Chapter 19: Biomolecules

Carbohydrates

Introduction

• Optically active polyhydroxy aldehydes or ketones, or compounds yielding these on hydrolysis.
• General formula: C_x(H₂O)_y, considered hydrates of carbon.
• Examples: glucose, sucrose, starch, lactose.
• 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 (e.g., sucrose, maltose).
  • Polysaccharides: Many monosaccharide units (100+), e.g., starch, cellulose.
• Based on reducing properties:
  • Reducing sugars: Contain free aldehyde or ketone groups; reduce Fehling's and Tollens' reagents (e.g., maltose, lactose).
  • Non-reducing sugars: No free aldehyde/ketone groups; do not reduce Fehling's or Tollens' reagents (e.g., sucrose).

Monosaccharides

• Classified by number of carbons and functional group:
  • Aldoses: Contain aldehyde group.
  • Ketoses: Contain ketone group.
  • Examples: trioses, pentoses, hexoses.
• Glucose (C₆H₁₂O₆)
  • Aldohexose: 6 carbons, aldehyde group.
  • Found in fruits, honey, grapes.
  • Preparation:
    • From sucrose by acid hydrolysis (dilute HCl/H₂SO₄): glucose + fructose.
    • From starch by acid hydrolysis (dilute H₂SO₄, 393 K, pressure): glucose.
  • Structure:
    • Open chain confirmed by reactions with HI, NH₂OH, HCN, bromine water, acetic anhydride, nitric acid.
    • Fischer gave exact spatial arrangement of -OH groups.
    • Open chain fails to explain absence of 2,4-DNP, Schiff’s test, and hydrogensulphite addition.
    • Exists as cyclic hemiacetal (pyranose) forms: α and β anomers differ at anomeric carbon (C-1).
    • Haworth projection used to represent cyclic forms.
  • Other reactions:
    • Reduction to D-sorbitol (Na-Hg).
    • Oxidation to gluconic acid (Fehling’s, Tollens’).
    • Derivatives: glucose pentaacetate, pentachloroglucose, methylglucosides.
• Fructose (C₆H₁₂O₆)
  • Ketohexose: 6 carbons, ketone group at C-2.
  • Obtained with glucose by hydrolysis of sucrose.
  • Natural sweetener in fruits, honey, vegetables.
  • Exists in two cyclic furanose forms (five-membered rings): α and β anomers.
  • Haworth structures represent cyclic forms.

Disaccharides

• Two monosaccharides linked by glycosidic bond (oxygen bridge) formed by loss of water.
• Sucrose (C₁₂H₂₂O₁₁)
  • Hydrolysis yields equimolar D-(+) glucose and D-(-) fructose.
  • Hydrolysis changes optical rotation from dextrorotatory to laevorotatory (invert sugar).
  • Glycosidic linkage between C-1 of glucose and C-2 of fructose.
• Maltose (C₁₂H₂₂O₁₁)
  • Two α-D-glucose units linked by α-1,4 glycosidic bond.
  • Has free aldehyde group; reducing sugar.
• Lactose (C₁₂H₂₂O₁₁)
  • β-D-galactose and β-D-glucose linked by β-1,4 glycosidic bond.
  • Found in milk.

Polysaccharides

• Starch
  • Main plant storage polysaccharide.
  • Polymer of α-D-glucose units.
  • Two components: amylose (linear, α-1,4 linkages) and amylopectin (branched, α-1,4 and α-1,6 linkages).
  • Amylose is water soluble, 15-20% of starch.
• Cellulose (C₆H₁₀O₅)_n
  • Linear polymer of β-D-glucose units linked by β-1,4 glycosidic bonds.
  • Major component of plant cell walls, provides strength.
• Glycogen (C₆H₁₀O₅)_n
  • Animal storage polysaccharide, similar to amylopectin but more branched.
  • Found in liver, muscles, brain, yeast, fungi.
  • Broken down enzymatically to glucose on demand.

Importance of Carbohydrates

• Essential for life of plants and animals.
• Major food source and energy storage (starch in plants, glycogen in animals).
• Cell walls of plants and bacteria contain cellulose.
• Wood and cotton fibers are cellulose-based.
• Nucleic acids contain ribose sugars.
• Most natural sugars are D-form.
• Bacteria convert sucrose to dextran polysaccharide.
• Mercerisation of cotton with NaOH imparts lustre.

