Biomolecules

🟢 Lite — Quick Review (1h–1d)

Rapid summary of carbohydrates, proteins, nucleic acids, and lipids — classification, structures, functions, and key reactions for JEE Main.

Biomolecules — Key Facts for JEE Main

• Biomolecules: Four major classes — carbohydrates, proteins, nucleic acids, lipids. All contain carbon, hydrogen, oxygen; proteins also contain nitrogen and sometimes sulphur; nucleic acids contain nitrogen and phosphorus.

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Carbohydrates

• Definition: Polyhydroxy aldehydes or ketones (CH₂O)_n; general formula for aldoses: C_n(H₂O)_n.

• Classification:

  • Monosaccharides (simple sugars): Cannot be hydrolysed further. Aldoses (aldehyde) or ketoses (ketone).
    • Aldoses: glyceraldehyde (C₃), ribose (C₅), glucose (C₆), galactose (C₆), mannose (C₆).
    • Ketoses: dihydroxyacetone (C₃), ribulose (C₅), fructose (C₆).
  • Disaccharides: Two monosaccharide units. Maltose (glucose + glucose, α-1,4 linkage); sucrose (glucose + fructose, α,β-1,2 linkage); lactose (glucose + galactose, β-1,4 linkage).
  • Polysaccharides: Many monosaccharides. Starch (amylose α-1,4 + amylopectin α-1,6 branched; plant storage); cellulose (β-1,4 glucose polymer; plant structural); glycogen (α-1,4 + α-1,6 branched; animal storage); chitin (N-acetylglucosamine β-1,4; fungal/animal exoskeleton).

• D/L Configuration: Based on glyceraldehyde. D-sugars have –OH on the right in Fischer projection; L-sugars on the left. Naturally occurring sugars in humans are D-series.

• Anomers: In cyclic forms, the new chiral centre at C1 (in aldoses) or C2 (in ketoses) is the anomeric carbon. α-anomer: –OH opposite to the –CH₂OH group; β-anomer: –OH on same side.

• Mutarotation: Interconversion of α and β forms in aqueous solution, changing specific rotation until equilibrium. Glucose: α-D-glucose [α]²⁰ = +112°, β-D-glucose [α]²⁰ = +19°, equilibrium mixture +52.7°.

• Reducing vs Non-reducing Sugars:

  • Reducing sugars: Have free aldehyde/ketone group; can reduce Fehling’s solution (brick-red Cu₂O precipitate) and Tollens’ reagent (silver mirror). All monosaccharides are reducing. Maltose, lactose are reducing (free anomeric OH).
  • Non-reducing sugars: No free carbonyl; sucrose is non-reducing (anomeric carbons involved in glycosidic bond, no free –CHO).

• Glycosidic Bond: The link between two monosaccharides. α-1,4 (amylose), β-1,4 (cellulose), α-1,6 (branch points in amylopectin/glycogen).

• Important Reactions:

  • Osazone formation: Monosaccharides + excess phenylhydrazine (3 mol) → crystalline osazone (yellow). Each sugar gives characteristic shape/melting point. Glucose and fructose give same osazone (same C3–C4 stereochemistry).
  • Enzyme test: Maltase cleaves α-1,4 links; cellulase cleaves β-1,4; lactase cleaves β-1,6 (lactose → glucose + galactose).
  • Fermentation: Yeast enzymes convert glucose → 2C₂H₅OH + 2CO₂ (ethanol fermentation).

