Biology › Plant transport and mineral nutrition › Transpiration: the price of keeping the stomata open

Transpiration: the price of keeping the stomata open

A plant needs carbon dioxide, and the only way in is a hole. Every hole that lets carbon dioxide in lets water vapour out, at a rate set by a gradient the plant cannot switch off. Guard cells are how it negotiates the deal, minute by minute.

Before this Diffusion and the factors that affect its rate · Water potential and turgor in plant cells

Before you start

Transpiration is how a plant transports water and minerals to its leaves, so the pull is the point of it. Read that again with the plant's interests in mind. Somewhere between 95 and 99 per cent of the water a plant absorbs is lost to the air, and drought kills more crops than anything else does. No organism evolves a system whose main output is losing almost everything it collects. Transpiration is what happens because carbon dioxide has to get in and the same pores let water out; the transport of water and minerals is a side effect the plant has learned to exploit, and the moment water runs short the plant shuts the pores and accepts less photosynthesis instead.

What you should be able to do

Define transpiration and explain why a photosynthesising plant cannot avoid it.
Predict and explain the effect of light, temperature, humidity and air movement on the rate, in every case by naming what happens to the gradient.
Describe how a guard cell opens a stoma, from the proton pump to the turgor.
Explain why the shape and wall thickness of a guard cell turn turgor into an open pore rather than a fatter cell.
Account for xerophyte and hydrophyte adaptations by saying what each does to the water potential gradient.
Process potometer data: calculate rates, compare conditions and state what was controlled.

Why a plant leaks

A leaf is built for one job above all others: getting carbon dioxide to the chloroplasts. Carbon dioxide is 0.04 per cent of the atmosphere, so the leaf needs a large internal surface, an air space system that reaches every cell, and thin wet walls for the gas to dissolve in. Every one of those features is also an excellent evaporating surface, and the stomata that let carbon dioxide in are open holes into a chamber saturated with water vapour.

Transpiration: The loss of water vapour from the aerial parts of a plant, mainly through the stomata, by evaporation from the mesophyll cell walls followed by diffusion out of the leaf.
Transpiration stream: The continuous movement of water from the soil, through the root and up the xylem, to replace the water lost from the leaves.
Stoma: A pore in the epidermis, bounded by two guard cells, through which gases and water vapour pass.

The quantities are worth a moment. A single maize plant moves a couple of litres of water in a summer day. A mature oak in leaf can shift several hundred litres. All of it evaporates, and the plant gets from it only what comes along for the ride: mineral ions delivered to the leaves, turgor pressure that holds a non-woody stem up, and evaporative cooling on a hot day, which is real and measurable but not worth the water on its own.

The two-step description matters for exam wording. Water first evaporates from the wet cellulose walls of the mesophyll cells into the leaf air spaces, which is where the energy is spent. It then diffuses out through the open stomata down a water potential gradient. Anything that changes either step changes the rate, and every factor below acts on one or the other.

Four factors, and the same explanation each time

The practical is a potometer with one condition changed at a time. The readings below are idealised — clean numbers, chosen so the arithmetic is readable. A real class set does not look like this. Different shoots have different leaf areas, bubbles stick, and repeats from one bench vary by ten or twenty per cent, which is why you take repeats and compare means rather than single readings.

Compare the humid bar with the still-air bar rather than reading it on its own. The absolute numbers depend on which shoot was cut; the ratios between conditions are the result.

Factor	Effect on rate	Because
Light intensity up	Faster	Stomata open in the light for photosynthesis, so more pores are available for diffusion; the rate levels off once they are all open
Temperature up	Faster	Molecules have more kinetic energy, so evaporation from the walls is faster and diffusion is faster; warmer air also holds more vapour, which lowers relative humidity outside and steepens the gradient
Humidity up	Slower	The air outside is closer in water potential to the saturated air inside the leaf, so the gradient across the stoma is shallower
Air movement up	Faster	Wind sweeps away the humid boundary layer clinging to the leaf surface, so the steep part of the gradient is restored

Humidity is the factor most often explained badly, and a number fixes it. Air at 20 °C and 50 per cent relative humidity has a water potential of about −94 000 kPa. At 95 per cent relative humidity it is about −7000 kPa. Leaf mesophyll might be at −1500 kPa. Even very humid air is drier than the inside of a leaf, so transpiration does not stop in a greenhouse — it slows, because the gradient has gone from enormous to merely large.

