Photosynthesis

Question 1 (8)

(a) (3) $6CO_2 + 6H_2O \xrightarrow[\text{chlorophyll}]{\text{light}} C_6H_{12}O_6 + 6O_2$

• Correct reactants & products (1); balanced (1); light + chlorophyll condition shown (1).

(b) (3) $6CO_2 + 12H_2O \xrightarrow[\text{chlorophyll}]{\text{light}} C_6H_{12}O_6 + 6O_2 + 6H_2O$

• Correct 12/6 form (1). Why more accurate: isotope (¹⁸O) tracing shows the O₂ released comes from water, not CO₂; splitting 12 waters supplies the electrons/protons, and 6 new waters are re-formed during the reactions — so water is both consumed and produced (2).

(c) (2)

• Left: $CO_2$ gives $6\times2=12$; $H_2O$ gives $12\times1=12$ → 24 O atoms.
• Right: glucose $C_6H_{12}O_6$ gives 6; $O_2$ gives $6\times2=12$; $H_2O$ gives $6\times1=6$ → $6+12+6=$ 24 O atoms. Match confirmed (1 each side, 2 total).

Question 2 (14)

(a) (6) — order (1 each, max 6): H₂O → PSII (P680) → plastoquinone (PQ) → cytochrome b6f complex → plastocyanin (PC) → PSI (P700) → ferredoxin (Fd) → NADP⁺ reductase → NADPH (terminal acceptor NADP⁺).

(b) (5)

• Protons accumulate in the thylakoid lumen (1) from two sources: photolysis of water dumps H⁺ into the lumen (1) and the cytochrome b6f complex pumps H⁺ from stroma to lumen as electrons pass (1).
• The resulting proton-motive force / gradient drives H⁺ back into the stroma through ATP synthase (1), which uses the flow (chemiosmosis) to phosphorylate ADP → ATP (1).

(c) (3): ATP and NADPH → used in the Calvin cycle (stroma) (1+1 for correct destination); O₂ → released as waste/by-product (1).

Question 3 (8)

(a) (2) $2H_2O \rightarrow 4H^+ + 4e^- + O_2$ Balanced (must include electrons) (2); accept $H_2O \rightarrow 2H^+ + 2e^- + \tfrac12 O_2$.

(b) (3) — one mark each:

• electrons (e⁻) replace those lost from P680 in PSII;
• protons (H⁺) contribute to the lumen proton gradient / reduce NADP⁺;
• oxygen (O₂) is released as a waste gas.

(c) (3): P680 in PSII loses electrons when it is photo-excited and passes them down the chain; it is a strong enough oxidant to pull electrons from water (1). PSI (P700) is re-supplied with electrons arriving via plastocyanin from PSII, not from water (1). So only PSII creates the "electron hole" that water fills (1).

Question 4 (12)

(a) (6) — 2 marks per phase:

• Fixation: CO₂ + RuBP (5C) → 2 × GP/3-PGA (3C), catalysed by RuBisCO.
• Reduction: GP + ATP + NADPH → G3P/triose phosphate (3C); some G3P leaves to form glucose.
• Regeneration: remaining G3P + ATP regenerates RuBP (5C) to continue the cycle.

(b) (4)

• Per 6 CO₂ → 1 glucose: 18 ATP and 12 NADPH consumed (2).
• Justification: 6 turns fix 6 CO₂; reduction uses 12 ATP + 12 NADPH (2 per turn × 6); regeneration uses 6 more ATP (1 per turn × 6) → 12 + 6 = 18 ATP, 12 NADPH (2).

(c) (2): RuBisCO catalyses carboxylation of RuBP (fixes CO₂) (1); inefficient because it also binds O₂ (oxygenase activity → photorespiration) and/or has a very slow turnover rate (1).

Question 5 (12)

(a) (4): When O₂ concentration is high / CO₂ low (hot, closed stomata) (1), RuBisCO's oxygenase activity fixes O₂ instead of CO₂ (1), producing one 3C + one 2C (phosphoglycolate) instead of two 3C molecules (1). Recovering the 2C molecule wastes ATP and releases previously-fixed CO₂, lowering net carbon gain (1).

(b) (6) — 2 marks each:

• C3: no separation — RuBisCO fixes CO₂ directly in mesophyll; suffers photorespiration in heat.
• C4: spatial separation — CO₂ fixed by PEP carboxylase in mesophyll into a 4C acid, shuttled to bundle-sheath cells where CO₂ is concentrated around RuBisCO, suppressing oxygenase activity.
• CAM: temporal separation — stomata open at night, CO₂ fixed into 4C acid and stored; by day stomata close and CO₂ is released internally for the Calvin cycle.

(c) (2): CAM (1) — closing stomata by day minimises water loss in hot/dry conditions while still supplying CO₂, and the concentrated CO₂ suppresses photorespiration (1). (C4 acceptable with strong water-availability justification.)

Question 6 (6)

(a) (3): any three of — light intensity, CO₂ concentration, temperature (also acceptable: water, chlorophyll amount). 1 each.

(b) (3): The plateau means light is no longer limiting (rate independent of increasing light) (1); another factor now caps the rate (1). Identify it by holding light high and varying one factor at a time — the factor whose increase raises the plateau is the current limiting factor (1).

[
  {"claim":"Overall equation O-atoms balance: 24 = 24 (12/6 water form)","code":"left = 6*2 + 12*1\nright = 6 + 6*2 + 6*1\nresult = (left==24) and (right==24) and (left==right)"},
  {"claim":"Photolysis half-equation electron balance: 2 H2O gives 4 e-, 4 H+, 1 O2","code":"H_left=2*2\nO_left=2*1\nH_right=4\nO_right=2\nresult = (H_left==H_right) and (O_left==O_right)"},
  {"claim":"Calvin cycle per glucose consumes 18 ATP and 12 NADPH","code":"turns=6\nATP=turns*2 + turns*1\nNADPH=turns*2\nresult = (ATP==18) and (NADPH==12)"},
  {"claim":"Simple photosynthesis equation carbon and hydrogen balance","code":"C_left=6\nC_right=6\nH_left=6*2\nH_right=12\nresult = (C_left==C_right) and (H_left==H_right)"}
]