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The Calvin cycle: three steps, and the arithmetic behind them questions
The light-independent stage in the stroma: RuBP and carbon dioxide combined by rubisco to give two molecules of GP, GP reduced to TP using reduced NADP and ATP, TP either leaving the cycle or regenerating RuBP, the arithmetic of six turns per hexose, and predicting changes in GP, TP and RuBP when light or carbon dioxide is removed.
6 original questions · 21 marks · the calvin cycle: three steps, and the arithmetic behind them notes · Photosynthesis and primary productivity
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Calculate the number of turns of the Calvin cycle, the number of molecules of triose phosphate made, the number of molecules of ATP used and the number of molecules of reduced NADP used in producing two molecules of glucose.
Mark scheme
- M1 each turn fixes one carbon dioxide and a hexose holds six carbons, so six turns are needed per hexose and 6 × 2 = 12 turns in all
- M1 each turn makes two GP and each GP is reduced to one triose phosphate, so 2 × 12 = 24 molecules of triose phosphate
- A1 reducing 24 molecules of GP uses 24 ATP and 24 reduced NADP, one of each per molecule
- M1 regenerating RuBP costs one further ATP per turn, so 12 more ATP
- A1 36 molecules of ATP and 24 molecules of reduced NADP in total, which is eighteen ATP per hexose
Describe the three stages of the Calvin cycle, naming the compounds involved and saying what reduced NADP and ATP each contribute.
Mark scheme
- B1 fixation: rubisco joins one molecule of carbon dioxide to the five-carbon RuBP, and the six-carbon product is so unstable that it splits immediately into two molecules of GP
- B1 reduction: GP is reduced to triose phosphate, with reduced NADP supplying the hydrogen and ATP supplying the energy for the reaction
- B1 the oxidised NADP and the ADP go straight back to the thylakoid membranes to be used again
- B1 regeneration: most of the triose phosphate is rearranged, at a further cost in ATP, back into RuBP, while the rest leaves the cycle to be built into hexose sugars, lipids or amino acids
A plant photosynthesising steadily in bright light is placed in complete darkness, with its carbon dioxide supply unchanged. Explain what happens to the concentrations of GP, of triose phosphate and of RuBP in the stroma over the following minute.
Mark scheme
- B1 ATP and reduced NADP are made only in the light and are not stored in quantity, so within seconds the blocked step is the reduction of GP to triose phosphate
- B1 GP rises, because it lies immediately before the block: rubisco carries on fixing carbon dioxide onto the RuBP still present, so GP is still being made and is no longer being used
- B1 triose phosphate falls, because it lies immediately after the block: it is still being used and is no longer being made
- B1 RuBP falls as well, because it is still being consumed by fixation but is regenerated from triose phosphate, and there is no longer any triose phosphate to regenerate it from
Six turns of the Calvin cycle make twelve molecules of triose phosphate, and only two of them leave the cycle. Explain why the other ten cannot be allowed to leave.
On a hot, dry, bright day the productivity of a wheat crop falls even though light intensity is high. Suggest how the properties of rubisco account for this.
State the name of the enzyme that catalyses the fixation of carbon dioxide in the Calvin cycle, and state how many molecules of GP one turn of the cycle produces.
Practise the calvin cycle: three steps, and the arithmetic behind them one question at a time
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