Biomolecules — Class 12 Chemistry NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Biomolecules" — 5 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Biomolecules" — 5 important questions with detailed answers for CBSE board exam prep…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Sep 17, 2026

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NCERT Solutions for Class 12 Chemistry — Biomolecules. Step-by-step answers to all 5 textbook questions from this chapter, written for CBSE board preparation and free to use.

Biomolecules — All 5 Questions Solved

Q1. What are carbohydrates? Classify them and give examples of monosaccharides, disaccharides, and polysaccharides.

Carbohydrates: Organic compounds composed of carbon, hydrogen, and oxygen with the general formula Cn(H2O)m. They are polyhydroxy aldehydes or ketones and their derivatives.

General characteristics:
- Soluble in water (multiple OH groups)
- Taste sweet (especially sugars)
- Provide energy (4 kcal/gram)
- Contain C=O and multiple C-OH groups

Classification:

1. MONOSACCHARIDES (simple sugars)
Definition: Carbohydrates that cannot be hydrolyzed into smaller sugars
General formula: CnH2nOn (where n = 3 to 8 typically)

Types based on functional group:
a) Aldoses (contain aldehyde group -CHO)
- Aldopentose: Ribose (C5H10O5)
- Aldohexose: Glucose (C6H12O6)

b) Ketoses (contain ketone group C=O)
- Ketohexose: Fructose (C6H12O6)

Examples of monosaccharides:
- Trioses: Glyceraldehyde (C3H6O3)
- Pentoses: Glucose (C6H12O6), Ribose (C5H10O5), Deoxyribose (C5H10O4)
- Hexoses: Fructose (C6H12O6), Galactose (C6H12O6)

2. DISACCHARIDES
Definition: Carbohydrates composed of two monosaccharide units joined by a glycosidic bond
General formula: C12H22O11
Formation: Monosaccharide1 + Monosaccharide2 → Disaccharide + H2O

Examples:
a) Sucrose (table sugar):
Glucose + Fructose → Sucrose + H2O
Used as food sweetener, non-reducing sugar

b) Maltose (malt sugar):
Glucose + Glucose → Maltose + H2O
Formed from starch hydrolysis, reducing sugar

c) Lactose (milk sugar):
Glucose + Galactose → Lactose + H2O
Found in milk, reducing sugar

3. POLYSACCHARIDES
Definition: Carbohydrates composed of many monosaccharide units joined by glycosidic bonds
General formula: (C6H10O5)n where n is large (100-1000s)
Nature: Non-reducing, insoluble or partially soluble

Examples:
a) Starch (α-glucose polymer):
- Plant energy storage
- Two components: amylose (linear) and amylopectin (branched)
- (C6H10O5)n where n ≈ 250-4000

b) Glycogen (α-glucose polymer):
- Animal energy storage (liver and muscles)
- Highly branched structure
- (C6H10O5)n where n ≈ 1000-10000

c) Cellulose (β-glucose polymer):
- Plant structural material
- Linear chain, insoluble
- (C6H10O5)n where n ≈ 1000-10000
- Not digestible by humans (β-linkage)

Structural differences:
- Starch: α-(1→4)-glycosidic bonds with α-(1→6) branches in amylopectin
- Glycogen: α-(1→4)-glycosidic bonds with α-(1→6) branches (more branched than starch)
- Cellulose: β-(1→4)-glycosidic bonds, linear, forms hydrogen bonds

Other carbohydrate derivatives:
- Sugar alcohols: Glycerol, Sorbitol
- Sugar acids: Gluconic acid, Ascorbic acid (Vitamin C)
- Amino sugars: Glucosamine, N-acetylglucosamine (in chitin)

Q2. Explain the structure of glucose. Draw the Fischer projection and the Haworth structure. Discuss mutarotation.

