OxfordAQA International AS and A-level Biology: Plant growth substances, tropisms, ethene and ABA (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.

The questions below follow OxfordAQA International AS and A-level Biology (9610), Version 5.1, section 3.4.5 and Required practical 10, in Unit 4, an International A-level only unit. Read the study guide and the revision notes first. Track progress with the course hub, the printable checklist and a free diagnostic.

Questions

1. State three features of plant growth substances. [3]

2. (a) Define a tropism. [1] (b) Name the response of a root to gravity and state its direction. [2] (c) Suggest an advantage of this response to the plant. [1]

3. A young pea plant is placed horizontally in a dark cupboard. Explain why the tip of its shoot turns upwards. [5]

4. Cress seedlings were grown for 24 hours with different concentrations of indoleacetic acid (IAA). Mean growth was measured.

IAA (µmol dm⁻³)	0	0.0002	0.002	0.02	0.2
Root growth (mm)	20.0	21.0	16.0	9.0	3.0
Shoot growth (mm)	10.0	11.0	14.0	17.0	18.0

(a) Calculate the percentage change in root growth at 0.02 µmol dm⁻³ compared with the control. [2] (b) Describe the effect of IAA on shoot growth. [2] (c) Contrast the effect of higher IAA concentrations on roots and shoots. [2] (d) Use the data to explain why a horizontal root curves downwards. [3]

5. A shoot was lit from one side for 6 hours. The shaded side elongated by 3.6 mm and the lit side by 2.4 mm.

(a) Calculate the mean rate of elongation of each side. [2] (b) Calculate by what percentage the shaded-side elongation exceeded the lit-side elongation. [1] (c) Explain how the difference in elongation arises. [3]

6. A student investigates the effect of light direction on the growth of seedling shoots (Required practical 10).

(a) Describe a method that would give valid results. [5] (b) Angles of curvature for five seedlings were 12°, 15°, 18°, 15° and 10°. Calculate the mean. [1] (c) Explain the purpose of a control group in even light. [1] (d) State one limitation of the investigation. [1]

7. Explain how bananas can be picked green, shipped and then be ripe when they reach the shops. [4]

8. Green mangoes took 10 days to ripen in open air. A matched batch sealed in a bag with one ripe banana took 4 days.

(a) Calculate the percentage reduction in ripening time. [2] (b) Explain the result. [2] (c) Suggest why the bag was sealed. [1]

9. Explain how abscisic acid (ABA) causes stomata to close in a plant whose roots are deficient in water. [5]

10. Five leaves had stomatal apertures (µm) of 6.2, 5.8, 6.6, 6.0 and 6.4. After ABA was applied, the same leaves measured 2.4, 2.0, 2.8, 2.2 and 2.6. Mass lost in 2 hours by a leaf fell from 0.84 g to 0.36 g.

(a) Calculate the mean aperture before and after ABA. [2] (b) Calculate the percentage decrease in mean aperture. [2] (c) Calculate the percentage decrease in water loss. [1] (d) Explain the link between your answers to (b) and (c). [2]

11. Water loss and stomatal aperture were measured in a wild-type plant and a mutant that makes very little ABA.

Plant and treatment	Mean aperture (µm)	Water loss (g per hour)
Wild type, well watered	7.0	0.50
Wild type, water-stressed	2.0	0.15
Mutant, water-stressed	6.5	0.46
Mutant, water-stressed, sprayed with ABA	2.2	0.16

(a) Calculate the percentage decrease in aperture of the wild type when water-stressed. [2] (b) Compare the response of the mutant with that of the wild type. [2] (c) Explain what the spraying result shows. [2] (d) Suggest why the mutant is at a disadvantage in dry soil. [2] (e) Suggest one cost to the wild type of closing its stomata. [2]

Answers

1. They are signal molecules produced in plants [1]; present in very low concentrations [1]; they act on target cells close to where they are made, or travel to act in other tissues [1]. [3] Examiner insight: “State” asks for one clear point each, so listing three short separate features is safer than one long sentence.

2. (a) A growth response to a directional stimulus [1]. (b) Positive gravitropism [1]; growth downwards [1]. (c) It carries the root into soil, where water and mineral ions are available (or anchors the plant) [1]. [4] Examiner insight: Give the name and the direction; “gravitropism” alone does not say whether it is positive or negative.

3. IAA collects on the lower side of the horizontal shoot [1]. In a shoot a higher IAA concentration stimulates cell elongation [1]. The cells underneath lengthen more than those above [1]. The tip turns upwards [1]. This is negative gravitropism [1]. [5] Examiner insight: A chain from distribution to concentration to elongation to bending earns the marks; “IAA makes it grow up” skips the chain.