The boundary layer deserves the same treatment. Still air next to a leaf becomes saturated with the vapour coming out of it, and that saturated layer, not the room, is what the leaf's interior is actually exchanging with. Hairs on a leaf surface hold that layer in place; a fan strips it away. This is why wind is such a powerful factor and why xerophytes are hairy.

Comparing two conditions properly

Using the readings in the figure: the bubble moved 24 mm in 10 minutes in still air and 51 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 each condition in mm³ min⁻¹, and the percentage increase caused by the fan.

Rates first: 24 ÷ 10 = 2.4 mm min⁻¹ and 51 ÷ 10 = 5.1 mm min⁻¹ of bubble travel.

Cross-sectional area = πr² = π × 0.50² = 0.785 mm². So still air gives 0.785 × 2.4 = 1.9 mm³ min⁻¹ and moving air gives 0.785 × 5.1 = 4.0 mm³ min⁻¹.

Percentage increase = (5.1 − 2.4) ÷ 2.4 × 100 = 112.5%, which is a rate slightly more than doubled.

Note what the percentage is calculated from. It is the change divided by the original, not by the new value, and the original here is the control condition. Working in bubble distance rather than volume gives the same percentage, because the conversion factor cancels — which is a useful check and a legitimate shortcut if the question only wants the comparison.

How a guard cell opens a hole

A stoma is a gap between two guard cells, and the gap opens because the cells change shape. What follows is a chain of causes, and the marks are distributed along it rather than concentrated at the end.

The asymmetry is the mechanism. Two cells that swelled evenly would press together and close the pore; these two can only bulge outwards, so swelling opens the gap.

In the light, and when carbon dioxide in the air spaces runs low, ATP from the guard cell's own chloroplasts and mitochondria drives proton pumps in its plasma membrane. Hydrogen ions are pumped out. That leaves the inside of the cell electrically negative relative to the outside, and potassium ions move in through channel proteins down the electrical gradient. Chloride ions follow, and stored starch is converted to malate, adding more solute still.

The consequences are osmotic and mechanical in turn. More solute means a lower solute potential, so a lower water potential; water enters from the neighbouring epidermal cells by osmosis; the guard cells become turgid. Now the wall does the rest. The wall on the side facing the pore is thicker and less elastic than the outer wall, and the cellulose microfibrils are wound in hoops around the cell, so it can lengthen but cannot fatten. A turgid guard cell therefore bows away from its partner, and the pore opens.

Closing runs the same chain backwards, and the plant can force it. Under water stress the roots and leaves produce abscisic acid, which binds receptors on the guard cell membrane and opens anion channels. Solutes leave, water follows, the cells go flaccid and the stoma shuts, often within minutes and regardless of how bright the light is. Guard cells have no plasmodesmata joining them to their neighbours, which is what lets them be controlled osmotically as an isolated pair.

Where this is examined varies. CAIE 9700 asks for the mechanism, abscisic acid included, inside the transport topic; OCR A examines abscisic acid under plant responses. It is the same chain of events either way, so learn it once.

Leaves built for the extremes

Plants in dry places are not doing anything new. Every xerophyte adaptation is a way of making the gradient out of the leaf less steep, or the path out of it longer, or the number of exits smaller.

Marram grass on the left is worth learning as a single named example, because it carries four adaptations at once and questions ask for them with the plant named.

Adaptation	How it slows water loss
Thick waxy cuticle	Cuticle is impermeable, so almost nothing evaporates through the epidermis itself
Stomata sunken in pits	The pit traps humid air outside the pore, so the gradient across the stoma is shallower
Hairs on the epidermis	They hold a still, humid boundary layer against the surface even when the wind blows
Rolled leaf, as in marram grass	The stomata end up enclosed inside the roll, in air that quickly becomes saturated
Leaves reduced to spines or needles	Less surface area from which to lose water
Stomata open only at night	Carbon dioxide is fixed into malate in the dark and released to the Calvin cycle by day, so the pores are shut when evaporation is fastest

A hydrophyte has the opposite problem and solves it by not bothering. A floating leaf has water under it and air above, so its stomata sit on the upper surface — the reverse of an ordinary leaf, and a favourite one-mark question. The cuticle is thin or missing, there is little lignified tissue because the water provides support, and large air spaces called aerenchyma give buoyancy and carry oxygen down to roots sitting in anaerobic mud. OCR A names hydrophytes alongside xerophytes; AQA and CAIE concentrate on the xerophyte side.

