Answer key · Unit 13

Ecology.

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 8

Lesson 1 of 8 · About 20 minutes

Food chains

Producers, consumers and trophic levels in a lake’s food chain, the arrows that show where the energy flows, and why energy flows one way and leaves as heat while matter goes round in a cycle.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Who eats whom
Practice · Choose
Question
The lake’s food chain: algae, zooplankton (water fleas and copepods), small fish, perch and herons. Which are the secondary consumers?

Correct answer
- Correct The small fish
Setup, choices and feedback
The Scene at this step
Energy passed on
10 %
Sunlight
125 W/m²
Build the energy pyramid
Pond
Choices
Shown in a random order.
- Correct The small fish
- Wrong The zooplankton
  Misconception: Counts the producer as the first consumer, so calls the primary consumer the secondary one.
  Feedback from the catalogue
  Consumers are counted from the first one that eats the producer. The algae are producers, not consumers at all. The zooplankton, which eat algae, are primary consumers; the small fish, which eat zooplankton, are secondary consumers; the perch are tertiary. A consumer’s number is one less than its trophic level.
- Wrong The perch
  Feedback
  The perch eat small fish, which eat zooplankton: they are tertiary consumers.
- Wrong The algae
  Feedback
  The algae are the producers: they make their own food by photosynthesis and eat nobody.
Feedback for a right answer
The small fish. The algae are the producers, the first trophic level; the zooplankton that graze them are the primary consumers, the second level; and the small fish that eat the zooplankton are the secondary consumers, the third level.

Hint
Start counting consumers from the first one that eats the algae.

Worked solution
1. The producers, the algae, make their own food from sunlight. They are not consumers.
2. The first consumer to eat them, the zooplankton, is the primary consumer.
3. Whatever eats the primary consumer is the secondary consumer: the small fish.
4. After them come the perch (tertiary) and the herons at the top.
If you chose the zooplankton, you counted the algae as the first consumer.
Step 2 of 5
Open this step
One way
Practice · Choose
Question
The heron breathes out carbon dioxide, and the algae can build sugar from it next spring. What happens to the energy a level gets and does not pass on?

Correct answer
- Correct It leaves as heat, and does not come back to the chain
Setup, choices and feedback
The Scene at this step
Energy passed on
10 %
Sunlight
125 W/m²
Build the energy pyramid
Pond
Choices
Shown in a random order.
- Correct It leaves as heat, and does not come back to the chain
- Wrong The decomposers return it to the algae, like the carbon
  Misconception: Thinks energy goes round an ecosystem like matter, back to the producers through the decomposers.
  Feedback from the catalogue
  Matter goes round in a cycle: the carbon a heron breathes out can be built into algae next spring. Energy does not go round. It comes in as sunlight and leaves as heat, through the decomposers too, which burn what they eat in cellular respiration just as everything else does. Algae cannot use heat, so every ecosystem needs fresh sunlight all the time.
- Wrong It is stored in the mud, and stays in the lake for ever
  Feedback
  A little energy is buried in the mud, in dead matter that does not rot, but most of the energy not passed on is burnt in respiration and leaves as heat.
Feedback for a right answer
It leaves the lake as heat and does not come back. Every creature burns most of its food in cellular respiration, and that energy warms the water and the air. Algae cannot use heat, so the lake needs fresh sunlight every year. Press Energy pyramid under Build the energy pyramid: the orange rims of each tier are that energy.

Hint
What leaves every living thing that respires, besides carbon dioxide?

Worked solution
1. Matter goes round: the same carbon atoms pass from the air to the algae, the fish, the heron and back to the air.
2. Energy flows one way: in as light, along as food, out as heat.
3. The decomposers burn what they eat too, so through them as well the energy leaves as heat.
4. Heat cannot drive photosynthesis, so it does not return to the chain.
If you chose the decomposers returning it to the algae, you mixed up the cycling of matter with the flow of energy.
Step 3 of 5
Open this step
A tenth
Practice · Calculate
Question
Each year the lake’s algae capture $20\,000$ ⁠ dots of energy, and each dot is a year of food for one heron. Each level passes on 10% of the energy it gets. How many dots reach the zooplankton?

Correct answer
2000 ± 1
Setup, choices and feedback
The Scene at this step
Energy passed on
10 %fixed
Sunlight
125 W/m²fixed
Build the energy pyramid
Pondfixed
Accepted within
± 1
Computed from the Scene at this step’s setting
Energy reaching the water fleas and copepods, in heron-years/nodes/pond/outputs/zooplankton
Wrong answers it recognises
- 18000 ± 1
  Misconception: Reads the 10% rule upside down: thinks 10% is lost at each link and 90% passed on.
  Feedback from the catalogue
  The 10% rule says the opposite: only about a tenth is passed on, and about nine tenths is lost. Most of the energy a level gets is burnt in cellular respiration and leaves as heat, and part of what is eaten is not digested or dies uneaten. So $20\,000$ ⁠ dots of algae become $2000$ ⁠ for the zooplankton, not $18\,000$ ⁠.
- 20000 ± 1
  Misconception: Thinks everything an animal eats becomes its body, so all the energy of one level passes to the next.
  Feedback from the catalogue
  An animal does not build all it eats into its body. Most of its food it burns in order to live: to swim, breathe, digest, repair its body and keep warm. Part passes out undigested, and part of the prey dies without being eaten. Only what is built into the body can pass to the next level, and in this lake that is a tenth.
Feedback for a right answer
$20\,000 \times 0.1 = 2000$ ⁠ dots. The other $18\,000$ ⁠ dots are not passed on: the algae burn most of them to stay alive, and some algae die and sink without being eaten.

Hint
10% is a tenth.

Worked solution
1. The algae’s energy: $20\,000$ ⁠ dots.
2. The share passed on: 10%, that is $0.1$ ⁠.
3. $20\,000 \times 0.1 = 2000$ ⁠.
4. $2000$ ⁠ dots for the zooplankton. In the next Step, press Energy pyramid under Build the energy pyramid to see it as the pyramid’s second tier.
If you got $18\,000$ ⁠, you turned the rule upside down: a tenth is passed on, nine tenths are lost. If you answered $20\,000$ ⁠, you assumed that everything eaten becomes the eater’s body.
Step 4 of 5
Open this step
Arrows
Practice · Choose
Question
You draw the lake’s food chain with arrows. Which way does the arrow between the algae and the zooplankton point?

Correct answer
- Correct From the algae to the zooplankton
Setup, choices and feedback
The Scene at this step
Energy passed on
10 %
Sunlight
125 W/m²
Build the energy pyramid
Pond
Choices
Shown in a random order.
- Correct From the algae to the zooplankton
- Wrong From the zooplankton to the algae
  Misconception: Draws food-chain arrows from the eater to the eaten, reading them as “eats”.
  Feedback from the catalogue
  In a food chain the arrow shows where the energy and matter flow: from the eaten to the eater. So it leaves the algae and points at the zooplankton, then from the zooplankton to the small fish. Read the arrow as “is eaten by”, not “eats”.
Feedback for a right answer
From the algae to the zooplankton: the arrow shows where the energy and matter flow, from the eaten to the eater. So the whole chain points upwards, from the algae to the heron, the same way the dots of energy fly to the pyramid.

Hint
What does the arrow show: who eats, or where the energy goes?

Worked solution
1. In a food chain an arrow is a flow, not an action.
2. The algae’s energy and matter pass to the zooplankton when it eats them.
3. So the arrow leaves the algae and points at the zooplankton: “are eaten by”.
If you chose an arrow from the zooplankton to the algae, you read the arrow as “eats”. It is a common slip, and it turns the whole web backwards.
Step 5 of 5
Open this step
Less sun
Practice · Predict, then look
Question
The lake feeds two herons. Now Sunlight drops from 125 W/m² to 50 W/m², as in a shaded lake. The herons eat perch, not algae. How many herons can the lake feed?

Correct answer
- Correct None
Setup, choices and feedback
The Scene at this step
Energy passed on
10 %fixed
Sunlight
125 W/m²fixed
Build the energy pyramid
Pond
Choices
- Wrong Still two: they eat perch, not algae
  Misconception: Thinks a top predator depends only on the prey it eats, not on the producers at the base of the chain.
  Feedback from the catalogue
  The heron eats perch, but all the energy in the perch began as sunlight the algae captured. Each level gets a fixed share of the level below, so a change at the base is passed all the way to the top. Less light, fewer algae, and less of everything above them, until there is not enough left for a heron.
- Wrong One
  Feedback
  Close, but work it out: $0.4$ ⁠ of the light gives $0.4$ ⁠ of the energy at every level, and at the top $0.4 \times 2 = 0.8$ ⁠, less than one heron.
- Correct None
Once the prediction is locked in, the Scene changes
Sunlight
50 W/m²fixed
The Scene then shows: 0 · Herons the lake can feed for a year/nodes/pond/outputs/herons
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
None. Less light, fewer algae: $8000$ ⁠ dots instead of $20\,000$ ⁠. A tenth of that to the zooplankton, a tenth of that to the fish, and so on, and only $0.8$ ⁠ of a dot reaches the top, less than a year of food for one heron. The chain ends at the perch, even though herons do not eat algae at all.

Worked solution
1. The algae grow in proportion to the light: 50 W/m² is $\tfrac{2}{5}$ ⁠ of 125 W/m², so $\tfrac{2}{5} \times 20\,000 = 8000$ ⁠ dots.
2. Four links to the herons: $8000 \times 0.1^4 = 0.8$ ⁠.
3. A heron needs a whole dot a year, and $0.8 < 1$ ⁠.
4. The lake feeds no heron.
If you predicted two herons, you took the heron to depend only on its prey. But each level gets a share of what is below it, and a change at the base reaches the top.

Lesson 2 of 8 · About 25 minutes

The ten per cent rule

How much energy reaches each level, when every link multiplies by the same share again. Why a small change in efficiency changes the top a great deal, why food chains are short, and which pyramid always stands upright.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Four links up
Practice · Calculate
Question
The algae capture $20\,000$ ⁠ dots a year, and each link passes on 10%. From the algae to the herons there are four links: algae to zooplankton, to small fish, to perch, to herons. How many dots reach the herons?

