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. Useful relationships: logistic growth dN/dt = rₘₐₓN(K − N)/K; percentage change = (new − original) ÷ original × 100. Assume no immigration or emigration. A calculator is assumed. All populations and data sets are fictional.
Question 1 (multiple choice · foundation)
Which of the following is most likely to act as a density-independent factor on a population of meadow plants?
- (A) Competition between neighbouring plants for light
- (B) A hailstorm that destroys about 30% of the plants, whether they grow sparsely or densely
- (C) A fungal disease that spreads from plant to plant by touching leaves
- (D) Grazing by rabbits that feed mostly where the plants are thickest
Answer and explanation
Answer: (B). The hailstorm removes about the same fraction of plants at any density, which is the test for a density-independent factor.
- (A) Competition gets stronger as plants crowd together, so it is density-dependent.
- (C) A disease that spreads by contact spreads faster when plants are close together: density-dependent.
- (D) Grazers concentrating where plants are densest take a larger share of dense patches: density-dependent.
Question 2 (multiple choice · core)
A population has rₘₐₓ = 0.2 per week, K = 1,000 and N = 800. What is dN/dt?
- (A) 32 individuals per week
- (B) 40 individuals per week
- (C) 128 individuals per week
- (D) 160 individuals per week
Answer and explanation
Answer: (A). (K − N)/K = (1,000 − 800) ÷ 1,000 = 0.2. dN/dt = 0.2 × 800 × 0.2 = 32 per week.
- (B) divides by N instead of K: 0.2 × 800 × (200 ÷ 800) = 40.
- (C) uses N/K (0.8) instead of (K − N)/K: 0.2 × 800 × 0.8 = 128. That is the share of K already used, not the share left.
- (D) leaves out the (K − N)/K term entirely, which is exponential growth: 0.2 × 800 = 160.
Question 3 (multiple choice · core)
A population grows logistically in an environment with K = 1,200. At which population size will it add the most individuals per year?
- (A) 120
- (B) 600
- (C) 1,080
- (D) 1,200
Answer and explanation
Answer: (B). dN/dt = rₘₐₓN(K − N)/K is largest when N = K/2 = 600. There the product of “many individuals” and “plenty of unused resources” is greatest.
- (A) At 120 each individual reproduces fastest (the per capita rate is highest), but there are few individuals, so the total added is small.
- (C) At 1,080 there are many individuals, but resources are nearly used up, so each one adds very little. (With any rₘₐₓ, N = 120 and N = 1,080 give the same dN/dt.)
- (D) At K, births balance deaths, so dN/dt = 0.
Question 4 (data analysis · core)
A fictional grasshopper population was introduced to a small island and counted each year.
| Year | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| N | 50 | 86 | 148 | 232 | 345 | 455 | 532 | 575 | 592 | 598 | 600 |
(a) Estimate the carrying capacity of the island for this population. (b) Calculate the increase in N for each year, and identify the year in which the population grew most. Explain how this fits the logistic model. (c) Calculate the per capita growth rate for year 0–1 and year 8–9 (use N at the start of each year), and the percentage decrease between them. (d) Explain, in terms of resources, why the per capita growth rate fell.
Worked solution and suggested Marlbridge rubric
(a) N levels off at about 600 grasshoppers, so K ≈ 600.
(b) Increases: 36, 62, 84, 113, 110, 77, 43, 17, 6, 2. The largest increase (113) is from year 3 to year 4, when N rose from 232 to 345. This range includes K/2 = 300, where the logistic model predicts the fastest growth. After that, increases shrink as N approaches K.
(c) Year 0–1: 36 ÷ 50 = 0.72 per year. Year 8–9: 6 ÷ 592 = 0.010 per year. Percentage decrease = (0.72 − 0.0101) ÷ 0.72 × 100 = 99% (98.6%).
(d) As the grasshoppers became denser, each one had less food (plants), space and shelter. With less energy and matter for reproduction, females produced fewer eggs (lower per capita birth rate), and more grasshoppers starved or were killed by predators and disease (higher per capita death rate). Near K, births roughly balance deaths.
| Point | What earns it |
|---|---|
| 1 | K about 600 |
| 1 | Correct increases, largest in year 3–4, linked to N near K/2 |
| 1 | Both per capita rates and a percentage decrease of about 99% |
| 1 | Explains the fall through less resource per individual affecting births and/or deaths |
Accept K from 595 to 610 if justified by the levelling off.
Question 5 (constructed response · core)
A student sows radish seeds in identical trays of compost at four densities, keeping light, water and temperature the same. After 6 weeks she counts surviving seedlings. Three trays were sown at 200 seeds; their counts were 80, 86 and 86. For the other densities, mean survivors are given.
| Seeds sown per tray | 25 | 50 | 100 | 200 |
|---|---|---|---|---|
| Mean survivors per tray | 22 | 41 | 66 | ? |
(a) Calculate the missing mean. (b) Calculate the percentage survival at each density. (c) Calculate the percentage change in survival between 25 and 200 seeds per tray. (d) Explain whether seedling death here is density-dependent, and identify the likely limiting resources. (e) Predict the number of survivors in a tray sown with 400 seeds, and justify your answer.
Model answer and suggested Marlbridge rubric
(a) (80 + 86 + 86) ÷ 3 = 252 ÷ 3 = 84.
