Before you start
Do the unit diagnostic first, then try these questions before opening any answer.
Big ideas of the unit
- Responses are shaped by fitness. Selection favours responses and signals that raise survival and reproduction (Topic 8.1).
- Energy flows; matter cycles. About 10% of energy passes to each next level; the rest leaves as heat. Carbon, nitrogen, phosphorus and water cycle (Topic 8.2).
- Energy budgets become births and deaths. A net gain funds growth and reproduction; a net loss cuts births and raises deaths. Ectotherms need far less food than endotherms.
- Growth depends on rates and size together. dN/dt = B − D = rN. With unlimited resources, r stays at rₘₐₓ and growth is exponential (Topic 8.3).
- Resources set a ceiling. The limiting resource sets K. Density-dependent factors lower the per capita rate as N rises (logistic growth); density-independent factors remove a fixed fraction (Topic 8.4).
- Populations change each other. Interactions move energy and matter, and a change at one level can cascade down a food chain (Topic 8.5).
- Diversity buys resilience. More species and genetic variety give back-ups; keystone species have effects far beyond their abundance (Topic 8.6).
- Disruptions filter existing variation. Invasions, pollution, climate change and natural events change which variants and species do well (Topic 8.7).
Key relationships and methods
| Idea | Rule or method | Where it appears |
|---|---|---|
| Energy transfer | about 10% per step; count arrows, not levels | food chains, pyramids, biomagnification |
| Energy budget | intake − use = energy for storage, growth and reproduction | endotherms, ectotherms, breeding timing |
| Population change | dN/dt = B − D; r = (B − D) ÷ N; exponential dN/dt = rₘₐₓN | closed populations, new habitats |
| Logistic growth | dN/dt = rₘₐₓN(K − N)/K; fastest at K/2; negative above K | resource-limited populations |
| Density test | does the fraction affected change with density? | competition, disease against storms, frost |
| Diversity | 1 − Σ(n/N)²; richness and evenness | comparing communities |
| Significance | χ² with df = categories − 1, or compare error bars | choice chambers, field surveys |
| Prediction chain | change → first population affected → what depends on it → community later | keystones, cascades, invasions |
These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric, not official scoring guidelines. All organisms and data sets are fictional.
Question 1 (multiple choice · mixed)
A fictional beetle has rₘₐₓ = 0.4 per year and, alone on an island, K = 500. A second species that eats the same leaves arrives, and the first species’ K falls to 300. What happens to its dN/dt at a population of 200?
- (A) It falls from 48 to about 27 beetles per year.
- (B) It stays at 48 beetles per year.
- (C) It stays at 80 beetles per year.
- (D) It rises from 48 to about 53 beetles per year.
Answer and explanation
Answer: (A). Alone: 0.4 × 200 × (500 − 200) ÷ 500 = 48. With the competitor: 0.4 × 200 × (300 − 200) ÷ 300 = 26.7, about 27. Competition between species (Topic 8.5) leaves less food per beetle, so K falls and the logistic brake (Topic 8.4) tightens.
- (B) assumes only rₘₐₓ matters. The brake (K − N)/K changes when K changes.
- (C) ignores the brake entirely: 0.4 × 200 = 80 is exponential growth.
- (D) uses N/K (200 ÷ 300) instead of (K − N)/K.
Question 2 (multiple choice · mixed)
Phosphate limits algal growth in a fictional lake. Which event is most likely to cause the lowest deep-water oxygen a few weeks later?
- (A) A dry summer slows rock weathering, so less phosphate enters the lake.
- (B) Rain carries extra nitrate, but no extra phosphate, into the lake.
- (C) Runoff carries phosphate fertiliser into the lake, so algae bloom, die and are broken down by aerobic decomposers.
- (D) A fish that eats zooplankton is removed, so zooplankton increase and graze more algae.
Answer and explanation
Answer: (C). Adding the limiting nutrient releases algal growth. When the bloom dies, decomposers respire and use up the oxygen: eutrophication (Topic 8.7) driven by the phosphorus cycle (Topic 8.2).
- (A) would reduce algal growth.
- (B) adds a nutrient that is not limiting, so it has little effect.
- (D) is a trophic cascade (Topic 8.5) that lowers algae, so less dead matter reaches decomposers.
Question 3 (multiple choice · mixed)
A fictional songbird starts laying when spring day length passes a set value, and feeds its chicks on caterpillars. After a run of warm springs, the caterpillar peak comes 12 days earlier, but the birds still lay on the same date. Which prediction is best supported?
- (A) dN/dt will rise, because there will be more caterpillars in total.
- (B) The birds will start laying earlier within one season because they need to.
- (C) The bird population will be unaffected, because adults can still eat caterpillars.
- (D) Chicks hatch after the food peak, so fewer fledge and dN/dt falls; females that happen to lay earlier will be favoured.
