OCR A Level Biology A: Photosynthesis (H420) – 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 learning outcomes 5.2.1(a) to 5.2.1(g) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. Read the study guide or the revision notes first. More practice skills are in Practical skills practice. Find the course on the OCR A Level Biology hub, use the printable checklist and try the free diagnostics.

Questions

1. (a) Write the balanced symbol equation for photosynthesis. [1] (b) A leaf fixes 11.0 g of carbon dioxide. Calculate the mass of glucose made and of oxygen released. (Relative molecular masses: CO₂ 44, glucose 180, O₂ 32.) [3] (c) Explain how photosynthesis and aerobic respiration depend on each other’s products. [2]

2. (a) State where in a chloroplast the light-dependent stage and the light-independent stage happen. [2] (b) An electron micrograph shows a chloroplast 63 mm long at a magnification of ×12 000. Calculate its actual length in µm. [2] (c) Suggest why a chloroplast contains DNA. [1]

3. Leaf pigments were separated by thin layer chromatography. The solvent front moved 8.0 cm. Four spots had centres 7.6, 5.2, 4.0 and 2.8 cm from the origin. (a) Calculate the Rf values of the first and fourth spots. [2] (b) The first spot is carotene. Explain why it travelled furthest. [2] (c) Explain why the origin is drawn in pencil. [1]

4. The rate of photosynthesis (arbitrary units) was measured at five wavelengths.

Wavelength (nm)	440	500	540	600	660
Rate	80	36	20	44	72

(a) Calculate the rate at 540 nm as a percentage of the rate at 440 nm. [1] (b) State what this type of graph is called. [1] (c) Chlorophyll a absorbs little green light. Explain why photosynthesis still occurs at 540 nm. [2] (d) Suggest why the rates at 440 and 660 nm are the highest. [1]

5. (a) Compare cyclic and non-cyclic photophosphorylation, giving four points of difference. [4] (b) Explain two roles of water in the light-dependent stage. [2]

6. A herbicide stops electrons leaving photosystem II. Explain the effect, shortly after application, on the output of oxygen and on the levels of ATP, reduced NADP, GP, TP and RuBP in the leaf. [5]

7. In a model of the Calvin cycle, 30 molecules of RuBP each combine with one carbon dioxide molecule. RuBP has five carbon atoms. (a) State how many GP molecules form. [1] (b) State how many TP molecules form from them. [1] (c) Calculate how many TP molecules leave the cycle, and explain your answer in terms of carbon atoms. [2] (d) Calculate how many RuBP molecules the remaining TP can regenerate. [1]

8. Algae were supplied with 0.20% carbon dioxide in bright light. The supply was reduced to 0.02%. The table shows the amounts of three compounds (µmol per g) before and 60 seconds after the change.

Compound	Before	After
GP	6.8	2.1
TP	5.0	1.8
RuBP	3.5	9.4

(a) Calculate the percentage change in GP. [1] (b) Explain the changes in GP and RuBP. [3] (c) Explain the change in TP. [1]

9. In one stage of an experiment 18 carbon dioxide molecules are fixed by RuBisCO. (a) Calculate the number of TP molecules produced, the number that leave the cycle and the percentage that are recycled. [2] (b) State two uses of the TP that leaves the cycle. [2]

10. A culture of the alga Chlorella was kept at 25 °C. The rate of photosynthesis (arbitrary units) was recorded at four carbon dioxide concentrations in dim light and in bright light.

CO₂ (%)	0.02	0.06	0.10	0.14
Dim light	6	10	11	11
Bright light	7	18	27	28

(a) Identify the limiting factor in dim light at 0.14% CO₂, with a reason. [1] (b) Explain why the rate in bright light rises from 0.02% to 0.10% CO₂ but hardly changes from 0.10% to 0.14%. [2] (c) Calculate the percentage increase in rate in bright light when CO₂ rises from 0.02% to 0.06%. [2] (d) Explain, in terms of GP and RuBP, why extra carbon dioxide cannot raise the rate in dim light. [2]

11. Gas released by pondweed collected in a capillary tube of internal radius 0.50 mm. In 10 minutes the gas column was 24 mm long. (Volume of a cylinder = πr²l.) (a) Calculate the volume of gas. [2] (b) Calculate the rate in mm³ min⁻¹. [1] (c) Explain why this underestimates the rate of oxygen production. [2] (d) State one way of keeping temperature constant while the lamp is moved. [1]

12. On a hot, dry afternoon the rate of photosynthesis in a wheat field falls although the light is bright. Explain why, with reference to GP, TP and RuBP, and suggest how a farmer could reduce the effect. [8]

Answers

1. (a) 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ [1] (b) Moles CO₂ = 11.0 ÷ 44 = 0.250 [1]; glucose = 0.250 ÷ 6 × 180 = 7.50 g [1]; oxygen = 0.250 × 32 = 8.00 g [1]. (c) Respiration uses glucose and oxygen made by photosynthesis [1]; photosynthesis uses carbon dioxide and water made by respiration [1]. [6] Examiner insight: Show the mole step even when the mass ratio looks obvious; it makes your method clear and any slip easy to find.

2. (a) Light-dependent: thylakoid membranes (grana) [1]. Light-independent: stroma [1]. (b) 63 mm ÷ 12 000 = 0.00525 mm [1] = 5.25 µm [1]. (c) It codes for some chloroplast proteins [1]. [5] Examiner insight: Unit conversion is an easy place to slip; write “× 1000” explicitly when going from mm to µm.

