Practice questions · Topic 8.5

Unit 8 · Practice for the Topic 8.5 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: Community ecology, summed up

Video coming soon

Population, community and ecosystem; species composition, richness, evenness and diversity; the pie chart; Simpson’s Diversity Index — what the formula does, how to calculate it and what a difference in it means; the five interactions and symbiosis; niche, competitive exclusion and niche partitioning; the dual-y graph and the predator–prey cycle; trophic cascades and the signed-arrow model; the ±2SE overlap rule.

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 community’s Simpson’s Diversity Index 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 Simpson’s Diversity Index, Diversity Index = 1 − Σ(n/N)², where n is the total number of organisms of a particular species and N is the total number of organisms of all species.

Question 1

A biologist lists every species of bird, plant and insect living in one gorge, together with the gorge’s rock, its stream and the sunlight that reaches its floor.

Which of the following is that group?

A. A population B. A community C. An ecosystem D. The gorge’s non-living surroundings

Question 2

The count of each plant species in the four ravines.

Suppose surveyors count the plants of four ravines, narrow steep-sided valleys, and list the count of each species in each ravine. The table below gives the four lists.

Which ravine has the greatest species diversity?

A. Ravine Q B. Ravine U C. Ravine Y D. Ravine Z

Question 3

The canyon’s shrubs as a pie chart, its slices lettered Q, U, Y and Z.

Suppose a survey of the shrubs of one canyon counts 60 plants of four species: 30 saltbush, 15 buckthorn, 9 wild rose and 6 dogwood. The pie chart below shows the four species as slices Q, U, Y and Z, with no other labels. The slices are drawn in a shuffled order, different from the order of the species in the list.

Which slice is the buckthorn’s?

A. Slice Q B. Slice U C. Slice Y D. Slice Z

Question 4

Suppose a student counts 80 insects in a dovecote, a tower built for pigeons, and finds that every one of them belongs to the same species.

Which of the following is Simpson’s Diversity Index for the dovecote’s insects?

A. 0 B. 0.5 C. 1 D. It cannot be worked out until the species is named

Question 5

Suppose a diver records the fish of one reef flat, the shallow shelf of a coral reef, each year for ten years. Six species of fish are present every year. Simpson’s Diversity Index for the fish rises from 0.44 to 0.82 over the ten years.

Which of the following happened to the reef flat’s fish over the ten years?

A. More species of fish came to live on the reef flat B. The fish became more equally shared among the six species C. One species came to hold most of the fish D. Every species’ count fell by the same fraction

Question 6

A kinkajou, a small tree-living mammal of the tropical Americas, drinks nectar from the flowers of the balsa tree at night. As it moves from flower to flower it carries pollen between them, and the flowers set seed. Write a plus for a population that gains, a minus for one that loses and a zero for one that is untouched.

Which of the following gives the sign each population gets?

A. Kinkajous +, balsa trees 0 B. Kinkajous +, balsa trees − C. Kinkajous 0, balsa trees + D. Kinkajous +, balsa trees +

Question 7

Suppose two species of pipistrelle, small bats, hunt the same small flies along one escarpment, a long steep slope. One species hunts at dusk and the other after midnight.

Compared with two species that hunt at the same time of night, how much do the two species of pipistrelle compete for the flies?

A. More B. The same C. Less D. Not at all: they never hunt at the same time

Question 8

Suppose redwings, thrushes that arrive in one wood each winter, and firecrests, tiny songbirds, are counted in the wood every winter for twelve winters. The redwing counts lie between 600 and 4 000 redwings. The firecrest counts lie between 9 and 45 firecrests. A student plots both series against the winters, one series against a left y-axis and the other against a right y-axis.

Which population takes the right y-axis, and why?

A. The firecrests, because their counts are the smaller and the right y-axis is numbered to fit them B. The firecrests, because the right y-axis always carries the species counted second C. The redwings, because the larger counts need the taller axis and the right y-axis is the taller D. Either population, because the two y-axes carry one scale

Question 9

Which of the following is a trophic cascade?

A. A dogfish, a small shark, catches a whiting and eats it B. Two species of skate hunt the same shellfish in one bay, and each leaves fewer for the other C. More dogfish eat more whiting, so more of the brown shrimps the whiting eat survive D. Booklice feed on the fungus growing under a tree’s bark, and the tree is untouched

Question 10

The mean number of bird species per visit in the two polders, each with its ±2SE.

Suppose a birdwatcher visits the ditches of two polders, flat land reclaimed from the sea behind a dike, twelve times each: an old polder and a new polder. On each visit the birdwatcher lists the bird species seen. The table below gives each polder’s mean number of bird species per visit with its ±2SE, so each error bar’s two ends can be worked out.

Which of the following do the two error bars show?

A. The two means differ B. The data cannot tell C. The two means are the same D. The new polder holds more bird species

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

Suppose a botanist counts the plants growing on one crag, a rocky outcrop, and finds 50 plants of four species. The table below gives the counts. The formula sheet gives

Diversity Index = 1 - \sum {(\frac{n}{N})}^{2}

, where n is the total number of organisms of a particular species and N is the total number of organisms of all species.

The plants counted on the crag, one row per species.

(a) Calculate the share of the crag’s plants held by the most numerous species, to two decimal places. (1 point)

Hint: n is the count of that one species; N is the count of every plant on the crag.

