Answer key · Unit 3

Transport in cells.

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 4

Lesson 1 of 4 · About 20 minutes

Diffusion

The random motion of molecules, net movement from high concentration to low, dynamic equilibrium, and the effect of temperature.

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The 5 steps and their answers

Step 1 of 5
Open this step
A random walk
Practice · Predict, then look
Question
The left side of the tank holds 300 mOsm/L of sugar, the right side 100 mOsm/L, and the membrane lets sugar through. Which way will the sugar move, overall?

Correct answer
- Correct From left to right
Setup, choices and feedback
The Scene at this step
Left side
300 mOsm/Lfixed
Right side
100 mOsm/Lfixed
Temperature
293 K
Liquids / Molecules
Liquidsfixed
Choices
- Correct From left to right
- Wrong From right to left
  Feedback
  Molecules wander every way at random, but there are more of them on the left to cross, so more cross to the right.
- Wrong Neither way: each molecule stays on its side
  Feedback
  The membrane is open to sugar, and the molecules do not know which side they belong on.
Once the prediction is locked in, the Scene changes
Liquids / Molecules
Moleculesfixed
The Scene then shows: -1 · Net sugar flow at the start: +1 into the left side, −1 out of it, 0 none (it cannot cross)/nodes/tank/outputs/soluteIn
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From left to right: from the high concentration to the low. The arrows show a net flow of sugar to the right. No molecule “decides” to cross; there are simply more molecules on the left bumping into the membrane.

Worked solution
1. Each molecule wanders at random, and when it hits the membrane it may cross.
2. The left holds three times as many sugar molecules in each volume, so three times as many hit the membrane and cross to the right.
3. The difference is a net flow from left to right: diffusion, from high concentration to low.
Step 2 of 5
Open this step
The end
Practice · Calculate
Question
Both sides hold the same volume: 300 mOsm/L on the left, 100 mOsm/L on the right. What will the concentration on the left be, in mOsm/L, once diffusion is done?

Correct answer
200 ± 2
Setup, choices and feedback
The Scene at this step
Left side
300 mOsm/Lfixed
Right side
100 mOsm/Lfixed
Temperature
293 K
Liquids / Molecules
Molecules
Accepted within
± 2
Computed from the Scene at this step’s setting
Concentration on the left, or inside the cell, once it settles, mOsm/L/nodes/tank/outputs/finalLeft
Wrong answers it recognises
- 100 ± 2
  Misconception: Thinks molecules stop moving once the concentrations are equal.
  Feedback
  The sugar does not “swap” sides. It crosses until there is no difference, then goes on moving with no net flow.
- 400 ± 2
  Misconception: Thinks a side whose solute cannot cross keeps its concentration.
  Feedback
  $400$ ⁠ is the sum of the concentrations. The total amount spreads over twice the volume, so take the average.
Feedback for a right answer
$\dfrac{300 + 100}{2} = 200$ ⁠: the same concentration on both sides. From then on molecules keep crossing both ways, at the same rate.

Hint
All the sugar spreads evenly over both sides.

Worked solution
1. The volumes are equal, so the sugar on each side is in proportion to its concentration: $300$ ⁠ and $100$ ⁠ units.
2. At the end all $400$ ⁠ units share two equal volumes.
3. On each side: $\dfrac{400}{2} = 200$ ⁠, that is 200 mOsm/L.
Step 3 of 5
Open this step
Still moving
Practice · Choose
Question
The concentrations on both sides are now equal. What are the sugar molecules doing?

Correct answer
- Correct Still crossing both ways, at the same rate.
Setup, choices and feedback
The Scene at this step
Left side
200 mOsm/L
Right side
200 mOsm/L
Temperature
293 K
Liquids / Molecules
Liquids
Choices
Shown in a random order.
- Correct Still crossing both ways, at the same rate.
- Wrong They have stopped, as there is no reason left to move.
  Misconception: Thinks molecules stop moving once the concentrations are equal.
  Feedback from the catalogue
  The molecules never stop: they keep wandering at random as long as there is any warmth. When the concentrations are equal, about as many cross each way every second, so there is no net movement. That is a dynamic equilibrium: everything moves, and nothing changes.
- Wrong Each stays on the side it reached.
  Misconception: Thinks molecules stop moving once the concentrations are equal.
  Feedback from the catalogue
  The molecules never stop: they keep wandering at random as long as there is any warmth. When the concentrations are equal, about as many cross each way every second, so there is no net movement. That is a dynamic equilibrium: everything moves, and nothing changes.
Feedback for a right answer
Right. Choose Molecules under Look inside: the molecules dart about and cross through the channels both ways, but the net-flow arrows are gone. That is a dynamic equilibrium.

