Photosynthesis

Question 1 (20 marks)

(a) (4 marks)

• 12 H₂O on the left because photolysis splits water to supply electrons/protons and release O₂ (1). Each O₂ requires 2 H₂O split → 6 O₂ needs 12 H₂O (1).
• Isotope labelling (⁠¹⁸O) shows the O₂ released comes entirely from water, not CO₂ (1).
• 6 H₂O reappear on the right because they are regenerated during the Calvin cycle / condensation reactions (e.g. ATP hydrolysis, sugar formation) (1). Net water consumed = 6.

(b)(i) (3 marks) $E_{\text{photon}} = hc/\lambda = \dfrac{(6.626\times10^{-34})(3.00\times10^{8})}{680\times10^{-9}}$ $= 2.923\times10^{-19}\ \text{J}$ (1) Per mole: $E = 2.923\times10^{-19}\times6.022\times10^{23} = 1.760\times10^{5}\ \text{J} = 176.0\ \text{kJ mol}^{-1}$ (2).

(b)(ii) (3 marks)

• 8 photons per O₂; 6 O₂ per glucose → $8\times6 = 48$ mol photons per mol glucose (1).
• Energy = $48 \times 176.0 = 8448\ \text{kJ}$ (≈ $8.45\times10^{3}$ kJ) (2).

(b)(iii) (3 marks) $\eta = \dfrac{2870}{8448} = 0.3397 = 34.0\%$ (2). Real efficiency much lower because: reflection/transmission of light, absorption by non-photosynthetic pigments, photorespiration, saturation/limiting factors, respiratory losses, and only certain wavelengths are usable (1, any valid reason).

(c) (4 marks)

• Similarities (any 2, 1 mark total): both are redox reactions involving electron transport chains and chemiosmotic ATP synthesis; both use similar electron carriers (NADP⁺/NAD⁺) and membrane-bound ATP synthase.
• Differences (2 marks, ½ each):
  • Organelle: chloroplast vs mitochondrion.
  • Energy flow: photosynthesis stores light energy (endergonic, +ΔG) vs respiration releases energy (exergonic, −ΔG).
  • Redox: photosynthesis reduces CO₂ to sugar (reduction/anabolic); respiration oxidises sugar to CO₂ (oxidation/catabolic).
  • Net ATP: respiration yields net ATP; light reactions produce ATP but Calvin cycle consumes it — net storage is in sugar not ATP.
• Not exact reverses (1 mark): different intermediates, enzymes, electron carriers (NADPH vs NADH), and pathways; oxygen source differs; energetically the pathways are not simply the reverse mechanism.

Question 2 (22 marks)

(a) (5 marks) Linear Z-scheme (1 mark per correct labelled element, max 5): PSII (P680) → photolysis supplies e⁻ → plastoquinone (PQ) → cytochrome b6f complex (proton pumping/ATP generation here) → plastocyanin (PC) → PSI (P700) → ferredoxin (Fd) → NADP⁺ reductase → NADPH. Electron gains energy at PSII and PSI (light absorption boosts it to higher energy). ATP generated via proton gradient built at the cytochrome b6f step.

(b)(i) (3 marks) Non-cyclic gives 3 ATP : 2 NADPH per O₂; per O₂ = 2 CO₂ fixed, so per CO₂ = 1.5 ATP : 1 NADPH. Calvin demand per CO₂ = 3 ATP : 2 NADPH = 1.5 ATP : 1 NADPH. These are equal ratios → non-cyclic alone exactly matches demand in this idealised case (3). (Award full marks for showing the per-CO₂ scaling and equality.)

(b)(ii) (4 marks) Pseudocode (2 marks):

INPUT demand_ATP = 3, demand_NADPH = 2   # per CO2
INPUT nc_ATP = 2.6, nc_NADPH = 2         # per O2 = per 2 CO2
# scale non-cyclic output to per CO2
nc_ATP_perCO2 = nc_ATP / 2
nc_NADPH_perCO2 = nc_NADPH / 2
extra_ATP = demand_ATP - nc_ATP_perCO2   # NADPH assumed sufficient
OUTPUT extra_ATP

Evaluation (2 marks): nc_ATP_perCO2 = 2.6/2 = 1.3; extra_ATP = 3 − 1.3 = 1.7 ATP per CO₂ must come from cyclic photophosphorylation.

