OxfordAQA International AS and A-level Biology: Photosynthesis and limiting factors (9610) – 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 OxfordAQA International AS and A-level Biology (9610), Version 5.1, section 3.3.2 (photosynthesis) and Required practical 7, in Unit 3 (International A-level only). Read the study guide first. The course page, the checklist and the free diagnostics help you plan.

Questions

1. Describe what happens when chlorophyll absorbs light, and what the electrons are then used for. [3]

2. Explain how ATP is made in the light-dependent reaction by the passage of protons across chloroplast membranes. [5]

3. A suspension of algae was given water containing a heavy isotope of oxygen, ¹⁸O, and ordinary carbon dioxide, and was kept in the light. The oxygen gas released contained ¹⁸O. A second suspension given ordinary water and carbon dioxide containing ¹⁸O released oxygen with no ¹⁸O.

(a) State the three products of photolysis. [3] (b) Explain what the results show. [2] (c) 36 µmol of water is split. Calculate the amount of oxygen released and the amount of electrons released. [2]

4. Algae were illuminated until the amounts of two compounds were steady. The light was then switched off. The table shows relative amounts.

Time after light off (min)	0	1	3
GP	100	160	210
RuBP	100	45	20

(a) Describe the changes, including the percentage change in GP between 0 and 3 minutes. [2] (b) Explain the change in GP. [2] (c) Explain the change in RuBP. [2]

5. Nine molecules of carbon dioxide are fixed in the Calvin cycle.

(a) How many molecules of GP are formed? [1] (b) How many molecules of ATP are used to reduce the GP? [1] (c) How many molecules of triose phosphate are needed to regenerate RuBP? Show your working. [2] (d) How many molecules of triose phosphate are available to leave the cycle? [1]

6. Describe how a plant uses triose phosphate to make (a) starch [2] and (b) a protein [2].

7. Rates of photosynthesis (arbitrary units) were recorded at two carbon dioxide concentrations over a range of light intensities, with temperature held constant.

Light intensity (arbitrary units)	0	15	30	45	60	75
Rate at 0.04% CO₂	0	2.5	5.0	7.5	8.0	8.0
Rate at 0.08% CO₂	0	2.5	5.0	7.5	10.0	11.5

(a) Identify the limiting factor at light intensity 15 and justify your answer. [2] (b) Identify the limiting factor at 0.04% CO₂ and light intensity 60, and justify your answer. [2] (c) Calculate the percentage increase in rate at light intensity 75 when CO₂ is raised from 0.04% to 0.08%. [2] (d) Suggest what limits the rate at 0.08% CO₂ and light intensity 75, and how you would test it. [2]

8. The rate of photosynthesis of a leaf (arbitrary units) at constant light intensity and CO₂ was: 2.5 at 15 °C, 5.4 at 25 °C, 9.6 at 35 °C, 4.3 at 45 °C and 0.8 at 55 °C.

(a) State how many times greater the rate is at 35 °C than at 15 °C. Give your answer to 2 significant figures. [1] (b) Explain why the rate increases from 15 °C to 35 °C. [2] (c) Calculate the percentage decrease in rate from 35 °C to 45 °C. [1] (d) Explain the decrease. [2]

9. Plan an investigation into how light intensity affects the rate of photosynthesis in a water plant. [8]

10. In a Required practical 7 investigation, a student collected the oxygen from a pondweed in a capillary tube and measured the length of the gas column after 5 minutes, with the lamp at different distances.

Distance (cm)	Length 1 (mm)	Length 2 (mm)	Length 3 (mm)
10	38	41	35
20	21	19	23
40	7	6	8

(a) Calculate the mean rate of gas collection at 40 cm in mm per minute. [1] (b) Light intensity is proportional to 1 ÷ distance². Calculate the intensity at 20 cm relative to that at 10 cm. [1] (c) Calculate how many times greater the mean length at 10 cm is than at 20 cm. Give your answer to 2 significant figures. [1] (d) Compare your answers to (b) and (c) and suggest an explanation. [2] (e) Room lighting was on during the experiment. Explain why this is a problem and how to avoid it. [2]

11. A grower compared tomato yields (kg per 100 plants) in four greenhouses. A: normal air and daylight, 52. B: extra CO₂, 64. C: extra lamps, 71. D: extra CO₂ and extra lamps, 98.

(a) Calculate the percentage increase in yield in D compared with A. [2] (b) The gains in B and C added together are less than the gain in D. Use the principle of limiting factors to explain this. [4] (c) Suggest why yield may not rise further if CO₂ and light are increased even more. [2]

Answers

1. Chlorophyll absorbs light energy [1]. This excites electrons, which leave the chlorophyll [1]. They pass along the electron transfer chain and the energy is used to make ATP and reduced NADP [1]. [3] Examiner insight: Say the electrons are excited by light energy and then say what their energy is used for. Answers that stop at “chlorophyll absorbs light” gain little.

2. Excited electrons pass along the electron transfer chain in the chloroplast membrane [1]. Energy is released at each transfer [1]. It is used to pump protons across the membrane [1], creating a proton gradient. The protons return through a channel in ATP synthase [1], and this enzyme catalyses ADP + inorganic phosphate → ATP [1]. [5] Examiner insight: Name the enzyme and say that protons pass through it. “Protons make ATP” does not show the role of the enzyme.

