Answers · Limits on photosynthesis

Measuring the rate of photosynthesis — answer key

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

1. RQ1 · Practice

Model answer. The bubbles are mainly oxygen, given out by photosynthesis. Collect the gas in an upside-down test tube of water over the plant, then put a glowing splint into it: the splint relights. Limewater tests for carbon dioxide, a squeaky pop with a lit splint is the test for hydrogen, and cobalt chloride paper tests for water.

A full answer includes

I chose: mainly oxygen, and a glowing splint relights in it.
I did not choose the squeaky pop: that is the test for hydrogen.

Also accept

oxygen; a glowing splint relights

Watch for

“Carbon dioxide: it turns limewater milky.” — In bright light photosynthesis takes in carbon dioxide faster than respiration gives it out, so the gas given out is mainly oxygen.
“Oxygen: a lit splint goes out with a squeaky pop.” — The gas is oxygen, but the squeaky pop is the test for hydrogen. Oxygen relights a glowing splint.
“Water vapour: it turns blue cobalt chloride paper pink.” — The bubbles rise through water, so they are not water vapour. They are the oxygen made by photosynthesis.
“The bubbles are pure oxygen.” — They are mainly oxygen: the oxygen left over after respiration, with a little nitrogen and carbon dioxide from the water.

2. RQ2 · Practice

Options as numbered on the pupil sheet: 1 — dependent variable; 2 — controlled variable; 3 — independent variable.

Model answer. The mass of sodium hydrogencarbonate is the independent variable: she changes it on purpose, to change the carbon dioxide concentration. The number of bubbles per minute is the dependent variable: she measures it. The distance from the lamp to the plant, the temperature of the water and the sprig of pondweed are controlled variables: they stay the same, so that only the carbon dioxide changes.

A full answer includes

I matched the mass of sodium hydrogencarbonate to the independent variable.
I matched the number of bubbles per minute to the dependent variable.
I matched the distance from the lamp to the plant to a controlled variable.
I matched the temperature of the water to a controlled variable.
I matched the sprig of pondweed to a controlled variable.
I did not match the lamp distance to the independent variable: in this investigation it stays the same.

Also accept

independent: mass of sodium hydrogencarbonate; dependent: bubbles per minute; controlled: lamp distance, temperature, the sprig

Watch for

“The lamp distance is the independent variable, because it was the one changed in the light test.” — In this investigation the lamp stays at one distance. Only the mass of sodium hydrogencarbonate is changed on purpose.
“The bubbles are the independent variable, because they change.” — The bubbles change because of what she did: they are what she measures, the dependent variable.

3. RQ3 · Practice

Model answer. (a) 22 is the anomalous count: it is far from the other three, which are all between 13 and 16. (b) Leave it out. Mean = (13 + 14 + 16) ÷ 3 = 43 ÷ 3 = 14.33…, which is 14.3 bubbles per minute to one decimal place.

Checked automatically

(a) Anomalous count: 22 (exact value)
- Common wrong answer 13 : That count fits the pattern of the others. Look for the one far from the rest.
- Common wrong answer 16 : That count is close to the others. Look for the one far from the rest.
(b) Mean rate without it: 14.3 min-1 (exact value)
- Common wrong answer 16.3 : That is the mean of all four counts. Leave out the anomalous count first.
- Common wrong answer 16.25 : That is the mean of all four counts. Leave out the anomalous count first.
- Common wrong answer 43 : That is the total of the three counts. Divide it by the number of counts.

A full answer includes

(a) I identified 22 as the anomalous count.
Method (b): I added the other three counts, 13 + 14 + 16 = 43, and divided by 3.
Answer (b): I wrote 14.3 bubbles per minute, to one decimal place.
I did not include the 22 in the mean.

Also accept

(a) 22; (b) 14.3 bubbles per minute

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 up.
“The mean is 43.” — That is the total of the three counts. Divide it by the number of counts, 3.

4. RQ4 · Practice

Model answer. As the lamp comes closer, the light on the plant gets brighter and the rate rises, from 3 bubbles per minute at 45 cm to 13 at 18 cm. From 18 cm to 9 cm it rises by only 1, to 14: the rate levels off, because something other than light is now holding it back.

