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OxfordAQA International AS and A-level Biology: Populations, communities and ecological succession (9610) – Practice Questions

Original questions with worked answers on niche, competition, predation, correlation, succession and seedling growth for OxfordAQA A-level Biology.

Subject
Biology
Level
A LEVEL
Topic
Populations, communities and ecological succession
Updated

Aligned to OxfordAQA A Level Biology (9610), Version 5.1, for International A-level exams May/June 2018 onwards. Official specification .

Syllabus page (what it covers and how it is assessed): OxfordAQA A Level Biology.

Syllabus points this page covers

9610 (A Level)

  • 3 Populations and genes (whole topic)

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These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.3.1 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, and required practical 9. Course pages: OxfordAQA A-level Biology hub and printable checklist.

Questions

1. Define each of the following terms: (a) population [1], (b) community [1], (c) ecosystem [1].

2. Two species of mayfly nymph live in the same stream. Species A lives on stones in fast, well-oxygenated water and scrapes algae from the surface. Species B lives in slow, silty water and feeds on dead leaves.

(a) Explain the difference between the habitat and the niche of species A. [2] (b) Name one abiotic and one biotic factor that form part of the niche of species A. [2]

3. Freshwater snails were kept in identical tanks at four starting densities. The mean eggs laid per snail each week were recorded.

Snails per tank 10 20 40 80
Eggs per snail per week 14 11 7 3

(a) Describe the trend shown. [1] (b) Explain the trend in terms of competition. [3] (c) Calculate the percentage decrease in eggs per snail between 10 and 80 snails per tank. [2]

4. Aphids and ladybird larvae were counted on the same plants each week.

Week 1 2 3 4 5 6 7 8
Aphids per plant 20 55 130 210 150 70 45 110
Larvae per plant 1 2 5 9 14 8 4 3

(a) Describe how the peak in larvae relates to the peak in aphids. [1] (b) Explain this relationship. [3] (c) Calculate the percentage change in aphids between weeks 4 and 6. [2] (d) Suggest why these data alone do not prove that the larvae caused the aphid decline. [1]

5. A survey of 12 moorland sites found heather beetles more abundant where soil pH was higher.

(a) Explain why this does not show that soil pH causes the difference in abundance. [2] (b) Describe an investigation that would test whether pH affects the beetles. [2]

6. Along a transect across a marsh, soil moisture (%) and the number of rush stems per m² were recorded in ten quadrats.

Quadrat 1 2 3 4 5 6 7 8 9 10
Soil moisture (%) 12 18 25 31 36 44 52 58 63 71
Rush stems per m² 3 5 4 9 12 10 18 20 16 27

Calculate the Spearman rank correlation coefficient using rₛ = 1 − 6Σd² / (n(n² − 1)). Give your answer to 2 significant figures. Using a critical value of 0.648 for n = 10 at p = 0.05, state and explain your conclusion. [6]

7. A student investigates the effect of competition on mustard seedling growth (required practical 9). Describe a method, including how growth is measured and how results are made reliable. [8]

8. Cress seedlings were grown in five replicate pots alone and in five replicate pots mixed with mustard. The mean dry mass per cress seedling in each pot (mg) was recorded.

Cress alone: 31, 35, 33, 30, 36. Cress with mustard: 27, 25, 29, 24, 28.

(a) The standard deviation for cress alone is 2.55 mg. Calculate the standard error and the 95% confidence limits (mean ± 2 × standard error). [3] (b) The mean for cress with mustard is 26.6 mg. Calculate the percentage decrease in mean dry mass compared with cress alone. [2] (c) The 95% limits for cress with mustard are 24.7 to 28.5 mg. State and explain what the data show about the effect of mustard on cress. [2]

9. Describe how primary succession proceeds on bare rock from the first colonisers to a climax community, explaining how the organisms at each stage affect the stages that follow. [6]

10. A lava flow was sampled at four sites of different ages.

Site age (years) 2 15 40 90
Plant species per 100 m² 2 9 17 13
Soil depth (cm) 0 1.5 8 20
Light at ground level (% of open) 100 98 50 10

(a) Calculate the percentage change in species number between 40 and 90 years. [2] (b) Explain the rise in species number up to 40 years. [2] (c) Suggest an explanation for the fall between 40 and 90 years. [2]

11. On a lowland heath reserve, plots were either cleared of tree seedlings each year or left alone. The mean number of heathland plant species per 10 m² was 12.0 in every plot at the start. After six years it was 15.0 in cleared plots and 8.4 in uncleared plots.

(a) Calculate the percentage change in each type of plot. [2] (b) Explain the difference in terms of succession. [3] (c) The local council wants to build houses on the reserve. Explain why the managers must weigh this against conservation. [2]

Answers

1. (a) Every organism of a single species living in one habitat [1]. (b) The populations of different species living in the same place [1]. (c) A community together with the non-living (abiotic) parts of its surroundings [1]. [3] Examiner insight: A definition of population that omits “one species” is a common way to lose the mark, because it then describes a community.

2. (a) The habitat is where species A lives, the stones in fast-flowing water [1]. The niche is its role, how it fits into the ecosystem, set by its adaptations to biotic and abiotic conditions, for example scraping algae [1]. (b) Abiotic: oxygen concentration or water flow speed (accept temperature) [1]. Biotic: algae as food, or competition from other grazers (accept predation) [1]. [4] Examiner insight: Give the niche as a role, and one factor of each type, as two different types are asked for.

