A single maize plant loses around 200 litres of water by transpiration over a growing season. What controls how fast it loses water, and how do we measure it? Start with the data and figure it out.
Pick up the worksheets, then commit to your own answer before you read.
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Which condition would make a leaf lose water fastest: wind, heat, bright light or dry air? Rank them and justify using gradients or stomata.
Set the intentions and clear the misconceptions that trip most students here.
Content from this lesson that appears directly in HSC Biology exams
Analysing potometer data (graph or table) to identify factor effects on transpiration rate, tested as a 3–4 mark working scientifically question in most HSC papers. Must identify trend, explain mechanism, evaluate the method, and calculate rate from data.
Explaining why each factor (temperature, humidity, wind, light) affects transpiration rate. Tested as 2–3 mark mechanism questions, must state the factor, explain the mechanism (water potential gradient, stomatal aperture, or diffusion gradient), and predict the effect direction.
Explaining how structural adaptations of xerophytes (plants in dry environments) reduce transpiration. Tested as 2–3 mark application questions, must name the adaptation, describe the structure, and explain how it reduces water loss.
Designing a valid potometer experiment (IV, DV, controlled variables) and evaluating its limitations (measures water uptake, not transpiration directly). Tested as 2–3 mark Working Scientifically questions in Section II.
Start with the Data
Wrong: Wind always increases transpiration rate.
Right: Moderate wind removes the boundary layer and increases transpiration. However, very high wind or hot, dry wind can cause stomata to close as a protective response, actually reducing transpiration. The relationship between wind speed and transpiration is not linear, stomatal regulation overrides physical effects at extreme conditions.
Four conditions, one plant. See what needs explaining before you explain it.
Look at the numbers. Which condition most affects the rate? Why?
A student uses a potometer to measure water uptake in a leafy shoot under four different conditions. In each trial, all other variables are held constant (same shoot, same time period, same starting conditions). The shoot is moved into each condition for 10 minutes and the distance the air bubble travels in the capillary tube is measured.
| Condition | Trial 1 (mm) | Trial 2 (mm) | Trial 3 (mm) | Mean (mm) | Rate (mm/min) |
|---|---|---|---|---|---|
| Still air, 22°C, 60% humidity, dim light | 12 | 14 | 13 | 13.0 | 1.3 |
| Still air, 22°C, 60% humidity, bright light | 19 | 21 | 20 | 20.0 | 2.0 |
| Fan (wind), 22°C, 60% humidity, dim light | 24 | 23 | 25 | 24.0 | 2.4 |
| Still air, 35°C, 30% humidity, dim light | 31 | 34 | 32 | 32.3 | 3.2 |
| Still air, 22°C, 90% humidity, dim light | 4 | 5 | 4 | 4.3 | 0.4 |
Your hypotheses, before reading Card 2: Rank the conditions fastest to slowest. For each, suggest why that condition affects water loss rate.
What trend or pattern would you expect to see in data about this topic?
The Mechanism
Light, temperature, humidity and wind, and the mechanism behind each.
Each factor acts on one of two things: the water potential gradient or stomatal aperture
Transpiration rate is determined by how steep the water potential gradient is between the leaf interior and the outside air, and by how open the stomata are. Every environmental factor acts through one of these two mechanisms, once you understand which, you can predict the direction of effect for any factor without memorising.
The five factors that affect transpiration rate, each acts on either the water potential gradient or stomatal aperture
High temperature → faster transpiration
Mechanism 1, Kinetic energy: Higher temperature increases the kinetic energy of water molecules, evaporation from mesophyll cell walls into leaf air spaces is faster, increasing the water vapour concentration in the leaf air spaces.
Mechanism 2, Atmospheric holding capacity: Warm air can hold more water vapour than cool air. This means at high temperature, the atmosphere is further from saturation, the water potential difference between the humid leaf air space and the dry outside air is larger, increasing the diffusion gradient driving water vapour out through stomata.
Mechanism 3, Stomatal response: High temperature also causes guard cells to open stomata wider (heat stress response), increasing stomatal conductance.
High humidity → slower transpiration
Mechanism, Gradient reduction: Transpiration is driven by the difference in water potential (water vapour concentration) between the leaf air spaces (nearly saturated, ~99% relative humidity) and the outside air.
When outside air is already highly humid (high relative humidity), the water potential difference between leaf interior and atmosphere is small, the gradient driving diffusion of water vapour through stomata is reduced. At 100% relative humidity, no net transpiration occurs because there is no gradient.
The potometer data confirms this dramatically: the 90% humidity condition (0.4 mm/min) is 8× slower than the high temperature/low humidity condition (3.2 mm/min), the single largest effect in the dataset.
