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Biomolecules

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Chapter Notes

Overview
This chapter explores the chemistry of life, focusing on the structure, classification, and function of biomolecules like carbohydrates, proteins, nucleic acids, enzymes, vitamins, and hormones. It explains how non-living atoms and molecules organize into complex systems that sustain life. The one big idea a student must walk away with is that the molecular logic of life processes is governed by specific structures and interactions of these biomolecules, and understanding their chemistry is essential for comprehending biology and medicine.
Key Concepts (20)
Carbohydrates
Optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis.
e.g. Glucose (C6H12O6) is a carbohydrate found in sweet fruits and honey.
Monosaccharides
A carbohydrate that cannot be hydrolysed further to give simpler unit of polyhydroxy aldehyde or ketone.
e.g. Glucose, fructose, ribose.
Oligosaccharides
Carbohydrates that yield two to ten monosaccharide units on hydrolysis.
e.g. Sucrose (disaccharide) yields glucose and fructose on hydrolysis.
Polysaccharides
Carbohydrates which yield a large number of monosaccharide units on hydrolysis.
e.g. Starch, cellulose, glycogen.
Reducing sugars
All carbohydrates which reduce Fehling’s solution and Tollens’ reagent.
e.g. All monosaccharides (e.g., glucose, fructose) are reducing sugars.
Glycosidic linkage
The oxide linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule.
e.g. In sucrose, C1 of α-D-glucose is linked to C2 of β-D-fructose.
Anomers
The two cyclic hemiacetal forms of a monosaccharide that differ only in the configuration of the hydroxyl group at the anomeric carbon (C1).
e.g. α-D-glucose and β-D-glucose are anomers.
Proteins
Polymers of α-amino acids linked by peptide bonds, with molecular mass higher than 10,000 u.
e.g. Keratin in hair, insulin in pancreas.
Amino acids
Compounds containing amino (–NH2) and carboxyl (–COOH) functional groups; only α-amino acids are obtained on hydrolysis of proteins.
e.g. Glycine (H2N–CH2–COOH) is the simplest amino acid.
Peptide bond
An amide bond formed between the –COOH group of one amino acid and the –NH2 group of another, with elimination of a water molecule.
e.g. Glycylalanine is a dipeptide formed from glycine and alanine.
Zwitter ion
A dipolar ion formed when the carboxyl group loses a proton and the amino group accepts a proton in aqueous solution, giving both positive and negative charges.
e.g. Glycine in water exists as +H3N–CH2–COO–.
Denaturation of proteins
Loss of biological activity of a protein due to disturbance of hydrogen bonds by physical or chemical changes, destroying secondary and tertiary structures but leaving primary structure intact.
e.g. Coagulation of egg white on boiling.
Enzymes
Biocatalysts, mostly globular proteins, that speed up biochemical reactions under mild conditions with high specificity.
e.g. Maltase catalyses hydrolysis of maltose into glucose.
Vitamins
Organic compounds required in small amounts in the diet to perform specific biological functions for normal maintenance of optimum growth and health.
e.g. Vitamin C (ascorbic acid) prevents scurvy.
Nucleic acids
Long chain polymers of nucleotides, also called polynucleotides, responsible for heredity and protein synthesis.
e.g. DNA and RNA.
Nucleoside
A unit formed by the attachment of a base to the 1' position of sugar.
e.g. Adenosine (adenine + ribose).
Nucleotide
A nucleoside linked to phosphoric acid at the 5'-position of the sugar moiety.
e.g. AMP (adenosine monophosphate).
Hormones
Molecules that act as intercellular messengers, produced by endocrine glands and poured directly into the bloodstream.
e.g. Insulin, thyroxine, testosterone.
Essential amino acids
Amino acids that cannot be synthesised in the body and must be obtained through diet.
e.g. Valine, leucine, lysine.
Non-essential amino acids
Amino acids that can be synthesised in the body.
e.g. Glycine, alanine, glutamic acid.
Formulas & Equations (8)
Cx(H2O)y
Reads as: C x times H2O y times
Variables: x and y are integers; this is the general formula for many carbohydrates.
Use when: Identifying compounds that may be carbohydrates based on empirical formula.
⚠ Assuming all compounds with this formula are carbohydrates (e.g., acetic acid fits but is not a carbohydrate).
C6H12O6
Reads as: C6 H12 O6
