Per molecule of ATP hydrolysed, the Na⁺/K⁺-ATPase

moves 2 Na⁺ out and 3 K⁺ in

moves 3 Na⁺ in and 2 K⁺ out

exchanges 1 Na⁺ for 1 K⁺, electroneutrally

Phosphofructokinase-1, the committed step of glycolysis, is allosterically activated by

fructose-2,6-bisphosphate and AMP

Which of the following are correct for one turn of the citric acid cycle starting from one acetyl-CoA?

Two molecules of CO₂ are released.

Succinate dehydrogenase is also complex II of the electron transport chain.

Two molecules of CO₂ are released.

O₂ is consumed directly by one of the cycle’s enzymes.

Succinate dehydrogenase is also complex II of the electron transport chain.

Cyanide poisoning stops respiration by inhibiting

cytochrome c oxidase (complex IV)

the F₀ proton channel of ATP synthase

the cytochrome bc₁ complex (complex III)

In the presence of an inhibitor, the Lineweaver–Burk plot of an enzyme gives a line parallel to the uninhibited one. The inhibition is

uncompetitive — Km and Vmax both decrease by the same factor

competitive — Km increases, Vmax unchanged

pure non-competitive — Vmax decreases, Km unchanged

irreversible — the enzyme concentration falls

For a pure non-competitive inhibitor, which of the following are true?

Lineweaver–Burk lines with and without inhibitor meet on the x-axis.

The inhibition is overcome by raising [S] sufficiently.

Biochemistry — Biomolecules, Membranes, Metabolism and Enzyme Kinetics

Section 2 of the GATE Biotechnology (BT) paper is General Biology, and it is three subjects under one heading: biochemistry, microbiology and immunology. This chapter is the biochemistry, the part with the most arithmetic in it, and the second chapter of the section covers microbiology and immunology. The order follows the syllabus: the structure and function of the four classes of biomolecule; biological membranes — channels, pumps, molecular motors and the action potential; the regulation of carbohydrate, lipid, amino acid and nucleotide metabolism, with glycolysis, the citric acid cycle and fatty acid oxidation in detail; photosynthesis, respiration and the electron transport chain; and enzymes — classification, catalytic and regulatory strategies, Michaelis–Menten kinetics and the three kinds of reversible inhibition. GATE numericals here are ATP counts, free energies from reduction potentials, Nernst potentials and inhibited rates, and each has a worked example below.

1. Biomolecules — structure and function

The four classes, their linkage and the structural numbers worth knowing
Class	Monomer and linkage	What GATE asks
Carbohydrates	Monosaccharides joined by glycosidic bonds (α or β)	Starch and glycogen are α(1→4) with α(1→6) branches; cellulose is β(1→4); sucrose (glucose α1→β2 fructose) is non-reducing
Lipids	Fatty acids esterified to glycerol; phospholipids; sterols	Amphipathic phospholipids form bilayers; cis double bonds lower the melting point and raise membrane fluidity; cholesterol buffers fluidity
Proteins	Twenty L-amino acids joined by peptide bonds	The peptide bond is planar and trans; α-helix has 3.6 residues per turn and a 0.54 nm pitch (0.15 nm per residue); β-sheets are parallel or antiparallel; quaternary structure is subunit assembly
Nucleic acids	Nucleotides joined by 3′→5′ phosphodiester bonds	B-DNA is right-handed, about 10.5 bp per turn, 0.34 nm rise per base pair; Chargaff: A = T, G = C; G·C has three hydrogen bonds, so higher GC raises the melting temperature

⚠️ Hyperchromicity runs the other way from what intuition suggests

Stacked bases in double-stranded DNA absorb less UV than free ones, so melting DNA raises its A260 — the hyperchromic effect. Tm is the temperature at the midpoint of that rise, and it increases with GC content and with ionic strength, because cations shield the repulsion between the two phosphate backbones.

