NCERT Solutions · Class 12 Chemistry · NCERT Chemistry Part I and II Part II · Chapter 10
Chapter 10: Biomolecules (Organic Chemistry)
Answers to all 5 intext questions and 25 exercises of Chapter 10, Biomolecules (NCERT Class 12 Chemistry Part II, 2026-27): monosaccharides, reducing sugars, glycosidic linkage, glycogen, starch and cellulose, reactions of glucose and its cyclic structure, amino acids and zwitterions, peptide bond and protein structure, denaturation, enzymes, vitamins, and the structure and functions of DNA and RNA. All 30 questions are answered, with the key answer highlighted.
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Glucose: CX6HX12OX6, an aldohexose; it exists mainly as the cyclic α- and β-pyranose forms. Sucrose (glucose + fructose) is non-reducing; maltose and lactose are reducing. Proteins: α-amino acids joined by peptide (–CO–NH–) bonds; structure levels: primary, secondary (α-helix, β-pleated sheet), tertiary, quaternary. DNA: 2-deoxyribose, bases A, G, C, T, double helix; RNA: ribose, bases A, G, C, U, single strand.
Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
Solution
Glucose and sucrose contain many –OH groups, which form hydrogen bonds with water, so they dissolve readily. Cyclohexane and benzene are non-polar hydrocarbons with no –OH groups; they cannot form hydrogen bonds with water and so are insoluble.
The many –OH groups of the sugars hydrogen-bond with water; the hydrocarbons cannot.
How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Solution
Glucose pentaacetate does not react with hydroxylamine (no oxime) or with NaHSOX3. In the pentaacetate, all five –OH groups are acetylated, including the –OH on C-1 of the cyclic form. With C-1 tied up in the ring as an acetylated hemiacetal, the ring cannot open to give a free –CHO group. This shows glucose exists mainly in a cyclic (hemiacetal) form, not as the open-chain aldehyde.
In the pentaacetate the cyclic C-1 –OH is acetylated, so the ring cannot open to free a –CHO group.
The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Solution
Amino acids exist as zwitterions (HX3NX+−CHR−COOX−), which are dipolar ionic species held together by strong electrostatic (salt-like) attractions. So they have high melting points, and their ionic nature makes them very soluble in water. Halo acids are ordinary covalent molecules without such ionic attractions.
Amino acids exist as zwitterions with strong ionic attractions, unlike covalent halo acids.
Where does the water present in the egg go after boiling the egg?
Solution
On boiling, the egg proteins are denatured: their globular structure unfolds and the chains coagulate, forming a solid network. The water present is trapped (absorbed and held) within this coagulated protein network, mainly by hydrogen bonding; it does not leave the egg.
It is trapped in the coagulated (denatured) protein network.
Monosaccharides are the simplest carbohydrates, which cannot be hydrolysed into simpler polyhydroxy aldehydes or ketones. Examples: glucose, fructose, ribose. They are classified by the number of carbon atoms and the carbonyl group (aldoses, ketoses).
Carbohydrates that cannot be hydrolysed further, e.g. glucose, fructose, ribose.
Sugars that reduce Fehling's solution and Tollens' reagent are reducing sugars. They have a free aldehyde or ketone group (or a hemiacetal that can open to one). All monosaccharides and disaccharides such as maltose and lactose are reducing sugars; sucrose is not.
Sugars that reduce Fehling's and Tollens' reagents (free carbonyl/hemiacetal group), e.g. glucose, maltose, lactose.
What do you understand by the term glycosidic linkage?
Solution
The glycosidic linkage is the oxide (–O–) bond that joins two monosaccharide units, formed with the loss of a water molecule between the hemiacetal –OH of one sugar and an –OH of the other. For example, in maltose the C-1 of one α-D-glucose is joined to C-4 of another glucose through oxygen.
The C–O–C link between two monosaccharide units, formed by loss of water.
What is the basic structural difference between starch and cellulose?
Solution
Starch is made of α-D-glucose units: amylose has C1–C4 α-glycosidic links (unbranched) and amylopectin also has C1–C6 branches. Cellulose is a straight-chain polymer of β-D-glucose units joined by C1–C4 β-glycosidic links.
Starch is built from α-D-glucose (with branching in amylopectin); cellulose is an unbranched chain of β-D-glucose.
Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Solution
Despite having an aldehyde group, glucose does not give Schiff's test and does not form the hydrogensulphite addition product with NaHSOX3.