Glucose (also called dextrose or grape sugar):
Molecular formula: C6H12O6
Type: Aldohexose (6-carbon aldose sugar)

Structural representations:

1. FISCHER PROJECTION (open-chain form):
Shows the 2D representation of the 3D structure

This is D-glucose (the naturally occurring form)
- Aldehyde group at C1
- Five chiral centers (C2, C3, C4, C5)
- Five hydroxyl groups

2. HAWORTH STRUCTURE (cyclic form):
Glucose exists primarily in cyclic form in aqueous solution

Formation: The aldehyde group (at C1) reacts with the hydroxyl group at C5 to form a hemiacetal
This forms a 6-membered ring (pyranose ring)

Haworth structure of α-D-glucose (most common form in solid state):

More clearly, the Haworth structure shows:
- A 6-membered ring (pyranose)
- Four CH and one O in the ring
- In α-glucose: The OH at C1 is below the ring (axial position)
- In β-glucose: The OH at C1 is above the ring (equatorial position)

3. CHAIR CONFORMATION:
More accurate representation (3D structure):

α-D-Glucose in chair form:
The pyranose ring adopts a chair conformation
- C1 (anomeric carbon) has -OH group
- C2, C3, C4, C5 are the ring carbons
- C6 is attached as -CH2OH
- All hydroxyl groups are in equatorial positions (except anomeric in α-form)

3. MUTAROTATION:

Definition: The gradual change in optical rotation when a solution of one anomeric form of glucose is allowed to stand, until it reaches a constant value.

Explanation:
When α-D-glucose is dissolved in water:
- Initial optical rotation: [α]D = +112.2°
- Over time, the solution changes as α-glucose converts to β-glucose
- Final equilibrium rotation: [α]D = +52.7°
- This constant value (52.7°) is the average of α and β forms

Mechanism of mutarotation:
1. The ring-chain equilibrium:
α-D-glucose (cyclic) ⇌ open-chain aldehyde form ⇌ β-D-glucose (cyclic)

2. In aqueous solution, a small amount of the ring opens
3. The free aldehyde can close to form either α or β anomeric form
4. The β form accumulates as the equilibrium shifts
5. At equilibrium: α:β ratio ≈ 36:64

Structural difference between α and β glucose:
Only the configuration at C1 (anomeric carbon) differs:
- α-D-glucose: C1-OH is axial (pointing down in Haworth)
- β-D-glucose: C1-OH is equatorial (pointing up in Haworth)

At equilibrium at 25°C:
- α-glucose: 36%
- β-glucose: 64%
- Open-chain form: <0.1%

Optical rotation values:
- α-D-glucose: [α]D = +112.2°
- β-D-glucose: [α]D = +18.7°
- Equilibrium mixture: [α]D = +52.7°

Importance:
- Explains why glucose crystallizes in different forms
- Important in biochemistry (enzymes recognize specific anomers)
- Relevant in food industry (different sweetness levels)
- Shows the dynamic nature of ring-chain equilibrium

Other points:
- Mutarotation is catalyzed by acids or bases
- Mutarotation stops when equilibrium is reached
- Each anomer, when pure, shows mutarotation
- The open-chain aldehyde form is present at very low concentration but is key to reactions

Q3. Explain amino acids. What is a peptide bond? Write the general structure of an amino acid and describe how proteins are formed from amino acids.

Amino acids: Organic compounds that contain both an amino group (-NH2) and a carboxyl group (-COOH) bonded to the same carbon atom (α-carbon).

General structure of amino acid:

Carboxyl group
|
v
COOH
|
H2N-C-H (α-carbon)
|
R (side chain)
^
|
Amino group

Structure details:
- Central carbon (α-carbon): sp3 hybridized
- Four groups bonded to α-carbon:
1. Carboxyl group (-COOH)
2. Amino group (-NH2)
3. Hydrogen atom (-H)
4. Side chain (R group, varies with amino acid)

Generalized formula: NH2-CHR-COOH

Zwitterionic form (in aqueous solution at physiological pH):
In water, the amino acid exists as a zwitterion:

COO- (carboxyl ionized)
|
H3N+-C-H (amino group protonated)
|
R

The -COOH loses H+, and -NH2 gains H+ to form -NH3+

Side chains (R groups):
There are 20 common amino acids with different R groups:

1. Nonpolar R groups:
- Alanine: R = -CH3
- Valine: R = -CH(CH3)2
- Leucine: R = -CH2CH(CH3)2
- Glycine: R = -H (simplest, no chiral center)

2. Polar R groups:
- Serine: R = -CH2OH
- Threonine: R = -CH(OH)CH3
- Cysteine: R = -CH2SH
- Tyrosine: R = -CH2-C6H4-OH

3. Charged R groups:
- Aspartic acid: R = -CH2COOH (acidic)
- Glutamic acid: R = -CH2CH2COOH (acidic)
- Lysine: R = -(CH2)4NH2 (basic)
- Arginine: R = -(CH2)3NH-C(=NH)NH2 (basic)

PEPTIDE BOND:

Definition: A covalent bond formed between the carboxyl group of one amino acid and the amino group of another amino acid, with the elimination of water.

