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Transpiration: the price of keeping the stomata open questions
Transpiration defined and explained as a consequence of gas exchange, the effect of light, temperature, humidity and air movement on the rate, the guard cell mechanism from the proton pump to the opening of the pore, the part played by abscisic acid, and the structural adaptations of xerophytes and hydrophytes.
6 original questions · 22 marks · the transpiration: the price of keeping the stomata open notes · Plant transport and mineral nutrition
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In a potometer investigation the bubble travelled 18 mm in 10 minutes in still air and 47 mm in 10 minutes with a fan running. The capillary tube has an internal diameter of 1.0 mm. Calculate the rate of water uptake in moving air in mm³ min⁻¹, and calculate the percentage increase in the rate of bubble travel caused by the fan.
Mark scheme
- M1 cross-sectional area = πr² = π × 0.50² = 0.785 mm²
- M1 rate of bubble travel in moving air = 47 ÷ 10 = 4.7 mm min⁻¹
- A1 rate of water uptake = 0.785 × 4.7 = 3.7 mm³ min⁻¹
- M1 percentage increase = (47 − 18) ÷ 18 × 100, dividing the change by the original still-air value
- A1 161 per cent, accepting 160 to two significant figures
Explain how a guard cell opens a stoma in bright light, beginning with the proton pump in its plasma membrane.
Mark scheme
- B1 ATP drives proton pumps that move hydrogen ions out of the guard cell across its plasma membrane
- B1 the inside of the cell becomes electrically negative relative to the outside, so potassium ions enter through channel proteins down the electrical gradient
- B1 chloride ions follow and stored starch is converted to malate, adding further solute to the guard cell
- B1 the extra solute lowers the water potential of the guard cell, so water enters from the neighbouring epidermal cells by osmosis and the cell becomes turgid
- A1 the wall facing the pore is thicker and less elastic and the cellulose microfibrils are wound in hoops, so the turgid cell lengthens by bowing away from its partner and the pore opens
Explain why moving air over a leaf raises the rate of transpiration while raising the humidity of the air lowers it.
Mark scheme
- B1 water vapour leaving the stomata saturates a boundary layer of still air held against the leaf surface
- B1 moving air sweeps that saturated layer away, so the air just outside the pore has a much lower water potential and the steep part of the gradient is restored
- B1 raising the humidity raises the water potential of the air outside the leaf, towards that of the saturated air in the leaf air spaces
- A1 the water potential gradient across the stoma is therefore shallower, so water vapour diffuses out more slowly
Compare the effect of a rise in temperature on the rate of transpiration with the effect of a rise in humidity.
A water lily has leaves that float on the surface of a pond, with their stomata on the upper surface rather than the lower. Suggest why the stomata are on that surface, and suggest one other feature you would expect such a leaf to show.
State the two steps by which water is lost from a leaf during transpiration, in the order in which they occur.
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