AQA GCSE Biology 8461: Ecology – 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 Topic 7, Ecology (sections 4.7.1 to 4.7.5), of the AQA GCSE Biology (8461) specification, for first teaching 2016 with exams from 2018 (version 1.0). Ecology is examined on Paper 2 at Foundation and Higher tier every May/June. Question 8 tests section 4.7.2.4 and is labelled “Higher tier only”; everything else is for both tiers.

Learn the content first in the study guide and the revision notes. The course hub is AQA GCSE Biology and the printable checklist lists every statement.

Questions

1.

(a) State what is meant by biodiversity. [1] (b) Explain why a community with high biodiversity is more stable. [2]

2. Bluebells grow on the floor of an oak wood. They flower in spring, before the oak leaves open.

(a) Suggest two factors that the bluebells and oak trees compete for. [2] (b) Jays eat acorns and bury some, which later grow into oak trees. Explain how this shows interdependence. [2]

3. A student estimates the number of buttercups in a meadow 30 m × 20 m. She places twelve 0.25 m² quadrats and counts: 3, 6, 0, 4, 2, 5, 1, 3, 4, 2, 3, 3.

(a) Calculate the mean number per quadrat and estimate the buttercup population of the meadow. [3] (b) State the median. [1] (c) Explain why the quadrats must be placed at random. [1] (d) Describe how she could investigate whether buttercup distribution changes with distance from a hedge. [3]

4. The dissolved oxygen in a stream is 9.0 mg/dm³ above a sewage outlet and 3.6 mg/dm³ below it.

(a) Calculate the percentage decrease in dissolved oxygen. [2] (b) Explain why fewer mayfly nymphs (aquatic insects) are found below the outlet. [2] (c) Suggest one biotic factor that could also reduce mayfly numbers. [1]

5. A scientist counts voles and owls in a forest.

Year	1	2	3	4	5	6	7	8
Voles	200	450	700	400	150	300	650	420
Owls	10	12	20	26	18	11	14	22

(a) Compare the years in which each population peaks. [1] (b) Explain the pattern in the two populations. [3]

6. Explain the role of microorganisms in cycling materials through an ecosystem. [4]

7. A student measures the pH of fresh milk (starting pH 6.8) after 60 minutes at three temperatures.

Temperature (°C)	15	25	35
pH after 60 min	6.5	6.0	5.3

(a) Calculate the rate of pH change at 35 °C. Give the unit. [2] (b) Explain the trend in the results. [2] (c) Predict and explain the result at 70 °C. [2] (d) Give one variable the student should keep the same. [1]

8. (Higher tier only) Over 30 years, the northern edge of a butterfly species’ range moved 90 km north. Mean summer temperature in the region rose by 0.8 °C. Evaluate the claim that the temperature rise caused the change in distribution. [4]

9. In a pond, producers have a biomass of 2000 g/m², primary consumers 160 g/m² and secondary consumers 14 g/m².

(a) Calculate the efficiency of biomass transfer from producers to primary consumers. [2] (b) Give this efficiency as a fraction in its simplest form. [1] (c) Explain two reasons why biomass is lost between primary and secondary consumers. [2] (d) Explain why there are fewer secondary consumers than primary consumers. [2]

10. Many gardeners buy compost made from peat. Evaluate the use of peat compost, including its effect on biodiversity and the atmosphere. [6]

11. The stock of a fish species in a sea area fell from 120 thousand tonnes to 84 thousand tonnes.

(a) Calculate the percentage decrease. [2] (b) Explain how two measures could return the stock to a sustainable level. [4]

12.

(a) Describe how mycoprotein is produced. [3] (b) Explain why keeping chickens in small, warm sheds increases the efficiency of food production. [2] (c) Give one reason why some people object to this method. [1]

Answers

1. (a) The variety of all the different species in an ecosystem (or on Earth) [1] (b) Species depend less on any one other species for food or shelter [1], so losing one species has less effect on the rest of the community [1]. Examiner insight: “Lots of animals” is not biodiversity – the mark needs the idea of different species.

2. (a) Any two of: light [1], water [1], mineral ions, space. (b) Jays depend on oaks for food [1]; oaks depend on jays for seed dispersal [1]. Examiner insight: Two marks need dependence in both directions, each named.

3. (a) Mean = 36 ÷ 12 = 3.0 per quadrat [1]; 3.0 ÷ 0.25 = 12 per m² [1]; 12 × 600 = 7200 buttercups [1] (b) 3 [1] (c) To avoid bias, so the sample represents the whole meadow [1]. (d) Lay a tape (transect) from the hedge into the meadow [1]; place quadrats at regular intervals along it [1]; count buttercups in each and repeat with more transects [1]. Examiner insight: An area calculation earns method marks even if the mean is wrong – allow error carried forward, so show each step.