3. (a) First spot: 7.6 ÷ 8.0 = 0.95 [1]. Fourth: 2.8 ÷ 8.0 = 0.35 [1]. (b) Carotene is the most soluble in the solvent [1] and is least attracted to the plate, so it moves with the solvent front [1]. (c) Pencil does not dissolve in the solvent, so it does not separate and mix with the pigments [1]. [5] Examiner insight: Explanations of separation are strongest when they give both sides: solubility in the mobile phase and attraction to the stationary phase.

4. (a) 20 ÷ 80 × 100 = 25.0% [1]. (b) An action spectrum [1]. (c) Accessory pigments such as carotenoids absorb some green light [1] and pass the energy to a reaction centre [1]. (d) These are the wavelengths (blue and red) that chlorophyll absorbs most strongly [1]. [5] Examiner insight: Naming the absorbing pigments and saying the energy is passed to the reaction centre is stronger than “other pigments help”.

5. (a) Any four: photosystems involved (PSI alone, or both PSII and PSI) [1]; destination of the electrons (back to PSI, or onto NADP) [1]; products (ATP only, or ATP plus reduced NADP) [1]; photolysis and oxygen release (absent in cyclic, present in non-cyclic) [1]. Each point must state both processes. (b) Photolysis supplies electrons to replace those lost from PSII [1]; it supplies protons that reduce NADP (or build the gradient) [1]. [6] Examiner insight: A comparison needs both sides of each point stated; describing only one process earns little credit.

6. Oxygen output falls because no electrons are passed from water, so photolysis stops [1]. Almost no reduced NADP is made, and ATP from the non-cyclic route falls (cyclic flow may still make some) [1]. GP cannot be reduced, so GP initially rises or stays high [1] while TP falls [1]. RuBP falls because less TP is available to regenerate it [1]. [5] Examiner insight: Link each change in a Calvin cycle compound to the missing product that causes it.

7. (a) 60 [1]. (b) 60 [1]. (c) 30 carbons were fixed, so 30 ÷ 3 = 10 TP leave [1] because carbon leaving equals the carbon from the fixed carbon dioxide [1]. (d) 50 TP × 3 = 150 carbons; 150 ÷ 5 = 30 RuBP [1]. [5] Examiner insight: Counting carbon atoms at each stage is a safe check, because an unbalanced answer shows up at once.

8. (a) (2.1 − 6.8) ÷ 6.8 × 100 = −69.1% [1]. (b) Less carbon dioxide is available to combine with RuBP, so less GP forms [1]; RuBP is used more slowly so it accumulates [1], while regeneration from TP continues [1]. (c) TP falls because less GP is available to be reduced [1]. [5] Examiner insight: Give the sign with percentage change; “69.1% decrease” or “−69.1%” both show the direction.

9. (a) 36 GP and 36 TP, so 6 leave [1] and 30 are recycled: 30 ÷ 36 × 100 = 83.3% [1]. (b) Any two: carbohydrates, lipids, amino acids [1][1]. [4] Examiner insight: “Used for respiration” is not one of the uses listed for TP; name the molecule types.

10. (a) Light intensity, because the rate is higher in bright light at the same CO₂ [1]. (b) Up to 0.10% carbon dioxide is limiting, so more allows faster fixation by RuBisCO [1]. By 0.14% another factor, such as temperature or light, is limiting (or RuBisCO is working at its maximum) [1]. (c) (18 − 7) ÷ 7 × 100 [1] = 157% [1]. (d) Little ATP and reduced NADP is made in dim light, so GP is reduced to TP slowly [1] and RuBP is regenerated slowly, so less RuBP is available to fix extra carbon dioxide [1]. [7] Examiner insight: Quote values from the table when naming a limiting factor, for example comparing the two light conditions at the same carbon dioxide level.

11. (a) πr²l = π × 0.50² × 24 [1] = 18.8 mm³ [1]. (b) 18.85 ÷ 10 = 1.88 mm³ min⁻¹ [1]. (c) The plant’s own respiring cells take up part of the oxygen made [1]; some oxygen stays dissolved in the water [1]. (d) Place a tank of water between the lamp and plant, or use a water bath [1]. [6] Examiner insight: Keep unrounded values until the end: 18.8496 divided by 10 gives 1.88, whereas rounding early to 18.85 or 18.8 can shift the last figure.

12. Hot, dry conditions cause water stress [1]. Guard cells close the stomata to reduce water loss [1]. Less carbon dioxide enters [1], so it becomes the limiting factor [1]. Less fixation means GP falls [1] and TP falls [1] while RuBP rises because less is used [1]. The light-dependent stage still makes ATP and reduced NADP, but they are not used up. A farmer could irrigate [1]. [8] Examiner insight: A full answer chains the cause (stomatal closure) to the effect on each of the three compounds; naming the limiting factor alone is not enough.

Where marks are usually lost

• Giving the right equation but not showing the mole step in a mass calculation.
• Writing the site of the light-dependent stage as “the chloroplast” rather than the thylakoid membranes.
• Measuring Rf to the edge of the spot instead of its centre.
• Saying that pigments “reflect” rather than absorb, or ignoring accessory pigments.
• Stating that water is “split for energy” instead of supplying electrons and protons.
• Reversing GP and RuBP changes when light or carbon dioxide falls.
• Forgetting that RuBP must be regenerated from TP.
• Naming a limiting factor without referring to the data.
• Leaving out units or rounding too early.

Next steps

• Re-read the revision notes and the study guide.
• Go to the OCR A Level Biology hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025, published by OCR. Section 5.2.1 Photosynthesis.