Carbohydrates are primarily produced by plants and form a very large group of naturally occurring organic compounds. They have a general formula Cx(H2O)y and were originally considered as "hydrates of carbon." Some common examples are glucose, sucrose, starch, cellulose, and glycogen.
Chemical Definition
Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds which produce such units on hydrolysis.
They are also called saccharides (Greek: sakcharon = sugar). Sweet-tasting ones are called sugars.
Note: Acetic acid (CH₃COOH) fits Cx(H2O)y but is NOT a carbohydrate. Rhamnose (C6H12O5) is a carbohydrate but doesn't fit the formula. So the formula alone doesn't define carbohydrates.
10.1.1 Classification of Carbohydrates
Classification of Carbohydrates based on behaviour on hydrolysis
Reducing vs Non-Reducing Sugars
Reducing sugars: Carbohydrates that reduce Fehling's solution (blue → brick-red Cu2O) and Tollens' reagent (silver mirror test). They have a free aldehydic or ketonic group. All monosaccharides (aldose or ketose) are reducing sugars. Examples: glucose, fructose, maltose, lactose.
Non-reducing sugars: Reducing groups (–CHO or C=O) are involved in glycosidic bond — no free group available. Example: sucrose.
Types of Monosaccharides
Carbon Atoms
General Term
Aldehyde Type
Ketone Type
3
Triose
Aldotriose
Ketotriose
4
Tetrose
Aldotetrose
Ketotetrose
5
Pentose
Aldopentose
Ketopentose
6
Hexose
Aldohexose
Ketohexose
7
Heptose
Aldoheptose
Ketoheptose
10.1.2 Monosaccharides
10.1.2.1 Glucose (Dextrose)
Glucose is an aldohexose, also called dextrose. It is the most abundant organic compound on Earth and the monomer of starch and cellulose. Molecular formula: C6H12O6.
Preparation of Glucose
1. From Sucrose (cane sugar):
C12H22O11 + H2O —dil. HCl or H₂SO₄→ C6H12O6 + C6H12O6 Glucose Fructose
Glucose reacts with hydroxylamine (NH2OH) to form an oxime and adds HCN to give a cyanohydrin → confirms the presence of a carbonyl group (C=O).
Evidence 4 – Aldehyde Group
Glucose is oxidised by bromine water (mild oxidising agent) to a six-carbon carboxylic acid (gluconic acid) → confirms the carbonyl is an aldehyde group (–CHO).
Acetylation with acetic anhydride gives glucose pentaacetate → confirms presence of 5 –OH groups, each on a different carbon (since the compound is stable).
Evidence 6 – Primary –OH Group
Oxidation with nitric acid (HNO3) converts both glucose and gluconic acid to a dicarboxylic acid (saccharic acid) → confirms a primary alcoholic –CH2OH group at C6.
Open Chain (Fischer) Structure of Glucose
D-(+)-Glucose (Structure I)
D-(+)-Glucose
–OH at C5 on right → D-config
Key Points — Open Chain
Aldehyde at C1 (top)
4 chiral carbons: C2, C3, C4, C5
Primary –OH at C6 (bottom)
–OH at C5 on right → D series
Rotates plane of polarised light to right → (+) dextrorotatory
Full name: D-(+)-glucose
D and L Notation
D/L Configuration
D and L describe relative configuration compared to glyceraldehyde. They have NO relation to optical rotation (d/l or +/–).
The lowest asymmetric carbon (C5 in glucose) is compared to glyceraldehyde:
–OH on right → D-configuration (like D-(+)-glyceraldehyde)
–OH on left → L-configuration (like L-(–)-glyceraldehyde)
Most naturally occurring monosaccharides are in the D-series.
Cyclic (Haworth) Structure of Glucose
The open-chain structure could not explain three anomalous facts:
Glucose does not give Schiff's test and does not form NaHSO3 addition product (despite having –CHO)
Glucose pentaacetate does not react with hydroxylamine → no free –CHO group
Glucose exists in two crystalline forms: α (m.p. 419 K, from concentrated solution at 303 K) and β (m.p. 423 K, from hot saturated solution at 371 K)
These are explained by a six-membered cyclic hemiacetal structure formed when the –OH at C5 adds to the –CHO group at C1.
KEY CONCEPT
Anomeric Carbon: C1 (the aldehyde carbon before ring closure) becomes a new chiral centre after ring formation. The two forms (α and β) are called anomers.
α-D-glucopyranose: –OH at C1 is on the same side as the ring oxygen (axial/below the ring in Haworth)
β-D-glucopyranose: –OH at C1 is on the opposite side from the ring oxygen (equatorial/above in Haworth)
The ring is called pyranose (6-membered, analogous to pyran). The two anomers exist in equilibrium with the open-chain form in solution (mutarotation).