Formation equation:
Amino acid 1 + Amino acid 2 → Dipeptide + H2O

Specific example:
Glycine + Alanine → Glycylalanine + H2O

NH2-CH2-COOH + NH2-CH(CH3)-COOH → NH2-CH2-CO-NH-CH(CH3)-COOH + H2O

Structure of peptide bond:
The peptide bond is a C-N bond (covalent) formed between:
- The carboxyl carbon (C=O) of one amino acid
- The amino nitrogen (-NH2) of the next amino acid
- Loss of one water molecule

Peptide bond structure:
O
||
...C-N-H...
|

The C-N bond has partial double bond character (resonance):
- C=N partial double bond (C-N bond is shorter than typical single bond)
- Restricted rotation around the C-N bond
- The six atoms in the peptide bond are approximately planar

Sequence formation:
Multiple amino acids join via peptide bonds to form a polypeptide chain:

General dipeptide structure:
H O H O
| || | ||
NH3+-C-C-N-C-COOH
| H | H
R1 R2

The chain has:
- N-terminus: free amino group (-NH2 or -NH3+ in physiological conditions)
- C-terminus: free carboxyl group (-COOH or -COO- in physiological conditions)
- Internal residues: amino acids linked by peptide bonds

PROTEIN FORMATION:

Dipeptide: 2 amino acids linked
Tripeptide: 3 amino acids linked
Oligopeptide: 2-10 amino acids
Polypeptide: 10-100 amino acids
Protein: >100 amino acids

Example reaction for forming a tripeptide (Gly-Ala-Ser):
Glycine + Alanine → Glycylalanine + H2O
Glycylalanine + Serine → Glycylalanylserine + H2O

Levels of protein structure:

1. Primary structure: Sequence of amino acids in the polypeptide chain
- Determined by the order of amino acids
- Unique for each protein

2. Secondary structure: Regular folding patterns
- α-helix: spiral structure stabilized by hydrogen bonds between C=O and N-H
- β-pleated sheet: extended structure with hydrogen bonds between chains
- Random coil: no regular structure

3. Tertiary structure: 3D shape of the entire polypeptide
- Stabilized by:
* Hydrogen bonds
* Disulfide bonds (between cysteine residues: Cys-S-S-Cys)
* Hydrophobic interactions (nonpolar R groups)
* Ionic interactions (charged R groups)

4. Quaternary structure: Multiple polypeptide chains in a protein
- Example: Hemoglobin has 4 subunits

Protein functions:
- Enzymes: Catalyze biochemical reactions
- Antibodies: Immune defense
- Hormones: Regulation and signaling
- Structural proteins: Support and strength
- Transport: Carry molecules (e.g., hemoglobin carries O2)

Properties of peptide bond:
- Planar geometry
- Partial double bond character
- Cannot rotate freely
- Shorter than typical C-N single bond (~1.32 Å vs ~1.47 Å)
- Stable under physiological conditions
- Can be hydrolyzed by acids, bases, or enzymes (proteases)

Q4. What are nucleic acids? Describe the structure of DNA, including the role of nucleotides and the double helix structure.

Nucleic acids: Polymers of nucleotides that store and transmit genetic information. Two main types: DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).

NUCLEOTIDES (building blocks):

A nucleotide consists of three components:
1. A five-carbon sugar (pentose)
- DNA: Deoxyribose (C5H10O4) - lacks one oxygen at C2
- RNA: Ribose (C5H10O5) - has OH at C2

2. A nitrogenous base (purine or pyrimidine)
- DNA bases: Adenine (A), Guanine (G), Cytosine (C), Thymine (T)
- RNA bases: Adenine (A), Guanine (G), Cytosine (C), Uracil (U)

3. A phosphate group (PO4^3-)

Structure of a nucleotide:

Nucleotide formation:
The three components are bonded by:
- Glycosidic bond: between the sugar (C1') and the base
- Phosphodiester bond: between the phosphate and the sugar

DNA STRUCTURE:

1. PRIMARY STRUCTURE (nucleotide sequence):
DNA is a polymer of nucleotides linked by phosphodiester bonds

Phosphodiester bond formation:
The phosphate group of one nucleotide bonds to the 3'-OH of the next nucleotide