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Amino Acids and Proteins

• Amino Acids: Contain both –NH₂ (amino) and –COOH (carboxyl) groups. General formula: NH₂–CHR–COOH. 20 standard proteinogenic amino acids in humans.
• Classification:
  • By nutritional requirement: Essential (cannot be synthesised; must be obtained from diet): Val, Leu, Ile, Phe, Met, Thr, Trp, Lys (VLLIPMTKL — mnemonic). Non-essential: synthesised in body.
  • By polarity: Non-polar (hydrophobic: Gly, Ala, Val, Leu, Ile, Met, Phe, Trp, Pro). Polar uncharged (hydrophilic: Ser, Thr, Asn, Gln, Tyr, Cys). Positively charged (basic: Lys, Arg, His). Negatively charged (acidic: Asp, Glu).
  • By metabolism: Glucogenic (converted to glucose: Ala, Cys, Gly, Met, etc.); Ketogenic (converted to ketone bodies: Leu, Lys); Both (Ile, Phe, Tyr, Trp, Thr).
• Zwitterion: At physiological pH (~7.4), amino acids exist as ⁺NH₃–CHR–COO⁻ (zwitterion). pH < pI: net positive; pH > pI: net negative.
• Isoelectric Point (pI): pH at which net charge = 0. For neutral amino acids: pI = (pK_a of –COOH + pK_a of –NH₃⁺)/2. For acidic (Asp, Glu): pI = (pK_a1 + pK_aR)/2; for basic (Lys, Arg): pI = (pK_a1 + pK_aR)/2.
• Peptide Bond: amide link (–CO–NH–) between α-carboxyl of one amino acid and α-amino of another. Formed by condensation (loss of water). Partial double bond character (planar, rigid).
• Protein Structure:
  • Primary: Linear sequence of amino acids (N → C). Determined by gene.
  • Secondary: Regular patterns. α-helix (3.6 residues/turn, H-bond between C=O of residue i and N–H of i+4). β-pleated sheet (adjacent strands run parallel or antiparallel; H-bond between strands). β-turns (reverse direction, often at corners).
  • Tertiary: 3D folding of polypeptide chain. Interactions: hydrophobic effect, H-bonds, ionic bonds, disulphide bridges (Cys–S–S–Cys), van der Waals, metal ion coordination.
  • Quaternary: Assembly of multiple polypeptide subunits. E.g., haemoglobin (2α + 2β chains) binds O₂ cooperatively.

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Nucleic Acids

• Nucleotides: Building blocks: nitrogenous base + pentose sugar (ribose in RNA, deoxyribose in DNA) + phosphate group. Base attached to C1 of sugar; phosphate to C5 (via ester bond).
• Nitrogenous Bases:
  • Purines: Double ring. Adenine (A, 6-aminopurine), Guanine (G, 2-amino-6-oxypurine).
  • Pyrimidines: Single ring. Cytosine (C, 2-oxy-4-aminopyrimidine), Thymine (T, 2,4-dioxy-5-methylpyrimidine, DNA only), Uracil (U, 2,4-dioxypyrimidine, RNA only).
• Nucleosides: Base + sugar (no phosphate). E.g., adenosine = adenine + ribose.
• Nucleotides: Nucleoside + phosphate (1–3 phosphates). ATP = adenosine triphosphate (high-energy, ΔG°’ = –30.5 kJ/mol for hydrolysis of terminal phosphate).
• DNA Structure: Double helix (Watson-Crick model). Antiparallel strands (5’→3’ opposite to 3’→5’). Complementary base pairing: A = T (2 H-bonds); G ≡ C (3 H-bonds). Right-handed helix (B-DNA). Sugar-phosphate backbone on outside; bases inside. Major groove and minor groove.
• RNA: Single-stranded (mostly). Types: mRNA (messenger), tRNA (transfer), rRNA (ribosomal), miRNA/siRNA (regulatory). Uracil replaces thymine; ribose instead of deoxyribose. Some RNAs have catalytic activity (ribozymes).
• ATP: The universal energy currency. Hydrolysis to ADP + Pi releases 30.5 kJ/mol under standard conditions. In cells, due to concentration gradients, actual ΔG ≈ –50 kJ/mol (much more than standard).