🔄
α-D-Glucopyranose ⇌ Open Chain ⇌ β-D-Glucopyranose
NCERT p.285
Three structures in equilibrium: α-D-(+)-Glucose (ring, –OH at C1 pointing down/right), open chain structure (with free CHO), and β-D-(+)-Glucose (ring, –OH at C1 pointing up/left). Also shows the Haworth projections of both pyranose forms with all carbons numbered 1–6.
📌 Image Source
NCERT Chemistry Class 12, Chapter 10, Page 285. Cyclic (Haworth) structures of α-D-(+)-Glucopyranose and β-D-(+)-Glucopyranose in equilibrium with open chain.
Pyran (reference ring)
6-membered ring Glucose → pyranose
Furan (reference ring)
5-membered ring Fructose → furanose
10.1.2.2 Fructose
Fructose is an important ketohexose (D-(–)-fructose). Molecular formula: C6H12O6. It contains a ketonic group at C2 and belongs to the D-series. It is laevorotatory (rotates polarised light to the left).
Found in fruits, honey, and vegetables. Obtained along with glucose by hydrolysis of sucrose.
Fructose vs Glucose
Glucose: aldehyde at C1 → aldohexose; D-(+)-glucose (dextrorotatory)
Fructose: ketone at C2 → ketohexose; D-(–)-fructose (laevorotatory)
Fructose forms a five-membered ring (furanose) by C5–OH adding to C2 carbonyl
Both are monosaccharides with same molecular formula C6H12O6 → isomers
Both are reducing sugars (Fehling's and Tollens' positive)
10.1.3 Disaccharides
Disaccharides are formed when two monosaccharide units are joined by a glycosidic linkage — an oxide linkage formed by loss of one water molecule.
Glycosidic Linkage
The linkage between two monosaccharide units through an oxygen atom. If the reducing groups (–CHO or C=O) of both units are involved → non-reducing sugar. If only one is involved → reducing sugar.
Disaccharide
Units
Linkage
Reducing?
Key Fact
Sucrose (cane sugar)
α-D-Glucose + β-D-Fructose
C1 of glucose – C2 of fructose (1,2-glycosidic)
Non-reducing
Both reducing groups involved; dextrorotatory but hydrolysis gives laevorotatory mixture → invert sugar
Maltose (malt sugar)
Two α-D-Glucose units
C1 of glucose(I) – C4 of glucose(II) (α-1,4 glycosidic)
Reducing
Free –CHO can form at C1 of second glucose unit; obtained from starch by enzyme amylase
Lactose (milk sugar)
β-D-Galactose + β-D-Glucose
C1 of galactose – C4 of glucose (β-1,4 glycosidic)
Reducing
Free –CHO can form at C1 of glucose unit; found in milk
⚠️ Invert Sugar
Sucrose is dextrorotatory ([α] = +66.5°). On hydrolysis it gives glucose ([α] = +52.5°) and fructose ([α] = –92.4°). Since fructose's laevorotation is stronger, the mixture is laevorotatory. This change in sign of rotation is called inversion, and the product is called invert sugar. Honey is mainly invert sugar.
🔗
Haworth Structures of Sucrose, Maltose and Lactose
NCERT p.287–288
Three Haworth projection diagrams: (1) Sucrose showing α-D-glucose ring linked via C1–O–C2 to β-D-fructose ring with dashed box around the glycosidic linkage; (2) Maltose showing two α-D-glucose pyranose rings joined by α-1,4 glycosidic linkage, with free –OH at C1 of second unit; (3) Lactose showing β-D-galactose and β-D-glucose rings with β-1,4 linkage.
📌 Image Source
NCERT Chemistry Class 12, Chapter 10, Pages 287–288. Haworth structures of sucrose, maltose, and lactose.
10.1.4 Polysaccharides
Polysaccharides contain a large number of monosaccharide units joined by glycosidic linkages. They are the most commonly encountered carbohydrates in nature, serving as food storage or structural materials. They are not sweet (non-sugars).
Polysaccharide
Source/Location
Monomer
Linkage
Structure
Solubility
Amylose
Starch (15–20%)
α-D-Glucose
C1–C4
Long unbranched chain, 200–1000 units
Water soluble
Amylopectin
Starch (80–85%)
α-D-Glucose
C1–C4 (main chain) + C1–C6 (branch)
Branched chain
Water insoluble
Cellulose
Plants — cell wall (most abundant organic compound on Earth)
β-D-Glucose
C1–C4 (β-glycosidic)
Straight chain, unbranched
Insoluble
Glycogen
Animal body — liver, muscles, brain (animal starch)
α-D-Glucose
C1–C4 (main) + C1–C6 (branch)
Highly branched — more than amylopectin
—
KEY CONCEPT
Why Can't Humans Digest Cellulose?
