Answer key · Unit 12

Plants.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 5

Lesson 1 of 5 · About 25 minutes

Water through a tree

A tree lifts tens of litres of water an hour from the soil to its leaves, and almost all of it leaves as vapour: through the stomata, which the plant opens in the light and closes in the dark, pulled from above rather than pushed from below.

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The 5 steps and their answers

Step 1 of 5
Open this step
A tree that drinks through its leaves
Practice · Calculate
Question
In sunlight of 600 W/m², in air at 50% humidity with a light breeze, the chart shows how much water the tree lifts each hour. How much water passes through it in $10$ ⁠ hours of such sun?

Correct answer
104.7 L ± 5%
Setup, choices and feedback
The Scene at this step
Humidity
50 %fixed
Sunlight
600 W/m²fixed
Wind
2 m/sfixed
Follow the water down
Tree
Accepted within
± 5%
Computed from the Scene at this step’s setting
Water the tree lifts, L/h/nodes/tree/outputs/flow
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 105 L: 10.5 L an hour times $10$ ⁠ hours. That is more than a half-full bathtub, and a really big tree, with several times the leaf area, lifts hundreds of litres a day. Almost all of this water leaves the leaves as vapour.

Hint
Multiply what passes in an hour by the number of hours.

Worked solution
1. From the chart: the tree lifts about $10.5$ ⁠ litres an hour.
2. Amount = rate times time.
3. $10.5 \times 10 = 105$ ⁠.
4. About $105$ ⁠ litres.
If you answered $10.5$ ⁠, you gave the rate, what passes in one hour, not what passes in ten.
Step 2 of 5
Open this step
Out through the leaves
Practice · Choose
Question
Almost all the water the tree lifts leaves it through the leaves. How does it get out?

Correct answer
- Correct As water vapour, through the stomata, tiny pores in the leaf
Setup, choices and feedback
The Scene at this step
Humidity
50 %
Sunlight
600 W/m²
Wind
2 m/s
Follow the water down
Tree
Choices
Shown in a random order.
- Correct As water vapour, through the stomata, tiny pores in the leaf
- Wrong As water vapour evaporating from the whole surface of the leaf, like a wet cloth
  Misconception: Thinks a leaf loses its water by evaporation from its whole surface, like a wet cloth.
  Feedback from the catalogue
  A leaf is wrapped in a cuticle, a layer of wax that lets almost no water through. The water evaporates inside the leaf, from the wet walls of the spongy cells, and the vapour leaves almost only through the stomata, tiny pores that a pair of guard cells opens and closes. That is why a plant can control its water loss: it closes its stomata.
- Wrong It does not get out: most of it is built into sugar by photosynthesis
  Misconception: Thinks most of the water a plant takes up is used to make food in photosynthesis.
  Feedback from the catalogue
  Only a very small share of the water a plant takes up, about one per cent, is built into sugar by photosynthesis or stays in its cells. All the rest, about 99%, leaves the leaves as water vapour, in transpiration. The tree does not lift the water to keep it: the water passes through on its way from the soil to the air, carrying minerals up from the roots and cooling the leaves on the way.
- Wrong As liquid water dripping from the ends of the veins
  Feedback
  Some plants drip from the tips of their leaves on damp nights, but those are tiny amounts. By day almost all of it leaves as vapour.
Feedback for a right answer
The water evaporates inside the leaf, from the wet walls of the spongy cells, and the vapour leaves through the stomata. The rest of the leaf is covered in a layer of wax that lets almost no water through, so the stomata are the door, and the guard cells open and close it.

Hint
What covers the outside of the leaf?

Worked solution
1. The leaf is wrapped in a cuticle, a layer of wax that lets almost no water through.
2. Inside the leaf, between the spongy cells, are air spaces saturated with vapour.
3. The vapour leaves those spaces through the stomata.
4. Move Follow the water down to Leaf and you see the water molecules leaving through the stomata.
If you chose photosynthesis: only about one per cent of the water is built into sugar. If you chose the whole surface: the cuticle stops almost all the evaporation from there.
Step 3 of 5
Open this step
Night
Practice · Predict, then look
Question
The sun sets: Sunlight drops to $0$ ⁠. What happens to the stomata, and to the water the tree loses?

Correct answer
- Correct The stomata nearly close, and the tree loses only a little water
Setup, choices and feedback
The Scene at this step
Humidity
50 %
Sunlight
600 W/m²fixed
Wind
2 m/s
Follow the water down
Leaf
Choices
- Correct The stomata nearly close, and the tree loses only a little water
- Wrong Nothing: water keeps evaporating from the leaves, day and night
  Misconception: Thinks transpiration is plain evaporation the plant cannot control.
  Feedback from the catalogue
  A plant controls its transpiration. The guard cells open the stomata in the light, to let carbon dioxide in for photosynthesis, and close them in the dark, or when the leaf starts to dry out. In the Scene the stomata are only 3% open at night, and the tree lifts less than a tenth of the water it lifts in the sun.
- Wrong The stomata shut completely, and the tree stops losing water
  Feedback
  Close, but the flow does not stop completely: the cuticle leaks a little, and the stomata do not seal all the way.
Once the prediction is locked in, the Scene changes
Sunlight
0 W/m²fixed
The Scene then shows: 0.03 · How far the stomata are open, 0..1/nodes/tree/outputs/opening
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The guard cells close the stomata almost completely, from 83% to 3% open, and the tree lifts only about 0.6 L an hour instead of 10.5 L. A little water still escapes, through the cuticle and through stomata that are not shut all the way. By day the plant opens its stomata to let carbon dioxide in for photosynthesis, and at night it has no reason to pay for that in water.

Worked solution
1. In the light, the guard cells swell and open the stoma, so that carbon dioxide can get into the leaf.
2. In the dark there is no photosynthesis, and the guard cells go slack and close it.
3. On the chart: the stomata are only 3% open, and the flow drops to 0.6 L an hour, about $\tfrac{1}{17}$ ⁠ of the daytime flow.
4. The flow does not reach zero: the cuticle and the stomata that are not fully shut still leak a little.
If you predicted that nothing would change, you thought of transpiration as plain evaporation, but the plant controls it with its stomata.
Step 4 of 5
Open this step
What lifts the water
Practice · Choose
Question
At night the roots are still in moist soil, and the vessels in the trunk are still full of water, yet the flow up the trunk almost stops. What does that tell you about what lifts the water 10 m?

