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Biomolecules — NCERT Solutions

CBSE · Class 11 · Biotechnology

NCERT Solutions for Biomolecules, CBSE Class 11 Biotechnology: 30 textbook questions solved step by step. Covers Exercises.

63 questions116 flashcards4 formulas & key relations5 concepts

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Exercises

1Describe the classification of carbohydrates.Show solution

Given/Concept: Carbohydrates are polyhydroxy aldehydes or ketones. They are classified on the basis of the number of sugar units they contain.

Classification of Carbohydrates:

1. Monosaccharides (Simple sugars):

  • Cannot be hydrolysed further into simpler sugars.
  • General formula: (CH2O)n(CH_2O)_n where n=3n = 3 to 77.
  • Classified by number of carbon atoms:
  • Trioses (n=3n=3): e.g., Glyceraldehyde, Dihydroxyacetone
  • Tetroses (n=4n=4): e.g., Erythrose
  • Pentoses (n=5n=5): e.g., Ribose, Deoxyribose
  • Hexoses (n=6n=6): e.g., Glucose, Fructose, Galactose
  • Heptoses (n=7n=7): e.g., Sedoheptulose
  • Further classified as aldoses (contain aldehyde group, −CHO-CHO) or ketoses (contain ketone group, C=OC=O).

2. Oligosaccharides:

  • Contain 2–10 monosaccharide units joined by glycosidic bonds.
  • Sub-classified as:
  • Disaccharides (2 units): e.g., Sucrose (glucose + fructose), Maltose (glucose + glucose), Lactose (glucose + galactose)
  • Trisaccharides (3 units): e.g., Raffinose
  • Tetrasaccharides (4 units): e.g., Stachyose

3. Polysaccharides:

  • Contain more than 10 (often hundreds to thousands) monosaccharide units.
  • Two types:
  • Homopolysaccharides: Made of one type of monosaccharide. e.g., Starch, Glycogen, Cellulose (all made of glucose)
  • Heteropolysaccharides: Made of two or more types of monosaccharides. e.g., Hyaluronic acid, Heparin

Conclusion: Carbohydrates are broadly classified into monosaccharides, oligosaccharides, and polysaccharides based on the degree of polymerisation.

2Differentiate between D- and L-forms of glucose.Show solution

Concept: The D- and L- designation of glucose is based on the configuration of the asymmetric carbon atom farthest from the carbonyl (aldehyde) group, i.e., C-5 in glucose. This is compared to the reference molecule glyceraldehyde.

FeatureD-GlucoseL-Glucose
Configuration at C-5The −OH-OH group on C-5 is on the right side (same as D-glyceraldehyde)The −OH-OH group on C-5 is on the left side (same as L-glyceraldehyde)
ReferenceBased on D-glyceraldehydeBased on L-glyceraldehyde
Occurrence in natureNaturally occurring form; found in plants and animalsRarely found in nature
Biological activityMetabolically active; can be utilised by cellsNot metabolised by most organisms
Optical rotationDextrorotatory (+) — rotates plane-polarised light to the rightLevorotatory (−) — rotates plane-polarised light to the left
Mirror imageD and L forms are mirror images (enantiomers) of each otherMirror image of D-glucose

Note: The D/L designation refers to the spatial configuration at the reference carbon, not to the direction of optical rotation.

Conclusion: D- and L-glucose are enantiomers differing in the orientation of the −OH-OH group at C-5. D-glucose is the biologically important form.

3Draw the structure of a disaccharide made up of two monosaccharides glucose and fructose.Show solution

Given: The disaccharide made up of glucose and fructose is Sucrose.

Concept: Sucrose is formed by a glycosidic bond between C-1 of α\alpha-D-glucose and C-2 of β\beta-D-fructose. This is an α\alpha-1,2-glycosidic bond (also written as α,β\alpha,\beta-1,2 linkage). Since both anomeric carbons are involved in the bond, sucrose is a non-reducing sugar.

