CBSE Class 12 Chemistry Biomolecules Questions Answers

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Class 12 Chemistry Biomolecules questions and answers for CBSE board exam practice. This set covers carbohydrates such as glucose, sucrose, lactose, starch and cellulose, proteins with their structures and denaturation, amino acids, nucleic acids, enzymes and vitamins, with one mark, two mark and three mark questions and a set of quick recall questions.

Biomolecules

Section A (One Mark Question)

1. Name the sugar present in milk.

Answer: Lactose.

Teacher's Note:
a) Lactose is the only sugar naturally found in milk, so it is the answer for any milk question.

2. How many monosaccharide units are present in it?

Answer: Two monosaccharide units are present, glucose and galactose.

Teacher's Note:
a) Lactose is a disaccharide, so it always gives two monosaccharide units on hydrolysis.

3. What are such oligosaccharides called?

Answer: Such oligosaccharides are called disaccharides.

Teacher's Note:
a) Oligosaccharides with two monosaccharide units are called disaccharides.

4. How do you explain the presence of all the six carbon atoms in glucose in a Straight chain?

Answer: On prolonged heating with HI, glucose gives n-hexane (CH₃-CH₂-CH₂-CH₂-CH₂-CH₃), which shows that all six carbon atoms are linked in a straight chain.

Teacher's Note:
a) Reduction with HI removes the oxygen atoms, and a straight chain of six carbons is left as n-hexane.

3. Name the linkage connecting monosaccharide units in polysaccharides.

Answer: Glycosidic linkage.

Teacher's Note:
a) A glycosidic linkage is an oxygen bridge formed between two monosaccharide units with loss of water.

4. Under what conditions glucose is converted to gluconic and saccharic acid?

Answer: Glucose is converted to gluconic acid by bromine water and to saccharic acid by conc. HNO₃.

Teacher's Note:
a) Bromine water oxidises only the aldehyde group, while concentrated nitric acid oxidises both the aldehyde and the primary alcohol group.

5. Which sugar is called invert sugar?

Answer: Sucrose.

Teacher's Note:
a) Sucrose on hydrolysis gives an equimolar mixture of glucose and fructose, which is called invert sugar.

6. During curdling of milk, what happens to sugar present in it?

Answer: It converts into lactic acid.

Teacher's Note:
a) Bacteria convert the lactose of milk into lactic acid, and the acid causes the milk protein to curdle.

7. Monosaccharide contain carbonyl group hence are classified, as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule are also considered for classification. In which class of monosaccharide will you place fructose?

Answer: Fructose is a ketohexose.

Teacher's Note:
a) Fructose has a ketone group, so it is a ketose, and it has six carbon atoms, so it is a hexose.

8. The letters 'D' or 'L' before the name of a stereoisomer of a compound indicate the correlation of configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde. Predict whether the following compound has 'D' or 'L' configuration.

[Figure: Fischer projection with CHO at the top and CH₂OH at the bottom. The second carbon from the top has HO on the left and H on the right. The third carbon has H on the left and OH on the right. The fourth carbon has HO on the left and H on the right. The fifth carbon has HO on the left and H on the right.]

Answer: 'L' configuration.

Teacher's Note:
a) The OH group on the carbon next to CH₂OH is on the left, so the compound has the 'L' configuration.

9. What are constituents of Starch?

Answer: Amylose and amylopectin.

Teacher's Note:
a) Amylose is the water soluble straight chain part, and amylopectin is the branched part of starch.

10. What D N A & R N A Stand for?

Answer: Deoxyribonucleic acid and ribonucleic acid.

Teacher's Note:
a) DNA contains deoxyribose sugar, while RNA contains ribose sugar.

11. What are Zwitter ions?

Answer: A zwitter ion is a dipolar ion formed by neutralization of acidic and basic centres present within the molecule.

Teacher's Note:
a) Amino acids exist as zwitter ions, with a positive ammonium group and a negative carboxylate group in the same molecule.

12. What is non reducing sugar? Give example.

Answer: Sugars in which the groups like CHO and C=O are not freely available in the molecule, so they do not answer Tollens' or Fehling's test, are called non reducing sugars. Example: sucrose.

Teacher's Note:
a) The printed key gives maltose and lactose as examples; these are reducing sugars, so the correct example is sucrose.
b) Maltose and lactose have a free hemiacetal group, which is why they reduce Tollens' and Fehling's reagents.

