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NCERT Exercises and Solutions: Biomolecules

🎓 Class 11 Biology CBSE Theory Ch 9 – Biomolecules ⏱ ~8 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: NCERT Exercises and Solutions: Biomolecules

આ મૂલ્યાંકન આના પર આધારિત હશે: NCERT Exercises and Solutions: Biomolecules

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NCERT Exercises and Solutions: Biomolecules

Chapter 9 — Summary

This chapter took us from extraction of biomolecules to the molecular machines (enzymes) that drive metabolism. Here are the essentials:

  • Acid soluble pool (small molecules) is separated from acid insoluble pool (macromolecules) by trichloroacetic acid extraction.
  • Primary metabolites have known physiological roles; secondary metabolites are species-specific products like alkaloids, pigments, gums, antibiotics.
  • Carbohydrates (CH₂O)ₙ — monosaccharides (glucose, fructose, ribose), disaccharides (sucrose, lactose, maltose), polysaccharides (starch, glycogen, cellulose, chitin).
  • Proteins are polymers of 20 amino acids linked by peptide bonds. Four levels of structure: 1° (sequence), 2° (α-helix, β-sheet), 3° (3-D fold), 4° (multi-subunit assembly).
  • Nucleic acids are polymers of nucleotides (base + pentose + phosphate). DNA uses deoxyribose + T; RNA uses ribose + U. Double helix held by A=T and G≡C complementary pairing (Chargaff's rule).
  • Lipids are insoluble in water. Simple lipids (fats, oils, waxes); compound lipids (phospholipids, glycolipids); derived lipids (steroids, sterols). Phospholipids form membrane bilayers.
  • Enzymes are biological catalysts (mostly proteins, some RNA) that lower activation energy. Show specificity via active site; obey Michaelis-Menten kinetics. Six IUBMB classes — Oxidoreductase, Transferase, Hydrolase, Lyase, Isomerase, Ligase.
  • Enzyme activity depends on temperature, pH, and substrate concentration. Many need cofactors — coenzymes (NAD⁺), prosthetic groups (haem), or metal ions (Zn²⁺).

Key Terms — Click Each

  • Macromolecule: MW > 10,000; polymer of small monomers.
  • Glycosidic bond: sugar-sugar bond formed by condensation.
  • Peptide bond: protein bond.
  • Phosphodiester bond: nucleic acid backbone bond.
  • Zwitterion: net-neutral but dipolar molecule.
  • Denaturation: loss of native folded structure.
  • Holoenzyme: full enzyme (apoenzyme + cofactor).

Chapter 9 — NCERT Exercises (Worked Solutions)

Q1. What are macromolecules? Give examples.

Macromolecules are large biomolecules with molecular weights greater than ~10,000 Daltons. They are formed by polymerization of small monomers.

Four major classes:
  • Proteins — polymers of amino acids. e.g., albumin, haemoglobin, collagen.
  • Polysaccharides — polymers of monosaccharides. e.g., starch, cellulose, glycogen, chitin.
  • Nucleic acids — polymers of nucleotides. e.g., DNA, RNA.
  • Lipids (operationally) — in the acid-insoluble pool though molecularly small; e.g., membrane lipids in vesicles.

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

Glycosidic bond: Formed between —OH of one sugar's anomeric carbon and —OH of another sugar (or another molecule). Loss of one H₂O.

Sugar1—O—Sugar2 (with loss of H₂O)
e.g., Glucose + Fructose → Sucrose + H₂O (α-1,2 glycosidic bond)

Peptide bond: Formed between —COOH of one amino acid and —NH₂ of next. Loss of one H₂O.

R₁—CH(NH₂)—CO—NH—CH(R₂)—COOH
The —C(=O)—NH— group is the peptide bond.

Phosphodiester bond: Phosphate links 3'-OH of one nucleotide sugar to 5'-OH of next. Two ester linkages share one phosphate.

Nucleotide-1 (3'-OH) — O — PO₂⁻ — O — (5') Nucleotide-2
Forms the sugar-phosphate backbone of DNA/RNA.

Q3. What is meant by tertiary structure of proteins?

Tertiary structure is the overall three-dimensional folding of a single polypeptide chain — how the entire chain (its α-helices, β-sheets, and connecting loops) packs in space to form a compact globular or fibrous shape.