Correct answer
2 ± 0.05
Setup, choices and feedback
The Scene at this step
Energy passed on
10 %fixed
Sunlight
125 W/m²fixed
Build the energy pyramid
Pondfixed
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Energy reaching the herons, in heron-years/nodes/pond/outputs/heronFood
Wrong answers it recognises
- 8000 ± 1
  Misconception: Takes a tenth per link as one subtraction or one multiplication by 0.1 × the number of links, instead of multiplying by 0.1 again at each link.
  Feedback from the catalogue
  At each link you multiply by $0.1$ ⁠ again, because each level gets a tenth of what the level below it got, not of the algae. After four links: $0.1 \times 0.1 \times 0.1 \times 0.1 = 0.1^4 = \tfrac{1}{10\,000}$ ⁠. It is exponential decay: $E_n = E_1 \cdot f^{\,n-1}$ ⁠.
- 13120 ± 2
  Misconception: Reads the 10% rule upside down: thinks 10% is lost at each link and 90% passed on.
  Feedback from the catalogue
  The 10% rule says the opposite: only about a tenth is passed on, and about nine tenths is lost. Most of the energy a level gets is burnt in cellular respiration and leaves as heat, and part of what is eaten is not digested or dies uneaten. So $20\,000$ ⁠ dots of algae become $2000$ ⁠ for the zooplankton, not $18\,000$ ⁠.
Feedback for a right answer
$20\,000 \times 0.1^4 = 2$ ⁠ dots: a year’s food for two herons, from a lake about a square kilometre in size. A tenth of a tenth of a tenth of a tenth is a ten-thousandth.

Hint
Multiply by $0.1$ ⁠ four times, once for each link.

Worked solution
1. Each level gets a tenth of the level below it: $E_n = E_1 \cdot 0.1^{\,n-1}$ ⁠.
2. The herons are the fifth level: $n - 1 = 4$ ⁠.
3. $0.1^4 = 0.0001$ ⁠.
4. $20\,000 \times 0.0001 = 2$ ⁠.
5. Press Energy pyramid under Build the energy pyramid and check on the chart: $20\,000$ ⁠, $2000$ ⁠, $200$ ⁠, $20$ ⁠, $2$ ⁠.
If you got $8000$ ⁠, you multiplied by $0.1 \times 4$ ⁠ instead of $0.1^4$ ⁠. If you got about $13\,000$ ⁠, you kept 90% at each link instead of 10%.
Step 2 of 6
Open this step
One heron
Practice · Reach the target
Question
Sunlight is fixed at 125 W/m². Change Energy passed on until the lake feeds exactly one heron, then press Check.

Goal
- 1 ± 0
  At the start of the step: 2
Setup, choices and feedback
The Scene at this step
Energy passed on
10 %
Sunlight
125 W/m²fixed
Build the energy pyramid
Energy pyramid
Read from the Scene at
Herons the lake can feed for a year/nodes/pond/outputs/herons
One setting that reaches it
Energy passed on
9 %
Feedback for a right answer
At 9%: $20\,000 \times 0.09^4 \approx 1.3$ ⁠ dots reach the herons, a year of food for one heron and a little more. At 8% only $0.8$ ⁠ arrive, and there is no heron at all. One per cent less at every link takes a third off what reaches the top.

Feedback for a wrong answer
Not yet. Check on the chart how many herons the lake feeds, and change Energy passed on one per cent at a time.

Hint
At 10% there are two herons. Try a little less.

Worked solution
1. To feed one heron and not two, you need $1 \le 20\,000 \cdot f^4 < 2$ ⁠.
2. $f^4 \ge \tfrac{1}{20\,000} = 0.00005$ ⁠, so $f \ge 0.00005^{1/4} \approx 0.084$ ⁠.
3. $f^4 < 0.0001$ ⁠, so $f < 0.1$ ⁠.
4. The only whole percentage in that range: 9%. On the chart: the lake feeds one heron.
Step 3 of 6
Open this step
Twice the share
Practice · Calculate
Question
Now each link passes on 20%, twice as much as before, and the algae still capture $20\,000$ ⁠ dots. How many herons can the lake feed?

Correct answer
32 ± 0
Setup, choices and feedback
The Scene at this step
Energy passed on
20 %fixed
Sunlight
125 W/m²fixed
Build the energy pyramid
Pondfixed
Accepted within
± 0
Computed from the Scene at this step’s setting
Herons the lake can feed for a year/nodes/pond/outputs/herons
Wrong answers it recognises
- 4 ± 0
  Misconception: Thinks doubling the share passed on at each link doubles what reaches the top.
  Feedback from the catalogue
  The share passed on appears as a power, once for every link. When it doubles, what reaches the top after four links goes up $2^4 = 16$ ⁠ times, not twice: $0.2^4 = 16 \times 0.1^4$ ⁠. That is why a small change in each link’s efficiency changes a great deal what is left for the top predators.
Feedback for a right answer
$20\,000 \times 0.2^4 = 32$ ⁠ herons, $16$ ⁠ times as many as at 10%, not twice. The share passed on appears once for every link, so doubling it multiplies what reaches the top by $2^4$ ⁠.

Hint
$0.2^4 = 0.2 \times 0.2 \times 0.2 \times 0.2$ ⁠.

Worked solution
1. $E_5 = 20\,000 \times 0.2^4$ ⁠.
2. $0.2^4 = 0.0016$ ⁠.
3. $20\,000 \times 0.0016 = 32$ ⁠.
4. $32$ ⁠ dots, that is $32$ ⁠ herons. The pond draws only three, but the chart counts them all.
If you answered $4$ ⁠, you doubled the number of herons, as the efficiency doubled. But the efficiency is doubled at each of the four links.
Step 4 of 6
Open this step
A sixth level
Practice · Calculate
Question
Suppose the lake had a predator that ate only herons, a sixth level. At 10% a link, how many dots of energy would reach it in a year?

Correct answer
0.2 ± 0.01
Setup, choices and feedback
The Scene at this step
Energy passed on
10 %fixed
Sunlight
125 W/m²fixed
Build the energy pyramid
Pondfixed
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Energy reaching the herons, in heron-years/nodes/pond/outputs/heronFood
Feedback for a right answer
$2 \times 0.1 = 0.2$ ⁠ of a dot: a fifth of a year’s food for an animal the size of a heron. Not enough for even one predator, which is why food chains are hardly ever longer than four or five links.

Hint
How much reaches the herons, and what is a tenth of that?

Worked solution
1. The herons get $2$ ⁠ dots.
2. The sixth level gets a tenth of that: $2 \times 0.1 = 0.2$ ⁠.
3. Or straight from the algae: $20\,000 \times 0.1^5 = 0.2$ ⁠.
4. $0.2 < 1$ ⁠: no such predator could live in the lake.
Step 5 of 6
Open this step
How long a chain
Practice · Calculate
Question
At 25% a link the algae still capture $20\,000$ ⁠ dots. If the chain went on, with an animal needing one dot a year at the top of each level, how many levels in all, counting the algae, could get at least one dot?

Correct answer
8 ± 0
Setup, choices and feedback
The Scene at this step
Energy passed on
25 %fixed
Sunlight
125 W/m²fixed
Build the energy pyramid
Energy pyramid
Accepted within
± 0
A fixed value in the question
Feedback for a right answer
$8$ ⁠ levels. Level $n$ ⁠ gets $20\,000 \times 0.25^{\,n-1}$ ⁠, and that is at least $1$ ⁠ as long as $n - 1 \le \log_4 20\,000 \approx 7.1$ ⁠. Even at a very high efficiency the chain comes to an end, because the decay is exponential.

Hint
How many times can you divide $20\,000$ ⁠ by $4$ ⁠ and still stay above $1$ ⁠?

Worked solution
1. You want $20\,000 \times 0.25^{\,n-1} \ge 1$ ⁠, that is $4^{\,n-1} \le 20\,000$ ⁠.
2. Take logarithms: $n - 1 \le \log_4 20\,000 = \dfrac{\log 20\,000}{\log 4} \approx \dfrac{4.30}{0.602} \approx 7.14$ ⁠.
3. $n - 1$ ⁠ is whole, so at most $7$ ⁠, and $n = 8$ ⁠.
4. Check: $4^7 = 16\,384 \le 20\,000$ ⁠, but $4^8 = 65\,536 > 20\,000$ ⁠.
If you answered $7$ ⁠, you counted the links, the steps between levels, not the levels themselves.
Step 6 of 6
Open this step
Three pyramids
Practice · Choose
Question
You can build a pyramid of trophic levels three ways: by the energy flowing through each level in a year, by the number of individuals, or by the weight of what is alive at one moment (biomass). Which one always stands with each tier smaller than the one below?

Correct answer
- Correct The pyramid of energy
Setup, choices and feedback
The Scene at this step
Energy passed on
10 %
Sunlight
125 W/m²
Build the energy pyramid
Energy pyramid
Choices
Shown in a random order.
- Correct The pyramid of energy
- Wrong The pyramid of numbers
  Misconception: Thinks pyramids of numbers and of biomass always stand upright, like the pyramid of energy.
  Feedback from the catalogue
  Only the pyramid of energy always stands upright, because energy can only be lost along the chain. A pyramid of numbers can stand on its head: one oak tree feeds thousands of caterpillars. A pyramid of biomass can flip over in a lake: algae are eaten almost as fast as they grow, so at any moment they can weigh less than the zooplankton grazing them, even though over a year they make far more.
- Wrong The pyramid of biomass
  Misconception: Thinks pyramids of numbers and of biomass always stand upright, like the pyramid of energy.
  Feedback from the catalogue
  Only the pyramid of energy always stands upright, because energy can only be lost along the chain. A pyramid of numbers can stand on its head: one oak tree feeds thousands of caterpillars. A pyramid of biomass can flip over in a lake: algae are eaten almost as fast as they grow, so at any moment they can weigh less than the zooplankton grazing them, even though over a year they make far more.
- Wrong All three
  Misconception: Thinks pyramids of numbers and of biomass always stand upright, like the pyramid of energy.
  Feedback from the catalogue
  Only the pyramid of energy always stands upright, because energy can only be lost along the chain. A pyramid of numbers can stand on its head: one oak tree feeds thousands of caterpillars. A pyramid of biomass can flip over in a lake: algae are eaten almost as fast as they grow, so at any moment they can weigh less than the zooplankton grazing them, even though over a year they make far more.
Feedback for a right answer
The pyramid of energy. Energy can only be lost along the chain, never added, so each tier is smaller than the one it stands on. In the Scene each tier stands exactly on the part of the tier below that was passed up.

Hint
Think of one tree full of caterpillars.

Worked solution
1. Energy: each level gets only a share of what the level below got. Always less.
2. Numbers: one oak tree feeds thousands of caterpillars, so the bottom tier can be narrower than the second.
3. Biomass: in a lake the algae are eaten almost as fast as they grow, so at a given moment they can weigh less than the zooplankton.
4. Only the pyramid of energy always stands upright.

