Questions on transpiration and stomatal control

Question 1

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.

Accepted Answer

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 160 per cent to two significant figures, accepting 161

Question 2

Explain how a guard cell opens a stoma in bright light, beginning with the proton pump in its plasma membrane.

Accepted Answer

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

Question 3

Explain why moving air over a leaf raises the rate of transpiration while raising the humidity of the air lowers it.

Accepted Answer

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

Question 4

A plant kept in a normal cycle of day and night is moved into constant dim light at a constant temperature and humidity. Its stomata continue to open and close on a cycle of about twenty-four hours for several days. Explain what this result shows about the control of stomatal opening, and explain why opening before dawn is useful to the plant.

Accepted Answer

B1 the environmental cues have been held constant, so nothing outside the plant is changing to drive the cycle, yet the cycle continues B1 the timing must therefore be generated inside the plant, by an internal clock, so this is a circadian rhythm rather than a simple response to light B1 the daily change in light normally acts as a cue that keeps that clock set to local time, which is why the rhythm eventually damps out once the cue is removed A1 opening in anticipation means the pores are already wide when light arrives, so carbon dioxide is available for photosynthesis from first light rather than some minutes later, and at a time of day when the air is cool and humid so relatively little water is lost

Question 5

Abscisic acid is produced when a plant is short of water. Describe how it brings about the closure of a stoma, naming the second messenger involved.

Accepted Answer

B1 abscisic acid binds to specific receptors of the guard cell; accept receptors on the guard-cell plasma membrane B1 the concentration of calcium ions in the cytoplasm rises, some entering through channels in the plasma membrane and some released from the vacuole, so calcium ions are acting as the second messenger B1 the raised calcium concentration opens anion channels, so chloride and malate leave the cell, which makes the inside less negative and opens potassium channels so potassium leaves too A1 with the solutes gone the water potential of the guard cells rises, water leaves by osmosis, the cells become flaccid and the pore closes

Question 6

Compare the effect of a rise in temperature on the rate of transpiration with the effect of a rise in humidity.

Accepted Answer

B1 a rise in temperature raises the rate whereas a rise in humidity lowers it B1 temperature acts on both steps, since molecules gain kinetic energy so evaporation from the mesophyll walls and diffusion out of the stomata are both faster, whereas humidity acts only on the diffusion step A1 warmer air also holds more vapour, so the air outside is further from saturation and the gradient steepens, while raising the humidity brings the water potential outside closer to that inside the leaf and makes the gradient shallower

Question 7

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.

Accepted Answer

B1 the lower surface of a floating leaf is in contact with water, so gases could not diffuse in or out through stomata placed there B1 stomata on the upper surface are in contact with the air, so carbon dioxide can diffuse in and oxygen and water vapour can diffuse out B1 any one further feature with a reason: a thin or absent cuticle since water loss is not a threat, little lignified tissue since the water supports the leaf, or large air spaces called aerenchyma giving buoyancy and carrying oxygen down to roots in anaerobic mud

Question 8

The capillary tube of the potometer has a cross-sectional area of 1.0 mm². Calculate the rate of water uptake at a wind speed of 2.0 m s⁻¹, in mm³ min⁻¹.

Accepted Answer

M1 volume taken up = distance moved × cross-sectional area = 48 × 1.0 M1 rate = volume ÷ time = 48 ÷ 10 A1 4.8 mm³ min⁻¹, with the unit

Question 9

The distance moved rises with wind speed but by less at the highest speed. Explain why moving air increases the rate, and explain why the increase gets smaller.

Accepted Answer

B1 moving air removes the layer of humid air held at the leaf surface, so the water potential gradient between the air spaces and the outside air is steeper B1 water vapour therefore diffuses out through the open stomata faster, and more water is drawn up the xylem to replace it B1 once the humid layer has been largely stripped away, more wind thins it only a little further, so the gradient steepens less and diffusion through the stomatal pores becomes what limits the rate; accept that the stomata begin to close as the shoot loses water faster than the cut stem can replace it

Question 10

Explain why an increase in light intensity increases the rate of transpiration, up to a point beyond which further increases in light intensity have little additional effect.

Accepted Answer

B1 stomata open wider in the light, because photosynthesis needs a supply of carbon dioxide, and a greater aperture gives a larger area available for the diffusion of water vapour out of the leaf B1 once the stomata are fully open, no further increase in light intensity can open them any wider, so the rate of transpiration levels off B1 the effect of light on transpiration is therefore indirect, acting through stomatal aperture rather than directly on evaporation or diffusion

Question 11

A xerophyte has its stomata sunk into pits in the leaf surface, each pit lined with hairs. Explain how each of these two features reduces the rate of water loss.

Accepted Answer

B1 a sunken pit traps a pocket of humid air immediately outside the pore, raising the water potential of the air the stoma is exchanging with and making the gradient across the stoma shallower B1 hairs lining the pit hold that trapped humid air in place, stopping wind from sweeping it away and restoring a steep gradient B1 together the two features keep the boundary layer both humid and still, which a stoma on a flat, exposed leaf surface could not do

Question 12

Explain why transpiration is an unavoidable consequence of gas exchange in a photosynthesising leaf, rather than a process that benefits the plant in its own right.

Accepted Answer

B1 a leaf must keep its stomata open to let carbon dioxide diffuse in for photosynthesis B1 the same open stomata, leading to air spaces saturated with water vapour, inevitably also let water vapour diffuse out, since both movements use the same pores B1 water loss is therefore a side effect of the stomata being open for carbon dioxide, not something the plant does deliberately, which is why the stomata close, and reduce photosynthesis, when water is in short supply

Question 13

A succulent desert plant opens its stomata only at night and fixes carbon dioxide into malate in the dark, releasing it to the Calvin cycle during the day. Suggest why opening the stomata at night, rather than during the day, reduces the plant's water loss.

Accepted Answer

B1 night air is cooler and more humid than daytime air, so the water potential gradient between the leaf air spaces and the outside air is shallower at night B1 a shallower gradient means a slower rate of diffusion of water vapour out of the stomata for a given aperture B1 the plant can therefore take in the carbon dioxide it needs for the day's photosynthesis while losing much less water than it would opening its stomata under the hotter, drier conditions of the day

Question 14

State the two steps by which water is lost from a leaf during transpiration, in the order in which they occur.

Accepted Answer

B1 water evaporates from the wet cellulose walls of the mesophyll cells into the air spaces of the leaf B1 water vapour then diffuses out through the open stomata down a water potential gradient

Question 15

State the approximate percentage of the water a plant absorbs that is lost again by transpiration, and state what happens to the stomata, and to photosynthesis, when water is short.

Accepted Answer

B1 between 95 and 99 per cent of the water absorbed is lost through transpiration B1 the stomata close, which reduces water loss but also reduces carbon dioxide uptake and so the rate of photosynthesis

Wind speed / m s⁻¹	Distance moved by the meniscus in 10 min / mm
0.0	24
1.0	36
2.0	48
3.0	54

Wind speed / m s⁻¹	Distance moved by the meniscus in 10 min / mm	Increase on the row above / mm
0.0	24	no earlier row
1.0	36	12
2.0	48	12
3.0	54	6