Practice questions · Topic 8.3

Unit 8 · Practice for the Topic 8.3 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: Population ecology, summed up

Video coming soon

A population sharpened: one species, one place, able to mix; births minus deaths as dN/dt = B − D; per head, rmax; the more there are, the more are added, dN/dt = rmax N; the J on an ordinary axis and the straight line on a log axis; drawing the graph.

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 two growth equations, dN/dt = B − D and dN/dt = rmax N.

Question 1

Each of the following is a group of armadillos, or of armadillos and skunks.

Which of the following is one population?

A. All the armadillos of one wood B. The armadillos and the skunks of one wood C. The armadillos of two islands 80 km apart, between which no armadillo swims D. All the armadillos of the species, wherever they live

Question 2

Nuthatches nest in a wood. Ecologists count how many hatch and how many die each year. This year more nuthatches hatched than died.

Which of the following would make the nuthatch population smaller next year than it is this year?

A. More nuthatches hatch next year than this year, and the same number die B. The same number hatch and the same number die next year as this year C. The same number hatch next year as this year, more die, and the deaths are still fewer than the births D. Next year the deaths outnumber the births

Question 3

Ecologists write four numbers about a colony of guillemots: B is 310 guillemots per year, D is 190 guillemots per year, N is 2 400 guillemots, and dN/dt is 120 guillemots per year.

Which of the four numbers is a count at one moment?

A. B B. D C. N D. dN/dt

Question 4

Partridges live on a farm. In one year 130 chicks hatch and 74 partridges die.

Which of the following is dN/dt for the partridges?

A. −56 partridges per year B. 56 partridges per year C. 74 partridges per year D. 204 partridges per year

Question 5

Tench live in a lake. The population holds 620 tench now, and its rate of change of population size is 45 tench per year.

Which of the following is the size of the population after 6 years, if the rate stays the same?

A. 270 tench B. 350 tench C. 665 tench D. 890 tench

Question 6

Suppose 250 houseflies live in a barn with food to spare and nothing that eats them. Their maximum per capita growth rate, rmax, is 0.2 per day.

Which of the following is dN/dt for the houseflies?

A. 0.2 houseflies per day B. 50 houseflies per day C. 250.2 houseflies per day D. 1 250 houseflies per day

Question 7

Krill in a large tank have food to spare. The tank held 400 krill a year ago and holds 500 krill now. Each krill adds the same number of young per year as before.

Which of the following is the tank’s gain over the next year?

A. 25 krill B. 100 krill C. 125 krill D. 625 krill

Question 8

The anemone counts on the four rocks.

Ecologists count the sea anemones on four rocks once a year for three years. The table below shows the counts.

Which rock’s anemone population is growing exponentially?

A. The north rock’s B. The east rock’s C. The south rock’s D. The west rock’s

Question 9

The leech counts on a log-scale y-axis.

A student counts the leeches in a pond once a week. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.

Which of the following is the count at 2 weeks?

A. 15 leeches B. 20 leeches C. 30 leeches D. 200 leeches

Question 10

A student counts the swans on a lake once a month for a year. Her counts run from 6 swans to 43 swans.

Which y-axis lets every one of her counts be read?

A. An ordinary axis from 0 to 50 with a gridline every 10 B. An ordinary axis from 10 to 50 with a gridline every 10 C. An ordinary axis from 0 to 500 with a gridline every 100 D. A log scale from 10 to 1 000

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

Razorbills nest on a small island off a rocky coast. The colony holds 260 razorbills now. In the past year 72 chicks hatched and 33 razorbills died. The birds have fish to spare, nothing on the island eats them, and no disease has reached them.

(a) Calculate dN/dt for the colony over the past year. (1 point)

Hint: Which of the two counts is the birth rate, B, and which is the death rate, D?

razorbills per year

Write down the values in the question:

B = 72 razorbills per year

D = 33 razorbills per year

Write down the equation:

\frac{d N}{d t} = B - D

Substitute the values into the equation, and calculate:

\frac{d N}{d t} = 72 - 33

\frac{d N}{d t} = 39 razorbills per year

A full-credit answer: dN/dt = B − D = 72 − 33 = 39 razorbills per year.

(b) Calculate the size of the colony after 4 years, using the rate of change from part (a). (1 point)

Hint: How many razorbills does the colony gain in one year, and how many years pass?

razorbills

Write down the values in the question:

N = 260 razorbills

\frac{d N}{d t} = 39 razorbills per year

time = 4 years

Write down the equation:

later size = present size + rate of change \times time

Substitute the values into the equation, and calculate:

later size = 260 + 39 \times 4

later size = 416 razorbills

A full-credit answer: later size = present size + rate of change × time = 260 + 39 × 4 = 416 razorbills.