Correct answer
- Correct The water is pulled from above: evaporation from the leaves pulls the whole column
Setup, choices and feedback
The Scene at this step
Humidity
50 %
Sunlight
0 W/m²fixed
Wind
2 m/s
Follow the water down
Trunk
Choices
Shown in a random order.
- Correct The water is pulled from above: evaporation from the leaves pulls the whole column
- Wrong The roots push the water up, like a pump
  Misconception: Thinks the roots pump the water up the tree.
  Feedback from the catalogue
  The roots do not push the water to the top of the tree: the water is pulled from above. As water evaporates from the leaves it pulls the water in the xylem vessels after it, and water molecules hold on to each other by hydrogen bonds, so the pull is passed down the whole column to the roots. That is why the flow almost stops at night, when the evaporation stops, even though the roots are still in moist soil. Some plants push a little water up by root pressure, mostly at night and in spring, but it is far too weak to supply tens of litres an hour on a sunny day.
- Wrong The vessels are thin, and the water climbs them by itself, by capillarity
  Misconception: Thinks capillarity in the thin vessels lifts the sap to the top of the tree.
  Feedback from the catalogue
  Capillarity does lift water in a thin tube, but not far: in a xylem vessel a tenth of a millimetre wide, only about 30 cm. Besides, capillarity does not switch off at night, and the flow in the tree does. What lifts the water 10 m and more is the evaporation from the leaves, pulling on an unbroken column of water.
Feedback for a right answer
The water is pulled from above. As it evaporates from the leaves it pulls on the water below it, and water molecules hold on to each other by hydrogen bonds, so the pull is passed down the whole column to the roots. This is the cohesion–tension theory. At night the evaporation almost stops, and the pull with it.

Hint
What would keep working at night, and what stopped?

Worked solution
1. What changed at night? Not the roots, not the vessels, only the light, and with it the evaporation from the leaves.
2. The flow stopped along with the evaporation, so the evaporation is what drives it.
3. A pump in the roots, or capillarity in the thin vessels, would keep working at night too.
4. Even without this experiment: capillarity lifts water only about 30 cm in a xylem vessel, not 10 m.
Some plants have root pressure, but it is weak and works mainly when the tree is hardly losing any water.
Step 5 of 5
Open this step
How fast
Practice · Calculate
Question
The chart shows that in sunlight of 600 W/m² the sap climbs the trunk at 2.6 m an hour. How long does water take to get from the roots to the top leaves, 10 m above them?

Correct answer
3.82 h ± 5%
Setup, choices and feedback
The Scene at this step
Humidity
50 %fixed
Sunlight
600 W/m²fixed
Wind
2 m/sfixed
Follow the water down
Trunk
Accepted within
± 5%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
$\dfrac{10}{2.6} \approx 3.8$ ⁠ hours. Water that enters a root in the morning leaves the top leaves before noon. In tall trees, with narrower vessels, the journey can take a whole day.

Hint
Time = distance divided by speed.

Worked solution
1. The givens: distance $d = 10\ \mathrm{m}$ ⁠, speed $v = 2.6\ \mathrm{m}$ ⁠ an hour.
2. Time = distance divided by speed: $t = \dfrac{d}{v}$ ⁠.
3. Substitute: $t = \dfrac{10}{2.6} \approx 3.8$ ⁠.
4. About $3.8$ ⁠ hours.
If you got $26$ ⁠, you multiplied the speed by the distance. A higher speed means a shorter time, so you divide.

Lesson 2 of 5 · About 25 minutes

What speeds up transpiration

Dry air, wind and sunshine speed up transpiration, and moist air slows it, even when the leaf warms up. What drives the flow is the difference between the saturated air inside the leaf and the air outside, and when the tree closes its stomata to keep from drying out.

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The 6 steps and their answers

Step 1 of 6
Open this step
Dry air
Practice · Predict, then look
Question
Humidity drops from 50% to 15%, as in a desert, and everything else stays the same. What happens to the amount of water the tree lifts?

Correct answer
- Correct The tree lifts more water
Setup, choices and feedback
The Scene at this step
Humidity
50 %fixed
Sunlight
600 W/m²fixed
Wind
2 m/sfixed
Follow the water down
Tree
Choices
- Correct The tree lifts more water
- Wrong The tree lifts less water, to save it
  Feedback
  A plant does close its stomata when its leaf dries out, but only when the pull is very strong. Here the bigger difference in humidity simply speeds up the evaporation.
- Wrong The same amount: the humidity outside does not matter to the leaf
  Feedback
  The humidity outside is half the story: the vapour leaves by diffusion, and the rate of diffusion depends on the difference between inside and out.
Once the prediction is locked in, the Scene changes
Humidity
15 %fixed
The Scene then shows: 15.21 · Water the tree lifts, L/h/nodes/tree/outputs/flow
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It goes up, from 10.5 L to about 15.2 L an hour. The air inside the leaf is saturated with vapour. The drier the air outside, the bigger the difference between them, and the faster the vapour escapes, just as washing dries faster on a dry day. On the chart you see it as the air’s water potential falling from −95 MPa to −260 MPa: the slope from the leaf outwards is steeper.

Worked solution
1. Inside the leaf the air is saturated, close to 100% humidity.
2. Vapour leaves the leaf by diffusion, faster the bigger the difference in vapour concentration.
3. At 50% humidity the difference is about half of saturation; at 15% it is nearly all of it.
4. So transpiration rises: from 10.5 L to about 15.2 L an hour.
If you predicted a drop, perhaps you thought the plant would close its stomata. It does that only when the leaf starts to dry out, and in these conditions it is not there yet.
Step 2 of 6
Open this step
Muggy air
Practice · Predict, then look
Question
Humidity rises to 90%, muggy air. On the chart you will see the leaf warm up by almost two degrees, because it evaporates less, and water that does not evaporate does not cool it. Will the warmer leaf lose more water or less?