Starch (α-glycosidic C1–C4 bonds) can be hydrolysed by human digestive enzymes (amylase). Cellulose has β-glycosidic C1–C4 bonds — humans lack the enzyme (cellulase) to break these bonds. Herbivores have bacteria in their gut that can digest cellulose. This structural difference (α vs β glucose) has huge biological consequences.
🌿
Amylose, Amylopectin and Cellulose Chain Structures
NCERT p.288–289
Three structural diagrams: (1) Amylose — repeating glucose units in unbranched chain with C1–C4 α-linkages; (2) Amylopectin — branched structure showing the C1–C4 main chain and C1–C6 branch points; (3) Cellulose — repeating β-D-glucose units with β-1,4 links, showing the alternating orientation of glucose units.
📌 Image Source
NCERT Chemistry Class 12, Chapter 10, Pages 288–289. Structures of amylose, amylopectin, and cellulose.
10.1.5 Importance of Carbohydrates
Energy source: Glucose is the primary metabolic fuel; honey is an instant energy source
Storage: Starch in plants, glycogen in animals
Structural: Cellulose forms plant cell walls; used as wood, cotton, paper
Industrial: Raw material for textiles, paper, lacquers, breweries
Nucleic acids: D-ribose and 2-deoxy-D-ribose are present in RNA and DNA respectively
Combination: Carbohydrates are found combined with proteins (glycoproteins) and lipids (glycolipids) in biosystems
10.2 Proteins
Proteins are the most abundant biomolecules of the living system. The word protein comes from the Greek proteios meaning "primary" or "of prime importance." All proteins are polymers of α-amino acids linked by peptide bonds.
10.2.1 Amino Acids
Amino acids contain both –NH2 (amino) and –COOH (carboxyl) groups. Only α-amino acids (amino group on the α-carbon, adjacent to –COOH) are obtained on hydrolysis of proteins.
General structure of an α-amino acid — α-carbon bears the –NH₂, –COOH, –H, and the variable R (side chain)
Classification of Amino Acids
By Charge (at neutral pH)
Neutral: Equal –NH₂ and –COOH groups (e.g., glycine, alanine)
Basic: More –NH₂ than –COOH (e.g., lysine, arginine)
Acidic: More –COOH than –NH₂ (e.g., aspartic acid, glutamic acid)
By Dietary Requirement
Non-essential: Can be synthesised by the body (e.g., glycine, alanine, serine)
Essential: Cannot be synthesised; must be obtained through diet (e.g., valine, leucine, isoleucine, methionine, phenylalanine, tryptophan, threonine, lysine — 10 total)
Zwitter Ion (Dipolar Ion)
In aqueous solution, the –COOH group loses a proton and the –NH₂ group accepts a proton, giving a dipolar ion called a zwitter ion. This explains why amino acids:
Are colourless crystalline solids with high melting points
Behave like salts rather than amines or acids
Are soluble in water
Show amphoteric behaviour (react with both acids and bases)
R–CH(NH₂)–COOH ⇌ R–CH(NH₃⁺)–COO⁻ (Zwitter ion) Neutral but carries both + and – charges → amphoteric
Optical Activity of Amino Acids
Except glycine (R = H), all other naturally occurring α-amino acids are optically active because the α-carbon is asymmetric. Most naturally occurring amino acids have L-configuration (–NH₂ on the left in Fischer projection).
Some Important Amino Acids
#
Name
R group (side chain)
Symbol
Type
1
Glycine
–H
Gly (G)
Neutral; only non-optically active
2
Alanine
–CH₃
Ala (A)
Neutral
3
Valine*
–CH(CH₃)₂
Val (V)
Neutral, essential
4
Leucine*
–CH₂CH(CH₃)₂
Leu (L)
Neutral, essential
5
Isoleucine*
–CH(CH₃)CH₂CH₃
Ile (I)
Neutral, essential
7
Lysine*
–(CH₂)₄NH₂
Lys (K)
Basic, essential
8
Glutamic acid
–CH₂CH₂COOH
Glu (E)
Acidic
14
Cysteine
–CH₂SH
Cys (C)
Contains –SH; forms disulphide bonds
16
Phenylalanine*
–CH₂C₆H₅
Phe (F)
Aromatic, essential
18
Tryptophan*
indole-CH₂–
Trp (W)
Aromatic, essential
* = essential amino acid (must be obtained from diet). There are 20 standard amino acids total.
Peptide Bond Formation
Two amino acids combine by the reaction between –COOH of one and –NH₂ of another with elimination of water → forming a peptide bond (–CO–NH–).
