Chapter 19: Biomolecules

Carbohydrates

Definition and General Properties

• Optically active polyhydroxy aldehydes or ketones, or compounds yielding these on hydrolysis.

• General formula: Cx(H2O)y, considered hydrates of carbon.

• Examples: glucose, sucrose, starch.

• Sugars are sweet carbohydrates; sucrose (table sugar) and lactose (milk sugar) are common examples.

• Also called saccharides (Greek: sakcharon = sugar).

Classification Based on Hydrolysis

• Monosaccharides: Simple sugars, cannot be hydrolyzed further (e.g., glucose, fructose).

• Oligosaccharides: 2-10 monosaccharide units; hydrolyze to monosaccharides (e.g., sucrose, maltose).

• Polysaccharides: Long chains of monosaccharides (100+ units); hydrolyze to many monosaccharides (e.g., starch, cellulose).

Classification Based on Reducing Properties

• Reducing sugars: Contain free aldehyde or ketone groups; reduce Fehling's and Tollens' reagents (e.g., glucose, maltose, lactose).

• Non-reducing sugars: No free aldehyde or ketone groups; do not reduce Fehling's or Tollens' reagents (e.g., sucrose).

Solved Examples

Ex 1: Identify if maltose is a reducing sugar.
- Maltose has a free aldehyde group at C-1 of one glucose unit.
- It reduces Fehling's solution.
- Therefore, maltose is a reducing sugar.

Ex 2: Hydrolysis of sucrose.
- Sucrose + H2O (acid/enzyme) → glucose + fructose.
- Both monosaccharides are reducing sugars, but sucrose itself is non-reducing.

Practice Set

• Q1: Which of the following is a non-reducing sugar?
  A. Glucose
  B. Maltose
  C. Sucrose
  D. Lactose

• Q2: Oligosaccharides contain how many monosaccharide units?
  A. 1
  B. 2-10
  C. 10-100
  D. More than 100

• Q3: Which reagent is used to test reducing sugars?
  A. Benedict's solution
  B. Fehling's solution
  C. Tollens' reagent
  D. All of the above

Answer Key

• Q1: C. Sucrose - It lacks free aldehyde or ketone groups and does not reduce Fehling's or Tollens' reagents.

• Q2: B. 2-10 - Oligosaccharides consist of 2 to 10 monosaccharide units.

• Q3: D. All of the above - Benedict's, Fehling's, and Tollens' reagents all test for reducing sugars.

Memory Aids

• "Mono" = single sugar unit; "Oligo" = few units; "Poly" = many units.

• Reducing sugars have free aldehyde/ketone groups; non-reducing sugars do not.

• Sucrose is non-reducing despite being a disaccharide.

Monosaccharides

Classification and Properties

• Classified by carbon number: trioses (3C), pentoses (5C), hexoses (6C), etc.

• Based on functional group: Aldoses (aldehyde group), Ketoses (ketone group).

• Glucose: aldohexose (6C, aldehyde), naturally occurring, found in fruits and honey.

• Fructose: ketohexose (6C, ketone), found in fruits and honey, sweeter than glucose.

Preparation of Glucose

• From sucrose: Boiling sucrose with dilute HCl or H2SO4 in alcoholic solution yields equal glucose and fructose.

• From starch: Hydrolysis of starch with dilute H2SO4 at 393 K under pressure produces glucose.

Structure of Glucose

• Open chain form confirmed by reactions with HI, NH2OH, HCN, bromine water.

• >

• 

  Fischer gave exact spatial arrangement of —OH groups.

• Glucose exists as cyclic hemiacetal (pyranose) forms: α and β anomers differ at C-1 (anomeric carbon).

• 

  Haworth projection represents cyclic glucose.

Other Reactions of Glucose

• Reduction: Glucose → D-Sorbitol (using Na-Hg).

• Oxidation: Glucose → Gluconic acid (Fehling's or Tollens' test positive).

• 

  Derivatives: Glucose pentaacetate (acetic anhydride), pentachloroglucose (PCl5), methylglucosides (CH3OH + HCl).

Solved Examples

Ex 1: Hydrolysis of sucrose.
- Sucrose + H2O (acid) → glucose + fructose (1:1 ratio).

