Gas exchange in flowering plants – IGCSE Biology exam questions with mark scheme

1Grade 1–3Triple only3 marks

(a) Gases enter and leave a leaf through small pores in the leaf surface.

Which of these is the name of these pores?

A  alveoli
B  stomata
C  villi
D  xylem [1]

(b) Complete the table to show the gas that is used and the gas that is produced by each process in a leaf cell. [2]

Process	Gas used	Gas produced
aerobic respiration
photosynthesis

(a) B (stomata)(1)
(b) aerobic respiration: uses oxygen and produces carbon dioxide; accept O$_2$ / CO$_2$; both needed for the mark(1)
(b) photosynthesis: uses carbon dioxide and produces oxygen; both needed for the mark; ignore water / glucose added correctly; reject 'air'(1)
Answer: (a) B. (b) Respiration: uses O₂, produces CO₂. Photosynthesis: uses CO₂, produces O₂.
Examiner tip: Each row needs both gases right for the mark, so learn the two processes as exact opposites.

2Grade 4–5Triple only5 marks

The diagram shows a section through part of a leaf.

(a) Name the structures labelled X and Y. [2]

(b) Z is a stoma. Describe the role of Z in gas exchange when the leaf is in bright light. [2]

(c) The cells of the spongy mesophyll have a thin layer of water on their surfaces. Suggest why this is important for gas exchange. [1]

(a) X: air space; accept intercellular space / air gap(1)
(a) Y: guard cell; accept guard cells(1)
(b) (allows) carbon dioxide to diffuse into the leaf; accept CO$_2$ enters through the stoma(1)
(b) (allows) oxygen to diffuse out of the leaf; ignore water vapour leaves; reject oxygen enters(1)
(c) gases (carbon dioxide / oxygen) dissolve in the water before diffusing into / out of the cells; accept so gases can diffuse across cell membrane in solution(1)
Answer: (a) X air space, Y guard cell. (b) CO₂ diffuses in, O₂ diffuses out through the stoma. (c) Gases dissolve in the water so they can diffuse into/out of cells.
Examiner tip: In bright light the net movement is CO₂ in and O₂ out; writing the reverse loses both marks in (b).

3Grade 4–5Triple only4 marks

Oxygen is produced by photosynthesis in the palisade mesophyll cells of a leaf.

Describe the pathway taken by this oxygen as it leaves the leaf, and explain how it moves along this pathway. [4]

oxygen diffuses out of the palisade (mesophyll) cell (across the cell membrane / cell wall)(1)
into the air spaces (in the spongy mesophyll); accept intercellular spaces(1)
out through the stomata / stoma (into the air outside the leaf); accept between guard cells(1)
by diffusion down a concentration gradient / from a high concentration (in the leaf) to a lower concentration (outside); the word 'diffusion' alone does not score this mark(1)
Answer: Diffuses out of palisade cell into air spaces of the spongy mesophyll, then out through stomata, down its concentration gradient.
Examiner tip: Name the air spaces AND the stomata; many answers jump straight from 'cell' to 'out of the leaf'.

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4Grade 6–7Triple only6 marks

A potted plant was sealed inside a clear chamber. A sensor recorded the carbon dioxide concentration of the air in the chamber over 24 hours. The plant was in darkness from 00:00 to 06:00 and from 18:00 to 24:00 (shaded on the graph), and in light from 06:00 to 18:00.

(a) Calculate the mean rate of increase in carbon dioxide concentration between 00:00 and 06:00.
Give your answer in ppm per hour. [2]

(b) Explain why the carbon dioxide concentration increased between 00:00 and 06:00. [2]

(a) increase = 520 − 400 = 120 (ppm)(1)
(a) 120 ÷ 6 = 20 (ppm per hour); allow 19 to 21 from a 6-hour reading of 515–525; correct answer with no working scores 2(1)
(b) no light so no photosynthesis (takes place); accept photosynthesis stops(1)
(b) (but) the plant (still) respires, releasing carbon dioxide (into the chamber); reject 'the plant only respires at night' if it implies no respiration in the light(1)
(c) (in light) rate of photosynthesis is greater than rate of respiration; accept photosynthesis takes place faster than respiration(1)
(c) so more carbon dioxide is taken in / used (in photosynthesis) than is released (by respiration); accept net uptake of CO$_2$(1)
Answer: (a) 20 ppm per hour. (b) Dark: no photosynthesis, but respiration releases CO₂. (c) Light: photosynthesis faster than respiration, so net CO₂ uptake.
Examiner tip: Never say plants only respire at night: respiration continues all the time, and in (c) you must compare the two rates.

5Grade 8–9Triple only6 marks

A scientist measured the net carbon dioxide exchange of a leaf at different light intensities at a constant temperature. Positive values show net uptake of carbon dioxide by the leaf; negative values show net release.

