Answers · Limits on photosynthesis

Limits on photosynthesis — final check key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. RK1 · Final check

Model answer. (a) Put a sprig of pondweed upside down in a boiling tube of water, with the cut end of its stem uppermost. Place a lamp at a measured distance, for example 55 cm, in a darkened room. Wait about 2 minutes for the plant to adjust, then count the bubbles given off in 1 minute; count again and work out the mean. Repeat with the lamp at 35, 15 and 5 cm: a closer lamp gives brighter light. (b) (i) Stand the tube in a water bath with a thermometer, and put a beaker of water between the lamp and the tube to soak up the heat of the lamp. (ii) Use the same water, with the same mass of sodium hydrogencarbonate in it, at every distance.

A full answer includes

(a) I put the pondweed upside down in a tube of water, cut end up.
(a) I set the lamp at several measured distances from the plant.
(a) I waited about 2 minutes after each move before counting.
(a) I counted the bubbles for a fixed time, such as 1 minute.
(a) I counted at least twice at each distance.
(a) I worked out the mean of the counts.
(b)(i) I wrote: stand the tube in a water bath with a thermometer.
(b)(i) I wrote: put a beaker of water between the lamp and the tube to soak up its heat.
(b)(ii) I wrote: use the same water, with the same sodium hydrogencarbonate, at every distance.
I did not count only once at each distance.

Also accept

(a) pondweed in water, lamp at measured distances, wait, count bubbles per minute twice, mean; (b) (i) water bath and a beaker of water as a heat filter; (ii) the same water and sodium hydrogencarbonate

Watch for

“Measure the light by how far away the lamp is, and calculate the intensity from it.” — Measure the distance, and read the results as "closer lamp, brighter light".
“Count once at each distance.” — Two counts, with a mean, give a more reliable result.
“The water bath alone keeps the temperature the same.” — A close lamp can still warm the tube. The beaker of water between them soaks up much of the heat.

2. RK2 · Final check

Model answer. (a) The rate rises as the temperature rises, from 12 bubbles per minute at 15 °C to 26 at 35 °C, the fastest of the temperatures tested. Then it falls to 9 at 45 °C. (b) The reactions of photosynthesis are controlled by enzymes. As the temperature rises towards their optimum, they work faster. Above the optimum, the enzymes start to be denatured: their shape changes, they work more slowly, and the rate falls.

A full answer includes

(a) I wrote that the rate rises from 12 at 15 °C to 26 at 35 °C.
(a) I wrote that it then falls to 9 at 45 °C.
(b) I explained the rise: the enzymes work faster when warmer, up to their optimum.
(b) I explained the fall: above the optimum the enzymes are denatured, so they work more slowly.
I did not write that the enzymes are killed.

Also accept

(a) rises to 35 °C, then falls; (b) enzymes faster when warmer, then denatured above the optimum

Watch for

“The enzymes are killed at 45 °C.” — Enzymes are not alive. They are denatured: their shape changes.
“The optimum is exactly 35 °C.” — It is near 35 °C; with only these temperatures tested, it could be a little higher or lower.

3. RK3 · Final check

Options as numbered on the pupil sheet: 1 — light; 2 — temperature; 3 — carbon dioxide.

Model answer. P at light intensity 1: all four conditions give 4, so only more light can raise the rate: light is limiting. P at light intensity 8: the rate stayed at 8 from light intensity 4 to 8, so light is not limiting; Q, with extra carbon dioxide at the same 15 °C, is also 8, so carbon dioxide is not limiting; R, at 25 °C, reaches 14, so temperature is limiting. R at light intensity 8: S, at the same 25 °C but with extra carbon dioxide, reaches 22, so carbon dioxide is limiting. S at light intensity 2: doubling the light to 4 doubles the rate from 8 to 16, so light is limiting.

A full answer includes

I matched P at light intensity 1 to light.
I matched P at light intensity 8 to temperature.
I matched R at light intensity 8 to carbon dioxide.
I matched S at light intensity 2 to light.
For P at 8, I wrote that more light did not raise the rate: it is 8 at light intensity 4 and at 8.
For P at 8, I wrote that extra carbon dioxide did not raise it: Q is also 8.
For P at 8, I wrote that a higher temperature raised it: R, at 25 °C, is 14.
I did not match R at light intensity 8 to temperature: S, at the same 25 °C, is higher.