Variables: Molecular formula of glucose and fructose.
Use when: Writing molecular formulas for monosaccharides like glucose and fructose.
⚠ Forgetting that fructose has the same molecular formula as glucose but is a ketose.
C12H22O11 + H2O → C6H12O6 + C6H12O6
Reads as: C12 H22 O11 plus water yields C6 H12 O6 plus C6 H12 O6
Variables: Sucrose (C12H22O11) hydrolyses to glucose and fructose.
Use when: Writing hydrolysis reaction of sucrose.
⚠ Writing the products as two molecules of glucose instead of one glucose and one fructose.
(C6H10O5)n + nH2O → nC6H12O6
Reads as: C6 H10 O5 subscript n plus n water yields n C6 H12 O6
Variables: Starch or cellulose (C6H10O5)n hydrolyses to n molecules of glucose.
Use when: Writing hydrolysis of starch or cellulose to glucose.
⚠ Forgetting the condition: dilute H2SO4, 393 K, 2-3 atm for starch.
R–CH(NH2)–COOH
Reads as: R C H NH2 C O O H
Variables: R is the side chain; general structure of an α-amino acid.
Use when: Drawing general structure of α-amino acids.
⚠ Writing the amino group on the wrong carbon (not α-carbon).
–CO–NH–
Reads as: C O N H
Variables: Peptide bond (amide linkage).
Use when: Identifying peptide linkage in protein structure.
⚠ Writing it as –CONH– without the dash indicating bond.
C5H10O4 (2-deoxyribose)
Reads as: C5 H10 O4
Variables: Pentose sugar in DNA.
Use when: Writing the sugar component of DNA nucleotides.
⚠ Confusing it with ribose (C5H10O5) which has one more oxygen.
C5H10O5 (ribose)
Reads as: C5 H10 O5
Variables: Pentose sugar in RNA.
Use when: Writing the sugar component of RNA nucleotides.
⚠ Forgetting that ribose has an –OH group at C2, unlike deoxyribose.
How It Works
Hydrolysis of sucrose
  1. Sucrose is boiled with dilute HCl or H2SO4 in alcoholic solution.
  2. Glycosidic linkage between C1 of α-D-glucose and C2 of β-D-fructose is broken.
  3. Equimolar mixture of D-(+)-glucose and D-(-)-fructose is obtained.
  4. The product mixture is laevorotatory because laevorotation of fructose (-92.4°) exceeds dextrorotation of glucose (+52.5°).
💡 Sucrose is dextrorotatory but the hydrolysed mixture (invert sugar) is laevorotatory.
Formation of peptide bond
  1. Amino group of one amino acid reacts with carboxyl group of another amino acid.
  2. A water molecule is eliminated.
  3. An amide bond (–CO–NH–) called peptide bond is formed.
  4. The product is a dipeptide (e.g., glycylalanine).
💡 Peptide bond formation is a condensation reaction (dehydration synthesis).
Denaturation of proteins
  1. Physical change (e.g., temperature) or chemical change (e.g., pH) disturbs hydrogen bonds.
  2. Globules unfold and helix uncoils.
  3. Secondary and tertiary structures are destroyed.
  4. Primary structure (sequence of amino acids) remains intact.
  5. Protein loses its biological activity.
💡 Denaturation does not break peptide bonds; it only disrupts higher-order structures.
Formation of cyclic structure of glucose
  1. The –OH group at C5 adds to the –CHO group (carbonyl carbon).
  2. A six-membered cyclic hemiacetal ring (pyranose) is formed.
  3. Two anomers (α and β) are produced, differing in configuration at C1 (anomeric carbon).
  4. The open chain and cyclic forms exist in equilibrium.
💡 The cyclic structure explains why glucose does not give Schiff's test or form NaHSO3 addition product.
DNA double helix structure (Watson-Crick model)
  1. Two nucleic acid chains are wound about each other.
  2. The two strands are held together by hydrogen bonds between specific base pairs.
  3. Adenine (A) pairs with thymine (T) via two hydrogen bonds.
  4. Cytosine (C) pairs with guanine (G) via three hydrogen bonds.
  5. The strands are complementary to each other.
💡 Base pairing is specific: A-T and C-G. This complementarity is key to DNA replication.
Enzyme action (mechanism)
  1. Enzyme binds to specific substrate at active site.
  2. Enzyme reduces activation energy of the reaction.
  3. Reaction proceeds under mild conditions.
  4. Product is released; enzyme is unchanged and can be reused.
💡 Enzymes are highly specific and named after the substrate or reaction (e.g., maltase, oxidoreductase).
Exam Traps — Don't Make These Mistakes
⚠ Confusing D and L configuration with optical rotation
✗ D-glucose is dextrorotatory because it has D-configuration.
→ D and L refer to relative configuration compared to glyceraldehyde, not to the direction of optical rotation. D-glucose happens to be dextrorotatory, but D-fructose is laevorotatory.