2. Biological membranes — channels, pumps, motors and the action potential

The fluid-mosaic membrane is a phospholipid bilayer carrying integral and peripheral proteins. Small nonpolar molecules cross by simple diffusion; ions and polar solutes need proteins. Channels (such as voltage-gated Na⁺ and K⁺ channels and aquaporins) let solutes run down their electrochemical gradient fast and without binding saturation in the usual sense; carriers bind and saturate, like enzymes (the GLUT glucose transporters). Primary active transport spends ATP directly: the Na⁺/K⁺-ATPase moves 3 Na⁺ out and 2 K⁺ in per ATP, which makes it electrogenic. Secondary active transport spends a gradient another pump built — the Na⁺–glucose symporter of intestinal epithelium moves glucose uphill on the Na⁺ gradient; antiporters move two solutes in opposite directions. Molecular motors convert ATP hydrolysis into movement along a track: myosin on actin filaments, kinesin toward the plus end of microtubules and dynein toward the minus end.

The Nernst equation gives the equilibrium potential of one ion: E = (RT/zF) ln(C_out/C_in), which at 37 °C is about (61.5/z) log₁₀(C_out/C_in) mV. For K⁺ with 5 mM outside and 140 mM inside, E_K = 61.5 × log₁₀(5/140) = 61.5 × (−1.447) ≈ −89 mV. The resting potential of a neuron (about −70 mV) sits near E_K because the resting membrane is most permeable to K⁺. In an action potential, voltage-gated Na⁺ channels open and Na⁺ rushes in (depolarisation toward E_Na, which is positive); Na⁺ channels then inactivate and delayed K⁺ channels open, so K⁺ leaves (repolarisation, often overshooting into hyperpolarisation). Inactivation of the Na⁺ channels is what produces the refractory period and makes conduction one-way.

⚠️ The sign of the Nernst potential follows the ratio outside-over-inside

Write C_out/C_in, never the reverse, and the sign takes care of itself: K⁺ is concentrated inside, so the log is negative and E_K is negative; Na⁺ and Ca²⁺ are concentrated outside, so their potentials are positive. For Ca²⁺ remember z = 2, which halves the 61.5 mV factor.

3. Metabolism and its regulation — glycolysis, the citric acid cycle and fatty acid oxidation

Glycolysis splits glucose into 2 pyruvate in the cytosol, with a net yield of 2 ATP and 2 NADH. Its three irreversible steps are its control points — hexokinase, phosphofructokinase-1 (PFK-1) and pyruvate kinase — and PFK-1 is the committed, most tightly regulated one: activated by AMP and fructose-2,6-bisphosphate, inhibited by ATP and citrate. Gluconeogenesis reverses glycolysis by bypassing exactly those three steps (pyruvate carboxylase with PEP carboxykinase, fructose-1,6-bisphosphatase, glucose-6-phosphatase), and fructose-2,6-bisphosphate regulates the pair reciprocally so the two pathways do not run at once. Pyruvate dehydrogenase converts pyruvate to acetyl-CoA, CO₂ and NADH in the mitochondrion. Each acetyl-CoA through the citric acid cycle yields 3 NADH, 1 FADH₂, 1 GTP (or ATP) and 2 CO₂, with control at citrate synthase, isocitrate dehydrogenase and α-ketoglutarate dehydrogenase. Complete oxidation of one glucose gives about 30–32 ATP, depending on the shuttle that carries cytosolic NADH into the mitochondrion.

ℹ️ Worked: the ATP yield of palmitate

Palmitate (C16) is activated to palmitoyl-CoA at a cost of 2 ATP equivalents (ATP → AMP + PPᵢ), enters the mitochondrion on the carnitine shuttle, and undergoes 7 rounds of β-oxidation, giving 8 acetyl-CoA, 7 FADH₂ and 7 NADH. With 10 ATP per acetyl-CoA through the cycle, 1.5 per FADH₂ and 2.5 per NADH: 80 + 10.5 + 17.5 = 108, minus 2 for activation = 106 ATP. The trap is counting 8 rounds for 8 acetyl-CoA — the last cleavage releases two acetyl-CoA at once.