Stabilising bonds:
  • Hydrogen bonds (between R-groups)
  • Ionic / salt bridges (between oppositely charged side chains)
  • Disulphide bridges (covalent —S—S— between cysteine residues)
  • Hydrophobic interactions (non-polar R-groups cluster inside, away from water)
  • Van der Waals forces
Importance: The tertiary structure positions the active-site residues of an enzyme — without correct folding, the protein has no function. Example: myoglobin's tertiary fold creates the haem-binding pocket that lets it carry oxygen.

Q4. Find and write the structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which makes use of any of these by isolating them.

MoleculeBrief StructureIndustry Use
Citric acidHOOC-CH₂-C(OH)(COOH)-CH₂-COOHSoft-drink, food preservation
Lactic acidCH₃-CHOH-COOHDairy (curd), biopolymer PLA
Acetic acidCH₃-COOHVinegar, textile
EthanolCH₃-CH₂-OHBeverages, fuel, sanitizer
GlucoseC₆H₁₂O₆ (aldohexose)IV drips, fermentation
CaffeineMethylated xanthine alkaloidBeverages, pharma
NicotinePyridine + pyrrolidine alkaloidPesticides, tobacco
AdrenalineCatecholamineAnaphylaxis drug
Penicillinβ-lactam ring + side chainAntibiotic industry
Vitamin C (ascorbate)C₆H₈O₆ lactoneFood/supplement industry
Industries: Pharmaceutical, food, beverage, cosmetics, fermentation — all rely on isolating and purifying these small biomolecules.

Q5. Proteins have primary structure. If you are given a method to know which amino acid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?

YES — knowing terminal amino acids is a powerful test for protein purity.

A pure protein is a homogeneous population — all molecules have identical sequences, hence the SAME amino acid at N-terminus and the SAME amino acid at C-terminus.

Test logic:
  • If N-terminus analysis (e.g., Edman degradation) gives ONE unique amino acid → likely pure.
  • If you find MULTIPLE different amino acids at "the" N-terminus → the sample contains multiple different proteins → impure / heterogeneous.
  • Same logic for C-terminus (by carboxypeptidase digestion).
Caveat: Some proteins have N-terminal blocking modifications (acetylation, pyroglutamate) — sequencing might fail even though pure. Modern techniques use mass spectrometry which sidesteps this.

This is exactly how Frederick Sanger first proved insulin was a pure protein, then sequenced it (Nobel 1958).

Q6. Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e.g., cosmetics etc.).

Therapeutic Proteins:
  • Insulin — diabetes treatment (recombinant human insulin since 1982)
  • Growth hormone (GH) — for growth disorders
  • Erythropoietin (EPO) — stimulates RBC production in anaemia, kidney disease
  • Interferons — antiviral and cancer treatment
  • Monoclonal antibodies — Herceptin (breast cancer), Rituximab (lymphoma), Humira (rheumatoid arthritis)
  • Clotting factors VIII, IX — haemophilia treatment
  • Tissue plasminogen activator (tPA) — for stroke
  • Vaccines — hepatitis B surface antigen, COVID spike protein-based vaccines
Cosmetic / Industrial proteins:
  • Collagen — anti-aging cream, wound dressings
  • Keratin — hair care products
  • Silk fibroin — luxury textiles, biomedical sutures
  • Casein — adhesives, food fortifier
  • Enzymes — protease in detergents, rennet in cheese, amylase in baking
  • Botox — botulinum toxin (a neurotoxic protein) for cosmetic wrinkle removal

Q7. Explain the composition of triglyceride.

A triglyceride = 1 molecule of glycerol + 3 molecules of fatty acids, joined by three ester bonds (with release of 3 water molecules).

Structure:
H₂C—O—CO—R₁
 |
HC—O—CO—R₂
 |
H₂C—O—CO—R₃
where R₁, R₂, R₃ = long hydrocarbon chains of fatty acids (often C₁₄–C₂₀).

Saturated: all R chains have only single C—C bonds → solid at room temperature (animal fats like butter, ghee).
Unsaturated: one or more C=C double bonds → liquid at room temperature (plant oils like mustard oil, olive oil).

Function: Most concentrated form of energy storage in animals (adipose tissue) — yields ~9 kcal/g vs 4 kcal/g for carbs/proteins.