Ex 2: Identify the anomeric carbon in glucose.
- Carbon 1 (C-1) is the anomeric carbon where α and β forms differ.

Practice Set

• Q1: Glucose is classified as:
  A. Ketohexose
  B. Aldohexose
  C. Aldopentose
  D. Ketopentose

• Q2: Which reagent confirms the aldehyde group in glucose?
  A. HI and heat
  B. NH2OH
  C. Bromine water
  D. All of the above

• Q3: The cyclic form of glucose is called:
  A. Furanose
  B. Pyranose
  C. Glucopyranose
  D. Both B and C

Answer Key

• Q1: B. Aldohexose - Glucose has six carbons and an aldehyde group.

• Q2: D. All of the above - These reagents confirm the aldehyde group in glucose.

• Q3: D. Both B and C - The six-membered cyclic form is pyranose, also called glucopyranose.

Memory Aids

• Glucose = aldohexose; fructose = ketohexose.

• Anomeric carbon = C-1 in glucose; α and β differ in OH position here.

• Pyranose = six-membered ring; furanose = five-membered ring.

Fructose

Structure and Properties

• Ketohexose: 6 carbons, ketone group at C-2.

• Found in fruits, honey, vegetables; used as sweetener.

• Belongs to D-series; laevorotatory (D-(-)-fructose).

  

• Exists in two cyclic forms (furanose rings) formed by OH at C-5 attacking ketone at C-2.

• α and β anomers differ in OH position at C-2.

Haworth Structures and Isomerism

• Cyclic forms represented by Haworth projections.

• Fructose anomers: α-D-(−)-Fructofuranose and β-D-(−)-Fructofuranose.

• Isomer types: Enantiomers, Anomers, Epimers, Diastereomers (differences in OH group positions).

Solved Examples

Ex 1: Identify the ring size of fructose cyclic form.
- Fructose forms a five-membered furanose ring.

Ex 2: Difference between α and β fructofuranose.
- Position of OH on C-2: down in α, up in β.

Practice Set

• Q1: Fructose is classified as:
  A. Aldohexose
  B. Ketohexose
  C. Aldopentose
  D. Ketopentose

• Q2: The cyclic form of fructose is:
  A. Pyranose
  B. Furanose
  C. Glucopyranose
  D. None of the above

• Q3: The difference between α and β fructofuranose is:
  A. Position of OH on C-1
  B. Position of OH on C-2
  C. Number of carbons
  D. Type of sugar

Answer Key

• Q1: B. Ketohexose - Fructose has a ketone group at C-2.

• Q2: B. Furanose - Fructose forms a five-membered ring.

• Q3: B. Position of OH on C-2 - α and β differ at C-2 OH position.

Memory Aids

• Fructose = ketohexose with furanose ring.

• α and β anomers differ at C-2 OH position.

Disaccharides

Structure and Properties

• Formed by two monosaccharides linked via glycosidic (oxide) linkage.

• Glycosidic bond formed by loss of water between OH groups.

• Examples:

  • Sucrose: Glucose + Fructose; non-reducing sugar; hydrolyzes to invert sugar.

  • 

    Maltose: Two α-D-glucose units; reducing sugar.

  • 

    Lactose: β-D-galactose + β-D-glucose; reducing sugar.

Hydrolysis Reactions

• Sucrose + H2O (invertase or acid) → glucose + fructose (invert sugar).

• Maltose + H2O (maltase) → 2 glucose molecules.

• Lactose + H2O (lactase) → galactose + glucose.

Solved Examples

Ex 1: Hydrolysis of sucrose.
- Sucrose + H2O → glucose + fructose (invert sugar).

Ex 2: Identify reducing sugar among disaccharides.
- Maltose and lactose are reducing sugars; sucrose is non-reducing.

Practice Set

• Q1: Sucrose is a disaccharide of:
  A. Glucose + Glucose
  B. Glucose + Fructose
  C. Galactose + Glucose
  D. Fructose + Galactose

• Q2: Which disaccharide is non-reducing?
  A. Maltose
  B. Lactose
  C. Sucrose
  D. Cellobiose

• Q3: Glycosidic linkage in maltose is between:
  A. C-1 and C-4 of two glucose units
  B. C-1 and C-6 of two glucose units
  C. C-1 and C-4 of glucose and fructose
  D. C-1 and C-4 of galactose and glucose

Answer Key

• Q1: B. Glucose + Fructose - Sucrose is composed of these monosaccharides.