Lipid metabolism is regulated at acetyl-CoA carboxylase, the committed step of fatty acid synthesis: citrate activates it, palmitoyl-CoA and AMP-kinase phosphorylation inhibit it. Its product malonyl-CoA inhibits carnitine acyltransferase I, so synthesis and oxidation do not run together.
Amino acids lose their nitrogen by transamination (pyridoxal phosphate enzymes) and oxidative deamination of glutamate; the nitrogen leaves as urea, and the urea cycle is controlled at carbamoyl phosphate synthetase I, which needs N-acetylglutamate. Carbon skeletons are glucogenic or ketogenic.
Nucleotides are made de novo (purines built on PRPP up to IMP; pyrimidines as a ring then attached to PRPP) or by salvage. In bacteria aspartate transcarbamoylase is the classic allosteric enzyme — CTP inhibits it, ATP activates it — and ribonucleotide reductase makes all four deoxyribonucleotides under allosteric control.

4. Photosynthesis, respiration and the electron transport chain

In the light reactions of the thylakoid, photosystem II (reaction centre P680) oxidises water at its oxygen-evolving complex, releasing O₂; electrons pass through plastoquinone, the cytochrome b₆f complex and plastocyanin to photosystem I (P700), then to ferredoxin and ferredoxin–NADP⁺ reductase. This non-cyclic Z-scheme makes NADPH and, through the proton gradient, ATP; cyclic flow around PSI alone makes ATP only, with no NADPH and no O₂. The Calvin cycle in the stroma fixes CO₂ with RuBisCO at a cost of 3 ATP and 2 NADPH per CO₂, so 18 ATP and 12 NADPH per hexose. RuBisCO also accepts O₂ (photorespiration); C4 plants concentrate CO₂ around it with PEP carboxylase in mesophyll cells and decarboxylation in bundle-sheath cells, and CAM plants separate the two steps in time instead of space.

The mitochondrial electron transport chain and its classic inhibitors
Component	Role	Inhibitor
Complex I (NADH dehydrogenase)	NADH → ubiquinone; pumps protons	Rotenone
Complex II (succinate dehydrogenase)	FADH₂ from succinate → ubiquinone; pumps no protons	Malonate (competitive, at the succinate site)
Complex III (cytochrome bc₁)	Ubiquinol → cytochrome c; pumps protons (Q cycle)	Antimycin A
Complex IV (cytochrome c oxidase)	Cytochrome c → O₂, the terminal acceptor; pumps protons	Cyanide, azide, carbon monoxide
ATP synthase (F₀F₁)	Proton flow back through F₀ drives ATP synthesis in F₁	Oligomycin; uncouplers such as 2,4-dinitrophenol dissipate the gradient instead

The free energy of an electron transfer follows from the difference in standard reduction potentials: ΔG°′ = −nFΔE°′. From NADH (E°′ = −0.320 V) to O₂ (E°′ for ½O₂/H₂O = +0.816 V), ΔE°′ = 1.136 V and n = 2, so ΔG°′ = −2 × 96.485 kJ mol⁻¹ V⁻¹ × 1.136 V ≈ −219 kJ/mol — enough, when coupled through the proton gradient, for about 2.5 ATP (the P/O ratio of NADH; 1.5 for FADH₂). Uncouplers let electron flow and O₂ consumption continue while ATP synthesis stops, releasing the energy as heat — the principle of thermogenin in brown fat.