(a) Give the rate at which the leaf releases carbon dioxide by respiration. Explain how you used the graph. [2]

(b) Use the graph to estimate the light intensity at which there is no net exchange of carbon dioxide. [1]

(c) Assume the rate of respiration stays the same at all light intensities.
Calculate the total rate at which carbon dioxide is used in photosynthesis at a light intensity of 4 kilolux. [2]

(d) Explain what is happening in the leaf at the light intensity you gave in (b). [1]

(a) 2.0 mg per hour; accept 2 mg/h; ignore negative sign(1)
(a) value read at 0 kilolux / in darkness, as no photosynthesis takes place so all CO$_2$ exchange is due to respiration(1)
(b) 1.3 kilolux; allow any value from 1.2 to 1.4(1)
(c) 3.5 + 2.0 (net uptake + respiration); allow ecf from (a)(1)
(c) 5.5 (mg per hour); correct answer with no working scores 2; 1.5 (3.5 − 2.0) scores 0(1)
(d) rate of photosynthesis equals rate of respiration; accept CO$_2$ used in photosynthesis = CO$_2$ released in respiration; reject 'no photosynthesis or respiration takes place'(1)
Answer: (a) 2.0 mg/h, from the value at 0 kilolux. (b) ≈1.3 kilolux. (c) 5.5 mg per hour. (d) Photosynthesis rate = respiration rate.
Examiner tip: Net uptake hides the CO₂ made by respiration: add the respiration rate back on, don't subtract it.

6Grade 4–5Triple only7 marks

A student set up four boiling tubes, each containing 10 cm$^3$ of hydrogen-carbonate indicator. The indicator was orange-red at the start. A leaf was placed in tubes A, B and C, held above the indicator on a platform of gauze. Each tube was sealed with a bung.

Tube	Contents	Conditions	Colour after 3 hours
A	leaf	bright light	purple
B	leaf	wrapped in foil (dark)	yellow
C	leaf	covered in thin muslin cloth (dim light)	orange-red
D	no leaf	bright light	orange-red

(a) Explain the result in tube A. [2]

(b) Explain the result in tube B. [2]

(d) Give the reason for setting up tube D. [1]

(a) rate of photosynthesis greater than rate of respiration; accept photosynthesis faster than respiration(1)
(a) (so) carbon dioxide (concentration) decreased / net uptake of CO$_2$ (so indicator turns purple)(1)
(b) no photosynthesis (in the dark) but the leaf respires(1)
(b) (so) carbon dioxide (concentration) increased (so indicator turns yellow)(1)
(c) rate of photosynthesis equals rate of respiration(1)
(c) (so) no net change in carbon dioxide concentration; accept CO$_2$ used = CO$_2$ released; reject 'no gas exchange'(1)
(d) control: to show that the colour change is caused by the leaf / not by light (alone); accept to compare with(1)
Answer: (a) Photosynthesis > respiration, CO₂ falls (purple). (b) Dark: respiration only, CO₂ rises (yellow). (c) Photosynthesis = respiration, no net CO₂ change. (d) Control: shows the leaf causes the change.
Examiner tip: Colour links to CO₂ level: yellow = more CO₂, purple = less; always say what happened to CO₂, not just which process occurred.

7Grade 6–7Triple only6 marks

A student placed leaf discs in sealed tubes of hydrogen-carbonate indicator at different distances from a lamp. The indicator started at pH 8.4. A tank of water was placed between the lamp and the tubes. After 2 hours the pH of the indicator in each tube was measured with a pH meter.

Distance from lamp / cm	10	20	30	40	50
pH after 2 hours	9.0	8.7	8.5	8.4	8.2

Carbon dioxide forms an acid when it dissolves, so a decrease in carbon dioxide concentration causes the pH to rise.

(a) Calculate the mean rate of change in pH per hour for the tube at 10 cm. [2]

(b) Give the distance at which there was no net gas exchange. Explain your answer. [2]

(d) Suggest why the tank of water was placed between the lamp and the tubes. [1]

(a) change in pH = 9.0 − 8.4 = 0.6(1)
(a) 0.6 ÷ 2 = 0.3 (per hour); correct answer with no working scores 2(1)
(b) 40 cm(1)
(b) pH did not change (from 8.4) so carbon dioxide used in photosynthesis = carbon dioxide released by respiration; accept rate of photosynthesis = rate of respiration(1)
(c) (light too dim so) rate of respiration greater than rate of photosynthesis so carbon dioxide concentration increased(1)
(d) to absorb heat (from the lamp) / act as a heat shield so temperature is kept constant; ignore 'to make it fair' alone(1)
Answer: (a) 0.3 per hour. (b) 40 cm: pH unchanged, photosynthesis = respiration. (c) Respiration > photosynthesis, CO₂ rises. (d) Heat shield to keep temperature constant.
Examiner tip: In (d) say what the water absorbs (heat) and why that matters (temperature affects rate), not just 'fair test'.