✓ D and L indicate the configuration of the lowest asymmetric carbon relative to glyceraldehyde, while (+) and (-) indicate the direction of optical rotation. They are independent.
⚠ Calling all compounds with formula Cx(H2O)y as carbohydrates
✗ Acetic acid (CH3COOH) is a carbohydrate because its formula fits C2(H2O)2.
→ Carbohydrates are defined as optically active polyhydroxy aldehydes or ketones, not just by formula. Acetic acid lacks polyhydroxy and aldehyde/ketone groups.
✓ Carbohydrates are optically active polyhydroxy aldehydes or ketones. Acetic acid does not meet this definition.
⚠ Thinking all disaccharides are reducing sugars
✗ Sucrose is a reducing sugar because it is a disaccharide.
→ Sucrose has its aldehyde and ketone groups involved in the glycosidic linkage, so no free reducing group is available. It is a non-reducing sugar.
✓ Sucrose is a non-reducing sugar because the reducing groups of glucose and fructose are bonded in the glycosidic linkage.
⚠ Confusing nucleoside and nucleotide
✗ A nucleotide consists of a base and a sugar.
→ A base + sugar is a nucleoside. A nucleotide includes a base, sugar, and phosphate group.
✓ A nucleoside is base + sugar; a nucleotide is nucleoside + phosphate.
⚠ Assuming all amino acids are optically active
✗ All naturally occurring α-amino acids are optically active.
→ Glycine has no asymmetric carbon (R = H), so it is optically inactive.
✓ Except glycine, all naturally occurring α-amino acids are optically active because the α-carbon is asymmetric.
Compare & Contrast
Monomer unit and glycosidic linkage
Starch
Cellulose
Starch is a polymer of α-D-glucose with C1–C4 and C1–C6 glycosidic linkages; cellulose is a polymer of β-D-glucose with only C1–C4 glycosidic linkages.
Sugar and bases
DNA
RNA
DNA contains 2-deoxyribose sugar and thymine base; RNA contains ribose sugar and uracil base instead of thymine.
Type of secondary protein structure
α-helix
β-pleated sheet
α-helix is a right-handed screw with hydrogen bonds within the same chain; β-pleated sheet has stretched chains held together by intermolecular hydrogen bonds between adjacent chains.
Solubility and storage
Fat-soluble vitamins
Water-soluble vitamins
Fat-soluble vitamins (A, D, E, K) are stored in liver and adipose tissues; water-soluble vitamins (B group, C) are excreted in urine and cannot be stored (except B12).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 11-12 marks
Most asked: Structure of glucose (open chain and cyclic) · Classification of carbohydrates (mono, oligo, poly) · Peptide bond and protein structure (primary, secondary, tertiary, quaternary) · DNA vs RNA differences · Vitamins and their deficiency diseases · Essential and non-essential amino acids
Don't skip: Explain the cyclic structure of glucose and why it explains the absence of aldehyde group reactions.; Differentiate between DNA and RNA with at least four points.; Describe the primary, secondary, tertiary, and quaternary structure of proteins.; Write the hydrolysis products of sucrose and lactose, and explain why sucrose is a non-reducing sugar.
One-Line Revision (read aloud before the exam)
  1. Carbohydrates are optically active polyhydroxy aldehydes or ketones.
  2. Monosaccharides cannot be hydrolysed further; glucose and fructose are examples.
  3. Sucrose is a non-reducing disaccharide because its aldehyde and ketone groups are bonded.
  4. Starch is a polymer of α-D-glucose; cellulose is a polymer of β-D-glucose.
  5. Proteins are polymers of α-amino acids linked by peptide bonds.
  6. Essential amino acids cannot be synthesised in the body and must be obtained from diet.
  7. Denaturation of proteins destroys secondary and tertiary structures but not primary structure.
  8. Enzymes are biocatalysts that reduce activation energy and are highly specific.
  9. Vitamins are classified as fat-soluble (A, D, E, K) and water-soluble (B group, C).
  10. DNA contains deoxyribose sugar and thymine; RNA contains ribose sugar and uracil.
  11. A nucleoside is base + sugar; a nucleotide is nucleoside + phosphate.
  12. In DNA, adenine pairs with thymine and cytosine pairs with guanine via hydrogen bonds.
  13. Hormones are intercellular messengers produced by endocrine glands.
  14. Glucose exists in two anomeric forms (α and β) due to cyclisation forming a hemiacetal.
  15. Hydrolysis of sucrose yields glucose and fructose, and the mixture is called invert sugar.

