Current Affairs updated daily
KalviGavaniKalviGavani

Biomolecules — NCERT Class 11 Biology

Complete NCERT Solutions for CBSE Class 11 Biology Chapter 9 — all intext and chapter-end exercise questions on proteins, carbohydrates, lipids, nucleic acids, enzymes, and metabolic pathways answered clearly.

Updated 11 June 20269 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

How to Use This Page

These are complete NCERT Solutions for CBSE Class 11 Biology Chapter 9 — Biomolecules. Attempt each question before checking. Use precise biochemical terms — marks are given for terms like 'peptide bond', 'glycosidic bond', 'phosphodiester bond', 'activation energy', etc.


Key Concepts (Quick Recap)

  • Chemical analysis: Living organisms contain C, H, O, N, S, P, Na, K, Ca, Mg, Fe, and trace elements.
  • Amino acids: 20 standard; α-amino acids with amino group, carboxyl group, hydrogen, and R-group on same carbon; zwitterion at neutral pH.
  • Proteins: Polymers of amino acids; peptide bonds; 4 structural levels (primary, secondary, tertiary, quaternary).
  • Monosaccharides: Glucose, fructose, ribose, deoxyribose; Disaccharides: sucrose, lactose, maltose; Polysaccharides: starch, glycogen, cellulose, chitin.
  • Lipids: Fats (triglycerides), phospholipids, steroids (cholesterol); fats = glycerol + 3 fatty acids.
  • Nucleic acids: DNA (double helix, deoxyribose, A-T, G-C) and RNA (single strand, ribose, A-U, G-C).
  • Enzymes: Biological catalysts; lower activation energy; competitive and non-competitive inhibition; cofactors and coenzymes.
  • Metabolic pathways: Anabolic (building up, consume energy) and catabolic (breaking down, release energy).
  • The Living State: Metabolism provides the dynamic organisation; steady state with energy input; living organisms are in non-equilibrium steady state.

Intext Questions

Page 141

Q1. What are macromolecules? Give examples.
Answer

Macromolecules are large, complex molecules with high molecular weight (more than 1000 Daltons) that are formed by polymerisation of micromolecules. Examples:

  • Proteins (polymer of amino acids)
  • Nucleic acids — DNA and RNA (polymer of nucleotides)
  • Polysaccharides (polymer of monosaccharides) — starch, cellulose, glycogen
Q2. Illustrate a glycosidic, peptide, and phosphodiester bond.
Answer

1. Glycosidic bond: A bond formed between two monosaccharide units by the loss of a water molecule (dehydration). For example, in sucrose: C₁ of glucose bonds with C₂ of fructose via an oxygen bridge, releasing H₂O.

2. Peptide bond: A bond formed between the carboxyl group (–COOH) of one amino acid and the amino group (–NH₂) of another amino acid, with the release of H₂O. Structure: –CO–NH–

3. Phosphodiester bond: A bond formed between the 3′-OH group of one nucleotide and the 5′-phosphate group of the next nucleotide, releasing H₂O. This forms the sugar-phosphate backbone of DNA and RNA.

Q3. What is meant by tertiary structure of proteins?
Answer

The tertiary structure of a protein is the three-dimensional folding of the entire polypeptide chain. It is stabilised by four types of bonds: 1. Hydrogen bonds — between polar side chains 2. Ionic bonds (salt bridges) — between charged side chains 3. Disulphide bonds — between cysteine residues (–S–S–) 4. Hydrophobic interactions — between non-polar side chains (they cluster inside the protein)

The tertiary structure gives the protein its functional shape and determines its biological activity.

Page 143

Q1. Find out who had written the book 'Genera Plantarum'?
Answer

The book Genera Plantarum was written by George Bentham and Joseph Dalton Hooker (1862–1883). It was a significant work on plant classification.


Chapter-End Exercise

Q1. What are macromolecules? Give examples.
Answer

(See Intext Page 141 Q1 — answered above.)