Lesson 3 of 8 · About 30 minutes

Predator and prey

Foxes and rabbits on one meadow rise and fall in a cycle: four quarters of a loop round a balance point, the foxes lagging behind the rabbits, and the surprise in the averages: good years for rabbits raise foxes, and better hunters stay few.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Four quarters
Practice · Predict, then look
Question
Year $2.8$ ⁠: the meadow has nearly $300$ ⁠ rabbits, more than the balance point of $120$ ⁠, and only $7$ ⁠ foxes, fewer than $15$ ⁠. The note in the Scene says the foxes are raising more cubs. What happens next?

Correct answer
- Correct The rabbits crash, because there are too many foxes
Setup, choices and feedback
The Scene at this step
Rabbit birth rate
1 per yearfixed
Hunting success
20 %fixed
Years
0.07 (Start → 40 years on)fixed
Choices
- Wrong The rabbits keep multiplying without end
  Feedback
  They could, if the foxes did not multiply. But plenty of rabbits means more fox cubs, and in the end too many foxes.
- Correct The rabbits crash, because there are too many foxes
- Wrong The foxes go hungry and dwindle
  Feedback
  That happens, but later. Right now there are plenty of rabbits, and the foxes are still rising.
Once the prediction is locked in, the Scene changes
Years
0.1 (Start → 40 years on)fixed
The Scene then shows: 3 · 0 the start, 1 rabbits multiply, 2 foxes multiply, 3 rabbits crash, 4 foxes starve, 5 two foxes or fewer/nodes/meadow/outputs/regime
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The rabbits crash. The foxes keep multiplying, and once there are more than $15$ ⁠ of them they eat rabbits faster than rabbits are born. By year $4$ ⁠ there are already $33$ ⁠ foxes, and the rabbits are falling fast, from $340$ ⁠ to $240$ ⁠, even though there are still plenty of them.

Worked solution
1. As long as there are more than $120$ ⁠ rabbits, each fox catches enough to raise more cubs than die: the foxes multiply.
2. As long as there are fewer than $15$ ⁠ foxes, more rabbits are born than eaten: the rabbits multiply too.
3. The foxes pass $15$ ⁠, and from then on more rabbits are eaten than born: the rabbits start to crash while the foxes are still rising.
4. That is the third quarter of the cycle; after it the foxes go hungry.
Step 2 of 6
Open this step
The foxes lag
Practice · Choose
Question
The chart shows the foxes peaking about a year after the rabbits: the rabbits are already falling, and the foxes are still rising. Why?

Correct answer
- Correct The foxes multiply as long as there are more than $120$ ⁠ rabbits, even while the rabbits are falling
Setup, choices and feedback
The Scene at this step
Rabbit birth rate
1 per yearfixed
Hunting success
20 %fixed
Years
0.125 (Start → 40 years on)
Choices
Shown in a random order.
- Correct The foxes multiply as long as there are more than $120$ ⁠ rabbits, even while the rabbits are falling
- Wrong The rabbits crash because the grass runs out, and the foxes only respond later
  Misconception: Thinks the prey crash because they run out of food, not because of the predators.
  Feedback from the catalogue
  On the meadow in the Scene the grass never runs out: the rabbits have unlimited food. What brings them down is the foxes. When there are more than $15$ ⁠ foxes, they eat rabbits faster than rabbits are born. In the wild food limits them too, but the predator–prey cycle happens without it.
- Wrong Fox cubs take a year to grow up before they hunt
  Feedback
  In the wild that does add a delay, but the model in the Scene has no ages at all, and still lags. The delay comes from the numbers alone.
Feedback for a right answer
Because the foxes keep multiplying as long as there are more rabbits than the balance point, $120$ ⁠. Even once the rabbits are falling from their peak, there are still plenty of them, and each fox catches enough to raise more cubs than die. Only when the rabbits drop below $120$ ⁠ do the foxes start to decline.

Hint
When do the foxes stop multiplying: when there are fewer rabbits than what?

Worked solution
1. What makes foxes multiply: enough rabbits to catch. The line is $R^* = 120$ ⁠.
2. The rabbits peak in year $3.4$ ⁠ and start to fall, but stay above $120$ ⁠ until about year $4.4$ ⁠.
3. All that time the foxes are still rising, and they peak when the rabbits cross $120$ ⁠: around year $4.5$ ⁠.
4. So the foxes’ wave lags behind the rabbits’.
If you chose the rabbits crashing because the grass runs out: on this meadow the grass never runs out. The foxes alone bring them down.
Step 3 of 6
Open this step
The balance point
Practice · Predict, then look
Question
The rabbits’ average over a whole turn is the balance point, $R^* = \dfrac{m}{kb}$ ⁠, with $m = 0.8$ ⁠ and $k = 0.1$ ⁠, and $b$ ⁠ is Hunting success as a fraction, times $\tfrac13$ ⁠. Now $b = 0.2 \times \tfrac13$ ⁠, and $R^* = 120$ ⁠. Predict: how many rabbits will there be on average when Hunting success rises to 30%?

Correct answer
80 ± 3
A number prediction counts as right within ± 3
Setup, choices and feedback
The Scene at this step
Rabbit birth rate
1 per yearfixed
Hunting success
20 %fixed
Years
40 years on
Once the prediction is locked in, the Scene changes
Hunting success
30 %fixed
The Scene then shows: 80 · Rabbits on average over one turn of the loop (the balance point)/nodes/meadow/outputs/meanRabbits
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$80$ ⁠: $R^* = \dfrac{0.8}{0.1 \times 0.3 / 3} = 80$ ⁠. Foxes that hunt better need fewer rabbits to raise as many cubs as they must, so they bring the prey’s average down. $R^*$ ⁠ shrinks in inverse proportion to the hunting success: $1.5$ ⁠ times the success, $\tfrac{2}{3}$ ⁠ of the rabbits.

Worked solution
1. $b = 0.3 \times \tfrac13 = 0.1$ ⁠.
2. $kb = 0.1 \times 0.1 = 0.01$ ⁠.
3. $R^* = \dfrac{0.8}{0.01} = 80$ ⁠.
4. On the chart the average drops from $120$ ⁠ to $80$ ⁠.
Or without recomputing: $R^*$ ⁠ is inversely proportional to $b$ ⁠, so $120 \times \dfrac{20}{30} = 80$ ⁠.
Step 4 of 6
Open this step
Breaking even
Practice · Calculate
Question
In the model each fox dies at a rate of $0.8$ ⁠ a year, and each rabbit it catches raises $0.1$ ⁠ of a cub. How many rabbits must a fox catch a year for the foxes to raise exactly enough cubs to replace those that die?

Correct answer
8 ± 0.1
Setup, choices and feedback
The Scene at this step
Rabbit birth rate
1 per yearfixed
Hunting success
20 %fixed
Years
40 years on
Accepted within
± 0.1
A fixed value in the question
Feedback for a right answer
$8$ ⁠: $\dfrac{0.8}{0.1} = 8$ ⁠ rabbits per fox a year. That is exactly what a fox catches at the balance point, however good a hunter it is: a better hunter reaches $8$ ⁠ catches with fewer rabbits on the meadow, which is why $R^*$ ⁠ goes down.

Hint
Births equal deaths.

Worked solution
1. Fox births a year, per fox: $0.1$ ⁠ of a cub times the number of catches $c$ ⁠.
2. Deaths per fox: $0.8$ ⁠ a year.
3. Balance: $0.1\,c = 0.8$ ⁠.
4. $c = \dfrac{0.8}{0.1} = 8$ ⁠.
At the balance point each fox catches $b R^* = \dfrac{m}{k}$ ⁠ rabbits: the same answer.
Step 5 of 6
Open this step
Bumper years
Practice · Predict, then look
Question
Under Rabbit birth rate, bumper years: the rate rises from $1$ ⁠ to $1.5$ ⁠ a year. What goes up on average over a whole turn?

Correct answer
- Correct The foxes’ average
Setup, choices and feedback
The Scene at this step
Rabbit birth rate
1 per yearfixed
Hunting success
20 %fixed
Years
40 years on
Choices
- Wrong The rabbits’ average
  Misconception: Thinks prey that breed faster are more numerous on average.
  Feedback from the catalogue
  The balance-point formula says otherwise: $R^* = \dfrac{m}{kb}$ ⁠ depends only on the foxes, their death rate and how well they hunt. The extra rabbits born in bumper years feed extra foxes, so the foxes’ average rises, $F^* = \dfrac{a}{b}$ ⁠, and the rabbits’ average stays at $120$ ⁠.
- Correct The foxes’ average
- Wrong Both
  Feedback
  Both peaks rise, but the rabbits’ average, $R^* = \dfrac{m}{kb}$ ⁠, does not depend on their birth rate.
Once the prediction is locked in, the Scene changes
Rabbit birth rate
1.5 per yearfixed
The Scene then shows: 22.5 · Foxes on average over one turn of the loop (the balance point)/nodes/meadow/outputs/meanFoxes
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The foxes’ average, from $15$ ⁠ to $22.5$ ⁠, and the rabbits’ average stays at $120$ ⁠. The extra rabbits feed extra foxes. $F^* = \dfrac{a}{b}$ ⁠ grows with the rabbits’ birth rate, and $R^* = \dfrac{m}{kb}$ ⁠ does not depend on it at all.

Worked solution
1. The balance point: $R^* = \dfrac{m}{kb}$ ⁠ and $F^* = \dfrac{a}{b}$ ⁠.
2. The birth rate $a$ ⁠ appears only in $F^*$ ⁠.
3. $F^* = 15 \times 1.5 = 22.5$ ⁠; $R^*$ ⁠ stays at $120$ ⁠.
4. On the chart, the averages: $120$ ⁠ rabbits and $22.5$ ⁠ foxes. The loop is wider and faster, but goes round the same number of rabbits.
If you predicted that the rabbits would go up, that is the model’s surprise: what feeds the prey ends up raising the predators.
Step 6 of 6
Open this step
Better hunters
Practice · Reach the target
Question
Rabbit birth rate is fixed at $1$ ⁠, and on average the meadow has $15$ ⁠ foxes. Change Hunting success until the foxes’ average falls to $7.5$ ⁠, half, then press Check.