A full answer includes

I chose: the rate rises from 3 to 13 between 45 cm and 18 cm, then only to 14 at 9 cm.
I did not choose "the same amount each time": the rises were 2, 3.5, 4.5 and then only 1.

Also accept

the rate rises from 3 to 13 between 45 cm and 18 cm, then only to 14 at 9 cm

Watch for

“The rate rises by about the same amount each time the lamp comes 9 cm closer.” — The rises were 2, 3.5 and 4.5, but from 18 cm to 9 cm only 1.
“The rate falls from 14 to 3 as the lamp comes closer.” — Read the table from the far lamp to the near lamp: 45 cm is the farthest, and the rate there is the lowest.
“The rate rises most between 18 cm and 9 cm, where the lamp is closest.” — Between 18 cm and 9 cm it rose by only 1, the smallest rise of all.

5. RQ13 · Practice

Model answer. (a) As the light intensity rises, the rate rises: from 2 bubbles per minute at 500 lux to 19 at 4000 lux. From 4000 to 8000 lux it rises by only 0.5, to 19.5: the rate levels off. (b) Light supplies the energy for photosynthesis. Brighter light gives the chloroplasts more energy, so the plant photosynthesises faster and gives off more oxygen, as more bubbles. (c) Doubling the light now hardly changes the rate, so light is no longer what holds it back: something else, such as the carbon dioxide in the water or the temperature, does.

A full answer includes

(a) I wrote that the rate rises from 2 to 19 bubbles per minute between 500 and 4000 lux.
(a) I wrote that it then levels off: it rises by only 0.5 from 4000 to 8000 lux.
(b) I wrote that light supplies the energy for photosynthesis.
(b) I wrote that brighter light gives more energy, so the plant photosynthesises faster.
(c) I wrote that something other than light, such as carbon dioxide or temperature, now holds the rate back.
I did not write that the bright light damages the plant: the rate did not fall.

Also accept

(a) the rate rises with light intensity, then levels off; (b) more light energy, so faster photosynthesis; (c) something else, such as carbon dioxide or temperature, holds the rate back

Watch for

“The rate rises by the same amount every time the light intensity doubles.” — From 500 to 1000 lux it rose by 3, but from 4000 to 8000 lux by only 0.5.
“At 8000 lux the light is so bright that it damages the plant.” — The rate did not fall: it levelled off, because something else was holding it back.

6. RQ5 · Practice

Model answer. (a) A lamp gives out heat as well as light. Without the beaker of water to soak up the heat, the closer lamp also warmed the water by 6 °C. Two things changed at once, the light and the temperature, so she cannot tell which one changed the rate. (b) The rate does not change at once when the conditions change. In the first minute the plant was still adjusting, so the count was a mixture of the old and new rates. She should wait about 2 minutes after moving the lamp, then count.

A full answer includes

(a) I wrote that the lamp also heated the water: the temperature changed as well as the light.
(a) I wrote that she cannot tell whether the light or the temperature changed the rate.
(b) I wrote that the plant had not yet adjusted to its new conditions in the first minute.
(b) I wrote that she should wait about 2 minutes before counting.
I did not write that the extra heat gives the plant more light.

Also accept

(a) the temperature rose too, so two variables changed; (b) the plant had not adjusted: wait about 2 minutes first

Watch for

“The warmer water does not matter, because temperature does not affect photosynthesis.” — Photosynthesis is controlled by enzymes, which work faster when warmer. A rise of 6 °C can change the rate.
“The second count is higher because she counted more carefully.” — Both counts are at the same distance. The plant was still adjusting to its new light in the first minute.

7. RQ6 · Practice

Model answer. The bubbles are not all the same size, so the number of bubbles does not show the volume of gas. Y gives fewer bubbles, but each is larger, so it gives off more gas in the same time. Counting bubbles is only a rough measure of the rate; measuring the volume of gas, as the syringe does, is a better one.

A full answer includes

I chose: the bubbles are not all the same size, so a count does not show the volume of gas.
I did not choose the statement that Y stops respiring: plants respire all the time.