Formation of a dipeptide by condensation of two amino acids with loss of water — the –CO–NH– bond formed is the peptide (amide) bond
🔗
Peptide Chain Size Terminology
Dipeptide (2 AA) → Tripeptide (3 AA) → Tetrapeptide (4 AA) → Pentapeptide (5 AA) → Polypeptide (>10 AA) → Protein (>100 AA residues, MW > 10,000 u). Note: Insulin (51 AA) is considered a protein because it has a well-defined 3D conformation.
Specific sequence of amino acids in the polypeptide chain. Any change creates a different protein.
Peptide bonds (covalent)
Sequence: Gly–Ala–Val–...
Secondary
Shape of the polypeptide chain. Two types: α-helix (right-handed screw, H-bonds within chain) and β-pleated sheet (extended chains side by side, intermolecular H-bonds)
Hydrogen bonds between C=O and N–H of peptide bonds
α-Keratin (hair) = α-helix; silk = β-sheet
Tertiary
Overall 3D folding of the secondary structure. Gives fibrous or globular shape.
H-bonds, disulphide (–S–S–) links, van der Waals forces, electrostatic forces
Myoglobin
Quaternary
Spatial arrangement of two or more polypeptide chains (subunits) relative to each other.
Same as tertiary
Haemoglobin (4 subunits)
🧬
Four Levels of Protein Structure — Primary, Secondary, Tertiary, Quaternary
NCERT p.293–294 (Figs. 10.1–10.4)
Fig. 10.1: α-Helix structure with dotted H-bonds between C=O and N–H of peptide bonds in adjacent turns. Fig. 10.2: β-Pleated sheet structure with side-by-side polypeptide chains and intermolecular H-bonds. Fig. 10.3: Diagrammatic representation of all four levels (ball-and-stick model). Fig. 10.4: Primary, secondary, tertiary, and quaternary structure of haemoglobin showing 4 subunits.
📌 Image Source
NCERT Chemistry Class 12, Chapter 10, Pages 293–294, Figures 10.1–10.4. All four protein structure levels including α-helix, β-pleated sheet, and haemoglobin quaternary structure.
10.2.4 Denaturation of Proteins
Definition
Denaturation: When a native protein is subjected to physical change (temperature) or chemical change (pH), the hydrogen bonds are disturbed → globules unfold and helices uncoil → protein loses its biological activity.
During denaturation: Secondary and tertiary structures are destroyed. Primary structure remains intact.
📝 Examples of Denaturation
Coagulation of egg white on boiling — irreversible denaturation by heat
Curdling of milk — lactic acid produced by bacteria lowers pH → denaturation of milk proteins
Enzymes are biocatalysts — biological catalysts produced by living cells. Almost all enzymes are globular proteins. They are highly specific for a particular reaction and substrate.
Key Properties of Enzymes
Needed only in small quantities
Act at mild conditions (body temperature, physiological pH)
Highly specific — one enzyme, one reaction/substrate (lock and key model)
Reduce activation energy like chemical catalysts
Named after the substrate or reaction with suffix –ase
📝 Example — Activation Energy Reduction
Acid hydrolysis of sucrose: activation energy = 6.22 kJ mol–1
Enzyme (sucrase) hydrolysis: activation energy = 2.15 kJ mol–1
Mechanism of enzyme action: enzyme has an active site (lock), substrate fits into active site (key), enzyme-substrate complex forms, reaction proceeds at lowered activation energy, products released, and enzyme is regenerated.
🤖 AI Image Prompt
Professional chemistry diagram, fully white background, clean and minimalistic. Lock-and-key model of enzyme action showing: (1) Enzyme with active site (highlighted), (2) Substrate approaching active site, (3) Enzyme-substrate complex formation, (4) Products released, (5) Enzyme regenerated. Labelled arrows for each step. No background patterns, no explanatory paragraphs, just the core mechanism diagram with clean labels.
10.4 Vitamins
Definition
Vitamins are organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health. Their deficiency causes specific diseases.
The term "Vitamine" was coined from vital + amine (early compounds had amino groups). When it was found most do not have amino groups, the 'e' was dropped → vitamin.