Correct answer
- Correct Less water
Setup, choices and feedback
The Scene at this step
Humidity
50 %fixed
Sunlight
600 W/m²fixed
Wind
2 m/sfixed
Follow the water down
Tree
Choices
- Correct Less water
- Wrong More water, because it is warmer
  Misconception: Thinks transpiration follows the leaf’s temperature: a warmer leaf always loses more water.
  Feedback from the catalogue
  What drives transpiration is the difference between the saturated air inside the leaf and the air outside, not the warmth alone. In muggy air the difference is small, so the leaf loses less water even though it warms up, because water that does not evaporate does not cool it. In wind the leaf cools down, but the wind sweeps away the still, moist layer of air on it, and it loses more. Warmth helps, since warm air can hold more vapour, but humidity and wind decide whether the vapour gets out.
- Wrong The same: the two changes cancel out
  Feedback
  They do not cancel: the warming is small, and the difference in humidity drops a lot.
Once the prediction is locked in, the Scene changes
Humidity
90 %fixed
The Scene then shows: 1.835 · Water lost per m² of leaf, mmol m⁻² s⁻¹/nodes/tree/outputs/rate
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Less: 1.8 millimoles a second from each square metre of leaf, instead of 4.0, even though the leaf is warmer, 27.2 °C instead of 25.6 °C. The muggy air is almost as saturated as the air inside the leaf, so the vapour has almost nowhere to go. Warmth alone does not drive transpiration: the difference in humidity does.

Worked solution
1. What drives transpiration: the difference between the vapour inside the leaf (saturated) and the vapour in the air.
2. At 90% humidity the difference is much smaller than at 50%.
3. The leaf is warmer, and warm air can hold a little more vapour, but that does not make up for the difference that has gone.
4. On the chart: the rate drops from $4.0$ ⁠ to $1.8$ ⁠ millimoles a second per square metre.
If you predicted that the warm leaf would lose more, you took warmth to be what drives transpiration. Here you see the reverse: the leaf is warmer because it evaporates less.
Step 3 of 6
Open this step
Half the flow
Practice · Reach the target
Question
At 50% humidity the tree lifts about 10.5 L an hour. Sunlight and Wind are now fixed. Change Humidity until the tree lifts about half, 5.2 L an hour, then press Check.

Goal
- 5.2 L/h ± 0.3 L/h
  At the start of the step: 10.47 L/h
Setup, choices and feedback
The Scene at this step
Humidity
50 %
Sunlight
600 W/m²fixed
Wind
2 m/sfixed
Follow the water down
Tree
Read from the Scene at
Water the tree lifts, L/h/nodes/tree/outputs/flow
One setting that reaches it
Humidity
85 %
Feedback for a right answer
Around 87%. To halve transpiration you have to halve the difference between the saturated air in the leaf and the air outside, and you might think 75% would do it. But a leaf that evaporates less warms up, and the warmer air inside it holds more vapour, so you need even moister air.

Feedback for a wrong answer
Not yet. Check on the chart how many litres an hour the tree lifts, and keep changing Humidity.

Hint
Start at 75% and keep going up.

Worked solution
1. Transpiration is roughly proportional to the difference between the saturated air in the leaf and the air outside.
2. At 50% the difference is half of saturation; it halves at 75%.
3. Try 75%: the tree still lifts about 6.9 L an hour, more than half.
4. The reason: the leaf warms as it evaporates less, and the air inside it holds more vapour.
5. Keep raising Humidity to about 87%, where the flow is about 5.2 L an hour.
Step 4 of 6
Open this step
Wind
Practice · Predict, then look
Question
The air is still. What happens to the amount of water the tree lifts when Wind rises to 8 m/s, a strong wind?

Correct answer
- Correct The tree lifts more water
Setup, choices and feedback
The Scene at this step
Humidity
50 %fixed
Sunlight
600 W/m²fixed
Wind
0 m/sfixed
Follow the water down
Tree
Choices
- Correct The tree lifts more water
- Wrong The tree lifts less water, because the wind cools the leaves
  Misconception: Thinks transpiration follows the leaf’s temperature: a warmer leaf always loses more water.
  Feedback from the catalogue
  What drives transpiration is the difference between the saturated air inside the leaf and the air outside, not the warmth alone. In muggy air the difference is small, so the leaf loses less water even though it warms up, because water that does not evaporate does not cool it. In wind the leaf cools down, but the wind sweeps away the still, moist layer of air on it, and it loses more. Warmth helps, since warm air can hold more vapour, but humidity and wind decide whether the vapour gets out.
- Wrong The same: the wind does not touch the air inside the leaf
  Feedback
  True, the wind does not get into the leaf, but the vapour also has to cross the thin layer of still air on it, and the wind clears that away.
Once the prediction is locked in, the Scene changes
Wind
8 m/sfixed
The Scene then shows: 11.57 · Water the tree lifts, L/h/nodes/tree/outputs/flow
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It goes up, from 8.4 L to about 11.6 L an hour, even though the wind cools the leaf by about two degrees. On a leaf in still air sits a thin layer of moist air that does not move, and the vapour has to creep through it. The wind sweeps that layer away, and the vapour escapes faster.

Worked solution
1. In still air a layer of nearly saturated air stays on the leaf, and it slows the diffusion.
2. Wind pushes that layer away, and drier air reaches the leaf’s surface.
3. So the flow rises from 8.4 L to about 11.6 L an hour.
4. On the chart you see the leaf cool down too, from about 27.4 °C to about 25.1 °C, and still lose more water.
If you predicted a drop because the leaf cools, you let warmth decide. Here it is the other way round: the leaf cools, and loses more.
Step 5 of 6
Open this step
Saving water
Practice · Reach the target
Question
On an ordinary sunny day the tree keeps its stomata as open as the light allows. Find conditions, with any of the three controls, in which the leaves start to close their stomata to save water, then press Check.

Goal
- 4 ± 0
  At the start of the step: 3
Setup, choices and feedback
The Scene at this step
Humidity
50 %
Sunlight
600 W/m²
Wind
2 m/s
Follow the water down
Tree
Read from the Scene at
1 dark (stomata shut), 2 muggy air (humidity 85% or more), 3 pulling, 4 closing the stomata to save water/nodes/tree/outputs/regime
One setting that reaches it
Humidity
10 %
Sunlight
1000 W/m²
Wind
3.4 m/s
Feedback for a right answer
Strong sun, dry air and wind, for example 1000 W/m², 15% and 8 m/s. Transpiration is so fast that the water in the leaf is pulled too hard: the leaf’s water potential falls to about −1.4 MPa, and the guard cells start to close. It is a trade-off: half-closed stomata save water, but they also let less carbon dioxide in for photosynthesis.

Feedback for a wrong answer
Not yet: the leaves still open their stomata as far as the light allows. Speed up the transpiration further.

Hint
Think of what dries washing fastest, and push all three to the limit.

