth projection of D-glucopyranose.
Answer:

Teacher's Note:
1. Check the D-glucose pattern in a Haworth ring: –OH down at C2, up at C3, down at C4, with CH₂OH up; only C1 differs between α (down) and β (up).

Question 14. Name the vitamin responsible for coagulation of blood.
Answer:
Vitamin K.

Teacher's Note:
1. Deficiency of vitamin K increases blood clotting time.

Question 15. Where does the water present in the egg go after boiling the egg?
Answer:
On boiling, during denaturation process water gets absorbed in denatured proteins.

Teacher's Note:
1. The water is held (bound) by the coagulated, denatured protein, which is why boiled egg white becomes solid but not dry.

Question 16. What is native state of protein?
Answer:
The energetically most stable shape of the protein at normal pH and temperature is called native state.

Teacher's Note:
1. A protein is biologically active only in its native state; denaturation destroys it.

Question 17. Why is cellulose not digested in human body?
Answer:
It is due to the fact that human beings do not have enzyme to digest cellulose.

Teacher's Note:
1. The missing enzyme is cellulase, which breaks the β-glycosidic links; ruminants get it from microbes in their gut.

Question 18. Name the enzyme that is used to dissolve blood clots.
Answer:
Streptokinase.

Teacher's Note:
1. Streptokinase is used medically as a clot-dissolving (thrombolytic) agent.

Short Answer Questions (2 Marks)

Question 19. Define the following terms in relation to proteins:
(i) Peptide linkage   (ii) Denaturation
Answer:
(i) Peptide linkage: A link between two amino acids with loss of water, –CO–NH– peptide linkage.
(ii) A process that changes the three dimensional structure of native protein is called denaturation of protein. It results into breaking of hydrogen bonds and disulphide linkages. Thus, a completely denatured protein has a shape of random coil.

Teacher's Note:
1. Denaturation can be caused by heat or change in pH; the primary structure is not affected.

Question 20. Explain what is meant by: Biocatalyst
Answer:
Biocatalysts are the catalysts which increase the rate of metabolism/biochemical reactions.

Teacher's Note:
1. Enzymes are biocatalysts – mostly globular proteins that are highly specific for their substrate.

Question 21. Explain the following terms:
(i) Invert sugar   (ii) Polypeptides
Answer:
(i) An equimolar mixture of glucose and fructose produced on hydrolysis of sucrose is called invert sugar. It is called so because sucrose is dextrorotatory whereas its hydrolysis product is laevorotatory.
(ii) Polypeptides are polymers of amino acids containing less than 100 amino acids. For example, oxytocin, vasopressin, etc.

Teacher's Note:
1. Chains of more than about 100 amino acid residues (molecular mass above 10,000 u) are called proteins.

Question 22. Explain what is meant by: Glycosidic linkage
Answer:
The linkage between the monosaccharide units through oxygen is called glycosidic linkage.

Teacher's Note:
1. It forms between the hemiacetal –OH of one sugar and an –OH of another with loss of water.

Question 23. Name the product of hydrolysis of sucrose. Why is sucrose not a reducing sugar?
Answer:
On hydrolysis, sucrose gives equimolar mixture of D-(+)-glucose and D-(−)-fructose. Sucrose is not a reducing sugar as glucose and fructose are linked through their reducing centres in structure of sucrose.

Teacher's Note:
1. The link is between C1 of α-glucose and C2 of β-fructose – both anomeric carbons are used up.

Question 24. State clearly what are known as nucleotides and nucleosides.
Answer:
A nucleoside contains only two basic components of nucleic acids i.e., pentose sugar and nitrogenous base.
A nucleotide contains all the three basic components of nucleic acids i.e., a phosphoric acid group, pentose sugar and nitrogenous base.

Teacher's Note:
1. Quick check: nucleoside + phosphate = nucleotide. The base is joined at C1′ and the phosphate at C5′ of the sugar.

