OCR A Level Biology A: Plant responses and plant hormones (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 5.1.5(a) to 5.1.5(f) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. Learn the content from the study guide and revision notes first. The course hub, the checklist and the free diagnostics show the wider course.

Questions

1. Name the type of response shown by each of the following. (a) A pea tendril coils round a cane. (b) The leaflets of Mimosa fold when touched. (c) A root grows towards a patch of moist soil. [3]

2. A seedling is laid on its side in the dark. After two days the shoot curves upwards and the root curves downwards. Explain these observations. [4]

3. A student investigates phototropism in cress seedlings. Five seedlings were lit from one side for 24 hours. Five others were rotated slowly on a clinostat in the same light. The angles of curvature were: one-sided light 62°, 70°, 58°, 66°, 64°; clinostat 3°, 5°, 2°, 6°, 4°.

(a) Calculate the mean angle for the seedlings lit from one side. [1] (b) State two variables the student should control. [2] (c) Explain the purpose of the clinostat treatment. [2]

4. A plant is deprived of water. Its leaf stomatal conductance falls from 220 to 70 mmol m⁻² s⁻¹.

(a) Calculate the percentage decrease in stomatal conductance. [2] (b) Explain how abscisic acid (ABA) causes stomata to close. [3]

5. Explain how plant hormones bring about leaf loss in a deciduous tree in autumn, and suggest one advantage of losing leaves. [5]

6. Barley seeds (50 per treatment) were soaked in different solutions and the number that germinated was recorded. Water: 18. Gibberellin (GA): 47. ABA: 3. GA and ABA together: 21.

(a) Calculate the percentage germination with water and with GA. [2] (b) Describe what the results show about the effect of ABA. [2] (c) Explain how gibberellin promotes germination in a barley seed. [3]

7. An investigator removed the apical bud from pea plants and put agar on the cut stump. Mean lateral shoot growth after 6 days is shown.

Treatment	Lateral growth / mm
Apex intact	2
Apex removed, plain agar	22
Apex removed, agar with 0.05% auxin	15
Apex removed, agar with 0.5% auxin	7
Apex removed, agar with 5% auxin	2

(a) Calculate the percentage inhibition caused by 0.5% auxin compared with plain agar. [2] (b) What do the data show about the effect of auxin concentration? [2] (c) Explain why the treatment with plain agar was included. [2]

8. Three types of pea plant were grown for 12 days with and without gibberellin (GA). Mean stem growth in mm is shown.

Plant type	No GA	With GA
Tall	44	47
Dwarf A	10	43
Dwarf B	9	10

(a) Calculate the percentage increase in growth of dwarf A when GA was applied. [2] (b) Explain what the results suggest about the cause of dwarfism in A and in B. [3] (c) Suggest why GA had little effect on the tall plants. [2]

9. A student investigates the effect of gibberellin (GA) on hypocotyl length in cress seedlings. The stock solution is 1 mg dm⁻³.

(a) Describe how to make solutions of 0.1, 0.01 and 0.001 mg dm⁻³ by serial dilution. [3] (b) Hypocotyl lengths in mm after 4 days for 8 seedlings each were: control 14, 12, 15, 13, 16, 13, 14, 15; 0.01 mg dm⁻³ GA 18, 20, 17, 21, 19, 18, 22, 19. State a null hypothesis for a t-test and the degrees of freedom. [2] (c) The calculated value of t is 7.00 and the critical value at p = 0.05 is 2.14. State the conclusion. [1] (d) Suggest why the student should plot the concentrations on a logarithmic axis. [2]

10. Explain how alkaloids and pheromones help plants survive attack by herbivores. [5]

11. Plant hormones have commercial uses.

(a) Explain why a synthetic auxin can be used as a selective weedkiller on a cereal crop. [3] (b) Explain the advantage to a grower of dipping cuttings in rooting powder. [2] (c) Bananas are picked green and transported cool, then exposed to ethene. Explain why, and why the fruit ripen together once started. [3]

Answers

1. (a) Thigmotropism, a response to touch or contact [1]. (b) A nastic response (not directional), since it does not depend on the direction of touch [1]. (c) Hydrotropism, a positive response to water [1]. [3] Examiner insight: For a nastic response, add that the direction of the stimulus does not decide the direction of movement.

2. Auxin (IAA) moves to the lower side of both organs [1]. In the shoot, the higher concentration stimulates cell elongation on the lower side, so the shoot curves up [1]. In the root, the same higher concentration inhibits elongation on the lower side [1]. The upper side grows faster, so the root curves down [1]. [4] Examiner insight: Saying auxin simply “makes the root grow down” loses credit; the contrast between stimulation and inhibition is the point.

3. (a) (62 + 70 + 58 + 66 + 64) ÷ 5 = 320 ÷ 5 = 64° [1]. (b) Any two from: species, seedling age, light intensity, distance from the light, temperature, duration, moisture [2]. (c) Rotation exposes all sides equally to light, so no side is more strongly lit [1]. This shows that the curvature is caused by the one-sided light and not by another factor such as handling or the container [1]. [5] Examiner insight: “Same conditions” is too vague; name each variable and say how it is held constant.

4. (a) (220 − 70) ÷ 220 [1] × 100 = 68.2% [1]. (b) Water stress raises ABA, which binds to receptors on the guard cells [1]. Calcium ions enter and channels open so potassium ions leave the guard cells [1]. Water potential of the guard cells rises, water leaves by osmosis, turgor falls and the stomata close [1]. [5] Examiner insight: Divide the change by the starting value. The osmosis step is the one most often left out.