• Q2: C. Sucrose - It lacks free aldehyde/ketone groups.

• Q3: A. C-1 and C-4 of two glucose units - Maltose linkage is α-1,4 glycosidic bond.

Memory Aids

• Sucrose = glucose + fructose, non-reducing.

• Maltose = two glucose units, reducing sugar.

• Lactose = galactose + glucose, reducing sugar.

Polysaccharides

Types and Structure

• Starch: Plant storage polysaccharide; polymer of α-D-glucose.

• Components:

  • Amylose: linear, 200-1000 α-D-glucose units linked by α-1,4 glycosidic bonds; water soluble.

  • 

    Amylopectin: branched polymer with α-1,4 linkages and α-1,6 branches at C-6.

• 

  Cellulose: Structural polysaccharide in plants; linear polymer of β-D-glucose linked by β-1,4 glycosidic bonds.

• 

  Glycogen: Animal storage polysaccharide; highly branched α-D-glucose polymer similar to amylopectin; stored in liver, muscles, brain.

Properties and Importance

• Cellulose forms plant cell walls; provides strength and rigidity.

• Starch and glycogen serve as energy storage in plants and animals respectively.

• Glycogen is broken down enzymatically to glucose on demand.

• Cellulose is used in cotton fibers, wood, and bacterial cell walls.

• Mercerisation: treatment of cellulose fibers with NaOH to impart lustre.

Solved Examples

Ex 1: Identify the glycosidic linkage in cellulose.
- β-1,4 glycosidic bonds between β-D-glucose units.

Ex 2: Difference between amylose and amylopectin.
- Amylose: linear α-1,4 linkages.
- Amylopectin: branched with α-1,4 and α-1,6 linkages.

Practice Set

• Q1: The main storage polysaccharide in plants is:
  A. Cellulose
  B. Glycogen
  C. Starch
  D. Dextran

• Q2: Glycogen is stored mainly in:
  A. Liver and muscles
  B. Brain and bones
  C. Kidneys
  D. Lungs

• Q3: The glycosidic linkage in cellulose is:
  A. α-1,4
  B. β-1,4
  C. α-1,6
  D. β-1,6

Answer Key

• Q1: C. Starch - Main plant storage polysaccharide.

• Q2: A. Liver and muscles - Glycogen storage sites.

• Q3: B. β-1,4 - Cellulose linkage type.

Memory Aids

• Starch = plant storage; glycogen = animal storage.

• Amylose = linear; amylopectin = branched.

• Cellulose = β-1,4 linkages; indigestible by humans.

Importance of Carbohydrates

Biological Roles

• Primary energy source for plants and animals.

• Storage molecules: starch in plants, glycogen in animals.

• Structural components: cellulose in plant cell walls, bacterial cell walls.

• Constituents of nucleic acids: D-ribose and 2-deoxy-D-ribose sugars.

• Natural sugars mostly exist in D-form.

• Bacterial enzyme converts sucrose to dextran, a polysaccharide.

• Mercerisation of cotton (cellulose) with NaOH improves fabric lustre.

Solved Examples

Ex 1: Name the sugar in nucleic acids.
- D-ribose and 2-deoxy-D-ribose.

Ex 2: What is mercerisation?
- Treatment of cellulose fibers with concentrated NaOH to enhance lustre.

Practice Set

• Q1: The sugar present in nucleic acids is:
  A. Glucose
  B. D-ribose
  C. Fructose
  D. Lactose

• Q2: Mercerisation is a treatment of:
  A. Starch
  B. Glycogen
  C. Cellulose
  D. Dextran

• Q3: Dextran is a polysaccharide produced by:
  A. Plants
  B. Animals
  C. Bacteria
  D. Fungi

Answer Key

• Q1: B. D-ribose - Sugar in nucleic acids.

• Q2: C. Cellulose - Mercerisation treats cellulose fibers.

• Q3: C. Bacteria - Bacterial enzyme converts sucrose to dextran.

Memory Aids

• Nucleic acids contain ribose sugars.

• Mercerisation = NaOH treatment of cotton (cellulose).