Also accept

P at 1: light; P at 8: temperature; R at 8: carbon dioxide; S at 2: light

Watch for

“P at light intensity 8 is limited by light.” — From light intensity 4 to 8 the rate under P stayed at 8, so more light does not help.
“P at light intensity 8 is limited by carbon dioxide.” — Q has extra carbon dioxide at the same 15 °C and is also 8. Warming the plant to 25 °C (R) raised the rate to 14.
“R at light intensity 8 is limited by temperature.” — Compare R with S, both at 25 °C: extra carbon dioxide raised the rate from 14 to 22.
“R at light intensity 8 is limited by light.” — From light intensity 4 to 8 the rate under R rose only from 13 to 14, but extra carbon dioxide (S) took it to 22.
“At low light, temperature or carbon dioxide is limiting.” — At light intensity 1 all four conditions give 4, and under S the rate doubles from 8 to 16 when the light doubles from 2 to 4. Only more light raises it there.

4. RK4 · Final check

Model answer. (a) At 0.4 g the counts are 17, 19, 7 and 19. The 7 is far from the other three, so it is the anomalous count. (b) Leave it out. Mean = (17 + 19 + 19) ÷ 3 = 55 ÷ 3 = 18.33…, which is 18.3 bubbles per minute to 3 significant figures. (c) Sodium hydrogencarbonate is a source of carbon dioxide. The mean rate rises from 3.5 bubbles per minute with none added to 9 at 0.2 g, 18.3 at 0.4 g and 23 at 0.8 g. From 0.8 g to 1.2 g it rises only from 23 to 24: it levels off, because something else, such as the light, now holds the rate back.

Checked automatically

(a) Anomalous count: 7 (exact value)
- Common wrong answer 17 : That count is close to the other counts at 0.4 g. Look for the one far from the rest.
- Common wrong answer 3 : That count fits the other counts at 0 g. Look along each row for the one far from the rest.
(b) Mean rate at 0.4 g without it: 18.3 min-1 (exact value)
- Common wrong answer 15.5 : That is the mean of all four counts. Leave out the anomalous count first.
- Common wrong answer 55 : That is the total of the three counts. Divide it by the number of counts.

A full answer includes

(a) I identified 7, at 0.4 g, as the anomalous count.
Method (b): I added the other three counts at 0.4 g, 17 + 19 + 19 = 55, and divided by 3.
Answer (b): I wrote 18.3 bubbles per minute, to 3 significant figures.
(c) I wrote that the rate rises as the mass of sodium hydrogencarbonate rises, up to 0.8 g.
(c) I wrote that from 0.8 g to 1.2 g the rate hardly changes (23 to 24): it levels off.
I did not include the 7 in the mean.
I did not write that the rate rises steadily all the way to 1.2 g.

Also accept

(a) 7; (b) 18.3 bubbles per minute; (c) the rate rises with the carbon dioxide concentration up to 0.8 g, then levels off

Watch for

“Every count must go into the mean.” — An anomalous count does not fit the pattern, so it is left out. Including it would pull the mean down to 15.5.
“The rate rises steadily all the way to 1.2 g.” — From 0.8 g to 1.2 g the mean rises only from 23 to 24. The rate has levelled off.

5. GK1 · Final check

Model answer. (a) After 7 hours in the light the indicator is purple. In the dark cupboard it changes back through red towards orange, and then yellow. (b) In the light the pondweed photosynthesises faster than it respires, so it takes in more carbon dioxide than it gives out, and the carbon dioxide in the indicator falls. In the dark there is no photosynthesis; the pondweed only respires, giving out carbon dioxide, so the carbon dioxide in the indicator rises again, past the starting level.

A full answer includes

(a) I predicted purple after the 7 hours in the light.
(a) I predicted that in the cupboard it changes back through red towards orange and yellow.
(b) I wrote that in the light photosynthesis is faster than respiration, so carbon dioxide is taken out of the indicator.
(b) I wrote that in the dark there is no photosynthesis.
(b) I wrote that respiration goes on giving out carbon dioxide, so the carbon dioxide rises again.
I did not write that the plant stops respiring in the light.

Also accept

(a) purple, then back through red to orange or yellow; (b) light: photosynthesis faster than respiration, less carbon dioxide; dark: respiration only, more carbon dioxide

Watch for

“In the light the plant stops respiring.” — Respiration goes on all the time; photosynthesis is faster.
“Purple shows the oxygen given out.” — The indicator shows carbon dioxide only.

