NCERT Solutions for Class 12 Chemistry Chapter 10: Biomolecules
These Class 12 Chemistry Chapter 10 solutions cover Biomolecules from the NCERT textbook (session 2026–27). Every Intext Question and every numbered Exercise question is reproduced exactly as in the book and solved fully — carbohydrates, proteins, enzymes, vitamins, nucleic acids and hormones — in clear, exam-ready language for the CBSE board exam.
Class 12 Chemistry Chapter 10 Solutions – Overview
Chapter 10, Biomolecules, studies the complex organic molecules that build and run living systems. It begins with carbohydrates (optically active polyhydroxy aldehydes or ketones, classified as mono-, oligo- and polysaccharides), explains the open-chain and cyclic structures of glucose and fructose, and the disaccharides sucrose, maltose and lactose. It then moves to proteins — polymers of α-amino acids joined by peptide bonds, with primary, secondary, tertiary and quaternary structure and denaturation — followed by enzymes as biocatalysts. The chapter ends with vitamins (fat-soluble A, D, E, K and water-soluble B-group and C), nucleic acids (DNA and RNA, nucleosides, nucleotides and base pairing) and hormones. The big idea is that all of life’s chemistry is carried out by a small set of well-organised biomolecules.
Key Concepts & Definitions
Carbohydrates: optically active polyhydroxy aldehydes or ketones, or compounds that yield such units on hydrolysis (general formula Cx(H2O)y).
Monosaccharide: a carbohydrate that cannot be hydrolysed to a simpler sugar (e.g. glucose, fructose, ribose). An aldose has a –CHO group; a ketose has a >C=O group.
Reducing sugar: a carbohydrate that reduces Fehling’s solution and Tollens’ reagent (all monosaccharides, plus maltose and lactose). Non-reducing sugar: e.g. sucrose.
Glycosidic linkage: the C–O–C oxide bridge joining two monosaccharide units, formed by loss of one water molecule.
α-Amino acid: a molecule with –NH2 and –COOH on the same (α) carbon; in water it exists as a dipolar zwitterion and is amphoteric.
Peptide bond: the amide –CO–NH– link between the –COOH of one amino acid and the –NH2 of the next.
Enzyme: a globular-protein biocatalyst that is highly specific and lowers the activation energy of a biochemical reaction.
Nucleotide: base + pentose sugar + phosphate; a nucleoside is base + sugar only.
Glucose + HNO3: both –CHO and 1° –CH2OH oxidise ⟶ saccharic (glucaric) acid, HOOC(CHOH)4COOH.
Base pairing (DNA): Adenine ≡ Thymine (2 H-bonds); Guanine ≡ Cytosine (3 H-bonds). In RNA, thymine is replaced by uracil.
Intext Questions — Solutions
10.1 Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
ANSWERGlucose and sucrose carry many polar –OH groups (5 in glucose, 8 in sucrose). These groups form strong hydrogen bonds with water molecules, so both dissolve readily in water.Cyclohexane and benzene are non-polar hydrocarbons with no –OH (or other polar) groups, so they cannot form hydrogen bonds with water. Hence they are insoluble in water but dissolve in non-polar solvents.
10.2 What are the expected products of hydrolysis of lactose?
ANSWERLactose (milk sugar) is a disaccharide made of β-D-galactose and β-D-glucose joined by a C1–C4 glycosidic linkage.On hydrolysis it gives one molecule of D-(+)-galactose and one molecule of D-(+)-glucose.
10.3 How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
ANSWERIn aqueous solution glucose exists mainly in its cyclic (pyranose) hemiacetal form. The C1 carbon (anomeric carbon) is no longer a free –CHO group but is locked in the ring as a hemiacetal –OH.On acetylation, this hemiacetal –OH is also acetylated along with the other four –OH groups, giving glucose pentaacetate in which the ring is fixed. As there is no free or potential –CHO group available, the pentaacetate does not react with hydroxylamine (it gives no oxime), confirming the absence of a free aldehyde group.
10.4 The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
ANSWERAmino acids exist as dipolar zwitterions (+H3N–CHR–COO−), so they behave like ionic salts. Strong electrostatic (ion–ion) forces hold the molecules together, which requires a lot of energy to break — hence very high melting points.The charged zwitterion also interacts strongly with polar water through ion–dipole attraction, giving high water solubility. The corresponding halo acids are not dipolar/ionic, so their inter-molecular forces are weaker, giving lower melting points and lower water solubility.