Question 25. What is essentially the difference between α-form of glucose and β-form of glucose? Explain.
Answer:
α-form of glucose and β-form of glucose differ only in the configuration of the hydroxyl group at C₁ in cyclic structure of glucose/hemiacetal form of glucose.

Teacher's Note:
1. Such isomers differing only at the anomeric carbon (C1) are called anomers.

Question 26. What are anomers? Give the structures of two anomers of glucose.
Answer:
Monosaccharides which differ in configuration at the functional group carbon atom (C₁ in aldoses and C₂ in ketoses), e.g., α-glucose and β-glucose.

Teacher's Note:
1. In α-glucose the C1–OH is on the same side as the ring oxygen bond (right) in this projection; in β-glucose it is on the opposite side.

Question 27. Write the hydrolysed product of:
(i) Maltose   (ii) Cellulose
Answer:
(i) α-D-glucose
(ii) β-D-glucose

Teacher's Note:
1. Maltose gives two molecules of α-D-glucose; cellulose is a linear polymer of β-D-glucose.

Question 28. (i) Acetylation of glucose with acetic anhydride gives glucose pentaacetate. Write the structure of pentaacetate.
(ii) Explain why glucose pentaacetate does not react with hydroxylamine?
Answer:
(i) CHO–(CH–O–CO–CH₃)₄–CH₂–O–CO–CH₃ (Glucose pentaacetate)
(ii) The molecule of glucose pentaacetate has a cyclic structure in which –CHO is involved in ring formation.

Teacher's Note:
1. Formation of a pentaacetate shows that glucose has five –OH groups, each on a different carbon.

Question 29. What are vitamins? How are they classified?
Answer:
Vitamins are a group of biomolecules (other than carbohydrates, fats and proteins) most of which cannot be produced by body but must be supplied in small amount to perform specific biological functions of the body.
Types:
(i) Water soluble vitamins: Vitamin B and C.
(ii) Fat soluble vitamins: Vitamin A, D, E and K.

Teacher's Note:
1. Fat soluble vitamins are stored in liver and adipose tissue; water soluble ones (except B₁₂) must be supplied regularly.

Question 30. Write the products of oxidation of glucose with:
(i) Bromine water   (ii) Nitric acid
Answer:
(i) CHO–(CHOH)₄–CH₂OH (Glucose) + [O] →^{Br₂–H₂O} COOH–(CHOH)₄–CH₂OH (Gluconic acid)
(ii) CHO–(CHOH)₄–CH₂OH (Glucose) →^HNO₃ COOH–(CHOH)₄–COOH (Saccharic acid)

Teacher's Note:
1. Bromine water is a mild oxidant and attacks only –CHO; nitric acid also oxidises the primary –CH₂OH, giving a dicarboxylic acid.

Teacher's Note:
1. Examples help: insulin and albumin are globular; keratin and myosin are fibrous.

Short Answer Questions (3 Marks)

Question 32. Define the following with an example of each:
(a) Polysaccharides
(b) Denatured protein
(c) Essential amino acids [CBSE Delhi/OD 2018]
Answer:
(a) Carbohydrates that give large number of monosaccharide units on hydrolysis / large number of monosaccharides units joined together by glycosidic linkage. Example: Starch/ glycogen/ cellulose (or any other)
(b) Proteins that lose their biological activity / proteins in which secondary and tertiary structures are destroyed. Example: Curdling of milk (or any other)
(c) Amino acids which cannot be synthesised in the body. Example: Valine / Leucine (or any other)

Teacher's Note:
1. For essential amino acids, keep a few ready: valine, leucine, isoleucine, lysine, threonine, methionine, phenylalanine, tryptophan, histidine, arginine.

