Transpiration and stomatal density

Organisms (Form and function) · Gas exchange · note 5 of 6

Spec B3.1.9, B3.1.10Download all Gas exchange notes (PDF)Download all notes (PDF)10 pages

Transpiration and stomatal densitySpec B3.1.9, B3.1.10

In short

Transpiration is the loss of water vapour from the leaves and stems of plants. It is a consequence of gas exchange because stomata must open to let carbon dioxide in, and water vapour then diffuses out. Its rate rises with light, temperature and wind, and falls with humidity. Stomatal density is stomata per unit area.

Water evaporates from the moist walls of mesophyll cells into the air spaces, then water vapour diffuses out through open stomata. Because stomata must be open for CO₂ to enter, transpiration is a consequence of gas exchange.

Factors affecting the rate of transpiration
Factor	Effect on rate	Reason
Light intensity	Increases	Stomata open in the light for photosynthesis.
Temperature	Increases	More evaporation from cell walls and faster diffusion of water vapour.
Humidity	Decreases	Moist air outside lowers the water vapour concentration gradient.
Wind speed	Increases	Moving air removes water vapour from around the stomata, keeping the gradient steep.
Soil water availability	Decreases if water is short	Guard cells lose turgor and stomata close.

The rate of transpiration can be estimated with a potometer, which measures water uptake by a cut shoot.

Stomatal density

stomatal density = number of stomata ÷ area of field of view (mm²)

Practical skill:

To make a leaf cast, paint clear nail varnish on the lower epidermis, let it dry, lift it off with clear tape and stick it to a slide. Count stomata at high power in several fields of view, or count them on a micrograph with a known scale.

Calculating stomatal density

At high power the field of view has a diameter of 0.40 mm. Counts of stomata in five fields of view are 24, 27, 22, 26 and 26. Calculate the mean stomatal density. (Practice data.)

Mean count = (24 + 27 + 22 + 26 + 26) ÷ 5 = 125 ÷ 5 = 25
Radius = 0.40 ÷ 2 = 0.20 mm
Area = πr² = 3.14 × 0.20² = 0.126 mm²
Density = 25 ÷ 0.126 = 198.4

Answer: about 200 stomata per mm² (2.0 × 10² mm⁻²)

Maths skill:

Repeated counts vary because biological material is variable. Replicate counts and a mean increase the reliability of the data; the range or standard deviation shows the spread.

Written and checked against the IB Biology HL specification · Updated October 2026

Frequently asked questions

Why do larger organisms need specialised gas exchange surfaces?

Larger organisms need specialised gas exchange surfaces because their surface area-to-volume ratio is smaller and the distance from their centre to the outside is greater. Diffusion across the body surface alone would be far too slow, so they need large, thin, moist, permeable surfaces such as lungs or gills and a transport system.

How are alveoli adapted for gas exchange?

Alveoli are adapted for gas exchange by their huge total surface area, walls one thin layer of cells thick, a moist lining and a dense capillary network around each one. Surfactant lowers surface tension so alveoli do not collapse, and ventilation with continuous blood flow keeps the concentration gradients steep.

What happens to the diaphragm and intercostal muscles when you breathe in?

When you breathe in, the diaphragm contracts and flattens and the external intercostal muscles contract, pulling the ribs up and out. The volume of the thorax increases, so the pressure inside falls below atmospheric pressure and air flows into the lungs.

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