Structure of Sucrose:

The Haworth projection of sucrose is represented as:

α-D-Glucose⏟(pyranose ring)→α(1→2)β-glycosidic bondβ-D-Fructose⏟(furanose ring)\underbrace{\alpha\text{-D-Glucose}}_{\text{(pyranose ring)}} \xrightarrow{\alpha(1\to2)\beta\text{-glycosidic bond}} \underbrace{\beta\text{-D-Fructose}}_{\text{(furanose ring)}}

The structural formula can be described as:

  • α\alpha-D-Glucopyranose ring (6-membered) linked at its C-1 anomeric carbon
  • to β\beta-D-Fructofuranose ring (5-membered) at its C-2 anomeric carbon
  • via an α\alpha-1,2-glycosidic bond with elimination of one water molecule (H2OH_2O).

Molecular formula of sucrose: C12H22O11C_{12}H_{22}O_{11}

Reaction:
Glucose (C6H12O6)+Fructose (C6H12O6)→−H2OSucrose (C12H22O11)\text{Glucose } (C_6H_{12}O_6) + \text{Fructose } (C_6H_{12}O_6) \xrightarrow{-H_2O} \text{Sucrose } (C_{12}H_{22}O_{11})

Key feature: Both anomeric −OH-OH groups are involved in the glycosidic bond, so sucrose has no free anomeric carbon and is a non-reducing disaccharide.

4Draw the partial structure of starch and glycogen.Show solution

Concept: Both starch and glycogen are homopolysaccharides made of α\alpha-D-glucose units. They differ in the degree of branching.


Partial Structure of Starch:

Starch has two components:

  • Amylose (~20%): Unbranched chain of α\alpha-D-glucose units linked by α\alpha(1→4) glycosidic bonds. Forms a helical structure.
  • Amylopectin (~80%): Branched chain with α\alpha(1→4) linkages in the main chain and α\alpha(1→6) linkages at branch points (branching every 24–30 glucose units).

Partial structure of amylose (linear portion):
…−[α-D-Glc]→α(1→4)−[α-D-Glc]→α(1→4)−[α-D-Glc]−…\ldots -[\alpha\text{-D-Glc}]\xrightarrow{\alpha(1\to4)}-[\alpha\text{-D-Glc}]\xrightarrow{\alpha(1\to4)}-[\alpha\text{-D-Glc}]-\ldots

At branch point in amylopectin:
…−[Glc]→α(1→4)−[Glc]→α(1→4)−[Glc]−…\ldots -[\text{Glc}]\xrightarrow{\alpha(1\to4)}-[\text{Glc}]\xrightarrow{\alpha(1\to4)}-[\text{Glc}]-\ldots
↓α(1→6)\hspace{5.5cm}\downarrow \alpha(1\to6)
[Glc]→α(1→4)−[Glc]−…\hspace{5.5cm}[\text{Glc}]\xrightarrow{\alpha(1\to4)}-[\text{Glc}]-\ldots


Partial Structure of Glycogen:

  • Glycogen is the animal storage polysaccharide (found in liver and muscle).
  • Structure is similar to amylopectin but more highly branched — branching occurs every 8–12 glucose units (compared to 24–30 in amylopectin).
  • Main chain: α\alpha(1→4) glycosidic bonds.
  • Branch points: α\alpha(1→6) glycosidic bonds.

Partial structure:
…−[Glc]→α(1→4)−[Glc]→α(1→4)−[Glc]−…\ldots -[\text{Glc}]\xrightarrow{\alpha(1\to4)}-[\text{Glc}]\xrightarrow{\alpha(1\to4)}-[\text{Glc}]-\ldots
↓α(1→6)↓α(1→6)\hspace{3.5cm}\downarrow \alpha(1\to6) \hspace{1.5cm}\downarrow \alpha(1\to6)
[Glc]−…[Glc]−…\hspace{3.5cm}[\text{Glc}]-\ldots \hspace{1.5cm}[\text{Glc}]-\ldots

Key difference: Glycogen is more extensively branched than starch (amylopectin), allowing faster mobilisation of glucose.

