Questions on the Calvin cycle

Question 1

Each turn of the Calvin cycle fixes one molecule of carbon dioxide and produces two molecules of GP. Reducing each GP to triose phosphate uses one ATP and one reduced NADP, and regenerating the RuBP used in each turn uses one further ATP. Calculate the number of turns of the 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.

Accepted Answer

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

Question 2

Two identical flasks of an algal suspension are photosynthesising steadily in bright light, each bubbled with air containing carbon dioxide. At time zero the gas stream to one flask is switched to carbon dioxide-free air, the light being left on; the second flask is left on the original gas stream. Samples from both are run into boiling methanol at intervals and their GP and RuBP measured, giving the constructed readings in the table. Explain what the control flask contributes to this investigation, and explain what the readings from the treated flask show about the step at which carbon dioxide acts.

Accepted Answer

B1 the control flask is set up, lit, warmed, bubbled and sampled in exactly the same way and at the same times, and differs in one thing only, the carbon dioxide in its gas stream, so it isolates that one variable B1 its readings stay level throughout, GP between 118 and 121 and RuBP between 59 and 61 arbitrary units, which shows that neither time spent in the flask nor the sampling and killing procedure changes the concentrations by itself; without that comparison the fall in GP in the treated flask could have been caused by the method rather than by the treatment B1 in the treated flask GP falls from 120 to 6 arbitrary units, because GP is produced only by fixation, which has stopped for want of carbon dioxide, while ATP and reduced NADP are still arriving in the light so GP goes on being reduced to triose phosphate B1 RuBP rises from 60 to 130 arbitrary units over the same period, because it is still being regenerated from triose phosphate using the ATP the light-dependent stage is still supplying, and is no longer being consumed by fixation A1 RuBP therefore lies immediately before the blocked step and GP immediately after it, which places the action of carbon dioxide at fixation onto RuBP by rubisco and nowhere later in the cycle. The sizes agree as well: GP falls by 114 arbitrary units while RuBP rises by only about 70, and five triose phosphate are needed to rebuild three RuBP, so 114 × 3 ÷ 5 = 68 is what regeneration predicts. The reasoning is set out in the lesson section 'Predicting the effect of removing light or carbon dioxide'

Question 3

A wheat leaf is held in a lit chamber at a constant high light intensity, and its temperature raised in steps, the composition of the air supplied to the chamber unchanged. The active site of rubisco can bind oxygen instead of carbon dioxide, which gas it binds depends on the ratio of the two around it, and a reaction with oxygen fixes no carbon. The table gives constructed measurements taken once the leaf had settled at each temperature. Explain how the first three columns of measurements account for the fall in the net uptake of carbon dioxide, and explain why the net uptake falls proportionally very much further than the share of rubisco's reactions that fix carbon dioxide does.

Accepted Answer

B1 as the leaf warms its stomata close to conserve water, so the carbon dioxide in the air spaces falls steeply, from 270 to 120 µmol mol⁻¹, while the oxygen released by the light-dependent stage builds up a little in a leaf that can vent it less freely, from 21.0 to 21.8 % B1 the active site of rubisco accepts either gas, and which one it binds depends on the ratio of the two around it; that ratio falls to well under half its starting value and the falling carbon dioxide does nearly all of the work, since the oxygen figure moves by less than a twentieth, so the share of reactions that fix carbon falls from 78 % to 60 %; accept also that rubisco's preference for carbon dioxide over oxygen falls as the leaf warms B1 the two falls are of quite different sizes: net uptake falls to 3.4 ÷ 18.4 = 0.18, that is to 18 % of its starting value, whereas the fixing share falls only to 60 ÷ 78 = 0.77, or 77 %, so the shift in the ratio cannot on its own account for the collapse in uptake B1 rubisco is nowhere near saturated at these concentrations, so the falling carbon dioxide also slows every carboxylation that does still happen, as well as making carboxylations a smaller share of the total; fewer of them, and each one slower, multiply together rather than adding A1 what the last column measures is a net figure, gross fixation minus the carbon dioxide the leaf's own respiration releases, and respiration rises steeply with temperature across this range while gross fixation is falling; a growing loss subtracted from a shrinking gain falls faster than either does alone, which is why the net uptake drops away ahead of every other column

Question 4

Describe the three stages of the Calvin cycle, naming the compounds involved and saying what reduced NADP and ATP each contribute.

Accepted Answer

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

Question 5

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.

Accepted Answer

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

Question 6

A plant photosynthesising steadily in bright light has its carbon dioxide supply cut off completely, with the light kept on. Explain what happens to the concentrations of RuBP and of GP in the stroma over the following minute, and state which step has become blocked.