Goal
- 7.5 ± 0.1
  At the start of the step: 15
Setup, choices and feedback
The Scene at this step
Rabbit birth rate
1 per yearfixed
Hunting success
20 %
Years
40 years on
Read from the Scene at
Foxes on average over one turn of the loop (the balance point)/nodes/meadow/outputs/meanFoxes
One setting that reaches it
Hunting success
40 %
Feedback for a right answer
At 40%, twice 20%: $F^* = \dfrac{a}{b}$ ⁠ is inversely proportional to the hunting success. Foxes that hunt twice as well end up half as many, because they thin out their prey to a level that can support only a few of them.

Feedback for a wrong answer
Not yet. Watch the foxes’ average on the chart. Perhaps it is the opposite way from what seems sensible?

Hint
If halving the foxes means changing $b$ ⁠, which way?

Worked solution
1. $F^* = \dfrac{a}{b}$ ⁠, and $b$ ⁠ is proportional to Hunting success.
2. To halve $F^*$ ⁠, double $b$ ⁠.
3. From 20% to 40%.
4. On the chart: $7.5$ ⁠ foxes and $60$ ⁠ rabbits on average.

Lesson 4 of 8 · About 25 minutes

The carbon cycle

Carbon passes from the air to the grass, the rabbit, the dead leaves and the soil, and back to the air. What plants build themselves from, how much they breathe back, how the decomposers close the cycle, and what happens when they are too slow or too fast.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Where the carbon comes from
Practice · Predict, then look
Question
From 1 March to the end of August the meadow’s plants grow from about $180$ ⁠ to about $480$ ⁠ grams of carbon a square metre. Where does the extra carbon come from?

Correct answer
- Correct From the carbon dioxide in the air
Setup, choices and feedback
The Scene at this step
Follow
Carbonfixed
Decomposers
100 %fixed
Labels
offdefault
Through the year
1 Marchfixed
Choices
- Correct From the carbon dioxide in the air
- Wrong From the soil, through the roots
  Misconception: Thinks plants get the material they grow from out of the soil.
  Feedback from the catalogue
  If plants built themselves out of the soil, it would empty. In the Scene the humus in the soil barely changes while the grass takes up hundreds of grams of carbon. The carbon, about half a plant’s dry weight, comes from the carbon dioxide in the air, in through the stomata, by photosynthesis. From the soil the plant takes water and a little of some minerals, such as nitrogen.
- Wrong From the water the roots take up
  Misconception: Thinks the plant’s carbon comes from the water its roots take up.
  Feedback from the catalogue
  Water holds no carbon: a water molecule is two hydrogens and one oxygen, $\mathrm{H_2O}$ ⁠. In photosynthesis the water gives its hydrogen to the sugar, and its oxygen goes out to the air. The sugar’s carbon comes from carbon dioxide, $\mathrm{CO_2}$ ⁠, which enters the leaves from the air.
Once the prediction is locked in, the Scene changes
Through the year
0.5 (1 March → A year on)fixed
The Scene then shows: 467.8 · G/m² in the humus that turns over within a few years/nodes/meadow/outputs/soil
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From the air: from the carbon dioxide the leaves take in through their stomata and build into sugar by photosynthesis. On the chart the humus in the soil has barely changed, about $475$ ⁠ grams, while the grass took up $300$ ⁠ grams. The soil gives a plant water and a few minerals, not most of its substance.

Worked solution
1. About half a dry plant’s weight is carbon.
2. In photosynthesis: $\mathrm{6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2}$ ⁠. The sugar’s carbon comes from the $\mathrm{CO_2}$ ⁠.
3. Water holds no carbon, and the humus on the chart has hardly dropped.
4. So the extra carbon comes from the air.
In 1648 the Flemish chemist Jan van Helmont weighed the soil in a pot with a willow tree. Five years later the tree weighed $74$ ⁠ kilograms more, and the soil had lost only $60$ ⁠ grams. He concluded that the tree was made of water; he never suspected the air.
Step 2 of 6
Open this step
Breathing it back
Practice · Calculate
Question
In a year the meadow’s plants take up about $1000$ ⁠ grams of carbon over each square metre by photosynthesis. The note in the Scene says that in summer they breathe half the carbon they take in straight back out. How many grams of carbon are built into new leaves and roots in a year?

Correct answer
500 ± 5
Setup, choices and feedback
The Scene at this step
Follow
Carbonfixed
Decomposers
100 %fixed
Labels
offdefault
Through the year
0.2 (1 March → A year on)
Accepted within
± 5
A fixed value in the question
Wrong answers it recognises
- 1000 ± 5
  Misconception: Thinks plants only photosynthesise and do not respire, so all the carbon they take in builds new growth.
  Feedback from the catalogue
  Plants respire, day and night, like every living thing: they burn sugar in cellular respiration to run their cells, and breathe its carbon straight out as carbon dioxide. In the Scene they breathe out half of what they take in, and only the other half is built into leaves and roots.
Feedback for a right answer
$500$ ⁠ grams. The plants burn half the sugar they make in cellular respiration, to run their cells, and its carbon goes straight back to the air. Only the other half is built into the plant’s body, and from there it passes to the rabbit, the dead leaves and the decomposers.

Hint
Plants respire too.

Worked solution
1. Photosynthesis: about $1000$ ⁠ grams of carbon taken up.
2. The plants’ own respiration: half, about $500$ ⁠ grams, back to the air.
3. What is left for growth: $1000 - 500 = 500$ ⁠ grams.
If you answered $1000$ ⁠, you forgot that plants respire, day and night, like every living thing.
Step 3 of 6
Open this step
No decomposers
Practice · Predict, then look
Question
Decomposers is at 0%: nothing rots. The grass goes on growing and dying back. Where does the carbon pile up by the end of the year?

Correct answer
- Correct In the dead leaves on the ground
Setup, choices and feedback
The Scene at this step
Follow
Carbonfixed
Decomposers
0 %fixed
Labels
offdefault
Through the year
1 Marchfixed
Choices
- Correct In the dead leaves on the ground
- Wrong In the humus in the soil
  Feedback
  Humus is what the decomposers leave of the leaves they eat. With no decomposers no new humus forms.
- Wrong In the living plants, which keep growing
  Feedback
  The plants die back in autumn as usual. With no decomposers they even grow less, because they run short of nitrogen.
Once the prediction is locked in, the Scene changes
Through the year
A year onfixed
The Scene then shows: 771.8 · G/m² in dead leaves and dung on the ground/nodes/meadow/outputs/litter
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
In the dead leaves, which almost double, from about $410$ ⁠ to about $770$ ⁠ grams of carbon a square metre. With no decomposers there is nothing to breathe their carbon back into the air, and no new humus forms. About $320$ ⁠ grams taken from the air do not go back to it.

Worked solution
1. The grass takes up carbon and sheds dead leaves, with decomposers or without.
2. The decomposers are what eat dead leaves and breathe out most of their carbon, making humus from the rest.
3. With no decomposers the dead leaves stay, and the humus neither grows nor is eaten.
4. On the chart: the dead leaves rise to about $770$ ⁠ grams, the humus stays at about $477$ ⁠.
That is how peat builds up in cold, wet bogs, where rot is slower than growth.
Step 4 of 6
Open this step
Closing the cycle
Practice · Reach the target
Question
Through the year is at A year on. With Decomposers at half pace, the meadow ends the year with more carbon than it started with. Change Decomposers until the cycle closes: every store ends the year with what it held at the start. Press Check.

Goal
- 0 ± 5
  At the start of the step: 176.2
Setup, choices and feedback
The Scene at this step
Follow
Carbonfixed
Decomposers
50 %
Labels
offdefault
Through the year
A year on
Read from the Scene at
G/m² by which the stores end the year above where they began (0 when the cycle closes)/nodes/meadow/outputs/yearGap
One setting that reaches it
Decomposers
100 %
Labels
off
Feedback for a right answer
At 100%, the normal pace: the decomposers return to the air exactly what the grass took from it, net. The same stores, different atoms: the carbon in the grass now came out of the air this year, and last year’s carbon has already gone back.

Feedback for a wrong answer
Not yet. Read on the chart how far above or below its start the meadow ends the year.

Hint
At half pace there is a surplus. What do the decomposers need?

Worked solution
1. At half pace the dead leaves rot more slowly than they fall, and the meadow gains about $175$ ⁠ grams a year.
2. At double pace the rot eats into the humus too, and the meadow loses carbon.
3. Between the two is a pace at which rot equals fall: 100%.
4. On the chart: “A year on, every store holds what it held a year before”.
Step 5 of 6
Open this step
A peat bog
Practice · Calculate
Question
Decomposers are at half pace, as in cold, wet soil. The chart shows how much more carbon the meadow ends the year with than it started with. If that went on for a hundred years, how many kilograms of carbon would build up on each square metre?

Correct answer
17.62 kg ± 3%
Setup, choices and feedback
The Scene at this step
Follow
Carbonfixed
Decomposers
50 %fixed
Labels
offdefault
Through the year
A year on
Accepted within
± 3%
Computed from the Scene at this step’s setting
G/m² by which the stores end the year above where they began (0 when the cycle closes)/nodes/meadow/outputs/yearGap
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 17.6 kg: about $176$ ⁠ grams a year times $100$ ⁠ years. That is ten times all the carbon in the column of air over that square metre, about 1.8 kg. That is how peat bogs store carbon over thousands of years, and when they are drained it goes back to the air.

Hint
Multiply by $100$ ⁠, then convert grams to kilograms.

Worked solution
1. From the chart: about $176$ ⁠ grams of extra carbon each year.
2. In a hundred years: $176 \times 100 = 17\,600$ ⁠ grams.
3. In kilograms: $17\,600 \div 1000 = 17.6$ ⁠.
If you got $1.76$ ⁠, you divided by $10\,000$ ⁠ instead of $1000$ ⁠.
Step 6 of 6
Open this step
Faster rot
Practice · Predict, then look
Question
Warmer soil speeds up the decomposers. Decomposers will rise to 200%, double pace. What happens to the meadow over a year?

Correct answer
- Correct It loses carbon to the air
Setup, choices and feedback
The Scene at this step
Follow
Carbonfixed
Decomposers
100 %fixed
Labels
offdefault
Through the year
A year on
Choices
- Wrong It gains carbon
  Feedback
  Fast rot builds nothing: it breathes carbon out. What builds is the grass, and that has not changed.
- Wrong The books balance, as in a normal year
  Feedback
  In a normal year rot equals fall. Doubled, it eats faster than things fall.
- Correct It loses carbon to the air
Once the prediction is locked in, the Scene changes
Decomposers
200 %fixed
The Scene then shows: -227.5 · G/m² by which the stores end the year above where they began (0 when the cycle closes)/nodes/meadow/outputs/yearGap
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 carbon, about $230$ ⁠ grams a square metre in a year. The fast rot eats not only the year’s dead leaves but also the humus in the soil, and the meadow gives back to the air more carbon than the grass took from it. It is a real concern: warming that speeds up rot in soils adds more carbon dioxide to the air, which warms it further.