13. Define mutarotation? Give example.

Answer: Mutarotation is the change in specific rotation of an optically active compound when its anomers are left in solution. Example: alpha-D-glucose (+111°) and beta-D-glucose (+19.2°) change to an equilibrium mixture with a rotation of +52.4°.

Teacher's Note:
a) The change happens because the open chain form of glucose forms both anomers again in solution.
b) Mutarotation is a change in rotation, not a change in the structure of the compound as a whole.

14. Amino acids are amphoteric in behavior? Explain.

Answer: Amino acids form a zwitter ion (dipolar ion) and act as both acid and base. At the isoelectric point they exist as neutral zwitter ions.

Teacher's Note:
a) The isoelectric point is not always pH 7; it depends on the amino acid, so the printed key's pH 7 is only a special case.

Section B (2 Mark Questions.)

1. Define native state and denaturation of protein. What happens when?
a. Protein is cooled to zero degree C?
b. Protein is heated to 80° C

Answer:
1. A protein in the native state has a definite configuration and biological activity.
2. Denaturation is the change in the higher structure of a protein, without disturbing the primary structure, which destroys its biological activity.
a. No change.
b. The coagulation of the protein takes place.

Teacher's Note:
a) Low temperature usually does not change the folded structure, so the protein stays in its native state.
b) Heating to about 80 °C breaks the hydrogen bonds and the protein coagulates.

2. Which forces are responsible for stability of alpha Helix of protein? Why it is called 3.6₁₃ helix?

Answer:
1. Hydrogen bonding is responsible for the stability of the alpha helix.
2. There are 3.6 amino acids in one complete turn of the helix.
3. A 13 membered ring is formed by hydrogen bonding, so it is called 3.6₁₃ helix.

Teacher's Note:
a) Hydrogen bonds form between the -NH group of one residue and the C=O group of another, four residues along the chain.
b) The 3.6 residues per turn means the helix rises about 5.4 Å for each turn.

3. What are essential amino acids? Give example and what happens when it is polymerized?

Answer:
1. Essential amino acids are amino acids required by the body which cannot be synthesized in the body. Example: lysine.
2. When they are polymerized, polypeptide chains are formed.

Teacher's Note:
a) Essential amino acids must come from food, because the body cannot make them.
b) Polymerising amino acids through peptide bonds gives the polypeptide chains that form proteins.

4. Glucose and sucrose are soluble in water but Cyclohexane and benzene are not soluble. Why?

Answer: Glucose and sucrose form hydrogen bonds with water, which makes them soluble. Cyclohexane and benzene are non polar and cannot form hydrogen bonds with water, so they are not soluble.

Teacher's Note:
a) Like dissolves like: polar solutes dissolve in polar water, and non polar hydrocarbons do not.
b) Many -OH groups in glucose and sucrose give them strong hydrogen bonding with water.

5. (i) Write the sequence of base on mRNA molecule synthesized on the following strand of DNA:
TATCTACCTGGA
(ii) Name a powerful antioxidant which is a water soluble vitamin.

Answer:
(i) AUAGAUGGACCU
(ii) Vitamin C.

Teacher's Note:
a) Each base of the DNA strand pairs with a complementary base in mRNA: A pairs with U, T with A, C with G and G with C.
b) Vitamin C is a powerful water soluble antioxidant.

Section C (THREE MARKS QUESTIONS)

1. (i) Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form is subjected to a physical change like change in temperature or a chemical change like change in pH, denaturation of protein takes place. Explain the cause.
(ii) Structures of glycine and alanine are given below. Show the peptide linkage in glycylalanine.

[Figure: Glycine is H₂N-CH₂-COOH. Alanine is H₂N-CH(CH₃)-COOH, with CH₃ drawn below the second carbon. An arrow with -H₂O shows them combining to give glycylalanine, H₂N-CH₂-CO-NH-CH(CH₃)-COOH, with the peptide linkage -CO-NH- boxed.]

Answer:
(i) Due to a physical or chemical change, the hydrogen bonds in the protein are disturbed, the globules unfold and the helix uncoils. Therefore the protein loses its biological activity. This is called denaturation of proteins.
(ii) The carboxyl group (-COOH) of glycine joins the amino group (-NH₂) of alanine with the loss of one water molecule. The linkage formed is -CO-NH-, which is the peptide linkage: H₂N-CH₂-CO-NH-CH(CH₃)-COOH.