Increased air movement → faster transpiration
Mechanism, Removal of boundary layer: In still air, a thin layer of humid air (the boundary layer) accumulates immediately outside the stomata, partially saturated with water vapour from previous transpiration. This boundary layer reduces the effective water potential gradient, water vapour must diffuse through this humid layer before reaching drier bulk air.
Wind constantly removes the boundary layer, replacing humid air around stomata with drier bulk air. This maintains a steep water potential gradient at the stomatal pore, increasing transpiration rate. This is why clothes dry faster on windy days, the same physics.
Note: very strong wind can cause stomatal closure as a water-conservation response, which would reduce transpiration.
Bright light → faster transpiration (indirectly)
Mechanism, Stomatal opening: Light does not directly accelerate water evaporation. Instead, it acts through guard cell biology. In bright light, guard cells photosynthesise, producing ATP and accumulating K⁺ ions by active transport, this lowers guard cell water potential, causing osmotic water entry and inflation, which opens stomatal pores (the mechanism from L09).
Wider stomatal aperture = more pathway for water vapour to diffuse out. In darkness, stomata close (most species) and transpiration drops to near zero, only cuticular transpiration remains through the waxy cuticle.
The potometer data shows light is a moderate factor (1.3 → 2.0 mm/min), it changes stomatal aperture, but the gradient itself (determined by temperature and humidity) determines maximum possible rate.
Figure: Four factors affecting transpiration rate. Coloured left strip identifies the factor. Temperature and humidity act on the water potential gradient; light acts on stomatal aperture; wind maintains the gradient by removing the boundary layer.
A student wants to use a potometer to test the hypothesis: "Increasing wind speed increases the rate of water uptake by a leafy shoot."
Type your written response below or write in your book.
Water uptake, not transpiration, and why the difference matters.
A critical distinction that examiners test every year
Variables in a valid potometer experiment:
Write one sentence summarising the main idea of this section.
Structural answers to a water problem.
Every adaptation acts on one of the four factors, or physically blocks the pathway
Xerophytes are plants adapted to environments with limited water availability, deserts, rocky slopes, coastal dunes. Their structural adaptations are not random; each one acts on a specific transpiration factor or directly reduces the water loss pathway. Understanding the mechanism is what separates Band 6 from Band 3 responses.
A thick layer of waxy cutin deposited on the leaf epidermis. Cutin is highly hydrophobic, essentially waterproof. Reduces cuticular transpiration (water diffusing through the cuticle rather than stomata).
Stomata positioned in crypts or pits below the leaf surface rather than flush with the epidermis. Still air accumulates in the crypt, creating a high-humidity boundary layer immediately outside the stomatal pore.
Leaves reduced to spines (cacti), needles (pines), or small scale-like structures. Fewer and smaller leaves mean fewer stomata and less total surface area for evaporation. Some have leaves that roll up in dry conditions, trapping humid air.
Dense covering of fine epidermal hairs (trichomes) on the leaf surface. Hairs trap a still, humid boundary layer around stomata, reducing the water potential gradient between the stomatal pore and bulk air. Hairs also reflect light, reducing leaf temperature.
Large vacuoles and parenchyma cells in leaves or stems that store water. Not a transpiration reduction adaptation directly, but provides a water reservoir that sustains the plant through dry periods when transpiration cannot be met by soil water uptake.
Crassulacean Acid Metabolism (CAM) plants open stomata only at night to fix CO₂ into organic acids. During the hot, dry day, stomata remain closed, CO₂ stored overnight is released internally for daytime photosynthesis. Eliminates daytime transpiration entirely.
Write one sentence summarising the main idea of this section.
Re-read the opening numbers with everything you now know.
Reread the potometer results through the lens of the mechanisms you now understand
Return to the data table in Card 1. With the mechanisms from Cards 2 and 3, you can now explain every result, not just describe the trend. This is the difference between Band 4 and Band 6.
What trend or pattern would you expect to see in data about this topic?
Activities
Multiple choice first, then the written answers. No notes.
Assessment
5 random review questions from a replayable lesson bank
6. Explain why a plant wilts faster on a hot, dry, windy day compared to a cool, humid, still day. Refer to the effect of each factor on transpiration rate in your answer. 4 MARKS
7. A student claims "a potometer directly measures how much water a plant transpires." Evaluate this claim, identifying one inaccuracy and explaining what the potometer actually measures. 3 MARKS
8. Explain how sunken stomata reduce transpiration in a xerophyte. In your answer, refer to the water potential gradient and boundary layer. 3 MARKS
1. C Wind's mechanism is boundary layer removal. In still air, water vapour transpired through stomata partially saturates the thin air layer immediately outside, reducing the effective gradient. Wind constantly sweeps this away, maintaining drier air at the stomatal pore. This is analogous to how a fan dries laundry faster, it removes the humid air immediately adjacent to the wet surface.