Answer and explanation
Answer: (D). The response to day length (Topic 8.1) no longer matches the food supply. Less energy reaches the chicks (Topic 8.2), so fewer survive, B falls and dN/dt falls (Topic 8.3). If laying date varies heritably, selection favours early layers (Topic 8.7).
- (A) confuses the timing of the peak with its size.
- (B) assumes change happens because it is needed. Selection acts on existing variation.
- (C) ignores the chicks, which need the most energy.
Question 4 (constructed response · mixed)
In a fictional pond, phytoplankton make 60 000 kJ of net production a year. Zooplankton eat them, and a small fish eats the zooplankton. In a simple model, the fish can use all the zooplankton production, each fish needs 20 kJ of food a year, and rₘₐₓ = 0.5 per year.
(a) Use the 10% rule to calculate the zooplankton production and the fish’s carrying capacity. (b) Calculate dN/dt and the per capita growth rate when there are 120 fish. (c) A storm washes silt into the pond. The cloudy water cuts phytoplankton production by 40% for a year. Calculate the new K and the new dN/dt at 120 fish. (d) Use a calculation to describe what would happen to 240 fish after the storm. (e) Explain, in terms of energy, why K fell, and name the type of factor that slows the fish population.
Worked solution and suggested Marlbridge rubric
(a) Zooplankton: 60 000 × 0.10 = 6000 kJ a year. K = 6000 ÷ 20 = 300 fish.
(b) dN/dt = 0.5 × 120 × (300 − 120) ÷ 300 = 0.5 × 120 × 0.6 = 36 fish per year. Per capita: 36 ÷ 120 = 0.30 per year.
(c) Phytoplankton: 60 000 × 0.60 = 36 000. Zooplankton: 3600. K = 3600 ÷ 20 = 180 fish. dN/dt = 0.5 × 120 × (180 − 120) ÷ 180 = 20 fish per year.
(d) dN/dt = 0.5 × 240 × (180 − 240) ÷ 180 = −40 fish per year. Above K, deaths exceed births, so N falls towards 180.
(e) Cloudy water means less light and photosynthesis, so less energy enters the chain; with 10% passed on, zooplankton fall by the same proportion, so fewer fish can each get 20 kJ. Food competition among the fish is density-dependent: the more fish, the less each one gets.
| Point | What earns it |
|---|---|
| 1 | 6000 kJ and K = 300 |
| 1 | 36 fish per year and 0.30 per year |
| 1 | K = 180 and 20 fish per year |
| 1 | −40 fish per year, with the population falling to K |
| 1 | Less light → less photosynthesis → less energy at each level → lower K |
| 1 | Food competition named as density-dependent, with the reason |
Total: 6 points. Accept (c) and (d) carried forward from an error in (a).
Question 5 (constructed response · mixed)
A fictional garden plant, X, was brought to a new region. Students recorded the plant at 100 random points in a meadow before X arrived and 8 years later.
| Species | X | A | B | C | D | E | F |
|---|---|---|---|---|---|---|---|
| Before | 0 | 24 | 20 | 18 | 16 | 12 | 10 |
| After 8 years | 70 | 12 | 8 | 6 | 4 | 0 | 0 |
Insects eat 35% of X’s leaf area in its native range. In the meadow they eat 4% of X’s and 22% of the natives’ leaf area.
(a) Calculate Simpson’s diversity index before and after, and state the species richness each time. (b) Use the leaf-damage data to explain how X became invasive. (c) The plant species differ in drought tolerance. Predict whether the meadow would keep more plant cover in a severe drought before or after X spread. Explain. (d) Describe a controlled field test of removing X to restore diversity, and the result that would support it.
Worked solution and suggested Marlbridge rubric
(a) Before: Σ(n/N)² = (24² + 20² + 18² + 16² + 12² + 10²) ÷ 100² = 1800 ÷ 10 000 = 0.180. Index = 0.82. Richness 6.
After: Σ(n/N)² = (70² + 12² + 8² + 6² + 4²) ÷ 10 000 = 5160 ÷ 10 000 = 0.516. Index = 0.484, so 0.48. Richness 5 (4 natives plus X).
(b) X escaped from its enemies: insects eat 35% of its leaves at home but only 4% here, against 22% of native leaves. With more leaf area for photosynthesis, X gains more energy, grows faster and outcompetes the natives for light and space.
(c) Before. Six fairly even species with different drought responses meant tolerant ones could keep production going (functional redundancy). Now 70% of plants are X: if X is drought-sensitive, most cover is lost at once.
(d) Remove X from randomly chosen plots; disturb the soil of similar control plots in the same way. After a few years, compare richness and Simpson’s index as means ± 2 SE. Support: higher values in removal plots, with non-overlapping error bars.
| Point | What earns it |
|---|---|
| 1 | Both indices correct |
| 1 | Both richness values (6 and 5) |
| 1 | Escape from enemies, using 35%, 4% and 22% |
| 1 | Link from less damage to more energy and outcompeting natives |
| 1 | More cover kept before, explained by redundancy or evenness |
| 1 | Replicated removal and control plots, with a similar disturbance in the controls |
| 1 | Result that would support the proposal, with error bars or repeats |
Total: 7 points.
