Answers · Limits on photosynthesis

Limits on photosynthesis — return questions key

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

1. RR1 · Return

Model answer. (a) Rate = volume ÷ time = 2.6 cm³ ÷ 4 min = 0.65 cm³/min. (b) Time = volume ÷ rate = 3.0 cm³ ÷ 0.65 cm³/min = 4.615… min, which is 4.6 min to one decimal place.

Checked automatically

(a) Rate of gas production: 0.65 cm³/min (exact value)
- Common wrong answer 10.4 : You multiplied. The rate is the volume divided by the time.
- Common wrong answer 1.54 : You divided the time by the volume. Divide the volume by the time.
- Common wrong answer 2.6 : That is the total volume. Divide it by the 4 minutes.
(b) Time to collect 3.0 cm³: 4.6 min (exact value)
- Common wrong answer 1.95 : You multiplied. Time = volume ÷ rate.
- Common wrong answer 0.22 : You divided the rate by the volume. Divide the volume by the rate.

A full answer includes

Method (a): I divided the volume by the time: 2.6 cm³ ÷ 4 min.
Answer (a): I wrote 0.65 cm³/min, to two decimal places.
Method (b): I divided the volume by the rate: 3.0 cm³ ÷ 0.65 cm³/min.
Answer (b): I wrote 4.6 min, to one decimal place.
I did not multiply the volume by the time.

Also accept

(a) 0.65 cm³/min; (b) 4.6 min

Watch for

“The rate is 2.6 × 4 = 10.4 cm³/min.” — A rate is the volume in each minute, so divide the volume by the time.
“The time is 3.0 × 0.65 = 1.95 min.” — At 0.65 cm³ each minute, 3.0 cm³ takes more than 4 minutes. Time = volume ÷ rate.

2. RR2 · Return

Model answer. (a) A closer lamp gives brighter light and raises the rate only until something else, such as the carbon dioxide or the temperature, holds it back. Then the rate levels off. (b) Sodium hydrogencarbonate is not food. It is a source of carbon dioxide, a raw material of photosynthesis; the plant makes its own food, glucose, with it. (c) Enzymes are proteins, not living things, so they cannot be killed. Above their optimum temperature they are denatured: their shape changes, so they work more slowly and the rate falls.

A full answer includes

(a) I wrote that the rate rises only until another factor, such as carbon dioxide or temperature, holds it back.
(a) I wrote that then the rate levels off, however close the lamp is.
(b) I wrote that sodium hydrogencarbonate supplies carbon dioxide, a raw material.
(b) I wrote that the plant makes its own food by photosynthesis.
(c) I wrote that enzymes are not alive, so they cannot be killed.
(c) I wrote that above the optimum the enzymes are denatured: their shape changes.
I did not write that the rate falls when the lamp is very close.

Also accept

(a) the rate levels off when another factor holds it back; (b) it supplies carbon dioxide, a raw material, not food; (c) enzymes are denatured, not killed

Watch for

“Statement (a) is wrong because a close lamp makes the rate fall.” — With a beaker of water to soak up the heat, the rate does not fall: it levels off.
“Statement (c) is wrong because the plant, not its enzymes, is killed.” — The plant still gives off bubbles at 45 °C. Its enzymes are denatured, and nothing is killed at once.

3. RR3 · Return

Model answer. Row B. Adding carbon dioxide raised the rate from 10 to 17, so carbon dioxide was the factor in shortest supply: the limiting factor. Moving the lamp closer gave more light, but light was not the limiting factor, so the rate hardly changed. Increasing a factor that is not limiting does not raise the rate.

A full answer includes

I chose row B.
I checked that carbon dioxide is the limiting factor: more of it raised the rate a lot.
I checked that more light hardly helped, because light was not in shortest supply.
I did not choose row A: if light were limiting, the closer lamp would raise the rate.

Also accept

Watch for

“Row A.” — If light were limiting, the closer lamp would have raised the rate. It hardly changed it.
“Row C.” — The beaker of water soaked up the heat, and the rate did not fall. The light simply was not the limiting factor.
“Row D.” — Adding carbon dioxide raised the rate a lot, at the same temperature, so temperature was not the limiting factor.

4. GR1 · Return

Options as numbered on the pupil sheet: 1 — red, as at the start; 2 — yellow; 3 — purple.

Model answer. Germinating peas respire and give out carbon dioxide, and in a sealed tube for a day the carbon dioxide rises a lot: yellow. Soda lime absorbs carbon dioxide, so air bubbled through after it has less carbon dioxide than ordinary air: purple. Outdoor air has the ordinary amount of carbon dioxide, and an empty sealed tube has nothing to change it, so both stay red.

A full answer includes

I matched the germinating peas to yellow.
I matched the air from the soda lime to purple.
I matched the outdoor air to red.
I matched the empty sealed tube to red.
I did not match the peas to purple: germinating peas have no green leaves yet, so they only respire.

Also accept

peas: yellow; air from soda lime: purple; outdoor air: red; empty tube: red

Watch for

“Air from the soda lime turns it yellow.” — Soda lime takes carbon dioxide out of the air, so the indicator gets less carbon dioxide and turns purple.
“The empty sealed tube turns purple over a day.” — With nothing living in it, nothing changes the carbon dioxide, so it stays red.

5. GR2 · Return

Model answer. Row B. In bright light the pondweed photosynthesises much faster than it respires and takes in more carbon dioxide than the snail gives out, so there is less carbon dioxide overall, but not as much less as with the pondweed alone: magenta, not purple. In the dim light the pondweed only balances its own respiration, so the carbon dioxide from the snail is not taken back in: more carbon dioxide, orange. In the dark nothing photosynthesises and both organisms respire, so the carbon dioxide rises more than with either alone: yellow.