10.5 Where does the water present in the egg go after boiling the egg?
ANSWEREgg white is mostly water held within a network of globular proteins (chiefly albumin). On boiling, the proteins are denatured — their secondary and tertiary structures unfold and the chains coagulate into a solid mesh.The water is not lost; it gets trapped (absorbed) within the pockets of this coagulated protein network, which is why the boiled egg appears solid although the water is still present inside it.
10.6 Why cannot vitamin C be stored in our body?
ANSWERVitamin C (ascorbic acid) is a water-soluble vitamin. Such vitamins are readily excreted from the body in urine and are not stored in fat or other tissues.Therefore it cannot be stored and must be supplied regularly through the diet.
10.7 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
ANSWEROn complete hydrolysis the DNA nucleotide containing thymine gives three products:(i) the nitrogenous base thymine, (ii) the pentose sugar β-D-2-deoxyribose, and (iii) phosphoric acid (H3PO4).
10.8 When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
ANSWERIn DNA the bases pair specifically (A with T, G with C), so the amounts of A & T are equal and of G & C are equal — a fixed relationship arising from its double-stranded structure.Since the bases of RNA show no such fixed relationship, RNA does not have complementary base pairing throughout. This suggests that RNA is normally a single-stranded molecule, not a double helix like DNA.
NCERT Exercises — Solutions
10.1 What are monosaccharides?
ANSWERMonosaccharides are the simplest carbohydrates that cannot be hydrolysed further into smaller carbohydrate (polyhydroxy aldehyde or ketone) units.About 20 occur in nature; common examples are glucose, fructose and ribose. They are classified by the number of carbon atoms (triose, tetrose, pentose, hexose) and by the functional group present (aldose or ketose).
10.2 What are reducing sugars?
ANSWERReducing sugars are carbohydrates that can reduce Fehling’s solution and Tollens’ reagent because they contain a free (or potentially free) aldehydic or ketonic group.All monosaccharides (aldoses and ketoses) are reducing sugars; among disaccharides, maltose and lactose are reducing, whereas sucrose is non-reducing.
10.3 Write two main functions of carbohydrates in plants.
ANSWER(i) Energy storage: carbohydrates are stored as starch, which serves as the reserve food and an instant source of energy for the plant.(ii) Structural support:cellulose forms the cell wall of plant cells, giving rigidity and mechanical strength to the plant body.
10.4 Classify the following into monosaccharides and disaccharides.
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
ANSWERMonosaccharides: ribose, 2-deoxyribose, galactose and fructose (each cannot be hydrolysed further).Disaccharides: maltose and lactose (each yields two monosaccharide units on hydrolysis).
10.5 What do you understand by the term glycosidic linkage?
ANSWERA glycosidic linkage is the linkage between two monosaccharide units through an oxygen atom, formed by the loss of one water molecule between them.It is the bond that joins the monosaccharide units in disaccharides (e.g. C1–C2 in sucrose) and polysaccharides (e.g. C1–C4 and C1–C6 linkages in starch).
10.6 What is glycogen? How is it different from starch?
ANSWERGlycogen is the polysaccharide in which carbohydrates are stored in the animal body (also called animal starch); it is present in liver, muscles and brain and is broken down to glucose when the body needs energy.Difference: Starch is the storage polysaccharide of plants and has two components — unbranched amylose and branched amylopectin. Glycogen is the storage polysaccharide of animals; its structure resembles amylopectin but it is much more highly branched.
10.7 What are the hydrolysis products of
(i) sucrose and (ii) lactose?
ANSWER(i) Sucrose ⟶ one molecule of D-(+)-glucose + one molecule of D-(−)-fructose.(ii) Lactose ⟶ one molecule of D-(+)-galactose + one molecule of D-(+)-glucose.
10.8 What is the basic structural difference between starch and cellulose?
ANSWERStarch is a polymer of α-D-glucose. It consists of amylose (a long unbranched C1–C4-linked chain) and amylopectin (a branched chain with C1–C4 chains and C1–C6 branch points).Cellulose is a straight-chain polymer made only of β-D-glucose units joined by C1–C4 glycosidic linkages, with no branching. The key difference is therefore α-glucose (starch) versus β-glucose (cellulose).
10.9 What happens when D-glucose is treated with the following reagents?
(i) HI (ii) Bromine water (iii) HNO3
ANSWER(i) HI (on prolonged heating): glucose is reduced and gives n-hexane, showing that the six carbon atoms are joined in a straight chain.(ii) Bromine water (a mild oxidising agent) oxidises only the –CHO group to –COOH, giving gluconic acid (a six-carbon monocarboxylic acid), CH2OH(CHOH)4COOH. This shows the carbonyl group is an aldehyde.(iii) HNO3 (a strong oxidising agent) oxidises both the –CHO group and the terminal 1° –CH2OH group to –COOH, giving the dicarboxylic saccharic acid (glucaric acid), HOOC(CHOH)4COOH. This shows the presence of a primary –OH group.
