AQA A-level Biology: Photosynthesis (7402) – 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 section 3.5.1 (Photosynthesis) of the AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015, including required practicals 7 and 8. Section 3.5 is A-level only. Learn the content first from the study guide or the revision notes. Course pages: AQA A-level Biology hub and printable checklist.

Questions

1. Describe what happens to chlorophyll during photoionisation and state where its replacement electrons come from. [3]

2. In a chloroplast the stroma is pH 8.0 and the thylakoid space is pH 5.0.

(a) Calculate how many times greater the hydrogen ion concentration is in the thylakoid space. [2] (b) Explain how the difference is formed and how it leads to the production of ATP. [4]

3. Photolysis of water produces three products. Name them and give one fate of each. [3]

4. Nine carbon dioxide molecules are fixed in the Calvin cycle.

(a) State how many molecules of GP are formed. [1] (b) Three molecules of triose phosphate leave the cycle. Calculate how many RuBP molecules the rest can regenerate. [2] (c) Explain why the number of carbon atoms leaving the cycle must equal the number entering. [1]

5. A herbicide blocks the electron transfer chain in chloroplasts. Explain the effect on the concentrations of GP, triose phosphate and RuBP shortly after treatment. [4]

6. The table shows the rate of oxygen release (arbitrary units) by a plant under four sets of conditions.

Treatment	Light	CO₂	Temperature	Rate
W	low	0.04%	20 °C	5
X	high	0.04%	20 °C	5
Y	high	0.20%	20 °C	12
Z	high	0.20%	30 °C	24

(a) Identify the limiting factor in W and in X, with evidence. [3] (b) Explain why raising the temperature from 20 °C to 30 °C increased the rate. [2] (c) Suggest why the rate would not keep rising as temperature is raised further. [1]

7. Pigments from a leaf were separated by chromatography. The solvent front moved 9.2 cm. Four spots had centres 8.7, 6.4, 5.9 and 4.1 cm from the origin.

(a) Calculate the Rf of the second and fourth spots. [2] (b) An extract from a shade-tolerant plant gave a larger spot than a sun plant at the fourth position. Suggest what this shows. [1] (c) Give two precautions for getting reliable measurements. [2]

8. Chloroplast extract was mixed with DCPIP and absorbance was read at the start and after 10 minutes.

Tube	Start	After 10 min
20 °C	0.62	0.38
30 °C	0.62	0.20
45 °C	0.62	0.55
Boiled extract, 30 °C	0.62	0.62
30 °C in foil	0.62	0.61

(a) Calculate the rate at 20 °C and at 30 °C, with units. [2] (b) Explain the result at 45 °C. [2] (c) State what the boiled and foil tubes show. [2]

9. In a 300 m² polytunnel, 1200 kg of strawberries were grown in ambient conditions. Carbon dioxide enrichment (cost £240) gave 1380 kg. Adding artificial light as well (extra cost £520) gave 1470 kg. Strawberries sell at £3.20 per kg.

(a) Calculate the percentage increase in yield from carbon dioxide enrichment. [1] (b) Calculate the net gain from carbon dioxide enrichment. [2] (c) Evaluate the addition of artificial light. [2] (d) State one limitation of the data. [1]

10. A lamp was placed 10 cm and then 20 cm from pondweed. The pondweed released 1.8 cm³ and 0.72 cm³ of gas in 12 minutes respectively.

(a) Calculate the rate at each distance. [2] (b) Light intensity is roughly proportional to 1 ÷ distance². State the expected ratio of rates, 10 cm to 20 cm, if light were the only limiting factor. [1] (c) The measured ratio is lower. Suggest two reasons. [2] (d) Suggest why a tank of water between lamp and plant improves the investigation. [1]

11. Describe and explain how light energy is conserved in the light-dependent reaction and used to make a simple sugar in the light-independent reaction. [8]

Answers

1. Chlorophyll absorbs light [1], which excites electrons so that they leave the chlorophyll (it is ionised) [1]. The electrons are replaced by electrons from the photolysis of water [1]. [3] Examiner insight: An answer that says electrons are “lost” but not that light caused it tends to miss the idea that makes it photoionisation.

2. (a) The difference is 8.0 − 5.0 = 3.0 pH units [1], so 10³ = 1000 times greater [1]. [2] (b) Energy released as electrons move along the electron transfer chain is used to move protons from the stroma into the thylakoid space [1]. The membrane is impermeable to protons, so a gradient builds up [1]. Protons flow back down the gradient through ATP synthase [1], which makes ATP from ADP and phosphate (chemiosmosis) [1]. [4] Examiner insight: Calculations of pH differences gain credit for converting units to a tenfold factor; stating only “3” does not answer “how many times”.

3. Electrons, which replace those lost from chlorophyll [1]. Protons, which add to the gradient or reduce NADP [1]. Oxygen, which is released or used in respiration [1]. [3] Examiner insight: Naming three products with no fate for each is a common way to lose credit when the question asks for both.