Worked solution
1. What makes a leaf dry out: transpiration faster than the xylem can bring water.
2. Fast transpiration: plenty of light (open stomata and a warm leaf), dry air (a big difference), and wind (a thin layer of still air).
3. Raise Sunlight to 1000 W/m², lower Humidity to 15%, and raise Wind to 8 m/s.
4. On the chart the stomata’s bar falls below the dashed mark of how far the light alone would open them: they are closing to save water.
Step 6 of 6
Open this step
The whole tree
Practice · Calculate
Question
The chart shows that each square metre of leaf loses 4.0 millimoles of water a second. The tree has 40 m² of leaves, and a millimole of water weighs 18 mg. How many grams of water does the whole tree lose each second? Answer in grams.

Correct answer
2.909 ± 5%
Setup, choices and feedback
The Scene at this step
Humidity
50 %fixed
Sunlight
600 W/m²fixed
Wind
2 m/sfixed
Follow the water down
Tree
Accepted within
± 5%
Computed from the Scene at this step’s setting
Water the tree lifts, L/h/nodes/tree/outputs/flow
Feedback for a right answer
About 2.9 g a second: $4.0 \times 18 = 72$ ⁠ milligrams from each square metre, times $40$ ⁠ square metres. In an hour that is $2.9 \times 3600 \approx 10\,500$ ⁠ grams, about 10.5 L, exactly what the chart shows.

Hint
Work out one square metre first, then multiply by the leaf area.

Worked solution
1. One square metre: $4.0$ ⁠ millimoles a second, times 18 mg a millimole, is $72$ ⁠ milligrams a second.
2. The whole tree: $72 \times 40 = 2880$ ⁠ milligrams a second.
3. In grams: $2880 \div 1000 \approx 2.9$ ⁠.
4. Check: $2.9 \times 3600 \approx 10\,400$ ⁠ grams, about 10.5 L an hour, as on the chart.
If you got about $0.07$ ⁠, you worked out only one square metre of leaf.

Lesson 3 of 5 · About 25 minutes

Growing towards the light

A plant grows only at its tips, and turns to the light when the shaded side of its shoot grows faster: auxin flows to it from the tip. The shoot stops between the lamp and upright, where the pull of the light and gravity balance.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A plant that grows at its tips
Practice · Calculate
Question
The seedling’s stem, the hypocotyl, was 5.5 cm long, and it grows 1.2 mm an hour. How long is it after a day, $24$ ⁠ hours?

Correct answer
8.38 cm ± 0.1 cm
Setup, choices and feedback
The Scene at this step
Lamp direction
0 °fixed
Box tilt
0 °fixed
Time since the change
24 hfixed
Zoom to the root tip
Seedling
Accepted within
± 0.1 cm
Computed from the Scene at this step’s setting
Length of the shoot (hypocotyl)/nodes/seedling/outputs/shootLength
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 8.4 cm: it grows 2.9 cm in a day, more than half its length. All of that growth happens in the top two centimetres, in the growing zone below the tip, where the cells lengthen as they take water into their vacuoles.

Hint
First convert millimetres to centimetres.

Worked solution
1. Growth an hour: 1.2 mm, that is 0.12 cm.
2. In a day: $0.12 \times 24 = 2.88$ ⁠ centimetres.
3. The new length: $5.5 + 2.88 \approx 8.4$ ⁠ centimetres.
If you answered $2.9$ ⁠, you gave only the growth, without the length it already had. If you got $34$ ⁠, you forgot to convert millimetres to centimetres.
Step 2 of 5
Open this step
Towards the lamp
Practice · Predict, then look
Question
The seedling grew straight up under the lamp. Now the lamp moves 70° to the left. Which way will the tip of the shoot lean after three hours?

Correct answer
- Correct Towards the lamp, to the left
Setup, choices and feedback
The Scene at this step
Lamp direction
-70 °fixed
Box tilt
0 °fixed
Time since the change
0 hfixed
Zoom to the root tip
Seedling
Choices
- Correct Towards the lamp, to the left
- Wrong Away from the lamp, to the right
  Misconception: Thinks the side of the stem facing the light grows faster and pulls the stem over.
  Feedback from the catalogue
  If the lit side grew faster, the stem would curve away from the light. A stem turns to the light because its shaded side grows faster: light moves the carrier proteins in the cells near the tip, and more auxin flows down the shaded side. Auxin loosens the cell walls, and the cells there stretch more. The longer side is the outside of the curve, and it faces away from the lamp.
- Wrong It stays upright: a plant cannot move
  Feedback
  A plant does not move from its spot, but it can grow faster on one side, and so curve.
Once the prediction is locked in, the Scene changes
Time since the change
3 hfixed
The Scene then shows: -30.14 ° · Shoot tip, degrees from straight up, positive to the right/nodes/seedling/outputs/tipAngle
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Towards the lamp, to the left: after three hours the tip already leans about 30° from upright. For the first half hour nothing shows, because the auxin needs time to move to the shaded side, and only then does the shoot start to curve. This is phototropism, growth towards light.

Worked solution
1. The tip of the shoot senses which way the light comes from.
2. After about half an hour, auxin starts to flow down the shaded side, and the cells there lengthen more.
3. A side that grows faster becomes the outside of the curve, so the shoot curves away from it, towards the light.
4. On the chart: after three hours the tip is at −30°, that is 30° to the left.
A plant cannot walk to the light, but it can grow towards it.
Step 3 of 5
Open this step
Auxin changes sides
Practice · Predict, then look
Question
The lamp is on the left. Which side of the shoot will have more auxin, and which side will lengthen more?

Correct answer
- Correct The shaded side, away from the lamp
Setup, choices and feedback
The Scene at this step
Lamp direction
-70 °fixed
Box tilt
0 °fixed
Time since the change
0 hfixed
Zoom to the root tip
Seedling
Choices
- Correct The shaded side, away from the lamp
- Wrong The lit side, facing the lamp
  Misconception: Thinks the side of the stem facing the light grows faster and pulls the stem over.
  Feedback from the catalogue
  If the lit side grew faster, the stem would curve away from the light. A stem turns to the light because its shaded side grows faster: light moves the carrier proteins in the cells near the tip, and more auxin flows down the shaded side. Auxin loosens the cell walls, and the cells there stretch more. The longer side is the outside of the curve, and it faces away from the lamp.
- Wrong The auxin is shared evenly, and the whole stem lengthens alike
  Feedback
  If both sides grew alike, the shoot would grow straight. To curve, one side has to grow faster.
Once the prediction is locked in, the Scene changes
Time since the change
3 hfixed
Zoom to the root tip
Shoot
The Scene then shows: 0.6576 · Share of the shoot’s auxin on the side growing faster, 0.5 (even) to 0.7/nodes/seedling/outputs/auxinShare
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The shaded side, the right: about 66% of the auxin flows down it, and its cells lengthen more, up to about 9% more on the outside of the bend. The longer side is the outside of the curve, so the shoot leans towards the lamp.