Teacher's Note:
a) Denaturation does not break the primary structure, only the hydrogen bonds that hold the higher structure.
b) In the peptide linkage the C=O belongs to glycine and the N-H belongs to alanine.
c) The sequence of amino acids is written from the free -NH₂ end to the free -COOH end.

2. (i) What are the expected products of hydrolysis of lactose?
(ii) How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?

Answer:
(i) Lactose is composed of beta-D-galactose and beta-D-glucose. The linkage is between C1 of galactose and C4 of glucose. On hydrolysis, it gives beta-D-galactose and beta-D-glucose.
(ii) The pentaacetate of glucose does not react with hydroxylamine, which indicates the absence of a free -CHO group.

Teacher's Note:
a) The C1 of galactose is linked to the C4 of glucose, which is why lactose is a reducing sugar through its glucose unit only.
b) All five -OH groups are acetylated, and the aldehyde group does not appear free in the pentaacetate.
c) This observation supports the cyclic structure of glucose.

3. (i) What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
(ii) How will you distinguish 1° and 2° hydroxyl groups present in glucose?

Answer:
(i) Complete hydrolysis of the nucleotide gives a pentose sugar (deoxyribose), phosphoric acid and thymine.
(ii) On oxidation with nitric acid, both glucose and gluconic acid give a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (-OH) group in glucose, because the -CH₂OH group is oxidised to -COOH.

Teacher's Note:
a) Thymine is a pyrimidine base that is found in DNA but not in RNA.
b) Saccharic acid has a carboxyl group at each end, which is why it is called dicarboxylic.
c) The primary alcohol at C6 is the only group that oxidises to a carboxyl group in addition to the aldehyde.

4. Write the reactions of D-glucose which can't be explained by its open-chain structure. How can cyclic structure of glucose explain these reactions?

Answer:
1. Despite having the aldehyde group, glucose does not give Schiff's test and does not form the hydrogensulphite addition product with NaHSO₃.
2. The pentaacetate of glucose does not react with hydroxylamine, which indicates the absence of a free -CHO group.
3. Glucose exists in two different crystalline forms, alpha and beta. The alpha-form (m.p. 419 K) is obtained by crystallisation from a concentrated solution of glucose at 303 K. The beta-form (m.p. 423 K) is obtained by crystallisation from a hot and saturated aqueous solution at 371 K.
4. In the cyclic (hemiacetal) structure the C1 aldehyde carbon is bonded to the ring oxygen, so the free -CHO group is absent and these reactions do not occur.
5. The alpha and beta forms differ only in the configuration at C1, the anomeric carbon, which explains the two crystalline forms.

Teacher's Note:
a) The cyclic structure is formed when the C5 -OH adds to the C1 aldehyde group.
b) Mutarotation happens because the ring opens and closes to give both anomers.
c) Students should remember that the free aldehyde group is needed for Schiff's test and the NaHSO₃ addition.

5. Write the evidences for the following on the basis open chain structure of Glucose.
(i) all the six carbon atoms are linked in a straight chain.
(ii) the presence of a carbonyl group (>C = O) in glucose.
(iii) five -OH groups are attached to different carbon atoms.

[Figure: Open chain glucose drawn as CHO at the top, a chain of four CHOH groups, and CH₂OH at the bottom. Three reaction schemes are shown: HI with heat gives n-hexane; NH₂OH gives the oxime CH=N-OH on the first carbon; HCN gives the cyanohydrin with CN and OH on the first carbon; acetic anhydride gives the pentaacetate with five -O-CO-CH₃ groups.]

Answer:
(i) On prolonged heating with HI, glucose gives n-hexane, which shows that all six carbon atoms are linked in a straight chain.
(ii) Glucose reacts with hydroxylamine (NH₂OH) to form an oxime, and with HCN to form a cyanohydrin. Both reactions show the presence of a carbonyl group.
(iii) Glucose reacts with acetic anhydride to form a pentaacetate, which shows that five -OH groups are present.

Teacher's Note:
a) The n-hexane test proves the straight chain but not the position of the groups.
b) Oxime and cyanohydrin formation both need a free carbonyl group.
c) Pentaacetate formation is the test for the number of -OH groups.