10.10 Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
ANSWER(i) Despite having an aldehyde group, glucose does not give the Schiff’s test and does not form the hydrogensulphite addition product with NaHSO3.(ii) The pentaacetate of glucose does not react with hydroxylamine, indicating the absence of a free –CHO group.(iii) Glucose exists in two crystalline forms, α and β (mutarotation), which the single open-chain structure cannot explain.These facts are explained by the cyclic (pyranose) hemiacetal structure of glucose.
10.11 What are essential and non-essential amino acids? Give two examples of each type.
ANSWEREssential amino acids cannot be synthesised by our body and must be obtained through diet. Examples: valine and leucine (also isoleucine, lysine, etc.).Non-essential amino acids can be synthesised within the body. Examples: glycine and alanine (also serine, aspartic acid, etc.).
10.12 Define the following as related to proteins
(i) Peptide linkage (ii) Primary structure (iii) Denaturation.
ANSWER(i) Peptide linkage: the amide bond –CO–NH– formed between the –COOH group of one amino acid and the –NH2 group of another, with elimination of a water molecule.(ii) Primary structure: the specific sequence (order) in which the amino acids are linked in a polypeptide chain. Any change in this sequence creates a different protein.(iii) Denaturation: the process in which a native protein loses its biological activity due to a change in temperature or pH, which disturbs hydrogen bonds and destroys the secondary and tertiary structures (the primary structure remains intact), e.g. coagulation of egg white on boiling.
10.13 What are the common types of secondary structure of proteins?
ANSWERThe two common types of secondary structure are the α-helix and the β-pleated sheet.Both arise from regular folding of the polypeptide backbone, stabilised by hydrogen bonding between the >C=O and –NH– groups of the peptide bonds.
10.14 What type of bonding helps in stabilising the α-helix structure of proteins?
ANSWERThe α-helix is stabilised by intramolecular hydrogen bonds.The –NH– group of each amino acid residue forms a hydrogen bond with the >C=O group of an adjacent turn of the helix, holding the right-handed coil in shape.
10.15 Differentiate between globular and fibrous proteins.
ANSWER
Fibrous proteins
Globular proteins
Polypeptide chains run parallel and are held by hydrogen and disulphide bonds, giving a fibre-like structure.
Polypeptide chains coil around to give a spherical (globular) shape.
10.16 How do you explain the amphoteric behaviour of amino acids?
ANSWERIn aqueous solution an amino acid exists as a zwitterion (+H3N–CHR–COO−), which contains both an acidic (–NH3+) and a basic (–COO−) centre.With an acid, the –COO− end accepts a proton; with a base, the –NH3+ end donates a proton. Because it can react both with acids and with bases, an amino acid is amphoteric.
10.17 What are enzymes?
ANSWEREnzymes are biocatalysts — almost all of them are globular proteins — that speed up the biochemical reactions of living organisms by lowering the activation energy.They are highly specific for a particular substrate and reaction, are required only in small amounts, and are usually named after the substrate or reaction with the ending -ase (e.g. maltase, sucrase).
10.18 What is the effect of denaturation on the structure of proteins?
ANSWEROn denaturation the hydrogen bonds are disturbed; the globules unfold and the helix uncoils, so the protein loses its biological activity.The secondary and tertiary structures are destroyed, but the primary structure (the sequence of amino acids) remains intact.
10.19 How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
ANSWERVitamins are classified by their solubility into (i) fat-soluble vitamins — A, D, E and K (stored in liver and adipose tissues) — and (ii) water-soluble vitamins — the B-group and vitamin C (must be supplied regularly as they are excreted in urine).The vitamin responsible for the coagulation (clotting) of blood is vitamin K.
10.20 Why are vitamin A and vitamin C essential to us? Give their important sources.
ANSWERVitamin A is essential for healthy eyes and vision; its deficiency causes xerophthalmia (hardening of the cornea) and night blindness. Sources: fish liver oil, carrots, butter and milk.Vitamin C (ascorbic acid) is essential for healthy gums, connective tissue and resistance to infection; its deficiency causes scurvy (bleeding gums). Sources: citrus fruits, amla and green leafy vegetables.
10.21 What are nucleic acids? Mention their two important functions.
ANSWERNucleic acids are long-chain polymers of nucleotides (polynucleotides) present in the nucleus of living cells; they are of two types, DNA and RNA.Functions: (i) DNA is the chemical basis of heredity — it stores and transfers genetic information from one generation to the next and is capable of self-duplication during cell division. (ii) Nucleic acids are responsible for protein synthesis in the cell (the message in DNA is carried out by RNA molecules).