The chain has directionality:
- 5' end: free phosphate group on C5' of first sugar
- 3' end: free hydroxyl group on C3' of last sugar
- Chain direction written as 5' → 3'

2. SECONDARY STRUCTURE (double helix):

Key features of DNA double helix (Watson-Crick model):
- Two polypeptide chains wound around each other
- Right-handed spiral (helical)
- Antiparallel strands (5' → 3' runs opposite to 3' → 5')
- Major and minor grooves formed between strands

Base pairing:
Complementary bases form hydrogen bonds between strands:
- Adenine (A) pairs with Thymine (T): 2 hydrogen bonds
A==T

- Guanine (G) pairs with Cytosine (C): 3 hydrogen bonds
G≡C

This is Chargaff's rule:
- Amount of A = Amount of T
- Amount of G = Amount of C
- A + G = T + C

Double helix parameters:
- Diameter: ~2 nm
- Rise per base pair: 0.34 nm
- Pitch (one complete turn): ~3.4 nm
- Base pairs per turn: 10
- Total base pairs in human genome: ~3 billion

Structure diagram (simplified):

Strand 1 (5' → 3'):
5'-A---T-3'
| |
G≡C
| |
T---A
| |
C≡G
| |
3'-G---C-5'
Strand 2 (3' → 5')

3. TERTIARY STRUCTURE:
Supercoiling: The double helix is further coiled into a superhelix
- Negative supercoiling in eukaryotes (underwound)
- Stabilized by histone proteins in eukaryotes
- Forms chromosomes

FUNCTIONS OF DNA:

1. Storage of genetic information: The sequence of bases encodes genetic information
2. Replication: Semi-conservative replication produces two identical DNA molecules
3. Transcription: DNA is transcribed into RNA
4. Gene expression: Genes are expressed as proteins

COMPARISON: DNA vs RNA

RNA types:
- mRNA (messenger RNA): Carries genetic code from DNA
- tRNA (transfer RNA): Brings amino acids during translation
- rRNA (ribosomal RNA): Component of ribosomes

DNA packaging in eukaryotes:
- DNA wraps around histone proteins forming nucleosomes
- Nucleosomes condense into chromatin
- Chromatin further condenses into chromosomes
- Human genome: 46 chromosomes, 3 billion base pairs

Key points:
- Nucleotides are the monomers
- Phosphodiester bonds link nucleotides
- Double helix provides stability and allows replication
- Complementary base pairing ensures accurate replication
- DNA stores the instructions for all living organisms

Q5. Explain the structure and properties of proteins. What are enzymes, and how do they catalyze biochemical reactions?

PROTEINS:

Definition: Large polypeptides composed of amino acids linked by peptide bonds, with specific 3D structures that determine their functions.

Composition:
- Primarily C, H, O, N
- Some contain S (from cysteine)
- Some contain metal ions (Fe, Zn, Mg, Cu)
- Molecular weight: 5,000 to >100,000 Da

Structural properties of proteins:

1. PRIMARY STRUCTURE:
Linear sequence of amino acids in the polypeptide chain
- Determined by the genetic code (DNA)
- Unique sequence for each protein
- Example: Insulin has 51 amino acids
- Direction: N-terminus (NH3+) to C-terminus (COO-)

2. SECONDARY STRUCTURE:
Regular folding patterns stabilized by hydrogen bonds between C=O and N-H of the backbone

α-helix:
- Right-handed spiral structure
- Hydrogen bond between C=O of residue n and N-H of residue (n+4)
- Rise per turn: 5.4 Å
- Residues per turn: 3.6
- Example: Keratin (hair protein)

β-pleated sheet:
- Extended structure with multiple strands
- Hydrogen bonds between adjacent strands (parallel or antiparallel)
- R groups alternate above and below the plane
- Example: Silk protein (fibroin)

Random coil:
- Irregular structure with no regular pattern
- Can exist in unstructured regions

Other patterns:
- β-turns: Change direction of polypeptide
- Loops: Connect secondary structures