Question 33. (a) Write the product when D-glucose reacts with conc. HNO₃.
(b) Amino acids show amphoteric behaviour. Why?
(c) Write one difference between α-helix and β-pleated structures of proteins. [CBSE Delhi/OD 2018]
Answer:
(a) Saccharic acid / HOOC–(CHOH)₄–COOH
(b) Due to the presence of carboxyl and amino group in the same molecule / due to formation of zwitter ion or dipolar ion.
(c) α-helix has intramolecular hydrogen bonding while β pleated has intermolecular hydrogen bonding / α-helix results due to regular coiling of polypeptide chains while in β pleated all polypeptide chains are stretched and arranged side by side.

Teacher's Note:
1. In the zwitter ion H₃N⁺–CHR–COO⁻, the –COO⁻ accepts a proton (base) and the –NH₃⁺ donates one (acid).

Question 34. (i) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar.
(ii) Why vitamin C cannot be stored in our body?
(iii) What is the difference between a nucleoside and nucleotide? [CBSE Delhi 2016; DDE]
Answer:
(i) β-D-Glucose and β-D-Galactose.
(ii) It is water soluble, hence it is excreted through urine.
(iii) Nucleoside: It is formed when pentose sugar combines with nitrogen base. Nucleotide: When nucleoside bonds with phosphate group.

Teacher's Note:
1. Remember: nucleoside = base + sugar; nucleotide = base + sugar + phosphate (nucleoside + phosphate at C-5′).

Question 35. (i) Write the structural difference between starch and cellulose.
(ii) What type of linkage is present in Nucleic acids?
(iii) Give one example each for fibrous protein and globular protein. [CBSE OD 2016]
Answer:
(i) In starch, the glucose monomers are in alpha configuration while in cellulose the glucose monomers are in beta configuration. Starch is a polymer consisting of amylose and amylopectin while cellulose is a long chain composed only of β-D-glucose units.
(ii) Phosphodiester linkage between the 5′ and 3′ atoms is present in nucleic acids.
(iii) Example of fibrous protein: Collagen, keratin, myosin. (Any one) Example of globular protein: Insulin, haemoglobin, egg albumin. (Any one)

Teacher's Note:
1. α- versus β-glycosidic links is why humans can digest starch but not cellulose – we lack the enzyme for β-links.

Question 36. How are vitamins classified? Name the vitamin responsible for the coagulation of blood. [CBSE Comptt. Delhi 2015; NCERT]
Answer:
Vitamins are classified into:
(i) Water soluble vitamins: Vitamins which are soluble in water but insoluble in fat and oils. e.g., vitamin B and C.
(ii) Fat soluble vitamins: Vitamins are soluble in fat and oils but insoluble in water. e.g., vitamin A, D, E and K.
Vitamin K is responsible for coagulation of blood.

Teacher's Note:
1. A quick memory aid for fat soluble vitamins is ‘A, D, E, K’; everything else (B group and C) is water soluble.

Question 37. (i) Which one of the following is a polysaccharide: starch, maltose, fructose, glucose
(ii) Write one difference between α-helix and β-pleated sheet structures of protein.
(iii) Write the name of the disease caused by the deficiency of vitamin B₁₂. [CBSE OD 2015]
Answer:
(i) Starch.
(ii) α-Helix polypeptide chains are stabilized by intramolecular H-bonding whereas β-pleated sheet is stabilized by intermolecular H-bonding. (or any other difference)
(iii) Pernicious anaemia.

Teacher's Note:
1. Understand the deficiency diseases of different vitamins in detail: A – night blindness, B₁ – beri-beri, B₁₂ – pernicious anaemia, C – scurvy, D – rickets, K – delayed blood clotting.

Question 38. (i) Deficiency of which vitamin causes night blindness?
(ii) Name the base that is found in nucleotide of RNA only.
(iii) Glucose on reaction with HI gives n-hexane. What does it suggest about the structure of glucose? [CBSE Delhi 2014; DDE]
Answer:
(i) Vitamin A
(ii) Uracil
(iii) It suggests that six carbon atoms are in straight chain/CHO—(CHOH)₄—CH₂OH

Teacher's Note:
1. Students often confuse the bases of RNA and DNA: uracil is only in RNA and thymine is only in DNA.

