NCERT Solutions for Class 12 Chemistry Chapter 14 - Biomolecules

1. What are monosaccharides?

Ans. Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone. Monosaccharides are classified on the basis of number of carbon atoms and the functional group present in them. Monosaccharides containing an aldehyde group are known as aldoses and those containing a keto group are known as ketoses. Monosaccharides are further classified as trioses, tetroses, pentoses, hexoses, and heptoses according to the number of carbon atoms they contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3 carbon atoms is called aldotriose. There are about 20 monosaccharides found in nature.

2. What are reducing sugars?

Ans. Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars. It is believed that monosaccharide in form of ketoses in alkaline medium convert into aldoses and reduces Fehling’s solution and Tollen’s reagent.

3. Write two main functions of carbohydrates in plants.

Ans. Two major functions of carbohydrates in plants are following

(i) Structural material for plant cell walls: The polysaccharide cellulose acts as the chief structural material of the plant cell walls.

(ii) Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and act as the reserve food material for the tiny plant till it is capable of making its own food by photosynthesis.

4. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.

Ans. Monosaccharides: Ribose, 2-deoxyribose, galactose and fructose.
Disaccharides: Maltose and lactose.

5. What do you understand by the term glycosidic linkage?

Ans. Glycosidic linkage is the etheral linkage formed between two monosaccharide units through an oxygen atom by the loss of water molecule.

For example, in a sucrose molecule, two monosaccharide units, α-glucose and β-fructose, are joined together by a glycosidic linkage.


6. What is glycogen? How is it different from starch?

Ans. Glycogen is a carbohydrate (polysaccharide) found in animals, carbohydrates are stored as glycogen. It is insoluble in water.

Starch is a carbohydrate consisting of two components—amylose (15-20%) water soluble and amylopectin
(80-85%), water insoluble.

However, glycogen consists of only one component whose structure is similar to amylopectin. Amylose is a linear polymer while glycogen and amylopectin are branched polymers of a-D-glucose. However glycogen is more highly branched compared to amylopectin.

7. What are the hydrolysis products of (i) sucrose, and (ii) lactose?

Ans. Both sucrose and lactose are disaccharides.

(i) On hydrolysis, sucrose gives one molecule of α-D glucose and one molecule of β-D-fructose.

$$\underset{\text{Sucrose}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}} + \text{H}_{2}\text{O}\xrightarrow[\text{H}^{\normalsize+}]{\text{Invertase}}\underset{\text{D(+) Glucose}}{\text{C}_{6}\text{H}_{12}\text{O}_{6}} + \underset{\text{D(-) Fructose}}{\text{C}_{6}\text{H}_{12}\text{O}_{6}}$$

(ii) The hydrolysis of lactose gives β-D galactose and β-D-glucose.

$$\underset{\text{Lactose}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}} + \text{H}_{2}\text{O}\xrightarrow[\text{H}^{\normalsize+}]{\text{Lactase}}\underset{\text{D-(+) Glucose}}{\text{C}_{6}\text{H}_{12}\text{O}_{6}} + \underset{\text{D-(+)-Galactose}}{\text{C}_{6}\text{H}_{12}\text{O}_{6}}$$

8. What is the basic structural difference between starch and cellulose?

Ans. Starch consists of two components-amylose and amylopectin. Amylose is a long linear chain of 200 – 1000 units of α-D-(+)-glucose units joined by C1-C4 glycosidic linkage (a-link).

Amylopectin is a branched-chain polymer of α-D-glucose units, in which the chain is formed by C1-C4 glycosidic linkage and the branching occurs by C1-C6 glycosidic linkage.

On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C1-C4 glycosidic linkage (β-link).

9. What happens when D-glucose is treated with the following reagents.

(i) HI

(ii) Bromine water

(iii) HNO3

Ans. (i) When D-glucose is heated with HI for a long time, n-hexane is formed as a result of reduction reaction.