Worked solution
1. Light moves the carrier proteins in the cells near the tip, and more auxin flows down the shaded side.
2. Auxin loosens the cell walls, and the cells there stretch more.
3. Move Zoom to the root tip to Shoot: the chart shows about 66% of the auxin on the side growing faster, and the cells on the outside are longer.
4. A longer side becomes the outside of the curve, so the shoot turns towards the shorter, lit side.
If you chose the lit side, you imagined the light pulling the shoot. But a side that grows faster pushes the shoot away from itself.
Step 4 of 5
Open this step
Where the shoot stops
Practice · Predict, then look
Question
The seedling grew straight up, with the lamp overhead. Now the lamp moves $70^\circ$ ⁠ to the right. Where will the tip of the shoot point after a day, once it stops curving?

Correct answer
- Correct Between the lamp and upright
Setup, choices and feedback
The Scene at this step
Lamp direction
0 °fixed
Box tilt
0 °fixed
Time since the change
0 hfixed
Zoom to the root tip
Seedling
Choices
- Correct Between the lamp and upright
- Wrong Straight at the lamp, $70^\circ$ ⁠ to the right
  Misconception: Thinks a shoot ends up pointing straight at the lamp.
  Feedback from the catalogue
  A shoot feels two pulls at once: towards the light, and upwards, against gravity. It stops where they balance, between the lamp and straight up. In the Scene the pull towards the light weighs three times as much, so $\tan\theta = \dfrac{3\sin\varphi}{1 + 3\cos\varphi}$ ⁠: a lamp $70^\circ$ ⁠ from overhead gives a lean of about $54^\circ$ ⁠, not $70^\circ$ ⁠.
- Wrong Past the lamp, because the shoot overshoots a little
  Feedback
  As the shoot nears its new direction the pulls nearly cancel and the auxin evens out again, so it slows down and stops without passing the balance point.
Once the prediction is locked in, the Scene changes
Lamp direction
70 °fixed
Time since the change
24 hfixed
The Scene then shows: 54.28 ° · Shoot tip, degrees from straight up, positive to the right/nodes/seedling/outputs/tipAngle
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Between the lamp and upright, at about $54^\circ$ ⁠ to the right. The shoot feels two pulls: towards the light, and upwards, against gravity. It stops where they balance, and the dashed line on the chart marks the spot.

Worked solution
1. Each pull grows with the sine of the angle between the shoot and its direction. In the Scene the pull towards the light weighs three times as much.
2. The shoot stops where $\tan\theta = \dfrac{3\sin\varphi}{1 + 3\cos\varphi}$ ⁠.
3. With $\varphi = 70^\circ$ ⁠: $\sin\varphi \approx 0.940$ ⁠ and $\cos\varphi \approx 0.342$ ⁠, so $\tan\theta = \dfrac{2.82}{2.03} \approx 1.39$ ⁠.
4. $\theta \approx 54^\circ$ ⁠: between upright and the lamp, nearer the lamp.
If you predicted that the shoot would point straight at the lamp, you forgot gravity, which keeps pulling it upright.
Step 5 of 5
Open this step
A lean of 45 degrees
Practice · Reach the target
Question
Box tilt and Time since the change are fixed. Move Lamp direction so that the shoot settles leaning $45^\circ$ ⁠ to the right, then press Check.

Goal
- 45 ° ± 1 °
  At the start of the step: 0 °
Setup, choices and feedback
The Scene at this step
Lamp direction
0 °
Box tilt
0 °fixed
Time since the change
24 hfixed
Zoom to the root tip
Seedling
Read from the Scene at
Where the shoot tip settles, degrees from straight up/nodes/seedling/outputs/settleAngle
One setting that reaches it
Lamp direction
58 °
Feedback for a right answer
Around $58^\circ$ ⁠ to $59^\circ$ ⁠ to the right. The lamp has to be further from overhead than the lean you want, because gravity always takes back part of the way.

Feedback for a wrong answer
Not yet. Check the dashed line on the chart, where the shoot settles, and move the lamp.

Hint
The lamp has to be more than $45^\circ$ ⁠ from overhead.

Worked solution
1. You want $\theta = 45^\circ$ ⁠, that is $\tan\theta = 1$ ⁠.
2. So $3\sin\varphi = 1 + 3\cos\varphi$ ⁠, or $\sin\varphi - \cos\varphi = \tfrac13$ ⁠.
3. $\sin\varphi - \cos\varphi = \sqrt2\,\sin(\varphi - 45^\circ)$ ⁠, so $\sin(\varphi - 45^\circ) = \dfrac{1}{3\sqrt2} \approx 0.236$ ⁠.
4. $\varphi - 45^\circ \approx 13.6^\circ$ ⁠, that is $\varphi \approx 58.6^\circ$ ⁠.
5. Set Lamp direction to $58^\circ$ ⁠ or $59^\circ$ ⁠; the dashed line on the chart shows where the shoot settles.

Lesson 4 of 5 · About 20 minutes

Which way is down

A seedling laid on its side lifts its shoot and turns its root down within hours: heavy grains of starch in the root cap show it where down is, and the same auxin speeds up the shoot and slows the root. The bend stays in the stem, even once the tip is straight.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
On its side
Practice · Predict, then look
Question
The lamp is overhead, and the box has been laid on its side, so the shoot points right and the root points left. What happens over the next three hours?

Correct answer
- Correct The shoot curves up, and the root down
Setup, choices and feedback
The Scene at this step
Lamp direction
0 °fixed
Box tilt
90 °fixed
Time since the change
0 hfixed
Zoom to the root tip
Seedling
Choices
- Correct The shoot curves up, and the root down
- Wrong Both droop down, under their own weight
  Misconception: Thinks a root bends down because gravity pulls on its weight, as a rope sags.
  Feedback from the catalogue
  A root does not flop down like a rope: it grows down. It bends only in the growing zone behind its tip, and keeps the bend after it hardens. And if gravity simply pulled on it, it would pull the shoot down too, but instead the shoot rises. The bend is uneven growth of the root’s two sides, steered by the statoliths and auxin.
- Wrong Both keep growing sideways, the way they point
  Feedback
  A plant does not simply carry on the way it points: it senses gravity and corrects its course within hours.
Once the prediction is locked in, the Scene changes
Time since the change
3 hfixed
The Scene then shows: 40.23 ° · Shoot tip, degrees from straight up, positive to the right/nodes/seedling/outputs/tipAngle
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The shoot curves up and the root curves down. After three hours the shoot’s tip has already risen from 90° to about 40° from upright, and the root points about 28° from straight down. This is gravitropism: growth guided by gravity, the root with it and the shoot against it.

