Question 1
Paper 1A style
In a capture–mark–release–recapture study, 40 snails were caught, marked and released. In a second sample of 50 snails, 10 were marked. What is the estimated population size?
- 100
- 200
- 500
- 2000
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Answer: B [1]: (40 × 50) ÷ 10 = 200
Question 2
Paper 1A style
Which of the following is an example of allelopathy?
- A fungus secreting a chemical that kills bacteria around it
- A tree releasing a chemical from its roots that inhibits the growth of other plants nearby
- Bacteria in root nodules fixing nitrogen for a legume
- A plant producing toxins in its leaves that deter herbivores
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Answer: B [1]. A is antibiotic secretion; C is mutualism; D is a defence against herbivory, not against competing plants.
Question 3
Paper 1A style
What do hard corals obtain from the zooxanthellae living in their cells?
- Carbon dioxide and ammonia
- Fixed nitrogen from the atmosphere
- Sugars and other products of photosynthesis
- Protection from predators
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Answer: C [1]. A is what the zooxanthellae receive from the coral.
Question 4
Paper 1B style
Students estimated the population of dandelion plants on a school field of area 600 m². They placed ten 1 m² quadrats at random coordinates and counted the dandelions in each. (a) Calculate the mean number of dandelions per quadrat. [1] (b) Estimate the population size on the field. [1] (c) The standard deviation of the counts is 2.6. State what this indicates about the distribution of the dandelions. [1] (d) Explain why the quadrats were placed at random coordinates. [2]
| Quadrat | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Dandelions | 4 | 7 | 2 | 0 | 5 | 8 | 3 | 6 | 1 | 4 |
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- (a) 40 ÷ 10 = 4.0 (dandelions per quadrat / per m²) [1]
- (b) 4.0 × 600 = 2400 dandelions [1] ECF from (a)
- (c) the counts vary widely around the mean / the dandelions are unevenly spread / clumped OWTTE [1]
- (d) to avoid (sampler) bias / so every part of the field has an equal chance of being sampled [1]
- (d) so the sample is representative / the estimate is closer to the true population size / reduces sampling error [1]
Question 5
Paper 1B style
An ecologist recorded the presence or absence of two moss species, X and Y, in 60 random quadrats on a woodland floor. (a) State a null hypothesis for a chi-squared test on these data. [1] (b) Calculate the expected number of quadrats with both species present. [1] (c) The chi-squared value for the data is 10.0. The critical value at p = 0.05 with 1 degree of freedom is 3.84. Deduce what the result shows about the two species. [2] (d) Suggest why this result does not prove that the two species compete, and outline one way to test the hypothesis of competition. [2]
| Y present | Y absent | Total | |
|---|---|---|---|
| X present | 6 | 24 | 30 |
| X absent | 18 | 12 | 30 |
| Total | 24 | 36 | 60 |
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- (a) there is no association between the distributions of species X and Y / the two species are distributed independently [1]
- (b) 30 × 24 ÷ 60 = 12 [1]
- (c) 10.0 > 3.84, so the null hypothesis is rejected / the association is significant [1]
- (c) negative association: the species occur together less often than expected (6 observed, 12 expected) [1]
- (d) association/correlation does not show cause / the species may need different conditions (e.g. light, moisture, soil pH) OWTTE [1]
- (d) remove one species from some plots and compare growth/abundance of the other with control plots / grow them alone and together in the laboratory [1]
Question 6
Paper 2A style
Outline how both partners benefit in the mutualistic relationship between plants of the family Fabaceae and the bacteria in their root nodules.
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- bacteria / Rhizobium fix nitrogen (gas) / convert N₂ to ammonia/ammonium [1]
- plant uses fixed nitrogen to make amino acids/proteins/nucleic acids / can grow in nitrogen-poor soil [1]
- bacteria receive sugars/carbon compounds/products of photosynthesis from the plant [1]
- bacteria gain a protected / low-oxygen environment in the nodule [1]
- max 3
Question 7
Paper 2A style
Duckweed plants were grown in a beaker of pond water and the number of fronds counted every three days. The numbers rose slowly, then very rapidly, then levelled off. (a) Explain the reasons for the rapid increase in the early part of the experiment. [2] (b) Suggest two factors that caused the numbers to level off. [2]
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- (a) resources (light, mineral ions, space) are plentiful / little competition [1]
- (a) birth/reproduction rate greatly exceeds death rate [1]
- (a) no/few pathogens/pests/consumers in the beaker [1]
- (b) competition for light as the surface becomes covered / shading [1]
- (b) mineral ions (e.g. nitrate/phosphate) in the water used up [1]
- (b) carrying capacity of the beaker reached / space on the water surface limited [1]
- max 4
Question 8
Paper 2B style
Explain how density-dependent factors control the size of animal populations, using a predator–prey relationship as an example.
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- density-dependent factors have a greater effect as population density increases [1]
- they act by negative feedback / push the population back towards the carrying capacity [1]
- competition for limited resources (food, space) increases in dense populations [1]
- pathogens / pests / parasites are transferred more easily in dense populations [1]
- dense prey populations are easier for predators to find/catch [1]
- named real example, e.g. Canada lynx and snowshoe hare [1]
- more prey → more predators survive and breed → predator numbers rise [1]
- increased predation reduces prey numbers [1]
- fewer prey → predators starve / breed less → predator numbers fall [1]
- prey numbers then recover / cycles repeat with predator peaks lagging behind prey peaks [1]
- this is top-down control; bottom-up control by food supply may also act [1]
- density-independent factors (e.g. weather) cause fluctuations but do not regulate [1]
- max 7