(b) 22 ÷ 25 = 88%; 41 ÷ 50 = 82%; 66 ÷ 100 = 66%; 84 ÷ 200 = 42%.
(c) (42 − 88) ÷ 88 × 100 = −52% (survival fell by about half). Note this is a 46 percentage-point drop, but a 52% decrease.
(d) Density-dependent: the fraction that died rose as density rose (from 12% to 58%). Crowded seedlings compete for light (they shade each other), water and mineral nutrients in the compost; with less of each per plant, more die.
(e) About 85–100 survivors, not double the 84. The number of survivors rose by 25 from 50 to 100 seeds but only by 18 from 100 to 200, so it is levelling off. Each tray’s resources can support only a limited number of seedlings (a carrying capacity for the tray), so extra seeds mainly add to the deaths. Survival would fall to roughly 21–25%.
| Point | What earns it |
|---|---|
| 1 | Mean of 84 and all four percentages correct |
| 1 | Percentage change about −52% (not −46) |
| 1 | Density-dependent, justified by the rising fraction dying, with a named resource |
| 1 | Prediction that survivors level off near the tray’s limit, justified by the shrinking increases |
Accept any prediction from about 84 to about 100 with this reasoning.
Question 6 (constructed response · core)
A fictional antelope population on a grassland has rₘₐₓ = 0.4 per year and K = 2,000. The population is currently 1,500. A long drought then reduces plant growth, lowering K to 1,200.
(a) Calculate dN/dt before the drought. (b) Calculate dN/dt after the drought, assuming N is still 1,500. (c) Explain the sign of your answer to (b) in biological terms. (d) Describe how the population size would change over the following years if K stays at 1,200.
Model answer and suggested Marlbridge rubric
(a) (2,000 − 1,500) ÷ 2,000 = 0.25. dN/dt = 0.4 × 1,500 × 0.25 = +150 antelope per year.
(b) (1,200 − 1,500) ÷ 1,200 = −0.25. dN/dt = 0.4 × 1,500 × (−0.25) = −150 antelope per year.
(c) The population is now larger than the new carrying capacity. There is not enough grass for 1,500 antelope, so many are underfed: fewer calves are born and survive, and more adults die. Deaths now exceed births, so dN/dt is negative.
(d) The population would fall, quickly at first and then more slowly as it nears 1,200, then level off around 1,200 (perhaps fluctuating around it). The drought acts through resources: it changes K, and density-dependent effects (competition for less grass) bring N down to match.
| Point | What earns it |
|---|---|
| 1 | +150 per year |
| 1 | −150 per year |
| 1 | N above the new K, linked to too little food per individual and deaths exceeding births |
| 1 | Decline that slows and levels off near 1,200 |
Question 7 (data analysis · stretch)
Ecologists studied a fictional ground squirrel in four 1-hectare meadow plots at different densities. In one spring a flood swept all four plots. Later in the year, a parasitic disease was also recorded.
| Squirrels per hectare | 50 | 100 | 200 | 400 |
|---|---|---|---|---|
| Killed by flood / % | 30 | 29 | 31 | 30 |
| Infected with parasite / % | 4 | 11 | 23 | 41 |
(a) Calculate the number of squirrels killed by the flood and the number infected in the 50 and 400 plots. (b) Calculate the mean percentage killed by the flood across the four plots. (c) Classify each factor as density-dependent or density-independent, and justify using the data. (d) A student argues: “The flood killed far more squirrels than the parasite, so the flood is what keeps this population near its carrying capacity.” Evaluate this claim.
Worked solution
(a) 50 plot: flood 50 × 0.30 = 15; infected 50 × 0.04 = 2. 400 plot: flood 400 × 0.30 = 120; infected 400 × 0.41 = 164.
(b) (30 + 29 + 31 + 30) ÷ 4 = 120 ÷ 4 = 30%.
(c) Flood: density-independent. It killed about 30% at every density (29–31%), even though the number killed rose from 15 to 120. Parasite: density-dependent. The fraction infected rose about tenfold (4% to 41%; 41 ÷ 4 ≈ 10) as density rose eightfold, probably because the parasite passes more easily between squirrels that live close together.
(d) The claim is not supported. The flood did kill many squirrels, but it removed the same fraction at every density, so it does not respond to how crowded the population is. It cannot push a large population down harder than a small one, so it cannot hold N near K. The parasite does exactly that: the denser the population, the larger the share infected (and, if infection lowers survival or breeding, the stronger the brake). Also, the flood was a single event, whereas a parasite can keep acting for as long as the population is crowded. Finally, in the 400 plot the parasite affected more squirrels (164) than the flood (120), so even the numbers do not fully support the claim.
Suggested mark points (4): 1 for all four counts; 1 for 30%; 1 for both classifications justified by the fraction affected; 1 for explaining that only a density-dependent factor can regulate N around K.
How did you do?
- Q1 or Q7 wrong: re-read “What limits growth: two kinds of factor” in the study guide; focus on the word “fraction”.
- Q2, Q3 or Q6 wrong: rework Worked example 1 and look again at Figures 1 and 2.
- Q4 or Q5 calculations wrong: rework Worked example 2; check percentage change uses the original value as the denominator.
- Q6 explanation incomplete: re-read “Carrying capacity”, especially how K changes with resources.
Then tick off the topic checklist.