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OCR GCSE Biology A: Photosynthesis: process, investigation and limiting factors (J247) – Practice Questions

Original practice questions with worked answers on OCR GCSE (9-1) Biology A (J247) Topic B1.4: photosynthesis, practicals and limiting factors.

Subject
Biology
Level
GCSE
Topic
Photosynthesis: process, investigation and limiting factors
Updated

Aligned to OCR GCSE Biology (J247), Version 4.0 (August 2026), for first assessment in 2018. Official specification .

Syllabus page (what it covers and how it is assessed): OCR GCSE Biology.

Syllabus points this page covers

J247

  • B1.4 Photosynthesis

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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 OCR GCSE (9-1) Biology A, Gateway Science (J247), Topic B1.4 Photosynthesis, statements B1.4a to B1.4f, from the specification Version 4.0 (August 2026). Questions 9 and 10 test B1.4f, which is printed in bold in the specification and is tested only in the Higher Tier papers; they are labelled. Revise first with the study guide and the revision notes. See also the OCR GCSE Biology hub, the printable checklist and the free diagnostics.

Questions

1. Write the balanced symbol equation for photosynthesis, and name the organelle in which it takes place. [3]

2. In the sea, a food chain runs: phytoplankton → small fish → large fish.

(a) Explain why phytoplankton are described as producers. [2] (b) Explain why a large fall in phytoplankton would reduce the biomass of large fish. [2]

3. A geranium plant is kept in the dark for 48 hours. Strips of black paper are then fixed over part of one leaf and the plant is put in bright light for six hours. The leaf is removed and tested for starch.

(a) Give the reason for keeping the plant in the dark first. [1] (b) The leaf is boiled in water and then warmed in hot ethanol. State what the ethanol does. [1] (c) Iodine solution is added. Describe the colours seen on the covered and uncovered parts. [2] (d) State the conclusion. [1]

4. A student puts a Cabomba stem in sodium hydrogencarbonate solution with a lamp 15 cm away. It gives off 42 bubbles in 3 minutes.

(a) Calculate the rate in bubbles per minute. [1] (b) The lamp is moved to 30 cm. State by what factor the light intensity changes, and show your working. [2] (c) Predict the rate at 30 cm, assuming light is the only limiting factor. [1] (d) At 7.5 cm the student measured 34 bubbles per minute, although the same reasoning predicts 56. Suggest why. [2]

5. Photosynthesis is an endothermic reaction. Explain what this means for the energy in the reaction. [2]

6. A pondweed was kept in bright light with plenty of carbon dioxide. Oxygen release (cm³ per hour) was measured at five temperatures.

Temperature (°C) 10 20 30 40 50
Oxygen released (cm³ per hour) 1.5 3.9 7.2 5.4 0.6

(a) Describe the pattern of the results. [2] (b) Explain the fall in rate between 30 °C and 50 °C. [2] (c) Calculate the percentage increase in rate from 20 °C to 30 °C. [2]

7. Three tubes each contain 20 alginate beads with algae and hydrogencarbonate indicator, which is red at normal carbon dioxide levels, yellow when carbon dioxide is higher and purple when lower. After 30 minutes, tube A (bright light) is purple, tube B (50 cm from the lamp) is red and tube C (wrapped in foil) is yellow. Explain these results. [4]

8. Plan an investigation into the effect of carbon dioxide concentration on the rate of photosynthesis in Cabomba. Include the independent variable, the measurement, three controlled variables and how you would improve reliability. [6]

9. (Higher Tier only) Rates of photosynthesis (arbitrary units) at 25 °C are shown at two carbon dioxide concentrations.

Light intensity (au) 20 40 60 80 100
0.04% carbon dioxide 12 22 24 24 24
0.12% carbon dioxide 12 23 33 40 40

(a) Identify the limiting factor at 20 au and give the evidence. [2] (b) Identify the limiting factor for the 0.04% results at 80 au and give the evidence. [2] (c) Calculate the percentage increase in rate at 80 au caused by the extra carbon dioxide. [1] (d) Suggest what limits the 0.12% results at 100 au. [1]

10. (Higher Tier only) A grower measures the rate (arbitrary units) at a constant carbon dioxide concentration of 0.10%.

Light intensity (au) 10 20 30 40 50
15 °C 6 12 15 15 15
25 °C 6 12 18 24 24

(a) State the limiting factor at 30 au and 15 °C, and justify it. [2] (b) Calculate the percentage increase in rate at 40 au when the temperature rises from 15 °C to 25 °C. [1] (c) At 25 °C and 50 au the grower can raise the temperature, add light or add carbon dioxide. Evaluate these options. [3] (d) Explain why a further rise to 45 °C would not be sensible. [2]

11. Twenty seedlings, each of equal starting dry mass, were grown in conditions that differed only in light. Mean dry mass per seedling: start 0.42 g; after 21 days in light 1.26 g; after 21 days in darkness 0.31 g.

(a) Calculate the percentage change in dry mass in the light. [2] (b) Calculate the percentage change in dry mass in darkness, to 2 significant figures. [2] (c) Explain the result in darkness. [1]

Answers

1. 6CO2 + 6H2O → C6H12O6 + 6O2: correct formulae [1]; balanced with 6, 6, 1, 6 [1]; chloroplasts [1] [3] Examiner insight: Balance an equation only by changing the numbers in front of the formulae; changing a subscript alters the substance and means the formula is no longer glucose or carbon dioxide.

