Transport in Plants

Lessons

Senior 5 · Biology

Transport in Plants

4 lessons

Lesson 3 of 4

Transpiration and the factors that affect its rate

25 min

Where does the water go?

Weigh a potted maize plant in the morning and again in the afternoon, and it will be lighter, although it has grown. Almost all the water a plant takes from the soil, often well over nine-tenths of it, is not used in photosynthesis or growth but evaporates from the leaves into the air. This loss is transpiration. It is the "engine" of the cohesion–tension mechanism you met in the last lesson. This lesson shows how to measure it and why its rate changes from hour to hour.

Definition

Transpiration: the loss of water vapour from the aerial parts of a plant, mainly through the stomata of the leaves.

Transpiration stream: the continuous movement of water from the roots, up the xylem, to the leaves, where it evaporates.

Stoma (plural stomata): a pore in the epidermis, mainly of the leaf, bounded by two guard cells that open and close it.

Potometer: an instrument that measures the rate at which a cut shoot takes up water, which is used to estimate its rate of transpiration.

Three routes out of the plant

Stomatal transpiration, through open stomata: by far the largest route, often around 90% of the total.
Cuticular transpiration, straight through the waxy cuticle: small, and smaller still where the cuticle is thick.
Lenticular transpiration, through lenticels, loose patches of cells in the bark of woody stems: very small.

Inside the leaf, water evaporates from the wet walls of the spongy mesophyll cells into the air spaces. The air spaces become nearly saturated with water vapour, so water vapour diffuses out through the stomata to the drier air outside, down a water vapour potential gradient.

Why transpire at all?

Transpiration is often called an unavoidable consequence of gas exchange. To photosynthesise, a leaf must open its stomata to let carbon dioxide in, and whenever they are open water vapour escapes. Yet it also brings benefits:

It drives the transpiration stream, which carries mineral ions from the roots to the leaves.
Evaporation absorbs heat (the latent heat of vaporisation), so it cools leaves in strong sun.
It supplies water to keep the leaf cells turgid, which supports the leaf, and to replace water used in photosynthesis.

The cost is that a plant that loses water faster than its roots can absorb it will wilt, close its stomata and stop photosynthesising.

Measuring transpiration with a potometer

Method (Figure 6):

Cut a leafy shoot under water, cutting the stem at a slant, so that no air enters the xylem and breaks the water columns.
Still under water, fit the shoot into the potometer, so that the apparatus is completely full of water. Seal every joint with petroleum jelly so that it is airtight and watertight.
Dry the leaves gently with tissue paper, since wet leaves cannot transpire normally.
Let a single air bubble into the capillary tube by lifting its end out of the beaker for a moment.
Leave the plant to settle for about 5 minutes in the conditions to be tested. Then record how far the bubble moves in a fixed time.
Open the tap of the reservoir to push the bubble back to the start, and repeat at least three times. Calculate a mean.

A potometer really measures water uptake, not water loss. The two are nearly equal, but not exactly: a little water is used in photosynthesis and in keeping cells turgid. A mass potometer, in which the whole plant and its pot stand on a balance, measures the water actually lost, provided the soil surface is covered so that no water evaporates from it.

Worked example 1: from bubble distance to a rate

In a potometer, the bubble moves 36 mm in 5.0 minutes. The capillary tube has an internal radius of 0.50 mm. The shoot's leaves have a total area of 48 cm². Calculate (a) the volume of water taken up, (b) the rate of uptake, (c) the rate per unit leaf area.

(a) The water taken up is a cylinder of length 36 mm: $V = \pi r^2 l = \pi \times 0.50^2 \times 36 = 28.3\,\mathrm{mm^3}$
(b) $\text{rate} = \frac{28.3\,\mathrm{mm^3}}{5.0\,\mathrm{min}} = 5.65\,\mathrm{mm^3\,min^{-1}} \approx 5.7\,\mathrm{mm^3\,min^{-1}}$
(c) $\frac{5.65}{48} = 0.118\,\mathrm{mm^3\,min^{-1}\,cm^{-2}} \approx 0.12\,\mathrm{mm^3\,min^{-1}\,cm^{-2}}$

Why (c) matters: two shoots of different sizes can only be compared fairly per unit of leaf area. Leaf area can be found by drawing round the leaves on squared paper.

Factors that change the rate

Factor	Effect on rate	Explanation
Light intensity	Rises in light	Stomata open in the light for photosynthesis, so more water vapour diffuses out. In the dark most stomata close.
Temperature	Rises with temperature	Water molecules have more kinetic energy, so evaporation and diffusion are faster; warm air can also hold more water vapour, so the gradient out of the leaf is steeper.
Humidity	Falls as humidity rises	Moist air outside reduces the water vapour gradient between the air spaces and the air.
Wind (air movement)	Rises with wind, up to a point	Wind blows away the layer of still, humid air next to the leaf surface, keeping the gradient steep. In very strong wind some plants close their stomata.
Soil water	Falls when soil is dry	The plant cannot replace water fast enough; the hormone abscisic acid (ABA) makes the stomata close.

The structure of the plant matters too: the leaf area, the thickness of the cuticle, and the number and position of the stomata. Many plants have more stomata on the lower surface of the leaf, which is cooler and shaded. The next lesson shows how plants of dry places use these features.