(ii) When D-glucose is treated with Br2 water, D-gluconic acid is produced, as a result of oxidation of –CHO group to –COOH group.
(iii) On being treated with HNO3, D-glucose gets oxidised to give saccharic acid, as a result of oxidation of both —CHO and 1° alcohol group –CH2OH to —COOH.

10. Enumerate the reactions of D-glucose which cannot be explained with open chain structure.

Ans. D-glucose cannot be explained open chain structure with following reactions:

(i) Aldehydes give 2, 4-DNP test, Schiff’s test, and react with NaHSO4 to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.

(ii) The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free −CHO group is absent from glucose.

(iii) Glucose exists in two crystalline forms − and β. The -form (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the β- form (m.p. = 423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

11. What are essential and non-essential amino acids? Give two examples of each type.

Ans. α-Amino acids which are needed for good health and proper growth of human beings but are not synthesized by the human body are called essential amino acids. For example, valine, leucine, phenylalanine, etc. On the other hand, α-amino acids which are needed for health and growth of human beings and are synthesized by the human body are called non-essential amino acids. For example, glycine, alanine, aspartic acid etc.

12. Define the following as related to proteins:

(i) Peptide linkage

(ii) Primary structure

(iii) Denaturation

Ans. (i) Peptide linkage: The amide linkage formed between –COOH group of one molecule of an amino acid and –NH2 group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage. These are the basic linkages present in proteins.

(ii) Primary structure: The primary structure of protein refers to the specific sequence in which various amino acids are present in it. The sequence in which amino acids arranged is different in each protein. A change in the sequence creates a different protein.

(iii) Denaturation: In a biological system, a protein is found to have a unique 3-dimensional structure and a unique biological activity. In such a situation, the protein is called native protein. However, when the native protein is subjected to physical changes such as change in temperature or chemical changes such as change in pH, its H-bonds are disturbed. This disturbance unfolds the globules and uncoils the helix. As a result, the protein loses its biological activity. This loss of biological activity by the protein is called denaturation. During denaturation, the secondary (2°) and the tertiary (3°) structures of the protein gets destroyed, but the primary structure remains unaltered.

One of the examples of denaturation of proteins is the curdling of milk due to change in pH.

13. What are the common types of secondary structure of proteins?

Ans. There are two common types of secondary structure of proteins:

(i) α-helix structure

(ii) β-pleated sheet structure.

(i) α-Helix structure: In this structure, the –NH group of an amino acid residue forms H-bond with the group of the adjacent turn of the right-handed screw (α-helix). The most common way of twisting of a polypeptide chain. Generally, proteins having larger alkyl (R) group in amino acid chain forms this structure.

(ii) β-pleated sheet structure: This structure is called so because it looks like the pleated folds of drapery. In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds. Generally proteins with smaller (R) groups in amino acids forms such structures.


14. What types of bonding helps in stabilising the α-helix structure of proteins?

Ans. α-helix structure of proteins is stabilised through hydrogen bonding. (a) α-Helix structure. If the size of the R-groups are quite large, the hydrogen bonding occurs between > C = O group of one amino acid unit and the > N-H group of the fourth amino acid unit within the same chain. As such the polypeptide chain coils up into a spiral structure called right handed a—helix structure. This type of structure is adopted by most of the fibrous structural proteins such as those present in wool, hair and muscles. These proteins are elastic i.e., they can be stretched. During this process, the weak hydrogen bonds causing the α-helix are broken. This tends to increase the length of the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.

15. Differentiate between globular and fibrous proteins.


S.No.  Fibrous Proteins Globular Proteins
  1. These consist of linear molecules which lie side by side to form fibres. For example: keratin.
These consist of polypeptide chains which are folder into compact units forming spheroidal shapes. For example: albumin
  1. These are insoluble in water.
These are soluble in water.
  1. These are the main structural units of animal tissues.
These acts as enzymes, antibodies, etc.
  1. Polypeptides are held through intermolecular hydrogen bonding.
Polypeptides are held through intramolecular hydrogen bonding.

16. How do you explain the amphoteric behaviour of amino acids?

Ans. In aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as Zwitter ion.


Thus, amino acids show amphoteric behaviour .

17. What are enzymes?

Ans. Enzymes are biological catalyst. Each biological reaction requires a different enzyme. Thus, as compared to conventional catalyst enzymes are very specific and efficient in their action. Each type of enzyme has its own specific optimum conditions of concentration, pH and temperature at which it works best.

18. What is the effect of denaturation on the structure of proteins?

Ans. As a result of denaturation, globules get unfolded and helices get uncoiled. Secondary and tertiary structures of protein are destroyed, but the primary structure remains unaltered. As the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity and becomes inactive as a bio-catalyst.

19. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.

Ans. Vitamins are classified into two groups on the basis of their solubility in water or fat:

(i) Fat soluble vitamins: Vitamins, which are soluble in fat but insoluble in water are called fat soluble vitamins. For example vitamins A, D, E and K.

(ii) Water-soluble vitamins: Vitamins, which are soluble in water and insoluble in fat. For example, vitamins B-complex and vitamin C.

Vitamin K is responsible for the coagulation of blood.

20. Why are vitamin A and vitamin C essential to us? Give their important sources.

Ans. The deficiency of vitamin A leads to xerophthalmia (hardening of the cornea of the eye) and night blindness. The deficiency of vitamin C leads to scurvy (bleeding gums). The sources of vitamin A are fish liver oil, carrots, butter and milk. The sources of vitamin C are citrus fruits, amla and green leafy vegetables.

21. What are nucleic acids? Mention their two important functions.

Ans. Nucleic acids are biomolecules which are found in the nuclei of all living cell in form of nucleoproteins or chromosomes (proteins contains nucleic acids as the prosthetic group).


Nucleic acids are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid.(RNA).

The two main functions of nucleic acids are:

(a) DNA is responsible for transmission of hereditary effects from one generation to another. This is due to its unique property of replication, during cell division and two identical DNA strands are transferred to the daughter cells.

(b) DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually the proteins are synthesized by various RNA molecules (r-RNA, m-RNA) and t-RNA) in the cell but the message for the synthesis of a particular protein is coded in DNA.

22. What is the difference between a nucleoside and a nucleotide?

Ans. A nucleoside is formed by the attachment of a base to C–1 position of ribose or deoxyribose sugar.

Nucleoside = Sugar + Base


On the other hand, all the three basic components of nucleic acids (i.e., pentose sugar, phosphoric acid, and base) are present in a nucleotide. Nucelotide = Sugar + Base + Phosphoric acid.


23. The two strands in DNA are not identical but are complementary. Explain.

Ans. The two strands in DNA molecule are held together by hydrogen bonds between purine base of one strand and pyrimidine base of the other and vice versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H-bonds, (i.e.,
C = G) and between A (adenine) and T (thiamine) through two H-bonds (i.e., A = T). Due to this base -pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complimentary and not identical.

24. Write the important structural and functional differences between DNA and RNA.


  1. The sugar present in DNA is 2-deoxy-D-(–)-ribose.
The sugar present in RNA is D-(–)-ribose.
  1. DNA contains cytosine and thymine as pyrimidine bases and, guanine and adenine as purine bases.
RNA contains cytosine and uracil as pyrimidine bases and guanine and adenine as purine bases.
  1. DNA has double stranded a-helix structure.
RNA has a single stranded α-helix structure.
  1. DNA molecules are very large, their molecular mass may vary from 6 × 106 – 16 × 106 u.
RNA molecules are comparatively much smaller with molecular mass ranging from 20,000 – 40,000 u.
Functional differences
  1. DNA has unique property of replication.
The sugar present in RNA is D-(–)-ribose.
  1. DNA controls the transmission of hereditary effects.
RNA controls the synthesis of proteins.

25. What are the different types of RNA found in the cell?

Ans. There are three types of RNA:
(a) Ribosomal RNA (r-RNA)
(b) Messenger RNA (m-RNA)
(c) Transfer RNA (t-RNA)