Solved Examples

• Ex 1: Hydrolysis of sucrose with dilute acid yields glucose and fructose in equal amounts.
• Ex 2: Hydrolysis of starch with dilute H₂SO₄ at 393 K under pressure yields glucose.
• Ex 3: Identify reducing sugar: Maltose has free aldehyde group, so it is reducing; sucrose is non-reducing.

Practice Set

• Q1: Which of the following is a reducing sugar?
  • A. Sucrose
  • B. Maltose
  • C. Sucrose and Maltose
  • D. None
• Q2: Glucose is classified as:
  • A. Ketohexose
  • B. Aldohexose
  • C. Aldopentose
  • D. Ketopentose
• Q3: Which glycosidic linkage is present in cellulose?
  • A. α-1,4
  • B. β-1,4
  • C. α-1,6
  • D. β-1,6

Answer Key

• Q1: B - Maltose has free aldehyde group, so it is a reducing sugar; sucrose is non-reducing.
• Q2: B - Glucose is an aldohexose (6 carbons, aldehyde group).
• Q3: B - Cellulose has β-1,4 glycosidic linkages.

Memory Aids

• Carbohydrates = Carbon + Water (C_x(H₂O)_y).
• Reducing sugars have free aldehyde/ketone groups; non-reducing sugars do not.
• Glucose = Aldohexose; Fructose = Ketohexose.
• Starch = Amylose (linear α-1,4) + Amylopectin (branched α-1,4 & α-1,6).
• Cellulose = β-1,4 linked glucose units.
• Glycogen = Highly branched α-D-glucose polymer.

Proteins

Overview

• Proteins are polymers of -amino acids linked by peptide bonds.
• Derived from Greek word "Proteins" meaning primary or of prime importance.
• Main sources: milk, cheese, pulses, peanuts, fish, meat.
• Essential for structure, function, growth, and maintenance of the body.

Solved Examples

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

Practice Set

• Q1: What is the monomeric unit of proteins?
  A. Fatty acids
  B. -amino acids
  C. Monosaccharides
  D. Nucleotides
• Q2: Which of the following is NOT a source of protein?
  A. Milk
  B. Cheese
  C. Glucose
  D. Meat
• Q3: Proteins are primarily responsible for:
  A. Energy storage
  B. Structure and function of life
  C. Genetic information
  D. Photosynthesis

Answer Key

• Q1: B - Proteins are polymers of -amino acids.
• Q2: C - Glucose is a carbohydrate, not a protein source.
• Q3: B - Proteins provide structure and function in living organisms.

Memory Aids

• Protein = Primary importance (Greek origin).
• Sources: Milk, Meat, Pulses, Fish.
• Monomer = -amino acids.

Amino Acids

Structure and Classification

• Bifunctional molecules with acidic carboxyl (—COOH) and basic amino (—NH₂) groups attached to the same carbon (alpha carbon).
• General structure:
• Side chain (R) varies, defining different amino acids.
• Classification by groups:
  • Neutral: Equal —COOH and —NH₂ groups (e.g., glycine, alanine).
  • Acidic: More —COOH than —NH₂ (e.g., aspartic acid).
  • Basic: More —NH₂ than —COOH (e.g., lysine).
• Classification by synthesis:
  • Essential: Not synthesized in body, must be dietary (e.g., valine, leucine).
  • Non-essential: Synthesized in body (e.g., glycine, alanine).

Properties

• Colorless, crystalline, water-soluble, high melting solids.
• Behave like salts due to acidic and basic groups.
• Form salts with acids and bases; reactions similar to primary amines and carboxylic acids.
• Zwitterion form in aqueous solution: dipolar ion with both + and - charges but overall neutral.
• Isoelectric point: pH where zwitterion is neutral and amino acid has least solubility.
• Except glycine, all -amino acids are optically active due to chiral carbon.
• L-form amino acids have —NH₂ group on left side.

Solved Examples

• Ex 1: Classify alanine based on acidic/basic groups.
  Answer: Neutral amino acid (equal —COOH and —NH₂).
• Ex 2: What is the zwitterion form?
  Answer: Dipolar ion with + and - charges, overall neutral.