Accepted Answer

B1 the blocked step is fixation itself, since rubisco has no carbon dioxide to add to RuBP B1 RuBP rises: it is immediately before the block, so it is still being regenerated from triose phosphate, using ATP the light reactions are still supplying, but is no longer being used up by fixation B1 GP falls: it is immediately after the block, so it is no longer being made by fixation, but reduced NADP and ATP are still available and it continues to be reduced to triose phosphate A1 because GP falls, triose phosphate is affected as well: it continues to be made from what GP remains and is still used in regeneration, so once the GP supply is exhausted triose phosphate falls too, a little later than GP

Question 7

Describe two of the biosynthetic uses for triose phosphate or GP that leave the Calvin cycle, other than forming glucose.

Accepted Answer

B1 triose phosphate can be converted into the glycerol part of a lipid molecule B1 GP's carbon skeleton, combined with nitrogen from nitrate absorbed by the roots, can be converted into an amino acid B1 fatty acids for a lipid molecule can also be built from triose phosphate or GP, so a lipid's glycerol and its fatty acid chains can both trace back to the Calvin cycle A1 amino acid synthesis by this route needs nitrate taken up from the soil as well as fixed carbon, tying the Calvin cycle's output to mineral nutrition as well as to carbon dioxide

Question 8

In a C3 plant such as wheat, rubisco in the mesophyll cells fixes carbon dioxide that has diffused in from the air through the stomata. In a C4 plant such as maize, a different enzyme in the mesophyll cells, which does not react with oxygen, first fixes carbon dioxide into a four-carbon compound; this passes into cells around the leaf veins and releases carbon dioxide there, around rubisco, at several times the concentration in the air. Rubisco can react with oxygen in place of carbon dioxide, which wastes energy and fixes less carbon. Explain why maize outproduces wheat in a hot, dry, bright climate.

Accepted Answer

B1 in hot, dry conditions the stomata partly close to reduce water loss, so carbon dioxide inside the leaf falls while oxygen from photosynthesis builds up B1 in wheat, rubisco is exposed to this low carbon dioxide and high oxygen, so it reacts with oxygen more often and less carbon is fixed B1 in maize, the first enzyme goes on fixing carbon dioxide even at low concentration, because oxygen does not compete with it, and the carbon dioxide it delivers keeps the concentration around rubisco high B1 so maize's rubisco rarely reacts with oxygen and fixation stays fast even with the stomata partly closed, giving higher productivity

Question 9

In an experiment, a chloroplast suspension fixes 18 molecules of carbon dioxide in one minute. Each turn of the Calvin cycle fixes one molecule of carbon dioxide and produces two molecules of GP. Reducing each GP to triose phosphate uses one ATP and one reduced NADP, and regenerating the RuBP used in each turn uses one further ATP. Calculate the number of complete turns of the cycle this represents, the number of molecules of GP produced, and the total number of ATP molecules used inside the cycle, assuming exactly one RuBP is regenerated for each one used.

Accepted Answer

M1 one turn of the cycle fixes one molecule of carbon dioxide, so 18 molecules of carbon dioxide fixed is 18 turns of the cycle A1 each turn produces two molecules of GP, so 18 × 2 = 36 molecules of GP M1 reducing 36 GP uses 36 ATP, and regenerating 18 RuBP uses a further 18 ATP A1 total ATP used = 36 + 18 = 54 molecules of ATP

Question 10

A suspension of algae photosynthesising steadily in bright light was switched into complete darkness at time zero, its carbon dioxide supply unchanged. The table gives the concentrations of GP, of triose phosphate and of RuBP in the stroma, all on the same scale, over the minute that followed. Calculate the change in the concentration of GP and the change in the concentration of RuBP between 0 s and 60 s, calculate the ratio of the two changes, and state what that ratio shows about the reaction still running. The readings are constructed for this question rather than measured.

Accepted Answer

M1 reads both compounds from the 0 s row and the 60 s row and takes the difference in each: GP 260 − 100, RuBP 0 − 80 A1 GP rises by 160 arbitrary units and RuBP falls by 80 arbitrary units; the stem states that the three compounds are measured on the same scale, which is what makes the two changes comparable at all A1 ratio of the rise in GP to the fall in RuBP = 160 ÷ 80 = 2 : 1, a ratio with no units B1 rubisco joins one carbon dioxide to one RuBP and the six-carbon product splits at once into two molecules of GP, so every RuBP consumed yields exactly two GP; a 2 : 1 ratio held over the whole minute is therefore what you would see if fixation is the only step still running, with the reduction of GP already stopped for want of ATP and reduced NADP. The one-carbon-in, two-molecules-out arithmetic is set out in the lesson section 'Three steps, and where the carbon goes'

Question 11

The table gives constructed concentrations of GP, triose phosphate and RuBP in the stroma of an alga that was photosynthesising steadily in bright light until the light was switched off at time zero, its carbon dioxide supply unchanged. Explain why the concentration of GP stops rising after 40 s although the alga is still in darkness, and explain why the concentration of triose phosphate falls rather than rising.