• Dextran produced by bacteria from sucrose.

Proteins

Overview

• Proteins are polymers of α-amino acids linked by peptide bonds.

• Essential for structure, function, growth, and maintenance of the body.

• Main sources: milk, cheese, pulses, peanuts, fish, meat.

Solved Examples

• Ex 1: Identify the monomer of proteins.
  Answer: α-amino acids.

Practice Set

• Q1: Proteins are polymers of:
  A. Monosaccharides
  B. Fatty acids
  C. α-amino acids
  D. Nucleotides

• Q2: Which of the following is a main source of protein?
  A. Glucose
  B. Milk
  C. Starch
  D. Cellulose

• Q3: Proteins are important for:
  A. Energy storage
  B. Growth and maintenance
  C. Photosynthesis
  D. Respiration

Answer Key

• Q1: C - Proteins are made of α-amino acids.

• Q2: B - Milk is a rich protein source.

• Q3: B - Proteins are essential for growth and maintenance.

Memory Aids

• Protein = Primary importance (Greek origin).

• Protein monomer = α-amino acid.

• Sources: Milk, Meat, Pulses (MMP).

Amino Acids

Structure and Properties

• Bifunctional molecules with acidic (—COOH) and basic (—NH2) groups attached to the same α-carbon.

• General formula: NH2–CH(R)–COOH, where R is side chain.

• Usually colorless, crystalline, water-soluble, high melting solids.

• Behave like salts due to acidic and basic groups.

• Exist as zwitterions in aqueous solution (dipolar ions with + and − charges).

Solved Examples

• Ex 1: Identify the functional groups in amino acids.
  Answer: Amino (—NH2) and carboxyl (—COOH).

Practice Set

• Q1: Amino acids contain which functional groups?
  A. Hydroxyl and aldehyde
  B. Amino and carboxyl
  C. Ketone and ester
  D. Ether and amide

• Q2: Zwitterion form of amino acid has:
  A. Only positive charge
  B. Only negative charge
  C. Both positive and negative charges
  D. No charge

• Q3: Amino acids are soluble in water because:
  A. They are nonpolar
  B. They form salts
  C. They are hydrocarbons
  D. They are gases

Answer Key

• Q1: B - Amino acids have amino and carboxyl groups.

• Q2: C - Zwitterions carry both positive and negative charges.

• Q3: B - Amino acids behave like salts, increasing solubility.

Memory Aids

• Remember: Amino + Acid = Amino acid.

• Zwitterion = Both charges present.

• Water solubility due to salt-like nature.

Classification of Amino Acids

Types Based on Functional Groups

• Neutral: Equal number of —COOH and —NH2 groups (e.g., glycine, alanine).

• Acidic: More —COOH groups than —NH2 (e.g., aspartic acid, glutamic acid).

• Basic: More —NH2 groups than —COOH (e.g., lysine, arginine).

Types Based on Synthesis

• Essential: Cannot be synthesized by body; must be obtained from diet (e.g., valine, leucine).

• Non-essential: Synthesized by body (e.g., glycine, alanine).

Solved Examples

• Ex 1: Classify lysine.
  Answer: Basic amino acid.

• Ex 2: Is valine essential or non-essential?
  Answer: Essential amino acid.

Practice Set

• Q1: Which amino acid is acidic?
  A. Glycine
  B. Aspartic acid
  C. Lysine
  D. Alanine

• Q2: Essential amino acids are:
  A. Synthesized in body
  B. Obtained from diet
  C. Nonpolar
  D. Acidic only

• Q3: Basic amino acid example is:
  A. Glutamic acid
  B. Arginine
  C. Valine
  D. Serine

Answer Key

• Q1: B - Aspartic acid is acidic.

• Q2: B - Essential amino acids must be obtained from diet.

• Q3: B - Arginine is basic.

Memory Aids

• Neutral = equal acidic and basic groups.

• Essential = Eat (must be in diet).

• Basic amino acids have extra amino groups.

Isoelectric Point

Concept

• pH at which amino acid exists as zwitterion with equal positive and negative charges.

• At this pH, amino acid does not migrate in electric field.

• Solubility in water is minimum at isoelectric point.

• Useful for amino acid separation.