Worked solution
1. The molecules’ motion comes from their heat, not from the difference in concentration.
2. As long as there is warmth, they move, and now and then cross the membrane.
3. At equal concentrations as many cross each way: no net flow, but plenty of crossing.
Step 4 of 5
Open this step
Warmer
Practice · Reach the target
Question
In warm water molecules move faster. Warm the tank with Temperature until diffusion is twice as fast as at Room temperature (20 °C), then press Check.

Goal
- 2 ± 3%
  At the start of the step: 0.9958
Setup, choices and feedback
The Scene at this step
Left side
300 mOsm/Lfixed
Right side
100 mOsm/Lfixed
Temperature
293 K
Liquids / Molecules
Molecules
Read from the Scene at
How fast molecules spread, times as fast as at 20 °C/nodes/tank/outputs/diffusionRate
One setting that reaches it
Temperature
321 K
Feedback for a right answer
Around 50 °C. Diffusion speeds up with absolute temperature, and also because warm water is less viscous: at 50 °C water’s viscosity is only about 55% of what it is at 20 °C.

Feedback for a wrong answer
Not yet. Warm it slowly and check again.

Hint
It is less hot than you might think: heat speeds the molecules up and also makes the water runnier.

Worked solution
1. By Stokes and Einstein, diffusion goes as $\dfrac{T}{\eta}$ ⁠: absolute temperature over the water’s viscosity.
2. From 20 °C to 50 °C the absolute temperature rises only about 10%, but the viscosity falls to about 55%.
3. Together: $1.1 \div 0.55 = 2$ ⁠, twice as fast, around 323 K.
Step 5 of 5
Open this step
In the lungs
Practice · Choose
Question
Why does oxygen pass from the air in the lungs’ air sacs into the blood?

Correct answer
- Correct Because the air in the sac holds more oxygen than the blood arriving there.
Setup, choices and feedback
The Scene at this step
Left side
1000 mOsm/L
Right side
0 mOsm/L
Temperature
310 K
Liquids / Molecules
Molecules
Choices
Shown in a random order.
- Correct Because the air in the sac holds more oxygen than the blood arriving there.
- Wrong Because lung cells pump it in using ATP.
  Feedback
  Oxygen needs no pump: it is a small molecule that diffuses through membranes, as long as there is a gradient.
- Wrong Because breathing pushes the air through the walls.
  Feedback
  Breathing brings fresh air into the sacs and keeps the concentration high there, but the crossing itself is diffusion.
Feedback for a right answer
Right: diffusion, from high concentration to low. Breathing keeps bringing oxygen-rich air, and the blood keeps carrying the oxygenated blood away, so the gradient never disappears. Carbon dioxide goes the other way, for the same reason.

Worked solution
1. The walls of the air sac and the capillary are very thin, one cell each.
2. The air in the sac holds more oxygen than the blood returning from the body.
3. So oxygen diffuses from the air into the blood, with no energy spent.

Lesson 2 of 4 · About 20 minutes

Osmosis

When only the water can cross: which way it flows, what happens to the concentrations and volumes, and when the flow stops.

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The 5 steps and their answers

Step 1 of 5
Open this step
Only water
Practice · Predict, then look
Question
Now the membrane lets only water through. The left holds 300 mOsm/L of sugar, the right 100 mOsm/L. What will happen to the liquid levels?

Correct answer
- Correct The left side will rise
Setup, choices and feedback
The Scene at this step
Left side
300 mOsm/Lfixed
Right side
100 mOsm/Lfixed
Temperature
293 K
Liquids / Molecules
Liquidsfixed
Choices
- Correct The left side will rise
- Wrong The right side will rise
  Misconception: Thinks water moves towards the weaker, more watery solution.
  Feedback from the catalogue
  Water moves from the dilute side to the concentrated side. One way to see it: the dilute side has more free water molecules in each volume, so more of them hit the membrane and cross; on the concentrated side some of the water is held by the solute. The result: a net flow of water into the side with more solute, which fills up.
- Wrong The levels will stay equal
  Misconception: Thinks osmosis moves the dissolved substance rather than the water.
  Feedback from the catalogue
  In osmosis it is water that crosses, not the solute. The membrane lets water through and not the sugar or salt, so the solute is stuck on its side, and the water is what moves, towards the more concentrated side. A solute moving from high to low concentration is diffusion, and that happens only when the membrane lets it through.
Once the prediction is locked in, the Scene changes
Liquids / Molecules
Moleculesfixed
The Scene then shows: 1 · Net water flow at the start: +1 into the left side or the cell, −1 out of it, 0 none/nodes/tank/outputs/waterIn
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The left side rises: water crosses from the right, the dilute side, into the concentrated side. The sugar does not move at all.