Accepted Answer

B1 GP is made only by fixation, and fixation needs an acceptor; the table shows RuBP has reached zero by 40 s, so from 40 s onwards there is no RuBP left for carbon dioxide to be fixed onto and no further GP can be made B1 the route by which nearly all GP is removed is its reduction to triose phosphate, and that needs ATP and reduced NADP, both of which are made only in the light-dependent stage and are not stored in quantity, so removal stopped within seconds of the switch-off; neither made nor used, GP holds level at 260 arbitrary units B1 triose phosphate lies immediately after the blocked step, so it is no longer being made from GP A1 it is still being drawn out of the cycle into the plant's products, so it is used and not replaced and falls almost to nothing, from 60 arbitrary units to 1. The rule is set out in the lesson section 'Predicting the effect of removing light or carbon dioxide'

Question 12

In an investigation of the Calvin cycle, a thin, flat, transparent flask holding a suspension of the alga Chlorella is lit from both sides, held at 25 °C in a water bath, stirred continuously and bubbled with a gas stream of known composition. At intervals a tap in the base of the flask is opened and a sample of the suspension runs straight down into boiling methanol. Suggest why the sample must run into boiling methanol rather than into an empty cold tube, and suggest why the flask is made thin and flat and the suspension is stirred.

Accepted Answer

B1 the intermediates of the Calvin cycle turn over in seconds, so a sample that goes on reacting while it is being collected gives the concentrations at some unknown later moment rather than at the moment the tap was opened B1 boiling methanol denatures the enzymes of the cycle, rubisco among them, the instant the sample meets it, so every reaction stops at a known time and the concentrations measured are the ones the cells were actually holding; it also dissolves the compounds out of the cells ready to be separated B1 light is absorbed as it passes through a suspension, so in a thick flask the cells at the back would be shaded by those in front of them; thin, flat and lit from both sides means every cell receives about the same light intensity A1 stirring keeps the dissolved carbon dioxide even throughout, so when the gas stream is changed every cell meets the new concentration at the same moment; without it a sample would be an average over cells in different conditions, and any change would look slower and less sharp than it really is

Question 13

Algae photosynthesising in bright light were given carbon dioxide labelled with carbon-14, and samples were killed at intervals and their compounds separated. The table gives the share of the labelled carbon found in each compound, constructed for this question rather than measured. Explain why the label appears in triose phosphate before it appears in RuBP, even though RuBP is the molecule that carbon dioxide is fixed onto.

Accepted Answer

B1 carbon dioxide is joined to RuBP, but the labelled carbon does not stay there: the six-carbon compound formed is unstable and splits at once into two molecules of GP, so the RuBP that accepted the label no longer exists and the label is already in GP, which is why 92 % of it is in GP after only 2 s B1 RuBP can only become labelled once labelled carbon is built back into it, and the one route to RuBP is regeneration from triose phosphate, since RuBP is not supplied to the chloroplast from outside B1 triose phosphate therefore lies between GP and RuBP along the path the label takes, which is why the table shows more label in triose phosphate than in RuBP at 5 s, 14 % against 6 %, and still at 15 s, 20 % against 17 % A1 the order in which a label appears is evidence for the order of the reactions, so this result is itself the evidence that RuBP is rebuilt from triose phosphate inside the cycle rather than fed into it; the half of the label sitting in sugars and starch by 120 s is the fraction that has left the cycle altogether. Regeneration is set out in the lesson section 'Three steps, and where the carbon goes'

Question 14

The table counts what the Calvin cycle uses and produces over nine turns and over twelve turns, on the assumption that the triose phosphate not needed for regeneration leaves the cycle. Four cells are marked with a question mark. Complete the table by working out those four values, giving your reasoning for each.