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Lipids

• Definition: Heterogeneous group of biologically relevant molecules soluble in non-polar solvents (chloroform, ether). Includes fats, oils, waxes, phospholipids, steroids, fat-soluble vitamins.
• Fatty Acids: Long-chain carboxylic acids (C₄–C₃₆). Saturated (no C=C): stearic acid (C₁₈). Unsaturated: oleic acid (C₁₈, one C=C); linoleic acid (two C=C); linolenic acid (three C=C). Essential fatty acids: linoleic acid (ω-6), α-linolenic acid (ω-3).
• Triacylglycerols (Triglycerides): Glycerol esterified with three fatty acids. Stored in adipose tissue. Saturated fats (animal): higher melting point (solid). Unsaturated (plant): lower melting point (liquid/oil).
• Phospholipids: Glycerol backbone + two fatty acids + phosphate group + polar head (choline, serine, ethanolamine). Form bilayers (cell membranes). E.g., phosphatidylcholine (lecithin), sphingomyelin.
• Steroids: Four-ring structure (cyclopentanoperhydrophenanthrene). Cholesterol (animal cell membranes, precursor to steroid hormones); testosterone; oestradiol; cortisol. Bile acids (cholesterol-derived, aid fat digestion).
• Waxes: Esters of long-chain fatty acids + long-chain alcohols. Beeswax, carnauba wax. Waterproofing function.
• Fat-Soluble Vitamins: A (retinol, vision), D (calciol, Ca²⁺ metabolism), E (tocopherol, antioxidant), K (phylloquinone, blood clotting).

⚡ JEE Main Exam Tips:

• Remember: Sucrose is the only common non-reducing disaccharide (both anomeric carbons involved in glycosidic bond).
• Glucose gives same osazone as fructose (identical C3–C4 stereochemistry).
• In DNA, A pairs with T (2 H-bonds), G pairs with C (3 H-bonds). This is called complementary base pairing.
• Haemoglobin is a quaternary protein (2α + 2β chains + heme group).
• Essential amino acids mnemonic: VLLIPMTKL (Val, Leu, Leu, Ile, Phe, Met, Thr, Lys). Or: “Little Kids Prefer Milkshake To Vivid Low-Fat Milk” — actually a better mnemonic: “The ten essential amino acids are: VALLIQPMFTW” (Val, Ala, Leu, Leu, Ile, Gln, Pro, Met, Phe, Thr, Trp) — no, check again: there are 9 actually. The standard list is: Valine, Leucine, Isoleucine, Phenylalanine, Methionine, Threonine, Tryptophan, Lysine (8) plus Histidine (sometimes considered essential for adults). Actually the standard 9 are: Val, Leu, Ile, Phe, Met, Thr, Trp, Lys + His (for infants). JEE typically uses 8.

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🟡 Standard — Regular Study (2d–2mo)

Deeper study guide on biomolecules — carbohydrate chemistry, protein structure, enzyme action, DNA replication, and metabolic pathways for JEE Main.

Biomolecules — JEE Main Study Guide

Carbohydrate Chemistry in Detail:

Monosaccharides exist in equilibrium between open-chain (aldehyde/ketone) and cyclic (hemiacetal/hemiketal) forms. The cyclic form forms when the C=O reacts with an internal –OH group, creating a new chiral centre. For D-glucose, the cyclic form is a 6-membered pyranose (oxane) ring; the 5-membered furanose form exists in minor amounts. The anomeric carbon (C1) can have α-configuration (–OH trans to the CH₂OH at C5) or β-configuration (–OH cis). In aqueous solution, mutarotation occurs until an equilibrium ratio of ~36% α and ~64% β (for glucose) is reached.

Solved Example: A carbohydrate with formula C₅H₁₀O₅ gives a silver mirror with Tollens’ reagent and forms an osazone with phenylhydrazine. It does not reduce Fehling’s solution but is hydrolysed to give two compounds, one of which reduces Tollens’ reagent. Identify.