5Write the major functions of carbohydrates.Show solution

Major Functions of Carbohydrates:

1. Energy Source:

  • Carbohydrates are the primary source of energy for living organisms.
  • Glucose is completely oxidised to release energy: C6H12O6+6O2→6CO2+6H2O+Energy (ATP)C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy (ATP)}
  • 1 gram of carbohydrate yields approximately 4 kcal of energy.

2. Energy Storage:

  • Starch (in plants) and glycogen (in animals) serve as stored forms of energy.
  • Glycogen is stored in liver and muscle cells and mobilised when energy is needed.

3. Structural Role:

  • Cellulose provides structural rigidity to plant cell walls.
  • Chitin (a polysaccharide) forms the exoskeleton of arthropods and cell walls of fungi.
  • Peptidoglycans (contain carbohydrate components) form bacterial cell walls.

4. Component of Nucleic Acids:

  • Ribose (in RNA) and deoxyribose (in DNA) are pentose sugars that form the backbone of nucleic acids.

5. Cell Recognition and Signalling:

  • Oligosaccharides on cell surfaces (as glycoproteins and glycolipids) act as cell recognition markers and are involved in cell–cell communication.
  • Blood group antigens (A, B, O) are determined by oligosaccharide chains.

6. Lubrication:

  • Mucopolysaccharides (e.g., hyaluronic acid) act as lubricants in joints and connective tissues.

7. Protein Sparing:

  • Adequate carbohydrate intake prevents the use of proteins as an energy source, thus sparing proteins for their structural and functional roles.

Conclusion: Carbohydrates serve as energy sources, structural components, informational molecules, and metabolic intermediates in living systems.

6Describe isomerisation in monosaccharides.Show solution

Concept: Isomers are compounds that have the same molecular formula but differ in the arrangement of atoms. In monosaccharides, several types of isomerism are observed.

Types of Isomerism in Monosaccharides:

1. Structural Isomers (Constitutional Isomers):

  • Compounds with the same molecular formula but different structural arrangements.
  • e.g., Glucose (C6H12O6C_6H_{12}O_6) is an aldohexose; Fructose (C6H12O6C_6H_{12}O_6) is a ketohexose — they are structural isomers.

2. Stereoisomers:
Compounds with the same structural formula but different spatial arrangement of atoms.

(a) D- and L-Isomers (Enantiomers):

  • Based on the configuration at the reference asymmetric carbon (farthest from carbonyl group).
  • D-glucose has −OH-OH on the right at C-5; L-glucose has −OH-OH on the left at C-5.
  • They are non-superimposable mirror images.

(b) Epimers:

  • Stereoisomers that differ in configuration at only one asymmetric carbon (other than the anomeric carbon).
  • e.g., D-Glucose and D-Galactose differ at C-4 → they are C-4 epimers.
  • e.g., D-Glucose and D-Mannose differ at C-2 → they are C-2 epimers.

(c) Anomers:

  • Stereoisomers that differ in configuration at the anomeric carbon (C-1 in aldoses, C-2 in ketoses) formed during ring closure.
  • α\alpha-D-Glucose: −OH-OH at C-1 is on the same side as the ring oxygen (axial/below in Haworth).
  • β\beta-D-Glucose: −OH-OH at C-1 is on the opposite side from the ring oxygen (equatorial/above in Haworth).
  • Interconversion between α\alpha and β\beta forms in solution is called mutarotation.

3. Mutarotation:

  • The spontaneous interconversion of α\alpha and β\beta anomers in aqueous solution through the open-chain form.
  • At equilibrium: ~36% α\alpha-D-glucose, ~64% β\beta-D-glucose.

Conclusion: Isomerisation in monosaccharides includes structural isomerism, D/L isomerism, epimerism, and anomerism, all arising from the presence of multiple asymmetric carbon atoms.

7Differentiate between sphingolipids and glycerolipids.Show solution

Concept: Both sphingolipids and glycerolipids are complex lipids found in biological membranes, but they differ in their backbone structure and composition.