4. (a) 9.0 − 3.6 = 5.4 [1]; 5.4 ÷ 9.0 × 100 = 60 % [1] (b) Less oxygen for aerobic respiration [1], so fewer nymphs can survive there [1]. (c) New predators, new pathogens, less food, or competition from another species [1]. Examiner insight: Percentage change divides by the original value; dividing by 3.6 loses the accuracy mark.

5. (a) Voles peak in years 3 and 7; owls peak one year later, in years 4 and 8 [1]. (b) More voles means more food, so more owls survive and breed [1]; more owls eat more voles, so vole numbers fall [1]; with less food, owl numbers then fall [1]. Examiner insight: Each mark is a separate cause-and-effect link; describing the numbers without “because” scores nothing in (b).

6. Microorganisms decompose dead organisms and waste [1] using enzymes [1]. They respire, returning carbon to the atmosphere as carbon dioxide [1], and return mineral ions to the soil [1]. Examiner insight: Both returns are separate marks – carbon dioxide to the air and mineral ions to the soil.

7. (a) (6.8 − 5.3) ÷ 60 [1] = 0.025 pH units per minute [1] (b) Higher temperature, faster fall in pH [1], because bacterial enzymes work faster, so acid is produced faster [1]. (c) pH would stay near 6.8 (little or no fall) [1]; enzymes denatured or bacteria killed [1]. (d) Volume of milk; type of milk; time; starting pH [1]. Examiner insight: The unit mark needs both parts: “pH units” and “per minute”.

8. Supports: the species moved north as temperature rose, so there is a correlation [1]; moving north is consistent with a species tracking cooler conditions [1]. Against: correlation does not prove cause – other changes, such as land use or water availability, could explain it [1]. Conclusion: only one species and one region, so the evidence supports but does not prove the claim [1]. Examiner insight: “Evaluate” needs points on both sides and a judgement; one-sided answers are capped.

9. (a) 160 ÷ 2000 × 100 [1] = 8 % [1] (b) 2/25 [1] (c) Some material is not absorbed and is egested as faeces [1]; some is lost as carbon dioxide and water from respiration, or as urea in urine [1]. (d) Only a small fraction of biomass (about 10 %) passes up [1], so less biomass is available to support organisms at the higher level [1]. Examiner insight: A fraction must be simplified when asked; 160/2000 alone loses the mark.

10. Benefits: peat compost is cheap and available [1] and helps increase food and plant production [1]. Drawbacks: extracting peat destroys the bog habitat [1], reducing the variety of plant, animal and microorganism species [1]; decay or burning of peat releases carbon dioxide [1], contributing to global warming. Judgement with reason, e.g. peat-free compost should be used because the losses outweigh the saving [1]. Examiner insight: A 6-mark evaluation is marked on levels; you need both sides and a reasoned conclusion for the top level.

11. (a) 120 − 84 = 36 [1]; 36 ÷ 120 × 100 = 30 % [1] (b) Fishing quotas limit the number or mass of fish caught [1], so enough fish remain to breed [1]. Larger net mesh lets young fish escape [1], so they survive to reproduce [1]. Examiner insight: Each measure needs the “what” and the “why it works” for two marks.

12. (a) The fungus Fusarium is grown on glucose syrup [1] in aerobic conditions [1]; the biomass is harvested and purified [1]. (b) Less movement and a warm environment mean less energy is transferred to the surroundings [1], so more biomass is used for growth [1]. (c) Ethical objection, e.g. animals cannot move freely or show natural behaviour [1]. Examiner insight: Name Fusarium as a fungus; “bacteria” loses the first mark.

Where marks are usually lost

• Not scaling from quadrat area to m² and then to the whole area.
• Dividing a percentage change by the new value instead of the original.
• Missing units on rates (pH units per minute, g per week).
• Describing predator–prey data without explaining cause and effect.
• Writing that decay always speeds up with temperature, ignoring denaturing.
• Naming only respiration as a biomass loss; faeces and urine count too.
• One-sided “evaluate” answers with no conclusion.
• Unsimplified fractions when the question asks for simplest form.

Next steps

• Ecology revision notes
• Ecology study guide
• Key ideas practice questions
• AQA GCSE Biology course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Biology (8461) specification, for first teaching 2016, exams from 2018, version 1.0, published by AQA – section 4.7 Ecology.