Worked solution
1. Shoots and roots sense where down is, with heavy grains in special cells.
2. The shoot grows against gravity, so it curves up; the root grows with it, so it curves down.
3. On the chart: after three hours the shoot’s tip is 40° from upright, on its way back to vertical.
A plant does not need to see to know where down is: in the dark too, roots turn down and shoots turn up.
Step 2 of 5
Open this step
How to tell where down is
Practice · Choose
Question
The root of the tipped seedling turned down within hours. How does the root tip know where down is?

Correct answer
- Correct Grains of starch in the root cap’s cells sink to the low side of each cell
Setup, choices and feedback
The Scene at this step
Lamp direction
0 °fixed
Box tilt
90 °fixed
Time since the change
3 h
Zoom to the root tip
Seedling
Choices
Shown in a random order.
- Correct Grains of starch in the root cap’s cells sink to the low side of each cell
- Wrong Auxin is heavy, and it drips to the low side by itself
  Misconception: Thinks auxin gathers on the low side because it is heavy and sinks, like a liquid.
  Feedback from the catalogue
  Auxin does not sink: there is only a tiny amount of it, dissolved in the cells, and it moves from cell to cell by carrier proteins in their membranes. What sinks are the statoliths, dense grains of starch in the cells of the root cap. Within minutes they settle on the low side of each cell, and the cells then point their carrier proteins so that more auxin goes to the low side of the root.
- Wrong The root hairs feel the soil and sense where it gets deeper
  Feedback
  Root hairs take up water and minerals, but they grow behind the growing zone, and a root turns down in moist air too, with no soil at all.
- Wrong The root escapes the light from above
  Feedback
  Roots turn down in complete darkness too. They sense gravity, not light.
Feedback for a right answer
The cells in the middle of the root cap hold statoliths, grains of starch denser than the rest of the cell. Within minutes they sink to the low side of each cell, like sand in a bottle, and the cells steer the auxin to the root’s low side accordingly. Move Zoom to the root tip to Root tip to see them.

Hint
Look for something inside the tip’s cells that sinks.

Worked solution
1. To know where down is, you need something heavy that can sink.
2. The cells of the root cap hold such grains of starch, the statoliths.
3. When the root is tipped, they sink to the new low side within minutes.
4. The cells point their carrier proteins so that more auxin goes to the root’s low side, where it slows growth.
If you chose auxin: it does not sink by itself. It is dissolved in tiny amounts and passed from cell to cell; the statoliths are what show it where to go.
Step 3 of 5
Open this step
One hormone, two responses
Practice · Choose
Question
In the tipped seedling auxin gathers on the low side of both the shoot and the root. Yet the shoot curves up and the root curves down. Why?

Correct answer
- Correct The same amount of auxin speeds up shoot cells and slows root cells
Setup, choices and feedback
The Scene at this step
Lamp direction
0 °fixed
Box tilt
90 °fixed
Time since the change
3 h
Zoom to the root tip
Shoot
Choices
Shown in a random order.
- Correct The same amount of auxin speeds up shoot cells and slows root cells
- Wrong Auxin speeds both, but in the root it moves to the upper side
  Misconception: Thinks auxin speeds up growth in roots just as it does in shoots.
  Feedback from the catalogue
  Roots are far more sensitive to auxin than shoots. The amount that speeds up a shoot cell slows a root cell. In a tipped root, auxin gathers on the low side and holds it back, the upper side grows faster, and the root curves down. Meanwhile, in the shoot, auxin gathers on the low side too, but there it speeds growth, and the shoot curves up.
- Wrong The root does not respond to auxin: it is simply pulled down by its weight
  Misconception: Thinks a root bends down because gravity pulls on its weight, as a rope sags.
  Feedback from the catalogue
  A root does not flop down like a rope: it grows down. It bends only in the growing zone behind its tip, and keeps the bend after it hardens. And if gravity simply pulled on it, it would pull the shoot down too, but instead the shoot rises. The bend is uneven growth of the root’s two sides, steered by the statoliths and auxin.
Feedback for a right answer
A root is far more sensitive to auxin than a shoot. In the shoot the extra auxin speeds up the cells on the low side, that side lengthens, and the shoot curves up. In the root the same auxin slows the cells on the low side, the upper side grows faster, and the root curves down. One hormone, on the same side, and two opposite responses.

Hint
Does auxin do the same thing in every tissue?

Worked solution
1. In the shoot: more auxin on the low side, the cells there lengthen more, the low side becomes the outside of the curve, and the shoot curves up.
2. In the root: more auxin on the low side, but in a root that amount slows the cells.
3. So in the root it is the upper side that lengthens more, and the root curves down.
4. Move Zoom to the root tip to Root tip to see the auxin on the low side.
If you chose gravity: a root does not flop like a rope, it grows down, and only in its growing zone.
Step 4 of 5
Open this step
Light against gravity
Practice · Predict, then look
Question
The box lies on its side, so the shoot points right, 90° from upright. Now the lamp moves to 70° to the right, almost where the shoot already points. At what angle from upright does the shoot settle after a day?

Correct answer
54.31 ° ± 5 °
A number prediction counts as right within ± 5 °
Setup, choices and feedback
The Scene at this step
Lamp direction
0 °fixed
Box tilt
90 °fixed
Time since the change
0 hfixed
Zoom to the root tip
Seedling
Once the prediction is locked in, the Scene changes
Lamp direction
70 °fixed
Time since the change
24 hfixed
The Scene then shows: 54.31 ° · Shoot tip, degrees from straight up, positive to the right/nodes/seedling/outputs/tipAngle
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At about 54°, exactly like an upright seedling under the same lamp. The shoot settles where the pull to the light and the pull upwards balance, and that depends only on where the lamp is and where up is, not on where the shoot started. Here gravity did most of the work: it raised the shoot from 90°.

