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9700 · 15.2

Control and coordination in plants — practice questions

Practice and worked examples for 9700 Control and coordination in plants. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

An experiment was set up to investigate the effect of unilateral light on the growth of an oat coleoptile. The initial length of the coleoptile was 15.0 mm. After 48 hours, the length of the shaded side was measured to be 19.5 mm, and the length of the illuminated side was 16.5 mm. Calculate the percentage elongation for each side and explain the difference.

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  1. Calculate the elongation of the shaded side:
    • Final length - Initial length = 19.5 mm - 15.0 mm = 4.5 mm
  2. Calculate the percentage elongation of the shaded side:
    • (Elongation / Initial length) × 100
    • (4.5 mm / 15.0 mm) × 100 = 30%
  3. Calculate the elongation of the illuminated side:
    • Final length - Initial length = 16.5 mm - 15.0 mm = 1.5 mm
  4. Calculate the percentage elongation of the illuminated side:
    • (Elongation / Initial length) × 100
    • (1.5 mm / 15.0 mm) × 100 = 10%
  5. Explanation: The difference in elongation (30% on the shaded side vs. 10% on the illuminated side) is due to the redistribution of auxin. Unilateral light causes auxin to move from the illuminated side to the shaded side. The higher concentration of auxin on the shaded side promotes greater cell elongation, causing it to grow more than the illuminated side. This differential growth results in the coleoptile bending towards the light source.

Worked example 2

A student set up an experiment where oat coleoptiles were placed horizontally in the dark. After 24 hours, the coleoptiles curved upwards. Explain the role of auxins in this observed response.

Show solution outline
  1. Identify the tropism: The coleoptile bending upwards in response to gravity, in the dark, is an example of negative gravitropism. This eliminates light as a factor, focusing purely on gravity.
  2. Auxin redistribution: When the coleoptile is placed horizontally, gravity causes auxins (produced in the apical meristem of the tip) to redistribute and accumulate on the lower side of the coleoptile.
  3. Differential growth in shoots: In shoots, an increased concentration of auxin promotes cell elongation. Therefore, the cells on the lower side of the coleoptile elongate more rapidly than those on the upper side, where auxin concentration is lower.
  4. Resulting bend: This differential growth means the lower side grows faster, causing the coleoptile to bend upwards, exhibiting negative gravitropism, orienting itself against the gravitational pull.
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