3. (a) Eggs per snail decrease as snail density rises [1]. (b) At higher density there is more intraspecific competition [1] for limited resources such as food or space [1], so each snail gets less, has less energy for egg production, and fewer eggs are laid per snail [1]. (c) (14 − 3) / 14 × 100 [1] = 78.6% (78.57) [1]. [6] Examiner insight: Name the type of competition and the resource. Percentage change is divided by the starting value.

4. (a) The larvae peak (week 5) comes one week after the aphid peak (week 4) [1]. (b) More aphids give the larvae more food [1], so larvae survive and increase in number after a delay, because they take time to grow [1]. More larvae eat more aphids, so aphid numbers fall [1]. (c) (70 − 210) / 210 × 100 [1] = −66.7% [1]. (d) Aphids could also have been reduced by weather, disease, a lack of food (plant growth) or other predators [1]. [7] Examiner insight: Quote the peak weeks and explain the delay, rather than just stating it.

5. (a) The data show only that the two variables move together [1]. Another factor, such as altitude, vegetation or soil moisture, may differ between sites and affect the beetles [1]. (b) Keep beetles in containers of the same soil at different pH, controlling temperature, moisture and food [1]. Compare survival or numbers after a fixed time, with replicates [1]. [4] Examiner insight: “Correlation is not causation” alone is not an explanation. Name a possible third factor.

6. Ranks of moisture: 1 to 10 in order [1]. Ranks of rushes: 1, 3, 2, 4, 6, 5, 8, 9, 7, 10 [1]. d: 0, −1, 1, 0, −1, 1, −1, −1, 2, 0 [1]. Σd² = 0 + 1 + 1 + 0 + 1 + 1 + 1 + 1 + 4 + 0 = 10 [1]. rₛ = 1 − (6 × 10) / (10 × 99) = 1 − 60/990 = 0.94 [1]. 0.94 is larger than 0.648, so there is a significant positive correlation: rush numbers rise with soil moisture, and the null hypothesis of no correlation is rejected [1]. [6] Examiner insight: Show d and d² values, then compare with the critical value and state the direction.

7. Fill identical trays with the same compost and sow mustard seeds at different densities, for example 5, 10, 20 and 40 per tray [1]. Use the same tray area and sowing depth [1]. Keep light, temperature and watering the same [1]. Use at least three replicate trays per density [1]. Grow for a fixed time such as 14 days [1]. Harvest the shoots and dry them until the mass stops changing [1]. Weigh them and divide total dry mass by the number of seedlings [1]. Calculate the mean per density and compare, or plot mean mass against density [1]. [8] Examiner insight: Specific controls and a named measurement are stronger than “keep it fair”.

8. (a) Mean = 33 mg (165/5). Standard error = 2.55 / √5 [1] = 1.14 mg [1]. 95% limits = 33 ± 2 × 1.14 = 30.7 to 35.3 mg [1]. (b) (33 − 26.6) / 33 × 100 [1] = 19.4% [1]. (c) The limits (30.7 to 35.3 and 24.7 to 28.5) do not overlap [1], so the difference is unlikely to be chance: interspecific competition from mustard reduces cress growth [1]. [7] Examiner insight: Refer to the overlap of the limits and give the biological conclusion as well as the statistical one.

9. Pioneers such as lichens, mosses and cyanobacteria colonise bare rock [1]. They weather it and add organic matter when they die, so thin soil forms [1]. The soil holds more water and mineral ions, so conditions are easier [1]. Grasses and shrubs, adapted differently (for example with deeper roots), then colonise and out-compete the shaded pioneers [1]. At every stage the resident species alter conditions in favour of the next group and against themselves [1]. A relatively stable climax community, such as woodland, is reached [1]. [6] Examiner insight: Name the environmental change at each stage and link it to the next group. A list of organisms does not.

10. (a) (13 − 17) / 17 × 100 [1] = −23.5% [1]. (b) Soil deepens as rock weathers and organic matter builds up, so the environment is less hostile [1]. More species can survive and more niches are available [1]. (c) Light at ground level falls to 10% as taller plants form a canopy [1], so light-demanding species are out-competed and fewer species survive [1]. [6] Examiner insight: Use the figures in the table to support the explanation, for example the 10% light at 90 years.

11. (a) Cleared: (15.0 − 12.0) / 12.0 × 100 = +25% [1]. Uncleared: (8.4 − 12.0) / 12.0 × 100 = −30% [1]. (b) Without clearing, succession continues towards scrub and woodland [1]. Birch and pine shade and out-compete the light-demanding heathland plants [1]. Removing the seedlings keeps the community at the heathland stage, so more species survive [1]. (c) People need housing and the council gains income [1], while the habitat needs the land managed for conservation, so the conflict between human needs and conservation must be managed [1]. [7] Examiner insight: Name the stage the management holds the habitat at. “Clearing is good” does not explain the data.

Where marks are usually lost

  • Leaving “one species” or “non-living” out of a definition.
  • Describing a niche as an address, or not naming the resource in competition.
  • Using the end figure as the divisor, or ignoring food supply in a predator-prey lag.
  • Treating a correlation as proof.
  • Stating rₛ without the critical value.
  • Writing a method for seedlings without tray area, replicates or a measure of growth.
  • Treating succession as species being added, not environments changing.

Next steps

Official syllabus

OxfordAQA International AS and A-level Biology (9610), specification Version 5.1, for International A-level exams May/June 2018 onwards, published by OxfordAQA. Section 3.3.1 and required practical 9 in Unit 3.

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