8Grade 8–9Triple only6 marks

Hydrogen-carbonate indicator changes colour when the carbon dioxide concentration around it changes.

Design an investigation to find out how light intensity affects the net gas exchange of leaves from a bean plant.

Include experimental details in your answer and write in full sentences. [6]

place leaves / equal-sized leaf discs (cut with a cork borer) from the same bean plant in tubes of hydrogen-carbonate indicator, held above the indicator on gauze(1)
same volume of indicator in each tube; tubes sealed with bungs(1)
vary light intensity by placing a lamp at different distances (e.g. 10, 20, 30, 40, 50 cm) or using different numbers of layers of muslin; include one tube in darkness (foil)(1)
control tube(s) with no leaf at each condition(1)
control temperature (water bath / heat shield of water between lamp and tubes); same leaf area / same age of leaf; same starting colour of indicator(1)
leave for the same time (e.g. 2 hours) then record colour against a colour chart / measure pH or absorbance with a colorimeter(1)
repeat for each light intensity and calculate a mean / check for anomalies(1)
interpretation: purple = net CO$_2$ uptake (photosynthesis > respiration), yellow = net CO$_2$ release, no change = compensation point(1)

Level	Marks	What it looks like
Level 0	0	Nothing worthy of credit.
Level 1	1–2	Basic outline: leaves in indicator under some light conditions, with at most one controlled variable or measurement mentioned. A colleague could not carry out the method from it.
Level 2	3–4	Workable method: light intensity varied in a stated way, colour change recorded after a set time, and at least one relevant controlled variable. Some details missing (e.g. no control tube or no repeats).
Level 3	5–6	Detailed, logically ordered method that could be followed: how light intensity is varied, how CO$_2$ change is measured (colour chart / pH), several controlled variables including temperature, a control without leaves, and repeats. May link colour to net gas exchange.

Answer: Equal leaf discs on gauze above equal volumes of indicator in sealed tubes; lamp at several distances (plus a dark tube); heat shield; no-leaf control; same time; compare colour with a chart or measure pH; repeat and take means.
Examiner tip: Top-level answers state HOW light intensity is changed and HOW temperature is kept constant; 'keep it a fair test' earns nothing.

9Grade 6–7Triple only3 marks

On a hot, dry afternoon the stomata of a plant close, even though light intensity is high.

Explain how closing the stomata affects gas exchange and the rate of photosynthesis in the leaf. [3]

carbon dioxide cannot diffuse into the leaf / less CO$_2$ enters the leaf; accept gas exchange (with the air) is reduced / stops(1)
carbon dioxide concentration in the air spaces (inside the leaf) falls as it is used in photosynthesis; accept oxygen builds up in the leaf(1)
(so) rate of photosynthesis decreases because carbon dioxide becomes a limiting factor; reject photosynthesis stops immediately(1)
Answer: CO₂ cannot diffuse in, so CO₂ in the air spaces is used up and falls; CO₂ becomes limiting, so photosynthesis slows.
Examiner tip: Link the stomata to CO₂ supply and then say CO₂ is the limiting factor; 'less water' is not what is being asked.

10Grade 8–9Triple only5 marks

A student painted nail varnish on the upper and lower surfaces of a leaf, peeled off the dried varnish and looked at it with a microscope. The circular field of view had a diameter of 0.40 mm.

The student counted 23 stomata in the field of view for the lower surface and 4 stomata for the upper surface.

(a) Calculate the number of stomata per mm$^2$ on the lower surface.
Area of a circle = $\pi r^2$ [3]

(b) Calculate the ratio of stomata on the lower surface to stomata on the upper surface. Give your answer in the form n : 1. [1]

(c) Suggest one reason why it is an advantage for gas exchange that most stomata are on the lower surface of the leaf. [1]

(a) radius = 0.20 mm(1)
(a) area = $\pi \times 0.20^2$ = 0.126 (mm$^2$); allow 0.13 / 0.1257(1)
(a) 23 ÷ 0.126 = 183 (stomata per mm$^2$); allow 177 to 184 (177 from area rounded to 0.13); correct answer with no working scores 3; using diameter as radius (23 ÷ 0.503 = 46) scores 1(1)
(b) 5.75 : 1; accept 5.8 : 1 / 6 : 1(1)
(c) lower surface is next to the spongy mesophyll / large air spaces so gases diffuse a short distance to and from the cells; accept less water is lost (as lower surface is shaded / cooler) so stomata can stay open (for CO$_2$ uptake) for longer; ignore 'to reduce transpiration' alone(1)
Answer: (a) 183 stomata per mm². (b) 5.75 : 1. (c) Lower surface is next to the spongy mesophyll air spaces (or: shaded, so less water lost and stomata stay open longer for gas exchange).
Examiner tip: Halve the diameter to get the radius before using $\pi r^2$; using 0.40 mm as the radius is the most common lost mark.

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