FeatureGlycerolipidsSphingolipids
BackboneGlycerol (3-carbon alcohol)Sphingosine (long-chain amino alcohol, 18 carbons)
Fatty acid linkageFatty acids are linked to glycerol via ester bonds (−COO−-COO-)One fatty acid is linked to the amino group of sphingosine via an amide bond (−CO−NH−-CO-NH-)
Number of fatty acidsTwo fatty acids (in phospholipids) or three (in triacylglycerols)One fatty acid
Head groupPhosphate + alcohol (e.g., choline, ethanolamine, serine) in phosphoglyceridesPhosphocholine (in sphingomyelin) or sugar residues (in glycosphingolipids)
ExamplesPhosphatidylcholine, Phosphatidylethanolamine, Phosphatidylserine, TriacylglycerolSphingomyelin, Cerebrosides, Gangliosides
LocationMajor component of all biological membranesAbundant in nervous tissue (myelin sheath), brain
Hydrolysis productsGlycerol + fatty acids + phosphate + alcoholSphingosine + fatty acid + phosphate/sugar

Conclusion: The key difference is the backbone: glycerolipids use glycerol while sphingolipids use sphingosine as the backbone molecule.

8Why are membrane lipids called amphipathic?Show solution

Given: Membrane lipids include phospholipids (glycerophospholipids and sphingomyelin) and glycolipids.

Concept: The term amphipathic (also called amphiphilic) refers to molecules that possess both a hydrophilic (water-loving) region and a hydrophobic (water-fearing) region within the same molecule.

Explanation:

Membrane lipids (e.g., phosphatidylcholine) have two distinct structural regions:

  1. Hydrophilic Head (Polar region):
  • Consists of the phosphate group and the attached polar alcohol (e.g., choline, ethanolamine, serine).
  • This region is charged/polar and interacts favourably with water molecules.
  • It faces the aqueous environment (cytoplasm or extracellular fluid).
  1. Hydrophobic Tail (Non-polar region):
  • Consists of two long fatty acid chains (hydrocarbon tails).
  • These are non-polar and do not interact with water.
  • They face the interior of the membrane, away from water.

Structural representation:
Polar head (phosphate + alcohol)⏟Hydrophilic−Glycerol−Two fatty acid tails⏟Hydrophobic\underbrace{\text{Polar head (phosphate + alcohol)}}_{\text{Hydrophilic}} - \text{Glycerol} - \underbrace{\text{Two fatty acid tails}}_{\text{Hydrophobic}}

Significance:

  • This amphipathic nature drives the spontaneous formation of lipid bilayers in aqueous environments.
  • In a bilayer, hydrophobic tails face inward (away from water) and hydrophilic heads face outward (toward water), forming a stable membrane structure.

Conclusion: Membrane lipids are called amphipathic because they contain both a polar hydrophilic head group and non-polar hydrophobic fatty acid tails in the same molecule.

9Differentiate between saturated and unsaturated fatty acids.Show solution

Concept: Fatty acids are long-chain carboxylic acids. They are classified as saturated or unsaturated based on the presence of double bonds in the hydrocarbon chain.

FeatureSaturated Fatty AcidsUnsaturated Fatty Acids
Double bondsNo carbon–carbon double bonds; all carbons are saturated with hydrogenContain one or more carbon–carbon double bonds (C=CC=C)
General formulaCH3−(CH2)n−COOHCH_3-(CH_2)_n-COOHContain −CH=CH−-CH=CH- in the chain
Types—Monounsaturated (one double bond); Polyunsaturated (two or more double bonds)
Physical stateSolid at room temperature (due to tight packing of straight chains)Liquid at room temperature (due to kinks/bends at double bonds preventing tight packing)
Melting pointHigher melting pointLower melting point
Chain geometryStraight, linear chainKinked/bent chain at the site of double bond (usually cis configuration)
ExamplesPalmitic acid (C16:0C_{16:0}), Stearic acid (C18:0C_{18:0})Oleic acid (C18:1C_{18:1}, one double bond), Linoleic acid (C18:2C_{18:2}), Arachidonic acid (C20:4C_{20:4})
SourcesAnimal fats (butter, lard), coconut oilVegetable oils (olive, sunflower), fish oils
Health effectsExcess intake associated with cardiovascular diseaseEssential fatty acids (linoleic, linolenic) are required in diet

Notation: Fatty acids are written as Cchain length:number of double bondsC_{\text{chain length}:\text{number of double bonds}}.

