These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: Simpson’s diversity index = 1 − Σ(n/N)², where n is the number of individuals of one species and N the total of all species; standard error SE = s ÷ √n; error bars are mean ± 2 SE. A calculator is assumed. Keep full calculator values until the last step. All species named by letter and all data sets are fictional.
Question 1 (multiple choice · foundation)
A small bird follows large grazing animals across a grassland. As the grazers walk, they disturb insects from the grass, and the bird catches them. Observations show that the grazers’ feeding, health and breeding are the same whether or not birds are present. Which interaction is this?
- (A) Mutualism (+/+)
- (B) Commensalism (+/0)
- (C) Parasitism (+/−)
- (D) Competition (−/−)
Answer and explanation
Answer: (B). The bird gains food (+). The data show no effect on the grazers (0). That is commensalism.
- (A) needs both species to benefit. The grazers gain nothing measurable here. (If the birds also removed ticks from the grazers, it could become mutualism.)
- (C) needs the grazers to be harmed, for example by losing food or blood. They are not.
- (D) needs both to use the same limited resource. The bird eats insects; the grazers eat grass.
Question 2 (multiple choice · core)
A sample from a stream contains three species of insect larvae: 6 of species J, 3 of species K and 1 of species L. What is Simpson’s diversity index?
- (A) 0.46
- (B) 0.54
- (C) 0.64
- (D) 3
Answer and explanation
Answer: (B). N = 10. Proportions: 0.6, 0.3, 0.1. Squares: 0.36, 0.09, 0.01. Sum = 0.46. Index = 1 − 0.46 = 0.54.
- (A) is Σ(n/N)², the step before subtracting from 1.
- (C) subtracts only the most common species’ square: 1 − 0.36 = 0.64.
- (D) is the species richness (3 species), not the index. Simpson’s index always lies between 0 and 1.
Question 3 (multiple choice · core)
In a greenhouse, a population of aphids (plant-sucking insects) reaches its peak in week 4. The population of ladybirds, which eat aphids, reaches its peak in week 6, and the aphids are then falling. Which statement best explains the pattern?
- (A) Ladybirds need time to convert the extra aphids they eat into extra offspring, so their numbers rise after the aphid peak; more ladybirds then eat more aphids, so aphids fall.
- (B) Aphids and ladybirds compete for the same plants, so when one rises the other must fall.
- (C) The ladybirds peak later because they are a larger species and grow more slowly; the aphid fall is unrelated to the ladybirds.
- (D) Aphid numbers fall because the aphids reached their carrying capacity in week 4; ladybirds increase because they no longer have to compete with aphids.
Answer and explanation
Answer: (A). This is the predator–prey lag. More prey means more food for predators, but turning food into offspring takes time, so predators peak later. Their larger numbers then push prey down.
- (B) treats predation (+/−) as competition. Ladybirds eat aphids, not plants.
- (C) ignores the link between the two populations. Size may affect breeding speed, but the timing and the aphid fall are explained by predation.
- (D) mixes up ideas: carrying capacity limits a population but does not explain why ladybirds rise; ladybirds do not compete with aphids.
Question 4 (constructed response · core)
A student nets aquatic snails from three ponds. Each sample contains 100 snails.
| Species | W | X | Y | Z | U | V |
|---|---|---|---|---|---|---|
| Pond P | 30 | 30 | 20 | 20 | 0 | 0 |
| Pond Q | 85 | 5 | 5 | 5 | 0 | 0 |
| Pond R | 20 | 20 | 0 | 0 | 30 | 30 |
(a) Calculate Simpson’s diversity index for each pond. (b) Ponds P and Q have the same species richness. Explain why their indices differ. (c) Ponds P and R have the same index. Explain why their communities are still different. (d) Suggest one interaction that could explain the pattern in Pond Q.
Worked solution and suggested Marlbridge rubric
(a) N = 100 in each pond.
- P: 0.30² + 0.30² + 0.20² + 0.20² = 0.09 + 0.09 + 0.04 + 0.04 = 0.26. Index = 1 − 0.26 = 0.74.
- Q: 0.85² + 3 × 0.05² = 0.7225 + 0.0075 = 0.73. Index = 1 − 0.73 = 0.27.
- R: 0.20² + 0.20² + 0.30² + 0.30² = 0.26. Index = 0.74.
(b) Both have four species, but Pond Q is very uneven: species W makes up 85% of the snails. Diversity depends on evenness as well as richness, so Q’s index is much lower.
(c) The index measures richness and evenness only. Pond R contains different species (U and V instead of Y and Z), so its species composition is different even though its diversity is the same.
(d) Any one with a reason, for example: species W out-competes the others for food (competition, −/−); a predator in Pond Q prefers to eat X, Y and Z (predation); a parasite affects the other species more than W.
| Point | What earns it |
|---|---|
| 1 | P and R = 0.74 with working |
| 1 | Q = 0.27 with working |
| 1 | (b) explains evenness, using the dominance of W |
| 1 | (c) distinguishes composition from diversity, naming the different species |
| 1 | (d) a named interaction linked to the dominance of W |
Question 5 (constructed response · stretch)
Two species of small insect-eating bird, A and B, live on the same wooded island. A researcher records the height at which individual birds were seen feeding (trees are up to 10 m tall). Six records for species A: 3.1, 2.6, 3.4, 2.9, 3.0 and 2.8 m. For species B, the mean is 7.6 m with 2 SE = 0.6 m. On a nearby island where species A is absent, species B feeds at a mean height of 5.2 m, with 2 SE = 0.7 m.
(a) Calculate the mean feeding height of species A and its mean ± 2 SE interval. (b) Use error bars to decide whether A and B feed at different heights on the shared island. (c) Compare species B on the two islands and explain the result in terms of niches. (d) State one limitation of comparing the two islands.