Worked solution
1. The grass takes up about as much carbon as in a normal year.
2. The decomposers breathe out twice as fast, and eat the humus built up in earlier years too.
3. What goes out to the air is more than what came in: the meadow becomes a source of carbon dioxide.
4. On the chart: “A year on, the meadow holds about $230$ ⁠ g less than it started with”.

Lesson 5 of 8 · About 25 minutes

The nitrogen cycle

The air is full of nitrogen that plants cannot use. Bacteria in clover’s nodules fix it, the decomposers free it from dead leaves, other bacteria turn it into nitrate, and the rain washes some away: a cycle that leaks, and what happens when rot is too slow or too fast.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Air full of nitrogen
Practice · Choose
Question
Four fifths of the air is nitrogen gas, about eight tonnes over each square metre. Yet grass is often short of nitrogen. Why?

Correct answer
- Correct Plants cannot break the triple bond of $\mathrm{N_2}$ ⁠; only certain bacteria can
Setup, choices and feedback
The Scene at this step
Follow
Nitrogenfixed
Decomposers
100 %
Labels
offdefault
Through the year
1 March
Choices
Shown in a random order.
- Correct Plants cannot break the triple bond of $\mathrm{N_2}$ ⁠; only certain bacteria can
- Wrong The grass does take the gas in through its leaves, but it is too dilute
  Misconception: Thinks plants take nitrogen gas straight from the air through their leaves.
  Feedback from the catalogue
  Plants cannot use nitrogen gas, even though it is four fifths of the air. Each molecule, $\mathrm{N_2}$ ⁠, is two nitrogen atoms held by a triple bond, one of the strongest in chemistry, and no plant or animal can break it. Only certain bacteria can, such as those in clover’s root nodules. Everything else takes its nitrogen from the soil, as nitrate or ammonium.
- Wrong Grass needs little nitrogen, only carbon and water
  Feedback
  Every protein and all DNA are built with nitrogen. Grass short of nitrogen turns pale and stops growing.
Feedback for a right answer
Because plants cannot break the triple bond that holds the two atoms of $\mathrm{N_2}$ ⁠ together. Only certain bacteria can, such as those in clover’s pink root nodules: they turn nitrogen gas into ammonia, by nitrogen fixation. The grass depends on the nitrate and ammonium in the soil.

Hint
What holds the two atoms of a nitrogen molecule together?

Worked solution
1. Nitrogen gas, $\mathrm{N_2}$ ⁠, is two atoms held by a triple bond, one of the strongest in chemistry.
2. Plants and animals have no enzyme that can break it.
3. Nitrogen-fixing bacteria can, with plenty of energy from sugar the clover gives them.
4. Everything else takes up nitrogen as nitrate, $\mathrm{NO_3^-}$ ⁠, and ammonium, $\mathrm{NH_4^+}$ ⁠, dissolved in the water in the soil.
That is why farmers sow clover, beans and peas to feed the soil.
Step 2 of 5
Open this step
In and out
Practice · Calculate
Question
A normal year, Decomposers at 100%: every store of nitrogen ends the year with what it held. Over the year the clover fixed from the air, less what bacteria returned to the air as nitrogen gas, about 2.8 g of nitrogen a square metre. How many grams of nitrogen were washed out of the meadow to the drain?

Correct answer
2.84 ± 0.15
Setup, choices and feedback
The Scene at this step
Follow
Nitrogenfixed
Decomposers
100 %fixed
Labels
offdefault
Through the year
A year on
Accepted within
± 0.15
Computed from the Scene at this step’s setting
G/m² of nitrogen washed out so far this year (0 for carbon)/nodes/meadow/outputs/washedOut
Wrong answers it recognises
- 0 ± 0.15
  Misconception: Thinks a meadow’s nitrogen cycle is closed: nothing leaves, so nothing has to come in.
  Feedback from the catalogue
  The meadow’s cycle leaks. Nitrate dissolves in water and the soil does not hold on to it, so the winter rain washes some of it down to the drain and the stream. In a steady year that leak equals exactly what the clover fixes from the air, less what bacteria return to the air as nitrogen gas.
Feedback for a right answer
About 2.8 g, exactly what came in. If the stores did not change, what came into the meadow equals what left it. Nitrate dissolves in water and the soil does not hold on to it, so the winter rain washes some of it to the stream.

Hint
If the stores did not change, how must what came in compare with what went out?

Worked solution
1. A store’s books: change = what comes in minus what goes out.
2. The stores are back where they started: change $= 0$ ⁠.
3. In, net, from the air: 2.8 g.
4. So out to the drain: 2.8 g.
If you answered $0$ ⁠, you thought the nitrogen never leaves the meadow. But the nitrogen cycle leaks, which is why the clover has to fix it again every year.
Step 3 of 5
Open this step
No rot
Practice · Predict, then look
Question
Follow is on Nitrogen, and Decomposers is at 0%. The grass has light, water and carbon dioxide as usual. What happens to it by the end of summer?

Correct answer
- Correct It is held back, because it runs out of nitrogen
Setup, choices and feedback
The Scene at this step
Follow
Nitrogenfixed
Decomposers
0 %fixed
Labels
offdefault
Through the year
1 Marchfixed
Choices
- Correct It is held back, because it runs out of nitrogen
- Wrong It grows more, because nothing is using up the material
  Misconception: Thinks decomposers only clear away waste, and that plants would do as well or better without them.
  Feedback from the catalogue
  The decomposers are what return to the soil the nitrogen locked up in dead leaves. When they eat a dead leaf, they free its nitrogen as ammonium, and bacteria turn it into nitrate, which the roots take up. Without decomposers the soil runs out of nitrogen by midsummer, and the grass turns pale and stops growing, even though it has light, water and carbon dioxide.
- Wrong Nothing changes: the grass has all it needs
  Misconception: Thinks decomposers only clear away waste, and that plants would do as well or better without them.
  Feedback from the catalogue
  The decomposers are what return to the soil the nitrogen locked up in dead leaves. When they eat a dead leaf, they free its nitrogen as ammonium, and bacteria turn it into nitrate, which the roots take up. Without decomposers the soil runs out of nitrogen by midsummer, and the grass turns pale and stops growing, even though it has light, water and carbon dioxide.
Once the prediction is locked in, the Scene changes
Through the year
0.5 (1 March → A year on)fixed
The Scene then shows: 0.33 · 0..1, how far the plants are held back by too little nitrogen/nodes/meadow/outputs/nitrogenShort
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It is held back: by midsummer the soil runs out of nitrate and ammonium, and growth falls to about half. The nitrogen the grass needs is locked up in last year’s dead leaves, and only decomposers can free it. The grass turns pale and stops growing early.

Worked solution
1. Most of the nitrogen the grass takes up each year does not come from the clover but from inside the meadow itself: from dead leaves that rotted.
2. Decomposers eat dead leaves and free their nitrogen as ammonium.
3. With no decomposers there is no new ammonium, and the nitrate there is gets taken up or washed out.
4. On the chart: ammonium and nitrate fall nearly to zero, and growth is held to about half.
If you predicted that the grass would grow more, you thought of the decomposers as cleaners only. They are a central part of the cycle.
Step 4 of 5
Open this step
From ammonium to nitrate
Practice · Choose
Question
The rabbit’s urine and the rotting dead leaves release nitrogen into the soil as ammonium, $\mathrm{NH_4^+}$ ⁠. What turns it into nitrate, $\mathrm{NO_3^-}$ ⁠?

Correct answer
- Correct Bacteria in the soil
Setup, choices and feedback
The Scene at this step
Follow
Nitrogenfixed
Decomposers
100 %
Labels
offdefault
Through the year
0.2 (1 March → A year on)
Choices
Shown in a random order.
- Correct Bacteria in the soil
- Wrong The bacteria in clover’s root nodules
  Feedback
  The bacteria in the nodules fix nitrogen: they turn nitrogen gas into ammonia. Ammonium is turned into nitrate by other bacteria, living free in the soil.
- Wrong The grass’s roots
  Feedback
  The roots take up both ammonium and nitrate, but do not turn one into the other in the soil.
- Wrong The rain water
  Feedback
  The rain washes nitrate down, but does not make it. The reaction needs oxygen and living things to carry it out.
Feedback for a right answer
Bacteria in the soil, nitrifying bacteria: they strip the hydrogens from the ammonium and add oxygens, and get energy from doing it. In wet soil another kind of bacteria does the opposite, and returns nitrate to the air as nitrogen gas.

Hint
Which living things in the soil get energy from chemical reactions of nitrogen?

Worked solution
1. From what to what: ammonium, $\mathrm{NH_4^+}$ ⁠, becomes nitrate, $\mathrm{NO_3^-}$ ⁠: the hydrogens go, oxygens come in.
2. That is an oxidation, and in the soil nitrifying bacteria do it, getting energy from it.
3. The bacteria in clover’s nodules do something else: they turn nitrogen gas from the air into ammonia.
4. Drag Through the year into the winter and watch the atoms move from the ammonium to the nitrate.
Step 5 of 5
Open this step
Too much rot
Practice · Reach the target
Question
Through the year is at A year on, and Follow is on Nitrogen. When the decomposers are too fast, they free more nitrate than the grass can take up, and the surplus washes away. Change Decomposers until the meadow ends the year about 2.1 g of nitrogen a square metre poorer, then press Check.

Goal
- -2.1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Follow
Nitrogenfixed
Decomposers
100 %
Labels
offdefault
Through the year
A year on
Read from the Scene at
G/m² by which the stores end the year above where they began (0 when the cycle closes)/nodes/meadow/outputs/yearGap
One setting that reaches it
Decomposers
150 %
Labels
off
Feedback for a right answer
At 150%. The fast rot frees nitrate in winter, when the grass takes up hardly any, and the rain washes it out: about 4 g a year instead of 2.8 g. A field given more fertiliser than the crop needs leaks the same way, and the nitrate in streams feeds blooms of algae.

Feedback for a wrong answer
Not yet. Read on the chart how far below its start the meadow ends the year.

Hint
Above 100% the meadow loses nitrogen. How far above?