10.4.1 Classification of Vitamins
Fat-Soluble Vitamins
Soluble in fat and oils; insoluble in water
Stored in liver and adipose (fat) tissues
Vitamins: A, D, E, K
Memory: ADEK
Water-Soluble Vitamins
Soluble in water; excreted in urine → must be supplied regularly
Cannot be stored in the body (except Vitamin B12)
Vitamins: B group (B₁, B₂, B₆, B₁₂) and Vitamin C
Vitamin
Chemical Name
Sources
Deficiency Disease
A
Retinol
Fish liver oil, carrots, butter, milk
Xerophthalmia (hardening of cornea), Night blindness
B1
Thiamine
Yeast, milk, green vegetables, cereals
Beri-beri (loss of appetite, retarded growth)
B2
Riboflavin
Milk, egg white, liver, kidney
Cheilosis (fissuring at corners of mouth and lips), digestive disorders, burning sensation of skin
B6
Pyridoxine
Yeast, milk, egg yolk, cereals, grams
Convulsions
B12
Cyanocobalamin
Meat, fish, egg, curd
Pernicious anaemia (RBC deficient in haemoglobin)
C
Ascorbic acid
Citrus fruits, amla, green leafy vegetables
Scurvy (bleeding gums)
D
Calciferol
Exposure to sunlight, fish, egg yolk
Rickets (bone deformities in children), Osteomalacia (soft bones, joint pain in adults)
E
Tocopherol
Vegetable oils (wheat germ oil, sunflower oil)
Increased fragility of RBCs, muscular weakness
K
Phylloquinone
Green leafy vegetables
Increased blood clotting time
💡
Quick Disease Mnemonics
A → Ability to see at Night (Night blindness) · B₁ → Beri-Beri · B₂ → Burning lips (Cheilosis) · B₆ → Convulsions · B₁₂ → Blood cells (Pernicious anaemia) · C → Cuts bleed (Scurvy) · D → Deformed bones (Rickets/Osteomalacia) · K → Koagulation (Blood clotting)
10.5 Nucleic Acids
Nucleic acids are long-chain polymers of nucleotides (polynucleotides). They are found in the nucleus of cells in the form of chromosomes and are responsible for heredity (transmission of characteristics from parents to offspring).
Two types: DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).
Function: Carries out protein synthesis in the cell
3 Types: mRNA, rRNA, tRNA
Nitrogenous Bases
Adenine (A)
PURINE
In both DNA & RNA Pairs with: T (DNA), U (RNA)
Guanine (G)
PURINE
In both DNA & RNA Pairs with: C (both)
Cytosine (C)
PYRIMIDINE
In both DNA & RNA Pairs with: G (both)
Thymine (T)
PYRIMIDINE
Only in DNA Pairs with: A (DNA)
Uracil (U)
PYRIMIDINE
Only in RNA Pairs with: A (RNA)
🧠
Memory: Purines are "Pure As Gold" — PURines have 2 rings (A and G). PYrimidines have 1 ring (C, T, U).
In DNA: A=T (2 H-bonds), G≡C (3 H-bonds). In RNA: Uracil (U) replaces Thymine (T). Remember: "DNA Thymine → RNA Uracil."
Nucleoside and Nucleotide
Term
Components
Linkage
Example
Nucleoside
Base + Pentose sugar
Base attached to C1′ of sugar (glycosidic bond)
Adenosine, thymidine, cytidine
Nucleotide
Base + Sugar + Phosphoric acid
Phosphate linked to C5′ of sugar
AMP, ADP, ATP (adenosine mono/di/triphosphate)
Nucleotides are joined together by phosphodiester linkages between the 5′ carbon of one sugar and the 3′ carbon of the next sugar.
🔬
Fig. 10.5 — Structure of a Nucleoside and a Nucleotide
NCERT p.298
Two diagrams: (a) Nucleoside — pentose sugar ring with base attached at C1′ position, showing 1′, 2′, 3′, 4′, 5′ numbering; (b) Nucleotide — same structure with phosphate group (–O–P–O–) attached at C5′ of sugar ring.
📌 Image Source
NCERT Chemistry Class 12, Chapter 10, Fig. 10.5, Page 298.
10.5.2 Structure of Nucleic Acids
KEY CONCEPT — Watson-Crick Double Helix (DNA)
Proposed by James Watson and Francis Crick (1953, Nobel Prize 1962)
Two nucleotide chains wound about each other → right-handed double helix
Held together by H-bonds between complementary base pairs
Base pairing rules: A pairs with T (2 H-bonds); G pairs with C (3 H-bonds)
The two strands are antiparallel (one runs 5′→3′, the other 3′→5′)
The sugar-phosphate backbone forms the rails; base pairs form the "rungs" of the ladder
The two strands are complementary, not identical
🧬
Fig. 10.6 — Formation of a Dinucleotide (Phosphodiester Linkage)
NCERT p.299
Two nucleotides joined: first nucleotide (5′ end, phosphate–sugar–base) links via phosphodiester bond to second nucleotide's 3′-OH forming a dinucleotide. Shows 5′ end at top and 3′ end at bottom, with the phosphodiester bridge clearly marked between 3′-OH of first sugar and 5′-phosphate of second sugar.
📌 Image Source
NCERT Chemistry Class 12, Chapter 10, Fig. 10.6, Page 299. Formation of dinucleotide with phosphodiester linkage.