4. (a) 18 GP [1]. [1] (b) 18 − 3 = 15 triose phosphate remain, which is 45 carbons [1]. 45 ÷ 5 = 9 RuBP [1]. [2] (c) If more carbon left than entered, the cycle could not regenerate enough RuBP to keep fixing carbon dioxide [1]. [1] Examiner insight: Showing the carbon count makes it clear how the answer was reached; a bare number leaves nothing to credit if it is wrong.

5. No electrons move along the chain, so less ATP is made [1] and NADP is not reduced [1]. GP cannot be reduced, so GP rises and triose phosphate falls [1]. Less TP and ATP means RuBP is not regenerated but is still used by rubisco, so RuBP falls [1]. [4] Examiner insight: Answers that give the direction of change for each compound, with a reason tied to the reaction that stopped, are easier to credit than a list of “decreases”.

6. (a) Raising light from W to X gave no change in rate [1]. Raising carbon dioxide from X to Y raised the rate [1]. Carbon dioxide is therefore the limiting factor in both W and X [1]. [3] (b) Higher temperature increases kinetic energy and collisions between enzymes (such as rubisco) and substrate [1], so more reactions occur per second [1]. [2] (c) Enzymes denature at high temperature [1]. [1] Examiner insight: Comparisons that change only one variable between treatments are what support a conclusion about limiting factors.

7. (a) Second: 6.4 ÷ 9.2 = 0.70 [1]. Fourth: 4.1 ÷ 9.2 = 0.45 [1]. [2] (b) The shade plant has more of that pigment [1]. [1] (c) Mark the solvent front as soon as the paper is removed [1]. Measure to the centre of each spot, and use pencil for the origin [1]. [2] Examiner insight: Rf has no unit; writing one, or quoting it to more decimal places than the measurements allow, can lose credit.

8. (a) 20 °C: (0.62 − 0.38) ÷ 10 = 0.024 absorbance units per minute [1]. 30 °C: (0.62 − 0.20) ÷ 10 = 0.042 absorbance units per minute [1]. [2] (b) The rate is low because enzymes have begun to denature [1], so fewer electrons reach DCPIP [1]. [2] (c) Boiled: no colour change shows enzymes need to be active [1]. Foil: very little change shows light is required for the reaction [1]. [2] Examiner insight: Rates need the unit; “per minute” is part of the answer.

9. (a) (1380 − 1200) ÷ 1200 × 100 = 15% [1]. [1] (b) Extra income = 180 × 3.20 = £576 [1]. Net gain = 576 − 240 = £336 [1]. [2] (c) Extra income from light is 90 × 3.20 = £288 [1], which is less than the cost of £520, a net loss of £232, so it is not worthwhile [1]. [2] (d) One season, one crop, no repeats, or prices vary [1]. [1] Examiner insight: Evaluation needs a judgement supported by figures; a percentage alone, with no cost, is unlikely to complete it.

10. (a) 1.8 ÷ 12 = 0.15 cm³ per minute [1]; 0.72 ÷ 12 = 0.060 cm³ per minute [1]. [2] (b) (20 ÷ 10)² = 4 [1]. [1] (c) The measured ratio is 2.5. Another factor, such as carbon dioxide concentration or temperature, is limiting [1]. The lamp is not a point source, or the gas collected includes oxygen used in respiration or dissolving [1]. [2] (d) It absorbs heat from the lamp, so temperature stays constant [1]. [1] Examiner insight: Suggestions that name a specific variable score better than “errors” or “mistakes”.

11. Chlorophyll absorbs light, which excites and releases electrons (photoionisation) [1]. Electrons pass along the electron transfer chain, losing energy [1]. Energy is used to move protons into the thylakoid space [1]. Protons flow through ATP synthase, driving ATP formation (chemiosmosis) [1]. Electrons and protons reduce NADP [1]. Photolysis replaces the electrons [1]. In the stroma, carbon dioxide reacts with RuBP, catalysed by rubisco, to form GP [1], reduced to triose phosphate using ATP and reduced NADP, then TP is converted to sugars and used to regenerate RuBP [1]. [8] Examiner insight: Extended answers earn credit for linking each stage to the next, such as ATP and reduced NADP being used in the reduction of GP.

Where marks are usually lost

• Stating that energy is made, rather than transferred.
• Forgetting that photolysis supplies electrons for chlorophyll.
• Saying that ATP synthase pumps protons.
• Not giving the direction of change for GP, triose phosphate and RuBP.
• Describing the light-independent reaction as needing darkness.
• Calling the x-axis factor limiting on a plateau.
• Giving a percentage without dividing by the original.
• Leaving out units, or giving Rf with units.
• Evaluating agricultural data with no mention of cost.

Next steps

• Review the revision notes and the study guide.
• Visit the AQA A-level Biology hub and the printable checklist.
• Try all free 10-minute diagnostics at diagnostics.
• Book a free trial class at trial.

Official syllabus

AQA AS and A-level Biology (7401 and 7402), specification for first teaching 2015, published by AQA. Section 3.5.1 and required practicals 7 and 8.