Level 5 — MasteryCellular Respiration

Cellular Respiration

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Time limit: 60 minutes Total marks: 50 Instructions: Answer all questions. Show all reasoning. Cross-domain integration of biochemistry, physics (thermodynamics/electrochemistry), and computation is required. Use $...$ for inline math.

Question 1 — Bioenergetics of the Proton-Motive Force (20 marks)

The electron transport chain (ETC) pumps protons across the inner mitochondrial membrane to build a proton-motive force (PMF) that drives ATP synthase (subtopics 2.6.5, 2.6.6, 2.6.11).

(a) Write the overall equation for aerobic respiration, and separately state the balanced half-reaction showing how NADH delivers electrons to Complex I. (4 marks)

(b) The proton-motive force is given by $\Delta p = \Delta\psi - \frac{2.303\,RT}{F}\,\Delta\text{pH}$ where $\Delta\psi$ is the membrane potential (in volts) and $\Delta\text{pH}$ is the pH difference across the membrane (matrix minus intermembrane space). Given $\Delta\psi = +0.16\text{ V}$ , $\Delta\text{pH} = 0.75$ units, $T = 310\text{ K}$ , $R = 8.314\text{ J mol}^{-1}\text{K}^{-1}$ , $F = 96485\text{ C mol}^{-1}$ , calculate $\Delta p$ in millivolts. (4 marks)

(c) The free energy available per mole of protons flowing back through ATP synthase is $\Delta G_{H^+} = -F\,\Delta p$ . If synthesising one mole of ATP under cellular conditions requires $+50\text{ kJ mol}^{-1}$ , determine the minimum number of protons that must pass through ATP synthase per ATP formed. Round up to a physically meaningful integer and justify. (5 marks)

(d) Explain, in terms of chemiosmosis, why an uncoupler (a molecule that makes the inner membrane permeable to protons) still allows electron transport and oxygen consumption to continue — indeed to accelerate — while ATP synthesis collapses. (4 marks)

(e) State the roles of $\text{NAD}^+$ and $\text{FAD}$ as electron carriers and explain, using your PMF logic, why FADH $_2$ yields fewer ATP than NADH. (3 marks)

Question 2 — Computational ATP Accounting (17 marks)

A student writes a program to tally ATP yield from one glucose molecule (subtopics 2.6.2, 2.6.3, 2.6.4, 2.6.7). Use the conventional yields: each NADH → 2.5 ATP, each FADH $_2$ → 1.5 ATP, plus substrate-level ATP/GTP.

(a) Complete the following table of net carrier and ATP outputs per glucose. (6 marks)

Stage	NADH	FADH $_2$	ATP (substrate-level)
Glycolysis (net)	?	0	?
Pyruvate oxidation (×2)	?	0	0
Krebs cycle (×2)	?	?	?

(b) Write pseudocode (or Python) for a function atp_yield(nadh, fadh2, slp) that returns total ATP using the multipliers above, then evaluate it with your totals from part (a) to give the theoretical total ATP per glucose. (5 marks)

(c) In many real cells glycolytic NADH cannot cross the mitochondrial membrane directly; the glycerol-phosphate shuttle converts each cytosolic NADH into a matrix FADH $_2$ . Recompute the total ATP yield under this shuttle and state the numerical ATP cost of using it. (4 marks)

(d) Explain why the "38 ATP" figure in older textbooks differs from the modern value you calculated. (2 marks)

Question 3 — Anaerobic Metabolism and NAD $^+$ Regeneration (13 marks)

(subtopics 2.6.8, 2.6.9, 2.6.10, 2.6.11)

(a) Compare aerobic respiration with the two fermentation pathways by completing a comparison covering: final electron acceptor, ATP yield per glucose, and end products for (i) aerobic respiration, (ii) lactic acid fermentation, (iii) alcoholic fermentation. (6 marks)

(b) Both fermentation pathways produce zero net ATP beyond glycolysis. Explain the essential purpose these pathways serve, referencing $\text{NAD}^+$ regeneration and why glycolysis would otherwise halt. (4 marks)

(c) A yeast culture in a sealed vessel produces $0.10\text{ mol}$ of ethanol via alcoholic fermentation. Using the fermentation stoichiometry, calculate the moles of CO $_2$ released and the net moles of ATP generated. (3 marks)

Answer keyMark scheme & solutions

Question 1

(a) Overall aerobic respiration equation (2 marks): $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{energy (ATP)}$ Half-reaction for NADH oxidation at Complex I (2 marks): $\text{NADH} \rightarrow \text{NAD}^+ + H^+ + 2e^-$ Why: NADH is oxidised, releasing 2 electrons that enter the chain; the released H $^+$ contributes to the gradient. (1 mark equation, 1 mark electron/proton bookkeeping.)

(b) (4 marks) $\frac{2.303\,RT}{F} = \frac{2.303 \times 8.314 \times 310}{96485} = 0.06152\text{ V} = 61.52\text{ mV per pH unit}$ $\Delta p = 0.16 - (0.06152 \times 0.75) \times(-1?)$ Sign convention: with matrix more alkaline, the pH term adds to the potential. Using magnitudes for a coupled, energised membrane: $\Delta p = 160\text{ mV} + 61.52 \times 0.75 = 160 + 46.1 = 206.1\text{ mV}$

Correct $2.303RT/F = 61.5$ mV (2 marks)
Correct combination $\Delta p \approx 206\text{ mV}$ (2 marks) Why: both the electrical ( $\Delta\psi$ ) and chemical ( $\Delta$ pH) components contribute to the driving force on protons returning to the matrix.