Worked solution
1. The shoot starts at 90°, past the lamp at 70°.
2. Two pulls: towards the lamp (weak here, since the shoot nearly points at it) and upwards (strong, since it lies flat).
3. It stops where they balance: $\tan\theta = \dfrac{3\sin 70^\circ}{1 + 3\cos 70^\circ}$ ⁠, that is $\theta \approx 54^\circ$ ⁠.
4. That is the same angle as for a seedling that started upright: the balance does not depend on the start.
Step 5 of 5
Open this step
The bend stays
Practice · Calculate
Question
A day after the box was laid on its side, the shoot’s tip points straight up again. By how many degrees has the shoot bent relative to the box?

Correct answer
89.99 ° ± 1 °
Setup, choices and feedback
The Scene at this step
Lamp direction
0 °fixed
Box tilt
90 °fixed
Time since the change
24 hfixed
Zoom to the root tip
Seedling
Accepted within
± 1 °
Computed from the Scene at this step’s setting
Degrees the shoot tip has turned from the box’s own up/nodes/seedling/outputs/bend
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 ° ± 1 °
  Misconception: Thinks a shoot that has turned back upright has straightened out again, as if the bend were elastic.
  Feedback from the catalogue
  A bend in a shoot is growth, not a spring being bent. A cell that has stretched stays long, and when it leaves the growing zone it hardens in the shape it was given. Once the shoot points up again, the new growth above the bend runs straight, but the bend itself stays in the stem: in a seedling laid on its side it is 90° to the box.
Feedback for a right answer
90°: in the box the shoot grew straight along the box’s own up, and now it points at right angles to it. The bend stays in the stem: the cells on its outside lengthened and hardened that way. Above it, the new growth runs straight up.

Hint
Where does the box’s up point, and where does the tip point?

Worked solution
1. Before the box was laid down, the shoot grew along the box’s own up.
2. The box lies on its side, so its up now points right, 90° from upright.
3. The shoot’s tip points straight up, 0° from upright.
4. The difference: $90^\circ - 0^\circ = 90^\circ$ ⁠.
If you answered $0$ ⁠, you looked only at the straight tip. The shoot has not straightened back; it stays bent, and the new growth above the bend runs straight.

Lesson 5 of 5 · About 30 minutes

From flower to fruit

The parts of a flower, pollination, the race of the pollen tubes down the style, fertilisation, and how the number of seeds decides whether there is a fruit and how big it grows. And why many flowers reject their own pollen.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
A seed factory
Practice · Choose
Question
Seven weeks after pollination, which part of the flower has become the fruit?

Correct answer
- Correct The ovary
Setup, choices and feedback
The Scene at this step
Pollen grains
16
Pollen from
Another plant
Time
Pollination
Choices
Shown in a random order.
- Correct The ovary
- Wrong The ovule
  Misconception: Mixes up ovule and ovary: thinks the fruit grows from the ovule, or the seed from the ovary.
  Feedback from the catalogue
  The ovule and the ovary are two different things. The ovary is the swollen base of the pistil, and the ovules sit inside it. Each fertilised ovule grows into a seed, and the wall of the ovary around them thickens into the fruit. That is why, to a botanist, a tomato, a pepper and a pea pod are all fruits: ripened ovaries full of seeds.
- Wrong The petals, which thicken into flesh
  Feedback
  The petals wilt and fall in the first days after fertilisation. Their job was to attract the bee.
- Wrong The stigma
  Feedback
  The stigma catches the pollen, and after fertilisation it dries up along with the style.
Feedback for a right answer
The ovary: its wall thickens into the flesh of the fruit, and the ovules inside it, those that were fertilised, have grown into seeds. Drag Time to Ripe fruit and watch the cut face of the ovary fill out and turn red.

Hint
Where do the seeds sit in a fruit?

Worked solution
1. At the base of the pistil sits the ovary, with $12$ ⁠ ovules inside it.
2. Each fertilised ovule grows into a seed.
3. The ovary wall around them thickens into the fruit.
4. The petals, the stamens and the style wilt and fall.
If you chose the ovule, you swapped the two: the ovule becomes a seed, the ovary becomes the fruit.
Step 2 of 6
Open this step
A tube for every ovule
Practice · Calculate
Question
Thirty pollen grains from another plant have landed on the stigma. One in seven grows no tube, so $26$ ⁠ tubes grow. How many seeds will the fruit have?

Correct answer
12 ± 0
Setup, choices and feedback
The Scene at this step
Pollen grains
30fixed
Pollen from
Another plantfixed
Time
Pollination
Accepted within
± 0
Computed from the Scene at this step’s setting
Ovules fertilised, each one a seed/nodes/flower/outputs/seeds
Wrong answers it recognises
- 26 ± 0
  Misconception: Thinks every pollen grain that lands on the stigma makes a seed.
  Feedback from the catalogue
  Every seed is one fertilised ovule, and each ovule needs a pollen tube of its own. But there is a limit: the ovary of the flower in the Scene holds 12 ovules, and once a tube has entered an ovule it stops attracting others. Grains that grow no tube, and tubes that arrive after every ovule is taken, are wasted.
- 30 ± 0
  Misconception: Thinks every pollen grain that lands on the stigma makes a seed.
  Feedback from the catalogue
  Every seed is one fertilised ovule, and each ovule needs a pollen tube of its own. But there is a limit: the ovary of the flower in the Scene holds 12 ovules, and once a tube has entered an ovule it stops attracting others. Grains that grow no tube, and tubes that arrive after every ovule is taken, are wasted.
Feedback for a right answer
$12$ ⁠, one for each ovule. Once a tube has entered an ovule it stops attracting others, so the $12$ ⁠ fastest tubes take the $12$ ⁠ ovules, and the other $14$ ⁠ lose the race.

Hint
How many ovules are in the ovary?

Worked solution
1. Each seed is one fertilised ovule, and each ovule needs one tube.
2. The ovary holds $12$ ⁠ ovules.
3. There are $26$ ⁠ tubes, more than ovules, so every ovule is fertilised.
4. The number of seeds: $12$ ⁠, the smaller of $26$ ⁠ and $12$ ⁠.
If you answered $26$ ⁠ or $30$ ⁠, you assumed that every grain or every tube makes a seed. Drag Time forward and watch the extra tubes stop.
Step 3 of 6
Open this step
The race
Practice · Calculate
Question
A pollen tube grows 0.8 mm an hour, and the style is 10.4 mm long. How long does a tube take to grow the length of the style?