Also accept

the bubbles are not all the same size, so a count does not show the volume of gas

Watch for

“Some of the oxygen from Y dissolves, so its count is too high.” — Oxygen that dissolves is never seen as bubbles, so dissolving makes a count too low, not too high.
“Sprig Y stops respiring in the light, so it gives off more oxygen.” — Every plant respires all the time, in light and in dark. That does not explain the syringe result.
“The gas from X is mostly carbon dioxide, not oxygen.” — In bright light both sprigs give off mainly oxygen. The difference is the size of the bubbles.

8. RQ11 · Practice

Model answer. Sodium hydrogencarbonate is a source of carbon dioxide. With the lamp at 8 cm, the light is bright, and as the carbon dioxide concentration rises the rate rises, from 9 bubbles per minute with none added to 21 with 0.2 g. From 0.2 g to 0.3 g it rises by only 1, to 22: it levels off. With the lamp at 60 cm, the light is dim, and the rate stays between 2 and 3 whatever is added: the light holds it back, so extra carbon dioxide hardly helps.

A full answer includes

I chose: at 8 cm the rate rises from 9 to 21, then hardly changes; at 60 cm it stays between 2 and 3.
I did not choose "rises steadily at both distances": at 60 cm it rose by only 1 in all.

Also accept

at 8 cm the rate rises from 9 to 21, then hardly changes; at 60 cm it stays between 2 and 3

Watch for

“At both distances the rate rises steadily with every extra 0.1 g.” — At 8 cm the rises get smaller (7, 5, then 1), and at 60 cm the rate rose by only 1 in all.
“Carbon dioxide has no effect, because at 60 cm the rate hardly changes.” — At 8 cm, in bright light, the rate more than doubled as carbon dioxide was added.
“At 60 cm carbon dioxide limits the rate, so adding it helps most there.” — At 60 cm it rose only from 2 to 3, but at 8 cm from 9 to 22. In the dim light at 60 cm, light holds the rate back, so extra carbon dioxide hardly helps.

9. RQ7 · Practice

Model answer. Sodium hydrogencarbonate is a source of carbon dioxide, so stirring it in raised the carbon dioxide concentration in the water around the plant. Carbon dioxide is a raw material of photosynthesis. With more of it, the plant photosynthesised faster, so it gave off more oxygen, as more bubbles. Only the carbon dioxide changed: the lamp stayed at 15 cm and the water at 25 °C.

A full answer includes

I wrote that sodium hydrogencarbonate is a source of carbon dioxide.
I wrote that it raises the carbon dioxide concentration in the water.
I wrote that carbon dioxide is a raw material of photosynthesis.
I wrote that with more carbon dioxide the plant photosynthesises faster, so it gives off more oxygen bubbles.
I did not write that sodium hydrogencarbonate is food for the plant.

Also accept

it adds carbon dioxide, a raw material, so photosynthesis is faster

Watch for

“Sodium hydrogencarbonate is food for the plant.” — The plant makes its own food. Sodium hydrogencarbonate only supplies carbon dioxide, a raw material.
“The sodium hydrogencarbonate gives off the bubbles itself.” — The bubbles come from the cut stem of the plant, and the gas is mainly oxygen made by photosynthesis.

10. RQ12 · Practice

Model answer. Row D. Water baths at 12, 22, 32 and 42 °C change the temperature, and only the temperature. The lamp stays at one distance, so the light on the plant stays the same; the beaker of water between the lamp and the tube, the heat filter, stops the lamp warming the water above the bath temperature; and the same mass of sodium hydrogencarbonate keeps the carbon dioxide concentration the same. The number of bubbles per minute is the dependent variable: it is what she measures.

A full answer includes

I chose row D.
I checked that water baths set the temperature.
I checked that the lamp stays at one distance, so the light stays the same.
I checked that the heat filter stops the lamp warming the water.
I checked that the same mass of sodium hydrogencarbonate keeps the carbon dioxide the same.
I did not choose row B: the bubbles per minute are measured, not kept the same.

Also accept

Watch for

“Row A.” — Moving the lamp closer warms the water, but it also makes the light brighter. Two variables change, so she cannot tell which one changed the rate.
“Row B.” — The number of bubbles per minute is the dependent variable: it is what she measures, so it cannot be kept the same.
“Row C.” — Without the heat filter, the lamp warms the water above the temperature of the bath, so the temperature is not the one she set.

