Metabolism Practice Questions

This set is for students who already know the outline of respiration and photosynthesis and want to test deeper understanding: carbon bookkeeping, redox, ATP yields, thermodynamics and the logic of metabolic control. It’s aimed at the end of high school (advanced courses) and first-year university. Most questions need some calculation. Work carefully, show units, and then compare with the answer key. For review, see glycolysis, the Krebs cycle and the electron transport chain.

Data: R = 8.314 J K⁻¹ mol⁻¹; F = 96,485 C mol⁻¹; ΔG°′(ATP hydrolysis) = −30.5 kJ mol⁻¹.

Questions

Q1. Glycolysis converts one glucose into two pyruvate. (a) How many ATP are used and how many are produced? What is the net gain? (b) How many NADH are produced? (c) Why is it described as “substrate-level phosphorylation”?

Q2. Trace the six carbon atoms of glucose through aerobic respiration. At which stages are they released as CO₂, and how many at each stage?

Q3. Per molecule of glucose, the Krebs cycle turns twice. Give the total yield per glucose of NADH, FADH₂ and ATP (or GTP) from the Krebs cycle alone.

Q4. Using modern estimates of about 2.5 ATP per NADH and 1.5 ATP per FADH₂ via oxidative phosphorylation, calculate the maximum ATP yield per glucose. Assume cytoplasmic NADH is transferred into mitochondria without loss. Show your working stage by stage.

Q5. The electron transport chain transfers electrons from NADH (E°′ = −0.32 V) to O₂ (E°′ = +0.82 V). (a) Calculate ΔE°′. (b) Calculate ΔG°′ for the transfer of two electrons, using ΔG°′ = −nFΔE°′. (c) If 2.5 ATP are made per NADH, what percentage of this energy is conserved in ATP (under standard conditions)?

Q6. Explain how the proton gradient across the inner mitochondrial membrane is used to make ATP. What would happen to ATP synthesis if the membrane became leaky to protons? Name a natural situation where cells do this deliberately.

Q7. The standard Gibbs energy of the phosphofructokinase reaction is about −14 kJ mol⁻¹, while that of the phosphoglucose isomerase reaction is +1.7 kJ mol⁻¹. Explain why the isomerase reaction still proceeds in the forward direction in cells, and why phosphofructokinase is a key control point.

Q8. For a reaction with ΔG°′ = +1.7 kJ mol⁻¹ at 310 K, calculate the equilibrium constant K′.

Q9. Palmitic acid (C₁₆H₃₂O₂) is completely oxidised: C₁₆H₃₂O₂ + 23O₂ → 16CO₂ + 16H₂O (a) Calculate the respiratory quotient (RQ = CO₂ produced / O₂ consumed). (b) Compare it with the RQ for glucose and explain the difference in terms of oxidation state. (c) β-Oxidation of palmitate gives 8 acetyl-CoA, 7 NADH and 7 FADH₂. Activation costs the equivalent of 2 ATP. Calculate the net ATP yield using the ratios in Q4 and 10 ATP per acetyl-CoA from the Krebs cycle and oxidative phosphorylation.

Q10. Compare ATP per carbon atom for glucose (use your answer to Q4) and palmitate (use Q9c). Explain why fat is the better long-term store per gram.

Q11. Explain why a person can’t make glucose from fatty acids (with an even number of carbons), but can make it from glycerol and from many amino acids.

Q12. In photosynthesis, the Calvin cycle needs 3 ATP and 2 NADPH per CO₂ fixed. (a) How many ATP and NADPH are needed to make one glucose? (b) Explain why the light-dependent reactions must include both cyclic and non-cyclic electron flow, given that non-cyclic flow produces only about 2.6 ATP per 2 NADPH.

Answer key

A1. (a) 2 ATP used (hexokinase and phosphofructokinase), 4 ATP produced (two per three-carbon half), net 2 ATP. (b) 2 NADH (from glyceraldehyde-3-phosphate dehydrogenase). (c) ATP is made by directly transferring a phosphate group from a high-energy substrate (1,3-bisphosphoglycerate and phosphoenolpyruvate) to ADP, not via a proton gradient.

A2. Glycolysis: 0 CO₂ (6 C → 2 × 3 C pyruvate). Link reaction: 2 CO₂ (one from each pyruvate, leaving 2 acetyl groups). Krebs cycle: 4 CO₂ (two per turn, two turns). Total = 6. Strictly, the two carbons lost in each turn aren’t the same carbons that entered in that turn’s acetyl group, but the accounting balances over time.

A3. Per turn: 3 NADH, 1 FADH₂, 1 ATP (GTP). Per glucose (two turns): 6 NADH, 2 FADH₂, 2 ATP.

A4.

