Cambridge O Level Biology 5090: Coordination and response in plants (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 15, Coordination and response in plants, of the Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028 (version 4): every learning outcome in section 15.1 (15.1.1 to 15.1.4). The syllabus is not tiered, so every question applies to every candidate.

Revise first with the study guide and the revision notes. The course hub and the printable checklist list the other topics, and the free 10-minute diagnostics help you decide what to revise next. A calculator is useful for questions 5, 6, 9, 10 and 11. All data in these questions are invented for practice.

Questions

1. Define the term gravitropism. [2]

2. Copy and complete the table using the words positive or negative.

Part of plant	Response to gravity	Response to light from one side
Shoot	……..	……..
Root	……..	not required

[3]

3. This question is about auxin.

(a) State where auxin is made. [1] (b) State the effect of auxin on the cells it reaches. [1] (c) A shoot is lit from the left only. State which side of the shoot has more auxin. [1]

4. Explain how auxin causes a shoot to grow towards light coming from one side. [5]

5. A student grew three groups of oat seedlings in light from one side for two days and measured the angle of curvature of each shoot towards the light.

Group	Treatment of shoot tip	Angles of curvature / °
A	No treatment	38, 42, 35, 45
B	Covered with a transparent cap	36, 41, 39, 40
C	Covered with a light-proof cap	0, 2, 0, 2

(a) Calculate the mean angle of curvature for group A. [1] (b) Explain the results for group C. [3] (c) Suggest why the student included group B. [2]

6. A student measured cells in the region just below the tip of a shoot that had bent. The mean cell length on one side was 45 µm and on the other side was 72 µm.

(a) Calculate the percentage by which the longer cells exceed the shorter cells in length. [2] (b) The shoot had bent towards a lamp. State which side, lit or shaded, had the 72 µm cells, and explain your answer. [2]

7. A potted seedling is placed on its side in a dark cupboard for two days.

(a) Describe what happens to the direction of growth of the shoot and of the root. [2] (b) Explain the response of the shoot in terms of auxin. [3] (c) Suggest why the seedling was kept in the dark. [1]

8. Plan an investigation to find out whether cress shoots show phototropism. You are provided with cress seedlings growing in small pots, cardboard boxes, scissors, a lamp and a protractor. [6]

9. Bean seeds were pinned to damp cotton wool in two dishes, with their young roots pointing horizontally. Both dishes were kept in the dark. Dish P was kept still. Dish Q was fixed to a clinostat.

(a) Describe the expected direction of root growth in dish P and in dish Q. [2] (b) Explain why dish Q was set up. [2] (c) One root in dish P grew from 18 mm to 42 mm long in 3 days. Calculate its mean growth rate. [2] (d) Suggest one source of error in measuring the length of a curved root. [1]

10. A student laid five bean seedlings on their sides in the dark. After two days the roots had curved downwards by these angles: 84°, 90°, 78°, 88°, 85°.

(a) Calculate the mean angle of curvature. [2] (b) Name this response and state whether it is positive or negative. [2] (c) Explain how this response benefits the plant. [2] (d) Suggest why the student used five seedlings rather than one. [1] (e) State one variable that should be kept constant for all five seedlings. [1]

11. Scientists measured the auxin in the two halves of a shoot tip that had been lit from one side. The lit half contained 35 arbitrary units of auxin and the shaded half contained 65 arbitrary units.

(a) Calculate the percentage of the total auxin that was in the shaded half. [2] (b) Explain how this difference leads to the shoot bending. [3] (c) The shoot is then given light evenly from all sides. Predict how the auxin would be distributed and how the shoot would grow. [2]

Answers

1. A response in which parts of a plant grow [1] towards or away from gravity [1]. [2] Examiner insight: “Towards or away from” is needed for the second mark; “the plant grows downwards” describes only a root and is not a definition.

2. Shoot, gravity: negative [1]. Shoot, light: positive [1]. Root, gravity: positive [1]. [3] Examiner insight: In this answer each cell carries its own mark, so writing the root and shoot gravity responses the wrong way round costs two marks.

3. (a) In the shoot tip [1]. (b) It stimulates cell elongation [1]. (c) The right side, which is shaded [1]. Examiner insight: “Makes the cells grow” is usually too vague for (b); the creditworthy idea is elongation, and “makes cells divide” scores nothing.

4.

• Auxin is made in the shoot tip; [1]
• light causes auxin to be unequally distributed, with more on the shaded side; [1]
• auxin spreads back from the tip to the region below; [1]
• cells on the shaded side elongate more than cells on the lit side; [1]
• so the shaded side grows longer and the shoot bends towards the light. [1] [5] Examiner insight: The bending point only scores when it follows from a stated difference between the two sides; “auxin makes the shoot bend” with no reference to the shaded side earns little.

