Cambridge O Level Biology 5090: Plant nutrition (5090) – 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 Topic 6, Plant nutrition, of the Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028 (version 4): sections 6.1 Photosynthesis, 6.2 Leaf structure and 6.3 Mineral nutrition. The syllabus is not tiered. Questions 4–7, 9 and 11 use practical contexts.

Learn the content first with the study guide and revision notes.

Questions

1. State the word equation and the balanced chemical equation for photosynthesis. [3]

2. Chlorophyll is needed for photosynthesis.

(a) State where chlorophyll is found in a plant cell. [1] (b) State the role of chlorophyll in photosynthesis. [1]

3. State how a plant uses each of these carbohydrates made in photosynthesis: starch, cellulose, glucose and sucrose. [4]

4. A plant with variegated (green and white) leaves is kept in the dark for two days, then placed in sunlight for six hours. A leaf is then tested for starch to show that chlorophyll is needed.

(a) Explain why the plant was kept in the dark for two days. [2] (b) During the starch test the leaf is put into boiling water and then into hot ethanol. Explain the purpose of each step. [2] (c) State one safety precaution when using ethanol, with a reason. [1] (d) Predict the colour of the green and the white areas after iodine solution is added, and explain the results. [3]

5. Two destarched leaves on the same plant are each sealed in a flask. Flask A contains sodium hydroxide solution, which absorbs carbon dioxide; flask B contains the same volume of water. After eight hours in bright light both leaves are tested for starch.

(a) State why flask B is included. [1] (b) Predict the result for each leaf. [2] (c) State the conclusion. [1] (d) Suggest one variable, other than light, that is kept the same for both leaves. [1]

6. Bubbles released by pondweed in one minute are counted three times at each lamp distance.

Lamp distance / cm	Count 1	Count 2	Count 3
5	38	41	35
15	22	26	41
25	13	15	14

(a) Identify the anomalous result and calculate the mean number of bubbles per minute at 15 cm, leaving it out. [2] (b) Describe and explain the effect of lamp distance on the rate of photosynthesis. [3] (c) A beaker of water is placed between the lamp and the pondweed. Explain why. [1] (d) Suggest why collecting the gas in a syringe gives more accurate results than counting bubbles. [1]

7. Pondweed is placed in different concentrations of sodium hydrogencarbonate, a source of carbon dioxide, at constant light and temperature.

Sodium hydrogencarbonate / %	0	0.25	0.5	1.0	2.0
Volume of gas in 8 min / cm³	0.4	1.2	2.0	2.8	2.8

(a) Calculate the rate of gas production at 0.5%, giving the unit. [2] (b) Explain the change in volume between 0 and 1.0%. [2] (c) Explain why the volume does not change between 1.0% and 2.0%, and suggest one factor that could be limiting. [2]

8. Describe and explain how the rate of photosynthesis of a plant changes as temperature rises from 10 °C to 45 °C, with light and carbon dioxide not limiting. [5]

9. Four sealed tubes contain the same volume of red hydrogencarbonate indicator. All are placed in bright light for three hours.

• Tube W: pondweed only
• Tube X: a water snail only
• Tube Y: pondweed and a water snail
• Tube Z: indicator only

(a) Predict the colour of the indicator in tubes W and X at the end. [2] (b) Explain your predictions for W and X. [3] (c) Explain the purpose of tube Z. [1]

10. Leaves are adapted for photosynthesis.

(a) Explain how being thin and having a large surface area help a leaf to photosynthesise. [2] (b) Explain how the palisade mesophyll and the spongy mesophyll with air spaces are each adapted for their role. [4] (c) Explain the role of guard cells in photosynthesis. [2]

11. Identical seedlings grow for four weeks in a complete mineral solution, a solution lacking nitrate ions, or a solution lacking magnesium ions. Those without magnesium have yellow leaves; those without nitrate are much smaller.

(a) Explain why lack of magnesium ions produces yellow leaves and reduces growth. [3] (b) Explain why lack of nitrate ions reduces growth. [2] (c) State which seedlings act as the control, and one variable that must be kept constant. [2]

12. A grower plots rate of photosynthesis against light intensity for tomato plants under three conditions. Each curve rises, then levels off.

• Curve P: 0.04% carbon dioxide, 15 °C; levels off at 20 arbitrary units
• Curve Q: 0.12% carbon dioxide, 15 °C; levels off at 35 arbitrary units
• Curve R: 0.12% carbon dioxide, 25 °C; levels off at 50 arbitrary units

(a) State the limiting factor on the rising part of all three curves. [1] (b) Identify the factor limiting the level part of curve P, and explain how the data shows this. [2] (c) Identify the factor limiting the level part of curve Q, and explain how the data shows this. [2] (d) The greenhouse is closed and unheated on a bright winter day. Suggest two changes to increase yield, and explain when each would work. [3]

Answers

1. carbon dioxide + water → glucose + oxygen [1]; correct formulae: CO₂ + H₂O → C₆H₁₂O₆ + O₂ [1]; balanced: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ (light and chlorophyll by the arrow) [1]. Examiner insight: the balancing mark needs the 6s on all three; light written as a reactant on the left side is not credited.

2. (a) In chloroplasts [1]. (b) It transfers light energy into chemical energy for making glucose [1]. Examiner insight: “absorbs light” alone is usually not enough; the mark is for transferring light energy into chemical energy.