Q2. Illustrate a glycosidic, peptide, and phosphodiester bond.
Answer

(See Intext Page 141 Q2 — answered above.)

Q3. What is meant by tertiary structure of proteins?
Answer

(See Intext Page 141 Q3 — answered above.)

Q4. Find out who had written the book 'Genera Plantarum'?
Answer

(See Intext Page 143 Q1 — answered above.)

Q5. Find out what makes the cholesterol molecule different from a steroid.
Answer

Cholesterol is a steroid — it belongs to the steroid family. The difference is:

  • Cholesterol is a specific steroid molecule with a complex ring structure (4 fused carbon rings: 3 six-membered and 1 five-membered) plus a hydrocarbon side chain (8 carbon atoms) and a hydroxyl group (–OH) at C₃. It has 27 carbon atoms.
  • Steroids are a broader class of compounds that all share the same four-ring structure (cyclopentanoperhydrophenanthrene ring) but differ in the functional groups, side chains, and number of carbon atoms attached. Examples: cholesterol, testosterone, oestrogen, cortisol, vitamin D.

So cholesterol is a type of steroid — the difference is that cholesterol is one specific molecule with a particular structure, while "steroid" refers to the entire class of compounds sharing the basic ring structure.

Q6. Write the difference between: (a) Reducing and non-reducing sugars (b) Saturated and unsaturated fatty acids
Answer

(a) Reducing vs Non-reducing Sugars:

FeatureReducing SugarsNon-reducing Sugars
Free aldehyde/ketone groupPresentAbsent (both anomeric carbons involved in glycosidic bond)
Benedict's/Fehling's testPositiveNegative (unless hydrolysed)
ExamplesGlucose, fructose, lactose, maltoseSucrose
StructureFree –CHO or –C=O availableBoth reducing ends bonded

(b) Saturated vs Unsaturated Fatty Acids:

FeatureSaturated Fatty AcidsUnsaturated Fatty Acids
Double bondsAbsent (single bonds only)Present (one or more)
State at room temperatureSolidLiquid (oil)
Melting pointHigherLower
SourcesAnimal fats (butter, ghee)Vegetable oils (olive oil, sunflower oil)
Health effectIncrease LDL cholesterolLower LDL, increase HDL
Q7. What do you understand by the following terms: (a) Primary structure of protein (b) Co-enzyme (c) Active site of enzyme
Answer

(a) Primary structure of protein: The linear sequence of amino acids in a polypeptide chain, connected by peptide bonds. The specific order of amino acids determines the protein's properties and higher-level folding. Any change in even one amino acid can alter the protein's function (e.g., sickle cell anaemia — glutamic acid replaced by valine at position 6 in haemoglobin).

(b) Co-enzyme: A non-protein organic molecule that binds to the enzyme and is required for the enzyme's activity. Co-enzymes are often derived from vitamins. Examples: NAD⁺ (from niacin), FAD (from riboflavin), coenzyme A (from pantothenic acid). They carry chemical groups or electrons from one reaction to another.

(c) Active site of enzyme: A specific region on the enzyme surface where the substrate binds and the catalytic reaction occurs. The active site has a unique three-dimensional shape that is complementary to the substrate (lock and key model), or adjusts its shape upon substrate binding (induced fit model). It is a small part of the total enzyme but determines its specificity.

Q8. Explain the composition of triglyceride.
Answer

A triglyceride (fat/oil) is composed of:

  • One molecule of glycerol (3-carbon alcohol with three –OH groups)
  • Three molecules of fatty acids (each with a long hydrocarbon chain and a –COOH group)

The three fatty acids undergo esterification with the three hydroxyl groups of glycerol, releasing three water molecules:

Glycerol + 3 Fatty acids → Triglyceride + 3 H₂O

The ester bonds formed are: –COO–

If the fatty acids are saturated (no double bonds), the triglyceride is a fat (solid at room temperature, e.g., butter). If unsaturated (one or more double bonds), it is an oil (liquid at room temperature, e.g., olive oil).