Solved Examples

• Ex 1: What happens to amino acid at isoelectric point?
  Answer: Exists as neutral zwitterion, no migration in electric field.

Practice Set

• Q1: Isoelectric point is the pH where amino acid:
  A. Is fully protonated
  B. Has no net charge
  C. Is fully deprotonated
  D. Is insoluble

• Q2: At isoelectric point, amino acid:
  A. Migrates to cathode
  B. Migrates to anode
  C. Does not migrate
  D. Decomposes

• Q3: Amino acid solubility at isoelectric point is:
  A. Maximum
  B. Minimum
  C. Zero
  D. Unchanged

Answer Key

• Q1: B - Net charge is zero at isoelectric point.

• Q2: C - No migration due to neutral charge.

• Q3: B - Solubility is minimum at isoelectric point.

Memory Aids

• Isoelectric = "iso" means equal charges.

• Neutral charge = no movement in electric field.

• Solubility dips at isoelectric point.

Peptide Bond

Formation and Structure

• Peptide bond is an amide bond formed between —COOH of one amino acid and —NH2 of another.

• Formed by condensation reaction with elimination of H2O.

• Links amino acids to form peptides and proteins.

Solved Examples

• Ex 1: Write the reaction for peptide bond formation.
  Answer: Amino acid 1 (—COOH) + Amino acid 2 (—NH2) → Peptide bond + H2O

Practice Set

• Q1: Peptide bond is formed between:
  A. Two amino groups
  B. Two carboxyl groups
  C. Amino and carboxyl groups
  D. Side chains

• Q2: Peptide bond formation involves:
  A. Hydrolysis
  B. Condensation
  C. Oxidation
  D. Reduction

• Q3: Peptides with 2-10 amino acids are called:
  A. Polypeptides
  B. Oligopeptides
  C. Proteins
  D. Monopeptides

Answer Key

• Q1: C - Peptide bond forms between amino and carboxyl groups.

• Q2: B - Condensation reaction with water loss.

• Q3: B - Oligopeptides contain 2-10 amino acids.

Memory Aids

• Peptide bond = Amide bond.

• Condensation = Water removed.

• Oligo = Few (2-10 amino acids).

Protein Structure

Levels of Structure

• Primary: Linear sequence of amino acids linked by peptide bonds.

• Secondary: α-helix and β-pleated sheet formed by hydrogen bonding.

• Tertiary: Overall 3D folding stabilized by hydrogen bonds, disulfide bridges, van der Waals, electrostatic forces.

• Quaternary: Arrangement of multiple polypeptide subunits.

Solved Examples

• Ex 1: Name two secondary structures.
  Answer: α-helix and β-pleated sheet.

• Ex 2: What stabilizes tertiary structure?
  Answer: Hydrogen bonds, disulfide linkages, van der Waals, electrostatic forces.

Practice Set

• Q1: Primary structure of protein is:
  A. 3D folding
  B. Sequence of amino acids
  C. Multiple subunits
  D. Hydrogen bonding pattern

• Q2: α-helix is stabilized by:
  A. Ionic bonds
  B. Hydrogen bonds
  C. Covalent bonds
  D. Metallic bonds

• Q3: Quaternary structure involves:
  A. Single polypeptide chain
  B. Multiple polypeptide subunits
  C. Peptide bond formation
  D. Amino acid sequence

Answer Key

• Q1: B - Primary structure is amino acid sequence.

• Q2: B - α-helix stabilized by hydrogen bonds.

• Q3: B - Quaternary structure is arrangement of subunits.

Memory Aids

• Primary = Sequence.

• Secondary = α-helix & β-sheet.

• Tertiary = 3D folding.

• Quaternary = Multiple subunits.

Denaturation of Proteins

Concept

• Loss of biological activity due to disruption of secondary and tertiary structures.

• Caused by physical (temperature) or chemical (pH) changes.

• Primary structure remains intact.

• Examples: Coagulation of egg white on boiling, curdling of milk.

Solved Examples

• Ex 1: What happens to protein on boiling?
  Answer: Denaturation; loss of structure and function.