Q8. Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins?

Milk → Curd: A case of bacterial fermentation + protein denaturation.
  1. Milk contains the protein casein (~80% of milk protein) suspended as micelles, stabilized at neutral pH.
  2. When a starter culture (Lactobacillus) is added, the bacteria ferment milk sugar (lactose) into lactic acid.
  3. Lactic acid lowers the pH of milk from ~6.7 to ~4.6 (the isoelectric pH of casein).
  4. At isoelectric pH, casein micelles lose their net surface charge → they no longer repel each other → they aggregate / coagulate.
  5. The resulting gel network traps water, fat globules, and other components → semi-solid curd.
Key insight: This is acid-induced denaturation/coagulation of casein. It's the same principle by which paneer is made (using lemon juice or vinegar — direct acid addition without bacterial fermentation).

Bonus: Yoghurt also has live Lactobacillus probiotics, and the fermentation generates flavour compounds and partially digests lactose — making yoghurt easier to digest than milk for lactose-intolerant individuals.

Q9. Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick models).

Yes! Recommended modelling activity:
  1. Glucose: 6 C atoms with H, OH groups; ring form has O in ring.
  2. Amino acid (glycine): central C, with —NH₂, —COOH, —H, —H. Then alanine: replace one H with —CH₃.
  3. Dipeptide: Link two amino acids by losing H₂O, forming —C(=O)—NH— bond.
  4. Watson-Crick base pair: Adenine paired with thymine using 2 H-bonds (dotted lines between N-H...N and N...N-H).
  5. Fatty acid: Long C chain (e.g., 16 C for palmitic) with —COOH at one end.
Online alternatives: Use JMol, RasMol, PyMOL, or AlphaFold for 3-D protein visualization. PDB (Protein Data Bank, rcsb.org) has free atomic models of thousands of proteins.

Q10. Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionisable) functional groups in the amino acid.

Acid-base titration of an amino acid:
  1. Start with glycine at low pH (~1.5) — fully protonated form: ⁺H₃N-CH₂-COOH.
  2. Add base (NaOH) slowly while measuring pH.
  3. Plot pH vs equivalents of base added.
Observation: Plateaus appear at two pH values for glycine:
  • pKa₁ ≈ 2.3 — the —COOH deprotonates to —COO⁻ (carboxyl group).
  • pKa₂ ≈ 9.6 — the —NH₃⁺ deprotonates to —NH₂ (amino group).
Between the two plateaus is the isoelectric point (pI) — for glycine, pI = (2.3 + 9.6)/2 ≈ 5.97 — where it exists as a zwitterion with net zero charge.

Number of ionisable groups in glycine = 2 (—COOH and —NH₃⁺). Glutamate or aspartate would show 3 plateaus (one extra side-chain —COOH); lysine would show 3 plateaus (one extra side-chain —NH₃⁺).

Q11. Draw the structure of the amino acid, alanine.

Alanine structure (L-alanine):
       H
       |
  H₂N—C—COOH
       |
      CH₃
- Central α-carbon (chiral)
- —NH₂ amino group
- —COOH carboxyl group
- —H
- —CH₃ (methyl R-group) — making alanine the simplest "non-trivial" amino acid (glycine has R=H).

At physiological pH 7.4, alanine exists as zwitterion: ⁺H₃N-CH(CH₃)-COO⁻. R-group character: non-polar (hydrophobic). Code: A (one-letter) or Ala (three-letter). Non-essential — synthesized from pyruvate by transamination.

Q12. What are gums made of? Is Fevicol different?

Natural gums are secondary metabolites — complex polysaccharides exuded by plants (often as defence/wound healing).
  • Gum arabic (from Acacia) — a branched polysaccharide of D-galactose, L-arabinose, L-rhamnose, and glucuronic acid.
  • Gum tragacanth (from Astragalus) — a mixture of polysaccharides including tragacanthic acid.
  • Guar gum (from guar bean) — galactomannan.
Fevicol is DIFFERENT. It is a synthetic adhesive — specifically a polyvinyl acetate (PVA) emulsion in water. PVA is a synthetic polymer made by polymerization of vinyl acetate monomers, not a natural polysaccharide. So:
  • Natural gums = biological polysaccharides (sugar polymers).
  • Fevicol = man-made plastic emulsion (vinyl polymer).
Both stick things, but their chemistry and origin are fundamentally different.