Worked solution
1. At 100% the cycle closes.
2. A faster pace frees more nitrate than is taken up, and the meadow loses nitrogen.
3. Raise Decomposers and read on the chart how far below its start the meadow ends.
4. At 150% it ends about 2.1 g below.

Lesson 6 of 8 · About 25 minutes

Biomes and climate

Two numbers, temperature and rain, predict what landscape will grow: rain against the air’s thirst, the growing season trees need, and a mountain climb that passes through every biome in a few thousand metres.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Same rain, different land
Practice · Predict, then look
Question
500 mm a year at Boreal grows a conifer forest. Predict what grows on the same rain when Mean temperature rises to Tropical.

Correct answer
- Correct A dry savanna: grass and scattered trees
Setup, choices and feedback
The Scene at this step
Mean temperature
271.1 K
Rainfall per year
500 mmfixed
Raise a mountain
Flat landfixed
Choices
- Correct A dry savanna: grass and scattered trees
- Wrong A rainforest: warmth and water are all it needs
  Feedback
  A tropical rainforest needs more than three times this rain.
- Wrong Still a forest: same rain, same trees
  Misconception: Thinks the yearly rainfall alone decides how wet a place is and what grows there.
  Feedback from the catalogue
  Rain is only half the budget. Warm air takes back far more water than cold air, about twice as much for every 10 °C, so 500 mm a year is enough for a conifer forest somewhere cold, a grassland somewhere temperate and a dry savanna in the tropics. What decides is the moisture ratio, the rain over the air’s thirst. And at the cold edge no amount of rain will grow trees: what is missing there is warmth, not water.
Once the prediction is locked in, the Scene changes
Mean temperature
299.1 K
The Scene then shows: 5 · The biome at the foot: 0 tundra, 1 taiga, 2 temperate forest, 3 grassland, 4 desert, 5 savanna, 6 tropical rainforest/nodes/land/outputs/biome
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A savanna. On the same rain, the warm air takes back nearly five times as much water: about 1,490 mm a year instead of about 320 mm. The moisture ratio falls from about $1.6$ ⁠ to about $0.34$ ⁠, too little for a closed forest. On the way, at Temperate, there is grassland.

Worked solution
1. At Boreal: $E \approx 320$ ⁠ mm, $M = 500 / 320 \approx 1.6$ ⁠, above $0.85$ ⁠: forest.
2. At Tropical: $E \approx 1490$ ⁠ mm, $M = 500 / 1490 \approx 0.34$ ⁠.
3. Between $0.3$ ⁠ and $1.25$ ⁠ in the tropics: savanna.
Step 2 of 6
Open this step
Rain against thirst
Practice · Calculate
Question
Somewhere temperate, with a mean of 12 °C, the air could take back about 618 mm a year, and the rain is 500 mm. What is the moisture ratio?

Correct answer
0.8088 ± 0.02
Setup, choices and feedback
The Scene at this step
Mean temperature
285.1 Kfixed
Rainfall per year
500 mmfixed
Raise a mountain
Flat land
Accepted within
± 0.02
Computed from the Scene at this step’s setting
Rainfall over that demand/nodes/land/outputs/moisture
Wrong answers it recognises
- 1.236 ± 0.02
  Misconception: Divides the air's thirst by the rain instead of the rain by the thirst.
  Feedback from the catalogue
  The moisture ratio is $\dfrac{P}{E}$ ⁠: the rain on top, how much water arrives, over what the air takes back. A large ratio means water to spare; a ratio below $0.3$ ⁠ means desert. Turn the fraction over, and a wet place looks dry and the other way round.
Feedback for a right answer
$\dfrac{500}{618} \approx 0.81$ ⁠. Above $0.3$ ⁠, so not desert, but below $0.85$ ⁠, so not water enough for a closed forest: grassland. About 25 mm more rain a year, and the trees would close a canopy.

Worked solution
1. $M = \dfrac{P}{E}$ ⁠, with $P = 500$ ⁠ and $E = 618$ ⁠.
2. $M = 500 / 618 \approx 0.81$ ⁠.
3. $0.3 < 0.81 < 0.85$ ⁠: grassland.
If you got $1.24$ ⁠, you divided the thirst by the rain instead of the rain by the thirst.
Step 3 of 6
Open this step
Enough for a rainforest
Practice · Reach the target
Question
Mean temperature is locked at Tropical. Find the smallest Rainfall per year at which a rainforest grows, and press Check.

Goal
- 1.255 ± 0.005
  At the start of the step: 0.6729
Setup, choices and feedback
The Scene at this step
Mean temperature
299.1 Kfixed
Rainfall per year
1000 mm
Raise a mountain
Flat landfixed
Read from the Scene at
Rainfall over that demand/nodes/land/outputs/moisture
One setting that reaches it
Rainfall per year
1860 mm
Feedback for a right answer
About 1,860 mm a year, more than twice the rain that grows a temperate forest. The warm tropical air takes back about 1,490 mm, and a forest green all year needs a quarter more on top. That is why rainforests grow only in the rainiest places on Earth.

Feedback for a wrong answer
Not yet. If the tile is still savanna, add rain; if you have added more than the least, take a little off.

Hint
A rainforest needs a moisture ratio of at least $1.25$ ⁠, and in the Scene’s tropics the air takes back about 1,490 mm a year.

Worked solution
1. The rainforest threshold: $M \ge 1.25$ ⁠.
2. $P = 1.25 \times E = 1.25 \times 1486 \approx 1858$ ⁠ mm.
3. On the slider, in steps of 10 mm: 1,860 mm.
Step 4 of 6
Open this step
The cold edge
Practice · Predict, then look
Question
At Arctic, with 1,000 mm a year, there is treeless tundra. Predict what happens when Rainfall per year rises to 3,000 mm.

Correct answer
- Correct Still tundra: what trees lack is warmth, not water
Setup, choices and feedback
The Scene at this step
Mean temperature
261.1 Kfixed
Rainfall per year
1000 mm
Raise a mountain
Flat landfixed
Choices
- Correct Still tundra: what trees lack is warmth, not water
- Wrong A conifer forest grows: with water enough, trees grow
  Misconception: Thinks the yearly rainfall alone decides how wet a place is and what grows there.
  Feedback from the catalogue
  Rain is only half the budget. Warm air takes back far more water than cold air, about twice as much for every 10 °C, so 500 mm a year is enough for a conifer forest somewhere cold, a grassland somewhere temperate and a dry savanna in the tropics. What decides is the moisture ratio, the rain over the air’s thirst. And at the cold edge no amount of rain will grow trees: what is missing there is warmth, not water.
- Wrong The tundra turns into a bog with no plants
  Feedback
  Wet tundra is still tundra, with mosses, sedges and dwarf shrubs.
Once the prediction is locked in, the Scene changes
Rainfall per year
3000 mm
The Scene then shows: 0 · The biome at the foot: 0 tundra, 1 taiga, 2 temperate forest, 3 grassland, 4 desert, 5 savanna, 6 tropical rainforest/nodes/land/outputs/biome
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Still tundra. At a mean of −12 °C the growing season is too short and cold: no month rises above 10 °C, and trees cannot build wood and leaves in time. At the cold end of the chart, rain changes almost nothing.

Worked solution
1. Trees need a growing season with a mean above about 6.4 °C that lasts more than about three months.
2. At a yearly mean of −12 °C that season hardly exists.
3. The limit here is warmth, so more rain does not help: it stays tundra.
Step 5 of 6
Open this step
Up a mountain
Practice · Predict, then look
Question
At the foot there is dry savanna, 500 mm a year in the tropics. Predict what grows on the slopes when you choose Mountain under Raise a mountain.

Correct answer
- Correct A belt of forest above the savanna, with tundra and snow above it
Setup, choices and feedback
The Scene at this step
Mean temperature
299.1 Kfixed
Rainfall per year
500 mmfixed
Raise a mountain
Flat land
Choices
- Correct A belt of forest above the savanna, with tundra and snow above it
- Wrong Bare slopes: higher is drier and harsher
  Misconception: Thinks a mountain's slopes are always drier and barer than the land at its foot.
  Feedback from the catalogue
  Up a mountain the air is colder, and cold air is less thirsty. The same rain then goes further: above a dry foot of savanna or desert, a belt of forest can grow. Only higher up, when the cold makes the growing season too short, do the trees stop, with tundra and snow above them.
- Wrong Savanna all the way up, since the rain does not change
  Misconception: Thinks the yearly rainfall alone decides how wet a place is and what grows there.
  Feedback from the catalogue
  Rain is only half the budget. Warm air takes back far more water than cold air, about twice as much for every 10 °C, so 500 mm a year is enough for a conifer forest somewhere cold, a grassland somewhere temperate and a dry savanna in the tropics. What decides is the moisture ratio, the rain over the air’s thirst. And at the cold edge no amount of rain will grow trees: what is missing there is warmth, not water.
Once the prediction is locked in, the Scene changes
Raise a mountain
Mountain
The Scene then shows: 1 · 1 when a forest grows on the slopes above dry grassland, desert or savanna at the foot, else 0/nodes/land/outputs/wetterUp
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A belt of forest. Up the mountain the air is colder and less thirsty, so the same 500 mm is enough for trees. Higher still the cold shortens the growing season, the trees stop, and tundra and snow lie above them. That is how Arizona’s “sky islands” wear belts of oak and pine above their cactus slopes.

Worked solution
1. Each 1,000 m cools by about 6.5 °C, and the rain stays at 500 mm.
2. Cold air takes back less water: $E$ ⁠ shrinks, and $M = P / E$ ⁠ rises above $0.85$ ⁠, to forest.
3. Higher still the growing season is too short for trees: tundra, and ice and snow at the summit.
Step 6 of 6
Open this step
Biome and habitat
Practice · Choose
Question
In a European temperate forest live a woodpecker, which lives in tree trunks, and a woodlouse, which lives under a rotting log on the ground. What do they share?