2. A Transpiration rate depends on the water potential gradient between leaf air spaces (near-saturated, ~99% humidity) and the outside atmosphere. At 95% humidity, the gradient is 99 − 95 = ~4 percentage points. At 30% humidity, the gradient is 99 − 30 = 69 percentage points, approximately 17 times larger. Transpiration rate reflects this proportional difference.
3. D The key limitation of a potometer is that it measures water uptake, not transpiration directly. At higher temperatures, photosynthesis rate also increases, meaning more water is incorporated into carbohydrates and other metabolic processes. This additional non-transpiratory water use would cause the potometer to slightly overestimate transpiration's specific contribution to water uptake. This is a valid methodological limitation specific to temperature experiments.
4. B Sunken stomata in crypts create a still-air micro-environment within the pit. Transpired water vapour accumulates there rather than being dispersed, forming a humid boundary layer immediately outside the pore. This reduces the effective water potential gradient, the driving gradient is between the pore and the humid crypt air, not between the pore and dry bulk air. The waxy cuticle blocks the cuticular pathway (not boundary layer); CAM changes timing (not boundary layer); succulents store water (not boundary layer).
5. A Elevated CO₂ signals guard cells to close stomata. CO₂ enters guard cells and is converted to bicarbonate, which triggers signalling pathways leading to K⁺ efflux from guard cells. Losing K⁺ raises guard cell water potential, water exits by osmosis, turgor falls, and the stomatal pore narrows or closes. This is the reverse of light-induced stomatal opening and is an adaptive response to high CO₂, when CO₂ is abundant, there is less need for open stomata to capture more.
On a hot, dry, windy day, all three environmental factors combine to dramatically increase transpiration rate relative to a cool, humid, still day.
High temperature increases the kinetic energy of water molecules in the leaf mesophyll, accelerating evaporation into leaf air spaces and increasing the water vapour concentration inside the leaf. Additionally, warm air has greater capacity to hold water vapour, so even at the same absolute humidity, warm air is further from saturation, the atmosphere can absorb much more water vapour, creating a larger gradient driving diffusion through stomata.
Low humidity means the atmosphere has very low water vapour concentration, the water potential difference between the near-saturated leaf interior (~99% humidity) and the dry outside air (~20–30% humidity) is very large, driving rapid diffusion of water vapour through open stomata.
Wind removes the humid boundary layer that would otherwise accumulate just outside the stomata in still conditions. By constantly replacing this humid air with dry bulk air, wind maintains the maximum possible gradient at the stomatal pore throughout the day.
Together, these factors push transpiration rate far above the rate of water uptake from soil via roots, causing a progressive water deficit in leaf and stem cells. As water leaves cells faster than it is replaced, turgor pressure falls, the cells become flaccid, and the plant wilts.
The claim is inaccurate. A potometer measures the rate of water uptake by the cut shoot, not transpiration directly.
Water uptake includes all water absorbed by the shoot, which is then used for: (1) transpiration, evaporation through stomata and cuticle (the dominant pathway, ~95%+ of uptake), (2) photosynthesis, water split in the light-dependent reactions, (3) cell expansion and growth, water retained in vacuoles of growing cells.
Therefore, the potometer slightly overestimates transpiration rate, particularly at conditions that stimulate photosynthesis (bright light, high temperature) where non-transpiratory water use increases. A more accurate statement is that the potometer measures water uptake as an indirect indicator of transpiration rate, valid because transpiration accounts for the vast majority of water uptake in a leafy shoot under normal conditions.
In a xerophyte with sunken stomata, the pores are positioned in recessed pits or crypts below the level of the leaf surface. Water vapour transpired through the stomata accumulates within the crypt rather than being swept away into bulk air. This creates a humid still-air boundary layer immediately outside the stomatal pore, with water vapour concentration significantly higher than the dry bulk atmosphere outside the leaf.
The water potential gradient driving diffusion of water vapour is determined by the difference in water vapour concentration between the leaf air spaces (near-saturated, ~99%) and the air immediately outside the stomatal pore. With sunken stomata, the relevant comparison is between the leaf interior and the humid crypt air, not between the leaf interior and the dry bulk atmosphere. This smaller gradient reduces the driving force for diffusion, slowing transpiration rate.
In effect, the crypt creates a microenvironment that mimics the effect of high humidity outside the leaf, reducing the apparent water potential gradient even when the bulk atmosphere is dry. This adaptation is particularly effective in calm conditions; strong wind may partially disrupt the crypt boundary layer.
Revisit your first answer, then bank the lesson.
Put your knowledge of transpiration, stomatal factors and measurement methods to the test. Answer correctly to deal damage, get it wrong and the boss hits back. Pool: lessons 1–17.
Look back at what you wrote at the start of this lesson. How has your thinking changed? What new connections can you make?
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