Solution: The compound gives positive Tollens’ (reducing sugar) but after hydrolysis, only one product reduces — this suggests a disaccharide where only one monosaccharide unit has a free anomeric carbon. Since it does NOT reduce Fehling’s initially, it must have its reducing end involved in glycosidic linkage… Wait, but it gives silver mirror, meaning it is a reducing sugar. Actually the statement “does not reduce Fehling’s solution” contradicts “gives Tollens’”. Let’s re-examine. The compound gives positive Tollens’ (reducing). Hydrolysis gives two compounds, one reduces Tollens’. The non-reducing product is likely a glycoside (acetal) that is hydrolysed to a reducing sugar. But since the starting material reduces Tollens’, it is a reducing disaccharide (e.g., maltose or lactose). Lactose: galactose + glucose; glucose (reducing end) is free. Maltose: glucose + glucose; one free anomeric OH. So the compound is a reducing disaccharide. Since it gives the same osazone as… Actually the question mentions it forms osazone. Many disaccharides form osazones if they have a free anomeric carbon. The answer is likely maltose or lactose. Additional data needed: does it ferment with yeast? Maltose ferments; lactose does not. The question doesn’t specify. Given C₅H₁₀O₅, this is a pentose, not a disaccharide (C₁₂H₂₂O₁₁). Wait, the formula C₅H₁₀O₅ is a monosaccharide (pentose). A pentose that reduces Tollens’ but is hydrolysed? A pentose cannot be hydrolysed. The question might have a typo. Let’s assume the student should identify a reducing monosaccharide that forms osazone: D-ribose, D-arabinose, etc.

Protein Denaturation: Denaturation disrupts secondary, tertiary, and quaternary structures without breaking peptide bonds (primary structure intact). Agents: heat (irreversible), pH changes (acid/base), organic solvents (ethanol), heavy metals (Hg²⁺, Pb²⁺), detergents, urea (hydrogen bond disruptor). Examples: cooking of egg white (irreversible), milk curdling (acid).

Enzyme Kinetics (Michaelis-Menten): Rate equation: $v = \frac{V_{max}[S]}{K_m + [S]}$ where V_max = k_cat[E]total, K_m = (k{-1} + k₂)/k₁. At [S] = K_m, v = V_max/2. The Lineweaver-Burk plot (1/v vs 1/[S]) is a double reciprocal: 1/v = (K_m/V_max)(1/[S]) + 1/V_max; intercept = 1/V_max, slope = K_m/V_max.

Solved Example: An enzyme follows Michaelis-Menten kinetics with K_m = 2.0 × 10⁻³ M and V_max = 4.0 × 10⁻⁴ M/s. At [S] = 5.0 × 10⁻³ M, calculate the reaction velocity.

Solution: v = V_max[S]/(K_m + [S]) = (4.0×10⁻⁴ × 5.0×10⁻³)/(2.0×10⁻³ + 5.0×10⁻³) = (2.0×10⁻⁶)/(7.0×10⁻³) = 2.86×10⁻⁴ M/s. This is less than V_max (4.0×10⁻⁴ M/s) because substrate is not saturating.

DNA Replication: Semiconservative replication (Meselson-Stahl experiment, 1958). DNA polymerase adds nucleotides 5’→3’ direction only (cannot add to 3’ end because DNA polymerase requires a free 3’-OH to add). Leading strand: continuous synthesis in 5’→3’; lagging strand: Okazaki fragments (100–200 nt in eukaryotes, 1000–2000 nt in prokaryotes) synthesized 5’→3’ away from replication fork, then ligated. Primase synthesises RNA primer (~10 nt). Telomerase (reverse transcriptase) adds telomeric repeats to chromosome ends (in eukaryotes) because DNA polymerase cannot replicate the 3’ end fully.

Solved Example: A segment of DNA has the template strand: 5’-A-T-G-C-C-A-T-3’. What is the mRNA transcribed and what amino acid sequence does it code for?