Correct answer
- Correct A biome, but not a habitat
Setup, choices and feedback
The Scene at this step
Mean temperature
285.1 K
Rainfall per year
1000 mm
Raise a mountain
Flat land
Choices
Shown in a random order.
- Correct A biome, but not a habitat
- Wrong A habitat, since they are in the same forest
  Misconception: Uses habitat and biome as if they meant the same thing.
  Feedback from the catalogue
  A biome is a large region defined by climate and the shape of its plants, such as taiga or savanna. A habitat is the particular place where one kind of living thing lives and finds what it needs: a rotting log, a stream bank, the treetops. One biome holds many habitats.
- Wrong Both a biome and a habitat; those are two words for one thing
  Misconception: Uses habitat and biome as if they meant the same thing.
  Feedback from the catalogue
  A biome is a large region defined by climate and the shape of its plants, such as taiga or savanna. A habitat is the particular place where one kind of living thing lives and finds what it needs: a rotting log, a stream bank, the treetops. One biome holds many habitats.
- Wrong Neither: one lives up high and one down low
  Feedback
  They are in the same forest, so the same biome, even if each is in a different corner of it.
Feedback for a right answer
A biome, temperate forest, but not a habitat. The biome is set by the climate and the shape of the plants; a habitat is the particular place where each finds what it needs: trunks with insects for the woodpecker, damp rotting wood for the woodlouse. One forest holds many habitats.

Worked solution
1. Biome: a large region, defined by climate and vegetation. Both are in temperate forest.
2. Habitat: the place where one kind of living thing lives. A tree trunk is not a rotting log on the ground.
3. So: the same biome, different habitats.

Lesson 7 of 8 · About 25 minutes

Islands of forest

Orangutans in a forest cut into patches: less forest feeds fewer apes, small isolated groups die out even while their forest stands, the risk of extinction is a chance rather than a sentence, and wildlife corridors join the patches into one population.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
A forest cut up
Practice · Calculate
Question
An untouched forest of 400 km² holds about $200$ ⁠ orangutans, one for every 2 km². How many orangutans can the forest left, 35% of it, feed, counting only patches that feed two or more?

Correct answer
66 ± 5
Setup, choices and feedback
The Scene at this step
Forest left
35 %fixed
Wildlife corridors
offfixed
Luck
Typical
Years
Year 0fixed
Accepted within
± 5
Computed from the Scene at this step’s setting
Orangutans at the start, as many as the forest left can feed/nodes/forest/outputs/start
Feedback for a right answer
$66$ ⁠. From 140 km² you would expect $70$ ⁠, but some scraps are too small to feed even two. The $66$ ⁠ are split between four patches of $39$ ⁠, $14$ ⁠, $7$ ⁠ and $6$ ⁠: clearing does two things at once, less forest and broken forest.

Hint
How many square kilometres are left? And how many orangutans per square kilometre? The small scraps take off a few more.

Worked solution
1. Forest left: $400 \times 0.35 = 140$ ⁠ km².
2. One per $2$ ⁠ km²: $140 / 2 = 70$ ⁠.
3. Scraps of forest feed fewer than two and do not count: $66$ ⁠ are left.
Step 2 of 6
Open this step
A century on
Practice · Predict, then look
Question
Four patches hold $39$ ⁠, $14$ ⁠, $7$ ⁠ and $6$ ⁠ orangutans, and no more trees will be felled. Predict what Years shows when you drag it to 100 years on.

Correct answer
- Correct The small patches empty, though their forest still stands
Setup, choices and feedback
The Scene at this step
Forest left
35 %fixed
Wildlife corridors
offfixed
Luck
Typicalfixed
Years
Year 0
Choices
- Correct The small patches empty, though their forest still stands
- Wrong Every group holds on: they have forest, so they have food
  Misconception: Thinks a population is safe as long as its habitat still stands.
  Feedback from the catalogue
  Habitat that still stands does not guarantee survival. A small, cut-off group can vanish by chance: a few deaths in a row, no mate, inbreeding or a bad year. In the Scene, the forest in the emptied patches was never felled; the groups simply ran out of luck. The size of the population and the links between patches matter as much as the trees themselves.
- Wrong Every group grows back, since the felling has stopped
  Feedback
  Each patch is already full: it feeds only as many as it feeds. And orangutans breed very slowly.
Once the prediction is locked in, the Scene changes
Years
100 years on
The Scene then shows: 2 · Patches with none left in the year shown/nodes/forest/outputs/emptied
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The two small patches empty for good, and $21$ ⁠ of the $23$ ⁠ orangutans left are in the big patch. Nobody felled the forest in the empty patches; groups of six and seven simply ran out of luck.

Worked solution
1. Each patch holds a cut-off group that must survive alone.
2. In a group of $6$ ⁠, a few deaths in a row or no mate ends it.
3. There is nowhere for newcomers to come from: the plantation blocks them.
4. So the small patches empty first, with no more felling at all.
Step 3 of 6
Open this step
Why small groups fail
Practice · Choose
Question
What makes a small, cut-off group fragile? Choose everything that is true.

Correct answer
- Correct Chance: a few deaths in a row can end a small group
- Correct Finding a mate: with no female or no male there are no young
- Correct Inbreeding: fewer young survive, and genetic variety is lost
- Correct Bad years of drought and fire, which hit every patch at once
Setup, choices and feedback
The Scene at this step
Forest left
35 %
Wildlife corridors
off
Luck
Typical
Years
100 years on
Choices
Shown in a random order. More than one choice is correct: every one must be selected.
- Correct Chance: a few deaths in a row can end a small group
- Correct Finding a mate: with no female or no male there are no young
- Correct Inbreeding: fewer young survive, and genetic variety is lost
- Correct Bad years of drought and fire, which hit every patch at once
- Wrong Running out of food, since a small group eats its whole forest
  Misconception: Thinks a population is safe as long as its habitat still stands.
  Feedback from the catalogue
  Habitat that still stands does not guarantee survival. A small, cut-off group can vanish by chance: a few deaths in a row, no mate, inbreeding or a bad year. In the Scene, the forest in the emptied patches was never felled; the groups simply ran out of luck. The size of the population and the links between patches matter as much as the trees themselves.
Feedback for a right answer
All four, and they feed one another. A group shrunk by a bad year is more inbred, raises fewer young and is more at the mercy of chance, so it shrinks further: the extinction vortex. Food is not the problem: each patch feeds exactly as many as it feeds.

Worked solution
1. Chance: in a group of $200$ ⁠ good and bad luck cancel out; in a group of $6$ ⁠ they do not.
2. A mate: a patch that has lost its last female breeds no more.
3. Inbreeding: in a cut-off group everyone becomes related.
4. Bad years: they hit every patch at once, so the groups cannot make up for one another.
5. Food: the group is no bigger than its patch feeds, so it is not the cause.
Step 4 of 6
Open this step
A matter of chance
Practice · Predict, then look
Question
At 35%, mostly cleared, all the orangutans vanish within a century in only about 1% of the runs. Predict the risk of extinction when Forest left drops to 15%, mere scraps.

Correct answer
- Correct About one run in six
Setup, choices and feedback
The Scene at this step
Forest left
35 %
Wildlife corridors
offfixed
Luck
Typical
Years
100 years on
Choices
- Correct About one run in six
- Wrong Still about 2%: less forest, but about the same risk
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Wrong 100%: with so little forest they will certainly vanish
  Misconception: Thinks whether a population dies out is fixed in advance, rather than a chance that can be worked out.
  Feedback from the catalogue
  A small population’s future cannot be known in advance: the same $66$ ⁠ orangutans can end a century with $7$ ⁠ or with $44$ ⁠. So the question is “what is the chance?”, not “will it?”. You run a model many times and count in what share of the runs none are left: that is the risk of extinction.
Once the prediction is locked in, the Scene changes
Forest left
15 %
The Scene then shows: 15.5 · Per cent of the 200 runs with no orangutans left after 100 years/nodes/forest/outputs/risk
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 16%, roughly one run in six. There is room for only $27$ ⁠ orangutans in three patches. The risk grows far faster than the forest shrinks: at 10% it is nearly one in three. And still, in most runs a few are left: extinction is a chance, not a sentence.

Worked solution
1. Less forest: room for $27$ ⁠ instead of $66$ ⁠.
2. Smaller groups in each patch: each more exposed to chance and bad years.
3. The model runs $200$ ⁠ times; in about $31$ ⁠ of them none are left: about 16%.
Step 5 of 6
Open this step
Corridors
Practice · Predict, then look
Question
Without corridors, $23$ ⁠ of the $66$ ⁠ orangutans are left after a hundred years. Predict how many are left in the same run when Wildlife corridors is switched on.

Correct answer
55 ± 12
A number prediction counts as right within ± 12
Setup, choices and feedback
The Scene at this step
Forest left
35 %fixed
Wildlife corridors
off
Luck
Typicalfixed
Years
100 years onfixed
Once the prediction is locked in, the Scene changes
Wildlife corridors
on
The Scene then shows: 55 · Orangutans in the run shown, in the year shown/nodes/forest/outputs/left
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$55$ ⁠, more than twice as many, and all four patches still have orangutans. In exactly the same forest. The corridors added hardly any food; they joined the groups into one population.

Worked solution
1. Without corridors: four cut-off groups, and the small ones empty.
2. With corridors: youngsters move between patches, an emptied patch is settled again, and genes flow.
3. The result in the typical run: $55$ ⁠ of $66$ ⁠.
Step 6 of 6
Open this step
How a corridor helps
Practice · Choose
Question
The corridors are narrow strips of planted forest. What is the main way they help the orangutans?

Correct answer
- Correct They connect the patches, so youngsters, new settlers and genes pass between them
Setup, choices and feedback
The Scene at this step
Forest left
35 %
Wildlife corridors
on
Luck
Typical
Years
100 years on
Choices
Shown in a random order.
- Correct They connect the patches, so youngsters, new settlers and genes pass between them
- Wrong They add forest that feeds more orangutans
  Misconception: Thinks wildlife corridors help mainly by adding habitat to feed more animals.
  Feedback from the catalogue
  A corridor is a narrow strip, and it adds little food. What it does is connect: youngsters looking for a home reach a patch with room instead of dying on the way, an emptied patch is settled again, and wanderers bring in new genes. So several small groups behave like one large population.
- Wrong They hide the orangutans from hunters
  Feedback
  Hunting is a real threat, but the model has none, and the corridors still help a lot.
Feedback for a right answer
They connect. Youngsters looking for a home reach a patch with room instead of dying in the plantation; an emptied patch is settled again; and wanderers bring new genes to each group. In the model, the smallest patch on its own ends the century empty in three runs out of four; joined to the others, almost never.

Worked solution
1. The corridors are narrow: they add very little food.
2. What they change is movement: the patches behave like one large population.
3. A large population is sturdier against chance, inbreeding and bad years.

Lesson 8 of 8 · About 25 minutes

Give and take

An ecosystem inside the large intestine: microbes that ferment fibre we cannot digest and give back fatty acids, a fence of mucus that wears thin when they go hungry, antibiotics that make room for opportunists, and the kinds of symbiosis, from mutualism to parasitism.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Food you cannot digest
Practice · Choose
Question
Why can the microbes in the large intestine break down dietary fibre, when we cannot?