3. TERTIARY STRUCTURE:
Overall 3D shape of the protein, stabilized by multiple interactions:

a) Hydrogen bonds:
- Between polar amino acid side chains
- Between backbone and side chains
- Most common stabilizing force

b) Disulfide bonds:
- Covalent bonds between two cysteine residues
- Cys-S-S-Cys (thioether bond)
- Provides stability and strength
- More common in extracellular proteins

c) Hydrophobic interactions:
- Nonpolar amino acids cluster in the interior
- Nonpolar R groups: Val, Leu, Ile, Phe, Pro
- Avoid contact with water
- Hydrophobic effect drives protein folding

d) Ionic interactions (salt bridges):
- Between charged amino acids
- Acidic (Asp, Glu) and basic (Lys, Arg) residues
- Example: -COO- with -NH3+

e) Van der Waals forces:
- Weak interactions between atoms
- Collectively important for stability

4. QUATERNARY STRUCTURE:
Association of multiple polypeptide subunits
- Example: Hemoglobin (4 subunits, each with heme group)
- Example: Collagen (3 polypeptide chains)
- Example: Chlorophyll protein complexes
- Stabilized by same forces as tertiary structure

FUNCTIONAL PROPERTIES OF PROTEINS:

1. Specificity: Each protein has a unique structure and function
2. Catalytic activity: Enzymes catalyze reactions
3. Transport: Carry molecules (hemoglobin carries O2)
4. Support: Structural proteins (collagen, keratin)
5. Movement: Muscle proteins (myosin, actin)
6. Immunity: Antibodies recognize pathogens
7. Regulation: Hormones regulate biological processes
8. Storage: Proteins store molecules (ferritin stores iron)

ENZYMES:

Definition: Biological catalysts (proteins) that increase the rate of biochemical reactions without being consumed in the reaction.

Characteristics of enzymes:
1. Protein nature: Most are proteins (some RNA enzymes exist)
2. Specificity: Each enzyme catalyzes a specific reaction or class of reactions
3. Efficiency: Can increase reaction rate by factors of 10^6 to 10^17
4. Reusability: Regenerated after reaction, unchanged
5. Regulation: Activity can be controlled by cells
6. Mild conditions: Work at body temperature, neutral pH

Enzyme structure:
- Active site: Specific region where substrate binds
- Catalytic residues: Amino acids that participate in catalysis
- Substrate binding site: Complementary to substrate shape
- Cofactors/coenzymes: Non-protein helpers (metal ions, vitamins)

Enzyme nomenclature:
Enzymes are named by adding -ase suffix to substrate name
- Amylase: Breaks down amylose (starch)
- Lactase: Breaks down lactose (milk sugar)
- Protease: Breaks down proteins
- DNA polymerase: Synthesizes DNA

CATALYTIC MECHANISM:

General reaction:
E + S → ES → EP → E + P
where E = enzyme, S = substrate, P = product, ES = enzyme-substrate complex

Lock-and-Key Model (outdated):
- Enzyme active site fits substrate like a lock and key
- Substrate fits perfectly into active site
- Overly simplistic (no induced fit)

Induced Fit Model (current):
- Enzyme active site changes shape upon substrate binding
- Enzyme molds around substrate
- Increases catalytic efficiency
- Explains activation and transition state stabilization

Mechanism steps:

1. Substrate binding:
E + S → ES (formation of enzyme-substrate complex)
- Weak interactions (H-bonds, hydrophobic, ionic)
- Specific orientation of substrate

2. Catalysis:
ES → EP
- Enzyme stabilizes transition state
- Lowers activation energy (Ea)
- Mechanism depends on enzyme type:
* Acid-base catalysis: H+ transfer
* Covalent catalysis: Temporary covalent bond formation
* Metal ion catalysis: Metal ion stabilizes transition state
* Coenzyme catalysis: Non-protein cofactor assists

3. Product release:
EP → E + P
- Product leaves active site
- Enzyme regenerated for next cycle

Factors affecting enzyme activity:

1. Temperature:
- Optimal temperature: ~37°C (body temperature)
- Below optimal: Activity increases with temperature
- Above optimal: Enzyme denatures, activity decreases

2. pH:
- Optimal pH: Varies by enzyme (pepsin 2.0, trypsin 8.0)
- Extreme pH: Denatures enzyme
- Charged amino acids affected by pH

3. Substrate concentration:
- Low [S]: Rate increases linearly with [S]
- High [S]: Rate plateaus (Vmax reached)
- Follows Michaelis-Menten kinetics

4. Enzyme concentration:
- Direct proportional relationship
- More enzyme = faster reaction (with excess substrate)

5. Enzyme inhibitors:
a) Competitive inhibitors: Compete with substrate for active site
- Reversible or irreversible
- Can be overcome by increasing substrate concentration

b) Noncompetitive inhibitors: Bind to site other than active site
- Cannot be overcome by increasing substrate concentration
- Reduce Vmax