Accepted Answer

M1 every row is proportional to the number of turns, so the value per turn found from the complete column gives the missing value in the other A1 GP made over twelve turns is 24, at 2 per turn, and triose phosphate made over nine turns is 18, also 2 per turn, one for each GP reduced A1 triose phosphate leaving over twelve turns is 4, since 3 over nine turns is one for every three turns A1 ATP used over nine turns is 27, since 36 over twelve turns is 3 per turn: one for each of the two GP reduced and one to regenerate the RuBP

Question 15

A grower raises the carbon dioxide concentration in a glasshouse in steps, with light intensity and temperature held constant, and records the constructed readings in the table. The grower concludes: 'Between 0.04 % and 0.10 % the GP concentration rose by 60 per cent, so the Calvin cycle is running 60 per cent faster.' Evaluate that conclusion.

Accepted Answer

B1 the direction of the conclusion is supported. The measured rate of carbon dioxide uptake did rise, from 12.0 to 16.5 µmol m⁻² s⁻¹, and the fall in RuBP from 120 to 66 arbitrary units agrees with it, since more carbon dioxide means RuBP is consumed by fixation faster than it is regenerated B1 the size of the conclusion is not supported, because it reads a level as though it were a rate. The rise in the measured rate is (16.5 − 12.0) ÷ 12.0 × 100 = 37.5 %, not 60 % B1 a concentration of an intermediate rises whenever it is made faster than it is used up. The reduction of GP did speed up, since at a steady rate of uptake GP is used as fast as it is made, but light intensity was held constant, so the supply of ATP and reduced NADP could rise only so far; GP had to build up before its reduction kept pace with the faster fixation, so part of the 60 % rise is GP queueing in front of that step, not carbon travelling round the cycle faster A1 so the judgement is that the grower's direction is right and the figure is wrong, and only the rate column could have supported a figure at all. The last row makes the same point again: from 0.10 % to 0.14 % the rate rises by only 0.4 µmol m⁻² s⁻¹, so carbon dioxide has nearly stopped being the limiting factor, while GP has still moved. Levels and rates are separated in the lesson section 'Predicting the effect of removing light or carbon dioxide'

Question 16

A plant photosynthesising in bright light with plenty of carbon dioxide is cooled from 25 °C to 10 °C. Explain the effect on the concentrations of RuBP and of GP in the stroma.

Accepted Answer

B1 at the lower temperature molecules have less kinetic energy, so fewer enzyme-substrate complexes form and the enzymes of the Calvin cycle, rubisco among them, work more slowly B1 states which steps are taken to slow: every step of the cycle together (fixation, reduction of GP and regeneration of RuBP), or fixation by rubisco, the slowest step, most of all B1 GP falls, because fixation of carbon dioxide onto RuBP is slower so GP is made more slowly; accept that it is still being reduced to triose phosphate B1 RuBP falls, because less triose phosphate is made and the regeneration of RuBP is slower; OR RuBP rises, because fixation slows most, so RuBP is still regenerated from triose phosphate but used more slowly; credit either direction only with its matching reason

Question 17

The table gives constructed concentrations of GP and triose phosphate in the stroma of a leaf at 25 °C and ten minutes after it was heated to 45 °C, in bright light with plenty of carbon dioxide throughout. Explain the changes, suggest the effect on the concentration of RuBP, and explain why the rate of photosynthesis stayed low when the leaf was cooled to 25 °C again.

Accepted Answer

B1 45 °C is above the optimum, so enzymes including rubisco denature: the active site changes shape and substrates no longer fit B1 fixation of carbon dioxide onto RuBP slows, so GP is made more slowly and falls from 100 to 55, and triose phosphate falls from 60 to 35 because less GP is reduced B1 RuBP falls, because the enzymes that regenerate it are denatured too and less triose phosphate is available; OR RuBP rises, because rubisco is the step most affected, so RuBP is still regenerated but used more slowly; credit either direction only with its matching reason B1 denaturation is permanent: the enzymes do not regain their shape when the leaf cools, so the rate stays low; accept that stomata close at high temperature, cutting carbon dioxide supply, as a further cause of the changes

Question 18

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.

Accepted Answer

B1 the ten that remain hold thirty carbons, which is exactly the number needed to rebuild the six molecules of RuBP the six turns started from B1 RuBP is not supplied from outside the chloroplast, so the only source of it is regeneration from triose phosphate B1 if all twelve left, there would be no acceptor for the next carbon dioxide and the cycle would stop after a single round, so no further carbon could be fixed

Question 19

Rubisco can bind oxygen at its active site in place of carbon dioxide. When it does, RuBP is split into one GP and a two-carbon compound, and the plant spends ATP converting the two-carbon compound back into GP, losing some of its carbon as carbon dioxide. 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.