🔬
Fig. 10.7 — Double Strand Helix Structure of DNA
NCERT p.299
Watson-Crick double helix: two antiparallel strands wound in a right-handed helix. Sugar-phosphate backbone on the outside (rails). Base pairs (A=T, G≡C) in the interior (rungs). H-bonds shown as dotted lines between base pairs. 5′ and 3′ ends labelled on each strand.
📌 Image Source
NCERT Chemistry Class 12, Chapter 10, Fig. 10.7, Page 299. Watson-Crick double helix structure of DNA.
Types of RNA
Type
Full Name
Function
mRNA
Messenger RNA
Carries the genetic message (code) from DNA to ribosomes for protein synthesis
rRNA
Ribosomal RNA
Structural and catalytic component of ribosomes; site of protein synthesis
tRNA
Transfer RNA
Brings specific amino acids to the ribosome during protein synthesis (adaptor molecule)
10.5.3 Biological Functions of Nucleic Acids
DNA: Chemical basis of heredity; stores and transmits genetic information; capable of self-duplication during cell division — identical strands transferred to daughter cells
RNA: Involved in protein synthesis; the message for a particular protein is encoded in DNA but proteins are actually synthesised by RNA molecules in the cell
DNA fingerprinting: Each individual has a unique base sequence in their DNA → used in forensics, paternity determination, identification of accident victims, and tracing racial/evolutionary lineages
10.6 Hormones
Definition
Hormones are molecules that act as intercellular messengers. They are produced by endocrine glands and transported via the bloodstream to the site of action.
Response to stress/external stimuli, metabolic rate
Important Hormone Examples
Key Hormones
Insulin: Released when blood glucose rises → promotes uptake of glucose by cells → lowers blood glucose. Deficiency → diabetes mellitus.
Glucagon: Antagonist to insulin → increases blood glucose level. Together, insulin and glucagon regulate blood glucose within a narrow range.
Thyroxine: Produced by thyroid gland; iodinated derivative of amino acid tyrosine. Low level → hypothyroidism (lethargy, obesity). High level → hyperthyroidism. Iodine deficiency → hypothyroidism + goitre → controlled by using iodised salt.
Epinephrine (adrenaline) and norepinephrine: "Fight or flight" hormones — mediate responses to external stimuli.
Testosterone: Major male sex hormone; responsible for secondary male characteristics (deep voice, facial hair).
Estradiol: Main female sex hormone; secondary female characteristics; controls menstrual cycle.
Progesterone: Prepares uterus for implantation of fertilised egg.
Glucocorticoids: From adrenal cortex; carbohydrate metabolism, modulate inflammation, stress response.
Mineralocorticoids: Control water and salt excretion by kidneys. Deficiency → Addison's disease (hypoglycaemia, weakness, stress susceptibility, fatal if untreated).
✏️ Practice Questions
Q1
Define carbohydrates chemically. Why is acetic acid (CH₃COOH) not classified as a carbohydrate even though it fits the formula Cx(H₂O)y?
Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds which produce such units on hydrolysis. Acetic acid (CH₃COOH) fits C₂(H₂O)₂ but it has neither a polyhydroxy structure nor an aldehyde/ketone group — it is simply a carboxylic acid. It does not show optical activity and does not produce polyhydroxy aldehyde/ketone on hydrolysis. Hence it is NOT a carbohydrate. Carbohydrates are defined by their chemical functional groups and optical activity, not just their molecular formula.
Q2
Write six chemical evidences that establish the open-chain structure of glucose. What are the three facts that this structure CANNOT explain?
Six evidences: (1) Molecular formula C₆H₁₂O₆; (2) HI heating → n-hexane (straight chain of 6 carbons); (3) NH₂OH and HCN reactions → carbonyl (C=O) group; (4) Br₂/H₂O oxidation → gluconic acid (aldehyde group); (5) Acetic anhydride → pentaacetate (5 –OH groups on different C atoms); (6) HNO₃ oxidation → saccharic acid (primary –OH at C6).
Three unexplained facts: (1) Glucose doesn't give Schiff's test and doesn't form NaHSO₃ addition product despite having –CHO; (2) Pentaacetate of glucose doesn't react with hydroxylamine (no free –CHO); (3) Glucose exists in two crystalline forms (α, m.p. 419 K and β, m.p. 423 K). These are explained by the cyclic hemiacetal (pyranose) structure.
Q3
Distinguish between sucrose, maltose and lactose with respect to (a) units they contain (b) type of linkage (c) reducing/non-reducing nature.
Sucrose: (a) α-D-glucose + β-D-fructose; (b) C1 of glucose–C2 of fructose (1,2-glycosidic); (c) Non-reducing — both reducing groups involved in bond.
Maltose: (a) Two α-D-glucose units; (b) C1 of glucose(I)–C4 of glucose(II), α-1,4 glycosidic; (c) Reducing — free –CHO can form at C1 of second glucose.