6. Cofactors and coenzymes:
- Required for some enzymes to function
- Metal ions (Zn, Fe, Cu, Mg)
- Vitamins (B vitamins act as coenzymes)
- Organic molecules (NAD+, NADP+, ATP)

Enzyme kinetics (Michaelis-Menten equation):
V = (Vmax × [S]) / (Km + [S])

where:
- V = reaction velocity
- Vmax = maximum velocity
- [S] = substrate concentration
- Km = Michaelis constant (inversely related to affinity)

Biological examples:

1. Amylase: Breaks down starch
Starch → Glucose (+ maltose)
Location: Saliva, pancreas

2. Pepsin: Protease in stomach
Proteins → Peptides
Optimal pH: 2.0

3. Trypsin: Protease in small intestine
Proteins → Peptides
Optimal pH: 8.0

4. DNA polymerase: Synthesizes DNA
dNTP → DNA
Requires template, primer, cofactors

5. Catalase: Breaks down hydrogen peroxide
2 H2O2 → 2 H2O + O2
Protects cells from oxidative damage

Regulation of enzyme activity:
1. Allosteric regulation: Binding at site other than active site
2. Covalent modification: Phosphorylation (often reversible)
3. Feedback inhibition: Product inhibits enzyme
4. Compartmentalization: Spatial separation of enzymes and substrates
5. Enzyme synthesis: Control at gene expression level

Biomolecules — Practice MCQs with Answers

Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.

MCQ 1. Which of the following is a non-reducing sugar?

A. Glucose
B. Fructose
C. Maltose
D. Sucrose

Answer: D. Sucrose — Sucrose is a disaccharide formed by α-glucose and β-fructose, where the reducing groups of both monosaccharides are involved in the glycosidic bond, thus it does not have a free hemiacetal or hemiketal group.

MCQ 2. What are the monomer units of starch?

A. Glucose
B. Fructose
C. Galactose
D. Ribose

Answer: A. Glucose — Starch is a polysaccharide composed of many glucose units linked by glycosidic bonds. It is a major energy storage polysaccharide in plants.

MCQ 3. The two cyclic hemiacetal forms of glucose, α-D-glucose and β-D-glucose, are called:

A. Epimers
B. Anomers
C. Enantiomers
D. Isomers

Answer: B. Anomers — Anomers are stereoisomers that differ in configuration at the anomeric carbon (C1 for aldoses, C2 for ketoses) of a cyclic saccharide.

MCQ 4. Hydrolysis of sucrose yields:

A. Glucose and Galactose
B. Glucose and Fructose
C. Fructose and Galactose
D. Only Glucose

Answer: B. Glucose and Fructose — Sucrose is a disaccharide that on hydrolysis breaks down into one molecule of α-D-glucose and one molecule of β-D-fructose.

MCQ 5. The change in optical rotation of a freshly prepared solution of glucose over time is known as:

A. Isomerism
B. Racemization
C. Mutarotation
D. Polymerization

Answer: C. Mutarotation — Mutarotation is the change in the optical rotation of a solution containing an anomer in equilibrium with its other anomer, typically occurring when cyclic forms interconvert via an open-chain intermediate.

MCQ 6. Which functional group is present in fructose?

A. Aldehyde
B. Ketone
C. Carboxylic acid
D. Amine

Answer: B. Ketone — Fructose is a ketohexose, meaning it contains a ketone functional group. Glucose, on the other hand, is an aldohexose with an aldehyde group.

MCQ 7. The bond formed between two monosaccharide units to form a disaccharide or polysaccharide is called a:

A. Ester bond
B. Peptide bond
C. Glycosidic bond
D. Phosphodiester bond

Answer: C. Glycosidic bond — Monosaccharides are joined together by an oxygen bridge formed by the removal of a water molecule, creating a glycosidic linkage.

MCQ 8. Ribose is a:

A. Pentose sugar
B. Hexose sugar
C. Tetrose sugar
D. Heptose sugar

Answer: A. Pentose sugar — Ribose is a five-carbon sugar (pentose) and is a crucial component of RNA and ATP. Deoxyribose, found in DNA, is also a pentose sugar.

