AQA A-level Biology: Populations in Ecosystems and Succession (7402) – Practice Questions

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 3.7.4, Populations in ecosystems (A-level only), and required practical 12 in the AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015. Revise first from the study guide and revision notes. Course pages: AQA A-level Biology hub and printable checklist. For a starting point, try the free diagnostics.

Questions

1. State what is meant by a community, an ecosystem and a niche. [3]

2. A woodland of 2400 m² is sampled for bluebells using twenty quadrats, each 0.5 m by 0.5 m. The counts per quadrat were 2, 5, 0, 3, 4, 1, 6, 2, 0, 3, 5, 4, 1, 2, 7, 3, 0, 4, 2 and 3.

(a) Estimate the number of bluebells in the woodland. [3] (b) Calculate the frequency of bluebells. [1] (c) Describe how the positions of the quadrats should be chosen. [2]

3. A population of snails in a hedge bank is estimated by mark-release-recapture. 48 snails are marked with a dot of paint on the shell and released. The next day, 60 snails are caught, of which 8 are marked.

(a) Calculate the estimated population. [2] (b) State two assumptions made in this method. [2] (c) Thrushes find brightly painted snails easier to see. Explain the effect on the estimate. [2]

4. The table shows mean numbers of field voles (per hectare) and stoats (per km²).

Year	1	2	3	4	5	6	7	8	9
Voles	40	110	70	25	30	95	80	30	35
Stoats	3	5	9	6	3	4	8	7	3

(a) Give the years between the vole peaks, and the delay from a vole peak to the next stoat peak. [2] (b) Explain the pattern in the data. [3] (c) Define carrying capacity and suggest one abiotic factor that could lower it for voles. [2]

5. Grass species A and B were grown in pots of five plants. The table shows the mean dry mass at eight weeks of one plant, by its four neighbours.

Plant	Neighbours	Mean dry mass (g)
A	four A	3.6
A	four B	4.8
B	four B	3.4
B	four A	2.2

(a) Calculate the percentage decrease in the mass of B when its neighbours are A instead of B. [1] (b) Use the data to decide which is the stronger competitor. [2] (c) Explain why A grows less with four A neighbours than with four B neighbours. [2]

6. Explain how bare rock covered with lichens can come to support grasses and shrubs, and why the lichens then decline. [6]

7. A glacier retreated and the land was studied at different times afterwards.

Years since retreat	0	10	40	100	200
Soil depth (mm)	0	5	30	90	150
Plant species per 100 m²	2	6	14	22	18
Light reaching the ground (%)	100	95	70	25	8

(a) Calculate the percentage increase in soil depth between 40 and 100 years. [1] (b) Give two ways in which organisms increase soil depth. [2] (c) Explain the fall in species number between 100 and 200 years. [2] (d) State what mainly determines the type of climax community. [1]

8. On a chalk hillside reserve, sheep graze some plots and others are fenced. After six years, grazed plots had 14.2 plant species per m² and scrub covered 3%. Ungrazed plots had 8.6 species per m² and scrub covered 41%.

(a) Explain why scrub cover is higher in ungrazed plots. [2] (b) Calculate the percentage more species in grazed plots. [1] (c) The farmer wants to stop grazing because it loses money. Suggest a compromise that meets both aims. [2]

9. A field-margin scheme was studied for skylarks. Study 1, run by a conservation charity on six reserves for ten years, found mean pairs per reserve rose from 40 to 52. Study 2, run by a farmers’ union on two farms for two years, found pairs fell from 15 to 12.

(a) Calculate the percentage change in each study. [2] (b) Evaluate the evidence and suggest why the studies conflict. [4]

10. A student investigated the effect of light on a fern along a transect from open ground into woodland. At seven quadrats the light intensity, as a percentage of full daylight, was 8, 15, 22, 35, 48, 66 and 90. Fern frequency (%) at the same points was 80, 75, 60, 50, 30, 25 and 5.

(a) Describe how the student should carry out the investigation. [6] (b) The correlation coefficient is −0.98. The critical value for seven pairs at the 5% level is 0.754. Interpret this result. [3]

11. Yeast cells per cm³ in a flask of broth were counted: 2.5 × 10⁴ at 0 hours, 1.0 × 10⁵ at 2, 4.0 × 10⁵ at 4, 1.6 × 10⁶ at 6, 2.4 × 10⁶ at 8 and 2.5 × 10⁶ at 10 hours.

(a) Calculate the doubling time between 0 and 6 hours. [2] (b) Explain why a logarithmic scale suits these data. [1] (c) Explain the change after 6 hours. [2]

Answers

1. A community is the populations of different species in one place [1]. An ecosystem is a community and the non-living components of its environment [1]. A niche is how a species fits into its habitat, governed by its adaptation to abiotic and biotic conditions [1]. [3] Examiner insight: Define niche by role and adaptation, not by where an organism lives.

2. (a) Mean = 57 ÷ 20 = 2.85 per quadrat [1]. Density = 2.85 ÷ 0.25 = 11.4 per m² [1]. Estimate = 11.4 × 2400 = 27 360 (about 27 000) [1]. (b) 17 of 20 quadrats contain bluebells, so frequency = 85% [1]. (c) Lay tapes as axes [1] and use random numbers as coordinates [1]. [6] Examiner insight: Convert the quadrat area to m² before scaling up.