Practice Set

• Q1: Which group is acidic in amino acids?
  A. —NH₂
  B. —COOH
  C. —R
  D. —H
• Q2: Amino acids that cannot be synthesized in the body are called:
  A. Neutral
  B. Essential
  C. Basic
  D. Non-essential
• Q3: At isoelectric point, amino acid:
  A. Has net positive charge
  B. Has net negative charge
  C. Exists as zwitterion with no net charge
  D. Is insoluble in organic solvents

Answer Key

• Q1: B - Carboxyl group is acidic.
• Q2: B - Essential amino acids must be obtained from diet.
• Q3: C - Zwitterion form with equal + and - charges.

Memory Aids

• Acidic group = —COOH, Basic group = —NH₂.
• Essential amino acids = must eat (dietary).
• Zwitterion = both charges, overall neutral.

Peptide Bond

Formation and Properties

• 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.
• Example:
• Peptides:
  • Dipeptides, tripeptides, tetrapeptides based on number of amino acids.
  • Oligopeptides: 2-10 amino acids.
  • Polypeptides: >10 amino acids.
  • Proteins: Polypeptides with >100 amino acids and molecular mass >10,000 u.

Solved Examples

• Ex 1: What is released during peptide bond formation?
  Answer: Water molecule (H₂O).
• Ex 2: Define polypeptide.
  Answer: Chain of >10 amino acids linked by peptide bonds.

Practice Set

• Q1: Peptide bond is formed between:
  A. Two amino groups
  B. Two carboxyl groups
  C. —COOH and —NH₂
  D. Side chains
• Q2: What is the product of two amino acids linked by peptide bond?
  A. Protein
  B. Peptide
  C. Monomer
  D. Salt
• Q3: Polypeptides contain how many amino acids?
  A. 2-10
  B. >10
  C. >100
  D. 1

Answer Key

• Q1: C - Peptide bond forms between —COOH and —NH₂.
• Q2: B - Peptides are products of amino acid linkage.
• Q3: B - Polypeptides have more than 10 amino acids.

Memory Aids

• Peptide bond = amide bond + water loss.
• Peptides: di-, tri-, oligo-, poly- based on amino acid count.
• Proteins = polypeptides >100 amino acids.

Protein Structure

Levels of Structure

• Primary: Linear sequence of amino acids linked by peptide bonds.
• Secondary: -helix and -pleated sheet formed by hydrogen bonding.
  • -helix: Right-handed coil stabilized by H-bonds within chain.
  • -pleated sheet: Chains stretched and aligned side-by-side, stabilized by intermolecular H-bonds.
• Tertiary: Overall 3D folding of polypeptide chain stabilized by H-bonds, disulphide, van der Waals, electrostatic forces.
• Quaternary: Spatial arrangement of multiple polypeptide subunits.
• Protein types:
  • Fibrous: Parallel chains, insoluble, e.g., keratin, myosin.
  • Globular: Coiled chains, soluble, e.g., insulin, albumin.

Solved Examples

• Ex 1: Name two types of secondary structures.
  Answer: -helix and -pleated sheet.
• Ex 2: What stabilizes tertiary structure?
  Answer: Hydrogen bonds, disulphide linkages, van der Waals, electrostatic forces.

Practice Set

• Q1: Primary structure of protein is:
  A. 3D folding
  B. Sequence of amino acids
  C. Multiple polypeptide chains
  D. -helix formation
• Q2: -helix is stabilized by:
  A. Ionic bonds
  B. Hydrogen bonds
  C. Covalent bonds
  D. Peptide bonds
• Q3: Fibrous proteins are:
  A. Soluble
  B. Insoluble
  C. Spherical
  D. Enzymes

Answer Key

• Q1: B - Primary structure is amino acid sequence.
• Q2: B - -helix stabilized by hydrogen bonds.
• Q3: B - Fibrous proteins are insoluble.

Memory Aids

• Primary = sequence, Secondary = -helix/-sheet, Tertiary = folding, Quaternary = subunits.
• Fibrous = insoluble, Globular = soluble.
• -helix = right-handed coil.