Worked solution
1. The sugar cannot cross, so there is no diffusion of sugar.
2. The water can, and water moves from the dilute side to the concentrated side: right to left.
3. The volume on the left grows, and the level there rises.
If you chose “right”, you thought water goes to water. It goes to the solute.
Step 2 of 5
Open this step
What osmosis is
Practice · Choose
Question
What is osmosis?

Correct answer
- Correct Water moving across a partially permeable membrane, from a dilute solution to a concentrated one.
Setup, choices and feedback
The Scene at this step
Left side
300 mOsm/L
Right side
100 mOsm/L
Temperature
293 K
Liquids / Molecules
Molecules
Choices
Shown in a random order.
- Correct Water moving across a partially permeable membrane, from a dilute solution to a concentrated one.
- Wrong A solute moving across a membrane, from high concentration to low.
  Misconception: Thinks osmosis moves the dissolved substance rather than the water.
  Feedback from the catalogue
  In osmosis it is water that crosses, not the solute. The membrane lets water through and not the sugar or salt, so the solute is stuck on its side, and the water is what moves, towards the more concentrated side. A solute moving from high to low concentration is diffusion, and that happens only when the membrane lets it through.
- Wrong Water moving from a concentrated solution to a dilute one.
  Misconception: Thinks water moves towards the weaker, more watery solution.
  Feedback from the catalogue
  Water moves from the dilute side to the concentrated side. One way to see it: the dilute side has more free water molecules in each volume, so more of them hit the membrane and cross; on the concentrated side some of the water is held by the solute. The result: a net flow of water into the side with more solute, which fills up.
- Wrong Pumping water into a cell using ATP.
  Misconception: Thinks substances can only ever move from high to low concentration.
  Feedback
  Osmosis uses no energy: the water flows by itself, down the gradient.
Feedback for a right answer
Right. Osmosis is a special case of diffusion: the diffusion of water across a membrane that stops the solute.

Worked solution
1. Osmosis needs a partially permeable membrane: open to water, closed to the solute.
2. The solute does not cross; the water does.
3. The direction: from the dilute side (more free water) to the concentrated side.
Step 3 of 5
Open this step
The concentration at the end
Practice · Calculate
Question
The left holds 300 mOsm/L, the right 100 mOsm/L, and only water crosses. What will the concentration on the left be, in mOsm/L, once the water stops flowing?

Correct answer
200 ± 2
Setup, choices and feedback
The Scene at this step
Left side
300 mOsm/Lfixed
Right side
100 mOsm/Lfixed
Temperature
293 K
Liquids / Molecules
Liquids
Accepted within
± 2
Computed from the Scene at this step’s setting
Concentration on the left, or inside the cell, once it settles, mOsm/L/nodes/tank/outputs/finalLeft
Wrong answers it recognises
- 300 ± 2
  Misconception: Thinks a side whose solute cannot cross keeps its concentration.
  Feedback from the catalogue
  Concentration is the amount of solute divided by the volume. Even if the sugar cannot cross, water coming in raises the volume, so the concentration falls. 300 mOsm/L whose volume grows 1.5 times becomes $300 \div 1.5 = 200$ ⁠.
- 100 ± 2
  Misconception: Thinks osmosis moves the dissolved substance rather than the water.
  Feedback
  The sugar did not cross, so the left side cannot lose it.
Feedback for a right answer
200 mOsm/L on both sides: the same result as diffusion, by another route. In diffusion the sugar moved and the volumes stayed; in osmosis the sugar stayed and the volumes changed.

Hint
The water flows until both sides have the same concentration, and the total sugar does not change.

Worked solution
1. At the start, in equal volumes, there are $300$ ⁠ units of sugar on the left and $100$ ⁠ on the right.
2. The flow stops when the concentrations are equal. The total volume does not change: $2$ ⁠ volumes.
3. The shared concentration: $\dfrac{300 + 100}{2} = 200$ ⁠.
4. The left side: 200 mOsm/L, because water came in and diluted it.
If you answered $300$ ⁠, you forgot that incoming water raises the volume and lowers the concentration.
Step 4 of 5
Open this step
How much water
Practice · Calculate
Question
In the same experiment: how many times bigger has the left side’s volume become by the time the water stops?

Correct answer
1.5 ± 0.02
Setup, choices and feedback
The Scene at this step
Left side
300 mOsm/Lfixed
Right side
100 mOsm/Lfixed
Temperature
293 K
Liquids / Molecules
Liquids
Accepted within
± 0.02
Computed from the Scene at this step’s setting
Volume of the left side, or of the cell, once it settles, times its start/nodes/tank/outputs/leftVolume
Wrong answers it recognises
- 3 ± 0.02
  Misconception: Thinks a side whose solute cannot cross keeps its concentration.
  Feedback
  $3$ ⁠ is the ratio of the starting concentrations. But the right side changes too: it loses water and grows stronger, so they meet in the middle, at $200$ ⁠.
Feedback for a right answer
1.5 times: $\dfrac{300}{200} = 1.5$ ⁠. And the right side shrank to half its volume, from $100$ ⁠ to $200$ ⁠. The total volume stayed the same.