3. (a) N = (48 × 60) ÷ 8 [1] = 360 snails [1]. (b) Any two from: no migration, births or deaths between samples; marked snails mix randomly; marking does not affect survival or behaviour; the mark is not lost [1] [1]. (c) More marked snails are eaten, so fewer are recaptured [1]. m is too low, so the estimate is too high [1]. [6] Examiner insight: Name the number that changes (m) and say which way the estimate moves.

4. (a) Vole peaks in years 2 and 6, so 4 years apart [1]. Stoat peaks follow one year later [1]. (b) More voles give stoats more food, so stoats increase after a delay while they breed [1]. More stoats eat more voles, so voles fall [1]. Stoats then have less food and decline, so voles recover [1]. (c) The size of population an ecosystem can support [1]. For example, a severe frost or drought reduces the vegetation voles eat [1]. [7] Examiner insight: Describe the delay, then explain it.

5. (a) (3.4 − 2.2) ÷ 3.4 × 100 = 35.3% [1]. (b) A is the stronger competitor [1]. A is heavier with B neighbours (4.8 g) than with A neighbours (3.6 g), while B is lighter with A neighbours (2.2 g) than with B neighbours (3.4 g) [1]. (c) Neighbours of the same species have identical requirements for light, water and mineral ions [1]. Intraspecific competition is stronger, so each plant gets fewer resources [1]. [5] Examiner insight: Use figures from both pairs of rows. One pair alone does not show which species is stronger.

6. Lichens are pioneers that survive on bare rock with little water or nutrients [1]. They weather the rock slightly and, when they die, decay and add organic matter [1]. A thin soil forms that holds water and nutrients, so the environment is less hostile [1]. Grasses and shrubs, with different adaptations such as deeper roots, colonise [1]. They shade the lichens and out-compete them for light [1]. The environment is now less suitable for lichens, so they decline as biodiversity increases [1]. [6] Examiner insight: Name a change in the abiotic environment at each stage. A list of species does not explain replacement.

7. (a) (90 − 30) ÷ 30 × 100 = 200% [1]. (b) Dead organisms decay and add organic matter [1]. Roots and pioneers break rock into fragments [1]. (c) Light reaching the ground falls from 25% to 8% [1]. Species that need light are out-competed under the canopy [1]. (d) Climate [1]. [6] Examiner insight: In “explain” parts based on a table, quote a figure and give the biological reason.

8. (a) Without grazing, succession continues [1]. Woody plants grow because no grazers remove seedlings, and they shade out grassland species [1]. (b) (14.2 − 8.6) ÷ 8.6 × 100 = 65% [1]. (c) For example, a payment for grazing at a lower stocking density [1] that keeps scrub down while cutting the farmer’s losses [1]. [5] Examiner insight: A compromise must meet both aims. “Fence the reserve” ignores the farmer.

9. (a) Study 1: (52 − 40) ÷ 40 × 100 = +30% [1]. Study 2: (12 − 15) ÷ 15 × 100 = −20% [1]. (b) Study 1 has more sites and a longer period, so is likely to be more reliable [1]. Study 2 has only two farms and two years, so chance could explain the fall [1]. Both funders have an interest in the result [1]. Site, weather or how the scheme was run may also explain the difference, so neither proves the scheme caused the change [1]. [6] Examiner insight: Tie each criticism to the study: “two farms is too small to rule out a chance fall” is stronger than “more repeats”.

10. (a) Lay a transect from open ground into woodland [1]. Place same-size quadrats at regular intervals, for example every 5 m [1]. Record fern frequency or percentage cover in each [1]. Measure light intensity with a light meter at each quadrat at the same time of day [1]. Control or record other factors, such as soil moisture [1]. Repeat on a second transect, take means and carry out a risk assessment [1]. (b) Fern frequency falls as light rises, a strong negative correlation [1]. 0.98 is greater than 0.754, so the correlation is significant and unlikely to be chance [1]. It does not prove light is the cause, because soil moisture or temperature may also vary with shade [1]. [9] Examiner insight: Name the variable measured at each quadrat and say how other factors are controlled; “use quadrats” alone is too general.

11. (a) The ratio is 64 = 2⁶, so 6 doublings [1]. Doubling time = 6 ÷ 6 = 1 hour [1]. (b) The counts span several orders of magnitude, and a log scale shows steady exponential growth as a straight line [1]. (c) Growth slows and levels off as nutrients become limiting and waste products build up [1]. The population has reached the carrying capacity of the flask [1]. [5] Examiner insight: Show how you reached six doublings, so that working can be credited.

Where marks are usually lost

• Writing “niche” as “habitat”.
• Reporting the quadrat mean as the population without scaling to the area.
• Stating assumptions without saying how a failure changes the estimate.
• Confusing frequency with percentage cover.
• Listing succession species without the abiotic change.
• Giving generic evaluation points that ignore the study given.
• Treating a significant correlation as proof of cause.

Next steps

• Re-read the study guide and revision notes.
• Track progress on the printable checklist and the hub.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA, AS and A-level Biology specification (7401 and 7402), for first teaching 2015: section 3.7.4, Populations in ecosystems (A-level only), and required practical 12.