Denaturation

Concept and Effects

• Denaturation: Loss of protein's native structure due to physical (temperature) or chemical (pH) changes.
• Disrupts hydrogen bonds, unfolds globules, uncoils helices.
• Primary structure remains intact.
• Examples: Coagulation of egg white on boiling, curdling of milk, cheese formation.
• Denatured proteins lose biological activity.
• Some proteins act as antibodies defending against foreign organisms.
• Change in amino acid sequence can cause diseases (e.g., sickle cell anemia: glutamic acid replaced by valine in hemoglobin).

Solved Examples

• Ex 1: What happens to protein on boiling?
  Answer: Denaturation, loss of structure and function.
• Ex 2: Disease caused by amino acid substitution in hemoglobin?
  Answer: Sickle cell anemia.

Practice Set

• Q1: Denaturation affects which protein structure?
  A. Primary
  B. Secondary and tertiary
  C. Quaternary only
  D. None
• Q2: Which causes denaturation?
  A. Cooling
  B. Boiling
  C. Neutral pH
  D. Normal conditions
• Q3: Sickle cell anemia is caused by:
  A. Deletion of amino acid
  B. Substitution of glutamic acid by valine
  C. Addition of amino acid
  D. Protein folding

Answer Key

• Q1: B - Denaturation changes secondary and tertiary structures.
• Q2: B - Boiling causes denaturation.
• Q3: B - Sickle cell anemia caused by glutamic acid to valine substitution.

Memory Aids

• Denaturation = loss of shape, primary intact.
• Heat and pH changes cause denaturation.
• Sickle cell = Glu  Val substitution.

Enzymes

Overview

• Biological catalysts that speed up biochemical reactions.
• Mostly globular proteins.
• Increase reaction rates significantly.
• Show high specificity for substrates and products.
• Named after the substrate or reaction they catalyze (e.g., maltase hydrolyzes maltose).

Explanation of Maltose Hydrolysis

• Maltose consists of two glucose units.
• Water (H₂O) is added during the reaction.
• Maltase enzyme catalyzes the breakdown.
• Maltose splits into two glucose molecules.
• This reaction is called hydrolysis (breaking with water).

Solved Examples

• Ex 1: Write the reaction catalyzed by maltase.
  • Maltose + H₂O 1865 glucose + glucose (catalyzed by maltase)

Practice Set

• Q1: What type of biomolecule are most enzymes?
  • A. Carbohydrates
  • B. Lipids
  • C. Globular proteins
  • D. Nucleic acids
• Q2: Which enzyme catalyzes the hydrolysis of maltose?
  • A. Amylase
  • B. Maltase
  • C. Lactase
  • D. Sucrase
• Q3: Hydrolysis of maltose produces:
  • A. Two fructose molecules
  • B. Two glucose molecules
  • C. Glucose and fructose
  • D. Glucose and galactose

Answer Key

• Q1: C - Enzymes are mostly globular proteins that act as biological catalysts.
• Q2: B - Maltase specifically catalyzes maltose hydrolysis.
• Q3: B - Maltose hydrolysis yields two glucose molecules.

Memory Aids

• Enzymes end with '-ase' named after their substrate (e.g., maltase for maltose).
• Hydrolysis means breaking bonds using water.
• Most enzymes are globular proteins.

Mechanism of Enzyme Action

Key Points

• Enzymes are required in small amounts.
• They lower activation energy of reactions.
• Example: Acid hydrolysis of sucrose requires 6.22 kJ mol^-1, but with sucrase enzyme, only 2.15 kJ mol^-1.
• Deficiency of enzymes causes diseases:
  • Phenylalanine hydroxylase deficiency causes phenylketonuria.
  • Tyrosinase deficiency causes albinism (reduced melanin pigment).