Question 6 (constructed response · mixed)
Tadpoles of a fictional frog hide when they detect chemicals from a native predatory fish. Another predatory fish has been introduced. In a tank with a refuge half and an open half, 60 tadpoles were tested in each type of water:
| Water | In refuge | In open water |
|---|---|---|
| Plain pond water | 33 | 27 |
| Water from native predator | 48 | 12 |
| Water from introduced predator | 35 | 25 |
In a separate trial, 33 of 60 tadpoles survived 3 days with the native predator, but only 12 of 60 with the introduced one.
(a) Calculate χ² for the two predator treatments (null hypothesis: no preference) and state each conclusion at p = 0.05. (b) Explain the purpose of the plain-water treatment. (c) Explain how hiding increases fitness. (d) Use the data to explain why the introduced predator could reduce the frog population more than the native one.
Worked solution and suggested Marlbridge rubric
(a) Expected: 30 and 30. Degrees of freedom = 2 − 1 = 1; critical value 3.84.
- Native: χ² = (48 − 30)² ÷ 30 + (12 − 30)² ÷ 30 = 10.8 + 10.8 = 21.6. 21.6 > 3.84: reject the null. The tadpoles hide.
- Introduced: χ² = (35 − 30)² ÷ 30 + (25 − 30)² ÷ 30 = 1.67. 1.67 < 3.84: fail to reject. No evidence that they respond.
(b) It is the control: it tests for a side preference unrelated to predators (χ² = 0.6, none found).
(c) Hiding lowers the chance of being eaten, so more tadpoles survive to reproduce.
(d) Tadpoles do not respond to the introduced predator’s chemicals, so they stay in the open. Survival is 20% against 55%. More deaths (higher D) lower dN/dt.
| Point | What earns it |
|---|---|
| 1 | 21.6 and 1.67 |
| 1 | Correct conclusions with df = 1 and 3.84 |
| 1 | Plain water as the control for a side or refuge bias |
| 1 | Hiding → higher survival → more reproduction |
| 1 | Lack of response and 20% against 55% survival linked to higher deaths and a falling population |
Total: 5 points.
Question 7 (constructed response · mixed)
A fictional lizard must bask before it is warm enough to forage. In a cool year, it could forage about 4 hours a day; in a warm year, 7 hours. The population was 400 at the start of each year.
| Year | Births | Deaths |
|---|---|---|
| Cool | 120 | 140 |
| Warm | 220 | 100 |
(a) Calculate dN/dt and the per capita growth rate for each year. (b) Explain, using the energy budget of an ectotherm, why births were higher in the warm year. (c) Explain why a vole of similar mass on the same hillside can forage in cool air but needs much more food. (d) Further warming makes midday too hot, and foraging falls to 3 hours a day. Predict the effect on the lizard population and explain one response that could reduce the harm.
Worked solution and suggested Marlbridge rubric
(a) Cool: dN/dt = 120 − 140 = −20 lizards per year; r = −20 ÷ 400 = −0.05 per year. Warm: dN/dt = 220 − 100 = +120 lizards per year; r = 120 ÷ 400 = 0.30 per year.
(b) A lizard’s body temperature follows its surroundings, so it hunts only when warm. In the warm year it foraged 75% longer and had a net energy gain, which went into eggs, so births rose.
(c) The vole is an endotherm: its metabolism keeps it warm, so it can forage in cool air. Making that heat costs a lot of energy, so it needs much more food.
(d) Foraging falls by about 57% from the warm year, so less energy means fewer births and more deaths: dN/dt is likely to turn negative. Lizards that shelter at midday and forage in cooler mornings and evenings keep more foraging time. If this behaviour or heat tolerance varies heritably, selection favours it.
| Point | What earns it |
|---|---|
| 1 | Both dN/dt values |
| 1 | Both per capita rates |
| 1 | Longer foraging → net energy gain → more reproduction |
| 1 | Vole as an endotherm, with the higher energy cost explained |
| 1 | Prediction of falling numbers linked to less energy |
| 1 | A behavioural response or selection on existing variation |
Total: 6 points.
How did you do?
Add up your points: 3 for multiple choice and 24 for constructed response. Use the total only to compare attempts; it does not predict an exam score.
- Lost points on growth equations (Questions 1, 4, 7(a)): rework the worked examples of Topic 8.3 and Topic 8.4.
- Lost points on energy (Questions 4(e), 7(b)–(c)): reread Topic 8.2.
- Lost points on communities and invasions (Questions 2, 5, 6): read Topic 8.5, Topic 8.6 and Topic 8.7.
- Quick recheck: retake the unit diagnostic.
Then tick off each topic checklist: