NCERT Solutions for Class 12 Chemistry Chapter 10 Biomolecules

Biomolecules class 12 chemistry Chapter 10 solutions can be access below. Biomolecules majorly deal with carbohydrates, proteins, vitamins and nucleic acids, and act as the foundation for many upcoming advanced topics of the subject. Students having confusions or doubts regarding the same can visit the answers provided here by SimplyAcad.

The solution tries to assist students in getting their fundamentals of biomolecules right, and simultaneously, boost their confidence while solving the questions from the chapter. Practising and revising the answers given make sure that students secure good marks in their chemistry paper. The questions are related to the topic of starch, saccharides, cellulose, glycogen, etc, which are briefly discussed in the solutions. Scroll below to find the answers.

NCERT Answers of Biomolecules Class 12 Chemistry Chapter 10 Questions 1 to 10

Question 1: What are monosaccharides?

Solution:
Monosaccharides are the simplest form of carbohydrates and cannot be broken down into smaller units of polyhydroxy aldehydes or ketones through hydrolysis. They are classified based on the number of carbon atoms they contain and the functional group present. Monosaccharides with an aldehyde group are called aldoses, while those with a keto group are called ketoses. Depending on the number of carbon atoms, monosaccharides are further categorized as trioses, tetroses, pentoses, hexoses, and heptoses. For example, a monosaccharide with three carbon atoms and a keto group is called a ketotriose, while one with three carbon atoms and an aldehyde group is known as an aldotriose.

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Question 2: What are reducing sugars?

Solution:
Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars.

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Question 3: Write two main functions of carbohydrates in plants.

Solution:
Two main functions of carbohydrates in plants are:

1. Polysaccharides such as starch serve as storage molecules.
2. Cellulose, a polysaccharide, is used to build the cell wall.

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

Solution:
Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
Disaccharides: Maltose, lactose

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Question 5: What do you understand by the term glycosidic linkage?

Solution:
Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule. For example, in a sucrose molecule, two monosaccharide units, α-glucose and β-fructose, are joined together by a glycosidic linkage.

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Question 6: What is glycogen? How is it different from starch?

Solution:
Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components − amylose (15 − 20%) and amylopectin (80 − 85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin.

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Question 7: What are the hydrolysis products of: (i) Sucrose (ii) Lactose?

Solution:

1. Sucrose:
   Sucrose is a disaccharide composed of glucose and fructose. When hydrolyzed, it yields: 

   $$\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{invertase}} \text{C}_6\text{H}_{12}\text{O}_6 (\text{β-D-glucose}) + \text{C}_6\text{H}_{12}\text{O}_6 (\text{D-fructose})$$
   The mixture of glucose and fructose resulting from sucrose hydrolysis is known as invert sugar.

2. Lactose:
   Lactose is a disaccharide composed of glucose and galactose. When hydrolyzed, it yields: 

   $$\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{lactase}} \text{C}_6\text{H}_{12}\text{O}_6 (\text{β-D-galactose}) + \text{C}_6\text{H}_{12}\text{O}_6 (\text{β-D-glucose})$$ 

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

Solution:
Starch and cellulose are both polysaccharides composed of glucose units, but they differ significantly in their structure and linkage:

1. Starch:
   • Starch consists of two main components: amylose and amylopectin.
   • Amylose is a linear polymer of glucose units connected by α-1,4-glycosidic linkages.
   • Amylopectin is a branched polymer of glucose units, with branches formed by α-1,6-glycosidic linkages.
2. Cellulose:
   • Cellulose is a linear polymer of glucose units connected by β-1,4-glycosidic linkages.
   • The β-1,4-glycosidic linkage in cellulose leads to a straight, rigid structure that forms strong fibers.

In summary, starch has α-1,4 and α-1,6 linkages (in amylopectin), while cellulose has β-1,4 linkages. Starch is primarily linear (amylose), while cellulose is a straight chain with a more rigid structure due to its β-linkages. Amylopectin, a component of starch, is highly branched compared to the linear structure of cellulose.

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Question 9: What happens when D-glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO₃

Solution:

1. HI: When D-glucose is heated with HI for a long time, n-hexane is formed.
2. Bromine water: When D-glucose is treated with Br₂ water, D-gluconic acid is produced.
3. HNO₃: On being treated with HNO₃, D-glucose gets oxidized to give saccharic acid.

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Question 10: Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.