Solved Examples

• Ex 1: Calculate the reduction in activation energy when sucrose hydrolysis is catalyzed by sucrase.
  • Activation energy without enzyme = 6.22 kJ mol^-1
  • With enzyme = 2.15 kJ mol^-1
  • Reduction = 6.22 - 2.15 = 4.07 kJ mol^-1

Practice Set

• Q1: What is the main role of enzymes in reactions?
  • A. Increase activation energy
  • B. Decrease activation energy
  • C. Change equilibrium constant
  • D. Get consumed in reaction
• Q2: Deficiency of which enzyme causes phenylketonuria?
  • A. Tyrosinase
  • B. Phenylalanine hydroxylase
  • C. Sucrase
  • D. Maltase
• Q3: Activation energy for acid hydrolysis of sucrose is 6.22 kJ mol^-1. What is it when catalyzed by sucrase?
  • A. 6.22 kJ mol^-1
  • B. 4.07 kJ mol^-1
  • C. 2.15 kJ mol^-1
  • D. 0 kJ mol^-1

Answer Key

• Q1: B - Enzymes lower activation energy to speed up reactions.
• Q2: B - Phenylalanine hydroxylase deficiency causes phenylketonuria.
• Q3: C - Sucrase reduces activation energy to 2.15 kJ mol^-1.

Memory Aids

• Enzymes lower activation energy, not change equilibrium.
• Phenylketonuria = phenylalanine hydroxylase deficiency.
• Albinism = tyrosinase deficiency.

Common Enzymes

Important Enzymes and Their Reactions

• Amylase: Starch 1865 glucose.
• Maltase: Maltose 1865 2 glucose molecules.
• Lactase: Lactose 1865 glucose + fructose.
• Pepsin and Trypsin: Proteins 1865 smaller peptides.
• Nucleases: DNA/RNA 1865 nucleotides.
• Urease: Urea 1865 NH₃ + CO₂.
• Carbonic anhydrase: H₂CO₃ 1865 CO₂ + H₂O.

Solved Examples

• Ex 1: Identify the enzyme that breaks down starch.
  • Answer: Amylase.
• Ex 2: What are the products of lactase action on lactose?
  • Answer: Glucose and fructose.

Practice Set

• Q1: Which enzyme breaks down proteins into smaller peptides?
  • A. Amylase
  • B. Pepsin
  • C. Lactase
  • D. Urease
• Q2: Urease catalyzes the breakdown of:
  • A. Urea
  • B. Starch
  • C. DNA
  • D. Lactose
• Q3: Carbonic anhydrase converts carbonic acid into:
  • A. Oxygen and water
  • B. Carbon dioxide and water
  • C. Ammonia and carbon dioxide
  • D. Glucose and fructose

Answer Key

• Q1: B - Pepsin breaks down proteins into peptides.
• Q2: A - Urease breaks down urea into NH₃ and CO₂.
• Q3: B - Carbonic anhydrase converts H₂CO₃ into CO₂ and H₂O.

Memory Aids

• Amylase for starch, maltase for maltose, lactase for lactose.
• Pepsin and trypsin digest proteins.
• Urease breaks urea; carbonic anhydrase acts on carbonic acid.

Vitamins

Definition and Importance

• Organic compounds needed in small amounts in diet.
• Essential for specific biological functions and normal growth.
• Maintain optimum health of the organism.

Solved Examples

• Ex 1: Identify the role of vitamins in the body.
  • Answer: Vitamins act as coenzymes or precursors for enzymes essential in metabolism.

Practice Set

• Q1: Vitamins are required in the diet in
  • A. Large amounts
  • B. Small amounts
  • C. No amounts
  • D. Variable amounts
• Q2: Vitamins mainly function as
  • A. Energy sources
  • B. Structural components
  • C. Coenzymes or precursors
  • D. Hormones
• Q3: Which of the following is NOT a vitamin?
  • A. Vitamin C
  • B. Vitamin B12
  • C. Glucose
  • D. Vitamin A

Answer Key

• Q1: B - Vitamins are needed in small amounts for biological functions.
• Q2: C - Vitamins mainly act as coenzymes or their precursors.
• Q3: C - Glucose is a carbohydrate, not a vitamin.

Memory Aids

• Remember: "Vitamins = Vital amines" (historical term).
• Small amounts, big impact.

Classification of Vitamins

Types and Characteristics

• Water Soluble Vitamins:
  • Soluble in water.
  • Not stored in body (except B₁₂).
  • Regular supply needed as excess excreted in urine.
  • Examples: Vitamin B₁, B₂, B₆, B₁₂, C.
• Fat Soluble Vitamins:
  • Soluble in fats and oils.
  • Stored in liver and adipose tissues.
  • Examples: Vitamin A, D, E, K.
• Vitamin H (Biotin) is neither water nor fat soluble.

