Hot, windy or humid — answer key

1. HQ1 · Practice

Model answer. Use the same shoot in the same potometer. Time three runs in still air, then switch on the fan, let the shoot settle for a few minutes, and time three more runs. Only the moving air differs, so any change in the rate was caused by it.

A full answer includes

• The same shoot, timed in still air and then with the fan on, everything else kept the same, with time to settle and three runs each.

Also accept

• same shoot; only the fan differs; settle; three runs each

Watch for

• “Use a fresh shoot for the fan test.” — A different shoot has a different size and number of leaves, so its rate would differ anyway.
• “Switch on the heater too, to make the change bigger.” — Then you could not tell whether the fan or the heat caused the change.

2. HQ2 · Practice

Options as numbered on the pupil sheet: 1 — The variable you measure (dependent variable); 2 — Kept the same; 3 — The variable you change (independent variable).

Model answer. Whether the fan is on is the independent variable: you change it. The distance the bubble moves in 2 minutes, which gives the rate, is the dependent variable: you measure it. The shoot, the light, the temperature and the time of each run are kept the same, so the test is fair.

A full answer includes

• Whether the fan is on: the variable you change (independent).
• The distance the bubble moves in 2 minutes: the variable you measure (dependent).
• The shoot, the light and the time of each run: kept the same.

Also accept

• fan: change; distance: measure; shoot, light, time: keep the same

Watch for

• “The distance moved is the independent variable.” — You do not choose it; you measure it. It is the dependent variable.
• “The shoot can be changed between runs.” — The shoot must stay the same, or its size would change the rate.

3. HQ3 · Practice

Model answer. With the fan on, the bubble moved 48 to 52 mm in 2 minutes, against 22 to 26 mm in still air: about twice as far. So moving air made the rate of water uptake, and so of transpiration, faster.

A full answer includes

• With the fan on, the bubble moved about twice as far in 2 minutes: in this test, moving air made the rate faster.

Also accept

• faster with the fan: about twice the rate

Watch for

• “The fan slowed the rate, because it cooled the leaves.” — The distances were larger with the fan on, so the rate was faster.
• “The fan made no difference.” — The fan runs were about twice as far as the still-air runs.

4. HQ4 · Practice

Model answer. (a) Mean distance = (38 + 40 + 42) ÷ 3 = 120 ÷ 3 = 40 mm. (b) Mean rate = 40 mm ÷ 2 minutes = 20 mm per minute.

Checked automatically

• (a) Mean distance: 40 mm (within ±2%)
  • Common wrong answer 120 : That is the total. Divide by the 3 runs.
• (b) Mean rate of water uptake: 20 mm per minute (within ±2%)
  • Common wrong answer 40 : That is the distance in 2 minutes. Divide by 2.
  • Common wrong answer 60 : Divide the total by 3 runs and by 2 minutes.

A full answer includes

• (a) (38 + 40 + 42) ÷ 3 = 40 mm.
• (b) 40 ÷ 2 = 20 mm per minute.

Also accept

• (a) 40 mm; (b) 20 mm per minute

Watch for

• “The mean distance is 120 mm.” — That is the total of three runs. Divide by 3 for the mean.
• “The mean rate is 40 mm per minute.” — 40 mm is the distance in 2 minutes. Divide by 2.

5. HQ5 · Practice

Model answer. The shoot takes a few minutes to settle at its new rate after a change. If the pupil timed straight away, part of the run would still be at the old rate, so the result would be too low. Waiting 5 minutes means every run measures the new rate.

A full answer includes

• The shoot takes a few minutes to settle at its new rate after a change.
• Timing straight away would give a rate between the old one and the new one.

Also accept

• the shoot needs time to settle at the new rate

Watch for

• “To let the fan warm up.” — The fan works at once; it is the shoot that takes time to reach its new rate.
• “To let the bubble rest.” — The bubble moves all the time the shoot takes up water. The wait is for the shoot.

6. HQ6 · Practice

Model answer. (a) A check run back in the starting conditions shows whether the shoot itself has changed during the experiment, for example by wilting or drying out. (b) The rate is back to the still-air rate, so the shoot has not changed, and the faster rate with the fan on was caused by the moving air. The next test can start from the same place.

A full answer includes

• (a) To see whether the shoot itself has changed (for example, damaged or drying out) during the experiment.
• (b) The shoot has not changed: the faster rate with the fan was caused by the fan.

Also accept

• (a) to check the shoot has not changed; (b) it has not, so the fan caused the faster rate

Watch for

• “The check run is a fourth repeat for the fan.” — It is in still air, not with the fan; it checks the shoot.
• “The check run shows the fan is broken.” — It shows that the shoot still behaves as it did at the start.

7. HQ11 · Practice

Model answer. Humid air already holds a lot of water vapour. The difference in water vapour concentration between the air spaces inside the leaf and the air outside is small, a shallow concentration gradient, so water vapour diffuses out through the stomata more slowly.

A full answer includes

• Humid air already holds a lot of water vapour, so the concentration gradient between the air spaces and the air outside is shallow and water vapour diffuses out more slowly.

Also accept

• shallow concentration gradient of water vapour, so slower diffusion out of the leaf

Watch for

• “Humid air pushes water back into the leaf.” — Nothing is pushed in. Water vapour still diffuses out, only more slowly, because the gradient is shallow.
• “Humid air is cold, so no water evaporates.” — Humid air can be warm. It slows diffusion because it already holds a lot of water vapour.

8. HQ7 · Practice

Model answer. In warmer air, water molecules have more kinetic energy. Water evaporates faster from the surfaces of the mesophyll cells, and the water vapour diffuses out through the stomata faster, so the rate of transpiration increases.