Q13. Find out a procedure to estimate all the carbohydrates, fats and proteins in a food item. Quantitatively analyse all of them.

Standard analytical procedures:
  1. Total carbohydrates: Anthrone test (anthrone + conc. H₂SO₄ → blue-green colour proportional to sugars). Measure absorbance at 620 nm. Compare with glucose standard curve.
  2. Reducing sugars: Benedict's test (quantitative if using DNS — dinitrosalicylic acid reagent). Read absorbance at 540 nm.
  3. Proteins:
    • Kjeldahl method — measures total nitrogen × 6.25 = crude protein.
    • Biuret method — Cu²⁺ in alkali binds peptide bonds → purple colour at 540 nm.
    • Lowry / Bradford — for higher sensitivity.
  4. Fats:
    • Soxhlet extraction — repeatedly extract food in petroleum ether/hexane → evaporate solvent → weigh residue (total lipid).
    • Sudan III / IV staining — qualitative.
Quantitative analysis example for rice (per 100 g):
  • Carbohydrates ≈ 78 g (mostly starch)
  • Proteins ≈ 7 g
  • Fats ≈ 1 g
  • Total calories = (78×4) + (7×4) + (1×9) = 312 + 28 + 9 ≈ 349 kcal

🎯 Mixed Competency-Based Questions

Q1. Which of these is NOT a polysaccharide? L1 Remember

  • (a) Cellulose
  • (b) Glycogen
  • (c) Sucrose
  • (d) Chitin
Answer: (c) Sucrose. It's a disaccharide (glucose + fructose). All others are long-chain polysaccharides.

Q2. Match the following: L2 Understand

  • (1) Insulin — (a) Antibody
  • (2) Catalase — (b) Hormone
  • (3) IgG — (c) Transport
  • (4) Haemoglobin — (d) Enzyme
1—(b) Hormone; 2—(d) Enzyme; 3—(a) Antibody; 4—(c) Transport. All are proteins illustrating the diverse functional repertoire of proteins.

Q3. Why are mRNA molecules generally short-lived in the cell? L3 Apply

mRNA carries the information from DNA to ribosome for protein synthesis. It is intentionally unstable because:
  • It has 2'-OH (in ribose) — makes RNA more chemically labile to hydrolysis than DNA (2'-deoxyribose has no such group).
  • Cells need to rapidly turn protein synthesis ON/OFF in response to changing conditions — short-lived mRNA allows quick switching.
  • Stable mRNA would lock the cell into producing one protein indefinitely.
DNA, by contrast, is the long-term archive → stable, double-stranded, and uses deoxyribose.

Q4. Evaluate: A scientist claims an enzyme she has discovered works at 100°C. What evidence would convince you? L5 Evaluate

Such claims are scientifically plausible but require strong evidence:
  1. Source organism: Was it isolated from a hyperthermophile (e.g., Pyrococcus, Thermus)? These bacteria genuinely live near 100°C.
  2. Activity assays: Show enzyme catalyses its reaction at 100°C with measurable rate, increasing from 60°C → 100°C.
  3. Structural stability: Circular dichroism or differential scanning calorimetry showing retained fold at 100°C.
  4. No artefacts: Control with boiled enzyme (should be inactive); buffer must be stable at 100°C; sealed tubes to prevent evaporation.
  5. Reproducibility: Same result independently.
Famous real example: Taq DNA polymerase from Thermus aquaticus (Yellowstone hot springs) works at 95°C — revolutionized molecular biology via PCR. So 100°C activity is real but rare.