Solved Examples

• Ex 1: Classify Vitamin C.
  • Answer: Water soluble vitamin.
• Ex 2: Why must water soluble vitamins be supplied regularly?
  • Answer: They are excreted in urine and not stored.

Practice Set

• Q1: Which vitamin is fat soluble?
  • A. Vitamin B₁
  • B. Vitamin C
  • C. Vitamin A
  • D. Vitamin B₁₂
• Q2: Water soluble vitamins are
  • A. Stored in liver
  • B. Excreted in urine
  • C. Stored in adipose tissue
  • D. Not required in diet
• Q3: Vitamin H (Biotin) is
  • A. Water soluble
  • B. Fat soluble
  • C. Neither water nor fat soluble
  • D. Both water and fat soluble

Answer Key

• Q1: C - Vitamin A is fat soluble.
• Q2: B - Water soluble vitamins are excreted in urine.
• Q3: C - Vitamin H is neither water nor fat soluble.

Memory Aids

• Water soluble = B-complex + C, excreted in urine.
• Fat soluble = A, D, E, K, stored in fat.
• Vitamin H exception: neither soluble.

Vitamin Table

Overview

• Vitamins listed with chemical names, solubility, sources, and deficiency diseases.
• Sources include milk, eggs, vegetables, cereals.
• Deficiency causes diseases like night blindness (A), scurvy (C), beriberi (B1).

Solved Examples

• Ex 1: Identify deficiency disease of Vitamin C.
  • Answer: Scurvy.
• Ex 2: Name source of Vitamin B₁.
  • Answer: Cereals and pulses.

Practice Set

• Q1: Deficiency of Vitamin A causes
  • A. Night blindness
  • B. Scurvy
  • C. Beriberi
  • D. Rickets
• Q2: Vitamin B₁ is also called
  • A. Riboflavin
  • B. Thiamine
  • C. Niacin
  • D. Ascorbic acid
• Q3: Which vitamin deficiency causes rickets?
  • A. Vitamin D
  • B. Vitamin E
  • C. Vitamin K
  • D. Vitamin C

Answer Key

• Q1: A - Vitamin A deficiency causes night blindness.
• Q2: B - Thiamine is Vitamin B₁.
• Q3: A - Vitamin D deficiency causes rickets.

Memory Aids

• Vitamin A = Night blindness; D = Rickets.
• B₁ = Thiamine, B₂ = Riboflavin.
• Scurvy = Vitamin C deficiency.

Vitamin Deficiency and Storage

Key Points

• Vitamin deficiency condition is called avitaminosis.
• Water soluble vitamins are not stored; excess excreted in urine.
• Excess fat soluble vitamins cause hypervitaminosis (toxicity).

Solved Examples

• Ex 1: What is avitaminosis?
  • Answer: Disease caused by vitamin deficiency.
• Ex 2: Why is hypervitaminosis caused by fat soluble vitamins?
  • Answer: Because they are stored in liver and fat tissues, excess accumulates.

Practice Set

• Q1: Avitaminosis means
  • A. Excess vitamins
  • B. Vitamin deficiency
  • C. Normal vitamin level
  • D. Vitamin overdose
• Q2: Which vitamins cause hypervitaminosis?
  • A. Water soluble
  • B. Fat soluble
  • C. Both
  • D. None
• Q3: Water soluble vitamins are
  • A. Stored in body
  • B. Excreted in urine
  • C. Cause toxicity on excess
  • D. Stored in fat

Answer Key

• Q1: B - Avitaminosis is vitamin deficiency.
• Q2: B - Fat soluble vitamins cause hypervitaminosis.
• Q3: B - Water soluble vitamins are excreted in urine.

Memory Aids

• Avitaminosis = "A" for absence of vitamins.
• Fat soluble vitamins stored = risk of toxicity.
• Water soluble vitamins = no storage, no toxicity.