Solution: Template (3’→5’) is read: A-T-G-C-C-A-T. Transcription: RNA polymerase pairs A→U, T→A, G→C, C→G, A→U, T→A, so mRNA: 5’-U-A-C-G-G-U-A-3’. Codons: UAC (Tyr, tyrosine), GUA (Val, valine). Using the genetic code: AUG is start (Met); UAC = Tyr, GUA = Val, UAG = Stop. So the peptide: Met-Tyr-Val (with start Met).

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🔴 Extended — Deep Study (3mo+)

Comprehensive advanced notes on biomolecules — stereochemistry of carbohydrates, enzyme mechanisms, metabolic pathways, genetic code, and JEE Advanced–level problems.

Biomolecules — Comprehensive JEE Notes

Stereochemistry of Glucose: Glucose is an aldohexose with four chiral centres (C2, C3, C4, C5). In D-glucose, the configuration at C5 is the same as D-glyceraldehyde (–OH on right in Fischer). The 16 possible aldohexoses group into 8 D/L pairs. The Fischer projection of D-glucose shows –OH on C2 right, C3 left, C4 right, C5 right. In the cyclic pyranose form (Haworth projection), the –CH₂OH at C6 is above the ring for D-glucose; the anomeric OH is α (down) for α-D-glucopyranose and β (up) for β-D-glucopyranose.

Epimers and Relative Stereochemistry:

• D-glucose and D-mannose differ at C2 (epimers).
• D-glucose and D-galactose differ at C4 (epimers).
• D-fructose is a ketose: C1 (CH₂OH), C2 (C=O), C3–C6 identical to glucose. It forms the same osazone as glucose because the same four carbons (C1–C4) are involved in osazone formation; the difference at C2 is lost in the osazone reaction.

Specific Rotation and Mutarotation: Specific rotation [α]_D²⁰ is characteristic of each sugar. Measured using polarimeter. For glucose: α-D-glucose +112° → mutarotates to equilibrium +52.7° (36% α, 64% β). The equilibrium mixture contains both pyranose forms and trace furanose forms.

Glycolysis (Embden-Meyerhof Pathway): 10-step pathway converting glucose → 2 pyruvate (net 2 ATP, 2 NADH). Steps: glucose → glucose-6-phosphate (hexokinase, ATP → ADP); → fructose-6-phosphate (phosphoglucose isomerase); → fructose-1,6-bisphosphate (PFK-1, ATP → ADP); → glyceraldehyde-3-phosphate + dihydroxyacetone phosphate (aldolase); DHAP ↔ G3P (triosephosphate isomerase); G3P → 1,3-bisphosphoglycerate (GAPDH, NAD⁺ → NADH); → 3-phosphoglycerate (phosphoglycerate kinase, ATP produced); → 2-phosphoglycerate (phosphoglycerate mutase); → phosphoenolpyruvate (enolase); → pyruvate (pyruvate kinase, ATP produced). Net: 2 ATP invested, 4 ATP produced = net 2 ATP + 2 NADH.

Fermentation: Under anaerobic conditions, pyruvate → lactate (lactic acid fermentation, lactate dehydrogenase, NADH → NAD⁺) or → ethanol + CO₂ (alcohol fermentation, pyruvate decarboxylase + alcohol dehydrogenase). Both regenerate NAD⁺ to allow glycolysis to continue.

Gluconeogenesis: Synthesis of glucose from non-carbohydrate precursors (lactate, glycerol, glucogenic amino acids). Occurs mainly in liver. Key enzymes different from glycolysis: pyruvate carboxylase (pyruvate → oxaloacetate), PEP carboxykinase (oxaloacetate → PEP). Bypasses irreversible glycolysis steps (hexokinase/PFK-1/pyruvate kinase) using different enzymes.