11. RQ8 · Practice

Model answer. Row C. Graph 1 rises as the carbon dioxide concentration rises, then levels off, because another factor, such as the light, holds the rate back. Graph 2 rises as the enzymes work faster when warmer, up to their optimum, then falls as the enzymes are denatured.

A full answer includes

I chose row C.
I checked that graph 1 rises, then levels off.
I checked that graph 2 rises to a peak, then falls.
I did not choose row B: above the optimum the rate falls, it does not just level off.

Also accept

Watch for

“Row A.” — More carbon dioxide raises the rate only until something else holds it back. Then graph 1 levels off.
“Row B.” — Above the optimum temperature the enzymes are denatured, so graph 2 falls, it does not stay level.
“Row D.” — Row D puts the peak on the carbon dioxide graph. A peak and a fall come from enzymes being denatured by heat, so they belong to the temperature graph.

12. RQ9 · Practice

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

Model answer. X (12 °C, light intensity 1): at light intensity 1 both temperatures give 3, so warming does not help, but doubling the light doubles the rate to 6: light is limiting. Y (12 °C, light intensity 8): from light intensity 6 to 8 the rate stays at 9.5, a plateau, so more light does not help, but at 26 °C and the same light the rate is 18: temperature is limiting. Z (26 °C, light intensity 2): doubling the light to 4 doubles the rate from 6 to 12, so light is limiting. Carbon dioxide is not limiting at any point, because the plant has plenty of it.

A full answer includes

I matched X to light.
I matched Y to temperature.
I matched Z to light.
I did not match Y to light: at 12 °C the rate is 9.5 at light intensity 6 and at 8.

Also accept

X: light; Y: temperature; Z: light

Watch for

“Y is limited by light, because light is the factor that changes along the row.” — From light intensity 6 to 8 the rate at 12 °C stays at 9.5, so more light does not raise it there.
“Y is limited by carbon dioxide, because the line has levelled off.” — The plant has plenty of carbon dioxide. Warming it to 26 °C at the same light raises the rate from 9.5 to 18.
“X is limited by temperature, because 12 °C is cold.” — At light intensity 1 the rate is 3 at both temperatures, so warming does not raise it.
“Z is limited by temperature, because the warmer row is already in use.” — At 26 °C, doubling the light from 2 to 4 doubles the rate. The light is holding it back.

13. RQ10 · Practice

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

Model answer. Moss in a sealed jar: the sun and the warm windowsill give plenty of light and warmth, but the moss has been taking carbon dioxide out of the air in the jar all day and nothing adds more, so by late afternoon carbon dioxide is limiting. Strawberry plants at 5 °C: the lamps give bright light and extra carbon dioxide is added, but the enzymes work slowly in the cold, so temperature is limiting. The lowest maize leaves: the sun is bright and the day is hot, but the leaves above absorb most of the light, so little reaches the lowest leaves and light is limiting. The fern in the office: 22 °C suits it and the room air has carbon dioxide, but office lights are far dimmer than sunlight, so light is limiting.

A full answer includes

I matched the moss in the sealed jar to carbon dioxide.
I matched the strawberry plants in the cold greenhouse to temperature.
I matched the lowest maize leaves to light.
I matched the fern under the office lights to light.
I did not match the maize leaves to carbon dioxide: the leaves above them take most of the sunlight.

Also accept

moss in sealed jar: carbon dioxide; strawberry plants at 5 °C: temperature; lowest maize leaves: light; fern under office lights: light

Watch for

“The lowest maize leaves are limited by carbon dioxide, because it is a hot, sunny day.” — The day is sunny, but the leaves above absorb most of the light. Little reaches the lowest leaves, so light is limiting there.
“The moss is limited by light, because it is late afternoon.” — The windowsill is still sunny. In a sealed jar nothing replaces the carbon dioxide the moss has used all day.
“The strawberry plants are limited by carbon dioxide or light, because it is night.” — The lamps give bright light and extra carbon dioxide is added. At 5 °C the enzymes work slowly, so temperature is limiting.
“The fern is limited by temperature, because it is indoors.” — At 22 °C the enzymes work well. Office lights are much dimmer than sunlight, so light is limiting.