Write down the values in the question:

n = 19

N = 50

Write down the equation:

share = \frac{n}{N}

Substitute the values into the equation, and calculate:

share = \frac{19}{50}

share = 0.38

A full-credit answer: share = n ÷ N = 19 ÷ 50 = 0.38.

(b) Calculate that species’ squared share, to four decimal places. (1 point)

Hint: Multiply the share by itself, not by 2.

Write down the value in the question:

\frac{n}{N} = 0.38

Square the share, and calculate:

{(\frac{n}{N})}^{2} = 0.38 \times 0.38

{(\frac{n}{N})}^{2} = 0.1444

A full-credit answer: squared share = 0.38 × 0.38 = 0.1444.

(c) Calculate the sum of the four species’ squared shares, to four decimal places. (1 point)

Hint: Square each of the other three shares the same way, then add all four squares up.

Square each species’ share, one line per species:

{(\frac{19}{50})}^{2} = 0.38 \times 0.38 = 0.1444

{(\frac{14}{50})}^{2} = 0.28 \times 0.28 = 0.0784

{(\frac{11}{50})}^{2} = 0.22 \times 0.22 = 0.0484

{(\frac{6}{50})}^{2} = 0.12 \times 0.12 = 0.0144

Add the squared shares up:

0.1444 + 0.0784 + 0.0484 + 0.0144 = 0.2856

A full-credit answer: sum of the squared shares = 0.1444 + 0.0784 + 0.0484 + 0.0144 = 0.2856.

(d) Calculate Simpson’s Diversity Index for the crag, to two decimal places. (1 point)

Hint: Take the sum of the squared shares away from 1, then round.

Write down the equation:

Diversity Index = 1 - \sum {(\frac{n}{N})}^{2}

Substitute the values into the equation, and calculate:

Diversity Index = 1 - 0.2856

Diversity Index = 0.71 (no unit)

A full-credit answer: Diversity Index = 1 − 0.2856 = 0.71.

(e) A second crag holds the same four species of plant, and its Simpson’s Diversity Index is 0.53. Describe what the difference between the two crags’ indices says about how the second crag’s plants are shared among its four species. (1 point)

Hint: Both crags hold the same four species, so which of richness and evenness is left for the two indices to differ by?

A full-credit answer: The second crag’s plants are shared less equally among the four species: one species holds a larger share of them than any species holds on the first crag.

Check the box for each point your answer earns

the second crag’s plants are shared less equally (accept: one species dominates the second crag; the second crag has the smaller species evenness). Accept a description consistent with the index calculated in part (d), provided it is compared with 0.53.

Common slip: Saying the second crag holds fewer species. Both crags hold the same four species; with the same richness a lower index comes from less equal shares.

Free-response score: 0 of 5

Free response 2 · Analyze Model or Visual Representation · 4 points

Suppose kookaburras, large birds of the kingfisher family, hunt dunnarts, mouse-sized marsupials, on one stretch of Australian grassland, and the dunnarts eat soldier beetles. The drawing below shows the three populations as a chain, with the arrows drawn but no signs. The sign at an arrow’s head is the effect on the population the arrow ends on.

The three populations drawn as a chain, with the arrows but no signs.

(a) Identify the sign that belongs at the head of the arrow from the kookaburras to the dunnarts. (1 point)

A full-credit answer: A minus (−).

Check the box for each point your answer earns

a minus (accept: −; negative).

Common slip: Writing a plus because the kookaburras gain a meal. The sign at the head is the effect on the dunnarts, which the arrow ends on: they lose members.

(b) Suppose kookaburras return to the grassland after years away. Predict what happens to the soldier beetles over the following years. (1 point)

A full-credit answer: More soldier beetles survive.

Check the box for each point your answer earns

more soldier beetles / the soldier beetle count rises (the direction must be stated). Do not award: ‘the soldier beetles are affected’ with no direction.

Common slip: Predicting fewer soldier beetles because a hunter has come back. The kookaburras never eat soldier beetles; the change reaches the beetles through the dunnarts.

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

A full-credit answer: With the kookaburras back, more dunnarts are eaten, so the dunnart count falls.
Fewer dunnarts eat fewer soldier beetles.
So more soldier beetles survive.

Check the box for each point your answer earns

more dunnarts are eaten so the dunnart count falls, AND fewer dunnarts eat fewer soldier beetles (both links, each with its direction).

Accept reasoning consistent with a wrong prediction in (b): award the point only for two links, each with its direction, that lead to the direction the student predicted.

Common slip: Justifying with the kookaburras alone. The point wants the dunnart link and the beetle link, each with its direction.

(d) A student says: “Removing the kookaburras will change the soldier beetle count, although no kookaburra ever eats a soldier beetle.” Evaluate the student’s claim. (1 point)

A full-credit answer: The claim is right.
Kookaburras eat dunnarts, and dunnarts eat soldier beetles.
So a change in the kookaburras passes down the chain through the dunnarts to the soldier beetles, which the kookaburras never eat.

Check the box for each point your answer earns

the judgement (the claim is right) AND the ground (the change passes down the chain level by level — kookaburras eat dunnarts and dunnarts eat soldier beetles — so it reaches a population the kookaburras never eat).

Common slip: Judging the claim wrong because the kookaburras never touch a soldier beetle. A change passes down the chain through the population in between.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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