Practice Set

• Q1: Denaturation affects which protein structure?
  A. Primary
  B. Secondary and tertiary
  C. Quaternary only
  D. None

• Q2: Denaturation can be caused by:
  A. Change in temperature
  B. Change in pH
  C. Both A and B
  D. None

• Q3: Which remains unchanged during denaturation?
  A. Primary structure
  B. Secondary structure
  C. Tertiary structure
  D. Biological activity

Answer Key

• Q1: B - Secondary and tertiary structures are affected.

• Q2: C - Both temperature and pH changes cause denaturation.

• Q3: A - Primary structure remains intact.

Memory Aids

• Denaturation = Unfolding, loss of function.

• Primary structure = Peptide bonds intact.

• Heat and pH = Common denaturants.

Enzymes

Overview

• Biological catalysts that speed up biochemical reactions.

• Mostly globular proteins with high specificity for substrates and products.

• Named after the substrate or reaction they catalyze (e.g., maltase hydrolyzes maltose).

• Increase reaction rates significantly without being consumed.

Solved Examples

• Ex 1: Hydrolysis of maltose by maltase

  • Maltose + H2O → 2 Glucose (catalyzed by maltase)

  • Water breaks the glycosidic bond with enzyme assistance.

Practice Set

• Q1: What type of biomolecule are most enzymes?

  • A. Lipids

  • B. Carbohydrates

  • C. Globular proteins

  • D. Nucleic acids

• Q2: The enzyme maltase catalyzes the breakdown of:

  • A. Glucose into maltose

  • B. Maltose into glucose

  • C. Lactose into glucose and galactose

  • D. Sucrose into fructose and glucose

• Q3: Which statement about enzymes is correct?

  • A. They are consumed in the reaction.

  • B. They decrease reaction specificity.

  • C. They increase activation energy.

  • D. They speed up reactions without being consumed.

Answer Key

• Q1: C - Enzymes are mostly globular proteins.

• Q2: B - Maltase hydrolyzes maltose into glucose.

• Q3: D - Enzymes speed up reactions and are not consumed.

Memory Aids

• Enzymes = Globular proteins acting as biological catalysts.

• "-ase" suffix indicates enzyme name related to substrate.

• Hydrolysis reactions often involve water and enzymes like maltase.

Mechanism of Enzyme Action

Key Points

• Enzymes lower activation energy, speeding up reactions.

• Only small amounts of enzyme are needed for reaction progress.

• Example: Activation energy for acid hydrolysis of sucrose is 6.22 kJ mol−1, reduced to 2.15 kJ mol−1 by sucrase.

• Enzyme deficiency can cause diseases (e.g., phenylalanine hydroxylase deficiency causes phenylketonuria).

• Tyrosinase deficiency leads to albinism (reduced melanin pigment).

Solved Examples

• Ex 1: Activation energy reduction by enzyme

  • Without enzyme: 6.22 kJ mol−1

  • With sucrase: 2.15 kJ mol−1

  • Enzyme lowers energy barrier, increasing reaction rate.

Practice Set

• Q1: What is the main effect of enzymes on activation energy?

  • A. Increase it

  • B. Decrease it

  • C. No effect

  • D. Eliminate it

• Q2: Phenylketonuria is caused by deficiency of which enzyme?

  • A. Tyrosinase

  • B. Phenylalanine hydroxylase

  • C. Sucrase

  • D. Maltase

• Q3: Which condition results from tyrosinase deficiency?

  • A. Phenylketonuria

  • B. Albinism

  • C. Diabetes

  • D. Scurvy

Answer Key

• Q1: B - Enzymes decrease activation energy.

• Q2: B - Phenylketonuria is due to phenylalanine hydroxylase deficiency.

• Q3: B - Tyrosinase deficiency causes albinism.

Memory Aids

• Enzymes lower activation energy, not consumed.

• Phenylketonuria = Phenylalanine hydroxylase deficiency.

• Albinism = Tyrosinase deficiency (melanin pigment loss).

Common Enzymes

Important Enzymes and Reactions

• Amylase: Starch → Glucose

• Maltase: Maltose → 2 Glucose

• Lactase: Lactose → Glucose + Fructose

• Pepsin & Trypsin: Proteins → Peptides

• Nucleases: DNA/RNA → Nucleotides

• Urease: Urea → NH3 + CO2

• Carbonic anhydrase: H2CO3 → CO2 + H2O

Solved Examples

• Ex 1: Lactase action

  • Lactose + H2O → Glucose + Fructose (catalyzed by lactase)

• Ex 2: Urease reaction

  • Urea + H2O → NH3 + CO2 (catalyzed by urease)

Practice Set

• Q1: Which enzyme breaks down starch into glucose?