Hint
The same amount of sugar, whose concentration fell from $300$ ⁠ to $200$ ⁠. How many times bigger is the volume?

Worked solution
1. The amount of sugar on the left does not change: concentration × volume is fixed.
2. $300 \times V_0 = 200 \times V$ ⁠, so $V = \dfrac{300}{200}\,V_0 = 1.5\,V_0$ ⁠.
3. The volume grows 1.5 times.
Step 5 of 5
Open this step
In balance
Practice · Reach the target
Question
The left holds 300 mOsm/L, and the right pure water. Change Right side until there is no net flow of water either way, then press Check.

Goal
- 0 ± 0.1
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Left side
300 mOsm/Lfixed
Right side
0 mOsm/L
Temperature
293 K
Liquids / Molecules
Molecules
Read from the Scene at
Net water flow at the start: +1 into the left side or the cell, −1 out of it, 0 none/nodes/tank/outputs/waterIn
One setting that reaches it
Right side
300 mOsm/L
Temperature
274 K
Feedback for a right answer
300 mOsm/L, like the left side: two such solutions are called isotonic. Water still crosses, but at the same rate both ways.

Feedback for a wrong answer
Not yet. Water is still flowing to the more concentrated side.

Hint
When does the water have no reason to favour either side?

Worked solution
1. Water flows net when the concentrations differ, towards the stronger side.
2. With no difference, there is no net flow.
3. Set Right side to 300 mOsm/L.

Lesson 3 of 4 · About 25 minutes

Cells in solutions

A red blood cell in hypotonic, isotonic and hypertonic solutions, why a plant cell does not burst, and active transport against the concentration gradient.

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The 6 steps and their answers

Step 1 of 6
Open this step
Pure water
Practice · Predict, then look
Question
Inside the red blood cell is 300 mOsm/L, as in plasma. We will move it into Pure water. What will happen to it?

Correct answer
- Correct It will swell until it bursts
Setup, choices and feedback
The Scene at this step
Outside the cell
300 mOsm/Lfixed
Temperature
293 K
Liquids / Molecules
Liquidsfixed
Choices
- Correct It will swell until it bursts
- Wrong It will shrivel
  Misconception: Thinks water moves towards the weaker, more watery solution.
  Feedback from the catalogue
  Water moves from the dilute side to the concentrated side. One way to see it: the dilute side has more free water molecules in each volume, so more of them hit the membrane and cross; on the concentrated side some of the water is held by the solute. The result: a net flow of water into the side with more solute, which fills up.
- Wrong It will not change: it keeps its shape
  Feedback
  A red blood cell has no wall, and a membrane that lets water through cannot stop the flow.
Once the prediction is locked in, the Scene changes
Outside the cell
0 mOsm/Lfixed
Liquids / Molecules
Moleculesfixed
The Scene then shows: 1 · 1 when the red blood cell swells until it bursts, else 0/nodes/tank/outputs/burst
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It swells until it is a sphere, then the membrane tears and the haemoglobin spills out. The salt inside cannot leave, and the water keeps coming in.

Worked solution
1. There is no solute outside the cell and 300 mOsm/L inside: the outside is hypotonic.
2. Water moves to the concentrated side, that is into the cell.
3. The cell swells. Its membrane can stretch only as far as a sphere; beyond that it tears.
Step 2 of 6
Open this step
Swelling
Practice · Calculate
Question
Outside the cell is 200 mOsm/L, inside it 300. Only 60% of the cell’s volume is water that can flow; the rest, mostly haemoglobin, does not change. How many times bigger will the cell become?

Correct answer
1.3 ± 0.02
Setup, choices and feedback
The Scene at this step
Outside the cell
200 mOsm/Lfixed
Temperature
293 K
Liquids / Molecules
Liquids
Accepted within
± 0.02
Computed from the Scene at this step’s setting
Volume of the left side, or of the cell, once it settles, times its start/nodes/tank/outputs/leftVolume
Wrong answers it recognises
- 1.5 ± 0.02
  Misconception: Treats the whole cell volume as water that follows the outside solution.
  Feedback from the catalogue
  Not all of a cell is water that can flow. About 40% of a red blood cell’s volume is haemoglobin and other material that does not change. Only the other 60% swells or shrinks with the solution outside (the Boyle–van ’t Hoff law): $V = 0.4 + 0.6 \times \dfrac{\text{inside}}{\text{outside}}$ ⁠.
Feedback for a right answer
$0.4 + 0.6 \times 1.5 = 1.3$ ⁠: 1.3 times. The cell swells but does not burst; it would burst only at 1.57 times, as a sphere.

