The shape of the lesson
By the end of the lesson, learners can
- Evaporate, then diffuse: Watch where the water vapour particles start and how they leave the leaf.
- Built to hold on to water: The lab opens on a marram grass leaf.
- Making an annotated drawing: An annotation is more than a label: it names the feature and explains how it reduces water loss by transpiration.
- Evaporation from mesophyll cell walls into air spaces.
- Diffusion of water vapour out through stomata, down a water potential gradient.
- Humidity down, temperature, wind and light up: rate up.
- Xerophytes: trap humid air near stomata or thicken the cuticle.
One picture every idea returns to
Transpiration is the loss of water vapour from the aerial parts of a plant, mainly the leaves. Water evaporates from the wet surfaces of mesophyll cell walls into the air spaces inside the leaf, and the water vapour then diffuses out through open stomata, down a water potential gradient, to the drier air outside. Low humidity, high temperature, wind and the light that opens stomata all increase the rate. The water lost is replaced from the xylem, which is what pulls the water column up the plant. Xerophytes are plants adapted to dry places. Marram grass reduces transpiration with rolled leaves (rolled by hinge cells), sunken stomata in grooves on the inner surface, hairs that trap moist air and a thick waxy cuticle on the exposed outer surface. Each adaptation either keeps humid air next to the stomata, so the water potential gradient is shallow, or makes the surface less permeable.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Losing water is the price of carbon dioxide.. Stomata must open to let carbon dioxide in, and water vapour escapes through the same pores. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Evaporate, then diffuse | Watch where the water vapour particles start and how they leave the leaf. Project tab 4 of the lab (Transport in plants lab, transpiration) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Built to hold on to water | The lab opens on a marram grass leaf. Project tab 6 of the lab (Transport in plants lab, xerophytic leaves) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Making an annotated drawing | An annotation is more than a label: it names the feature and explains how it reduces water loss by transpiration. | Connect-Extend-Challenge |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 18 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Making an annotated drawing, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Say that water evaporates from the internal surfaces (the mesophyll cell walls) and then water vapour diffuses out through the stomata. | Writing that "water evaporates out of the stomata" or that "water diffuses out of the leaf" merges two steps and loses marks. |
| Xerophyte adaptations reduce water loss; they do not stop it. | The stomata still have to open for carbon dioxide, so some transpiration always happens. |
Transpiration and xerophytic leaves: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
Explain what is meant by transpiration.
Explain why the rate of transpiration increases when the air temperature rises.
In a potometer, the air bubble moved 45 mm along a capillary tube of radius 0.50 mm in 5 minutes. Calculate the rate of water uptake in mm3 per minute. (Volume of a cylinder = πr2l)
Explain why a potometer measures the rate of water uptake rather than the exact rate of transpiration.
Marram grass grows on sand dunes. Describe and explain four features of a marram grass leaf, seen in transverse section, that reduce water loss by transpiration.
A leafy shoot transpired 12 mm3 min-1 in still, dry air. In still, humid air the rate fell to 3 mm3 min-1. Calculate the percentage decrease and explain the change.
Answer key and marking notes
- Water evaporates from the surfaces of mesophyll cell walls into the air spaces of the leaf.
- Water vapour diffuses out of the leaf through open stomata.
- It moves down a water potential gradient from the moist air spaces to the drier atmosphere.
- Water molecules gain kinetic energy, so more water evaporates from the mesophyll cell walls.
- Water vapour molecules diffuse faster.
- Warm air can hold more water vapour, so the air outside has a lower water potential and the gradient is steeper.
- Volume = π × 0.502 × 45 = 35.3 mm3.
- Rate = 35.3 ÷ 5 = 7.07 mm3 per minute.
- Answer: about 7.1 mm3 per minute.
- Some of the water taken up is used by the plant, for example in photosynthesis, or kept in cells to maintain turgor.
- So the volume taken up is not exactly equal to the volume of water vapour lost.
- Rolled leaf (rolled by hinge cells that lose water): traps humid air inside, so the water potential gradient is reduced.
- Sunken stomata in grooves on the inner surface: humid air is held in the pits above them.
- Hairs on the inner surface: trap a layer of still, moist air, reducing diffusion of water vapour.
- Thick waxy cuticle on the outer surface: reduces evaporation through the epidermis exposed to sun and wind.
- Percentage decrease = (12 - 3) ÷ 12 × 100 = 75%.
- Humid air outside the leaf has a higher water potential (more water vapour).
- The water potential gradient from the air spaces to the atmosphere is less steep, so water vapour diffuses out more slowly.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
What is the first step in the loss of water by transpiration from a leaf?
- Awater evaporates from the surfaces of mesophyll cell walls into the air spaces
- Bwater vapour diffuses out through the stomata into the atmosphere
- Cwater moves out of the xylem into the air spaces as liquid
- Dguard cells actively pump water vapour out of the leaf
Under which set of conditions would a well-watered plant transpire fastest?