Glucose pentaacetate does not react with hydroxylamine, showing no free –CHO group.
Glucose exists in two crystalline forms, α and β (different melting points and specific rotations), and each changes in solution to an equilibrium mixture (mutarotation).
These are explained by the cyclic (hemiacetal) structure.
No Schiff's test or NaHSO₃ addition, the pentaacetate does not react with NH₂OH, and the existence of α and β forms (mutarotation).
What are essential and non-essential amino acids? Give two examples of each type.
Solution
Essential amino acids: cannot be made in the body and must be supplied in the diet. Examples: valine, leucine (also lysine, isoleucine, phenylalanine).
Non-essential amino acids: can be synthesised in the body. Examples: glycine, alanine (also glutamic acid, aspartic acid).
Essential (from the diet): valine, leucine. Non-essential (made in the body): glycine, alanine.
Define the following as related to proteins (i) Peptide linkage (ii) Primary structure (iii) Denaturation.
Solution
(i) Peptide linkage: the amide bond (–CO–NH–) formed between the –COOH group of one amino acid and the –NH₂ group of another, with loss of water.
(ii) Primary structure: the specific sequence of amino acids in the polypeptide chain(s) of a protein.
(iii) Denaturation: the loss of biological activity of a protein when its secondary and tertiary structures are destroyed by heat, change in pH etc.; the hydrogen bonds break and the chains unfold, but the primary structure stays intact (e.g. coagulation of egg white on boiling).
(i) –CO–NH– bond between amino acids. (ii) The amino acid sequence. (iii) Loss of the 2° and 3° structure (and activity) by heat or pH, keeping the 1° structure.
What type of bonding helps in stabilising the α-helix structure of proteins?
Solution
Intramolecular hydrogen bonding between the –NH group of one amino acid residue and the C=O group of the residue four places along the same chain (the next turn of the helix).
Hydrogen bonds between N–H and C=O groups within the same chain.
How do you explain the amphoteric behaviour of amino acids?
Solution
In water, an amino acid exists as a zwitterion, HX3NX+−CHR−COOX−, which has both an acidic group (−NHX3X+, can give HX+) and a basic group (−COO−, can take HX+). So it reacts with both bases and acids:
with acid: HX3NX+−CHR−COOX−+HX+HX3NX+−CHR−COOH
with base: HX3NX+−CHR−COOX−+OHX−HX2N−CHR−COOX−+HX2O
As zwitterions, amino acids have both a proton-donating (–NH₃⁺) and a proton-accepting (–COO⁻) group.
Enzymes are biological catalysts: mostly globular proteins that speed up specific biochemical reactions in living organisms under mild conditions by lowering the activation energy. They are highly specific; for example, maltase catalyses the hydrolysis of maltose to glucose. Their names usually end in "-ase".
Protein biocatalysts that speed up specific reactions in living organisms.
What is the effect of denaturation on the structure of proteins?
Solution
Denaturation (by heat, change in pH, etc.) breaks the hydrogen bonds and other weak interactions, so the secondary and tertiary structures are destroyed: the globules unfold and helices uncoil. The primary structure (amino acid sequence) remains intact, but the protein loses its biological activity.
The 2° and 3° structures are lost (the protein unfolds and loses activity); the 1° structure stays.
Why are vitamin A and vitamin C essential to us? Give their important sources.
Solution
Vitamin A: needed for healthy eyes and skin; its deficiency causes night blindness and xerophthalmia (hardening of the cornea). Sources: fish liver oil, carrots, butter and milk.
Vitamin C: needed for healthy gums and skin and for resistance to infection; its deficiency causes scurvy (bleeding gums). Sources: citrus fruits (lemon, orange), amla and green leafy vegetables.
Vitamin A prevents night blindness (fish liver oil, carrots, milk); vitamin C prevents scurvy (citrus fruits, amla, green vegetables).
The two strands in DNA are not identical but are complementary. Explain.
Solution
In the DNA double helix, the bases of the two strands pair by hydrogen bonds in a fixed way: adenine pairs with thymine (two H-bonds) and guanine pairs with cytosine (three H-bonds). So the sequence of one strand decides the sequence of the other: where one strand has A, the other has T, and so on. The two strands are therefore not identical but complementary.
Because of specific base pairing (A with T, G with C), each strand's sequence fixes the other's.