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AS · Practice questions · Transpiration and xerophytic leaves

Evaporate, diffuse, then slow it down.

Six original Cambridge-style questions on transpiration and xerophytes, including a potometer calculation and an annotated marram grass leaf.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Two steps, then the gradient.

01
Recall
[3 marks]

Explain what is meant by transpiration.

  • Water evaporates from the surfaces of mesophyll cell walls into the air spaces of the leaf. ✓
  • Water vapour diffuses out of the leaf through open stomata. ✓
  • It moves down a water potential gradient from the moist air spaces to the drier atmosphere. ✓
02
Explain
[3 marks]

Explain why the rate of transpiration increases when the air temperature rises.

  • Water molecules gain kinetic energy, so more water evaporates from the mesophyll cell walls. ✓
  • Water vapour molecules diffuse faster. ✓
  • Warm air can hold more water vapour, so the air outside has a lower water potential and the gradient is steeper. ✓
03
Calculation
[3 marks]

In a potometer, the air bubble moved 45 mm along a capillary tube of radius 0.50 mm in 5 minutes. Calculate the rate of water uptake in mm3 per minute. (Volume of a cylinder = πr2l)

  • Volume = π × 0.502 × 45 = 35.3 mm3. ✓
  • Rate = 35.3 ÷ 5 = 7.07 mm3 per minute. ✓
  • Answer: about 7.1 mm3 per minute. ✓
04
Analysis
[2 marks]

Explain why a potometer measures the rate of water uptake rather than the exact rate of transpiration.

  • Some of the water taken up is used by the plant, for example in photosynthesis, or kept in cells to maintain turgor. ✓
  • So the volume taken up is not exactly equal to the volume of water vapour lost. ✓
05
Explain
[4 marks]

Marram grass grows on sand dunes. Describe and explain four features of a marram grass leaf, seen in transverse section, that reduce water loss by transpiration.

  • Rolled leaf (rolled by hinge cells that lose water): traps humid air inside, so the water potential gradient is reduced. ✓
  • Sunken stomata in grooves on the inner surface: humid air is held in the pits above them. ✓
  • Hairs on the inner surface: trap a layer of still, moist air, reducing diffusion of water vapour. ✓
  • Thick waxy cuticle on the outer surface: reduces evaporation through the epidermis exposed to sun and wind. ✓
06
Data
[3 marks]

A leafy shoot transpired 12 mm3 min-1 in still, dry air. In still, humid air the rate fell to 3 mm3 min-1. Calculate the percentage decrease and explain the change.

  • Percentage decrease = (12 - 3) ÷ 12 × 100 = 75%. ✓
  • Humid air outside the leaf has a higher water potential (more water vapour). ✓
  • The water potential gradient from the air spaces to the atmosphere is less steep, so water vapour diffuses out more slowly. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.