5. Enzymes — classification, strategies, Michaelis–Menten kinetics and inhibition

The Enzyme Commission numbers seven classes: 1 oxidoreductases, 2 transferases, 3 hydrolases, 4 lyases, 5 isomerases, 6 ligases and 7 translocases, the last added in 2018 for enzymes that move ions or molecules across membranes. Enzymes speed reactions by lowering the activation energy, never by changing ΔG or the equilibrium constant, and they use four catalytic strategies: covalent catalysis (the serine of chymotrypsin’s Ser–His–Asp triad), general acid–base catalysis, metal-ion catalysis and catalysis by approximation (holding substrates together in the right orientation). Their activity is regulated by four regulatory strategies: allosteric control, isozymes, reversible covalent modification (phosphorylation above all) and proteolytic activation of zymogens (trypsinogen to trypsin).

Michaelis–Menten kinetics, from the steady-state assumption on the ES complex, gives v = Vmax[S]/(Km + [S]). At [S] = Km, v = Vmax/2; at [S] = 3Km, v = 0.75 Vmax; at [S] ≫ Km the enzyme is saturated and the rate is zero order in substrate. kcat = Vmax/[E]t is the turnover number, and kcat/Km is the specificity constant, bounded above by diffusion at roughly 10⁸–10⁹ M⁻¹s⁻¹. The Lineweaver–Burk plot linearises the equation as 1/v = (Km/Vmax)(1/[S]) + 1/Vmax: slope Km/Vmax, y-intercept 1/Vmax, x-intercept −1/Km.

Reversible inhibition, with α = 1 + [I]/Kᵢ
Type	Inhibitor binds	Apparent Km and Vmax	Lineweaver–Burk lines
Competitive	Free enzyme, at the active site	Km rises to αKm; Vmax unchanged — overcome by high [S]	Meet on the y-axis
Uncompetitive	ES complex only	Both Km and Vmax fall by the same factor	Parallel
Non-competitive (pure)	E and ES with equal affinity, away from the active site	Vmax falls to Vmax/α; Km unchanged	Meet on the x-axis at −1/Km

ℹ️ Worked: a competitively inhibited rate

Vmax = 100 µmol/min, Km = 2 mM, [S] = 6 mM, a competitive inhibitor at [I] = 2 mM with Kᵢ = 1 mM. α = 1 + 2/1 = 3, so the apparent Km is 6 mM and v = 100 × 6/(6 + 6) = 50 µmol/min, against 75 µmol/min uninhibited. Applying α to Vmax instead would treat the inhibitor as non-competitive.

Key takeaways

B-DNA: about 10.5 bp per turn and 0.34 nm per base pair; α-helix: 3.6 residues per turn and 0.54 nm pitch; melting DNA raises A260.
The Na⁺/K⁺-ATPase moves 3 Na⁺ out and 2 K⁺ in per ATP; E = (61.5/z) log₁₀(C_out/C_in) mV at 37 °C, about −89 mV for K⁺ at 5/140 mM.
Glycolysis nets 2 ATP and 2 NADH with PFK-1 as the key control; each acetyl-CoA gives 3 NADH, 1 FADH₂, 1 GTP; palmitate nets 106 ATP after 7 rounds of β-oxidation.
ΔG°′ = −nFΔE°′: NADH to O₂ is about −219 kJ/mol; the Calvin cycle spends 3 ATP and 2 NADPH per CO₂; cyanide blocks complex IV, rotenone complex I, oligomycin ATP synthase.
Competitive inhibition raises Km only (lines meet on the y-axis); uncompetitive lowers both (parallel lines); pure non-competitive lowers Vmax only (lines meet on the x-axis).