A full answer includes

I chose row B.
I checked that in bright light the pondweed takes in more carbon dioxide than the snail and the pondweed give out: less carbon dioxide, magenta.
I checked that in dim light the pondweed only balances its own respiration, so the snail adds carbon dioxide: more carbon dioxide, orange.
I checked that in the dark both respire and nothing photosynthesises: more carbon dioxide, yellow.
I did not choose row A: it leaves out the carbon dioxide from the snail.

Also accept

Watch for

“Row A.” — Those are the colours for the pondweed on its own. The snail respires in every tube and adds carbon dioxide.
“Row C.” — In bright light the pondweed photosynthesises much faster than it respires, fast enough to take in the carbon dioxide from the snail as well.
“Row D.” — In the dim light the pondweed only just balances its own respiration. The light limits its photosynthesis, so it cannot take in the extra carbon dioxide from the snail.

6. GR3 · Return

Model answer. (a) First, her own result goes against her conclusion: magenta means less carbon dioxide than at the start, so in the light photosynthesis was faster than respiration. Second, the two tubes had different volumes of indicator. The same amount of carbon dioxide taken in or given out changes the colour of 5 cm³ of indicator much more than 20 cm³. So the light and the volume both changed, and she cannot compare the sizes of the two colour changes. She should use the same volume in both tubes. (b) For example, the size of the sprig, the temperature, or the time the tubes are left.

A full answer includes

(a) I wrote that the tube in the light turned magenta: less carbon dioxide than at the start.
(a) I wrote that so, in the light, photosynthesis was faster than respiration.
(a) I wrote that the volumes of indicator were different: 20 cm³ and 5 cm³.
(a) I wrote that the same amount of carbon dioxide changes the colour of 5 cm³ more than 20 cm³.
(b) I stated one more variable: for example the size of the sprig, the temperature or the time.
I did not write that the light tube was magenta because it had no photosynthesis.

Also accept

(a) magenta shows photosynthesis was faster than respiration in the light; the volumes differ, so the colours cannot be compared; (b) e.g. sprig size, temperature or time

Watch for

“The conclusion is right, because yellow is a bigger change than magenta.” — The tube in foil had a quarter of the volume, so the same carbon dioxide made a bigger change to its colour.
“The magenta tube shows that photosynthesis stopped.” — Magenta means less carbon dioxide than at the start, so photosynthesis was faster than respiration there.

7. LR1 · Return

Model answer. Row B. P is a stoma, the gap between two guard cells in the lower epidermis; S is one of the guard cells beside the other stoma. R is xylem, a wide, thick-walled tube on the upper side of the vascular bundle; M is the phloem, the small cells on its lower side. N is a palisade mesophyll cell, in the upper of the two rows of tall cells just under the upper epidermis: these cells have the most chloroplasts. Q is a spongy mesophyll cell, with fewer chloroplasts and air spaces around it.

A full answer includes

I chose row B.
I identified P as a stoma: the gap between two guard cells.
I identified R as xylem: a wide, thick-walled tube on the upper side of the bundle.
I identified N as a palisade mesophyll cell, the kind of cell with the most chloroplasts.
I did not choose row A: S is a guard cell, not the stoma.

Also accept

Watch for

“Row A.” — S is a cell, one of the two guard cells beside a gap. The stoma is the gap itself, like P.
“Row C.” — M marks the small cells in the lower half of the bundle: the phloem. The xylem is the wide, thick-walled tubes in the upper half, R.
“Row D.” — Q is a rounded spongy mesophyll cell with air spaces around it and fewer chloroplasts. The tall cells near the top, such as N, have the most.

8. LR2 · Return

Model answer. Most leaves have a large surface area, so they absorb more light and have a large surface for carbon dioxide to diffuse in. They are thin, so light reaches all the cells inside, and carbon dioxide and oxygen diffuse only a short distance.

A full answer includes

I stated a large surface area.
I explained that a large surface absorbs more light.
I stated that leaves are thin.
I explained that in a thin leaf light reaches all the cells.
I explained that in a thin leaf carbon dioxide diffuses only a short distance to the cells.
I did not write that a large surface area cuts down water loss.

Also accept

large surface area: more light; thin: light reaches all cells, short diffusion distance

Watch for

“A large surface area stops the leaf losing water.” — A large surface catches light; the waxy cuticle cuts down water loss.
“Leaves are thin to be light enough to float.” — Thinness lets light reach all the cells and keeps the diffusion distance short.

9. LR3 · Return

Model answer. Row C. The xylem brings water, a raw material of photosynthesis, and mineral ions into the leaf from the roots. The phloem carries the sucrose made from the glucose of photosynthesis out of the leaf to the rest of the plant. The air spaces let carbon dioxide diffuse from the stomata to the mesophyll cells.

A full answer includes

I chose row C.
I checked that the xylem brings water and mineral ions into the leaf.
I checked that the phloem carries sucrose out of the leaf.
I checked that the air spaces let carbon dioxide reach the mesophyll cells.
I did not choose row B: photosynthesis gives out oxygen, it does not need it.

Also accept

Watch for

“Row A.” — The xylem and the phloem are swapped. The xylem brings water in; the phloem carries sucrose out.
“Row B.” — Photosynthesis gives out oxygen. The gas it needs from the air spaces is carbon dioxide.
“Row D.” — The glucose made in photosynthesis is changed to sucrose before the phloem carries it out of the leaf.