5. Shorter days and cooler weather reduce the auxin produced by the leaf [1]. Ethene stimulates the abscission zone at the base of the leaf stalk [1]. Cells in the separation layer make enzymes that digest cell walls [1]. A protective layer forms on the stem side of the scar, preventing infection [1]. Advantage: less water is lost in winter when water is hard to absorb [1]. [5] Examiner insight: Name both hormones and say which is rising and which is falling; “hormones cause the leaf to fall” is too general.

6. (a) Water: 18 ÷ 50 × 100 = 36% [1]. GA: 47 ÷ 50 × 100 = 94% [1]. (b) ABA reduced germination to 6% compared with 36% in water [1]. With GA present, ABA lowered germination from 94% to 42%, so ABA opposes the effect of GA [1]. (c) The embryo makes gibberellin, which diffuses to the aleurone layer [1] and stimulates transcription of the amylase gene [1]. Amylase hydrolyses starch to maltose and then glucose, which the embryo respires to release energy for growth [1]. [7] Examiner insight: Quote figures from the data when describing results; “it went down” is weaker than 94% to 42%. In (c), name the enzyme and its substrate.

7. (a) (22 − 7) ÷ 22 × 100 [1] = 68.2% [1]. (b) Greater auxin concentration gave greater inhibition of lateral growth (22 to 15, 7 and 2 mm) [1]. At 5% the lateral growth was the same as with the apex intact (2 mm), so enough auxin replaces the apex [1]. (c) Plain agar shows the effect of removing the apex without hormone [1]. It also shows that agar itself does not inhibit lateral growth, so differences are due to auxin [1]. [6] Examiner insight: Compare each auxin treatment with plain agar, not the intact plant, when asked for inhibition by the hormone.

8. (a) (43 − 10) ÷ 10 [1] × 100 = 330% [1]. (b) A grew much taller with GA (10 to 43 mm), so it is short because it makes too little gibberellin [1]. B hardly responded (9 to 10 mm) even with GA applied [1], so its dwarfism is not due to lack of GA but to being unable to respond, for example a fault in receptors or in the response pathway [1]. (c) Tall plants already make enough GA [1], so the response is near its maximum and extra GA has little further effect (a 3 mm rise) [1]. [7] Examiner insight: Distinguish “not enough hormone” from “cannot respond”; only the response to added hormone separates them.

9. (a) Put 1 cm³ of the 1 mg dm⁻³ stock into 9 cm³ of distilled water and mix to give 0.1 mg dm⁻³ [1]. Take 1 cm³ of that into a fresh 9 cm³ of water to give 0.01 mg dm⁻³, then repeat to give 0.001 mg dm⁻³ [1]. Use a clean pipette each time and discard 1 cm³ from the last tube so every tube holds 9 cm³ [1]. (b) Null hypothesis: there is no significant difference in mean hypocotyl length between the control and 0.01 mg dm⁻³ GA [1]. Degrees of freedom = 8 + 8 − 2 = 14 [1]. (c) 7.00 is greater than 2.14, so reject the null hypothesis: GA significantly increased hypocotyl length [1]. (d) The concentrations differ by factors of ten, so on a linear axis most points crowd near zero [1]. A log axis spaces them evenly [1]. [8] Examiner insight: Write the null hypothesis as “no difference”, and compare t with the critical value in the conclusion.

10. Alkaloids such as caffeine or nicotine are bitter or toxic chemicals [1]. They deter or poison herbivores, so less tissue is eaten [1]. Pheromones are airborne signals released by a damaged plant [1]. Neighbouring plants detect them and raise their own defences before being attacked [1]. Less damage means more plants survive and reproduce [1]. [5] Examiner insight: Give a named alkaloid and say who detects the pheromone; “it warns others” is too thin.

11. (a) Synthetic auxins are not broken down by the plant like natural IAA [1]. In broad-leaved weeds they persist and cause fast, uncontrolled growth that kills the plant [1]. Cereals (narrow-leaved) are much less affected, so the crop survives [1]. (b) Auxin in the powder stimulates roots to form from the cut stem [1], so many genetically identical plants are produced quickly and reliably [1]. (c) Unripe fruit bruise less in transport, and ethene at the destination times ripening for sale [1]. Ethene stimulates starch breakdown to sugar and softening of cell walls [1]. Ripening fruit release more ethene, a positive feedback, so nearby fruit ripen together [1]. [8] Examiner insight: “Kills weeds” does not explain selectivity; say why the crop differs.

Where marks are usually lost

• Treating auxin as always stimulating; higher concentrations inhibit roots.
• Describing results without numbers.
• Calculating percentage change from the wrong baseline.
• Leaving out the control, or naming the wrong control.
• Not naming the enzyme (amylase) and the substrate (starch) in germination.
• Saying guard cells “shrink” without water potential, osmosis and turgor.
• Giving vague controlled variables such as “same conditions”.
• Omitting degrees of freedom or the null hypothesis.
• Calling a weedkiller a poison rather than a synthetic auxin.

Next steps

• Revisit the revision notes and study guide.
• Open the course hub and the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), first assessment 2025: learning outcomes 5.1.5(a) to 5.1.5(f).