TRY IT — Explaining an adaptation, not listing it

Marram grass grows on sand dunes, where water drains away within minutes of rain and the wind rarely stops. Its leaves roll into a tube with the stomata on the inside, and the inner surface carries dense hairs. Explain how these two features reduce water loss. (4 marks)

Check your answer

Rolling encloses the stomata in a small volume of trapped air. Water vapour diffusing out of the stomata quickly saturates that air, so its water potential rises towards the water potential inside the leaf.

The water potential gradient between the leaf air spaces and the air just outside the pore is therefore shallower, so the rate of diffusion out of the stomata falls.

The hairs hold that humid air still. Without them, air moving over the surface would sweep the saturated layer away and restore the steep gradient, which is exactly what wind does to an unprotected leaf.

The rolled tube also keeps the stomata out of the moving air altogether, so the boundary layer survives even in the wind that a dune is never without.

What loses marks here is stopping at 'traps water vapour'. The mark is for the gradient: trapped vapour raises the water potential outside the stoma, and a shallower gradient means slower diffusion.

In the exam

Every factor gets the same shape of answer: say what happens to the water potential gradient between the leaf air spaces and the air outside, then say what that does to the rate of diffusion.
Evaporation and diffusion are two different steps. Temperature speeds up both; humidity and wind act only on the second.
For guard cells, run the chain in order — protons out, potassium in, water potential falls, water in by osmosis, turgid, thicker inner wall makes the cell bow. Skipping to 'they become turgid so the stoma opens' throws away most of the marks.
Say which surface. Stomata are usually on the lower epidermis; a floating hydrophyte leaf has them on the upper. Questions are set on precisely that reversal.
Name the plant when you can. 'Marram grass rolls its leaves' is worth more than 'some plants roll their leaves', and examiners set questions around the named examples in the specification.
In an evaluation, say what was controlled. Temperature, humidity, light and air movement all affect the rate, so a potometer investigation of one of them is only valid if the other three were held constant.

Check yourself

A student measures water uptake by a leafy shoot in a potometer. She records 2.4 mm min⁻¹ of bubble travel in still air at 20 °C, and 0.9 mm min⁻¹ after sealing a clear plastic bag loosely around the shoot. She concludes that the bag 'stopped the plant transpiring'. Evaluate her conclusion and explain the result.

Answer

The conclusion is wrong on two counts. The rate fell to 0.9 mm min⁻¹, which is a reduction of 62.5%, not a stop — the shoot went on taking up water throughout.

And the potometer was measuring uptake, not transpiration, so no reading it produces can be described directly as a transpiration rate.

The explanation is the gradient. Water vapour leaving the leaves accumulated inside the bag, so the humidity of the air around the shoot rose and its water potential rose with it, towards the water potential of the saturated air inside the leaf. A shallower gradient across the stomata means slower diffusion out, so less water is lost and less is drawn in to replace it.

It does not fall to zero because the air in the bag never quite reaches saturation, and even air at 95% relative humidity sits far below the leaf in water potential.

A fair criticism of the method: the bag also cut the light reaching the leaves slightly and raised the temperature inside it, so humidity was not the only variable that changed. A better design would humidify the air around an unbagged shoot instead.

Questions

Written to the command words the boards use. Try them on paper before opening a scheme: the marks go to points made, not to length.

Question 15 marks

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

Question 25 marks

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

Question 34 marks

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

Question 43 marks

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

Mark scheme

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 53 marks

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.

Mark scheme

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 62 marks

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

Mark scheme

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

Worth remembering

Transpiration is the cost of opening stomata for carbon dioxide, not a purpose.
Evaporation from the mesophyll walls, then diffusion out of the stomata — two steps, and factors act on one or the other.
Light and temperature raise the rate; humidity lowers it; wind raises it by removing the boundary layer.
Protons out, potassium in, water in, turgid, thicker inner wall bows the cell and the pore opens.
Abscisic acid shuts stomata under water stress, whatever the light is doing.
Xerophyte adaptations make the gradient shallower or the exits fewer; a hydrophyte puts its stomata on top.

← Xylem and the transpiration stream: the pull comes from the top · Phloem and translocation: pumped at the ends, flowing in the middle →