3. Starch: energy store [1]. Cellulose: builds cell walls [1]. Glucose: respiration to provide energy [1]. Sucrose: transport through the plant [1]. Examiner insight: one mark per carbohydrate, so a use given for the wrong molecule (for example “starch for transport”) loses that mark.

4. (a) To remove starch from the leaves (destarch) [1], so any starch found must have been made during the experiment [1]. (b) Boiling water kills the cells / breaks membranes so iodine can enter [1]; ethanol removes chlorophyll so the colour change can be seen [1]. (c) Heat the ethanol in a water bath with no naked flame, because ethanol is flammable [1]. (d) Green areas blue-black; white areas orange-brown [1]. Only green areas contain chlorophyll [1], so only they photosynthesised and made starch; chlorophyll is needed [1]. Examiner insight: in (d) the colour mark needs both areas correct; “black” alone is not accepted for blue-black.

5. (a) As a control, to show any difference is caused by the lack of carbon dioxide [1]. (b) Leaf A: no starch (orange-brown) [1]. Leaf B: starch present (blue-black) [1]. (c) Carbon dioxide is needed for photosynthesis [1]. (d) Any one: same temperature, same leaf size or age, same time in light, same plant [1]. Examiner insight: “to compare” is too vague for a control mark; say what the control shows.

6. (a) Anomaly: 41 at 15 cm [1]. Mean = (22 + 26) ÷ 2 = 24 bubbles per minute [1]. (b) As lamp distance increases, the number of bubbles (rate) decreases [1]. Further away means lower light intensity [1], so less light energy is transferred by chlorophyll and light is the limiting factor [1]. (c) To absorb heat from the lamp so temperature stays constant [1]. (d) Bubbles vary in size, so volume measures the gas released more accurately [1]. Examiner insight: in (a) the mean is credited only if the anomaly is excluded; including it gives 29.7, which scores the anomaly mark only.

7. (a) 2.0 ÷ 8 [1] = 0.25 cm³ per minute [1]. (b) More sodium hydrogencarbonate gives more carbon dioxide [1]; carbon dioxide is a raw material / limiting factor, so the rate of photosynthesis rises and more oxygen is released [1]. (c) Carbon dioxide is no longer the limiting factor [1]; light intensity or temperature is now limiting [1]. Examiner insight: the unit is part of the accuracy mark in (a); 0.25 with no unit, or “cm³”, does not get it.

8. Rate increases from 10 °C to an optimum [1]; molecules gain kinetic energy so there are more effective collisions between enzymes and substrates [1]. Above the optimum the rate falls sharply [1]; enzymes are denatured [1], the active site changes shape so the substrate no longer fits [1]. Examiner insight: “enzymes are killed” is not credited; denaturation and the change in active-site shape earn separate marks.

9. (a) W: purple (magenta accepted) [1]. X: yellow [1]. (b) W: in bright light photosynthesis is faster than respiration [1], so carbon dioxide concentration falls [1]. X: the snail only respires, so carbon dioxide concentration rises [1]. (c) Control: shows light alone does not change the indicator colour [1]. Examiner insight: each explanation must link the process to a change in carbon dioxide concentration; a colour with “because of photosynthesis” alone is not enough.

10. (a) Large surface area absorbs more light [1]; thin gives a short diffusion distance for carbon dioxide [1]. (b) Palisade: many chloroplasts [1], tightly packed near the upper surface where light is strongest [1]. Spongy and air spaces: loosely packed with a large surface area for gas exchange [1]; air spaces let carbon dioxide diffuse to the mesophyll cells [1]. (c) Guard cells open the stomata [1] so carbon dioxide can diffuse in (and oxygen out) [1]. Examiner insight: “Explain” in (b) needs feature and effect; “has chloroplasts” alone scores one of the two palisade marks.

11. (a) Magnesium is needed to make chlorophyll [1]; less chlorophyll, so leaves look yellow and absorb less light [1], so less photosynthesis and less carbohydrate for growth [1]. (b) Nitrate is needed to make amino acids [1], which make proteins for growth [1]. (c) The seedlings in the complete solution [1]; any one: same volume of solution, same light, same temperature, same seedling type or age [1]. Examiner insight: “magnesium makes leaves green” does not score; the mark is for linking magnesium to chlorophyll.

12. (a) Light intensity [1]. (b) Carbon dioxide concentration [1]; curve Q at higher carbon dioxide but the same temperature levels off higher [1]. (c) Temperature [1]; curve R at the same carbon dioxide but 25 °C levels off higher [1]. (d) Heat the greenhouse [1]; add carbon dioxide (for example with a burner or supply) [1]; each helps only while it is the limiting factor; once another factor is in shortest supply, raising it has no effect [1]. Examiner insight: in (b) and (c) the explanation mark needs a comparison of two named curves in which only one factor differs.

Where marks are usually lost

• Leaving destarching out of experiment descriptions, or not saying why it is done.
• Giving a colour for the starch test without both outcomes (blue-black and orange-brown).
• Naming the x-axis factor as limiting on the level part of a curve.
• Calculating a mean that includes an obvious anomaly.
• Missing units on rates (cm³ per minute, bubbles per minute).
• Linking indicator colours to “oxygen” instead of carbon dioxide concentration.

Next steps

• Recap with the plant nutrition revision notes.
• Go back to the plant nutrition study guide for any outcome you missed.
• Related: Enzymes practice questions.
• See every topic on the Cambridge O Level Biology hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028, version 4 (Cambridge University Press & Assessment), section 6 Plant nutrition.