Q9. Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins?
Answer

When milk is converted into curd: 1. Lactobacillus bacteria (or a starter culture) are added to warm milk. 2. The bacteria ferment lactose (milk sugar) into lactic acid. 3. Lactic acid lowers the pH of milk, making it acidic. 4. The acidic environment denatures the milk proteins (primarily casein) — the tertiary and quaternary structures unfold. 5. The denatured casein molecules lose their solubility and coagulate (precipitate), forming a network that traps water and fat globules. 6. This coagulation gives curd its semi-solid texture.

This is a process of protein denaturation and coagulation — the primary structure (amino acid sequence) remains intact, but the secondary, tertiary, and quaternary structures are disrupted by the change in pH.

Q10. Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick models)?
Answer

Yes, you can build models using ball-and-stick kits available in chemistry labs:

  • Glucose (C₆H₁₂O₆): Use 6 black balls (C), 12 white balls (H), 6 red balls (O), and sticks for bonds.
  • Amino acid: Build a central carbon with an amino group (–NH₂), carboxyl group (–COOH), hydrogen, and R-group.
  • Nucleotide: Build a phosphate group, a pentose sugar (ribose/deoxyribose), and a nitrogenous base connected together.

These models help visualise the 3D structure and bonding patterns. The key learning is understanding how monomers connect via glycosidic, peptide, and phosphodiester bonds to form polymers.

Q11. What are gums? How are they formed?
Answer

Gums are complex polysaccharides (heteropolysaccharides) produced by plants. They are formed by the injury or infection of plant tissues — the gum is secreted as a protective response. Chemically, gums are made of sugars like arabinose, galactose, and glucuronic acid. They swell in water to form a viscous solution or gel. Examples: gum arabic (from Acacia senegal), gum tragacanth, gum ghatti.

Q12. Find out a qualitative test for proteins, fats, and oils, amino acids and starch, and explain the process.
Answer
BiomoleculeTestProcedurePositive Result
ProteinsBiuret testAdd NaOH and CuSO₄ to sampleViolet/purple colour
Amino acidsNinhydrin testAdd ninhydrin reagent, heatBlue/purple colour
Fats/OilsSudan III testAdd Sudan III dyeRed-stained oil droplets
Fats/OilsEmulsion testAdd ethanol, then waterWhite emulsion forms
StarchIodine testAdd iodine solutionBlue-black colour
Q13. Find out exactly how much cholesterol you have in your blood and what your doctor says about it.
Answer

(This is a practical observation question — students should get their lipid profile tested and report their values. Normal ranges:)

ParameterNormal Range
Total CholesterolLess than 200 mg/dL
LDL (bad cholesterol)Less than 100 mg/dL
HDL (good cholesterol)More than 40 mg/dL (men) / 50 mg/dL (women)
TriglyceridesLess than 150 mg/dL
VLDL2–30 mg/dL

High total cholesterol and LDL increase the risk of atherosclerosis and heart disease. High HDL is protective.


Internal Links

  • [Chapter 9 Guide — Biomolecules](/school/cbse/class/11/biology/cbse-biomolecules)
  • [All Class 11 Biology Solutions](/school/cbse/class/11/biology/solutions)
  • [Class 11 Biology Subject Guide](/school/cbse/class/11/biology)

Frequently Asked Questions

Reducing sugars have a free aldehyde (–CHO) or ketone (–C=O) group that can reduce Benedict's or Fehling's reagent — examples: glucose, fructose, lactose, maltose. Non-reducing sugars do not have a free aldehyde or ketone group because their glycosidic bond involves both anomeric carbons — example: sucrose (glucose + fructose bonded at both C1 positions). Sucrose must be hydrolysed first to test positive for reducing sugars.