Accepted Answer

B1 the stomata close to reduce water loss, so less carbon dioxide enters the leaf while oxygen from the light-dependent reaction builds up inside it B1 oxygen and carbon dioxide compete for the same active site, so with the ratio of oxygen to carbon dioxide higher, rubisco binds oxygen more often B1 each time it does, one GP is made instead of two, ATP is spent and carbon is lost in the recovery, so less carbon is fixed overall and productivity falls

Question 20

When light intensity is raised in an otherwise unchanged, already photosynthesising plant, the concentration of GP in the stroma is found to fall, even though the overall rate of the Calvin cycle has increased. Suggest why a fall in GP concentration does not mean the cycle has slowed down.

Accepted Answer

B1 raising light intensity increases the supply of ATP and reduced NADP, so GP is reduced to triose phosphate faster than before B1 fixation, the only source of GP, also speeds up as RuBP rises, but at first by less than reduction does, since the carbon dioxide supply has not changed, so for a time GP is used up faster than it is made and its concentration falls to a new, lower level A1 a falling concentration of an intermediate shows only that it is being consumed faster than it is produced at that moment; it does not mean the pathway making it has slowed

Question 21

Explain why calling the Calvin cycle the 'dark stage' is misleading, even though none of its steps directly absorbs a photon of light.

Accepted Answer

B1 it is true that no step in the Calvin cycle itself absorbs light directly, so in principle the reactions could run in a darkened flask if ATP and reduced NADP were supplied artificially B1 in an intact, real chloroplast, ATP and reduced NADP are made only in the light-dependent reactions and neither is stored in any quantity A1 within seconds of the light being switched off, the cycle stalls for lack of its own raw materials, meaning that in practice the cycle does depend on light, and 'light-independent' is the more accurate name

Question 22

Rubisco can account for half of all the soluble protein in a leaf. Each rubisco molecule fixes only about three molecules of carbon dioxide a second, whereas many enzymes act on thousands of substrate molecules a second. Explain why a plant needs to make so much of this one enzyme.

Accepted Answer

B1 when substrate is plentiful, the rate of an enzyme-catalysed reaction is proportional to the number of enzyme molecules, so with each rubisco working this slowly, a high rate of fixation needs a very large number of them B1 fixation by rubisco is the only route by which carbon enters the cycle, so it limits the rate at which the plant can make all of its organic compounds B1 plenty of rubisco lets fixation keep pace with the ATP and reduced NADP supplied by the light-dependent reaction in bright light, so that supply is not wasted

Question 23

Carbon dioxide concentration around an already photosynthesising plant is suddenly raised, with light intensity kept the same. Suggest what happens to the concentrations of RuBP and of GP in the stroma as a result, explaining your reasoning.

Accepted Answer

B1 with more carbon dioxide available, rubisco fixes it onto RuBP faster than before, so GP is produced faster and its concentration rises B1 RuBP is used up by fixation faster than before, and because reduced NADP and ATP have not increased, the reduction of GP and the regeneration of RuBP proceed at their previous rate, so RuBP falls as it is consumed faster than it is replaced A1 this shows fixation was the step limited by carbon dioxide before the increase; if reduction or regeneration had instead been the limiting step, raising carbon dioxide would have made little difference to either concentration

Question 24

A plant growing in soil that lacks nitrate ions and phosphate ions photosynthesises in bright light but grows very little. Explain why it cannot make nucleic acids from the products of the Calvin cycle alone.

Accepted Answer

B1 the pentose sugar of each nucleotide, ribose or deoxyribose, is made from the sugar phosphates the Calvin cycle produces, so only this part comes from photosynthesis B1 the nitrogenous bases contain nitrogen, which has to be taken up from the soil as nitrate (or ammonium) ions B1 the phosphate groups come from phosphate ions absorbed by the roots, so without both ions nucleotides cannot be assembled and DNA and RNA cannot be made for new cells

Question 25

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.

Accepted Answer

B1 the enzyme is rubisco, also called RuBP carboxylase B1 one turn produces two molecules of GP

Question 26

State what is meant by carbon fixation, and state where in the chloroplast fixation takes place.

Accepted Answer

B1 carbon fixation is the joining of carbon from an inorganic molecule, carbon dioxide, to an organic molecule already present in the cell, converting the inorganic carbon into an organic form B1 fixation takes place in the stroma, where rubisco and the other Calvin cycle enzymes are dissolved

Counted over	Nine turns	Twelve turns
Carbon dioxide fixed	9	12
GP made	18	?
Triose phosphate made	?	24
Triose phosphate leaving the cycle	3	?
ATP used	?	36
Reduced NADP used	18	24