MCQ 9. In α-D-glucose, the -OH group at C1 is:

A. Above the plane of the ring
B. Below the plane of the ring
C. Inside the ring
D. Outside the ring

Answer: B. Below the plane of the ring — In α-D-glucose, the hydroxyl group at the anomeric carbon (C1) is on the opposite side to the CH2OH group at C5 when drawn in a Haworth projection.

MCQ 10. Which of the following is structurally a highly branched-chain polymer of α-D-glucose?

A. Amylose
B. Cellulose
C. Glycogen
D. Chitin

Answer: C. Glycogen — Glycogen is a highly branched polysaccharide, primarily composed of α-D-glucose units linked by α-1,4-glycosidic bonds with α-1,6-glycosidic bonds at the branch points. It is the main storage polysaccharide in animals.

MCQ 11. The bond that links amino acids together to form a polypeptide chain is called a:

A. Glycosidic bond
B. Ester bond
C. Peptide bond
D. Phosphodiester bond

Answer: C. Peptide bond — A peptide bond is an amide linkage formed between the carboxyl group of one amino acid and the amino group of another amino acid with the elimination of a water molecule.

MCQ 12. At its isoelectric point, an amino acid exists as a:

A. Cation
B. Anion
C. Zwitterion
D. Neutral molecule

Answer: C. Zwitterion — At the isoelectric point, an amino acid exists as a zwitterion, which is a dipolar ion carrying both positive and negative charges, making the net charge zero.

MCQ 13. The loss of biological activity of a protein due to changes in pH, temperature, or chemical agents is known as:

A. Polymerization
B. Esterification
C. Denaturation
D. Saponification

Answer: C. Denaturation — Denaturation is the process by which proteins lose their secondary, tertiary, and quaternary structures due to disruption of forces holding them, leading to loss of biological activity.

MCQ 14. The sequence of amino acids in a protein chain defines its:

A. Primary structure
B. Secondary structure
C. Tertiary structure
D. Quaternary structure

Answer: A. Primary structure — The primary structure of a protein refers to the unique linear sequence of amino acids linked by peptide bonds. This sequence dictates the higher-order structures.

MCQ 15. Which of the following is an essential amino acid for humans?

A. Glycine
B. Alanine
C. Valine
D. Aspartate

Answer: C. Valine — Essential amino acids cannot be synthesized by the human body and must be obtained from the diet. Valine is one of the essential amino acids, while Glycine, Alanine, and Aspartate are non-essential.

MCQ 16. What is the primary function of enzymes in biological systems?

A. Store genetic information
B. Provide structural support
C. Catalyze biochemical reactions
D. Transport oxygen

Answer: C. Catalyze biochemical reactions — Enzymes are biological catalysts that significantly increase the rate of biochemical reactions without being consumed in the process. They are highly specific in their action.

MCQ 17. Keratin, found in hair and nails, is an example of a:

A. Globular protein
B. Fibrous protein
C. Conjugated protein
D. Simple carbohydrate

Answer: B. Fibrous protein — Fibrous proteins, like keratin, are elongated, water-insoluble, and provide structural support, whereas globular proteins are spherical and water-soluble.

MCQ 18. Which type of linkage is primarily responsible for the stability of tertiary and quaternary structures of proteins, especially between cysteine residues?

A. Hydrogen bonds
B. Ionic bonds
C. Disulfide linkages
D. Van der Waals forces

Answer: C. Disulfide linkages — Disulfide linkages (-S-S-) are strong covalent bonds formed between the sulfhydryl groups of two cysteine residues, playing a crucial role in stabilizing higher-order protein structures.

MCQ 19. A nucleotide is composed of:

A. Sugar, Phosphate, and Nitrogenous base
B. Sugar and Nitrogenous base
C. Phosphate and Nitrogenous base
D. Sugar and Phosphate

Answer: A. Sugar, Phosphate, and Nitrogenous base — A nucleotide is the basic structural unit of nucleic acids, consisting of a pentose sugar (ribose or deoxyribose), a phosphate group, and a nitrogenous base.

MCQ 20. Which of the following is a pyrimidine base?

A. Adenine
B. Guanine
C. Cytosine
D. None of these

Answer: C. Cytosine — Pyrimidines are single-ring nitrogenous bases, which include Cytosine, Thymine (in DNA), and Uracil (in RNA). Adenine and Guanine are purines (double-ring bases).

How to Use These Solutions

Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.

Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.

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