Q5. HOT (Create): Design a hypothetical "super-protein" that combines features of: (i) silk's strength, (ii) haemoglobin's gas-binding, (iii) antibody's specificity. L6 Create

Design Concept — "Spider-Hemo-AbProtein":
  • Backbone: Silk fibroin-like β-sheet regions (Gly-Ala-Gly-Ala-Gly-Ser repeats) for extreme tensile strength — the protein can be spun into a fiber.
  • Functional cluster 1: A heme-binding pocket adapted from haemoglobin's globin fold — embedded within the silk core. Heme bound here can carry O₂ or CO₂.
  • Functional cluster 2: Variable surface loops (CDR-like, from antibodies) at fixed positions along the fibre — engineered to recognize specific targets (toxins, viral spikes).
  • Applications:
    • Smart wound dressings — fibre catches pathogens (CDR), delivers O₂ (heme), provides scaffold (silk).
    • Biomedical sensors — colour change when target is bound (because heme shifts spectrum).
    • Self-strengthening textiles for soldiers/medical applications.
  • Engineering: Use modular DNA cloning — fuse silk gene + globin gene + scFv antibody gene; express in E. coli or transgenic silkworms.
This kind of fusion-protein engineering is already happening — e.g., spider silk + GFP, spider silk + bone-binding peptides.

🧠 Mixed Assertion-Reason Questions

Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.

A: Lipids are not technically macromolecules.

R: Individual lipid molecules have molecular weights below 1000 Da, much smaller than the 10,000 Da threshold for macromolecules.

Answer: (A). Both true; R explains A. Despite being grouped with macromolecules, lipids are individually small. They appear in the acid-insoluble pool only because they're trapped in membrane vesicles, which are large aggregates.

A: All enzymes are proteins.

R: Only proteins have the structural complexity to form active sites.

Answer: (D). A is FALSE — ribozymes (catalytic RNAs like the ribosome's peptidyl transferase center) are non-protein enzymes. R is FALSE in implication — RNA also folds to form catalytic active sites. So strictly D-style: A false, R weak. Best fit: (D) with caveat that R is also imperfect.

A: Eating only rice may lead to protein deficiency despite rice having protein.

R: Rice protein is deficient in the essential amino acid lysine.

Answer: (A). Both true; R explains A. A "complete protein" needs ALL essential amino acids in adequate amounts. Rice has methionine but is lysine-poor. Combining rice + dal (which is lysine-rich, methionine-poor) gives a balanced amino acid profile — the wisdom behind India's traditional rice-dal staple.

Frequently Asked Questions - NCERT Exercises and Solutions: Biomolecules

What are the most-asked NCERT exercise questions in Chapter 9 Biomolecules?
NCERT Class 11 Biology Chapter 9 Biomolecules exercises cover definitions, classification, structure-function relationships, labelled diagrams, and application-based questions. The MyAiSchool exercise set provides full step-by-step solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should master scientific terminology, diagram labelling, and concept comparison tables to score full marks.
How should students approach labelled diagram questions in Biomolecules?
For labelled diagram questions in NCERT Class 11 Biology Chapter 9 Biomolecules: (1) draw a clean, large, proportional diagram with sharp pencil lines, (2) label each part with horizontal lines on the right or left side, (3) use scientific terminology (Latin/Greek names where applicable), (4) write a 1-2 line description below if asked. The MyAiSchool solutions provide editable reference diagrams aligned with NCERT textbook figures.
What types of CBSE board questions come from Chapter 9?
CBSE Class 11 Biology board questions from Chapter 9 (Biomolecules) typically include: (1) 1-mark MCQs on definitions and classification, (2) 2-mark short-answers on structure-function or comparisons, (3) 3-mark labelled-diagram questions, (4) 5-mark long-answer essays combining diagram + description + significance. The MyAiSchool exercise set tags each question by mark weight and Bloom level (L1-L6).
How do I compare two biological concepts in 5-mark questions?
For 5-mark comparison questions in NCERT Class 11 Biology Chapter 9: (1) use a two-column table with feature headings down the left side, (2) compare on at least 5-6 features (structure, function, location, examples, significance), (3) include one labelled diagram if relevant, (4) end with one line on biological significance. The MyAiSchool solutions follow this CBSE-aligned tabular format consistently for full marks.
What are common mistakes in Chapter 9 exercises?
Common mistakes in NCERT Class 11 Biology Chapter 9 (Biomolecules) include: (1) confusing similar scientific names or terminology, (2) skipping diagram labels or drawing too small, (3) missing examples in classification questions, (4) writing essay-style answers when a table is expected, (5) forgetting to mention biological significance. The MyAiSchool solutions highlight these traps so students avoid losing marks unnecessarily.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Biology Chapter 9 Biomolecules solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with biological reasoning, fully labelled SVG diagrams, comparison tables, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT textbook and CBSE marking schemes.
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