Worked solution and suggested Marlbridge rubric
(a) Mean = 17.8 ÷ 6 = 2.97 m. Deviations: 0.13, −0.37, 0.43, −0.07, 0.03, −0.17; squares sum to 0.373. s = √(0.373 ÷ 5) = 0.273 m. SE = 0.273 ÷ √6 = 0.112 m; 2 SE = 0.22 m. Interval: 2.74 to 3.19 m.
(b) Species B: 7.6 ± 0.6, from 7.0 to 8.2 m. The intervals (2.74–3.19 and 7.0–8.2) are far apart, so the mean heights are probably different. On the shared island, A feeds low and B feeds high.
(c) Without A, species B feeds lower: 4.5–5.9 m, which does not overlap 7.0–8.2 m. So B can use the middle heights but does not do so when A is present. This is niche partitioning: competition with A pushes B into a narrower part of its possible niche, higher in the trees. Dividing the trees by height reduces competition for insects, so both species can live on the same island.
(d) Any one: the islands may differ in tree height, insect supply or other predators and competitors, so the difference might not be caused by A; only one island of each type was studied; the birds were not tested experimentally (A was not added or removed).
| Point | What earns it |
|---|---|
| 1 | Mean 2.97 m and 2 SE ≈ 0.22 m |
| 1 | Correct interval and correct conclusion from non-overlap in (b) |
| 1 | (c) states B feeds lower without A, using the intervals |
| 1 | (c) links the shift to competition and niche partitioning |
| 1 | (d) a valid limitation (confounding difference or no replication) |
Accept a 2 SE of 0.2 m if the working is shown.
Question 6 (constructed response · core)
Six outdoor tanks are set up with water, algae and zooplankton (tiny animals that graze on algae). Small fish that eat zooplankton are added to three tanks; three tanks have no fish. After four weeks:
| Tanks | Zooplankton / individuals L⁻¹ (mean ± 2 SE) | Chlorophyll (a measure of algae) / µg L⁻¹ (mean ± 2 SE) |
|---|---|---|
| No fish | 42 ± 6 | 11 ± 2 |
| Fish added | 9 ± 3 | 29 ± 4 |
(a) Calculate the percentage change in each variable when fish were added. (b) Use the error bars to decide whether each difference is likely to be real. (c) Explain the change in algae, even though the fish do not eat algae. (d) Predict the effect on algae of adding a larger fish that eats the small fish. Explain.
Worked solution and suggested Marlbridge rubric
(a) Zooplankton: (9 − 42) ÷ 42 × 100 = −79%. Chlorophyll: (29 − 11) ÷ 11 × 100 = +164%.
(b) Zooplankton: 36–48 against 6–12. Chlorophyll: 9–13 against 25–33. Neither pair overlaps, so both differences are probably real.
(c) This is a trophic cascade. The fish eat zooplankton, so zooplankton fall. With fewer grazers, less algae is eaten, so algae grow and chlorophyll rises. The fish affect the algae indirectly, through the zooplankton.
(d) Algae should decrease, back towards the no-fish level. The larger fish eat small fish, so fewer zooplankton are eaten; zooplankton recover and graze more algae. Each added level reverses the direction of the effect.
| Point | What earns it |
|---|---|
| 1 | Both percentage changes correct with signs |
| 1 | Both decisions justified with the intervals |
| 1 | Cascade explained in sequence: fish → fewer zooplankton → less grazing → more algae |
| 1 | Correct prediction for the added level, with the chain of reasoning |
Question 7 (constructed response · stretch)
Two species of single-celled protist, A and B, feed on the same bacteria. Each is grown alone in a flask and then together in one flask, with the same food supply. Densities on day 16 (cells mL⁻¹):
| Culture | Species A | Species B |
|---|---|---|
| Alone | 520 | 380 |
| Together | 470 | 15 |
(a) Calculate the percentage change in each species when grown together. (b) Name the interaction and use the data to justify its sign pattern. (c) Explain why species B nearly disappears. (d) Suggest one change to the mixed flask that might allow both species to persist, and explain how it would work.
Worked solution and suggested Marlbridge rubric
(a) A: (470 − 520) ÷ 520 × 100 = −9.6%. B: (15 − 380) ÷ 380 × 100 = −96%.
(b) Competition (−/−). Both species reach lower densities together than alone, so each harms the other, but the effect on B is about ten times larger.
(c) Both species need the same limited bacteria. Species A is the better competitor: it takes the bacteria faster or survives on less food. B gets too little energy and matter to grow, so its population collapses. This is competitive exclusion: two species with the same niche and a limited resource cannot both persist indefinitely.
(d) Any reasonable change with a mechanism, for example: add a second food (such as yeast) that B can eat but A cannot; add a structure, such as glass beads, that creates places only B can feed. Each gives B a resource or space A does not use, so their niches partly separate (niche partitioning) and competition falls.
| Point | What earns it |
|---|---|
| 1 | Both percentage changes correct |
| 1 | Competition named, with data showing both species decreased |
| 1 | Exclusion explained by a shared limited resource and A’s advantage |
| 1 | A change that separates the niches, with its mechanism |
Do not award (b) for “predation”: neither protist eats the other.
How did you do?
- Q1 or Q3 wrong: re-read “Interactions between populations” and Figure 1 in the study guide.
- Q2 or Q4 wrong: rework Worked example 1; square every proportion before adding, and remember composition is not the same as diversity.
- Q5 or Q6 error-bar decisions wrong: rework Worked example 2.
- Q6 or Q7 explanations incomplete: re-read “Trophic cascades” and “Competition and niche partitioning”.
Then tick off the topic checklist.