3. (a) Protons [1], electrons [1], oxygen [1]. [3] (b) The ¹⁸O appears in oxygen only when the water was labelled [1], so the oxygen released comes from water, not from carbon dioxide [1]. [2] (c) Oxygen = 36 ÷ 2 = 18 µmol [1]; electrons = 18 × 4 = 72 µmol [1]. [2] Examiner insight: In (b) refer to both samples. A conclusion quoted without the comparison is weaker than one that uses both results.

4. (a) GP rises from 100 to 210, an increase of 110 ÷ 100 × 100 = 110% [1]; RuBP falls from 100 to 20 [1]. [2] (b) With no light, no ATP and reduced NADP are made [1]. GP is not reduced to triose phosphate, but is still made from CO₂ and RuBP, so GP builds up [1]. [2] (c) Less triose phosphate is made, so RuBP is not regenerated [1]. RuBP is still used up in reacting with CO₂ [1]. [2] Examiner insight: Quote figures from the table. Link the change in each compound to the missing ATP and reduced NADP, not just to “no light”.

5. (a) 9 × 2 = 18 GP [1]. [1] (b) One ATP per GP, so 18 ATP [1]. [1] (c) Nine RuBP have 9 × 5 = 45 carbons [1]; 45 ÷ 3 = 15 triose phosphate [1]. [2] (d) 18 − 15 = 3 triose phosphate [1]. [1] Examiner insight: Show the carbon count so your reasoning can be followed.

6. (a) Two triose phosphate molecules join to make a hexose such as glucose [1], and hexose molecules are joined by condensation reactions into starch [1]. [2] (b) Triose phosphate is converted into amino acids [1], using nitrogen from nitrate ions absorbed by roots, and amino acids are joined by peptide bonds [1]. [2] Examiner insight: For protein, nitrogen must appear. Triose phosphate contains only carbon, hydrogen, oxygen and phosphate.

7. (a) Light [1]; both curves have the same rate, which rises with light intensity [1]. [2] (b) Carbon dioxide [1]; raising CO₂ to 0.08% increases the rate from 8.0 to 10.0 [1]. [2] (c) (11.5 − 8.0) ÷ 8.0 [1] × 100 = 43.8% [1]. [2] (d) Temperature [1]; repeat at a higher temperature and see whether the rate rises [1]. [2] Examiner insight: Justify a limiting factor with data from the table. Naming the factor without evidence rarely gains full credit.

8. (a) 9.6 ÷ 2.5 = 3.84, so 3.8 times [1]. [1] (b) A rise in temperature increases kinetic energy [1], so there are more frequent collisions between enzymes (such as rubisco) and substrates, forming more enzyme-substrate complexes [1]. [2] (c) (9.6 − 4.3) ÷ 9.6 × 100 = 55.2% [1]. [1] (d) Enzymes begin to denature as hydrogen bonds break [1], changing the shape of the active site, so substrates no longer fit [1]. [2] Examiner insight: Divide the change by the value you started from, not the one you ended on. For (d), “denature” must be explained in terms of active-site shape.

9. Any eight from: place pondweed in a boiling tube of sodium hydrogencarbonate solution as a source of CO₂ [1]; keep the same length of pondweed each time [1]; set a lamp a known distance from the tube, measured with a ruler [1]; use at least five distances [1]; stand the tube in a water bath or use a heat shield to keep temperature constant [1]; allow a few minutes for the plant to adjust [1]; count bubbles per minute, or collect and measure the gas [1]; repeat at least three times at each distance and calculate a mean [1]. [8] Examiner insight: State what you alter, what you record and what you keep fixed. A plan that leaves out temperature control usually loses credit.

10. (a) Mean length = (7 + 6 + 8) ÷ 3 = 7.0 mm; 7.0 ÷ 5 = 1.4 mm per minute [1]. [1] (b) (10 ÷ 20)² = 0.25 [1]. [1] (c) Means are 38.0 mm at 10 cm and 21.0 mm at 20 cm; 38.0 ÷ 21.0 = 1.8 times [1]. [1] (d) Intensity at 10 cm is four times that at 20 cm, but the mean length is only 1.8 times greater [1]. Another factor, such as CO₂ concentration or temperature, limits the rate at high light intensity [1]. [2] (e) Room light adds to the lamp light, so the light intensity at the plant is not the value calculated from the distance [1]. Darken the room, or shield the apparatus, so the lamp is the only light source [1]. [2] Examiner insight: In (d) quote both ratios. In (e) link the stray light to the independent variable.

11. (a) 98 − 52 = 46 [1]; 46 ÷ 52 × 100 = 88.5% [1]. [2] (b) The gain from B (12) plus C (19) is 31, less than the gain of 46 in D [1]. In B, extra CO₂ helps but light is still limiting [1]. In C, extra light helps but CO₂ soon becomes limiting [1]. In D, both are raised, so neither limits and the rate is far higher [1]. [4] (c) Temperature becomes limiting [1]; the amount of enzyme such as rubisco, or chlorophyll, is limiting [1]. [2] Examiner insight: Use the figures from the table and name which factor is limiting in each greenhouse.

Where marks are usually lost

• Saying the electrons in the chain make ATP, without protons and ATP synthase.
• Giving the source of oxygen as carbon dioxide.
• Swapping GP and RuBP, or the number of carbons.
• Dividing percentage change by the final value.
• Stating a limiting factor with no data.
• Saying enzymes are “killed” rather than denatured.

Next steps

• Review the revision notes and the study guide.
• Go to the OxfordAQA A-level Biology course page and the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Biology (9610), Version 5.1, for International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards, published by OxfordAQA. Section 3.3.2 and Required practical 7.