  • e.g., Palmitic acid = C16:0C_{16:0}; Oleic acid = C18:1C_{18:1}

Conclusion: The key difference is the presence of C=CC=C double bonds: saturated fatty acids have none (solid, high melting point) while unsaturated fatty acids have one or more (liquid, low melting point).

10Describe the various categories of amino acids.Show solution

Concept: Amino acids are the building blocks of proteins. They contain an amino group (−NH2-NH_2), a carboxyl group (−COOH-COOH), a hydrogen atom, and a variable side chain (R group) attached to the α\alpha-carbon. They are classified based on the chemical nature of the R group.

Categories of Amino Acids:

1. Non-polar (Hydrophobic) Amino Acids:

  • R group is non-polar/aliphatic or aromatic.
  • Tend to be located in the interior of proteins (away from water).
  • Examples: Glycine (Gly), Alanine (Ala), Valine (Val), Leucine (Leu), Isoleucine (Ile), Proline (Pro), Phenylalanine (Phe), Tryptophan (Trp), Methionine (Met).

2. Polar, Uncharged Amino Acids:

  • R group is polar but carries no net charge at physiological pH (7.0).
  • Can form hydrogen bonds with water.
  • Examples: Serine (Ser), Threonine (Thr), Cysteine (Cys), Tyrosine (Tyr), Asparagine (Asn), Glutamine (Gln).

3. Positively Charged (Basic) Amino Acids:

  • R group carries a positive charge at physiological pH.
  • Examples: Lysine (Lys) — ε\varepsilon-amino group; Arginine (Arg) — guanidinium group; Histidine (His) — imidazole group.

4. Negatively Charged (Acidic) Amino Acids:

  • R group carries a negative charge at physiological pH.
  • Examples: Aspartic acid/Aspartate (Asp), Glutamic acid/Glutamate (Glu).

5. Classification based on nutritional requirement:

  • Essential amino acids: Cannot be synthesised by the human body; must be obtained from diet. (9 in humans): His, Ile, Leu, Lys, Met, Phe, Thr, Trp, Val.
  • Non-essential amino acids: Can be synthesised by the body. e.g., Ala, Gly, Ser, Asp, Glu.

6. Classification based on R group structure:

  • Aliphatic: Gly, Ala, Val, Leu, Ile
  • Aromatic: Phe, Tyr, Trp
  • Sulphur-containing: Cys, Met
  • Hydroxyl-containing: Ser, Thr
  • Amide-containing: Asn, Gln
  • Imino acid: Pro (contains imino group, −NH−-NH-, not −NH2-NH_2)

Conclusion: Amino acids are categorised based on the polarity, charge, and chemical nature of their R groups into non-polar, polar uncharged, acidic, and basic types.

11What is zwitterion? Draw the structure of zwitterion.Show solution

Definition of Zwitterion:

A zwitterion (from German Zwitter = hybrid) is a molecule that carries both a positive charge and a negative charge simultaneously on different atoms, resulting in an overall net charge of zero. It is also called a dipolar ion or inner salt.

Concept in Amino Acids:

Amino acids contain both an acidic carboxyl group (−COOH-COOH) and a basic amino group (−NH2-NH_2). At physiological pH (around 7.0, i.e., at the isoelectric point), the carboxyl group loses a proton (becomes −COO−-COO^-) and the amino group gains a proton (becomes −NH3+-NH_3^+). This doubly charged form is the zwitterion.