10.22 What is the difference between a nucleoside and a nucleotide?
ANSWERA nucleoside is formed by the attachment of a nitrogenous base to the 1′ position of a pentose sugar (base + sugar).A nucleotide is formed when a nucleoside is further linked to phosphoric acid at the 5′ position of the sugar (base + sugar + phosphate). Thus a nucleotide = nucleoside + phosphate.
10.23 The two strands in DNA are not identical but are complementary. Explain.
ANSWERIn the DNA double helix the two strands are held together by hydrogen bonds between specific pairs of bases: adenine pairs only with thymine and cytosine pairs only with guanine.Because of this fixed base pairing, wherever one strand has A the other must have T, and wherever one has G the other must have C. So the strands are not identical but are complementary — the sequence of one strand fixes the sequence of the other.
10.24 Write the important structural and functional differences between DNA and RNA.
ANSWER
DNA
RNA
Sugar is β-D-2-deoxyribose.
Sugar is β-D-ribose.
Bases: adenine, guanine, cytosine and thymine.
Bases: adenine, guanine, cytosine and uracil.
Usually double-stranded (double helix).
Usually single-stranded.
Function: stores genetic information; chemical basis of heredity; capable of self-duplication.
Function: actually carries out protein synthesis (mRNA, rRNA, tRNA).
10.25 What are the different types of RNA found in the cell?
ANSWERThree types of RNA are found in the cell:(i) messenger RNA (m-RNA), (ii) ribosomal RNA (r-RNA) and (iii) transfer RNA (t-RNA). Together they carry out protein synthesis in the cell.
Extra Practice Questions
Short Answer Type Questions
Q1. Why is sucrose called invert sugar after hydrolysis?
ANSWERSucrose is dextrorotatory (+66.5°). On hydrolysis it gives dextrorotatory glucose (+52.5°) and laevorotatory fructose (−92.4°). As the laevorotation of fructose is greater, the mixture becomes laevorotatory, so the sign of rotation is inverted from (+) to (−); hence the product is called invert sugar.
Q2. Why is sucrose a non-reducing sugar while maltose is a reducing sugar?
ANSWERIn sucrose the reducing groups of both glucose (C1) and fructose (C2) are involved in the glycosidic bond, so no free reducing group is left — it is non-reducing. In maltose only one C1 is used in the linkage, so a free aldehyde group can form at the C1 of the other glucose unit, making it a reducing sugar.
Q3. What is a zwitterion?
ANSWERA zwitterion is the dipolar ion of an amino acid (+H3N–CHR–COO−) formed when the –COOH group loses a proton and the –NH2 group accepts it. It is overall neutral but carries both a positive and a negative charge, which explains the high melting points, water solubility and amphoteric behaviour of amino acids.
Q4. Distinguish between fat-soluble and water-soluble vitamins.
ANSWERFat-soluble vitamins (A, D, E, K) dissolve in fats/oils, are stored in the liver and adipose tissues, and need not be supplied daily. Water-soluble vitamins (B-group and C) dissolve in water, are not stored (except B12) and are excreted in urine, so they must be supplied regularly in the diet.
Q5. Name one disease each caused by deficiency of vitamin B1, vitamin C and vitamin D.
ANSWERVitamin B1 (thiamine) deficiency ⟶ beri-beri; vitamin C deficiency ⟶ scurvy; vitamin D deficiency ⟶ rickets (in children) / osteomalacia (in adults).
Long Answer Type Questions
Q1. Describe the four levels of structure of proteins.
ANSWERPrimary structure is the specific sequence of amino acids in a polypeptide chain; any change in the sequence gives a different protein. Secondary structure is the regular shape of the chain — the α-helix or β-pleated sheet — held by hydrogen bonds between the >C=O and –NH– groups. Tertiary structure is the overall three-dimensional folding of the chain (giving fibrous or globular shapes), stabilised by hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces. Quaternary structure is the spatial arrangement of two or more polypeptide sub-units in proteins made of several chains, e.g. haemoglobin.
Q2. Explain how the cyclic structure of glucose accounts for facts the open-chain structure cannot.
ANSWERThe open chain cannot explain why glucose fails the Schiff’s test, does not form a NaHSO3 addition product, gives a pentaacetate that does not react with hydroxylamine, and exists in two crystalline forms (α and β). These are explained by the cyclic structure: one of the –OH groups (at C5) adds across the –CHO group to form a six-membered hemiacetal (pyranose) ring. This converts the free aldehyde at C1 into a hemiacetal –OH, so the typical aldehyde reactions are suppressed. The new chiral centre at C1 (the anomeric carbon) gives two anomers, the α-form and β-form, which exist in equilibrium with the open chain — explaining the two crystalline forms and mutarotation.