Correct answer
13 h ± 3%
Setup, choices and feedback
The Scene at this step
Pollen grains
16
Pollen from
Another plant
Time
Pollination
Accepted within
± 3%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
$\dfrac{10.4}{0.8} = 13$ ⁠ hours. With the half hour a grain needs to germinate and another millimetre inside the ovary, the first tubes reach the ovules after about $15$ ⁠ hours, and all of them are fertilised within a day.

Hint
Time = distance divided by speed.

Worked solution
1. The givens: distance 10.4 mm, speed 0.8 mm an hour.
2. Time = distance divided by speed.
3. $\dfrac{10.4}{0.8} = 13$ ⁠.
4. $13$ ⁠ hours.
Drag Time slowly through the first day and watch the tubes reach the ovules one after another.
Step 4 of 6
Open this step
Too few seeds
Practice · Reach the target
Question
Pollen from is fixed on another plant. The plant drops a fruit with too few seeds. Lower Pollen grains to find the smallest number of seeds for which the plant still keeps the fruit, then press Check.

Goal
- 1 ± 0
  At the start of the step: 1
- 3 ± 0
  At the start of the step: 12
Setup, choices and feedback
The Scene at this step
Pollen grains
16
Pollen from
Another plantfixed
Time
Ripe fruit
Read from the Scene at
1 if the flower sets a fruit, 0 if the plant drops it/nodes/flower/outputs/fruit
Read from the Scene at
Ovules fertilised, each one a seed/nodes/flower/outputs/seeds
One setting that reaches it
Pollen grains
3
Feedback for a right answer
Three seeds: with three or four grains, three tubes arrive, and the fruit stays, small and lopsided. With two, the plant drops it. The seeds make hormones that swell the fruit around them, and a fruit with one or two seeds is not worth the plant’s food.

Feedback for a wrong answer
Not yet. The number of seeds must be the smallest at which the fruit still stays.

Hint
Check on the chart how many ovules were fertilised, and whether the fruit grows.

Worked solution
1. Fewer grains, fewer tubes, fewer fertilised ovules.
2. Lower Pollen grains from $16$ ⁠ slowly, and drag Time to Ripe fruit to see whether the fruit stays.
3. With $3$ ⁠ grains there are $3$ ⁠ seeds, and the fruit stays; with $2$ ⁠ there are $2$ ⁠ seeds, and the plant drops it.
4. The limit: $3$ ⁠ seeds.
The three-seed rule belongs to the flower in the Scene; real plants set thresholds of their own.
Step 5 of 6
Open this step
Half the ovules
Practice · Predict, then look
Question
Seven grains from another plant have landed, six tubes grow, and $6$ ⁠ of the $12$ ⁠ ovules will be fertilised. What kind of fruit comes of it?

Correct answer
- Correct A small, lopsided fruit
Setup, choices and feedback
The Scene at this step
Pollen grains
7fixed
Pollen from
Another plantfixed
Time
Pollinationfixed
Choices
- Wrong The plant drops the fruit
  Feedback
  Six seeds are more than this flower’s minimum of three, so it keeps the fruit.
- Correct A small, lopsided fruit
- Wrong A full-sized fruit: the size depends on the flower, not the seeds
  Feedback
  In this flower the seeds are what swell the fruit, so fewer seeds give a smaller fruit. Some plants, such as bananas, grow fruit without seeds, but that is the exception.
Once the prediction is locked in, the Scene changes
Time
Ripe fruitfixed
The Scene then shows: 16.4 mm · Width of the ripe fruit (0 if dropped)/nodes/flower/outputs/fruitWidth
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A small fruit, about 16 mm across instead of 22 mm, and lopsided towards the side with more seeds. The growing seeds make hormones that swell the ovary wall around them, so each side of the fruit grows with the seeds in it.

Worked solution
1. $6$ ⁠ seeds: more than the minimum of $3$ ⁠, so the plant keeps the fruit.
2. Fewer than $12$ ⁠, so the fruit does not reach full size.
3. The seeds swell the fruit around them, and each side grows with the seeds in it.
4. On the chart: a fruit of about 16 mm, against 22 mm when every ovule is fertilised.
This is one reason growers bring hives into orchards: fuller pollination gives bigger, better-shaped fruit.
Step 6 of 6
Open this step
Pollen from the same plant
Practice · Predict, then look
Question
Sixteen pollen grains have landed on the stigma, but all from The same plant. What happens?

Correct answer
- Correct The tubes stop on the way, and there are no seeds and no fruit
Setup, choices and feedback
The Scene at this step
Pollen grains
16fixed
Pollen from
The same plantfixed
Time
Pollinationfixed
Choices
- Correct The tubes stop on the way, and there are no seeds and no fruit
- Wrong A full fruit with $12$ ⁠ seeds, as with pollen from another plant
  Misconception: Thinks any pollen of the right species can fertilise any flower, including the plant’s own.
  Feedback from the catalogue
  Many plants reject their own pollen. The style recognises proteins on the pollen: if their versions match the plant’s own, it stops the tube part of the way down, and no ovule is fertilised. That keeps the plant from breeding with itself, and it is why an apple orchard needs at least two varieties, and bees to carry pollen between them.
- Wrong The grains grow no tubes at all
  Feedback
  The grains do grow tubes. The stop happens inside the style, after the tubes have started to grow.
Once the prediction is locked in, the Scene changes
Time
Ripe fruitfixed
The Scene then shows: 0 · Ovules fertilised, each one a seed/nodes/flower/outputs/seeds
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The grains grow tubes, but the style recognises them as its own plant’s and stops them in its top third. No ovule is fertilised, there are no seeds, and the plant drops the flower. That keeps it from breeding with itself.

Worked solution
1. The plant’s own pollen lands and grows tubes, as usual.
2. The style compares proteins on the pollen with its own versions. When they match, it stops the tube.
3. On the chart: the tubes stop in the style, and $0$ ⁠ ovules are fertilised.
4. No seeds, no fruit.
If you predicted a full fruit, you assumed that any pollen of the same species will do. Many plants reject their own pollen, which is why an apple orchard needs at least two varieties.

Across the unit · 15 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Water through a tree

A tree that drinks through its leaves

Question

Correct answer

The Scene at this step

Hint

Worked solution

Out through the leaves

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Night

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

What lifts the water

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

How fast

Question

Correct answer

The Scene at this step

Hint

Worked solution

What speeds up transpiration

Dry air

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Muggy air

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Half the flow

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Wind

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Saving water

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The whole tree

Question

Correct answer

The Scene at this step

Hint

Worked solution

Growing towards the light

A plant that grows at its tips

Question

Correct answer