14. RT1 · Apply

Model answer. (a) At 10 minutes the line for the lamp at 10 cm is at 7 discs floating, and the line for the lamp at 30 cm is at 2 discs floating. (b) Going across from 5 discs: the line for the lamp at 10 cm reaches 5 at 8 minutes, and the line for the lamp at 30 cm reaches 5 at 16 minutes. (c) The same number of discs floated in half the time, so with the lamp at 10 cm the discs floated 16 ÷ 8 = 2 times faster: the brighter light gave faster photosynthesis.

Checked automatically

(a) Discs floating at 10 minutes, lamp at 10 cm: 7 (exact value)
- Common wrong answer 8 : That is the reading at 12 minutes. Read at 10 minutes.
- Common wrong answer 5 : That is the reading at 8 minutes. Read at 10 minutes.
(a) Discs floating at 10 minutes, lamp at 30 cm: 2 (exact value)
- Common wrong answer 3 : That is the reading at 12 minutes. Read at 10 minutes.
- Common wrong answer 7 : That is the line for the lamp at 10 cm. Use the line with squares.
(b) Time for 5 discs to float, lamp at 10 cm: 8 min (exact value)
- Common wrong answer 16 : That is the line for the lamp at 30 cm. Use the line with circles.
- Common wrong answer 10 : At 10 minutes 7 discs were floating. Go across from 5 discs instead.
(b) Time for 5 discs to float, lamp at 30 cm: 16 min (exact value)
- Common wrong answer 8 : That is the line for the lamp at 10 cm. Use the line with squares.
- Common wrong answer 20 : At 20 minutes 6 discs were floating. Go across from 5 discs instead.
(c) How many times faster with the lamp at 10 cm: 2 (exact value)
- Common wrong answer 8 : You subtracted the times. Divide the longer time by the shorter time.
- Common wrong answer 0.5 : You divided the shorter time by the longer one. A shorter time means a faster rate.

A full answer includes

(a) I wrote 7 discs for the lamp at 10 cm.
(a) I wrote 2 discs for the lamp at 30 cm.
(b) I wrote 8 minutes for the lamp at 10 cm.
(b) I wrote 16 minutes for the lamp at 30 cm.
Method (c): I divided the longer time by the shorter time: 16 ÷ 8.
Answer (c): I wrote 2 times faster.
I did not read (a) across from 10 discs on the y-axis.
I did not subtract the two times in (c).

Also accept

(a) 10 cm: 7; 30 cm: 2; (b) 10 cm: 8 min; 30 cm: 16 min; (c) 2 times faster

Watch for

“Read the value at 10 discs instead of at 10 minutes.” — Start from 10 on the time axis (x-axis), then read the number of discs on the y-axis.
“At 30 cm the reading is 3 discs.” — Check the time you read at: 3 discs is a later reading on the line with squares.
“The discs at 10 cm were 8 times faster, because 16 − 8 = 8.” — A difference of 8 minutes is not a number of times. Divide the times: 16 ÷ 8 = 2.
“The discs at 10 cm were 0.5 times as fast.” — You divided the shorter time by the longer one. A shorter time means a faster rate, so the closer lamp is 2 times faster.

15. RT2 · Apply

Model answer. (a) In the light the discs photosynthesise and make oxygen faster than respiration uses it. The oxygen collects in their air spaces, which makes the discs less dense, so they float. (b) With the lamp at 10 cm the light on the discs is brighter, so they photosynthesise faster, fill up with oxygen sooner, and float sooner. (c) In the dark there is no photosynthesis, so no more oxygen is made. The cells of the discs go on respiring, using up the oxygen in their air spaces; the carbon dioxide they give out dissolves in the solution. With less gas in their air spaces, the discs become denser again and sink.

A full answer includes

(a) I wrote that photosynthesis makes oxygen faster than respiration uses it.
(a) I wrote that the oxygen collects in the air spaces of the discs, so they float.
(b) I wrote that the closer lamp gives brighter light on the discs.
(b) I wrote that brighter light means faster photosynthesis, so the discs fill with oxygen sooner.
(c) I wrote that in the dark the discs do not photosynthesise, so no more oxygen is made.
(c) I wrote that the discs go on respiring and use up the oxygen in their air spaces.
(c) I wrote that with less gas in their air spaces, the discs sink.
I did not write that the discs float because they take in carbon dioxide.
I did not write that the discs stop respiring in the light.

