Level 3 — ProductionCellular Respiration

Cellular Respiration

⏱ 45 minutes50 marksprintable — key stays hidden on paper

Level: 3 (Production — from-scratch derivations, explain-out-loud reasoning) Time limit: 45 minutes Total marks: 50

Answer all questions. Show full reasoning. Where a derivation or calculation is required, state each step and justify it.

Question 1 — The master equation (6 marks)

(a) From memory, write the balanced overall equation for the complete aerobic respiration of one molecule of glucose, including energy. Ensure atoms balance. (3)

(b) Explain, in your own words, why this single equation "hides" a great deal of biochemistry — name the four major stages it summarises and where in the cell each occurs. (3)

Question 2 — Glycolysis from scratch (9 marks)

(a) Glycolysis has an investment phase and a payoff phase. Explain what happens to ATP in each phase, and hence derive the net ATP yield of glycolysis. (4)

(b) Write down the full input/output balance sheet for glycolysis per glucose: list substrates consumed and every product formed (carbon products, reduced carriers, and net ATP). (3)

(c) Glycolysis can proceed briefly without oxygen. Explain why, despite this, it cannot continue indefinitely without either oxygen or fermentation. (2)

Question 3 — Tracing the carbons and carriers (10 marks)

Trace a single glucose molecule from the end of glycolysis through the link reaction and Krebs cycle.

(a) State the inputs and outputs of the link reaction per pyruvate, and then per glucose. (3)

(b) For one turn of the Krebs cycle, state the reduced carriers and ATP/GTP produced, and the CO $_2$ released. Then give the per glucose totals for the Krebs cycle alone. (4)

(c) Account for all 6 carbon atoms of glucose: show that they are fully released as CO $_2$ by the end of the Krebs cycle, stating at which stage each pair of carbons leaves. (3)

Question 4 — Chemiosmosis explained out loud (9 marks)

(a) Explain, step by step, how the electron transport chain builds a proton gradient. Reference NADH/FADH $_2$ , the carriers, and the final electron acceptor. (4)

(b) Describe how ATP synthase uses this gradient to make ATP. Name the process and explain why FADH $_2$ yields fewer ATP than NADH. (3)

(c) Predict and justify what happens to ATP production if a chemical makes the inner mitochondrial membrane freely permeable to protons. (2)

Question 5 — Full ATP yield derivation (10 marks)

Derive the theoretical maximum ATP yield from one glucose under aerobic conditions.

(a) Using the conversion factors NADH → 3 ATP and FADH $_2$ → 2 ATP, build a stage-by-stage table (glycolysis, link reaction, Krebs cycle) listing NADH, FADH $_2$ , and direct ATP for each, then compute the total. (6)

(b) Using the modern conversion factors NADH → 2.5 ATP and FADH $_2$ → 1.5 ATP, recompute the total. (2)

Question 6 — Fermentation, compared and contrasted (6 marks)

(a) Write the word/near-molecular scheme for lactic acid fermentation and for alcoholic fermentation, starting from pyruvate. (2)

(b) Both fermentations regenerate a key molecule. Name it and explain why regenerating it allows glycolysis to keep running. (2)

(c) In a table, contrast aerobic respiration and anaerobic fermentation on: final electron acceptor, ATP yield per glucose, and end products. (2)

Answer keyMark scheme & solutions

Question 1 (6 marks)

(a) (3 marks) $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{energy (ATP + heat)}$

Correct reactants glucose + O $_2$ (1)
Correct products CO $_2$ + H $_2$ O (1)
Correctly balanced (6:6:6:6) and energy noted (1)

Check: C: 6=6; H: 12 = 12 (6×2); O: 6+12 = 18 = 12+6. Balances.

(b) (3 marks)

The equation only shows net reactants/products; it omits the stepwise oxidation and the electron carriers. (1)
Four stages: Glycolysis (cytoplasm/cytosol); Link reaction / pyruvate oxidation (mitochondrial matrix); Krebs cycle (matrix); Electron transport chain + chemiosmosis (inner mitochondrial membrane). (1 for naming all four)
Correct locations paired to stages. (1)

Question 2 (9 marks)

(a) (4 marks)

Investment phase: 2 ATP consumed to phosphorylate glucose (to fructose-1,6-bisphosphate). (1)
Payoff phase: 4 ATP produced by substrate-level phosphorylation. (1)
Net = 4 − 2 = 2 ATP. (1)
Correct stated derivation/logic (gross vs net). (1)

(b) (3 marks) Balance sheet per glucose:

Inputs: 1 glucose, 2 ATP, 2 NAD $^+$ , 2 P $_i$ , 4 ADP. (1)
Outputs: 2 pyruvate (3C each), 2 NADH, 2 H $^+$ . (1)
Net 2 ATP (4 made − 2 used). (1)

(c) (2 marks)

Glycolysis needs a continuous supply of NAD $^+$ ; without O $_2$ /ETC the NADH is not reoxidised. (1)
Once NAD $^+$ is exhausted, glycolysis stalls unless fermentation regenerates NAD $^+$ . (1)

Question 3 (10 marks)

(a) (3 marks) Per pyruvate: inputs = 1 pyruvate + 1 NAD $^+$ + 1 CoA; outputs = 1 acetyl-CoA + 1 CO $_2$ + 1 NADH. (1) Per glucose (×2 pyruvate): 2 acetyl-CoA, 2 CO $_2$ , 2 NADH. (1 for CO2/NADH, 1 for acetyl-CoA doubling)