MCQs — 10 Questions with Answers

Q1. Which of the following compounds fits the general formula Cx(H2O)y but is NOT a carbohydrate?

  • A. Glucose (C₆H₁₂O₆)
  • B. Acetic acid (CH₃COOH) ✓
  • C. Sucrose (C₁₂H₂₂O₁₁)
  • D. Fructose (C₆H₁₂O₆)

Answer: B — Acetic acid fits Cx(H2O)y as C₂(H₂O)₂ but lacks polyhydroxy aldehyde/ketone functional groups, so it is not classified as a carbohydrate.

Q2. Glucose reacts with hydroxylamine to form an oxime and with HCN to form a cyanohydrin. This indicates the presence of which functional group in glucose?

  • A. Carboxylic acid (—COOH)
  • B. Ester (—COO—)
  • C. Carbonyl (>C=O) ✓
  • D. Ether (—O—)

Answer: C — Both oxime formation and HCN addition are characteristic reactions of aldehydes and ketones (carbonyl groups), confirming the presence of >C=O in glucose.

Q3. When glucose is oxidised with bromine water, gluconic acid is formed. What does this observation prove about glucose structure?

  • A. Glucose contains a ketone group at C1
  • B. Glucose contains an aldehyde group at C1 ✓
  • C. Glucose contains a primary alcohol at C1
  • D. Glucose contains an ester group at C1

Answer: B — Mild oxidation with Br₂ water selectively oxidises aldehydes (—CHO) to carboxylic acids (—COOH); ketones are not oxidised by Br₂ water, so the formation of gluconic acid proves —CHO exists in glucose.

Q4. Glucose pentaacetate is a stable compound formed by acetylation of glucose. How many hydroxyl (—OH) groups must glucose contain?

  • A. Three
  • B. Four
  • C. Five ✓
  • D. Six

Answer: C — Pentaacetate means five acetyl groups are attached; since each —OH group is converted to —O—CO—CH₃, glucose must contain exactly five separate hydroxyl groups.

Q5. When glucose is oxidised with nitric acid, saccharic acid (a dicarboxylic acid) is formed. This proves that glucose has which two functional groups at opposite ends of the carbon chain?

  • A. Two aldehyde groups (—CHO)
  • B. One aldehyde (—CHO) and one primary alcohol (—CH₂OH) ✓
  • C. Two ketone groups (C=O)
  • D. One aldehyde (—CHO) and one secondary alcohol

Answer: B — Strong oxidation by HNO₃ converts both —CHO (at C1) and —CH₂OH (at C6) to —COOH, forming saccharic acid with two carboxylic acid groups; this proves both functional groups are present in glucose.

Q6. Which statement about D/L and (+)/(−) notation in carbohydrate nomenclature is correct?

  • A. D means dextrorotatory (+) and L means laevorotatory (−)
  • B. D and L denote stereochemical configuration; (+) and (−) denote optical rotation direction ✓
  • C. D and L are related to letter 'd' and 'l' used for optical activity
  • D. D(+)-glucose and L(−)-glucose have identical chemical structures

Answer: B — D/L are stereochemical descriptors based on the highest-numbered chiral carbon; (+) and (−) indicate optical rotation; they are independent properties of glucose.

Q7. Which of the following is NOT correct regarding reducing and non-reducing sugars?

  • A. All monosaccharides are reducing sugars
  • B. Maltose is a non-reducing sugar because it lacks a free carbonyl group ✓
  • C. Polysaccharides like starch and cellulose are non-reducing sugars
  • D. Reducing sugars can reduce Fehling's and Tollens' reagents

Answer: B — Maltose IS a reducing sugar because it has a free anomeric carbon with a —CHO or —C(OH)— group that can exist as an aldehyde; starch and cellulose are non-reducing because no free carbonyl is available.