6. Explain the terms primary structure of proteins.

Answer: Primary structure of proteins: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence, and this sequence of amino acids is called the primary structure of that protein. Any change in this primary structure, that is in the sequence of amino acids, creates a different protein.

Teacher's Note:
a) The primary structure is held together by peptide bonds.
b) Changing one amino acid in the sequence can change the whole protein, as in sickle cell anaemia.
c) The primary structure is not disturbed by denaturation.

7. Explain the terms secondary structure of proteins. What is the difference between alpha-helix and beta-pleated sheet structure of proteins?

Answer: Secondary structure of proteins: The secondary structure of a protein refers to the shape in which a long polypeptide chain can exist. It exists in two types of structures, alpha-helix and beta-pleated sheet. These structures arise due to the regular folding of the backbone of the polypeptide chain, caused by hydrogen bonding between -CO- and -NH- groups of the peptide bonds.
In alpha-helix, the polypeptide chain twists into a right handed screw (helix). The -NH group of each amino acid residue is hydrogen bonded to the C=O of an adjacent turn of the helix, so all possible hydrogen bonds are formed.
In beta-pleated sheet, the polypeptide chains are stretched out to nearly maximum extension and laid side by side. They are held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery.

Teacher's Note:
a) Alpha-helix is a coiled structure held by intramolecular hydrogen bonds within one chain.
b) Beta-pleated sheet is an extended structure held by hydrogen bonds between neighbouring chains.
c) Hydrogen bonding is the common force behind both structures, which is why they are called secondary structures.

8. Explain tertiary structure of Protein.

Answer: Tertiary structure of proteins: The tertiary structure of proteins represents the overall folding of the polypeptide chains, that is further folding of the secondary structure. It gives rise to two major molecular shapes, fibrous and globular. The main forces which stabilise the secondary and tertiary structures of proteins are hydrogen bonds, disulphide linkages, van der Waals forces and electrostatic forces of attraction.

Teacher's Note:
a) Fibrous proteins are long and insoluble, like keratin, while globular proteins are spherical and often soluble.
b) Disulphide linkages form between two cysteine residues.
c) Tertiary structure is what gives an enzyme its active site shape.

9. What are structural difference between Cellulose and Starch?

Answer:
1. Starch is a polymer of alpha-glucose and consists of two components, amylose and amylopectin.
2. Amylose is the water soluble component which constitutes about 15 to 20% of starch. Chemically, amylose is a long unbranched chain with 200 to 1000 alpha-D-(+)-glucose units held by C1 to C4 glycosidic linkages.
3. Amylopectin is the water insoluble component. It is a branched chain polymer of alpha-D-glucose units, with C1 to C4 linkages in the chain and C1 to C6 linkages at the branch points.
4. Cellulose is a straight chain polysaccharide composed only of beta-D-glucose units, which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.

[Figure: Amylose drawn as glucose rings joined by alpha-links, with the glycosidic O bridge between C1 of one ring and C4 of the next. Cellulose drawn as glucose rings joined by beta-links, with the arrows marking each beta-link.]

Teacher's Note:
a) Amylopectin is the branched part of starch, and this branching is the main structural difference from the straight chain of cellulose.
b) The alpha-linkage in starch makes it digestible, while the beta-linkage in cellulose cannot be broken by human digestive enzymes.
c) Starch forms a helical shape because of its alpha-linkages, while cellulose forms straight, hydrogen bonded fibres.

BIOMOLECULES

Q1. What is invert sugar ?

Answer: A mixture of glucose and fructose.

Teacher's Note:
a) Invert sugar is formed when sucrose is hydrolysed.
b) The name comes from the change in the direction of rotation during hydrolysis.

Q2. Which disaccharide is present in milk ?

Answer: Lactose.

Teacher's Note:
a) Milk sugar is lactose, made of glucose and galactose.

Q3. Which Polysaccharide is present in Rice ?

Answer: Starch.

Teacher's Note:
a) Rice, wheat and potato are rich in starch, which the body stores as a food reserve in plants.
b) Starch is not a sugar, because it is a polymer of many glucose units.

Q4. Write hydrolysis product of Lactose.

Answer: Glucose and galactose.

Teacher's Note:
a) Lactose is split by the enzyme lactase or by acid into these two monosaccharides.

