Biology • Year 11 • Module 2 • Lesson 17
Transpiration: Factors and Measurement
Apply transpiration factors to real potometer data, experimental design, and a xerophyte adaptation scenario.
1. Interpret potometer data
A student uses a potometer to measure water uptake by a leafy shoot under five different conditions. Each condition was tested for 10 minutes and the distance the air bubble moved in the capillary tube was recorded across three trials. All other variables were kept constant in each trial. 8 marks
| Condition | Trial 1 (mm) | Trial 2 (mm) | Trial 3 (mm) | Mean (mm) | Rate (mm/min) |
|---|---|---|---|---|---|
| Still air, 22°C, 60% humidity, dim light | 12 | 14 | 13 | 13.0 | 1.3 |
| Still air, 22°C, 60% humidity, bright light | 19 | 21 | 20 | 20.0 | 2.0 |
| Fan (wind), 22°C, 60% humidity, dim light | 24 | 23 | 25 | 24.0 | 2.4 |
| Still air, 35°C, 30% humidity, dim light | 31 | 34 | 32 | 32.3 | 3.2 |
| Still air, 22°C, 90% humidity, dim light | 4 | 5 | 4 | 4.3 | 0.4 |
1.1 Describe the trend in transpiration rate from the lowest to the highest value in the dataset. Quote specific figures. 2 marks
1.2 The high-temperature/low-humidity condition produced the fastest rate (3.2 mm/min). Explain why both temperature and humidity contribute to this result. Refer to the water potential gradient in your answer. 3 marks
1.3 The high-humidity condition (90%) produced a rate of 0.4 mm/min, 8 times lower than the high-temperature/low-humidity condition. Explain this difference mechanistically. 3 marks
2. Interpret graph, wind speed and transpiration rate
The graph below shows the relationship between wind speed and transpiration rate in a broad-leaved plant measured in a laboratory wind tunnel. Note the non-linear pattern. 6 marks
Figure 2. Effect of wind speed on transpiration rate. Conditions: 25°C, 50% humidity, bright light. Hypothetical data.
2.1 Describe the shape of the curve from 0 to 8 m/s, identifying three phases. 2 marks
2.2 Using lesson content, explain why transpiration rate levels off at moderate wind speeds (roughly 2–5 m/s) rather than continuing to rise. 2 marks
2.3 The misconceptions box in the lesson states: “Wind always increases transpiration rate” is wrong. Using the graph, explain what happens at high wind speeds and why. 2 marks
3. Apply experimental design, potometer investigation
A student wants to investigate whether light intensity affects transpiration rate using a potometer. 7 marks
3.1 State the independent variable (IV) and dependent variable (DV) in this investigation. 2 marks
3.2 List three variables that must be controlled to ensure a fair test. For each, state why it must be controlled. 3 marks
3.3 A student claims: “The potometer proves that light increases transpiration rate.” Identify one limitation of the potometer method that makes this claim too strong. 2 marks
4. Apply to a new scenario, a greenhouse at night
A tomato grower keeps plants in a glass greenhouse. At night, the temperature drops to 12°C, humidity rises to 88%, and the lights are turned off. In the day, temperature reaches 28°C, humidity is 45%, and the lights are at full intensity. 5 marks
4.1 Predict the transpiration rate at night relative to daytime. Give three reasons using factor mechanisms. 3 marks
4.2 A student suggests opening ventilation fans during the day would further increase plant water loss. Using lesson content, evaluate this suggestion. 2 marks
Q1.1, Trend description (2 marks)
Transpiration rate ranges from 0.4 mm/min (still air, 90% humidity) to 3.2 mm/min (35°C, 30% humidity) [1]. The high-humidity condition is the lowest (0.4 mm/min) and the high-temperature/low-humidity condition is the highest (3.2 mm/min), an 8× difference; fan (wind) and bright light conditions are intermediate (2.4 and 2.0 mm/min respectively) [1].
