Practice questions · Topic 8.4

Unit 8 · Practice for the Topic 8.4 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: The effect of density on populations, summed up

Video coming soon

Population density; K read from the S-shaped curve; why growth slows near the top; density-dependent and density-independent limits; the braking term and dN/dt = r_max N ((K − N)/K); fastest halfway to K; past K, and a moving K; percent change.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one population’s growth one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. The formula sheet gives the growth equations, dN/dt = B − D, dN/dt = rmax N and dN/dt = rmax N ((K − N)/K), and percent change = (final − initial) / initial × 100.

Question 1

Suppose a net sample of 14 L of sea water holds 602 sprats, small fish.

Which of the following is the population density of the sprats in the sample?

A. 0.02 sprats per L B. 43 sprats per L C. 588 sprats per L D. 8 428 sprats per L

Question 2

The estuary’s godwit counts, years 0 to 12.

Godwits, large wading birds, begin to winter on a new estuary. The graph below shows the count every winter for twelve years.

Which of the following is the estuary’s carrying capacity, K, for godwits?

A. 40 godwits B. 315 godwits C. 630 godwits D. 900 godwits

Question 3

Yellowhammers, small seed-eating birds, nest along one hedge, and the hedge’s seeds and nest sites stay as they were. The count of yellowhammers climbs and nears the hedge’s K.

As the count nears K, what happens to the births per yellowhammer and the deaths per yellowhammer each year?

A. Births per bird rise and deaths per bird fall B. Births per bird and deaths per bird both keep their earlier values C. Births per bird fall and deaths per bird rise D. Births per bird fall and deaths per bird stay as they were

Question 4

Suppose greenfinches feed at a garden’s feeders all winter, and the gardener puts out the same weight of seed each day. By midwinter the seed is gone within an hour of being put out. The count of greenfinches at the feeders climbs through the autumn and then levels off.

Which of the following is the limiting factor for the greenfinches?

A. The seed B. The winter cold C. The perches on the feeders D. The greenfinches’ own nature: a flock stops growing at a set size

Question 5

The two hedges’ goldfinch nests, and how many the hail destroyed.

Two hedges of the same length hold goldfinch nests. A hailstorm crosses both hedges on the same afternoon. The table below shows the nests in each hedge and how many the hail destroyed.

Which kind of limiting factor is the hail, and why?

A. Density-dependent, because the hail destroyed more nests in the crowded hedge B. Density-dependent, because the hail destroyed a larger fraction of the crowded hedge’s nests C. Density-independent, because the hail destroyed the same fraction of the nests in both hedges D. Density-independent, because the crowded hedge held three times as many nests

Question 6

Redpolls, small finches, roost in two thickets of the same size on winter nights: a few dozen birds in one thicket and several hundred in the other. An infection that passes from bird to bird when they perch touching reaches both roosts. By spring it has reached a larger fraction of the crowded roost’s birds.

Which of the following explains why the infection reached a larger fraction of the crowded roost’s birds?

A. The infection is stronger where the air is warmer, and a crowded roost is warmer B. The crowded roost holds more birds, so more birds are there to catch the infection C. The birds in the sparse roost are fitter, so they fight the infection off D. Each bird in the crowded roost perches touching more other birds every night

Question 7

Suppose an estuary’s mud holds food for no more than 2 320 dunlin, a small wading bird: K is 2 320 dunlin. In October the estuary holds 580 dunlin.

Which of the following is the braking term for the dunlin?

A. 0.25 B. 0.75 C. 4 D. 1 740

Question 8

A flock of pintails, a duck, on a lake has reached the lake’s carrying capacity: N equals K.

Which of the following does the logistic equation give for the flock’s dN/dt?

A. 0 pintails per year B. r_max N C. r_max K D. A negative rate

Question 9

Suppose a shore’s seaweed and crevices can support no more than 2 120 sea slaters, shore-living relatives of woodlice: K is 2 120 sea slaters.

At which of the following counts does the sea slater population grow fastest?

A. 530 sea slaters B. 1 060 sea slaters C. 1 590 sea slaters D. 2 120 sea slaters

Question 10

Suppose a reservoir’s fish feed no more than 76 goosanders, a fish-eating duck: K is 76 goosanders, and the flock has stood at 76 for years. The reservoir’s owners then stock it with fish every month, and the fish now feed no more than 152 goosanders.

Predict what the flock’s count does over the following years.

A. Stays at 76 goosanders B. Falls below 76 goosanders C. Rises with no limit, for as long as the stocking continues D. Rises toward 152 goosanders, the new K

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Analyze Data · 5 points

Lugworms live buried in the sand of a beach. Suppose the food in one marked patch of sand can support no more than 2 040 lugworms: K is 2 040 lugworms. This spring the patch holds 510 lugworms, and the lugworms’ maximum per capita growth rate, r_max, is 0.8 per year.

(a) Calculate the braking term for the lugworms this spring. (1 point)

Hint: Which of the two counts is K, the most the patch can support, and which is N, the count now?

Write down the values in the question:

K = 2 040 lugworms

N = 510 lugworms

Write down the equation:

braking term = \frac{K - N}{K}

Substitute the values into the equation, and calculate:

braking term = \frac{2 040 - 510}{2 040}

braking term = 0.75

A full-credit answer: braking term = (K − N)/K = (2 040 − 510)/2 040 = 0.75.