Hint
The water in the cell, 0.6 of the volume, grows $\dfrac{300}{200}$ ⁠ times. Add the 0.4 that does not change.

Worked solution
1. Water comes in until the inside matches the outside: from $300$ ⁠ to $200$ ⁠.
2. The solute in the water is fixed, so the water’s volume grows $\dfrac{300}{200} = 1.5$ ⁠ times: from $0.6$ ⁠ to $0.6 \times 1.5 = 0.9$ ⁠.
3. The part that does not change: $0.4$ ⁠.
4. In all: $0.4 + 0.9 = 1.3$ ⁠.
If you answered $1.5$ ⁠, you enlarged the whole cell, haemoglobin included, as if it were all water.
Step 3 of 6
Open this step
Seawater
Practice · Predict, then look
Question
Now we move the cell into Seawater, 1000 mOsm/L. What will happen to it?

Correct answer
- Correct It will shrink and crinkle
Setup, choices and feedback
The Scene at this step
Outside the cell
300 mOsm/Lfixed
Temperature
293 K
Liquids / Molecules
Liquidsfixed
Choices
- Correct It will shrink and crinkle
- Wrong It will swell and burst, as the salt gets in
  Misconception: Thinks osmosis moves the dissolved substance rather than the water.
  Feedback from the catalogue
  In osmosis it is water that crosses, not the solute. The membrane lets water through and not the sugar or salt, so the solute is stuck on its side, and the water is what moves, towards the more concentrated side. A solute moving from high to low concentration is diffusion, and that happens only when the membrane lets it through.
- Wrong It will not change
  Feedback
  The outside is more than three times as strong as the cell: there is a steep gradient.
Once the prediction is locked in, the Scene changes
Outside the cell
1000 mOsm/Lfixed
Liquids / Molecules
Moleculesfixed
The Scene then shows: 0.58 · Volume of the left side, or of the cell, once it settles, times its start/nodes/tank/outputs/leftVolume
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It loses water and shrinks into a spiky ball, to about 58% of its volume. The outside is hypertonic. That is why drinking seawater dehydrates you: the kidneys need more water to get rid of the salt than you drank.

Worked solution
1. Outside is 1000 mOsm/L, inside 300: the concentrated side is outside.
2. Water leaves the cell for the solution.
3. The water in the cell shrinks $\dfrac{1000}{300}$ ⁠ times: $0.4 + 0.6 \times 0.3 = 0.58$ ⁠. The cell shrinks.
Step 4 of 6
Open this step
A safe drip
Practice · Reach the target
Question
A drip into a patient’s vein must not harm the blood cells. Change Outside the cell until the red blood cell keeps its volume, then press Check.

Goal
- 1 ± 0.015
  At the start of the step: 1.567
Setup, choices and feedback
The Scene at this step
Outside the cell
0 mOsm/L
Temperature
310 K
Liquids / Molecules
Liquids
Read from the Scene at
Volume of the left side, or of the cell, once it settles, times its start/nodes/tank/outputs/leftVolume
One setting that reaches it
Outside the cell
300 mOsm/L
Temperature
274 K
Feedback for a right answer
300 mOsm/L, like the inside of the cell: an isotonic solution. That is the hospital saline drip, 0.9% salt.

Feedback for a wrong answer
Not yet. The cell is still swelling or shrinking.

Hint
What concentration outside makes the water flow neither way?

Worked solution
1. The cell keeps its volume when there is no net flow of water.
2. That happens when the outside matches the inside, 300 mOsm/L.
3. Set Outside the cell to 300 mOsm/L, Like blood.
Step 5 of 6
Open this step
Plant cells
Practice · Choose
Question
What happens to a lettuce leaf cell placed in pure water?

Correct answer
- Correct It fills with water and presses against its wall, but does not burst.
Setup, choices and feedback
The Scene at this step
Outside the cell
0 mOsm/L
Temperature
293 K
Liquids / Molecules
Liquids
Choices
Shown in a random order.
- Correct It fills with water and presses against its wall, but does not burst.
- Wrong It bursts, like the red blood cell.
  Misconception: Thinks a plant cell in pure water bursts like a red blood cell.
  Feedback from the catalogue
  A plant cell has a strong cellulose wall outside its membrane. As water comes in, the vacuole swells and presses the cell against the wall, and the wall pushes back and stops the flow before the cell tears. The cell becomes turgid, and that is what holds a plant upright. A red blood cell has no wall, so it bursts.
- Wrong It shrinks away from its wall.
  Misconception: Thinks water moves towards the weaker, more watery solution.
  Feedback
  That happens in a strong solution, like salt water: the cell loses water and the membrane pulls away from the wall (plasmolysis).
Feedback for a right answer
Right: it becomes turgid. That is why limp lettuce perks up when you soak it in water, and why an unwatered plant wilts: the vacuoles empty and nothing presses on the walls.