Q3. Describe the chemical composition and biological functions of nucleic acids.
ANSWERComplete hydrolysis of a nucleic acid gives three components: a pentose sugar (β-D-2-deoxyribose in DNA, β-D-ribose in RNA), phosphoric acid, and nitrogenous bases. DNA contains adenine, guanine, cytosine and thymine; RNA contains the first three plus uracil. A base joined to the sugar forms a nucleoside, and a nucleoside joined to phosphate forms a nucleotide; nucleotides are linked by phosphodiester bonds between the 3′ and 5′ carbons to build the polynucleotide chain. DNA is a double helix with complementary base pairing (A–T, G–C). Functionally, DNA is the chemical basis of heredity — it stores genetic information, self-duplicates during cell division and carries the coded message for protein synthesis, while the three RNAs (mRNA, rRNA, tRNA) actually carry out protein synthesis in the cell.
MCQs & Assertion–Reason
1. Which of the following is a ketohexose?
(a) glucose (b) fructose (c) ribose (d) galactose
2. The glycosidic linkage in sucrose is between:
(a) C1 of glucose and C4 of glucose (b) C1 of glucose and C2 of fructose (c) C1 of galactose and C4 of glucose (d) C2 of fructose and C4 of glucose
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: Sucrose is a non-reducing sugar.
Reason: In sucrose the reducing groups of both glucose and fructose are involved in the glycosidic linkage.
A-R 2. Assertion: Amino acids have high melting points and high water solubility.
Reason: Amino acids exist as zwitterions and behave like ionic salts.
A-R 3. Assertion: On denaturation a protein loses its biological activity.
Reason: Denaturation destroys the primary structure of the protein.
A-R 4. Assertion: Glucose pentaacetate does not react with hydroxylamine.
Reason: In the pentaacetate there is no free aldehyde group available.
A-R 5. Assertion: Vitamin C must be taken regularly in the diet.
Reason: Vitamin C is water-soluble and is excreted in urine, so it cannot be stored in the body.
Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).
Common Mistakes & Exam Tips
Common mistakes to avoid
Calling sucrose a reducing sugar — it is non-reducing (both reducing groups are tied up in the glycosidic bond).
Confusing the products: bromine water ⟶ gluconic acid (mono-acid), HNO3 ⟶ saccharic acid (di-acid).
Saying denaturation destroys the primary structure — only the secondary and tertiary structures are destroyed.
Mixing up the sugars: DNA has 2-deoxyribose, RNA has ribose; DNA has thymine, RNA has uracil.
Writing α-glucose for cellulose — cellulose is made of β-D-glucose; starch of α-D-glucose.
Learn the reaction products of glucose (HI, Br2 water, HNO3) and the four cyclic-structure facts cold — they are repeated favourites. Always use subscripts and superscripts correctly in formulae (C6H12O6, –NH3+). Present comparison answers (DNA vs RNA, fibrous vs globular, starch vs cellulose) as neat two-column tables. Remember the base-pairing rule (A–T, G–C) and the “-ase” rule for enzyme names. State definitions exactly — peptide linkage, glycosidic linkage, denaturation, zwitterion.
Frequently Asked Questions
What is Class 12 Chemistry Chapter 10 Biomolecules about?
Chapter 10 studies the main biomolecules of living systems — carbohydrates, proteins, enzymes, vitamins, nucleic acids and hormones. It covers the structure and reactions of glucose and fructose, the disaccharides sucrose, maltose and lactose, α-amino acids and protein structure, denaturation, classification of vitamins, and the structure and functions of DNA and RNA.
What is the difference between DNA and RNA?
DNA contains the sugar 2-deoxyribose, the base thymine, and is normally double-stranded; it stores genetic information. RNA contains ribose, the base uracil instead of thymine, and is normally single-stranded; it carries out protein synthesis through mRNA, rRNA and tRNA.
Why is glucose pentaacetate unable to react with hydroxylamine?
In its cyclic form the C1 of glucose is a hemiacetal, not a free –CHO. On acetylation this hemiacetal –OH is also acetylated and the ring is locked, so there is no free aldehyde group left to react with hydroxylamine.
Are these Class 12 Chemistry Chapter 10 solutions free?
Yes. All solutions are free and follow the official NCERT Chemistry textbook for session 2026–27, with every Intext and Exercise question reproduced and solved.