Reaction:
H2N−CH(R)−COOH⏟Uncharged form⇌+H3N−CH(R)−COO−⏟Zwitterion (dipolar ion)\underbrace{H_2N-CH(R)-COOH}_{\text{Uncharged form}} \rightleftharpoons \underbrace{^+H_3N-CH(R)-COO^-}_{\text{Zwitterion (dipolar ion)}}

Structure of Zwitterion (using Alanine as example):

H∣+H3N−C−COO−∣CH3\begin{array}{c} H \\ | \\ ^+H_3N - C - COO^- \\ | \\ CH_3 \end{array}

Or in a more explicit representation:
+H3N−C∣∣(R)(H)−COO−^+H_3N-\underset{|}{\overset{|}{C}}(R)(H)-COO^-

Where:

  • +H3N−^+H_3N- = protonated amino group (positively charged)
  • −COO−-COO^- = deprotonated carboxyl group (negatively charged)
  • RR = side chain
  • Net charge = +1+(−1)=0+1 + (-1) = 0

Key point: In aqueous solution at the isoelectric point (pI), amino acids predominantly exist as zwitterions, not as uncharged molecules. This is the most stable form of an amino acid in solution.

12What are non-standard and non-protein amino acids?Show solution

1. Non-Standard Amino Acids:

  • The 20 amino acids commonly found in proteins are called standard or canonical amino acids.
  • Non-standard amino acids are amino acids that are derived from the 20 standard amino acids by post-translational modifications (chemical modifications after the protein is synthesised).
  • They are found within proteins but are not directly encoded by the genetic code.
  • They arise by enzymatic modification of standard amino acids after incorporation into the polypeptide chain.

Examples of non-standard amino acids:

  • 4-Hydroxyproline: Derived from proline by hydroxylation; found in collagen.
  • 5-Hydroxylysine: Derived from lysine by hydroxylation; found in collagen.
  • ε\varepsilon-N-Methyllysine: Found in histone proteins.
  • γ\gamma-Carboxyglutamate: Found in blood clotting proteins (e.g., prothrombin).
  • Selenocysteine: Sometimes called the 21st amino acid; contains selenium instead of sulphur.
  • Desmosine: Found in elastin (cross-linking amino acid).

2. Non-Protein Amino Acids:

  • These are amino acids that are NOT incorporated into proteins at all.
  • They are found as free molecules in cells and serve important metabolic or physiological functions.
  • They may be D-amino acids or β\beta-, γ\gamma-amino acids (not α\alpha-amino acids).

Examples of non-protein amino acids:

  • β\beta-Alanine: Found in coenzyme A and carnosine.
  • γ\gamma-Aminobutyric acid (GABA): A neurotransmitter in the brain.
  • Ornithine and Citrulline: Intermediates in the urea cycle.
  • Homocysteine: Intermediate in methionine metabolism.
  • D-Amino acids: Found in bacterial cell walls (e.g., D-glutamate, D-alanine in peptidoglycan).
  • Canavanine: Found in jack beans; toxic analogue of arginine.

Conclusion: Non-standard amino acids are modified forms of standard amino acids found in proteins, while non-protein amino acids are amino acids not incorporated into proteins but serve other biological roles.

13How the peptide bonds are formed?Show solution

Concept: A peptide bond is a covalent amide bond formed between the α\alpha-carboxyl group (−COOH-COOH) of one amino acid and the α\alpha-amino group (−NH2-NH_2) of another amino acid, with the elimination of a water molecule (H2OH_2O). This reaction is called a condensation reaction (or dehydration synthesis).

Mechanism of Peptide Bond Formation:

Step 1: The carboxyl group of amino acid 1 reacts with the amino group of amino acid 2.

Step 2: A water molecule is eliminated.

Step 3: A peptide (−CO−NH−-CO-NH-) bond is formed.

Reaction:
H2N−CH∣R1−COOH  +  H2N−CH∣R2−COOH→−H2OH2N−CH∣R1−CO−NH⏟Peptide bond−CH∣R2−COOHH_2N-\underset{R_1}{\underset{|}{CH}}-COOH \;+\; H_2N-\underset{R_2}{\underset{|}{CH}}-COOH \xrightarrow{-H_2O} H_2N-\underset{R_1}{\underset{|}{CH}}-\underbrace{CO-NH}_{\text{Peptide bond}}-\underset{R_2}{\underset{|}{CH}}-COOH

