Cellular Respiration

Question 1 (10 marks)

(a) Aerobic respiration (2): $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \;(+\text{ATP})$

• Correct reactants and products (1); balanced (1).

(b) Alcoholic fermentation (2): glucose → ethanol + carbon dioxide (+ 2 ATP).

• End products ethanol + CO₂ named (1); glucose as substrate/ATP (1).

(c) (3): Aerobic respiration in A yields ~30–32 ATP per glucose (1), whereas anaerobic fermentation in B yields only 2 ATP per glucose (1). To make the same amount of ATP, B must break down far more glucose — i.e. A uses less glucose per ATP (1).

(d) (3): Flask A — O₂ consumed and CO₂ released (aerobic respiration uses O₂ as final electron acceptor and releases CO₂ in link reaction + Krebs) (1½). Flask B — no O₂ used (none present); CO₂ still released by alcoholic fermentation (1½). Why: CO₂ comes from decarboxylation of pyruvate to ethanol, independent of O₂.

Question 2 (12 marks)

(a) (4): All 6 carbons leave the cell as carbon dioxide (1). Released at:

• Link reaction: 2 CO₂ (one C per pyruvate × 2 pyruvate) (1)
• Krebs cycle: 4 CO₂ (2 per turn × 2 turns) (1)
• (No CO₂ released in glycolysis) (1)

(b) (3): The link reaction / pyruvate oxidation (1). Each of the 2 pyruvate (3C) is decarboxylated, losing one CO₂ each (1), giving 2 CO₂ total and forming 2 acetyl-CoA (2C) (1).

(c) (2): Acetyl-CoA (2C) combines with oxaloacetate (4C) to form citrate (6C); two decarboxylations release 2 CO₂ per turn, regenerating oxaloacetate (1) — so the cycle intermediate is regenerated unchanged and can accept the next acetyl group (1).

(d) (3): The claim is incorrect (1). The carbon in exhaled CO₂ comes from the glucose carbon skeleton, released during link reaction and Krebs (1). The oxygen we breathe is the final electron acceptor in the ETC and is reduced to form water (H₂O), not CO₂ (1).

Question 3 (12 marks)

(a) (4): Totals: NADH = 2+2+6 = 10; FADH₂ = 2.

• From NADH: $10 \times 2.5 = 25$ ATP (1)
• From FADH₂: $2 \times 1.5 = 3$ ATP (1)
• Substrate-level: $2 + 2 = 4$ ATP (1)
• Total: $25 + 3 + 4 = 32$ ATP (1)

(b) (3): The 2 glycolytic NADH now count as FADH₂.

• New carriers: NADH = 8, FADH₂ = 4.
• NADH: $8 \times 2.5 = 20$; FADH₂: $4 \times 1.5 = 6$; SLP = 4.
• Total: $20 + 6 + 4 = 30$ ATP (1 for adjustment, 1 for oxidative calc, 1 for total).

(c) (3): Oxidative phosphorylation ATP production stops (falls to 0 from that source) (1). Protons pumped by the chain cannot flow back through blocked ATP synthase, so the proton gradient builds up (1); the back-pressure eventually halts electron transport and O₂ consumption (1). (Only substrate-level ATP = 4 remains.)

(d) (2): NADH donates electrons at Complex I (higher/earlier energy level), while FADH₂ enters at Complex II, bypassing Complex I (1). Fewer protons are pumped across the membrane per FADH₂, so less ATP is made per pair of electrons (1).

Question 4 (10 marks)

(a) (3): During intense sprinting O₂ supply is insufficient for the ETC (1). NADH cannot be reoxidised by the (oxygen-dependent) ETC, so NAD⁺ runs out (1). Pyruvate is reduced to lactate, oxidising NADH back to NAD⁺, allowing glycolysis to continue (1).

(b) (2): Fermentation regenerates NAD⁺ so glycolysis and substrate-level ATP production can continue rapidly without oxygen (1) — provides fast ATP when demand exceeds aerobic supply (1).

(c) (3): After exercise, lactate is transported (e.g. to the liver) and, with oxygen now available, is oxidised back to pyruvate / converted back toward glucose (1). Extra O₂ is needed to fully oxidise pyruvate via aerobic respiration / to power reconversion (1) — this repayment is the "oxygen debt," restoring normal metabolic state (1).

(d) (2): Lactic acid fermentation → lactate (no CO₂); alcoholic fermentation → ethanol + CO₂ (1). In both, NAD⁺ must be regenerated so glycolysis (and its ATP production) can continue in the absence of O₂ (1).

Question 5 (6 marks)

(a) (3): A leaky membrane dissipates the proton gradient (1). Protons flow back without passing through ATP synthase (1), so chemiosmosis is uncoupled and little/no ATP is synthesised by oxidative phosphorylation (1).

(b) (3): Oxygen consumption increases — the chain keeps pumping harder to try to build the gradient, so electron transport (and O₂ reduction) runs fast (1). Energy from the proton flow is released as heat instead of ATP (1). Why: uncoupling converts the chemiosmotic potential energy into thermal energy (basis of thermogenesis) (1).

[
  {"claim":"Standard aerobic ATP yield with NADH=2.5, FADH2=1.5 equals 32","code":"NADH=2+2+6; FADH2=2; total=NADH*Rational(5,2)+FADH2*Rational(3,2)+4; result=(total==32)"},
  {"claim":"Shuttle-as-FADH2 case yields 30 ATP","code":"NADH=8; FADH2=4; total=NADH*Rational(5,2)+FADH2*Rational(3,2)+4; result=(total==30)"},
  {"claim":"Total labelled carbons lost as CO2 equals 6 (2 link + 4 Krebs)","code":"link=2; krebs=2*2; result=(link+krebs==6)"},
  {"claim":"Difference between standard and shuttle yield is 2 ATP","code":"std=10*Rational(5,2)+2*Rational(3,2)+4; shu=8*Rational(5,2)+4*Rational(3,2)+4; result=(std-shu==2)"}
]