Solution:

1. Aldehydes give 2,4-DNP test, Schiff’s test, and react with NaHSO₃ to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
2. The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free −CHO group is absent from glucose.
3. Glucose exists in two crystalline forms − α and β. The α-form (m.p. = 419 K) crystallizes from a concentrated solution of glucose at 303 K, and the β-form (m.p = 423 K) crystallizes from a hot and saturated aqueous solution at 371 K. This behavior cannot be explained by the open chain structure of glucose.

NCERT Answers of Biomolecules Class 12 Chemistry Chapter 10 Questions 11 to 20

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

Solution:
Essential amino acids are required by the human body, but they cannot be synthesized in the body. They must be taken through food.
Examples: Valine and Leucine

Non-essential amino acids are also required by the human body, but they can be synthesized in the body.
Examples: Glycine and Alanine

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Question 12: Define the following as related to proteins: (i) Peptide linkage (ii) Primary structure (iii) Denaturation

Solution:

1. Peptide linkage: A peptide linkage (or peptide bond) is a covalent bond formed between the amino group of one amino acid and the carboxyl group of another amino acid. This bond is a key feature in the formation of proteins, linking the amino acids together in a chain.
2. Primary structure: The primary structure of a protein or peptide refers to the linear sequence of amino acid residues in its polypeptide chain. This sequence is read from the amino-terminal (N) end to the carboxyl-terminal (C) end and determines the unique sequence of amino acids in the protein.
3. Denaturation: Denaturation is the process by which proteins or nucleic acids lose their native quaternary, tertiary, and secondary structures due to external stress or chemical agents such as strong acids, bases, or concentrated salts. This alteration in structure often leads to a loss of biological function.

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Question 13: What are the common types of secondary structure of proteins?

Solution:
There are two common types of secondary structure of proteins:

1. α-Helix structure: In this structure, the −NH group of an amino acid residue forms an H-bond with the group of the adjacent turn of the right-handed screw (α-helix).
2. 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.

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Question 14: What type of bonding helps in stabilizing the α-helix structure of proteins?

Solution:
The H-bonds formed between the −NH group of each amino acid residue and the group of the adjacent turns of the α-helix help in stabilizing the helix.

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Question 15: Differentiate between globular and fibrous proteins.

Solution:

Fibrous Protein	Globular Protein
1. It is a fiber-like structure formed by the polypeptide chain. These proteins are held together by strong hydrogen and disulfide bonds.	1. The polypeptide chain in this protein is folded around itself, giving rise to a spherical structure.
2. It is usually insoluble in water.	2. It is usually soluble in water.
3. Fibrous proteins are usually used for structural purposes. For example, keratin is present in nails and hair; collagen in tendons; and myosin in muscles.	3. All enzymes are globular proteins. Some hormones such as insulin are also globular proteins.

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Question 16: How do you explain the amphoteric behavior of amino acids?

Solution:
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 a zwitterion. Therefore, in zwitterionic form, the amino acid can act both as an acid and as a base. Thus, amino acids show amphoteric behavior.

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Question 17: What are enzymes?

Solution:
Enzymes are macromolecular biological catalysts. They accelerate chemical reactions by lowering the activation energy required for the reaction to proceed. The molecules upon which enzymes act are called substrates. The enzyme converts these substrates into different molecules known as products. Examples of enzymes include peptidases, which break down proteins, and kinases, which transfer phosphate groups.

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Question 18: What is the effect of denaturation on the structure of proteins?

Solution:
As a result of denaturation, globules get unfolded, and helices get uncoiled. Secondary and tertiary structures of proteins are destroyed, but the primary structures remain unaltered. It can be said that during denaturation, secondary and tertiary-structured proteins get converted into primary-structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.

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

Solution:
On the basis of their solubility in water or fat, vitamins are classified into two groups:

1. Fat-soluble vitamins: Vitamins that are soluble in fat and oils, but not in water, belong to this group. Examples: Vitamins A, D, E, and K.
2. Water-soluble vitamins: Vitamins that are soluble in water belong to this group. Examples: B group vitamins (B1, B2, B6, B12, etc.) and vitamin C.

Vitamin K is responsible for the coagulation of blood.

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

Solution:
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.

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NCERT Answers of Biomolecules Class 12 Chemistry Chapter 10 Questions 21 to 33

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

Solution:
Nucleic acids are biomolecules found in the nuclei of all living cells, as one of the constituents of chromosomes. There are mainly two types of nucleic acids − deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Nucleic acids are also known as polynucleotides as they are long-chain polymers of nucleotides.