Q8. When sucrose is hydrolysed with dilute HCl, the products obtained are glucose and fructose in a 1:1 molar ratio. This indicates that sucrose is composed of:

  • A. One glucose and one galactose unit
  • B. One glucose and one fructose unit ✓
  • C. Two glucose units
  • D. Two fructose units

Answer: B — The 1:1 molar ratio of glucose to fructose in the hydrolysate directly shows sucrose = glucose + fructose; the hydrolysis reaction is C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose).

Q9. Which of the following reactions would NOT occur with glucose?

  • A. Oxidation with Br₂ water to form gluconic acid
  • B. Oxidation with HNO₃ to form saccharic acid
  • C. Acetylation with acetic anhydride to form pentaacetate
  • D. Reduction with Tollens' reagent to form alcohol ✓

Answer: D — Tollens' reagent is an oxidising agent that converts aldehydes to carboxylic acids (and gets reduced to Ag mirror); glucose is oxidised, not reduced, by Tollens' reagent.

Q10. In the Fischer projection of D(+)-glucose, if the aldehyde group (—CHO) is at C1 and the primary alcohol (—CH₂OH) is at C6, how are the carbon atoms arranged in the straight-chain structure?

  • A. C1(—CHO) — C2 — C3 — C4 — C5 — C6(—CH₂OH) with four chiral carbons
  • B. C1(—CHO) — C2 — C3 — C4 — C5 — C6(—CH₂OH) with five chiral carbons ✓
  • C. C1(—CHO) — C2 — C3 — C4 — C5 — C6(—CH₂OH) with three chiral carbons
  • D. Branched chain with —CHO at the central carbon

Answer: B — The straight chain has 6 carbons total; C1 (aldehyde carbon) and C6 (primary alcohol carbon) are not chiral; C2, C3, C4, C5 are chiral centres, giving five chiral carbons total.

Flashcards

What is the general molecular formula for carbohydrates, and why is it misleading?

Cx(H2O)y; misleading because non-carbohydrates like acetic acid fit the formula, and some true carbohydrates like rhamnose do not.

Define monosaccharide and give two examples.

A carbohydrate that cannot be hydrolysed further to give simpler polyhydroxy aldehyde or ketone units; examples: glucose and fructose.

What is the difference between aldose and ketose monosaccharides?

Aldose contains an aldehyde group (—CHO) while ketose contains a ketone group (C=O); glucose is aldohexose, fructose is ketohexose.

How are reducing sugars identified in the lab?

Reducing sugars (all monosaccharides and some disaccharides like maltose) reduce Fehling's solution and Tollens' reagent to give a precipitate or mirror.

Write the hydrolysis reaction of sucrose and name the products.

C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose); sucrose gives equal moles of glucose and fructose.

What evidence proves glucose has a straight-chain structure with 6 carbon atoms?

Prolonged heating with HI converts glucose to n-hexane, proving all six carbons are linked in a single continuous chain.

Why does glucose form a pentaacetate product, and what does this tell us?

Five —OH groups are present in glucose attached to different carbon atoms; the stable pentaacetate confirms five acetylable hydroxyl groups.

Distinguish between the Fischer projection of glucose and gluconic acid.

Glucose has —CHO at C1 and —CH₂OH at C6; gluconic acid has —COOH at C1 and —CH₂OH at C6 (C1 oxidised to carboxylic acid).

What do the prefixes D and L mean in carbohydrate nomenclature?

D and L denote stereochemical configuration at the highest-numbered chiral carbon, not optical rotation; they are unrelated to (+) or (−) activity.

How is glucose prepared commercially from starch?

Starch is hydrolysed by boiling with dilute H₂SO₄ at 393 K under 2–3 atm pressure to yield glucose as the major product.

Important Board Questions

Define monosaccharides and classify them based on the number of carbon atoms and the type of carbonyl group present. Give one example of an aldopentose. [2 marks]

State: monosaccharides cannot hydrolyse further; classify as triose, tetrose, pentose, hexose, etc., and as aldose (—CHO) or ketose (C=O); example: ribose or arabinose (aldopentose with 5 carbons and —CHO).