Nucleic Acids

Chemical Composition

• Nucleic acids are polymers of nucleotides found in the nucleus of living cells.
• Complete hydrolysis yields pentose sugar, phosphoric acid, and nitrogenous bases.
• DNA sugar: β-deoxyribose; RNA sugar: β-D-ribose.
• DNA bases: Adenine (A), Guanine (G), Cytosine (C), Thymine (T).
• RNA bases: Adenine (A), Guanine (G), Cytosine (C), Uracil (U) instead of Thymine.

Nucleosides and Nucleotides

• Nucleoside = Pentose sugar + Nitrogenous base.
• Nucleotide = Nucleoside + Phosphate group.
• Nucleotides are building blocks of DNA and RNA.

Structure of Nucleic Acids

• Primary structure: Nucleotides linked by phosphodiester bonds between 5' and 3' carbons.
• Secondary structure: Two polynucleotide strands form a double helix.
• Backbone: Sugar-phosphate units; bases face inward and pair via hydrogen bonds.
• Watson and Crick proposed double helix model for DNA.
• Base pairing: A pairs with T (DNA), A pairs with U (RNA), G pairs with C.
• RNA secondary structure: Single strands fold to form double helices.
• Chargaff's rule: Purines = Pyrimidines; A=T (2 H-bonds), G≡C (3 H-bonds).

Biological Functions

• DNA stores genetic information and transmits heredity.
• Maintains species identity.
• Self-duplicates during cell division (replication).
• Protein synthesis directed by RNA; DNA contains the message.
• Replication: DNA produces two identical copies.
• Protein synthesis steps: Transcription (RNA synthesis) and Translation (protein synthesis in cytoplasm).
• Genetic code: Triplet nucleotide sequences determine amino acid sequence in proteins.

Solved Examples

• Ex 1: Identify sugar in DNA and RNA.
  Answer: DNA has β-deoxyribose; RNA has β-D-ribose.
• Ex 2: State base pairing in DNA.
  Answer: A pairs with T (2 H-bonds), G pairs with C (3 H-bonds).
• Ex 3: Define replication.
  Answer: Process where DNA makes two identical copies.

Practice Set

• Q1: Which sugar is present in RNA?
  A. β-deoxyribose
  B. β-D-ribose
  C. Glucose
  D. Fructose
• Q2: Which base is found in RNA but not in DNA?
  A. Thymine
  B. Uracil
  C. Adenine
  D. Guanine
• Q3: How many hydrogen bonds form between guanine and cytosine?
  A. One
  B. Two
  C. Three
  D. Four

Answer Key

• Q1: B. β-D-ribose - RNA contains ribose sugar.
• Q2: B. Uracil - Uracil replaces thymine in RNA.
• Q3: C. Three - G and C form three hydrogen bonds.

Memory Aids

• RNA sugar has an extra -OH group at 2' carbon compared to DNA.
• Base pairs: A-T (2 H-bonds), G-C (3 H-bonds), A-U in RNA.
• Chargaff's rule: Purines = Pyrimidines.

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; derived from single amino acids.

Functions

• Act as intercellular messengers secreted directly into the bloodstream by endocrine glands.
• Maintain balance of biological activities in the body.
• Growth and sex hormones regulate growth and development.
• Insulin lowers blood glucose; glucagon raises blood glucose; together regulate blood sugar levels.
• Epinephrine and norepinephrine mediate responses to external stimuli.

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: What is the main function of glucagon?
  • A. Decrease blood glucose
  • B. Increase blood glucose
  • C. Mediate stress response
  • D. Promote growth
• Q3: Which hormones are amino acid derivatives?
  • A. Insulin and endorphins
  • B. Estrogens and androgens
  • C. Epinephrine and norepinephrine
  • D. Growth hormone and insulin

Answer Key

• Q1: B. Estrogen — It is a steroid hormone derived from cholesterol.
• Q2: B. Increase blood glucose — Glucagon raises blood glucose by glycogen breakdown.
• Q3: C. Epinephrine and norepinephrine — These are amino acid derivative hormones.

Memory Aids

• Steroids: Think "-one" endings like estrogen, androgen.
• Polypeptides: Insulin is a classic example, remember as protein-based.
• Amino acid derivatives: Epinephrine and norepinephrine respond to stress.
• Insulin and glucagon work as a pair to regulate blood glucose.