Correct answer
- Correct They have genes for hundreds of enzymes that cut the bonds in plant sugars; we have fewer than twenty
Setup, choices and feedback
The Scene at this step
Fibre in the diet
30 g a day
Just after antibiotics
off
Zoom to the give and take
Gut wall
Choices
Shown in a random order.
- Correct They have genes for hundreds of enzymes that cut the bonds in plant sugars; we have fewer than twenty
- Wrong The large intestine is more acidic than the stomach, and the acid breaks the fibre down
  Feedback
  The large intestine is only slightly acidic, and that acidity is a result of fermentation, not its cause.
- Wrong The fibre just stays there long enough
  Feedback
  Time will not cut a bond that no enzyme cuts.
- Wrong They do not really break it down: fibre leaves the body as it came in
  Misconception: Thinks dietary fibre passes through unused, adding only bulk.
  Feedback from the catalogue
  We do not digest fibre, but the microbes in the large intestine do. They ferment it to short-chain fatty acids, and about 95% of them are absorbed: butyrate feeds the lining cells, propionate goes to the liver and acetate to the rest of the body. Fibre also feeds the community so it does not turn to eating the mucus layer that guards the wall.
Feedback for a right answer
Enzymes. Our genes make fewer than twenty carbohydrate-digesting enzymes, enough for starch, sugar and milk sugar. A single gut bacterium, such as Bacteroides thetaiotaomicron, carries genes for about two hundred, each fitting a different bond in a different plant sugar. The microbes do what we cannot, and pay us back part of the profit.

Worked solution
1. Fibre is chains of sugars joined by special bonds.
2. To cut such a bond takes an enzyme that fits it.
3. Humans have very few such enzymes; gut microbes have hundreds.
4. So they break down and ferment the fibre in the large intestine, which it reaches whole.
Step 2 of 6
Open this step
What you get back
Practice · Calculate
Question
At Recommended, 30 g of fibre a day. 70% of the fibre is fermented, plus 6 g of mucus and starch, and each fermented gram gives about $9$ ⁠ millimoles of short-chain fatty acids. How many millimoles of them are made a day?

Correct answer
243 ± 3%
Setup, choices and feedback
The Scene at this step
Fibre in the diet
30 g a dayfixed
Just after antibiotics
offfixed
Zoom to the give and take
Give and take
Accepted within
± 3%
Computed from the Scene at this step’s setting
Short-chain fatty acids made, mmol a day/nodes/gut/outputs/fattyAcids
Feedback for a right answer
About $243$ ⁠ millimoles a day. About 95% of them are absorbed through the wall. Butyrate, about a fifth of them, gives the lining cells nearly two-thirds of their energy; propionate becomes glucose in the liver; acetate goes to the muscles. About $70$ ⁠ kilocalories a day in all, energy that would otherwise be lost.

Worked solution
1. Fermented: $0.7 \times 30 + 6 = 27$ ⁠ grams a day.
2. Fatty acids: $27 \times 9 = 243$ ⁠ millimoles a day.
If you got $270$ ⁠, you assumed all the fibre, and nothing else, is fermented.
Step 3 of 6
Open this step
A fence of mucus
Practice · Predict, then look
Question
Predict what happens to the inner mucus layer when Fibre in the diet drops from Typical Western to Very little.

Correct answer
- Correct It thins: hungry microbes eat the mucus and creep closer to the wall
Setup, choices and feedback
The Scene at this step
Fibre in the diet
18 g a day
Just after antibiotics
offfixed
Zoom to the give and take
Gut wallfixed
Choices
- Correct It thins: hungry microbes eat the mucus and creep closer to the wall
- Wrong Nothing: fibre only adds bulk
  Misconception: Thinks dietary fibre passes through unused, adding only bulk.
  Feedback from the catalogue
  We do not digest fibre, but the microbes in the large intestine do. They ferment it to short-chain fatty acids, and about 95% of them are absorbed: butyrate feeds the lining cells, propionate goes to the liver and acetate to the rest of the body. Fibre also feeds the community so it does not turn to eating the mucus layer that guards the wall.
- Wrong It thickens: less fibre, less rubbing on the wall
  Feedback
  The mucus is not worn by rubbing but eaten. Without fibre, more microbes turn to eating it.
Once the prediction is locked in, the Scene changes
Fibre in the diet
5 g a day
The Scene then shows: 0.5987 · Inner mucus thickness, relative to a 30 g a day diet/nodes/gut/outputs/mucus
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It wears down to about two-thirds of its thickness. With little fibre, more members of the community turn to the only carbohydrate left: the mucus. The grazers multiply, opportunists move in, and the immune cells in the wall gather on alert. In experiments on mice, a diet with no fibre made it easier for a disease-causing bacterium to settle in.

Worked solution
1. Fibre is the community’s main food.
2. When it runs short, the microbes look for another carbohydrate: the sugar chains of the mucus.
3. The grazers multiply and wear the inner layer away faster than goblet cells renew it.
4. The layer thins, and the microbes come close to the cells.
Step 4 of 6
Open this step
Burning butyrate
Practice · Reach the target
Question
At Very little the lining cells get too little butyrate, and oxygen leaks into the gut. Find the smallest Fibre in the diet at which oxygen no longer leaks, and press Check.

Goal
- 0 ± 0.001
  At the start of the step: 0.4829
- 0.553 ± 0.004
  At the start of the step: 0.4051
Setup, choices and feedback
The Scene at this step
Fibre in the diet
5 g a day
Just after antibiotics
offfixed
Zoom to the give and take
Give and takefixed
Read from the Scene at
Oxygen the lining leaves unused, 0..1/nodes/gut/outputs/oxygenLeak
Read from the Scene at
Share of the lining cells' energy met by butyrate, 0..1/nodes/gut/outputs/wallFed
One setting that reaches it
Fibre in the diet
16 g a day
Feedback for a right answer
16 g a day. There butyrate meets about 55% of the lining cells’ energy, and they burn all the oxygen arriving from the blood. Any less, and the leftover oxygen leaks out and feeds opportunists such as E. coli. A typical Western diet, 18 g, is close to the edge.

Feedback for a wrong answer
Not yet. If oxygen still leaks, add fibre; if it no longer does, try one gram less.

Hint
Add one gram at a time, and watch the red pairs of oxygen in the zoom.

Worked solution
1. The lining cells burn butyrate with oxygen from the blood.
2. When butyrate meets less than 55% of their energy, they switch to sugar without oxygen, and the spare oxygen leaks out.
3. More fibre, more butyrate makers and more butyrate.
4. At 15 g a little still leaks; at 16 g none does.
Step 5 of 6
Open this step
After antibiotics
Practice · Predict, then look
Question
Predict which group’s share of the community grows most when Just after antibiotics is switched on.

Correct answer
- Correct The opportunists, such as E. coli
Setup, choices and feedback
The Scene at this step
Fibre in the diet
30 g a dayfixed
Just after antibiotics
off
Zoom to the give and take
Gut wall
Choices
- Correct The opportunists, such as E. coli
- Wrong The butyrate makers
  Feedback
  They are the worst hit: strict anaerobes, especially sensitive.
- Wrong None: the antibiotic harms only harmful bacteria, and the community stays as it was
  Misconception: Thinks antibiotics kill only the harmful bacteria and leave the gut's own community alone.
  Feedback from the catalogue
  Most antibiotics kill a wide range of bacteria and do not tell “bad” from “good”. One course can cut the large intestine’s community to a small part of its size, the butyrate makers, which live without oxygen, worst of all. What is left and spreads are opportunists that tolerate oxygen, such as E. coli.
Once the prediction is locked in, the Scene changes
Just after antibiotics
on
The Scene then shows: 0.3788 · Share of opportunists such as E. coli, 0..1/nodes/gut/outputs/opportunists
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The opportunists, from about 6% to about 38% of a community shrunk to a quarter of its size. The butyrate makers are hit hardest, the lining gets little butyrate, and the oxygen it does not burn leaks out and feeds exactly those that tolerate oxygen. A gap like this also lets C. difficile take over.

Worked solution
1. A broad-spectrum antibiotic kills most bacteria, without telling them apart.
2. The butyrate makers are especially sensitive: few are left.
3. Less butyrate, lining cells that leave oxygen unburnt, and oxygen leaking into the gut.
4. Those that tolerate oxygen and survive well, the opportunists, take the space.
Step 6 of 6
Open this step
Kinds of symbiosis
Practice · Choose
Question
After antibiotics, C. difficile takes over the gut and causes severe diarrhoea. Is that symbiosis too?

Correct answer
- Correct Yes: parasitism, in which the bacterium gains at the host’s cost
Setup, choices and feedback
The Scene at this step
Fibre in the diet
30 g a day
Just after antibiotics
on
Zoom to the give and take
Give and take
Choices
Shown in a random order.
- Correct Yes: parasitism, in which the bacterium gains at the host’s cost
- Wrong No: symbiosis means both sides gain
  Misconception: Thinks symbiosis always means a partnership in which both sides gain.
  Feedback from the catalogue
  Symbiosis is any close living together; that is all the word means. It includes mutualism, where both gain, commensalism, where one gains and the other is hardly affected, and parasitism, where one gains at the other’s cost. The same microbe can move between them as conditions change.
- Wrong Yes: mutualism, since both bacterium and person live
  Feedback
  In mutualism both sides gain. Here the person is harmed.
- Wrong Yes: commensalism, since the person is hardly affected
  Feedback
  Severe diarrhoea is not “hardly affected”.
Feedback for a right answer
Yes, parasitism. Symbiosis only means “living together” closely, and it covers every kind. Butyrate makers in a fibre-fed gut are mutualism: you give food and a home, they give fuel for the lining. The same community can move between the kinds: whether a microbe is a partner or a pest depends on where it is and what is going on around it.

Worked solution
1. Symbiosis: two kinds of living thing living together closely. C. difficile lives in your gut: that is symbiosis.
2. Who gains? The bacterium, which multiplies.
3. Who loses? The person, the host.
4. One gains at the other’s cost: parasitism.

Across the unit · 27 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

Food chains

Who eats whom

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

One way

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

A tenth

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Arrows

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Less sun

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The ten per cent rule

Four links up

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

One heron

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Twice the share

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A sixth level

Question

Correct answer

The Scene at this step

Hint

Worked solution

How long a chain

Question

Correct answer

The Scene at this step

Hint

Worked solution

Three pyramids

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Predator and prey

Four quarters

Question