  • A. Maltase

  • B. Amylase

  • C. Lactase

  • D. Pepsin

• Q2: Carbonic anhydrase catalyzes the conversion of:

  • A. CO2 to H2CO3

  • B. H2CO3 to CO2 and H2O

  • C. Urea to NH3 and CO2

  • D. Proteins to peptides

• Q3: Which enzyme breaks down proteins into peptides?

  • A. Trypsin

  • B. Nuclease

  • C. Lactase

  • D. Urease

Answer Key

• Q1: B - Amylase breaks starch into glucose.

• Q2: B - Carbonic anhydrase converts H2CO3 to CO2 and H2O.

• Q3: A - Trypsin breaks proteins into peptides.

Memory Aids

• Amylase = starch; Maltase = maltose; Lactase = lactose.

• Pepsin & Trypsin = protein digestion.

• Urease = urea to ammonia and CO2.

• Carbonic anhydrase = carbonic acid to CO2 and water.

Vitamins

Definition and Importance

• Organic compounds needed in small amounts in diet.

• Essential for specific biological functions and normal growth.

• Cannot be synthesized sufficiently by the body.

Classification of Vitamins

• Water soluble vitamins: Dissolve in water; not stored; require regular intake; excreted in urine (except B12).

• Examples: Vitamin B1 (Thiamine), B2 (Riboflavin), B6 (Pyridoxine), B12 (Cobalamin), C (Ascorbic acid).

• Fat soluble vitamins: Dissolve in fats/oils; stored in liver and adipose tissue.

• Examples: Vitamins A (Retinoids), D, E, K.

Key Facts

• Vitamin deficiency causes avitaminosis.

• Excess fat soluble vitamins cause hypervitaminosis.

• Water soluble vitamins cannot be stored in the body.

• Vitamin H (Biotin) is neither water nor fat soluble.

Solved Examples

• Ex 1: Identify the vitamin deficient in night blindness.
  Solution: Vitamin A deficiency causes night blindness.

• Ex 2: Which vitamin deficiency causes scurvy?
  Solution: Vitamin C deficiency causes scurvy.

Practice Set

• Q1: Which vitamin is stored in the liver and adipose tissue?
  A. Vitamin B1
  B. Vitamin C
  C. Vitamin A
  D. Vitamin B12

• Q2: Which vitamin deficiency leads to pernicious anemia?
  A. Vitamin B12
  B. Vitamin D
  C. Vitamin K
  D. Vitamin E

• Q3: Excess intake of which vitamin can cause hypervitaminosis?
  A. Vitamin C
  B. Vitamin B6
  C. Vitamin A
  D. Vitamin B1

Answer Key

• Q1: C. Vitamin A - It is fat soluble and stored in liver and adipose tissue.

• Q2: A. Vitamin B12 - Deficiency causes pernicious anemia.

• Q3: C. Vitamin A - Excess fat soluble vitamins like A cause hypervitaminosis.

Memory Aids

• Water soluble vitamins: B-complex and C; need regular intake.

• Fat soluble vitamins: A, D, E, K; stored in body fat.

• Avitaminosis = vitamin deficiency diseases.

• Hypervitaminosis = excess fat soluble vitamin toxicity.

Nucleic Acids

Chemical Composition

• Nucleic acids are polymers of nucleotides found in the nucleus of all living cells.

• Complete hydrolysis yields pentose sugar, phosphoric acid, and nitrogenous bases.

• DNA contains β-deoxyribose; RNA contains β-D-ribose.

• DNA bases: Adenine (A), Guanine (G), Cytosine (C), Thymine (T).

• RNA bases: Adenine (A), Guanine (G), Cytosine (C), Uracil (U) replaces thymine.

Nucleosides and Nucleotides

• Nucleoside = Pentose sugar + nitrogenous base.

• Nucleotide = Nucleoside + phosphate group.

• Nucleotides are building blocks of DNA and RNA.

• 

  Phosphate group has one phosphorus atom bonded to four oxygen atoms, carrying a negative charge.

