Biology

1. 1.

   Explain why random quadrat placement is preferable to choosing visually species-rich patches when estimating mean plant density.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The population of interest is the whole defined habitat, not its most attractive patches. Randomisation limits observer choice, while sufficient replication addresses spatial variability.

   Marking points

   • Visual selection biases sampling towards conspicuous/rich areas.
   • Random coordinates give eligible locations a defined chance of sampling.
   • The resulting mean better represents the defined habitat when sampling is adequate.

   Examiner tip: Random does not guarantee representativeness with only one quadrat.

2. 2.

   Explain why decomposers matter for nutrient cycling but do not recycle energy back to producers.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Atoms can pass repeatedly through organisms and the environment. Energy flow requires new input, usually sunlight, because metabolic transfers degrade useful energy into dispersed heat.

   Marking points

   • They break down organic remains and waste.
   • Mineral nutrients are released and can be taken up by producers.
   • Energy is dissipated as heat through respiration rather than recycled as usable input energy.

   Examiner tip: Matter cycles; usable energy flows through the ecosystem.

3. 3.

   Five 0.50 m^2 quadrats contain 4, 6, 5, 3 and 7 plants. Estimate mean density per m^2 and total plants in a uniform 200 m^2 habitat.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Sum counts to 25 and divide by five. Each quadrat covers half a square metre, so divide five by 0.50 before multiplying density by the habitat area.

   Marking points

   • Mean count per quadrat = 5.
   • Density = 10 plants m^-2.
   • Estimated total = 2000 plants.

   Examiner tip: Extrapolation assumes the sampled density represents the whole stated habitat.

4. 4.

   Producers have net production 12000 kJ m^-2 year^-1 and herbivores 900 kJ m^-2 year^-1. Calculate transfer efficiency and explain two routes accounting for energy not entering herbivore production.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Compare production values at successive trophic levels in matching units. The missing fraction is not simply destroyed; it enters other biological pathways or becomes heat rather than new herbivore biomass.

   Marking points

   • Efficiency = 900/12000 x 100 = 7.5%.
   • Some producer biomass is not eaten or is egested without assimilation.
   • Assimilated energy used in herbivore respiration is dissipated as heat.

   Examiner tip: Use net production in the denominator specified, not an invented gross-production value.

5. 5.

   A mark-release-recapture study captures 40 animals, marks and releases them, then captures 50 animals including 10 marked. Estimate population size and explain three assumptions whose violation could invalidate the estimate.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The recapture fraction 10/50 estimates the marked fraction 40/N. This proportional reasoning fails if marks disappear, capture is biased or migration/births/deaths change the population between sampling events.

   Marking points

   • Estimate N = 40 x 50/10 = 200.
   • Marked animals must mix and have comparable capture probability to unmarked animals.
   • Marks must persist without substantially affecting survival or behaviour.
   • The population should be approximately closed between samples.

   Examiner tip: This simple estimate is uncertain; 200 is not an exact census.

6. 6.

   River sites with high nitrate also have low dissolved oxygen. Evaluate a eutrophication explanation and design evidence that distinguishes it from a temperature confound.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The correlation fits a mechanism but does not isolate it. Controlling temperature and tracing the predicted intermediate changes tests the nutrient-to-algae-to-decomposition pathway rather than only its endpoints.

   Marking points

   • Nutrients can promote algal growth, followed by increased microbial decomposition.
   • Respiring decomposers consume oxygen.
   • Higher temperature can independently lower oxygen solubility and alter respiration.
   • Measure/match temperature and use replicated controlled nutrient additions with algae/respiration measurements.

   Examiner tip: Algae also photosynthesise; oxygen varies with light and time of day, which must be standardised.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.