OxfordAQA International GCSE Biology: Photosynthesis, limiting factors and transport in plants (9201) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.2.1 and 3.2.2 of the OxfordAQA International GCSE Biology (9201) specification, Version 5.2, for exams May/June 2018 onwards. The qualification is untiered. Full explanations are in the study guide and the revision notes. Course page: OxfordAQA IGCSE Biology. Printable list: checklist. Wider topic practice: Bioenergetics practice.

Questions

1. Write the word equation and the balanced symbol equation for photosynthesis. [3]

2. A wheat plant uses glucose from photosynthesis in several ways.

(a) State three uses of glucose in the plant. [3] (b) The soil is low in nitrate ions. Explain why the plant grows slowly. [2]

3. A student counts bubbles released by pondweed in 2 minutes with a lamp 15 cm away. Three repeats give 34, 36 and 52 bubbles.

(a) One count does not fit the pattern. Pick it out and give a likely reason for it. [2] (b) Calculate the mean number of bubbles per minute, leaving out the anomaly. [2]

4. The table shows the rate of photosynthesis (mm³ of oxygen per minute) at two temperatures. Carbon dioxide concentration was constant.

Light intensity (units)	0	1	2	3	4	5	6
Rate at 20 °C	0	5	10	15	18	18	18
Rate at 30 °C	0	5	10	15	20	25	28

(a) Name the limiting factor at 20 °C and 2 units of light. [1] (b) Explain what limits the rate at 20 °C and 5 units of light. [2] (c) Calculate the percentage increase in rate at 6 units of light when the temperature is raised from 20 °C to 30 °C. Give your answer to the nearest whole number. [2]

5. A greenhouse grows 1800 kg of lettuce per season. Lettuce sells for £0.85 per kg. A grower compares three ways of changing conditions.

Option	Yield (kg per season)	Extra running cost (£)
Extra heating	2400	420
Extra lamps	2250	390
Carbon dioxide enrichment	2000	60

(a) Calculate the net gain or loss for each option. [3] (b) State which option you would recommend. [1] (c) Explain why carbon dioxide enrichment would not help if light were the limiting factor. [2]

6. A leaf is modelled as a cuboid 8 cm long, 3 cm wide and 0.08 cm thick.

(a) Calculate its surface area to volume ratio. [2] (b) Explain how the flat shape and the air spaces inside a leaf help photosynthesis. [2]

7. (a) Explain how root hairs help a plant to take in water and mineral ions. [2] (b) State where in a flowering plant the carbon dioxide for photosynthesis enters, and how it moves. [2]

8. The table shows the width of stomata in a leaf and the water lost from the leaf during one day.

Time	07:00	10:00	13:00	16:00	19:00
Stomatal width (µm)	0	7	11	4	0
Water lost (g per hour)	0.1	0.9	1.6	0.5	0.1

(a) Describe how the width of the stomata changes between 07:00 and 19:00. [2] (b) State the relationship between stomatal width and water loss. [1] (c) Explain how closing stomata can prevent wilting, and what effect it has on photosynthesis. [2]

9. Three identical shoots are kept for 3 hours in different conditions. Mass of water lost: A, cool and still, 1.5 g; B, warm and still, 3.6 g; C, warm and windy, 6.3 g.

(a) Calculate the percentage increase in water loss from A to C. [2] (b) Explain why shoot C loses the most water. [2]

10. (a) State what is carried by xylem and what is carried by phloem. [2] (b) Explain how the structure of xylem suits its function. [2] (c) Explain how the structure of phloem suits its function. [2]

11. Design an investigation to find out how carbon dioxide concentration affects the rate of photosynthesis of pondweed. Your answer should include how you would control other variables, how you would obtain reliable results and one hazard. [8]

Answers

1. Word equation: carbon dioxide + water → glucose + oxygen [1]. Symbol equation using the right formulae for all four substances [1]. Balanced: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ [1]. [3] Examiner insight: writing “light” over the arrow shows the energy source; light is not a reactant, so do not list it with the substances on the left.

2. (a) Respiration for energy [1]; converted to starch, or fat or oil, for storage [1]; made into cellulose for cell walls, or into protein [1]. [3] (b) Nitrate ions are needed to make proteins [1]. Less protein means less growth and repair, so growth is slow [1]. [2] Examiner insight: name the reason for each use, such as “storage” or “strengthens the wall”, since a bare list of products does not show use.