Worked solution
1. The cell holds solutes, and outside is pure water: water comes in.
2. The vacuole swells and presses the membrane against the wall.
3. The strong wall pushes back and stops the flow. The cell is turgid, not torn.
Step 6 of 6
Open this step
Against the gradient
Practice · Choose
Question
Inside a plant’s root hairs nitrate ions are a hundred times more concentrated than in the soil water, yet the root keeps taking in more. How?

Correct answer
- Correct Active transport: pump proteins in the membrane move them in using energy from ATP.
Setup, choices and feedback
The Scene at this step
Outside the cell
300 mOsm/L
Temperature
310 K
Liquids / Molecules
Molecules
Choices
Shown in a random order.
- Correct Active transport: pump proteins in the membrane move them in using energy from ATP.
- Wrong Diffusion, because the ions are small enough to cross the membrane.
  Misconception: Thinks substances can only ever move from high to low concentration.
  Feedback from the catalogue
  Diffusion and osmosis always go down the gradient, from high to low concentration, and cost no energy. But cells can also move substances against the gradient, with pump proteins in the membrane that use ATP: that is active transport. That is how roots take up minerals from dilute soil water, and how nerve cells keep a lot of potassium inside and little sodium.
- Wrong Osmosis: the water entering the root drags the ions in with it.
  Misconception: Thinks osmosis moves the dissolved substance rather than the water.
  Feedback from the catalogue
  In osmosis it is water that crosses, not the solute. The membrane lets water through and not the sugar or salt, so the solute is stuck on its side, and the water is what moves, towards the more concentrated side. A solute moving from high to low concentration is diffusion, and that happens only when the membrane lets it through.
Feedback for a right answer
Right. Diffusion would carry the ions out, from high concentration to low. To move them against the gradient the cell pays in energy: that is why root cells have many mitochondria, and why a root without oxygen, in waterlogged soil for instance, stops taking up minerals.

Worked solution
1. The concentration is high inside the root and low outside: diffusion would go outward, not in.
2. Moving against the gradient takes energy.
3. The cell uses pump proteins that use ATP: active transport.

Lesson 4 of 4 · About 25 minutes

Surface area and volume

The square–cube law: when area grows as the square and volume as the cube, the surface area to volume ratio shrinks, which is why cells are small and big animals need lungs.

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The 6 steps and their answers

Step 1 of 6
Open this step
Skin
Practice · Calculate
Question
The copy is 3 times the original in every length, with the same shape. How many times bigger is its surface area?

Correct answer
9 ± 0.05
Setup, choices and feedback
The Scene at this step
Size
3 ×fixed
Leg thickness
1 ×fixed
Slice and count
Slices and cubes
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Its whole outer surface, times the original/nodes/beast/outputs/surfaceRatio
Wrong answers it recognises
- 3 ± 0.05
  Misconception: Thinks area and volume grow in step with length: twice as tall, twice the area and twice the weight.
  Feedback from the catalogue
  When every length grows $k$ ⁠ times, an area, which has two lengths in it, grows $k^2$ ⁠ times, and a volume, which has three, grows $k^3$ ⁠ times. A cube whose edge grows 3 times gets 9 times the surface area and 27 times the volume.
- 27 ± 0.05
  Misconception: Mixes up how area and volume scale: cubes where it should square, or squares where it should cube.
  Feedback from the catalogue
  Area is measured in squared units (cm²) and volume in cubed units (cm³), and that is how they grow: area $k^2$ ⁠ times, volume (and weight) $k^3$ ⁠ times. To remember which is which, picture little cubes: one face of a cube three times bigger is covered by $9$ ⁠ squares, and $27$ ⁠ cubes fill it.
Feedback for a right answer
$3^2 = 9$ ⁠: the copy’s leg slice is gridded in squares the size of the original’s slice, and it holds $9$ ⁠.

Hint
An area is a length times a length.

Worked solution
1. Every area is the product of two lengths.
2. Each length grows 3 times, so each area grows $3 \times 3 = 9$ ⁠ times.
3. The surface area grows 9 times.
If you answered $3$ ⁠, you grew the area like a length. If $27$ ⁠, like a volume.
Step 2 of 6
Open this step
Body
Practice · Calculate
Question
And how many times bigger are the volume and weight of the copy, 3 times the size?