(b) Calculate r_max N for the lugworms this spring. (1 point)

Hint: r_max is the number each lugworm adds per year with nothing holding it back; N is how many lugworms are adding.

lugworms per year

Write down the values in the question:

r_{m a x} = 0.8 per year

N = 510 lugworms

Substitute the values into the equation, and calculate:

r_{m a x} N = 0.8 \times 510

r_{m a x} N = 408 lugworms per year

A full-credit answer: r_max N = 0.8 × 510 = 408 lugworms per year.

(c) Calculate dN/dt for the lugworms this spring, using your answers to parts (a) and (b). (1 point)

Hint: The logistic equation multiplies the exponential part by the braking term.

lugworms per year

Write down the equation:

\frac{d N}{d t} = r_{m a x} N (\frac{K - N}{K})

Substitute the values into the equation, and calculate:

\frac{d N}{d t} = 0.8 \times 510 \times 0.75

\frac{d N}{d t} = 306 lugworms per year

A full-credit answer: dN/dt = r_max N × braking term = 408 × 0.75 = 306 lugworms per year.

(d) Predict how the number of lugworms the patch adds each year changes as the count climbs from 510 toward K, and explain your prediction. (1 point)

Hint: Think of the two parts you multiplied in part (c): which one grows as the count climbs, and which one shrinks?

A full-credit answer: The yearly gain rises at first and then falls to 0 as the count reaches K, because r_max N grows with the count while the braking term shrinks to 0.
The gain peaks halfway to K, at 1 020 lugworms, so from 510 the gain rises before it falls.

Check the box for each point your answer earns

the number the patch adds each year rises at first, is largest halfway to K, at 1 020 lugworms, and then falls to 0 as the count reaches K, because r_max N grows with the count while the braking term shrinks to 0 (accept: fastest growth at half of K, then the gain falls to 0 at K). ‘The number added falls as the count nears K’ on its own is the second half of the answer — it is true only once the count passes 1 020 — and is worth half of this point; the point wants the rise first.

Common slip: Predicting a gain that only falls as the count climbs. From 510 the count is below half of K, so the gain rises first; it falls once the count passes 1 020.

(e) Next spring the patch holds 816 lugworms. Calculate the percent change in the count from this spring to the next, with its sign. (1 point)

Hint: Which count is the starting count, the one you divide by?

Write down the values in the question:

initial = 510 lugworms

final = 816 lugworms

Write down the equation:

percent change = \frac{final - initial}{initial} \times 100

Substitute the values into the equation, and calculate:

percent change = \frac{816 - 510}{510} \times 100

percent change = +60 %

A full-credit answer: percent change = (final − initial)/initial × 100 = (816 − 510)/510 × 100 = +60 %.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

Stonechats, small perching birds, nest in the thorn scrub of a moor. Suppose the moor’s scrub shelters and feeds no more than 68 stonechat pairs: K is 68 pairs, and 65 pairs nest there this spring. In late summer a fire burns half of the scrub. No stonechat dies in the fire, but the scrub left shelters and feeds no more than 34 pairs. The burnt half stays bare in the years that follow.

(a) Identify the limiting factor that the fire changed for the stonechats. (1 point)

A full-credit answer: The thorn scrub: the nest sites and food it gives the stonechats.

Check the box for each point your answer earns

the scrub (accept: the nest sites; the shelter and food the scrub provides).

Common slip: Naming the fire itself. The fire is the event; the limiting factor is the resource it cut.

(b) Predict how the count of nesting pairs changes over the following years. (1 point)

A full-credit answer: The count falls over the following years toward 34 pairs, the new K.

Check the box for each point your answer earns

the count falls (accept: falls toward the new K of 34 pairs).

Common slip: Predicting no change because no bird died in the fire. K fell, so the count is now above K and falls.

(c) Justify your prediction in part (b). (1 point)

A full-credit answer: K is set by the moor’s scrub, so when the fire burned half the scrub, K fell from 68 to 34 pairs.
The 65 pairs are now above the new K.
So each pair’s share of nest sites and food is too small, and deaths outnumber births until the count reaches the new K.

Check the box for each point your answer earns

the count is above the new K, so each pair’s share of the scrub is too small and deaths outnumber births (accept: the braking term is negative above K; K fell because the scrub sets it).

Common slip: Justifying with the birds killed by the fire. None died; the resource fell, so K fell.

(d) A student says: “A fire is a density-independent factor, so it leaves K where it was, and the stonechats will nest at 68 pairs again within a few years.” Evaluate the student’s prediction. (1 point)

A full-credit answer: The prediction is wrong.
Calling the fire a density-independent factor is fair: a fire kills the same fraction of birds whatever their density.
But this fire burned the scrub, the resource that sets K, so K fell from 68 to 34 pairs.
The scrub stays bare, so the count falls to about 34 pairs and settles there, not at 68.

Check the box for each point your answer earns

the judgement (the prediction is wrong) AND the ground (this fire destroyed the scrub that sets K, so K fell to 34 pairs and the count settles there while the scrub stays bare). Accept: the label is fair — a fire is density-independent — but the conclusion is wrong, because this fire took away the resource that sets K.

Common slip: Judging the prediction right because fires are listed as density-independent. Whether K moves depends on whether the hazard took a resource away.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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