Question 39. (i) Deficiency of which vitamin causes scurvy?
(ii) What type of linkage is responsible for the formation of proteins?
(iii) Write the product formed when glucose is treated with HI. [CBSE Delhi 2014; DDE]
Answer:
(i) Vitamin C.
(ii) Peptide linkage.
(iii) n-hexane or its structure, CH₃–CH₂–CH₂–CH₂–CH₂–CH₃.

Teacher's Note:
1. HI with prolonged heating reduces every carbon of glucose, so the product has no oxygen at all – it is the alkane C₆H₁₄.

Teacher's Note:
1. Bone deformity in children is rickets (vitamin D); in adults the same deficiency causes osteomalacia.

Question 41. (i) Why water soluble vitamins must be supplied regularly in the diet? Give one example of it.
(ii) Differentiate between the following:
(a) Essential and non-essential amino acids.
(b) Fibrous and globular proteins. [CBSE Comptt. OD 2017]
Answer:
(i) Because they are excreted in urine and cannot be stored in body; Vitamin C/B₁/B₆
(ii)(a) Essential amino acids are those which cannot be synthesized in the body and are supplied through diet whereas non-essential amino acid can be synthesized in the body.
(b) In fibrous proteins, the polypeptide chains run parallel and are held together by hydrogen or disulphide bonds while in globular, polypeptide chains coil around to give a spherical shape.

Teacher's Note:
1. Avoid giving vitamin B₁₂ as the example in part (i): it is water soluble but is stored in the liver.

Long Answer Questions (5 Marks)

Question 42. On the basis of which evidences D-glucose was assigned the following structure?
CHO–(CHOH)₄–CH₂OH [NCERT Exemplar]
Answer:
This structure was assigned on the basis of the following evidences:
(i) Molecular formula: C₆H₁₂O₆ is molecular formula of glucose.
(ii) Straight chain structure:
(a) When aqueous solution of glucose is treated with sodium amalgam (Na/Hg) or sodium borohydride, it is reduced to sorbitol (or glucitol) a hexahydric alcohol.
OHC–(CHOH)₄–CH₂OH + 2[H] \( \xrightarrow{\text{Na-amalgam}} \) HOH₂C–(CHOH)₄–CH₂OH (Sorbitol)
(b) Prolonged heating with hydriodic acid and red phosphorous at 100°C gives a mixture of n-hexane and 2-iodohexane.
CH₂OH(CHOH)₄CHO \( \xrightarrow[\text{red P, 100}^{\circ}\text{C}]{\text{HI}} \) CH₃(CH₂)₄CH₃ (n-hexane) + CH₃–CH(I)–(CH₂)₃CH₃ (2-Iodohexane)
The formation of n-hexane suggests that all the six carbon atoms in glucose are arranged in a straight chain structure.
(c) Presence of five –OH groups: On acetylation with acetic anhydride, glucose gives a pentaacetate. This confirms that glucose contains five –OH groups. We know that the presence of two or more –OH groups on the same carbon atom makes the molecules unstable. Since glucose is a stable compound, therefore, the five –OH groups must be present on different carbon atoms.
(d) Presence of one primary alcoholic group: On oxidation with conc. HNO₃, both glucose and gluconic acid give the same dicarboxylic acid and saccharic acid. The primary alcoholic group (–CH₂OH) is always present at the end of the carbon chain.
(e) Presence of an aldehyde (–CHO) group: Glucose reacts with hydroxylamine, NH₂OH to form oxime. This suggest that glucose contains a carbonyl (>C=O) group.
On the basis of above observations, the following open chain structure for glucose can be written as follows:
CHO–(CHOH)₄–CH₂OH

Teacher's Note:
1. Oxime formation only shows a carbonyl group; it is the oxidation by bromine water to gluconic acid (same number of carbons) that shows the carbonyl is an aldehyde.
2. Learn the evidence in this order – formula, chain, five OH, carbonyl, aldehyde, primary alcohol – so no point is missed in a 5-mark answer.