2. (a) They make their own food (organic compounds) [1] from carbon dioxide and water using light energy by photosynthesis [1] [2] (b) Less phytoplankton means less biomass and energy enters the chain [1]; so fewer small fish to eat, and less biomass passes to large fish [1] [2] Examiner insight: “Energy” and “biomass” both work, but each link in the chain should be named, otherwise the explanation reads as a single jump.

3. (a) So the leaf starts with no starch and any starch found was made during the six hours [1] (b) Removes chlorophyll [1] (c) Uncovered part blue-black [1]; covered part orange-brown (stays the colour of iodine solution) [1] (d) Light is needed for photosynthesis to make starch [1] [5] Examiner insight: A conclusion must link the result to the variable tested; “the leaf has starch” alone is only a result.

4. (a) 42 ÷ 3 = 14 bubbles per minute [1] (b) Distance doubles [1]; intensity = (1/2)² = 1/4 [1] (c) 14 × 1/4 = 3.5 bubbles per minute [1] (d) At 7.5 cm light is no longer the only limiting factor [1]; carbon dioxide or temperature (the lamp warms the water) limits the rate, so the rate does not quadruple [1] [6] Examiner insight: When predicting by calculation, write down the factor you used, so your reasoning is visible even if the arithmetic slips.

5. It takes in energy from the surroundings [1]; the energy is light, and is stored as chemical energy in glucose [1] [2] Examiner insight: State where the energy comes from and where it goes; “absorbs heat” alone is a weaker answer.

6. (a) Rate rises from 1.5 to a peak of 7.2 at 30 °C [1]; then falls to 0.6 at 50 °C [1] (b) Enzymes denature above the optimum [1]; the active site loses its shape, so substrate molecules cannot bind and fewer reactions happen [1] (c) (7.2 − 3.9) ÷ 3.9 × 100 [1] = 85% (84.6, to 2 significant figures) [1] [6] Examiner insight: Quote values with units when describing a graph; “it goes up then down” is too vague for two marks.

7. Tube A: bright light means fast photosynthesis, using up more carbon dioxide than respiration releases, so purple [1]. Tube B: photosynthesis is slow, so carbon dioxide used roughly balances carbon dioxide released by respiration, so red [1]. Tube C: no light, so no photosynthesis [1]; algae respire and release carbon dioxide, so yellow [1] [4] Examiner insight: Mention respiration for the dark tube; saying only “no photosynthesis” does not explain why the colour changes.

8. Independent variable: carbon dioxide concentration, by different amounts of sodium hydrogencarbonate in water [1]. Measure: bubbles per minute or gas volume per minute, counted for a fixed time [1]. Control, with how: lamp distance kept fixed [1]; temperature kept steady with a water bath [1]; same stem and same counting time [1]. Reliability: repeat each concentration and calculate a mean [1] [6] Examiner insight: Controlled variables must be named with how they are kept constant; “keep it fair” does not earn credit.

9. (a) Light [1]; both rates are 12 at 20 au even though carbon dioxide differs [1] (b) Carbon dioxide [1]; raising it to 0.12% increased the rate from 24 to 40 [1] (c) (40 − 24) ÷ 24 × 100 = 67% [1] (d) Temperature (or another factor) [1] [6] Examiner insight: A limiting factor claim needs its evidence; quote both numbers from the table and say what changed.

10. (a) Temperature [1]; the 25 °C rate is higher at 30 au (18 against 15) [1] (b) (24 − 15) ÷ 15 × 100 = 60% [1] (c) At 50 au the rate is flat, so extra light would not help [1]; carbon dioxide or temperature is limiting and the data cannot say which, so test one at a time [1]; a small rise in temperature is cheap but risks denaturing enzymes if the optimum is passed, while extra carbon dioxide costs money but carries no such risk, so weigh cost against gain and choose with a reason [1] (d) Enzymes denature above the optimum [1]; the rate would fall [1] [8] Examiner insight: An evaluation needs a decision plus a reason against the alternatives; listing the three options without choosing is not enough.

11. (a) (1.26 − 0.42) ÷ 0.42 × 100 [1] = +200% [1] (b) (0.31 − 0.42) ÷ 0.42 × 100 [1] = −26% [1] (c) No photosynthesis in the dark, and stored food is used in respiration [1] [5] Examiner insight: A percentage change needs the sign; a loss in mass must be written as negative.

Where marks are usually lost

  • Writing unbalanced equations, or leaving out the condition of light.
  • Describing graphs without values or units.
  • Giving “temperature” as the answer to a plateau at a different factor with no evidence.
  • Using the inverse square law the wrong way round.
  • Forgetting the sign on a percentage change, or dividing by the new figure when the original is needed.
  • Missing that algae and plants respire in the dark.
  • Naming controlled variables without saying how they are held constant.
  • Saying enzymes are “killed” rather than denatured.

Next steps

Official syllabus

OCR GCSE (9-1) Gateway Science Biology A (J247), Version 4.0 (August 2026), for first assessment in 2018, published by OCR (Oxford Cambridge and RSA Examinations). Topic B1.4 Photosynthesis, statements B1.4a to B1.4f.

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