6. GK2 · Final check

Model answer. Put the same volume of red hydrogencarbonate indicator into each of several boiling tubes. Add similar sprigs of pondweed to all but one. Leave one plant tube uncovered, wrap one in foil for darkness, and, if you like, shade one with gauze. The tube with indicator but no plant is the control. Close every tube with a bung, so no gas gets in or out, and stand them all at the same distance from the same lamp, at the same temperature, for a few hours. Then remove the covers and compare the colours against a white background.

A full answer includes

I put the same volume of red indicator in each tube.
I put similar sprigs of pondweed in the test tubes.
I left one plant tube in the light and wrapped one in foil for the dark.
I set up a control tube with indicator and no plant.
I closed every tube with a bung.
I stood every tube at the same distance from the same lamp.
I kept every tube at the same temperature.
I left every tube for the same time.
I took off the covers and compared the colours against a white background.
I did not use a tube of plain water as the control.

Also accept

same indicator volume and pondweed; light, dark (foil); control with no plant; bungs; same lamp, temperature, time; compare colours

Watch for

“Use pondweed in tap water as the control.” — Tap water cannot change colour, so it does not show whether the indicator changes on its own.
“Leave the tubes open so the plant can get air.” — Open tubes exchange gases with the room, so the colour would not show only what the plant did.

7. GK3 · Final check

Model answer. A living plant respires all the time and gives out carbon dioxide. In dim light it can also photosynthesise at about the same rate as it respires. Then the carbon dioxide taken in balances the carbon dioxide given out, and the indicator stays red. So a red tube does not show that the plant is dead: a dead plant and a living plant in dim light both leave the indicator red.

A full answer includes

I wrote that a living plant respires all the time and gives out carbon dioxide.
I wrote that in dim light it can photosynthesise about as fast as it respires.
I wrote that the carbon dioxide taken in then balances the carbon dioxide given out, so the colour does not change.
I did not write that a living plant always changes the colour of the indicator.

Also accept

in dim light photosynthesis and respiration may balance, so a living plant can also leave the indicator red

Watch for

“A red tube always means the plant is dead.” — Living pondweed in dim light also leaves the indicator red, because its two processes balance.
“A living plant in the dark would also leave the indicator red.” — In the dark there is no photosynthesis, so a living plant only respires and the indicator turns yellow. The storeroom is dim, not dark.

8. LK1 · Final check

Options as numbered on the pupil sheet: 1 — upper epidermis; 2 — palisade mesophyll; 3 — spongy mesophyll; 4 — lower epidermis; 5 — stoma; 6 — guard cell; 7 — air space; 8 — cuticle; 9 — xylem.

Model answer. Put the layers in order by depth. X, 0.01 mm down, is a single row of cells with no chloroplasts: the upper epidermis. Z, at 0.07 mm, has tall cells packed side by side with about 50 chloroplasts each: the palisade mesophyll. W, at 0.16 mm, has rounded, loosely packed cells with fewer chloroplasts and large gaps, the air spaces: the spongy mesophyll. Y, the deepest at 0.24 mm, is another single row of clear cells with pores in it: the lower epidermis. Each pore is a stoma, between a pair of guard cells, the bean-shaped cells with chloroplasts. Layer Z lies just under the upper surface, where the light is brightest; its many chloroplasts absorb as much of the light as possible, and packing the tall cells side by side fits many of them into that bright layer.

A full answer includes

I matched X to the upper epidermis.
I matched Z to the palisade mesophyll.
I matched W to the spongy mesophyll.
I matched Y to the lower epidermis.
I matched the pores in Y to a stoma.
I matched the bean-shaped cells in Y to a guard cell.
I matched the large gaps in W to an air space.
I explained that layer Z is just under the upper surface, where the light is brightest.
I explained that each cell of Z has many chloroplasts, about 50, to absorb the light.
I explained that the cells of Z are packed side by side, so many of them fit where the light is.
I did not match the pores in Y to air spaces.