General equation:
Amino acid1+Amino acid2→condensationDipeptide+H2O\text{Amino acid}_1 + \text{Amino acid}_2 \xrightarrow{\text{condensation}} \text{Dipeptide} + H_2O

Properties of the Peptide Bond:

  1. It has partial double bond character due to resonance between the C=OC=O and C−NC-N bonds.
  2. The peptide bond is planar (all four atoms CαC_\alpha, CC, OO, NN, HH, CαC_\alpha lie in the same plane).
  3. It is rigid and does not allow free rotation.
  4. It is predominantly in the trans configuration (R groups on opposite sides).
  5. The bond length of C−NC-N in peptide bond is 0.133 nm (shorter than a typical C−NC-N single bond of 0.149 nm).

Polypeptide chain: Multiple amino acids joined by peptide bonds form a polypeptide. The chain has a free −NH2-NH_2 at the N-terminus and a free −COOH-COOH at the C-terminus.

Conclusion: Peptide bonds are formed by condensation reactions between the carboxyl group of one amino acid and the amino group of the next, releasing water.

14Draw the structure of Lys-Glu-Lys.Show solution

Given: A tripeptide: Lys-Glu-Lys (Lysine–Glutamic acid–Lysine)

Concept: In a peptide, amino acids are joined by peptide bonds (−CO−NH−-CO-NH-). The sequence is written from N-terminus (left) to C-terminus (right).

Side chains (R groups):

  • Lysine (Lys, K): R = −(CH2)4−NH2-(CH_2)_4-NH_2 (positively charged at physiological pH: −(CH2)4−NH3+-(CH_2)_4-NH_3^+)
  • Glutamic acid (Glu, E): R = −(CH2)2−COOH-(CH_2)_2-COOH (negatively charged at physiological pH: −(CH2)2−COO−-(CH_2)_2-COO^-)

Structure of Lys-Glu-Lys:

H2N−CH∣∣(CH2)4NH2−CO⏟Lys (N-terminus)−NH−CH∣∣(CH2)2COOH−CO⏟Glu−NH−CH∣∣(CH2)4NH2−COOH⏟Lys (C-terminus)\underbrace{H_2N-\overset{\displaystyle|(CH_2)_4NH_2}{\underset{|}{CH}}-CO}_{\text{Lys (N-terminus)}}-\underbrace{NH-\overset{\displaystyle|(CH_2)_2COOH}{\underset{|}{CH}}-CO}_{\text{Glu}}-\underbrace{NH-\overset{\displaystyle|(CH_2)_4NH_2}{\underset{|}{CH}}-COOH}_{\text{Lys (C-terminus)}}

Expanded structural representation:

H2N−CH∣(CH2)4−NH2−CO−NH⏟peptide bond−CH∣(CH2)2−COOH−CO−NH⏟peptide bond−CH∣(CH2)4−NH2−COOHH_2N-\underset{\displaystyle(CH_2)_4-NH_2}{\underset{|}{CH}}-\overset{\text{peptide bond}}{\underbrace{CO-NH}}-\underset{\displaystyle(CH_2)_2-COOH}{\underset{|}{CH}}-\overset{\text{peptide bond}}{\underbrace{CO-NH}}-\underset{\displaystyle(CH_2)_4-NH_2}{\underset{|}{CH}}-COOH

Key features:

  • N-terminus: Free −NH2-NH_2 of the first Lys
  • C-terminus: Free −COOH-COOH of the second Lys
  • Two peptide bonds (−CO−NH−-CO-NH-) connecting the three amino acids
  • Net charge at physiological pH: +1+1 (Lys) −1-1 (Glu) +1+1 (Lys) =+1= +1 (overall positive)

Molecular formula: C16H32N4O6C_{16}H_{32}N_4O_6 (after loss of 2 water molecules from condensation)

15Describe the various secondary structures of protein.Show solution

Concept: The secondary structure of a protein refers to the local, regular, repeating folding patterns of the polypeptide backbone, stabilised primarily by hydrogen bonds between the carbonyl oxygen (C=OC=O) and the amide hydrogen (N−HN-H) of the peptide backbone.