(c) Explain what the calculation in part (b) assumes about the colony’s rate of change. (1 point)

Hint: In your working for part (b), did the number you multiplied by the years change from one year to the next?

A full-credit answer: The calculation assumes that the rate of change stays the same every year.
It multiplies one year’s rate by four years.
So it takes the colony to gain the same number of razorbills in each of the four years.

Check the box for each point your answer earns

the rate of change (the yearly gain) stays the same in every one of the four years.

Common slip: Saying the calculation assumes no deaths. Deaths are inside the rate; the assumption is that the rate does not change.

(d) The birds have fish to spare, no predators and no disease. Predict how the colony’s yearly gain changes over the four years, and explain your prediction. (1 point)

Hint: How many razorbills are adding chicks in year 2, compared with year 1?

A full-credit answer: The yearly gain grows because each razorbill keeps adding the same number of chicks per year.
Each year the colony holds more razorbills than the year before.
More razorbills, each adding the same number, add more chicks in all.

Check the box for each point your answer earns

the gain grows, because each bird adds the same number while more birds are adding each year.

Common slip: Predicting the same gain every year because nothing has changed for the birds. Nothing changes per bird; the number of birds adding changes.

(e) An ecologist plots the colony’s yearly counts, which climb from 260 to about 450 razorbills. Determine which y-axis scale, ordinary or log, fits these counts. (1 point)

Hint: How many powers of ten do the counts cross between 260 and 900?

A full-credit answer: An ordinary y-axis fits, because the counts stay within one power of ten.
On an ordinary axis from 0 to 500 razorbills, every count from 260 to 450 sits clear of the x-axis and can be read.

Check the box for each point your answer earns

an ordinary axis, because the counts span less than one power of ten (from the hundreds to the hundreds), so every count can be read on it.

Common slip: Choosing a log scale because the counts grow exponentially. The scale is chosen by the span of the counts, not by the shape of the growth.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

Gobies of one kind live in two rock pools on a shore. At every high tide the sea covers both pools, and the gobies swim between them. The two pools hold 310 gobies now. In the past year 140 gobies hatched in the pools and 95 died. The gobies have food to spare, no predator reaches the pools, and no disease has been seen.

(a) Determine whether the gobies of the two pools are one population or two. (1 point)

A full-credit answer: The gobies of the two pools are one population.
They are one species, and at every high tide they swim between the pools.
So the gobies of the two pools mix: they compete for the same food and breed with each other.

Check the box for each point your answer earns

one population, because the gobies are one species and swim between the pools at high tide, so they mix (compete and breed together).

Common slip: Answering two populations because there are two pools. Two places hold one population when the individuals move between them and mix.

(b) Calculate dN/dt for the gobies over the past year. (1 point)

gobies per year

Write down the values in the question:

B = 140 gobies per year

D = 95 gobies per year

Write down the equation:

\frac{d N}{d t} = B - D

Substitute the values into the equation, and calculate:

\frac{d N}{d t} = 140 - 95

\frac{d N}{d t} = 45 gobies per year

A full-credit answer: dN/dt = B − D = 140 − 95 = 45 gobies per year.

(c) A student predicts the count after three more years by adding three years of this year’s rate of change to the count now. Evaluate the student’s prediction. (1 point)

A full-credit answer: The prediction is too small.
The student’s method assumes the gobies gain 45 every year.
The gobies have food to spare, so each goby keeps adding the same number of young.
Each year more gobies are adding, so the yearly gain grows above 45.

Check the box for each point your answer earns

the judgement (too small) AND the ground (the method assumes a constant yearly gain, but with food to spare each goby adds the same number and more gobies are adding each year, so the gain grows).

Common slip: Judging the prediction right because the rate was measured. The measured rate is this year’s; next year more gobies are adding.

(d) The student plots the gobies’ yearly counts on a log-scale y-axis against time in years. Predict the shape of the plotted points while the pools stay as described, and justify your prediction. (1 point)

A full-credit answer: The points lie on a straight line sloping up.
With food to spare, the count multiplies by the same factor every year.
On a log-scale y-axis, equal distances along the y-axis are equal factors.
So each year the point climbs the same distance, and equal rises in equal steps make a straight line.

Check the box for each point your answer earns

a straight rising line, because the count multiplies by the same factor each year and on a log-scale axis equal factors are equal distances, so each yearly rise is the same.

Common slip: Predicting a curve that bends upward. That is the shape on an ordinary y-axis; a log-scale y-axis straightens it.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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