A full answer includes

• Water molecules have more kinetic energy, so water evaporates faster from the mesophyll cells and water vapour diffuses out faster.

Also accept

• more kinetic energy: faster evaporation and diffusion

Watch for

• “The shoot is thirsty in warm air, so it pulls harder.” — A shoot has no muscles. Faster evaporation from the leaves pulls more water up.
• “Warmth always opens the stomata wider.” — Very hot air can close stomata. In moderate warmth the main effect is faster evaporation and diffusion.

9. HQ8 · Practice

Model answer. (a) In still air, water vapour that has diffused out of the stomata collects just outside the leaf. Moving air carries it away. (b) The air outside the stomata stays drier, so the concentration gradient of water vapour between the air spaces and the air outside stays steep, and water vapour diffuses out faster.

A full answer includes

• (a) It carries away the water vapour that collects just outside the stomata.
• (b) The air outside stays drier, so the concentration gradient of water vapour stays steep and water vapour diffuses out faster.

Also accept

• (a) carries water vapour away; (b) keeps the concentration gradient steep, faster diffusion

Watch for

• “The wind cools the leaves, so they lose more water.” — Cooling would slow evaporation. The effect is on the water vapour outside the leaf.
• “The wind pushes water out of the leaves.” — Nothing is pushed. Water vapour diffuses out faster because the gradient is steeper.

10. HQ9 · Practice

Model answer. Humid air makes the concentration gradient of water vapour shallow, so diffusion is slower. Still air lets water vapour collect around the leaf, which also makes the gradient shallow. Cooler air gives the water molecules less kinetic energy, so evaporation and diffusion are slower. Warmer air, a breeze and drier air all speed transpiration up.

A full answer includes

• Slow it down: more humid air; still air instead of a breeze; cooler air.
• Speed it up: warmer air; a breeze; drier air.

Also accept

• more humid, still instead of a breeze, cooler

Watch for

• “Drier air slows transpiration.” — Drier air makes the concentration gradient steeper, so it speeds transpiration up.
• “A breeze slows transpiration by cooling the leaves.” — A breeze carries water vapour away and speeds transpiration up.

11. HQ10 · Practice

Model answer. (a) Water vapour from the leaves was trapped in the bag, so the air around the leaves became very humid. The concentration gradient of water vapour between the air spaces and the air outside became shallow, so water vapour diffused out much more slowly. (b) The bag also changes other things: it stops the air round the leaves from moving, and it may warm them or change the light a little. So it is a demonstration of humid air, not a fair test of humidity alone.

A full answer includes

• (a) Water vapour from the leaves was trapped, so the air in the bag became humid; the concentration gradient became shallow, so water vapour diffused out more slowly.
• (b) Any one: the bag also stops the air moving; it may warm the leaves; it may change the light a little.

Also accept

• (a) humid air, shallow gradient, slower diffusion; (b) the bag also stills the air, or warms, or changes the light

Watch for

• “The bag kept the light off the leaves.” — The bag is clear, so it changes the light only a little. The main change is the humid air inside it.
• “The leaves took the droplets back in.” — The rate fell because less water vapour was leaving, not because water came back.

12. HT1 · Apply

Model answer. (a) The higher the wind speed, the faster the rate of water uptake: 10 mm per minute with the fan off, 16 on low and 22 on high. (b) Any two of: the same shoot in the same potometer; the same air temperature; the same light; the same time for each run; the fan at the same distance from the leaves; reading the same edge of the bubble.

A full answer includes

• (a) The higher the wind speed, the faster the rate (10, 16, then 22 mm per minute).
• (b) Any two: the same shoot; the same temperature; the same light; the same time for each run; the same distance from the fan; reading the same edge of the bubble.

Also accept

• (a) higher wind speed, faster rate; (b) two of: same shoot, temperature, light, time, fan distance, bubble edge

Watch for

• “The rate doubles each time the speed goes up.” — It goes from 10 to 16 to 22: it rises by about the same amount each time, not doubling.
• “Keep the wind speed the same.” — Wind speed is what they are changing. Keep the other conditions the same.

13. HT2 · Apply

Model answer. A cutting has no roots yet, so it can take up only a little water. Under a cover, or in a mist, the air around the leaves becomes humid. The concentration gradient of water vapour between the air spaces and the air outside is then shallow, so water vapour diffuses out more slowly, and transpiration is slow. The cutting loses water no faster than it can take it in, so it does not wilt or dry out while its roots grow.

A full answer includes

• The air around the cuttings becomes humid.
• The concentration gradient of water vapour is shallow, so less water vapour diffuses out of the leaves: slower transpiration.
• So the cuttings do not lose water faster than they can take it up, and they do not wilt (or dry out).

Also accept

• humid air, shallow gradient, slower transpiration, so the cutting does not lose water faster than it takes it up

Watch for

• “The mist waters the leaves, which drink it.” — Leaves do not take in much water this way. The mist makes the air humid, so less water is lost.
• “The cover keeps the cuttings warm, so they transpire faster and grow faster.” — Faster transpiration would dry a rootless cutting out. The point is to slow water loss.

14. HT3 · Apply

Model answer. In the cool corner with still air. Lower temperature and still air both make transpiration slower, so the plants take up less water from the soil, and the soil dries out most slowly.

A full answer includes

• In a cool corner where the air is still.

Also accept

• cool and still

Watch for

• “Next to the heater, because there is no draught.” — Still air slows transpiration, but warmth speeds it up. Cool and still is slowest.
• “Under the cool vent, because cool air slows it.” — Cool air slows transpiration, but moving air speeds it up. Cool and still is slowest.