Stage	ATP directly	NADH	FADH₂
Glycolysis	2	2	0
Link reaction	0	2	0
Krebs cycle	2	6	2
Total	4	10	2

From oxidative phosphorylation: 10 × 2.5 + 2 × 1.5 = 25 + 3 = 28. Total = 4 + 28 = 32 ATP per glucose. (If cytoplasmic NADH enters via a shuttle that passes electrons to FAD, those two NADH give 1.5 ATP each, and the total falls to 30.) Older textbooks quote 36–38, based on older ratios of 3 and 2.

A5. (a) ΔE°′ = +0.82 − (−0.32) = +1.14 V. (b) ΔG°′ = −2 × 96,485 × 1.14 = −219,986 J mol⁻¹ ≈ −220 kJ mol⁻¹. (c) Energy in 2.5 ATP = 2.5 × 30.5 = 76.3 kJ mol⁻¹. Percentage = 76.3 / 220 × 100 ≈ 35 % (higher in real cellular conditions).

A6. Complexes I, III and IV use the energy of electron transfer to pump H⁺ from the matrix into the intermembrane space, creating a concentration and electrical gradient (the proton-motive force). Protons flow back through ATP synthase, whose rotating part drives the enzyme to join ADP and phosphate (chemiosmosis — Peter Mitchell, Nobel Prize 1978). If the membrane leaked protons, the gradient would collapse: electron transport and O₂ use would continue (even speed up), but ATP synthesis would fall and the energy would be released as heat. Brown fat does this deliberately using uncoupling protein 1 (thermogenin), to generate heat in newborns and hibernating animals.

A7. The actual Gibbs energy depends on concentrations: ΔG = ΔG°′ + RT ln Q. In cells, fructose-6-phosphate is removed quickly by the next enzyme, keeping Q small, so ΔG for the isomerase is close to zero or slightly negative — the reaction is near equilibrium and runs forwards. Phosphofructokinase has a large negative ΔG in cells, making it essentially irreversible; such steps are where control is exerted. PFK is inhibited by high ATP and citrate (signals of plenty) and activated by AMP (a signal of low energy) — a classic example of allosteric regulation (see enzyme inhibition).

A8. ln K′ = −ΔG°′ / RT = −1,700 / (8.314 × 310) = −0.660. K′ = e^(−0.660) ≈ 0.52. The equilibrium slightly favours reactants under standard conditions, which is why removing product matters.

A9. (a) RQ = 16/23 = 0.70. (b) For glucose, C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, RQ = 6/6 = 1.0. Fatty acids contain much less oxygen relative to carbon and hydrogen — their carbons are more reduced — so more O₂ is needed per CO₂ produced. Measuring RQ from breath gas tells physiologists what mix of fuels a person is burning. (c) 8 acetyl-CoA × 10 = 80; 7 NADH × 2.5 = 17.5; 7 FADH₂ × 1.5 = 10.5. Total = 108. Minus 2 for activation: 106 ATP.

A10. Glucose: 32 / 6 ≈ 5.3 ATP per carbon. Palmitate: 106 / 16 ≈ 6.6 ATP per carbon. Per gram, the difference is larger: glucose (180 g mol⁻¹) gives 32/180 ≈ 0.18 mol ATP per g; palmitate (256 g mol⁻¹) gives 106/256 ≈ 0.41 mol ATP per g — more than double. Fat is also stored nearly water-free, whereas glycogen holds about two to three times its own mass of water, so fat is far more compact (see lipids explained).

A11. Even-chain fatty acids are broken down entirely into acetyl-CoA. The step from pyruvate to acetyl-CoA (pyruvate dehydrogenase) is irreversible, and in the Krebs cycle the two carbons of each acetyl group are balanced by two carbons lost as CO₂, so there’s no net gain of oxaloacetate to make glucose. Glycerol enters glycolysis as a three-carbon intermediate (dihydroxyacetone phosphate), and many amino acids are converted to pyruvate or Krebs cycle intermediates, which can be turned into oxaloacetate and then glucose by gluconeogenesis (see ketosis).

A12. (a) Glucose has 6 carbons: 18 ATP and 12 NADPH. (b) The Calvin cycle needs ATP : NADPH = 3 : 2 = 1.5, but non-cyclic flow supplies only about 2.6 : 2 = 1.3. Without a top-up, NADPH would pile up while ATP ran short, and the cycle would stall; cells also need extra ATP for other processes. Cyclic electron flow, which returns electrons from photosystem I to the cytochrome complex, pumps protons and makes ATP without NADPH, letting the chloroplast make up the shortfall and adjust the ratio. See light-dependent reactions and the Calvin cycle.

Key takeaways

• Keep carbon and electron bookkeeping straight: CO₂ from the link reaction and Krebs cycle; electrons carried by NADH and FADH₂.
• Modern ATP yield: about 30–32 per glucose, 106 per palmitate.
• Use ΔG°′ = −nFΔE°′ and ΔG°′ = −RT ln K′ to connect redox potentials and equilibria to energy.
• Control points are irreversible steps with large negative ΔG, regulated allosterically.
• RQ reveals the fuel being burned: 1.0 for carbohydrate, about 0.7 for fat.