Two main functions of nucleic acids are:

1. DNA is responsible for the transmission of inherent characters from one generation to the next. This process of transmission is called heredity.
2. Nucleic acids (both DNA and RNA) are responsible for protein synthesis in a cell. Even though the proteins are actually synthesized by the various RNA molecules in a cell, the message for the synthesis of a particular protein is present in DNA.

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Question 22: What is the difference between a nucleoside and a nucleotide?

Solution:
A nucleoside is formed by the attachment of a base to position 1 of sugar.
Nucleoside = Sugar + Base

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

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

Solution:
In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms a hydrogen bond with guanine, while adenine forms a hydrogen bond with thymine. As a result, the two strands are complementary to each other.

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

Solution:
Structural Differences:

DNA	RNA
1. The sugar moiety in DNA molecules is β-D-2-deoxyribose.	1. The sugar moiety in RNA molecules is β-D-ribose.
2. DNA contains thymine (T). It does not contain uracil (U).	2. RNA contains uracil (U). It does not contain thymine (T).
3. The helical structure of DNA is double-stranded.	3. The helical structure of RNA is single-stranded.

Functional Differences:

DNA	RNA
1. DNA is the chemical basis of heredity.	1. RNA is not responsible for heredity.
2. DNA molecules do not synthesize proteins but transfer coded messages for the synthesis of proteins in the cells.	2. Proteins are synthesized by RNA molecules in the cells.

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

Solution:
(i) Messenger RNA (m-RNA)
(ii) Ribosomal RNA (r-RNA)
(iii) Transfer RNA (t-RNA)

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Question 26: Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.

Solution:
A glucose molecule contains five −OH groups, while a sucrose molecule contains eight −OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water. Hence, these are soluble in water. But cyclohexane and benzene do not contain −OH groups. Hence, they cannot undergo H-bonding with water and, as a result, are insoluble in water.

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Question 27: What are the products of hydrolysis of lactose?

Solution:
The hydrolysis of lactose produces glucose and galactose. The chemical reaction is as follows:

 

$$\text{C}_{12}\text{H}_{22}\text{O}_{11} \text{ (Lactose)} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{12}\text{O}_6 \text{ (Glucose)} + \text{C}_6\text{H}_{12}\text{O}_6 \text{ (Galactose)}$$

 

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Question 28: How do you explain the absence of the aldehyde group in the pentaacetate of D-glucose?

Solution:
D-glucose reacts with hydroxylamine (NH₂OH) to form an oxime because of the presence of an aldehydic (−CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH₂OH to give an oxime. But pentaacetate of D-glucose does not react with NH₂OH. This is because pentaacetate does not form an open chain structure.

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Question 29: The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.

Solution:
In amino acids, both acidic ($-\text{COOH}$)and basic ($-\text{NH}_2$) groups are present. In the presence of water, the carboxyl group loses a proton, and the amino group accepts a proton, forming a dipolar ion known as a zwitterion. This zwitterionic form imparts ionic salt-like character to amino acids, resulting in very strong dipole-dipole interactions and hence, very high melting points. For solubility, the principle “like dissolves like” applies. Amino acids, like water, can form hydrogen bonds due to their zwitterionic nature. This ability to form hydrogen bonds makes them soluble in water. On the other hand, halo acids do not exhibit the property of forming dipolar compounds. Only the carboxyl group of halo acids is involved in hydrogen bonding, not the halogen atom. Therefore, halo acids have lower melting points and less solubility compared to amino acids.

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Question 30: Where does the water present in the egg go after boiling the egg?

Solution:
When an egg is boiled, the proteins present inside the egg get denatured and coagulate. After boiling the egg, the water present in it is absorbed by the coagulated protein through H-bonding.

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Question 31: Why cannot vitamin C be stored in our body?

Solution:
Vitamin C cannot be stored in our body because it is water soluble. As a result, it is readily excreted in the urine.

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Question 32: What products would be formed when a nucleotide from DNA containing thymine is hydrolyzed?

Solution:
When a nucleotide from the DNA containing thymine is hydrolyzed, thymine β-D-2-deoxyribose and phosphoric acid are obtained as products.

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Question 33: When RNA is hydrolyzed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?

Solution:
A DNA molecule is double-stranded, in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine. Therefore, on hydrolysis of DNA, the quantity of adenine produced is equal to that of thymine, and similarly, the quantity of cytosine is equal to that of guanine. But when RNA is hydrolyzed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded.