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A2 · Practice questions · Sampling and measuring biodiversity

Random sample, fair number.

Six original Cambridge-style questions on measuring biodiversity in the field, including Simpson's index and Lincoln index calculations.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Random, square, divide.

01
Recall
[2 marks]

Define the terms ecosystem and niche.

  • Ecosystem: a relatively self-contained, interacting community of organisms and the abiotic environment in which they live. ✓
  • Niche: the role of a species within an ecosystem, including how it gains energy and nutrients, where it lives and its interactions with other organisms and the environment. ✓
02
Explain
[3 marks]

Explain the three different levels at which the biodiversity of an area can be assessed.

  • Ecosystem / habitat diversity: the number and range of different ecosystems and habitats in the area. ✓
  • Species diversity: the number of different species (richness) and their relative abundance (evenness). ✓
  • Genetic diversity: the variation within each species, the number of different alleles of each gene in the population. ✓
03
Analysis
[4 marks]

Describe how you would use random sampling with frame quadrats to estimate the percentage cover of a moss on a lawn, and explain why sampling must be random.

  • Lay two tape measures at right angles along two edges of the lawn to form a grid. ✓
  • Use random numbers (calculator or table) to generate coordinates and place the quadrat there; repeat for many quadrats (for example 20, or until the running mean levels off). ✓
  • In each quadrat estimate the percentage of the area covered by moss (for example using a 100-square gridded quadrat); calculate the mean. ✓
  • Random placement avoids bias from the investigator choosing sites, so the sample is representative of the whole lawn. ✓
04
Calculation
[3 marks]

Random quadrats in a hedge bottom contain: species A 24, species B 12, species C 8, species D 6. Calculate Simpson's index of diversity, D = 1 − Σ(n/N)2, to two decimal places.

  • N = 24 + 12 + 8 + 6 = 50. ✓
  • (n/N)2: 0.482 = 0.2304; 0.242 = 0.0576; 0.162 = 0.0256; 0.122 = 0.0144; sum = 0.328. ✓
  • D = 1 − 0.328 = 0.67. ✓
05
Calculation
[3 marks]

In a pond, 45 water boatmen were caught, marked and released. Two days later 60 were caught, of which 9 were marked. Use the Lincoln index, N = (n1 × n2) ÷ m2, to estimate the population, and state two assumptions of the method.

  • N = (45 × 60) ÷ 9 = 2700 ÷ 9 = 300. ✓
  • Assumption: marks are not lost and do not affect survival or the chance of being recaptured. ✓
  • Assumption: no births, deaths, immigration or emigration between samples, and the marked animals mix randomly with the population. ✓
06
Explain
[2 marks]

Habitat X has D = 0.21 and habitat Y has D = 0.78. Explain what these values show about the two habitats.

  • Y has a high D, close to 1: high biodiversity, with many species and individuals spread evenly among them. ✓
  • X has a low D: low biodiversity, dominated by one or a few species, so it is likely to be less stable and more easily affected by a change in conditions. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.