RP12 Effect of an environmental factor on species distribution

Free AQA A-level Biology RP12 questions on Effect of an environmental factor on species distribution, including practical reasoning, variables, data handling, clues and explanations.

Specification route
RP12
Question bank
12 questions
Course stage
Year 13 / A-level only

Sample questions

QUESTION 1 · RP12 · LEVEL 2

A student uses a randomly placed quadrat to estimate the percentage cover of moss in a field. What is the main advantage of measuring 'percentage cover' instead of counting individual plants?

  • It is much faster and more accurate for plant species that grow in dense clumps where individual organisms are impossible to distinguish.
  • Percentage cover is the only data type that can be used in a Student's t-test.
  • It physically prevents investigator bias during sampling.
  • It automatically calculates the species richness of the entire field without further math.
Show clue

How do you count one single piece of grass in a lawn?

Show answer and explanation

Answer: It is much faster and more accurate for plant species that grow in dense clumps where individual organisms are impossible to distinguish.

Required practical 12. For species like moss or grass that lack distinct individual boundaries, counting frequency is impossible. Percentage cover is an efficient alternative to estimate abundance.

QUESTION 2 · RP12 · LEVEL 2

A student is using an interrupted belt transect to survey a large sand dune system. What does 'interrupted' mean in this context?

  • Quadrats are placed at regular, set intervals (e.g., every $5$ metres) along the line, rather than continuously end-to-end.
  • The transect line is physically cut by environmental barriers like streams or fences.
  • The sampling is stopped halfway through and resumed the following day.
  • Quadrats are thrown randomly along the length of the tape measure.
Show clue

Doing it continuously over a $500m$ dune would take weeks.

Show answer and explanation

Answer: Quadrats are placed at regular, set intervals (e.g., every $5$ metres) along the line, rather than continuously end-to-end.

Required practical 12. An interrupted belt transect involves taking samples at defined intervals along the transect line. It is highly efficient for capturing gradual environmental changes over very long distances without sampling every single square meter.

QUESTION 3 · RP12 · LEVEL 2

When estimating the population of daisies in a uniform, flat field, why is 'random sampling' the most appropriate method?

  • Because the environment is uniform, random sampling removes investigator bias, ensuring every part of the field has an equal chance of being selected, providing a representative estimate.
  • Because daisies grow in distinct environmental gradients that require random analysis.
  • Because random sampling actively forces the daisies to grow in a perfectly even distribution.
  • Because random sampling is the only method that allows the use of a point quadrat.
Show clue

If the whole field is basically the same, you just need a fair, unbiased snapshot of the whole area.

Show answer and explanation

Answer: Because the environment is uniform, random sampling removes investigator bias, ensuring every part of the field has an equal chance of being selected, providing a representative estimate.

Required practical 12. Random sampling is used in uniform habitats. It eliminates human bias (e.g., throwing a quadrat towards the most flowers), ensuring the calculated mean is statistically valid for the whole area.

QUESTION 4 · RP12 · LEVEL 1

How should a student properly generate coordinates for random quadrat placement?

  • By laying out two tape measures at right angles to form a grid, and using a random number generator or table to select the $x$ and $y$ coordinates.
  • By spinning around with eyes closed and throwing the quadrat over their shoulder.
  • By walking strictly in a straight line and placing a quadrat exactly every two paces.
  • By dividing the field into equal squares and choosing the centre of each square.
Show clue

Throwing objects is not mathematically random.

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Answer: By laying out two tape measures at right angles to form a grid, and using a random number generator or table to select the $x$ and $y$ coordinates.

Required practical 12 / AT k. True randomness requires a grid and a mathematical generation of coordinates to completely remove subconscious human bias.

QUESTION 5 · RP12 · LEVEL 2

In a woodland, a student wants to see how the abundance of bluebells changes from the dark centre of the woods out into the bright open field. What is the correct sampling technique?

  • A belt transect (continuous or interrupted), systematically placing quadrats along a tape measure extending from the woods to the field.
  • Random sampling using randomly generated coordinates across both the woods and the field.
  • The mark-release-recapture method using non-toxic paint on the bluebells.
  • Using a sweep net to randomly collect all bluebells within a $50m$ radius.
Show clue

When investigating a change over a distance (an environmental gradient), you use a line.

Show answer and explanation

Answer: A belt transect (continuous or interrupted), systematically placing quadrats along a tape measure extending from the woods to the field.

Required practical 12 / AT k. Systematic sampling (transects) is used when investigating how species distribution correlates with a changing abiotic factor (like light intensity moving out of a forest).

QUESTION 6 · RP12 · LEVEL 3

When deciding how many quadrat samples to take in a large field, how does a student know they have taken 'enough' samples?

  • They calculate a running mean after each sample; when the running mean stabilises and stops fluctuating significantly, enough samples have been taken.
  • They stop when they have exactly $10\%$ of the field area covered.
  • They stop when they have found at least one of every single species in the field.
  • They must always take exactly $100$ samples, regardless of the field size.
Show clue

As the sample gets larger, extreme anomalies have less effect on the average.

Show answer and explanation

Answer: They calculate a running mean after each sample; when the running mean stabilises and stops fluctuating significantly, enough samples have been taken.

Required practical 12. A running mean tracks the average as new data is added. Once it plateaus, the sample size is large enough to be representative of the whole population, and further sampling will not change the mean significantly.

QUESTION 7 · RP12 · LEVEL 2

Why might a researcher choose to record the 'percentage cover' of a plant species in a quadrat rather than its 'frequency' (number of individuals)?

  • Percentage cover is much faster and more accurate for species that grow in dense, continuous mats (like grass or moss) where individual plants cannot be easily distinguished.
  • Percentage cover automatically calculates the biomass of the organisms.
  • Frequency counting is highly inaccurate for large, distinct organisms like oak trees.
  • Percentage cover is the only metric that can be used in a Student's t-test.
Show clue

Try counting exactly how many 'plants' of grass are in a $1m \times 1m$ patch of lawn.

Show answer and explanation

Answer: Percentage cover is much faster and more accurate for species that grow in dense, continuous mats (like grass or moss) where individual plants cannot be easily distinguished.

Required practical 12. Counting frequency requires distinct individuals. For creeping, mat-forming, or highly abundant species, estimating the percentage of the quadrat area they cover is much more practical.

QUESTION 8 · RP12 · LEVEL 4

A student uses a $0.5m \times 0.5m$ quadrat and takes $20$ random samples in a $500m^2$ field. They count a total of $100$ buttercups. What is the estimated total population in the field?

  • $10,000$ buttercups
  • $2,500$ buttercups
  • $50,000$ buttercups
  • $1,000$ buttercups
Show clue

Area of one quadrat = $0.25m^2$. Mean per quadrat = $100 / 20 = 5$. How many quadrats fit in the whole field? $500 / 0.25 = 2000$. Total = $2000 \times 5$.

Show answer and explanation

Answer: $10,000$ buttercups

MS 0.3. Area of quadrat = $0.25m^2$. Mean number per quadrat = $100 / 20 = 5$. Total area of field = $500m^2$. Number of quadrats that fit in field = $500 / 0.25 = 2000$. Total population = $2000 \times 5 = 10,000$.

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