Correct answer
27 ± 0.1
Setup, choices and feedback
The Scene at this step
Size
3 ×fixed
Leg thickness
1 ×fixed
Slice and count
Slices and cubes
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Its weight, times the original/nodes/beast/outputs/weightRatio
Wrong answers it recognises
- 3 ± 0.1
  Misconception: Thinks area and volume grow in step with length: twice as tall, twice the area and twice the weight.
  Feedback from the catalogue
  When every length grows $k$ ⁠ times, an area, which has two lengths in it, grows $k^2$ ⁠ times, and a volume, which has three, grows $k^3$ ⁠ times. A cube whose edge grows 3 times gets 9 times the surface area and 27 times the volume.
- 9 ± 0.1
  Misconception: Mixes up how area and volume scale: cubes where it should square, or squares where it should cube.
  Feedback from the catalogue
  Area is measured in squared units (cm²) and volume in cubed units (cm³), and that is how they grow: area $k^2$ ⁠ times, volume (and weight) $k^3$ ⁠ times. To remember which is which, picture little cubes: one face of a cube three times bigger is covered by $9$ ⁠ squares, and $27$ ⁠ cubes fill it.
Feedback for a right answer
$3^3 = 27$ ⁠: the cube lifted from the copy’s top holds $27$ ⁠ cubes the size of the original’s. The surface grew 9 times, the volume 27.

Hint
A volume is length times width times height.

Worked solution
1. A volume is the product of three lengths.
2. Each grows 3 times: $3 \times 3 \times 3 = 27$ ⁠.
3. The weight, density times volume, grows 27 times too.
Step 3 of 6
Open this step
The ratio
Practice · Calculate
Question
Now the copy is 2 times the size. By what factor has its surface area to volume ratio changed, compared with the original?

Correct answer
0.5 ± 0.01
Setup, choices and feedback
The Scene at this step
Size
2 ×fixed
Leg thickness
1 ×fixed
Slice and count
Slices and cubes
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Its surface area to volume ratio, times the original’s/nodes/beast/outputs/surfaceToVolume
Wrong answers it recognises
- 2 ± 0.01
  Misconception: Thinks a bigger body has more surface for each unit of its volume.
  Feedback from the catalogue
  A big body has more surface in total, but less surface for each unit of volume. Every cubic centimetre inside needs oxygen and food that come in through the surface, and in a big body each square centimetre of surface has to serve more volume. That is why big animals have lungs, a folded gut and a circulation, and why cells are small.
- 1 ± 0.01
  Misconception: Thinks a scaled-up copy of the same shape keeps the same surface area to volume ratio.
  Feedback from the catalogue
  The same shape does not mean the same ratio. Scaling up $k$ ⁠ times multiplies the area by $k^2$ ⁠ and the volume by $k^3$ ⁠, so the ratio $\dfrac{\text{area}}{\text{volume}}$ ⁠ falls $k$ ⁠ times. A 1 cm cube: $6 : 1$ ⁠. A 2 cm cube: $24 : 8$ ⁠, that is $3 : 1$ ⁠.
Feedback for a right answer
$\dfrac{4}{8} = 0.5$ ⁠ times: each cubic centimetre of the copy has half the surface the original had. $k$ ⁠ times bigger, a ratio $k$ ⁠ times smaller.

Hint
The area grew 4 times, the volume 8 times.

Worked solution
1. Area: $2^2 = 4$ ⁠ times.
2. Volume: $2^3 = 8$ ⁠ times.
3. The ratio $\dfrac{\text{area}}{\text{volume}}$ ⁠ changes $\dfrac{4}{8} = 0.5$ ⁠ times.
If you answered $2$ ⁠, you turned the ratio upside down. If $1$ ⁠, you thought the same shape keeps the same ratio.
Step 4 of 6
Open this step
A quarter
Practice · Reach the target
Question
Change the copy’s Size until its surface area to volume ratio is a quarter of the original’s, then press Check.

Goal
- 0.25 ± 0.005
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Size
1 ×
Leg thickness
1 ×fixed
Slice and count
Whole tables
Read from the Scene at
Its surface area to volume ratio, times the original’s/nodes/beast/outputs/surfaceToVolume
One setting that reaches it
Size
4 ×
Feedback for a right answer
4 times, a table 3 m high: area 16 times, volume 64 times, and the ratio $\dfrac{16}{64} = \dfrac14$ ⁠.

Feedback for a wrong answer
Not yet. The ratio is $\dfrac{1}{\text{size}}$ ⁠.

Hint
The ratio shrinks by the size.