5. (a) (38 + 42 + 35 + 45) ÷ 4 = 160 ÷ 4 = 40° [1] (b) The shoots grew straight / did not bend [1]. Light could not reach the tip, so auxin was not unequally distributed [1]. Auxin spread back evenly, so cells on all sides elongated equally [1]. (c) To show that the cap itself does not stop bending [1]; so the lack of bending in C is caused by keeping light from the tip [1]. Examiner insight: For (c), “as a control” alone earns at most one mark; say what the control rules out (the effect of the cap, not of the light).

6. (a) (72 − 45) ÷ 45 × 100 [1] = 60% [1] (b) The shaded side [1]. More auxin collects on the shaded side and auxin stimulates cell elongation, so cells there are longer [1]. Examiner insight: Dividing by 72 instead of 45 gives 37.5%, which is a different comparison; use the value the question compares against, and expect to lose the accuracy mark for the wrong base.

7. (a) The shoot curves upwards [1]; the root curves downwards [1]. (b) Auxin is made in the shoot tip and spreads back [1]. Gravity causes more auxin on the lower side [1]. Cells on the lower side elongate more, so the shoot curves upwards [1]. (c) So that light cannot cause a response; only gravity affects the direction of growth [1]. Examiner insight: In (b) the side must be named; an answer that says auxin is “unequally distributed” without saying which side has more cannot score the second or third mark.

8.

• Put one group of seedlings in a box with a slit cut in one side, with the lamp shining through the slit, so light comes from one side; [1]
• control: an identical group in a box with no slit, or lit evenly from all sides; [1]
• keep at least two named variables the same, for example temperature, watering, type and age of seedlings; [1]
• leave both groups for the same stated time, for example two or three days; [1]
• measure the angle of curvature of each shoot from the vertical with the protractor; [1]
• use several seedlings in each group and calculate a mean angle. [1] [6] Examiner insight: Each controlled variable must be named; “keep everything else the same” usually scores nothing, and the control group needs to be described, not just mentioned.

9. (a) Dish P: roots curve downwards [1]. Dish Q: roots keep growing horizontally / in the direction they started [1]. (b) The clinostat turns the seeds so gravity acts on all sides equally [1]; this shows that the downward curve in P is caused by gravity [1]. (c) (42 − 18) ÷ 3 [1] = 8 mm per day [1] (d) A ruler cannot follow the curve / thread laid along the root may stretch / the start of the root is hard to judge [1]. Examiner insight: The rate needs units; “8” alone loses the accuracy mark even when the working is correct.

10. (a) (84 + 90 + 78 + 88 + 85) ÷ 5 [1] = 425 ÷ 5 = 85° [1] (b) Gravitropism [1]; positive [1]. (c) The root grows down into the soil [1]; to anchor the plant / to absorb water and mineral ions [1]. (d) Seedlings vary, so a mean of several is more reliable [1]. (e) Temperature / type or age of seedling / time / moisture of the cotton wool [1]. [8] Examiner insight: Part (b) has two separate marks, one for the name and one for the direction; “gravitropism” with no direction scores only one.

11. (a) 65 ÷ (35 + 65) × 100 [1] = 65% [1] (b) Auxin spreads back from the tip to the region below [1]. More auxin on the shaded side stimulates more cell elongation there [1]. The shaded side grows longer, so the shoot bends towards the light [1]. (c) Auxin evenly distributed [1]; cells elongate equally on all sides, so the shoot grows straight up [1]. [7] Examiner insight: In (a), the divisor is the total (100 units), not the lit-side value; dividing 65 by 35 gives a ratio, not the percentage asked for.

Where marks are usually lost

• Definitions of a tropism that leave out “grow” or “towards or away from”.
• Writing “auxin is made in the tip” without the word shoot.
• Saying auxin makes cells divide, rather than stimulating cell elongation.
• Not naming the side (shaded, or lower) that has more auxin.
• Saying the shoot bends towards the side with more auxin; it bends away from it.
• Swapping positive and negative for roots and shoots.
• In planning answers, writing “keep everything the same” instead of naming each controlled variable.
• Forgetting a control (even light, or a clinostat), or not keeping a gravitropism test in the dark.
• Leaving units off a mean angle (°) or a growth rate (mm per day).

Next steps

• Recap with the revision notes.
• Reread any section you found hard in the study guide.
• Link this topic to light and photosynthesis in Plant nutrition and to water uptake in Transport in flowering plants.
• See every topic on the course hub and tick off outcomes on the printable checklist.
• Try the 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), Topic 15: Coordination and response in plants (15.1 Coordination and response in plants), published by Cambridge Assessment International Education.