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

In B-DNA, the axial rise per base pair is approximately
1. 0.34 nm
2. 3.4 nm
3. 0.15 nm
4. 0.54 nm
Show answer
Answer: A — 0.34 nm
Adjacent base pairs in B-DNA are stacked 0.34 nm apart, and with about 10.5 bp per turn one helical turn is roughly 3.4–3.6 nm — which is where the 3.4 nm distractor comes from. 0.15 nm per residue and 0.54 nm pitch belong to the protein α-helix.
At 37 °C, [K⁺] is 5 mM outside and 140 mM inside a cell. Taking E = 61.5 log₁₀(C_out/C_in) mV, the equilibrium potential for K⁺ is ______ mV (round off to the nearest integer).
Numerical answer — type the value.
Show answer
Answer: -89
log₁₀(5/140) = log₁₀(0.0357) = −1.447, and 61.5 × (−1.447) = −89.0 mV. The sign is negative because K⁺ is concentrated inside; inverting the ratio to C_in/C_out gives +89 mV, the most common slip.
Per molecule of ATP hydrolysed, the Na⁺/K⁺-ATPase
1. moves 3 Na⁺ out and 2 K⁺ in
2. moves 2 Na⁺ out and 3 K⁺ in
3. moves 3 Na⁺ in and 2 K⁺ out
4. exchanges 1 Na⁺ for 1 K⁺, electroneutrally
Show answer
Answer: A — moves 3 Na⁺ out and 2 K⁺ in
The pump exports 3 Na⁺ and imports 2 K⁺ per cycle, so it moves one net positive charge outward each time and is electrogenic. That Na⁺ gradient is what secondary active transporters such as the Na⁺–glucose symporter then spend; the reversed and electroneutral options misstate both direction and stoichiometry.
Taking P/O ratios of 2.5 for NADH and 1.5 for FADH₂, and 10 ATP per acetyl-CoA oxidised in the citric acid cycle, the net number of ATP formed from the complete oxidation of one molecule of palmitate (C16) is ______.
Numerical answer — type the value.
Show answer
Answer: 106
Seven rounds of β-oxidation give 8 acetyl-CoA, 7 FADH₂ and 7 NADH: 8 × 10 + 7 × 1.5 + 7 × 2.5 = 80 + 10.5 + 17.5 = 108 ATP. Activation to palmitoyl-CoA costs 2 ATP equivalents (ATP → AMP), leaving 106. Answering 108 forgets activation; counting 8 rounds overstates both FADH₂ and NADH.
Phosphofructokinase-1, the committed step of glycolysis, is allosterically activated by
1. fructose-2,6-bisphosphate and AMP
2. ATP and citrate
3. glucose-6-phosphate and NADH
4. acetyl-CoA and alanine
Show answer
Answer: A — fructose-2,6-bisphosphate and AMP
PFK-1 responds to energy charge: AMP (low energy) and fructose-2,6-bisphosphate (the hormonal signal to run glycolysis) activate it, while ATP and citrate (energy and biosynthetic precursors are plentiful) inhibit it. The ATP-and-citrate option lists the inhibitors, not the activators.
Which of the following are correct for one turn of the citric acid cycle starting from one acetyl-CoA?
1. Three NADH are produced.
2. Two molecules of CO₂ are released.
3. O₂ is consumed directly by one of the cycle’s enzymes.
4. Succinate dehydrogenase is also complex II of the electron transport chain.
Show answer
Answer: A — Three NADH are produced.; B — Two molecules of CO₂ are released.; D — Succinate dehydrogenase is also complex II of the electron transport chain.
Each turn yields 3 NADH, 1 FADH₂, 1 GTP and 2 CO₂, and succinate dehydrogenase is the one cycle enzyme embedded in the inner membrane as complex II. The cycle uses no O₂ itself — it needs O₂ only indirectly, because the ETC must reoxidise its NADH and FADH₂.
Given E°′ = −0.320 V for NAD⁺/NADH and E°′ = +0.816 V for ½O₂/H₂O, and F = 96.485 kJ mol⁻¹ V⁻¹, the magnitude of ΔG°′ for the oxidation of NADH by O₂ is ______ kJ/mol (round off to the nearest integer).
Numerical answer — type the value.
Show answer
Answer: 219