Worked solution
1. For a copy of the same shape the ratio changes $\dfrac{1}{k}$ ⁠ times.
2. You want $\dfrac{1}{k} = \dfrac{1}{4}$ ⁠, so $k = 4$ ⁠.
3. Set Size to 4.
Step 5 of 6
Open this step
Cubes
Practice · Calculate
Question
A cube with 1 cm edges has a surface area of 6 cm² and a volume of 1 cm³: a ratio of $6$ ⁠. What is the surface area to volume ratio of a cube with 3 cm edges, in cm⁻¹?

Correct answer
2 ± 0.01
Setup, choices and feedback
The Scene at this step
Size
3 ×
Leg thickness
1 ×
Slice and count
Slices and cubes
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 6 ± 0.01
  Misconception: Thinks a scaled-up copy of the same shape keeps the same surface area to volume ratio.
  Feedback from the catalogue
  The same shape does not mean the same ratio. Scaling up $k$ ⁠ times multiplies the area by $k^2$ ⁠ and the volume by $k^3$ ⁠, so the ratio $\dfrac{\text{area}}{\text{volume}}$ ⁠ falls $k$ ⁠ times. A 1 cm cube: $6 : 1$ ⁠. A 2 cm cube: $24 : 8$ ⁠, that is $3 : 1$ ⁠.
- 18 ± 0.01
  Misconception: Thinks a bigger body has more surface for each unit of its volume.
  Feedback from the catalogue
  A big body has more surface in total, but less surface for each unit of volume. Every cubic centimetre inside needs oxygen and food that come in through the surface, and in a big body each square centimetre of surface has to serve more volume. That is why big animals have lungs, a folded gut and a circulation, and why cells are small.
- 0.5 ± 0.01
  Misconception: Mixes up how area and volume scale: cubes where it should square, or squares where it should cube.
  Feedback
  $0.5$ ⁠ is volume over area. The ratio asked for is area over volume.
Feedback for a right answer
Area $6 \times 3^2 = 54$ ⁠, volume $3^3 = 27$ ⁠, ratio $\dfrac{54}{27} = 2$ ⁠: a third of the small cube’s. And for a 10 µm cube, the size of a cell, the ratio is a thousand times the centimetre cube’s.

Hint
A cube has six faces, each a square of edge times edge.

Worked solution
1. Surface area: $6$ ⁠ faces of $3 \times 3$ ⁠: $6 \times 9 = 54\ \mathrm{cm^2}$ ⁠.
2. Volume: $3 \times 3 \times 3 = 27\ \mathrm{cm^3}$ ⁠.
3. The ratio: $\dfrac{54}{27} = 2$ ⁠, that is 2 cm⁻¹.
If you answered $6$ ⁠, you assumed the big cube keeps the small one’s ratio.
Step 6 of 6
Open this step
Why cells are small
Practice · Choose
Question
Why are most cells small, a few tens of micrometres at most?

Correct answer
- Correct A big cell would not have enough membrane to bring oxygen and food into all its volume and get the waste out.
Setup, choices and feedback
The Scene at this step
Size
0.2 ×
Leg thickness
1 ×
Slice and count
Whole tables
Choices
Shown in a random order.
- Correct A big cell would not have enough membrane to bring oxygen and food into all its volume and get the waste out.
- Wrong Because a big cell has more surface per volume, which is dangerous.
  Misconception: Thinks a bigger body has more surface for each unit of its volume.
  Feedback from the catalogue
  A big body has more surface in total, but less surface for each unit of volume. Every cubic centimetre inside needs oxygen and food that come in through the surface, and in a big body each square centimetre of surface has to serve more volume. That is why big animals have lungs, a folded gut and a circulation, and why cells are small.
- Wrong Because there is no room in the body for big cells.
  Feedback
  A body could be made of fewer big cells instead of many small ones. The problem lies in each cell alone: its membrane has to serve all the volume inside it.
Feedback for a right answer
Right. Everything that enters or leaves a cell crosses its membrane, and every substance has to diffuse to the cell’s middle. The bigger the cell, the less membrane for each unit of volume, and the longer the way to the middle. That is why big creatures are built of many small cells, and why they have lungs, a gut and a circulation.

Worked solution
1. A cell uses oxygen and food throughout its volume, and takes them in through its surface.
2. When a cell grows $k$ ⁠ times, its needs grow $k^3$ ⁠ times and its membrane only $k^2$ ⁠.
3. Past a certain size the membrane is not enough, and the cell must divide.

Across the unit · 11 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

Diffusion

A random walk

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The end

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Still moving

Question

Correct answer

The Scene at this step

Choices

Worked solution

Warmer

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

In the lungs

Question

Correct answer

The Scene at this step

Choices

Worked solution

Osmosis

Only water

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

What osmosis is

Question

Correct answer

The Scene at this step

Choices

Worked solution

The concentration at the end

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

How much water

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

In balance

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Cells in solutions

Pure water

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Swelling

Question

Correct answer