- Awarm, dry, windy air in bright light
- Bcool, humid, still air in bright light
- Cwarm, humid, still air in darkness
- Dcool, dry, windy air in darkness
In many cacti the leaves are reduced to spines and photosynthesis happens in a thick, swollen stem. How does this reduce water loss?
- Athe plant has a small surface area to volume ratio, so less surface for water to evaporate from
- Bthe spines absorb water vapour from the air and pass it into the stem
- Cthe swollen stem has more stomata than leaves would, so it controls loss better
- Dspines trap a thick layer of wax that seals the stomata completely
In a rolled marram grass leaf, the thick waxy cuticle is on the outer surface, while the stomata are on the inner surface. Why is this arrangement effective?
- Athe outer surface faces dry, moving air, while the stomata open into humid air trapped inside the roll
- Bthe cuticle on the outer surface absorbs sunlight to warm the stomata
- Cthe inner surface needs no cuticle because it carries out no photosynthesis
- Dthe stomata on the inner surface take in water vapour from the trapped air
In a potometer, the air bubble moved 30 mm in 2 minutes along a capillary tube with a cross-sectional area of 0.80 mm2. What was the rate of water uptake?
- A12 mm3 per minute
- B24 mm3 per minute
- C48 mm3 per minute
- D15 mm3 per minute
Four similar leafy shoots were treated as shown and the mass of water each lost in 24 hours was recorded. Which conclusion is best supported?
- Amost water vapour is lost through stomata on the lower surface, and a little through the cuticle
- Bmost water vapour is lost through the upper surface of the leaves
- Cthe cuticle is completely waterproof, so all loss is through stomata
- Dgreasing the lower surface stopped all transpiration from the shoot
The rate of transpiration of a shoot was 4.0 mm3 min-1 in still air. When a fan was switched on, it rose to 10.0 mm3 min-1. What is the percentage increase?
- A150%
- B60%
- C250%
- D40%
Oleander is a shrub of dry regions. Its stomata are found only in deep pits (crypts) in the lower epidermis, and the crypts are lined with hairs. Which explanation of the advantage is correct?
- Ahumid, still air is held in the crypts, so the water vapour gradient across the stomata is reduced
- Bthe hairs absorb water from the air and supply it to the guard cells
- Cthe crypts keep the stomata permanently closed, so no gas exchange occurs
- Dthe hairs increase the surface area for evaporation, cooling the leaf
Quiz key
Transpiration begins with evaporation of water from the wet surfaces of mesophyll cell walls into the air spaces; the vapour then diffuses out through the stomata, which is the second step, not the first. Water leaves the xylem into the mesophyll cells and walls, not into the air spaces as liquid. Guard cells control the size of the stoma but do not pump water vapour.
Warmth increases evaporation and diffusion, dry air and wind keep the water vapour gradient steep, and light keeps the stomata open, so warm, dry, windy and bright gives the fastest rate. Humid, still air makes the gradient shallow. In darkness most stomata close, which cuts transpiration sharply even if the air is warm or windy.
Thin spines and a thick, rounded stem give the plant a small surface area relative to its volume, so less water evaporates from it. Spines do not absorb water vapour. Cactus stems have relatively few stomata, not more. Wax forms a cuticle on the epidermis, but stomata must still open for gas exchange, so they are not sealed completely.
The exposed outer surface meets dry, windy air, so a thick waterproof cuticle there cuts evaporation through the epidermis. The stomata open into the humid air held inside the roll, so the gradient across them is shallow. The cuticle does not warm the stomata, and warming would increase loss. The inner surface does photosynthesise. Stomata release water vapour; they do not take it in.
Volume = 0.80 × 30 = 24 mm3 in 2 minutes, so the rate is 24 ÷ 2 = 12 mm3 per minute. 24 mm3 per minute forgets to divide by the time. 48 multiplies by the time instead of dividing. 15 is the distance per minute (30 ÷ 2) without using the area.
Greasing the lower surface cut the loss from 410 mg to 80 mg, while greasing the upper surface only reduced it to 350 mg, so most vapour leaves through the lower surface, where most stomata are. The upper surface accounts for only about 60 mg. With both surfaces greased, 20 mg was still lost (through the stem and any gaps in the cuticle), so the cuticle is not perfectly waterproof; the 80 mg in C shows loss did not stop.
(10.0 - 4.0) ÷ 4.0 × 100 = 150%. 60% divides the increase by the new value (6 ÷ 10). 250% is the new value as a percentage of the original (10 ÷ 4), not the increase. 40% is the original as a percentage of the new value (4 ÷ 10).
Water vapour collects in the sheltered crypts and the hairs stop it being blown away, so the air next to the stomata stays humid and the gradient for diffusion is shallow, reducing transpiration. The hairs do not absorb water for the guard cells. Stomata must still open for carbon dioxide uptake, so gas exchange continues. Increasing evaporation would increase water loss, the opposite of the adaptation's purpose.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.