ΔE°′ = 0.816 − (−0.320) = 1.136 V and two electrons are transferred, so ΔG°′ = −nFΔE°′ = −2 × 96.485 × 1.136 = −219.2 kJ/mol, magnitude 219. Using n = 1 halves the answer to about 110, and subtracting the potentials the wrong way gives 0.496 V.
Cyanide poisoning stops respiration by inhibiting
1. cytochrome c oxidase (complex IV)
2. NADH dehydrogenase (complex I)
3. the F₀ proton channel of ATP synthase
4. the cytochrome bc₁ complex (complex III)
Show answer
Answer: A — cytochrome c oxidase (complex IV)
Cyanide binds the ferric haem a₃ of cytochrome c oxidase and blocks electron transfer to O₂, as do azide and carbon monoxide. Rotenone acts on complex I, antimycin A on complex III and oligomycin on the F₀ channel — each distractor is a real inhibitor attached to the wrong site.
In the presence of an inhibitor, the Lineweaver–Burk plot of an enzyme gives a line parallel to the uninhibited one. The inhibition is
1. uncompetitive — Km and Vmax both decrease by the same factor
2. competitive — Km increases, Vmax unchanged
3. pure non-competitive — Vmax decreases, Km unchanged
4. irreversible — the enzyme concentration falls
Show answer
Answer: A — uncompetitive — Km and Vmax both decrease by the same factor
The slope of the Lineweaver–Burk line is Km/Vmax. An uncompetitive inhibitor divides both Km and Vmax by α′, so the slope is unchanged and the line shifts up parallel. Competitive inhibition pivots the line about the y-intercept and pure non-competitive pivots it about the x-intercept, so neither keeps the slope.
An enzyme has Vmax = 100 µmol/min and Km = 2 mM. At [S] = 6 mM in the presence of 2 mM of a competitive inhibitor with Kᵢ = 1 mM, the reaction rate is ______ µmol/min.
Numerical answer — type the value.
Show answer
Answer: 50
α = 1 + [I]/Kᵢ = 1 + 2/1 = 3, so the apparent Km = 3 × 2 = 6 mM and Vmax is unchanged: v = 100 × 6/(6 + 6) = 50 µmol/min. Without inhibitor v = 100 × 6/8 = 75; dividing Vmax by α instead gives 25, which is the non-competitive answer.
For a pure non-competitive inhibitor, which of the following are true?
1. The apparent Vmax decreases.
2. The apparent Km is unchanged.
3. Lineweaver–Burk lines with and without inhibitor meet on the x-axis.
4. The inhibition is overcome by raising [S] sufficiently.
Show answer
Answer: A — The apparent Vmax decreases.; B — The apparent Km is unchanged.; C — Lineweaver–Burk lines with and without inhibitor meet on the x-axis.
Binding E and ES equally removes a fraction of active enzyme whatever the substrate level, so Vmax falls to Vmax/α while Km is unchanged, and the lines share the x-intercept −1/Km. Because the inhibitor does not compete for the active site, raising [S] cannot displace it — that property belongs to competitive inhibition.
To fix one molecule of CO₂ into carbohydrate, the Calvin cycle consumes
1. 3 ATP and 2 NADPH
2. 2 ATP and 3 NADPH
3. 18 ATP and 12 NADPH
4. 1 ATP and 1 NADPH
Show answer
Answer: A — 3 ATP and 2 NADPH
Per CO₂, reducing two 3-phosphoglycerates costs 2 ATP and 2 NADPH and regenerating ribulose-1,5-bisphosphate costs 1 more ATP, so 3 ATP and 2 NADPH. The 18 ATP and 12 NADPH option is the cost of a whole hexose, six CO₂, which is the usual confusion of per-CO₂ with per-glucose.
Enzymes that cleave bonds by the addition of water belong to EC class
1. 3 — hydrolases
2. 4 — lyases
3. 2 — transferases
4. 6 — ligases
Show answer
Answer: A — 3 — hydrolases
Hydrolases (EC 3) such as proteases, lipases and phosphatases use water to cleave a bond. Lyases (EC 4) also break bonds but without water or oxidation, often forming a double bond, which is why they are the tempting wrong choice; ligases (EC 6) join molecules using ATP.