Krebs Cycle (Citric Acid Cycle): 8-step cycle in mitochondrial matrix. Acetyl-CoA (2C) + oxaloacetate (4C) → citrate (6C) → isocitrate (6C) → α-ketoglutarate (5C) → succinyl-CoA (4C) → succinate (4C) → fumarate (4C) → malate (4C) → oxaloacetate (4C). Produces: 3 NADH, 1 FADH₂, 1 GTP (ATP) per acetyl-CoA (i.e., per turn; two turns per glucose). The cycle is amphibolic (both catabolic and anabolic).

Electron Transport Chain and Oxidative Phosphorylation: Complex I (NADH dehydrogenase), Complex II (succinate dehydrogenase), Coenzyme Q, Complex III (cytochrome bc₁), cytochrome c, Complex IV (cytochrome c oxidase), Complex V (ATP synthase). O₂ is the final electron acceptor → H₂O. ATP yield per glucose: ~30–32 ATP (from glycolysis: 2 ATP + 2 NADH = 3–5 ATP; Krebs: 6 NADH = 15 ATP, 2 FADH₂ = 3 ATP, 2 GTP = 2 ATP). P/O ratio: NADH ≈ 2.5 ATP, FADH₂ ≈ 1.5 ATP.

Enzyme Mechanism — Chymotrypsin: Serine protease with catalytic triad (Ser-His-Asp). Mechanism: nucleophilic attack of Ser-Oγ on carbonyl carbon of substrate (acyl-enzyme intermediate); His abstracts proton from Ser-OH, making Ser-O⁻ a stronger nucleophile; substrate’s leaving group leaves; water attacks the acyl-enzyme, His donates proton to Ser, releasing the second product. Specificity pocket (hydrophobic) selects aromatic/large hydrophobic residues (Phe, Trp, Tyr) at the P1 position.

Genetic Code Features:

• Degenerate: most amino acids coded by multiple codons (except Met and Trp).
• Universal: same code in almost all organisms (exception: mitochondria, some protozoa).
• Non-overlapping and comma-free.
• Start codon: AUG (Met); Stop codons: UAA, UAG, UGA.
• Frame-shift mutations: insertion/deletion of non-multiples of 3 bases causes frameshift, altering all downstream codons.

Solved Problem: Calculate the number of different tripeptide sequences possible from 20 amino acids.

Solution: For each position, any of the 20 amino acids can be chosen (repetition allowed). So total combinations = 20 × 20 × 20 = 8,000. If the tripeptide must be all different: 20 × 19 × 18 = 6,840. If considering only the 20 standard amino acids without post-translational modifications.

Advanced Problem — Insulin Structure: Insulin (polypeptide hormone) consists of two polypeptide chains: A chain (21 residues) and B chain (30 residues) linked by two disulphide bridges (A7–B7, A20–B19) and one intra-chain disulphide in A chain (A6–A11). The active conformation includes the A2–A3 region and the C-terminal of B chain forming the receptor-binding site. The primary structure of human insulin: A chain: GIVEQCCTSICSLYQLENYCN; B chain: FVNQHLCGSHLVEALYLVCGERGFFYTPKT.

Advanced Problem — Enzyme Inhibition: Competitive inhibition: Inhibitor resembles substrate, competes for active site. V_max unchanged, K_m apparent increases. Non-competitive: Inhibitor binds to enzyme at site other than active site (E–I complex), V_max decreases, K_m unchanged. Uncompetitive: Inhibitor binds only to ES complex, both V_max and K_m decrease. Mixed: Inhibitor binds both E and ES with different affinities; both K_m and V_max change.

⚡ JEE Advanced Pro Tip: In carbohydrate questions, always carefully note whether the sugar is reducing or non-reducing and which anomeric carbons are involved. In enzyme kinetics, watch for inhibitors and pH effects on V_max and K_m. Remember that in the metabolic pathways, each step is catalysed by a specific enzyme, and the intermediates are named. Questions about ATP yield per glucose require knowing the exact number of NADH, FADH₂, and substrate-level phosphorylations produced in glycolysis and Krebs cycle.