Lactose: (a) β-D-galactose + β-D-glucose; (b) C1 of galactose–C4 of glucose, β-1,4 glycosidic; (c) Reducing — free –CHO can form at C1 of glucose unit.
Q4
Compare amylose, amylopectin, cellulose and glycogen in terms of monomer, linkage type, branching and biological role.
Amylose: α-D-glucose; C1–C4 α-glycosidic; unbranched (200–1000 units); water-soluble fraction of starch (15–20%); energy storage.
Amylopectin: α-D-glucose; C1–C4 (main chain) + C1–C6 (branch); branched; water-insoluble fraction of starch (80–85%); energy storage.
Cellulose: β-D-glucose; C1–C4 β-glycosidic; unbranched straight chain; structural — cell wall of plants; humans cannot digest (no cellulase).
Glycogen: α-D-glucose; C1–C4 (main) + C1–C6 (branch); highly branched (more than amylopectin); animal starch; energy storage in liver, muscles, brain.
Q5
What is a zwitter ion? Explain the amphoteric behaviour of amino acids using the zwitter ionic form.
A zwitter ion (dipolar ion) is an internally neutralised species where the –COOH group has lost a proton (forms –COO⁻) and the –NH₂ group has accepted a proton (forms –NH₃⁺). The molecule is neutral overall but carries both positive and negative charges: R–CH(NH₃⁺)–COO⁻.
Amphoteric behaviour: In acidic medium (H⁺ added), the carboxylate group accepts H⁺ → R–CH(NH₃⁺)–COOH (cationic form); In basic medium (OH⁻ added), the ammonium group loses H⁺ → R–CH(NH₂)–COO⁻ (anionic form). This ability to react with both acids and bases is called amphoteric behaviour.
Q6
Describe the four levels of protein structure with examples and forces involved at each level.
Primary: Specific sequence of amino acids in polypeptide chain. Force: peptide (covalent) bonds. Change in sequence = different protein.
Secondary: Shape of polypeptide chain — α-helix (H-bonds between C=O and N–H within chain, right-handed screw) or β-pleated sheet (intermolecular H-bonds between extended chains). Force: H-bonds.
Tertiary: Overall 3D folding of secondary structure → fibrous or globular shapes. Forces: H-bonds, disulphide (–S–S–) bonds, van der Waals, electrostatic forces. Example: Myoglobin.
Quaternary: Spatial arrangement of two or more polypeptide chains (subunits). Same forces as tertiary. Example: Haemoglobin (4 subunits — 2α, 2β chains).
Q7
What is denaturation of proteins? How does it affect the different levels of protein structure? Give two common examples.
Denaturation is the loss of biological activity of a native protein when subjected to physical (heat) or chemical (pH change) changes. H-bonds are disrupted → globules unfold, helices uncoil.
Effect: Secondary and tertiary structures are DESTROYED. Primary structure (sequence of amino acids linked by peptide bonds) remains INTACT. The protein loses its specific 3D shape and thus its biological function.
Examples: (1) Coagulation of egg white on boiling — heat denatures egg albumin. (2) Curdling of milk — lactic acid (from bacteria) lowers pH, denaturing milk proteins (casein).
Q8
Tabulate the differences between DNA and RNA under: (a) Sugar present (b) Bases present (c) Structure (d) Function.
Sugar: DNA → β-D-2-deoxyribose (no –OH at C2); RNA → β-D-ribose (–OH at C2).
Bases: DNA → A, G, C, T (Thymine); RNA → A, G, C, U (Uracil; no Thymine).
Structure: DNA → double-stranded helix (Watson-Crick); RNA → single-stranded (sometimes folds).
Function: DNA → stores genetic information; chemical basis of heredity; self-replicates during cell division. RNA → carries out protein synthesis (mRNA: carries message; rRNA: ribosome component; tRNA: transfers amino acids).
Q9
Explain what is meant by "complementary strands" in DNA. Why is this important biologically?
Complementary strands means the two strands in DNA are NOT identical but are matched by specific base-pairing rules: A always pairs with T (2 H-bonds), and G always pairs with C (3 H-bonds). If one strand has the sequence 5′–ATGCTA–3′, the complementary strand reads 3′–TACGAT–5′.
Biological importance: During cell division, DNA unwinds and each strand serves as a template to synthesise a new complementary strand. This ensures that both daughter cells receive an exact copy of the parental DNA — the basis of heredity and faithful transmission of genetic information. This is called semi-conservative replication.
Q10
Classify the following vitamins as fat-soluble or water-soluble and name the deficiency disease: Vitamin A, B₁, C, D, K, B₁₂.
Vitamin A: Fat-soluble (stored in liver/adipose) → Xerophthalmia and Night blindness.