Q5. Name the carbohydrate which is not digested in human body ?

Answer: Cellulose.

Teacher's Note:
a) Humans lack the enzyme that breaks the beta-glycosidic linkage of cellulose.
b) Cellulose is still useful as dietary fibre, which helps digestion.

Q6. Which Functional groups are present in glucose ?

Answer: -OH and -CHO.

Teacher's Note:
a) Glucose has five -OH groups and one aldehyde group in its open chain form.
b) The aldehyde group is the reason glucose is a reducing sugar.

Q7. What are the constituents of Maltose ?

Answer: Two units of alpha-D-glucose.

Teacher's Note:
a) Maltose is formed from starch by the action of the enzyme diastase.
b) Maltose has a free hemiacetal group, so it is a reducing sugar.

Q8. What are the constituents of Starch ?

Answer: Amylose and amylopectin.

Teacher's Note:
a) Amylose makes up about 15 to 20% of starch.
b) Amylopectin makes up the rest and is branched.

Q9. Which Polysaccharide is stored in the Liver of Animals ?

Answer: Glycogen.

Teacher's Note:
a) Glycogen is the animal equivalent of starch.
b) Glycogen is also stored in muscles, and it is more highly branched than amylopectin.

Q10. Name two Carbohydrades which act as byfuels?

Answer: Starch and glycogen.

Teacher's Note:
a) Both are stored food reserves that are broken down to glucose to release energy.
b) Starch is the reserve in plants, and glycogen is the reserve in animals.

Q11. Write the water soluble vitamins .

Answer: Vitamin B (the B-complex group) and vitamin C.

Teacher's Note:
a) Water soluble vitamins are not stored much in the body, so they must be taken daily.
b) Vitamins A, D, E and K are fat soluble.

Q12. Give an example of Fibrous Protein .

Answer: Keratin.

Teacher's Note:
a) Keratin is found in hair, nails and wool.
b) Fibrous proteins are long, insoluble and held by hydrogen bonds.

Q13. Give two examples of essential amino acids .

Answer: Valine and lysine.

Teacher's Note:
a) The printed key spells the second example as Lycine; the correct name is lysine.
b) Other essential amino acids include leucine, isoleucine, methionine and phenylalanine.

Q14. Name the Enzymes present in the saliva of Human.

Answer: Amylase (ptyalin).

Teacher's Note:
a) Salivary amylase starts the digestion of starch in the mouth.
b) The enzyme acts best at a neutral pH.

Q15. What is Glycogen ?

Answer: Glycogen is a polysaccharide.

Teacher's Note:
a) It is a polymer of alpha-D-glucose units.
b) It is the storage carbohydrate of animals.

Q16. What type of bonding occurs in globular Proteins ?

Answer: Van der Waals interactions, dipolar interactions and hydrogen bonds.

Teacher's Note:
a) Globular proteins are folded into compact shapes, held together by these forces.
b) Hydrogen bonds also help to give the protein its secondary and tertiary structures.

Q17. Which bio-molecules act as catalyst ?

Answer: Enzymes.

Teacher's Note:
a) Enzymes are protein molecules that speed up biochemical reactions.
b) Most enzymes are highly specific for one substrate.

Q18. Name the enzyme which breaks proteins into peptides ?

Answer: Pepsin and trypsin.

Teacher's Note:
a) Pepsin acts in the stomach, and trypsin acts in the small intestine.
b) Both are proteolytic enzymes.

Q19. Name the purines present in DNA ?

Answer: Adenine and guanine.

Teacher's Note:
a) Purines have two fused rings, while cytosine and thymine are pyrimidines with one ring.

Q20. Name the base that is found in RNA only .

Answer: Uracil.

Teacher's Note:
a) Uracil replaces thymine in RNA.
b) Thymine is found in DNA, and it is absent from RNA.

A: ‘L’ configuration

A: The anomers of glucose i.e. alpha and beta are having specific rotation of + 1110and + 19.20 respectively. The mixtures of these two have a rotation of +52.40. this is called a mutarotation.

A: they form zwitter ion (dipolar ion) and behave as neutral molecule at pH 7(isoelectric point).

A: H-bonding. It has 3.6 amino acids in one single turn, and a 13 member ring is formed by H bonding.

A: AUAGAUGGACCU