3. (a) The count of 52 stands out [1]. Possible cause: lamp moved nearer, a larger piece of pondweed, or a miscounted result [1]. [2] (b) Mean = (34 + 36) ÷ 2 = 35 bubbles in 2 minutes [1]. Rate = 35 ÷ 2 = 17.5 bubbles per minute [1]. [2] Examiner insight: show which result you left out and divide by the time as well, because the question asks for a rate per minute.

4. (a) Light intensity [1]. [1] (b) The rate at 20 °C is 18 at 4, 5 and 6 units, so light is no longer limiting [1]. The rate is higher at 30 °C, so temperature is limiting [1]. [2] (c) (28 − 18) ÷ 18 × 100 [1] = 56% [1]. [2] Examiner insight: in (b) a data comparison with the 30 °C line gives a stronger answer than “something else is limiting”.

5. (a) Heating: 600 × 0.85 = £510, minus 420 = £90 gain [1]. Lamps: 450 × 0.85 = £382.50, minus 390 = £7.50 loss [1]. Carbon dioxide: 200 × 0.85 = £170, minus 60 = £110 gain [1]. [3] (b) Carbon dioxide enrichment, because it gives the highest net gain [1]. [1] (c) Light would still be in shortest supply [1], so adding carbon dioxide would not raise the rate and the money would be wasted [1]. [2] Examiner insight: economics answers gain credit when the net figure is compared with the cost, not just the extra yield.

6. (a) Surface area = 2(8×3 + 8×0.08 + 3×0.08) = 49.76 cm² [1]. Volume = 1.92 cm³, so ratio = 49.76 ÷ 1.92 = 25.9 : 1 [1]. [2] (b) A flat leaf has a large surface area to volume ratio, so more light and gas exchange per unit volume [1]. Air spaces let carbon dioxide diffuse to cells inside the leaf [1]. [2] Examiner insight: quote the ratio with units or a ratio form and tie it to a specific benefit, such as faster diffusion.

7. (a) Root hairs increase the surface area of the root [1], so more water and mineral ions are absorbed at once [1]. [2] (b) It enters through the stomata [1] by diffusion [1]. [2] Examiner insight: “absorbs” alone is weak for gas entry; naming diffusion and the stomata is what the specification asks for.

8. (a) Width increases from 0 at 07:00 to a maximum of 11 µm at 13:00 [1], then falls back to 0 by 19:00 [1]. [2] (b) The wider the stomata, the more water is lost [1]. [1] (c) If water is lost faster than roots replace it, closing stomata reduces loss and prevents wilting [1]. Closing also stops carbon dioxide entering, so photosynthesis slows [1]. [2] Examiner insight: cite values from the table, and include both sides of the trade-off in (c).

9. (a) (6.3 − 1.5) ÷ 1.5 × 100 [1] = 320% [1]. [2] (b) Warm conditions make evaporation faster [1]; wind carries away water vapour near the stomata so the concentration gradient stays steep [1]. [2] Examiner insight: give the effect of both warmth and wind; “windy” alone covers only half of the explanation.

10. (a) Xylem carries water and mineral ions [1]; phloem carries dissolved sugars [1]. [2] (b) Xylem is made of hollow dead tubes, so water flows through easily [1]. Strong walls with lignin stop the tubes collapsing [1]. [2] (c) Phloem has sieve tubes with pores in the end walls so sugar solution can pass [1]. Companion cells supply the energy [1]. [2] Examiner insight: link each structural feature to its function with “so” or “because”.

11. Independent variable: different concentrations of sodium hydrogencarbonate solution [1]. Dependent variable: volume of oxygen collected in a set time, or bubbles per minute [1]. Control light by keeping the lamp at a fixed distance [1]. Control temperature using a water bath or heat shield [1]. Same pondweed and same length of piece [1]. Allow the plant a few minutes to adjust to each concentration [1]. Repeat at least three times at each concentration and calculate a mean, ignoring anomalies [1]. Hazard: mains lamp near water, so keep sockets dry [1]. [8] Examiner insight: plans gain credit when each control variable is named with the way it is kept constant, rather than saying “keep everything the same”.

Where marks are usually lost

• Leaving out the units or the “per minute” in a rate question.
• Not showing which result was ignored when calculating a mean.
• Writing “light” as a reactant in the photosynthesis equation.
• Naming a limiting factor without referring to the data.
• Calculating the extra yield but not subtracting the cost.
• Saying “stomata open and close” without mentioning guard cells.
• Mixing up xylem and phloem functions.
• Listing control variables with no method of control.

Next steps

• Revision notes
• Study guide
• Course hub and checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International GCSE Biology (9201), Version 5.2, for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.2.1 and 3.2.2.