Glucose is oxidised with bromine water to give gluconic acid. Further oxidation of gluconic acid with nitric acid gives saccharic acid (a dicarboxylic acid). Explain what these oxidation reactions reveal about the structure of glucose, and write the Fischer projection of glucose. [5 marks]

Br₂ oxidises —CHO to —COOH (proving aldehyde at C1); HNO₃ oxidises both —CHO and —CH₂OH to —COOH (proving primary —OH at C6); Fischer projection shows C1(—CHO) at top, —CH₂OH at C6 at bottom, with chiral carbons C2–C5 in between with —OH groups; show all three structures (glucose, gluconic acid, saccharic acid).

Glucose pentaacetate is formed when glucose is treated with acetic anhydride. The molecular formula of glucose pentaacetate is C₁₆H₂₂O₁₁. Show that glucose must contain exactly five hydroxyl groups, and explain how this evidence combined with oxidation reactions proves the complete structural formula of glucose as C₆H₁₂O₆ with the structure given in the Fischer projection. [6 marks]

Glucose C₆H₁₂O₆ + 5 (CH₃CO)₂O → pentaacetate + 5 CH₃COOH; calculate: glucose (180 g/mol) + 5 acetyl groups (5 × 42 = 210) − 5 water molecules (5 × 18 = 90) = 300 g/mol, matching C₁₆H₂₂O₁₁; five separate —OH groups on different carbons; combined with Br₂ water (—CHO at C1), HNO₃ (—CH₂OH at C6), HI degradation (straight chain), oxime + HCN (carbonyl present), derive that glucose = open-chain aldohexose with structure CHO-CHOH-CHOH-CHOH-CHOH-CH₂OH; show Fischer projection with correct stereochemistry.

Assertion-Reason Questions for Biomolecules

Q1. Assertion (A):

Acetic acid (CH3COOH) is not classified as a carbohydrate, even though its molecular formula fits the general formula Cx(H2O)y.

Reason (R):

The chemical definition of carbohydrates specifically requires optically active polyhydroxy aldehydes or ketones, which acetic acid does not possess.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. Acetic acid, despite fitting C2(H2O)2, lacks the multiple hydroxyl groups and does not function as a polyhydroxy aldehyde or ketone. The reason correctly explains why the general formula alone is insufficient to define carbohydrates.

Q2. Assertion (A):

All monosaccharides are reducing sugars that give positive tests with Tollens' reagent.

Reason (R):

Monosaccharides contain free aldehyde or ketone groups that can be oxidized by the mild oxidizing agent in Tollens' solution.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are correct, and the reason fully explains the assertion. The presence of free or readily available carbonyl groups in monosaccharides is the chemical basis for their reducing sugar property.

Q3. Assertion (A):

Glucose pentaacetate is a stable compound, confirming the presence of five distinct hydroxyl groups in glucose.

Reason (R):

Multiple hydroxyl groups attached to the same carbon atom would prevent stable acetylation due to steric hindrance.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both statements are true, but the reason does not directly explain the assertion. While steric factors may play a role, the key point is that five separate acetylation sites indicate five distinct –OH groups on different carbons. The reason provides a supporting argument but is not the complete explanation.

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Case-Based Questions (CBQ) for Biomolecules

Case Study:

A food scientist in India analyzed a sample of honey collected from a local apiary. The sample was found to contain three sugars: glucose, fructose, and maltose. When 10 g of the honey sample was dissolved in water and tested with Tollens' reagent, a bright silver mirror formed immediately. However, when sucrose (table sugar) was added to the same honey solution and tested again, the mixture did not form a silver mirror even after heating, until dilute HCl was added to the mixture first. Based on these observations, the scientist concluded that the honey contains reducing sugars but the added sucrose does not participate in the initial reaction.

Q1. Why did the original honey sample show a positive Tollens' test while sucrose did not?

  • A. Glucose and fructose are ketoses; maltose is a disaccharide with free carbonyl groups
  • B. Glucose and fructose are monosaccharides with free aldehyde or ketone groups; maltose has a free anomeric carbon; sucrose lacks a free carbonyl group ✓
  • C. Honey contains an enzyme that activates sugars; sucrose is too large to react
  • D. All sugars in honey are reducing, but sucrose is a non-reducing sugar because it is plant-derived

Answer: B

Q2. What is the chemical explanation for the formation of a silver mirror only after HCl treatment of the honey-sucrose mixture?