Worked example 2: interpreting a potometer experiment

A class tests one shoot under four conditions, with a capillary of internal radius 0.40 mm, timing each run for 10 minutes. Illustrative results:

Condition	Mean distance moved by bubble / mm	Rate / mm³ min⁻¹
A: shade, still air	12	0.60
B: bright light, still air	30	1.51
C: bright light, fan blowing	55	2.76
D: bright light, shoot covered by a clear plastic bag	6	0.30

Rate for B: $V = \pi \times 0.40^2 \times 30 = 15.1\,\mathrm{mm^3}$ , so the rate $= \frac{15.1}{10} = 1.51\,\mathrm{mm^3\,min^{-1}}$ . The others are found the same way.
A to B: light opens the stomata, so the rate is about 2.5 times higher.
B to C: the fan removes the humid layer next to the leaves, keeping the gradient steep.
B to D: the bag traps water vapour, so the air around the leaves becomes almost saturated and the gradient nearly disappears, even though the stomata are open in the light.

Control variables: the same shoot, the same temperature, and a 5-minute settling time before each run.

Transpiration is not guttation

Drops of water on the edges of leaves early in the morning are not transpiration, which is loss of water vapour. They are guttation: liquid water forced out through special pores at the leaf edges, called hydathodes, by root pressure, at times when the soil is wet, the air is humid and transpiration is slow.

Warning

Common mistakes and safety

"A potometer measures transpiration." It measures water uptake, which is only approximately equal to the rate of transpiration.
"Water leaves the leaf as liquid through the stomata." It leaves as vapour, by diffusion; it evaporates inside the leaf.
Explaining humidity. Do not just write "it is humid". Say that humid air reduces the water vapour potential gradient between the inside and the outside of the leaf.
Safety: take care with the sharp blade when cutting the shoot, and with glass capillary tubing, which snaps easily. Wipe up spilt water at once.

Miss Estruch Biology: "Potometer Experiment: Measuring the rate of transpiration". Watch how the shoot is cut and fitted under water and the joints sealed, how the bubble is introduced and timed, and how the distance is turned into a volume, as in Worked example 1.

Tip

Exam technique: for every factor, link three things: the factor, what it does to the water vapour potential gradient or to the stomata, and the effect on the rate of diffusion of water vapour out of the leaf. For example: "Higher wind speed removes the humid air near the leaf, so the gradient stays steep and water vapour diffuses out faster."

✦

Practice: transpiration

Define transpiration and name the three routes by which a plant loses water vapour.
Give two benefits of transpiration to a plant.
Why is the shoot cut under water when a potometer is set up?
A bubble moves 42 mm in 3.0 minutes in a capillary of internal radius 0.25 mm. Calculate the rate of water uptake in mm³ min⁻¹ and in mm³ h⁻¹.
A potted plant with its soil covered by a plastic sheet loses mass from 412.6 g to 398.2 g in 6.0 hours. Calculate its mean rate of water loss in g h⁻¹.
Explain why the rate of transpiration of a plant is usually higher on a windy, sunny day than on a still, cloudy day.
Suggest why a potometer reading might be slightly higher than the true rate of transpiration.
Predict and explain the effect of covering the lower surface of every leaf with petroleum jelly on the rate measured by a potometer.

Show answersHide answers

The loss of water vapour from the aerial parts of a plant. Routes: stomata, the cuticle and lenticels.
Any two: it moves mineral ions up from the roots; it cools the leaves; it keeps a supply of water to keep cells turgid and for photosynthesis.
To stop air entering the xylem vessels, which would break the water columns and stop the uptake of water.
$V = \pi \times 0.25^2 \times 42 = 8.25\,\mathrm{mm^3}$ ; rate $= \frac{8.25}{3.0} = 2.75\,\mathrm{mm^3\,min^{-1}}$ , about 2.7 mm³ min⁻¹, or $2.75 \times 60 = 165\,\mathrm{mm^3\,h^{-1}}$ .
$\frac{412.6 - 398.2}{6.0} = \frac{14.4}{6.0} = 2.4\,\mathrm{g\,h^{-1}}$ .
In sunlight the stomata are open and it is warmer, so evaporation is faster; wind removes the layer of humid air next to the leaf, so the water vapour potential gradient stays steep.
Some of the water taken up is used in photosynthesis or stays in the cells to keep them turgid and is not lost.
The rate falls sharply, because most stomata are on the lower surface and the jelly blocks them; a small uptake continues through the stomata on the upper surface and the cuticle.

Key points

Transpiration is the loss of water vapour, mainly through stomata; it is the unavoidable cost of letting CO₂ in.
It drives the transpiration stream, carries mineral ions and cools the leaf.
A potometer measures water uptake; rate $= \frac{\pi r^2 l}{t}$ , best compared per unit leaf area.
Rate rises with light, temperature and wind, and falls with humidity and dry soil.

Sources and credits

Written by Ganzaa. Facts were checked against 30.5 Transport of Water and Solutes in Plants, from Biology 2e by Mary Ann Clark, Matthew Douglas and Jung Choi, OpenStax, licensed under CC BY 4.0. No text was copied.
Photo: Crucifer Guttation droplets, Scot Nelson, CC0, via Wikimedia Commons.
Video: Potometer Experiment: Measuring the rate of transpiration, Miss Estruch Biology, embedded from YouTube.

Figure 6, the tables, the worked examples and the practice questions are by Ganzaa. The results in Worked example 2 are illustrative.

1Where does the water go?

2Three routes out of the plant

3Why transpire at all?

4Measuring transpiration with a potometer

5Factors that change the rate

6Transpiration is not guttation