(c) (5 marks)

• Protons accumulate in the thylakoid lumen creating an electrochemical gradient (1).
• Three sources (1 each): (i) photolysis of water releases H⁺ into the lumen; (ii) the cytochrome b6f complex pumps H⁺ from stroma into lumen as electrons pass; (iii) reduction of NADP⁺ in the stroma consumes H⁺, lowering stromal [H⁺] and steepening the gradient.
• Coupling (1): protons flow back through ATP synthase (CF₀CF₁); the energy released drives phosphorylation of ADP + Pi → ATP (chemiosmosis).

(d) (5 marks) (1 mark each, +1 for reasoning quality)

• (i) Proton gradient collapses/dissipates because H⁺ leaks back freely across the membrane.
• (ii) ATP synthesis stops/falls — no proton-motive force to drive ATP synthase.
• (iii) Electron transport rate INCREASES (or stays high) — without a back-pressure gradient, electron flow is no longer restricted ("uncoupled").
• (iv) O₂ evolution continues (may increase) since photolysis/electron flow proceeds; O₂ production is decoupled from ATP output.

Question 3 (18 marks)

(a)(i) (4 marks)

• Carboxylation: RuBP + CO₂ → 2 × 3-phosphoglycerate (2×3C) (1).
• Oxygenation: RuBP + O₂ → 1 × 3-phosphoglycerate + 1 × 2-phosphoglycolate (1).
• High temperature decreases CO₂ solubility relative to O₂ and reduces RuBisCO's specificity for CO₂; stomatal closure in heat lowers internal CO₂ raising O₂:CO₂ ratio → favours oxygenase (2).

(a)(ii) (3 marks)

• Photorespiration consumes fixed carbon: 2-phosphoglycolate is salvaged in a pathway (peroxisome/mitochondrion) that releases previously fixed CO₂ (1).
• It consumes ATP and reducing power (energy cost) without net sugar gain (1).
• Net effect: reduces photosynthetic efficiency/yield by up to ~25% in C3 plants (1).

(b) (6 marks) — ½ mark per correct cell (12 cells):

(c) (5 marks)

• (i) Both curves plateau because another factor becomes limiting once CO₂ is no longer limiting (light, temperature, or enzyme/RuBisCO saturation) (1).
• (ii) High-light curve plateaus higher because light is not limiting; the plateau is set by CO₂-independent factors so more CO₂ can be used before saturation, and the light reactions supply more ATP/NADPH (2).
• (iii) On the low-light curve: at low CO₂ the initial rising region is CO₂-limited (1); at the plateau, light intensity is the limiting factor (1).

[
  {"claim": "One mole of 680nm photons carries ~176 kJ", "code": "h=6.626e-34; c=3.00e8; lam=680e-9; Na=6.022e23; E=h*c/lam*Na/1000; result = abs(E-176.0)<1.0"},
  {"claim": "Minimum photon energy per mole glucose is ~8448 kJ (48 mol photons)", "code": "h=6.626e-34; c=3.00e8; lam=680e-9; Na=6.022e23; Emol=h*c/lam*Na/1000; Etot=48*Emol; result = abs(Etot-8448)<50"},
  {"claim": "Max theoretical efficiency ~34%", "code": "eff=2870/8448*100; result = abs(eff-34.0)<1.0"},
  {"claim": "Extra ATP from cyclic per CO2 is 1.7 when non-cyclic gives 2.6 ATP:2 NADPH per O2", "code": "nc_ATP_perCO2=Rational(26,10)/2; extra=3-nc_ATP_perCO2; result = extra==Rational(17,10)"},
  {"claim": "Non-cyclic 3ATP:2NADPH per O2 equals Calvin demand per CO2", "code": "result = (Rational(3,2)==Rational(3,1)/2) and (Rational(2,2)==Rational(2,1)/2)"}
]