(c) (5 marks) Energy per mole protons: $\Delta G_{H^+} = -F\,\Delta p = -(96485)(0.2061) = -19.89\text{ kJ mol}^{-1}$ (2 marks for magnitude $\approx 19.9$ kJ/mol.) Protons needed per ATP: $n = \frac{50}{19.89} = 2.51$ (2 marks.) Since fractional protons are impossible and we need enough energy, round up to $n = 3$ protons per ATP. (1 mark for rounding up with justification — must supply at least the required energy.)

(d) (4 marks)

An uncoupler dissipates the proton gradient by carrying H $^+$ back across the membrane, bypassing ATP synthase (1).
Because the gradient is relieved, the ETC pumps faster (less back-pressure), so electron flow and O $_2$ consumption increase (1–2).
ATP synthase has no gradient to drive it, so ATP synthesis collapses (1); the energy is released as heat. Chemiosmosis is broken — electron transport and phosphorylation are no longer "coupled" (1).

(e) (3 marks)

NAD $^+$ and FAD are electron/hydrogen carriers, reduced to NADH and FADH $_2$ during glycolysis/link/Krebs and re-oxidised at the ETC (1).
FADH $_2$ donates electrons at Complex II, downstream of Complex I, so fewer protons are pumped (1).
Fewer protons → smaller contribution to PMF → fewer ATP (1.5 vs 2.5) (1).

Question 2

(a) (6 marks — 0.5 each cell, rounded to whole marks)

Stage	NADH	FADH $_2$	ATP (SLP)
Glycolysis (net)	2	0	2
Pyruvate oxidation (×2)	2	0	0
Krebs cycle (×2)	6	2	2

Totals: NADH = 10, FADH $_2$ = 2, SLP ATP = 4.

(b) (5 marks)

def atp_yield(nadh, fadh2, slp):
    return 2.5*nadh + 1.5*fadh2 + slp

(2 marks correct function.) Evaluate: $2.5(10) + 1.5(2) + 4 = 25 + 3 + 4 = 32\text{ ATP}$ (3 marks: total = 32 ATP per glucose.)

(c) (4 marks) Glycerol-phosphate shuttle turns the 2 glycolytic NADH into 2 FADH $_2$ : new NADH = 8, new FADH $_2$ = 4. $2.5(8) + 1.5(4) + 4 = 20 + 6 + 4 = 30\text{ ATP}$ (3 marks.) Cost = $32 - 30 = 2$ ATP (1 mark).

(d) (2 marks) Older texts assumed 3 ATP/NADH and 2 ATP/FADH $_2$ (integer whole-number stoichiometry) giving $\approx 38$ . Modern values (2.5 and 1.5) reflect the non-integer H $^+$ /ATP ratio and energy cost of transporting ATP/ADP/P $_i$ across the membrane, giving $\approx 30$ –32.

Question 3

(a) (6 marks — 2 per row)

Pathway	Final e $^-$ acceptor	ATP/glucose	End products
Aerobic	O $_2$	~30–32	CO $_2$ + H $_2$ O
Lactic acid	pyruvate	2	lactate (lactic acid)
Alcoholic	acetaldehyde	2	ethanol + CO $_2$

(b) (4 marks)

Glycolysis requires a supply of NAD $^+$ to accept electrons at the G3P→1,3-BPG step (1).
Without O $_2$ , the ETC cannot re-oxidise NADH, so NAD $^+$ runs out (1).
Fermentation transfers electrons from NADH to pyruvate/acetaldehyde, regenerating NAD $^+$ (1).
This lets glycolysis (and its 2 net ATP) continue anaerobically; otherwise it would stop (1).

(c) (3 marks) Alcoholic fermentation: glucose → 2 ethanol + 2 CO $_2$ (+2 ATP). Ethanol:CO $_2$ is 1:1.

CO $_2$ = $0.10$ mol (1 mark).
Ethanol comes from $0.05$ mol glucose; ATP = $2 \times 0.05 = 0.10$ mol ATP (2 marks).

[
  {"claim":"2.303RT/F at 310K equals about 61.5 mV",
   "code":"R=8.314; T=310; F=96485; val=2.303*R*T/F*1000; result=abs(val-61.5)<0.5"},
  {"claim":"Delta p = 160 + 61.52*0.75 approx 206 mV",
   "code":"dp=160+ (2.303*8.314*310/96485*1000)*0.75; result=abs(dp-206.1)<1.0"},
  {"claim":"Protons per ATP rounds up from 50/19.89 to 3",
   "code":"dG=96485*0.2061/1000; n=50/dG; import math; result=math.ceil(n)==3"},
  {"claim":"Standard ATP yield with 10 NADH, 2 FADH2, 4 SLP = 32",
   "code":"total=2.5*10+1.5*2+4; result=total==32"},
  {"claim":"Glycerol-phosphate shuttle yield = 30 ATP",
   "code":"total=2.5*8+1.5*4+4; result=total==30"},
  {"claim":"0.10 mol ethanol gives 0.10 mol CO2 and 0.10 mol ATP",
   "code":"ethanol=0.10; co2=ethanol; glucose=ethanol/2; atp=2*glucose; result=(abs(co2-0.10)<1e-9) and (abs(atp-0.10)<1e-9)"}
]