  • A. HCl hydrolyzes the glycosidic bond in sucrose, releasing glucose and fructose with free aldehyde groups ✓
  • B. HCl converts the primary alcohols in sucrose into aldehyde groups
  • C. HCl denatures the enzyme that was inhibiting the Tollens' reaction
  • D. HCl oxidizes the –OH groups in sucrose to carbonyl groups

Answer: A

Q3. If the honey sample contains 2% glucose, 3% fructose, and 4% maltose by mass, approximately how many monosaccharide units are present in 100 g of honey?

  • A. Approximately 0.56 moles of monosaccharide units
  • B. Approximately 0.67 moles of monosaccharide units ✓
  • C. Approximately 0.50 moles of monosaccharide units
  • D. Approximately 0.45 moles of monosaccharide units

Answer: B

Q4. Why is maltose classified as a reducing sugar while sucrose is not?

  • A. Maltose is composed of two glucose units; sucrose contains both glucose and fructose
  • B. Maltose is sweeter than sucrose
  • C. Maltose has a free anomeric carbon from at least one monosaccharide unit; sucrose's glycosidic bond links both anomeric carbons ✓
  • D. Maltose is always produced from starch hydrolysis; sucrose is extracted from plants

Answer: C

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Previous Year Questions (PYQ) for Biomolecules

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 56(B) · 1 mark

Vitamin B12 is a complex compound of :

  • Fe2+
  • Mg2+
  • Co3+
  • Cu2+

CBSE 2026 · Set 56(B) · 1 mark

In the ring structure of fructose, the anomeric carbon is :

  • C2
  • C3
  • C4
  • C1

CBSE 2026 · Set 56(B) · 2 marks

What are essential amino acids ? How do you explain the amphoteric behaviour of amino acids ?

CBSE 2026 · Set 56(B) · 4 marks

Carbohydrates are optically active polyhydroxy aldehydes or ketones or molecules that provide such units on hydrolysis. They are broadly classified into three groups : monosaccharides, oligosaccharides and polysaccharides. Monosaccharide units are held together by glycosidic linkages to form disaccharides or polysaccharides. Starch and cellulose are polysaccharides. Starch is a polymer of α-glucose and consists of two components : Amylose and Amylopectin. Amylose is water soluble, having 15 – 20% of starch, whereas Amylopectin is insoluble in water, having 80 – 85% starch. Chemically, amylose is a long unbranched chain of α-glucose units held together by C1 − C4 glycosidic linkage. Amylopectin is a branched chain of α-glucose units held together by C1 − C4 and C1 − C6 glycosidic linkages.

Write two differences between Amylose and Amylopectin.

CBSE 2026 · Set 56/1/1 · 1 mark

The deficiency of which of the following vitamins causes increased fragility of red blood cells and muscular weakness:

  • Vitamin A
  • Vitamin K
  • Vitamin E
  • Vitamin D

CBSE 2026 · Set 56/1/1 · 1 mark

Assertion (A): The pentaacetate of glucose does not react with H2N – OH. Reason (R): It indicates the presence of free – CHO group in glucose.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/1 · 2 marks

Differentiate between the following: (i) Peptide linkage and Glycosidic linkage (ii) Essential amino acids and Non-essential amino acids

CBSE 2026 · Set 56/1/1 · 3 marks

Write the reactions of D-Glucose with the following: (a) HI (b) Br2 water (c) Conc. HNO3

CBSE 2026 · Set 56/1/2 · 1 mark

Which of the following is 'not' true about enantiomers?

  • They have the same chemical reactivity.
  • They have the same specific rotation.
  • They have the same melting or boiling point.
  • They have the same refractive index.

CBSE 2026 · Set 56/1/2 · 1 mark

An example of non-reducing sugar is

  • Glucose
  • Sucrose
  • Lactose
  • Maltose

CBSE 2026 · Set 56/1/2 · 1 mark

Assertion (A): The two strands of DNA are complementary to each other. Reason (R): The hydrogen bonds are formed between specific pairs of bases.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/2 · 2 marks

Differentiate between the following: (i) Acidic amino acids and basic amino acids (ii) Nucleotide and Nucleoside

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Class 12 Chemistry Formula Sheet

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Class 12 Chemistry Previous Year Questions, Chapter-wise

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