(b) (4 marks) Per turn: 3 NADH, 1 FADH $_2$ , 1 ATP (GTP), 2 CO $_2$ . (2) Per glucose (2 turns): 6 NADH, 2 FADH $_2$ , 2 ATP, 4 CO $_2$ . (2)

(c) (3 marks)

Glucose (6C) → 2 pyruvate (2×3C) — all 6 carbons still present. (1)
Link reaction: each pyruvate loses 1 C as CO $_2$ → 2 CO $_2$ (2 carbons gone). (1)
Krebs cycle: each acetyl-CoA (2C) contributes to release of 2 CO $_2$ per turn → 4 CO $_2$ (remaining 4 carbons gone). Total 2+4 = 6 CO $_2$ , all carbons accounted for. (1)

Question 4 (9 marks)

(a) (4 marks)

NADH and FADH $_2$ deliver electrons to the ETC (NADH at Complex I, FADH $_2$ at Complex II). (1)
Electrons pass down a chain of carriers (complexes, ubiquinone, cytochrome c) losing energy stepwise. (1)
Energy released pumps H $^+$ from matrix into the intermembrane space, creating a proton (electrochemical) gradient. (1)
Oxygen is the final electron acceptor, combining with electrons and H $^+$ to form water. (1)

(b) (3 marks)

Protons flow back down their gradient through ATP synthase — this is chemiosmosis. (1)
The flow drives rotation/conformational change that phosphorylates ADP → ATP (oxidative phosphorylation). (1)
FADH $_2$ enters "lower" (at Complex II), bypassing one proton-pumping complex, so fewer protons pumped → fewer ATP. (1)

(c) (2 marks)

Protons leak back without passing through ATP synthase, so the gradient collapses → ATP synthesis stops (or drops sharply). (1)
Energy from electron transport is released as heat instead; electron transport may continue but is uncoupled from ATP production. (1)

Question 5 (10 marks)

(a) (6 marks) Table (classic factors 3/2):

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

Correct table entries (3). Computation:

NADH: 10 × 3 = 30
FADH $_2$ : 2 × 2 = 4
Direct: 4
Total = 30 + 4 + 4 = 38 ATP (3)

(If glycolytic NADH costs 2 ATP for shuttle → 36; either 38 or the shuttle-corrected 36 accepted with justification.)

(b) (2 marks) Modern factors 2.5/1.5:

NADH: 10 × 2.5 = 25
FADH $_2$ : 2 × 1.5 = 3
Direct: 4
Total = 25 + 3 + 4 = 32 ATP (2)

(c) (2 marks)

Proton gradient is used for other work / some protons leak, so coupling is imperfect. (1)
Shuttling cytosolic NADH into mitochondria costs energy; membranes are not 100% efficient — actual yield ≈ 30–32. (1)

Question 6 (6 marks)

(a) (2 marks)

Lactic acid: pyruvate + NADH → lactate + NAD $^+$ . (1)
Alcoholic: pyruvate → CO $_2$ + acetaldehyde; acetaldehyde + NADH → ethanol + NAD $^+$ . (1)

(b) (2 marks)

The regenerated molecule is NAD $^+$ . (1)
NAD $^+$ is needed as the electron acceptor in glycolysis (at the GAP → 1,3-BPG step); regenerating it keeps glycolysis (and thus a small ATP supply) going without O $_2$ . (1)

(c) (2 marks) (½ each cell, round to 2)

Feature	Aerobic	Anaerobic fermentation
Final electron acceptor	Oxygen	Organic molecule (pyruvate/acetaldehyde)
ATP per glucose	~36–38 (≈32)	2
End products	CO₂ + H₂O	Lactate, OR ethanol + CO₂

Full correct table (2); partially correct (1).

[
  {"claim":"Aerobic equation carbon balance: 6 C on left equals 6 C in 6 CO2", "code":"result = (6 == 6*1)"},
  {"claim":"Aerobic equation oxygen balance: 6*2 (O2) + 6 (glucose) = 6*2 (CO2) + 6 (H2O)", "code":"left = 6*2 + 6; right = 6*2 + 6; result = (left == right)"},
  {"claim":"Aerobic equation hydrogen balance: glucose 12 H = 6 H2O = 12 H", "code":"result = (12 == 6*2)"},
  {"claim":"Net glycolysis ATP = 4 produced - 2 consumed = 2", "code":"result = (4 - 2 == 2)"},
  {"claim":"Classic total ATP = 10 NADH*3 + 2 FADH2*2 + 4 direct = 38", "code":"result = (10*3 + 2*2 + 4 == 38)"},
  {"claim":"Modern total ATP = 10 NADH*2.5 + 2 FADH2*1.5 + 4 direct = 32", "code":"result = (Rational(10)*Rational(5,2) + Rational(2)*Rational(3,2) + 4 == 32)"},
  {"claim":"Total CO2 released per glucose = 2 (link) + 4 (Krebs) = 6", "code":"result = (2 + 4 == 6)"},
  {"claim":"Krebs per glucose carriers: 2 turns give 6 NADH and 2 FADH2", "code":"result = (2*3 == 6 and 2*1 == 2)"}
]