Also accept

(a) oxygen from photosynthesis collects in the air spaces; (b) brighter light, faster photosynthesis; (c) no photosynthesis in the dark, and respiration uses up the oxygen, so they sink

Watch for

“The discs float because they take in carbon dioxide.” — Carbon dioxide is used up in photosynthesis. The discs float because the oxygen they make collects in their air spaces.
“The lamp at 10 cm warmed the discs, so they floated sooner.” — Both beakers were kept at 25 °C. The light changed, not the temperature.
“The discs sink in the dark because they have cooled down.” — Warmth does not hold a disc up: the gas in its air spaces does. In the dark respiration uses up the oxygen and none is made.

16. RT5 · Apply

Model answer. The light and the temperature were kept the same, so neither caused the fall. The plant takes carbon dioxide, a raw material of photosynthesis, from the water around it, and nothing was adding more, so as the hour went on the carbon dioxide concentration in the tube fell. With less carbon dioxide the plant photosynthesised more slowly and gave off fewer bubbles. Adding a little sodium hydrogencarbonate solution, a source of carbon dioxide, or changing the water for fresh water, would raise the carbon dioxide concentration and so the rate.

A full answer includes

I wrote that the plant uses up the carbon dioxide dissolved in the water.
I wrote that nothing adds more, so the carbon dioxide concentration falls.
I wrote that with less carbon dioxide, a raw material, photosynthesis is slower.
I suggested adding sodium hydrogencarbonate, or changing the water for fresh water.
I did not write that the plant is running out of water.

Also accept

the plant used up the dissolved carbon dioxide, so the rate fell; add sodium hydrogencarbonate or fresh water

Watch for

“The plant is getting tired.” — A plant does not tire like a muscle. Its rate fell because a raw material, carbon dioxide, was running short.
“The plant is running out of water.” — The plant stands in water the whole time. Water is not what runs short.

17. RT3 · Apply

Model answer. At 45 °C, above the optimum, some of the enzymes in the chloroplasts were denatured. Cooling the water does not change their shape back, so fewer enzymes are working than before the heating. The plant still photosynthesises at 28 °C, but more slowly than it did before it was overheated.

A full answer includes

I wrote that at 45 °C some of the enzymes that control photosynthesis were denatured.
I wrote that denatured enzymes do not go back to their working shape when cooled.
I wrote that fewer enzymes now work, so the rate at 28 °C is lower than before.
I did not write that the plant died: it still gives off 17 bubbles per minute.

Also accept

some enzymes were denatured at 45 °C and do not recover when cooled

Watch for

“Cooling the water back to 28 °C should give exactly the old rate.” — That would be true only if nothing in the plant had changed. Denatured enzymes stay denatured.
“The plant died at 45 °C.” — It still gives off 17 bubbles per minute, so it is still photosynthesising.

18. RT4 · Apply

Model answer. Switching on the lamps is most likely to raise the rate. At sunrise the light is dim, so light is the limiting factor; the heaters keep the temperature suitable. All night the plants respired and gave out carbon dioxide, there was no light for photosynthesis to take it back in, and the shut vents kept it inside, so it built up to more than twice the level outside. Carbon dioxide is plentiful, so adding more would not raise the rate while the light is dim.

A full answer includes

I chose: switch on the lamps.
I wrote that the light is still dim at sunrise.
I wrote that light is therefore the limiting factor.
I wrote that the plants respired all night and gave out carbon dioxide.
I wrote that in the dark there was no photosynthesis to take the carbon dioxide back in.
I wrote that the shut vents kept the carbon dioxide inside.
I did not choose extra carbon dioxide: there is already more than in the air outside.

Also accept

switch on the lamps: light is limiting at sunrise; the carbon dioxide built up from respiration all night with the vents shut

Watch for

“Release extra carbon dioxide into the greenhouse.” — There is already more than twice as much as outside. While the light is dim, more carbon dioxide cannot raise the rate.
“Turn the heaters up to 30 °C.” — At 22 °C the enzymes already work well. More warmth does not help while light is in shortest supply.
“Open the vents to let fresh air in.” — Fresh air holds less carbon dioxide than the greenhouse holds now, and the light would still be dim.