Major Secondary Structures:


1. α\alpha-Helix (Alpha Helix):

  • Proposed by Linus Pauling and Robert Corey (1951).
  • The polypeptide backbone coils into a right-handed helix (clockwise when viewed from the N-terminus).
  • Stabilisation: Hydrogen bonds form between the C=OC=O of residue nn and the N−HN-H of residue n+4n+4 (i.e., every 4th amino acid).
  • Parameters:
  • 3.6 amino acid residues per turn
  • Pitch (rise per turn) = 0.54 nm
  • Rise per residue = 0.15 nm
  • Diameter ≈ 0.5 nm
  • R groups project outward from the helix axis.
  • Helix breakers: Proline (Pro) disrupts the α\alpha-helix due to its rigid ring structure.
  • Example: Found abundantly in α\alpha-keratin (hair, nails), myoglobin.

2. β\beta-Pleated Sheet (Beta Sheet):

  • Also proposed by Pauling and Corey.
  • The polypeptide chain is almost fully extended and arranged side by side.
  • Stabilisation: Hydrogen bonds form between adjacent polypeptide strands (inter-strand H-bonds between C=OC=O and N−HN-H).
  • The backbone has a pleated/zigzag appearance; R groups alternate above and below the plane.
  • Types:
  • Parallel β\beta-sheet: Adjacent strands run in the same direction (N→C); H-bonds are slightly bent.
  • Antiparallel β\beta-sheet: Adjacent strands run in opposite directions (N→C and C→N); H-bonds are straight and stronger.
  • Example: Found in β\beta-keratin (silk fibroin), immunoglobulins.

3. β\beta-Turn (Beta Turn / Reverse Turn):

  • A short loop structure that allows the polypeptide chain to reverse direction (180°).
  • Involves 4 amino acid residues.
  • Stabilised by a hydrogen bond between the C=OC=O of residue 1 and the N−HN-H of residue 4.
  • Proline and Glycine are commonly found in β\beta-turns.
  • Important in connecting β\beta-strands in antiparallel β\beta-sheets.

4. Random Coil:

  • Regions of the polypeptide that do not have a regular, repeating secondary structure.
  • Flexible and irregular regions.

Conclusion: The major secondary structures are the α\alpha-helix and β\beta-pleated sheet, both stabilised by hydrogen bonds between backbone atoms. The β\beta-turn helps in chain reversal.

16Differentiate between tertiary and quaternary structure of proteins.

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17Differentiate between nucleosides and nucleotides.

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18Explain the primary structure of DNA.

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19Draw the structure of A-T-C-G oligonucleotide.

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20Explain Watson and Crick model of DNA.

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21Describe the various forms of DNA.

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22Describe the clover leaf model of tRNA.

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23In carbohydrates which of these functional groups are present
(a) Alcohol and carboxyl groups
(b) Aldehyde and ketone groups
(c) Hydroxyl and hydrogen groups
(d) Ether and ester groups

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24Which of the following is a non-reducing disaccharide?
(a) Maltose
(b) Lactose
(c) Sucrose
(d) Cellobiose

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25The repeating units of proteins are
(a) Glucose units
(b) Amino acids
(c) Fatty acids
(d) Nucleotides

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26Which of the following is the most common secondary structure of proteins
(a) α\alpha-helix
(b) β\beta-pleated sheet
(c) Both (a) and (b)
(d) None of the above

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27A nucleotide contains
(a) Nitrogenous base, sugar and phosphate
(b) Sugar and phosphate
(c) Nitrogenous base and sugar
(d) None of the above

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28The two strands in DNA double helix are joined by
(a) Covalent bond
(b) Hydrogen bond
(c) Glycosidic bond
(d) Phosphodiester bond

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29Which is an example of storage lipid?
(a) Fatty acids
(b) Triacylglycerol
(c) Sphingolipids
(d) Eicosanoids

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30In glycerolipids fatty acids are joined to glycerol through which bond?
(a) Phosphodiester bond
(b) Glycosidic bond
(c) Peptide bond
(d) Ester bond

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15 more solved questions in Biomolecules

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