Q1.2, High temperature / low humidity (3 marks)
High temperature (35°C) increases the kinetic energy of water molecules in the mesophyll, increasing evaporation into leaf air spaces and raising water vapour concentration inside the leaf. Warm air also has greater capacity to hold water vapour, so the atmosphere is further from saturation [1]. Low humidity (30%) means the outside air already has very low water vapour concentration. The leaf interior is near-saturated (~99% relative humidity), while the outside air is at 30%, the water potential difference is very large, driving rapid diffusion of water vapour through stomata [1]. These two factors act simultaneously on the same gradient, producing an additive effect that gives the highest rate in the dataset [1].
Q1.3, High humidity (3 marks)
At 90% humidity, the outside air is nearly as saturated with water vapour as the leaf interior (~99% relative humidity) [1]. The water potential difference between the leaf air spaces and the external atmosphere is therefore very small (approximately 9 percentage points), providing only a tiny driving force for diffusion through stomata [1]. This near-elimination of the gradient reduces transpiration to near zero, confirming that humidity is the primary limiting factor when taken to its extreme, more influential than any single other factor in this dataset [1].
Q2.1, Three phases of the curve (2 marks)
Phase 1 (0–~2 m/s): transpiration rate rises steeply as increasing wind removes the boundary layer and steepens the water potential gradient [1]. Phase 2 (~2–5 m/s): rate plateaus, the boundary layer is already fully removed so further wind speed provides diminishing returns [1]. Phase 3 (>5 m/s): rate declines as very strong wind triggers stomatal closure as a water-conservation response. Award 1 mark for two correct phases described; 2 marks for all three with appropriate direction.
Q2.2, Plateau (2 marks)
The boundary layer effect is the reason wind increases transpiration. Once wind speed is sufficient to completely remove the humid boundary layer from around the stomata, the air immediately outside the pore is already as dry as the bulk atmosphere [1]. Any further increase in wind speed cannot remove a boundary layer that no longer exists, so the rate cannot rise further, the gradient is already at its maximum determined by the ambient humidity [1].
Q2.3, High wind reduces rate (2 marks)
At wind speeds above approximately 5 m/s, the curve shows a declining transpiration rate [1]. Very high wind or hot, dry wind triggers a physiological response in guard cells, water stress causes guard cells to close stomata as a protective mechanism, reducing stomatal aperture and therefore reducing the pathway for water vapour to leave the leaf. This overrides the physical boundary-layer effect [1].
Q3.1, IV and DV (2 marks)
IV: light intensity (e.g. distance of lamp from shoot, or lux reading). DV: rate of water uptake by the shoot, measured as the distance the air bubble moves per unit time (mm/min) [1 each].
Q3.2, Controlled variables (3 marks: 1 per valid pair)
Accept any three from: temperature (changing temperature would alter kinetic energy and gradient independently of light); humidity (high/low humidity affects the gradient regardless of light); wind/air movement (would change boundary layer effect); same species/leaf area shoot (different leaf area changes total evaporative surface); same time period per measurement (needed to calculate rate); same starting bubble position (ensures comparable distances are measured).
Q3.3, Limitation (2 marks)
A potometer measures water uptake by the cut shoot, not transpiration directly [1]. Light also increases photosynthesis rate, meaning more water is incorporated into organic molecules, this extra non-transpiratory water use means the potometer slightly overestimates the transpiration contribution at high light intensity. Therefore the result does not prove light increases transpiration alone; it proves light increases water uptake, most of which is transpiration [1].
Q4.1, Night vs day transpiration (3 marks)
Night transpiration rate will be much lower than daytime. (1) Lower temperature (12 vs 28°C) reduces kinetic energy of water molecules and reduces evaporation into leaf air spaces, the water potential gradient is less steep. (2) High humidity (88%) means the outside air is close to saturation, the gradient driving diffusion through stomata is very small. (3) No light means guard cells do not receive the signal to accumulate K¹+; stomata close (or are minimally open), drastically reducing the pathway for water vapour to exit. One mark for each factor with mechanism; accept any three of these factors.
Q4.2, Ventilation fans (2 marks)
The suggestion is correct: opening ventilation fans during the day would increase transpiration rate [1]. Wind removes the humid boundary layer that accumulates around stomata, restoring a steeper water potential gradient between the leaf interior and the bulk greenhouse air, water vapour diffuses out faster. The grower should ensure adequate soil water is available to supply this increased demand, or wilting may occur [1].