Vitamin B₁ (Thiamine): Water-soluble → Beri-beri.
Vitamin C (Ascorbic acid): Water-soluble → Scurvy (bleeding gums).
Vitamin D (Calciferol): Fat-soluble → Rickets (children) and Osteomalacia (adults).
Vitamin K (Phylloquinone): Fat-soluble → Increased blood clotting time.
Vitamin B₁₂ (Cyanocobalamin): Water-soluble (exception: can be stored in body) → Pernicious anaemia.
Q11
What is the difference between a nucleoside and a nucleotide? What is a phosphodiester linkage?
Nucleoside = Base + Pentose sugar (base attached to C1′ of sugar via glycosidic bond). No phosphate. Examples: adenosine, thymidine.
Nucleotide = Nucleoside + Phosphoric acid (phosphate attached to C5′ of sugar). Examples: AMP, GMP, CMP, TMP.
Phosphodiester linkage: The bond that joins nucleotides in a nucleic acid chain. The phosphate group forms an ester bond with the 3′-OH of one sugar and another ester bond with the 5′-OH of the next sugar — creating a 3′–5′ phosphodiester bridge. This gives the nucleic acid chain directionality (5′ end and 3′ end).
Q12
What is "invert sugar"? Why is honey predominantly laevorotatory even though it contains glucose (dextrorotatory)?
Invert sugar is the equimolar mixture of D-(+)-glucose and D-(–)-fructose obtained by hydrolysis of sucrose. Sucrose is dextrorotatory ([α] = +66.5°). On hydrolysis, glucose ([α] = +52.5°) and fructose ([α] = –92.4°) are formed. Since the laevorotation of fructose (–92.4°) is greater in magnitude than the dextrorotation of glucose (+52.5°), the mixture is net laevorotatory (–39.9°). This change in sign from (+) to (–) is called inversion. Honey is mainly invert sugar, so it is laevorotatory despite containing dextrorotatory glucose.
🎯 Important Exam Points – Quick Reference
CONCEPTCarbohydrates = optically active polyhydroxy aldehydes or ketones or compounds that give such units on hydrolysis. General formula Cx(H₂O)y is necessary but not sufficient.
CONCEPTD/L has NO relation to optical activity (+/–). D vs L is relative configuration (reference: glyceraldehyde). (+) vs (–) is direction of rotation of polarised light.
CONCEPTα-D-Glucopyranose ⇌ Open chain ⇌ β-D-Glucopyranose (mutarotation). Anomeric carbon = C1 of glucose after ring formation. α: –OH same side as ring O; β: –OH opposite.
CONCEPTSucrose = non-reducing (both reducing groups involved in 1,2-glycosidic bond). Maltose, Lactose = reducing sugars (free aldehyde possible).
CONCEPTStarch = α-1,4 (main) + α-1,6 (branch). Cellulose = β-1,4 only (humans can't digest). Glycogen = highly branched, like amylopectin but more branches.
CONCEPTProtein structure: Primary (sequence, peptide bonds) → Secondary (α-helix or β-sheet, H-bonds) → Tertiary (overall 3D folding, disulphide + H-bonds + van der Waals) → Quaternary (subunits arrangement).
CONCEPTDenaturation destroys 2° and 3° structure only. Primary structure (sequence + peptide bonds) remains intact. Protein loses biological activity.
CONCEPTFat-soluble vitamins: A, D, E, K (stored in body). Water-soluble: B group and C (excreted in urine, must be supplied regularly; exception B₁₂ can be stored).
CONCEPTDNA = double-stranded, deoxyribose, bases A-G-C-T. RNA = single-stranded, ribose, bases A-G-C-U (Uracil replaces Thymine). Base pairing: A=T, G≡C (in DNA); A=U (in RNA).
REACTIONGlucose + Br₂/H₂O → Gluconic acid (–CHO oxidised; ketose fructose NOT oxidised by Br₂). Used to distinguish aldose from ketose.
REACTIONSucrose + H₂O (H⁺) → Glucose + Fructose (invert sugar; solution changes from + to – optical rotation).
REACTIONGlucose + HI (prolonged heat) → n-hexane. This proves all 6 carbons are in a straight chain (no branching, no ring permanent structure).
MCQWhich vitamin prevents scurvy? → Vitamin C (Ascorbic acid). Which vitamin is responsible for blood coagulation? → Vitamin K.
MCQPurines (double-ring): Adenine (A) and Guanine (G). Pyrimidines (single-ring): Cytosine (C), Thymine (T, DNA only), Uracil (U, RNA only).
MCQEssential amino acids cannot be synthesised by body → must come from diet. Non-essential amino acids can be synthesised. Glycine is the only achiral (non-optically active) α-amino acid (R = H).