4. (a) (9.0 - 20.0) / 20.0 × 100 [1] = -55 % [1]. (b) Shoot growth rises as IAA concentration rises [1], from 10.0 mm to 18.0 mm, with the rise levelling off at the highest concentrations [1]. (c) Higher IAA concentrations inhibit root growth [1] but stimulate shoot growth [1]. (d) The lower side of a horizontal root has the higher IAA concentration [1], which inhibits elongation there [1]. The upper cells elongate more, so the root curves down [1]. [9] Examiner insight: Show the subtraction and the division by the control value, so that a slip can be traced; keep the minus sign, because it shows you understand inhibition.

5. (a) Shaded: 3.6 / 6 = 0.60 mm per hour [1]. Lit: 2.4 / 6 = 0.40 mm per hour [1]. (b) (3.6 - 2.4) / 2.4 × 100 = 50 % [1]. (c) There is more IAA on the shaded side [1], which stimulates cell elongation in the shoot [1], so the shaded cells lengthen more and the shoot tip turns to face the light [1]. [6] Examiner insight: Divide by the lit-side value when the question says the shaded side “exceeded” it; dividing by the larger figure gives a different percentage.

6. (a) Use seedlings of the same species, age and size [1]. Light one group from one side only [1]. Keep temperature, water supply and light intensity the same [1]. After a set time, read the angle of curvature with a protractor on each seedling or photograph [1]. Use many seedlings and calculate a mean and range [1]. (b) (12 + 15 + 18 + 15 + 10) / 5 = 70 / 5 = 14° [1]. (c) It shows the seedlings do not curve without a directional stimulus [1]. (d) For example, seedlings grow at different rates, so results vary [1]. [8] Examiner insight: “Fair test” is too vague; name the variable you control and say how you would measure the response.

7. Green bananas are firmer and are damaged less in transport [1]. Ethene is a growth substance that stimulates ripening [1], so its level is limited while the fruit is shipped [1]. At the destination the bananas are exposed to ethene so that the batch ripens together [1]. [4] Examiner insight: Link the practice to the role of ethene; “they ripen on the ship” alone does not explain control.

8. (a) (10 - 4) / 10 × 100 [1] = 60 % [1]. (b) The ripe banana releases ethene [1], which stimulates ripening of the mangoes [1]. (c) To keep the ethene gas around the fruit rather than letting it escape [1]. [5] Examiner insight: State that the source is the ripe banana and the effect is on the mango; a one-word “ethene” gives no explanation.

9. ABA acts on the guard cells [1]. It stimulates K⁺ and Cl⁻ ions to move out of the guard cells [1]. The water potential in the guard cells rises [1]. Water then moves out by osmosis [1]. The guard cells shrink and the stomata close [1]. [5] Examiner insight: The order matters: ions leave first and water follows. Reversing them contradicts the sequence.

10. (a) Before: 31.0 / 5 = 6.2 µm [1]. After: 12.0 / 5 = 2.4 µm [1]. (b) (6.2 - 2.4) / 6.2 × 100 = 3.8 / 6.2 × 100 [1] = 61.3 % [1]. (c) (0.84 - 0.36) / 0.84 × 100 = 57.1 % [1]. (d) Smaller apertures mean less water vapour leaves by transpiration [1], so less mass is lost [1]. [7] Examiner insight: Both percentages are of the starting value; quote both with units or the percent sign and compare them in words.

11. (a) (7.0 - 2.0) / 7.0 × 100 [1] = 71.4 % [1]. (b) The wild type closed its stomata (7.0 to 2.0 µm) [1], but the mutant’s aperture barely changed (7.0 to 6.5 µm) [1]. (c) ABA restored closure (6.5 to 2.2 µm) [1], so the mutant fails to close because it lacks ABA [1]. (d) Its stomata stay open, so it loses water by transpiration [1] and may wilt as the soil dries [1]. (e) Less carbon dioxide enters the leaf [1], so photosynthesis may fall [1]. [10] Examiner insight: In a compare question, quote figures for both plants; a statement about one plant only gains limited credit.

Where marks are usually lost

• Stating a hormone effect without the organ: the effect of higher IAA concentration is opposite in roots and shoots.
• Omitting which side of the root or shoot has the higher IAA concentration.
• Using the new value rather than the control or starting value as the base of a percentage.
• Writing “ethene makes fruit ripe” without the link to green picking and artificial ripening.
• Reversing ions and water in the ABA sequence.
• Not saying that water potential in the guard cells rises when ions leave.
• Leaving out a control or the number of replicates in Required practical 10 answers.
• Rounding too early in multi-step calculations; round only the final value.

Next steps

• Re-read the study guide for any question you lost marks on.
• Condense with the revision notes.
• Open the OxfordAQA A-level Biology course page and tick off the checklist.
• Take one of 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 A-level exams May/June 2018 onwards, published by OxfordAQA. Sections 3.4.5.1 to 3.4.5.3 and Required practical 10.