Structure of Nucleic Acids

• Primary structure: Nucleotides linked by phosphodiester bonds between 5' and 3' carbons of sugar.

• Secondary structure: Two polynucleotide strands coil into a double helix.

• Backbone: Sugar-phosphate units; bases face inward and pair via hydrogen bonds.

• Watson and Crick proposed the double helix model for DNA.

• Base pairing: Adenine pairs with Thymine (2 H-bonds), Guanine pairs with Cytosine (3 H-bonds).

• RNA secondary structure: Single strand folds to form double helices.

• Chargaff's rule: Purine = Pyrimidine amount; A = T, G = C in DNA.

Biological Functions

• DNA stores genetic information and transmits heredity.

• Maintains species identity.

• Self-duplicates during cell division (replication).

• RNA synthesizes proteins using DNA's instructions.

• Protein synthesis involves transcription (RNA synthesis) and translation (protein assembly).

• Genetic code: Triplet nucleotide sequences determine amino acid sequence in proteins.

Solved Examples

Ex 1

• Identify sugar in DNA and RNA: DNA has deoxyribose; RNA has ribose.

• Explain base pairing: A pairs with T (DNA) or U (RNA), G pairs with C.

Practice Set

• Q1: Which sugar is present in RNA?

  • A. β-D-deoxyribose

  • B. β-D-ribose

  • C. Glucose

  • D. Fructose

• Q2: How many hydrogen bonds form between guanine and cytosine?

  • A. One

  • B. Two

  • C. Three

  • D. Four

• Q3: Which component is NOT part of a nucleotide?

  • A. Phosphate group

  • B. Nitrogenous base

  • C. Pentose sugar

  • D. Amino acid

Answer Key

• Q1: B. β-D-ribose - RNA contains ribose sugar.

• Q2: C. Three - G and C pair via three hydrogen bonds.

• Q3: D. Amino acid - Not a nucleotide component.

Memory Aids

• RNA has ribose; DNA has deoxyribose (missing oxygen at 2' carbon).

• A pairs with T (or U in RNA), G pairs with C.

• Purine bases (A, G) pair with pyrimidines (T/U, C).

• Nucleotide = sugar + base + phosphate.

Hormones

Types of Hormones

• Steroids: Estrogens and androgens; lipid-soluble molecules derived from cholesterol.

• Polypeptides: Insulin and endorphins; chains of amino acids, water-soluble.

• Amino Acid Derivatives: Epinephrine and norepinephrine; modified single amino acids.

Functions of Hormones

• Act as intercellular messengers secreted directly into the bloodstream by endocrine glands.

• Maintain balance of biological activities in the body.

• Growth hormones and sex hormones regulate growth and development.

• Insulin lowers blood glucose; glucagon raises blood glucose; together regulate glucose homeostasis.

• Epinephrine and norepinephrine mediate responses to external stimuli (fight or flight response).

Solved Examples

• Ex 1: Identify the hormone type: Insulin is a polypeptide hormone because it is a chain of amino acids.

• Ex 2: Explain glucose regulation: Insulin decreases blood glucose by promoting uptake; glucagon increases glucose by stimulating glycogen breakdown.

Practice Set

• Q1: Which hormone is a steroid?

  • A. Insulin

  • B. Estrogen

  • C. Epinephrine

  • D. Endorphin

• Q2: Which hormone increases blood glucose level?

  • A. Insulin

  • B. Glucagon

  • C. Estrogen

  • D. Endorphin

• Q3: Epinephrine and norepinephrine are derived from:

  • A. Cholesterol

  • B. Polypeptides

  • C. Amino acids

  • D. Fatty acids

Answer Key

• Q1: B. Estrogen — It is a steroid hormone derived from cholesterol.

• Q2: B. Glucagon — It raises blood glucose by stimulating glycogen breakdown.

• Q3: C. Amino acids — Epinephrine and norepinephrine are amino acid derivatives.

Memory Aids

• Steroid hormones end with '-one' or '-ol' (e.g., estrogen, androgen).

• Polypeptides include insulin; remember "Insulin is a Protein".

• Amino acid derivatives include epinephrine; think "Epi = amino acid derivative".

• Insulin lowers glucose; glucagon raises glucose — opposite actions maintain balance.