Also accept

W spongy mesophyll; X upper epidermis; Y lower epidermis; Z palisade mesophyll; pores: stoma; bean-shaped cells: guard cell; gaps in W: air space; Z near the top, many chloroplasts, tightly packed

Watch for

“Layer X is the cuticle, because it is the outermost layer.” — The cuticle is a thin waxy layer, not a row of cells. X is a single row of cells with no chloroplasts: the upper epidermis.
“The pores in Y are air spaces.” — Air spaces lie between the cells inside the leaf, as in W. A pore in the epidermis, between two bean-shaped cells, is a stoma.
“Layer W is the palisade mesophyll, because its cells have chloroplasts.” — W has about 15 chloroplasts in each cell and large gaps between its cells. The palisade, Z, is packed tight with about 50 in each cell.
“Layer Y is the upper epidermis, because it has no chloroplasts.” — Y is the deepest layer, 0.24 mm down, and it has the stomata: it is the lower epidermis.

9. LK2 · Final check

Model answer. (a) The middle of the needle is half its width from the surface: 1.0 ÷ 2 = 0.5 mm. The middle of the broad leaf is 0.2 ÷ 2 = 0.1 mm from its surface. 0.5 ÷ 0.1 = 5, so carbon dioxide must diffuse 5 times further. (b) In a leaf, carbon dioxide diffuses quickly through the air spaces to the mesophyll cells. With very few air spaces, it has to diffuse through the cells and the water in them, which is much slower. So the cells in the middle of the needle get less carbon dioxide, a raw material, and photosynthesise more slowly.

Checked automatically

(a) How many times further in the needle: 5 (exact value)
- Common wrong answer 0.2 : You divided the shorter distance by the longer one. Divide the longer distance by the shorter.
- Common wrong answer 0.4 : You subtracted the distances to the middle. Divide them to find how many times further.
- Common wrong answer 0.8 : You subtracted the widths. Divide them to find how many times further.

A full answer includes

Method (a): I found each distance to the middle, half the width: 0.5 mm for the needle and 0.1 mm for the leaf.
Method (a): I divided the longer distance by the shorter: 0.5 ÷ 0.1.
Answer (a): I wrote 5 times further.
(b) I wrote that carbon dioxide diffuses quickly through air spaces to the mesophyll cells.
(b) I wrote that with few air spaces it must diffuse through the cells instead, which is much slower.
(b) I wrote that the inner cells get less carbon dioxide, a raw material, so they photosynthesise more slowly.
I did not subtract the distances in (a).

Also accept

(a) 5 times further; (b) with few air spaces carbon dioxide diffuses slowly through cells, so the inner cells get less of it

Watch for

“The needle is 0.5 − 0.1 = 0.4 mm further, so the answer is 0.4.” — A difference in millimetres is not a number of times. Divide: 0.5 ÷ 0.1 = 5.
“Packed cells are better, because more cells fit into the needle.” — More cells photosynthesise faster only if carbon dioxide reaches them. With few air spaces it reaches the inner cells slowly.

10. LK3 · Final check

Model answer. (a) The bubbles come out of the stomata, the pores in the lower epidermis. The expanding air in the air spaces escapes through them. (b) The upper epidermis has few stomata, and the waxy cuticle over it lets very little gas through, so little air can get out of the upper surface. (c) The lower surface faces away from the sun, so it is shaded and cooler. Open stomata there lose less water vapour than they would on the hot upper surface, while carbon dioxide can still diffuse in through them for photosynthesis.

A full answer includes

(a) I identified the stomata, in the lower epidermis.
(b) I suggested that the upper epidermis has few stomata.
(b) I suggested that the waxy cuticle on the upper surface lets very little gas through.
(c) I suggested that the lower surface is shaded and cooler, so less water vapour is lost through its open stomata.
(c) I wrote that carbon dioxide can still diffuse in through these stomata for photosynthesis.
I did not identify the bubbles as oxygen from photosynthesis: they are air pushed out by the heat.

Also accept

(a) stomata in the lower epidermis; (b) few stomata in the upper epidermis, and the waxy cuticle lets little gas through; (c) the shaded lower surface loses less water, and carbon dioxide still gets in

Watch for

“The bubbles are oxygen from photosynthesis.” — The bubbles come at once in the hot water. They are air already in the air spaces, expanding with the heat; photosynthesis is far too slow to make so much gas.
“The air comes out through the cuticle of the lower surface.” — The waxy cuticle lets very little gas through. The air leaves through the pores, the stomata.
“The stomata are on the lower surface to take in water.” — Water reaches the leaf through the xylem. Stomata let gases in and out.