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Biology  ·  Year 11  ·  Module 4  ·  Lesson 10

HSC Exam Practice

Ecological Sampling: Quadrats, Transects and Mark-Recapture

7 questions / 3 sections / 26 marks total
Section 1

Short answer

1.Short answer

1.1

Explain why ecologists use sampling rather than counting every individual in a population.

2marks Band 3
1.2

Distinguish between a line transect and a belt transect. Include the type of data each produces.

3marks Band 3
1.3

State the Lincoln-Petersen formula and define each variable.

2marks Band 3
1.4

Explain why quadrat sampling is considered an inappropriate method for estimating the population size of a mobile species such as a butterfly.

2marks Band 3–4
1.5

Outline two assumptions that must hold for a mark-recapture study to produce a reliable population estimate.

2marks Band 3
Section 2

Data response

2.Data response — agile wallaby mark-recapture study

2.1

An ecologist studying agile wallabies in Kakadu National Park conducted three mark-recapture trials in consecutive months. The data are shown below.

0 10 20 30 40 50 Number of wallabies 45 45 60 15 Trial 1 45 60 9 Trial 2 45 60 25 Trial 3 M (marked & released) C (second sample total) R (marked recaptures) Trial
Figure 2.1. Mark-recapture data for three consecutive monthly trials on the same agile wallaby population. M and C were held constant across trials.

(a) Calculate the estimated wallaby population N for each trial. Show your working for Trial 1. (Trials 2 and 3: calculation only)

(b) The three estimates are different even though M and C were held constant. Identify the variable that changed and explain how its change affects N.

(c) Trial 2 was conducted one month after a flood that forced many wallabies to move out of the study area. Explain how this would affect the validity of the Trial 2 estimate, with reference to a specific assumption of mark-recapture.

8marks Band 4–5
Section 3

Extended response

3.Extended response

3.1

Describe how you would use quadrat sampling to estimate the population of Themeda triandra (kangaroo grass) in a 10-hectare paddock. In your answer, explain how random placement is achieved, how many quadrats to use and why, what measurements are taken, and how you would calculate the final population estimate. Identify one potential source of error and explain how it could be minimised.

7marks Band 4–6

Biology · Year 11 · Module 4 · Lesson 10

Answer Key & Marking Guidelines

1.1

Section 1 · Short answer · 2 marks · Band 3

Sample response. Most populations are too large, too widely distributed, or too mobile to count every individual directly. Sampling involves counting organisms in a representative subset of the habitat and using those data to estimate the total population, providing a statistically reliable estimate in a practical time frame.

Marking notes. 1 mark for identifying a reason why complete counting is impractical (too large / mobile / widely spread / time-prohibitive); 1 mark for explaining that sampling provides an estimate by extrapolating from representative subsets. Both required for full marks.

1.2

Section 1 · Short answer · 3 marks · Band 3

Sample response. A line transect involves laying a tape measure across the habitat and recording every species that touches the line, noting the position along the line where contact occurs. It produces semi-quantitative data — presence, absence, and relative abundance at different points along the gradient — but not actual counts or density. A belt transect surveys a strip of defined width (e.g. 1 m) on either side of the line, counting all individuals or estimating percentage cover within the strip. It produces quantitative data: actual counts, density, and biomass estimates at each section of the gradient.

Marking notes. 1 mark for correctly describing a line transect (touches the line only; presence/absence or relative abundance). 1 mark for correctly describing a belt transect (strip of defined width; all organisms within the strip counted or measured). 1 mark for identifying the key data difference: line transect = qualitative/semi-quantitative; belt transect = quantitative counts/density.

1.3

Section 1 · Short answer · 2 marks · Band 3

Sample response. N = (M × C) / R. N = estimated total population size; M = number of individuals marked and released in the first capture; C = total number of individuals captured in the second (recapture) sample; R = number of marked individuals found in the second (recapture) sample.

Marking notes. 1 mark for the correct formula written with variables; 1 mark for all four variables correctly defined. Deduct the variable-definition mark if any variable is defined incorrectly.

1.4

Section 1 · Short answer · 2 marks · Band 3–4

Sample response. Quadrat sampling requires organisms to remain within the frame long enough to be counted. Butterflies are highly mobile and fly continuously; they enter and leave the quadrat between the researcher setting it up and completing the count. As a result, a density figure derived from a quadrat would be meaningless — the number of butterflies touching the frame at any instant is not representative of their local population density. Mark-recapture is the appropriate method for mobile species like butterflies.

Marking notes. 1 mark for identifying that mobile organisms do not remain within the quadrat long enough to be counted reliably; 1 mark for explaining the consequence — density estimates from quadrats for mobile species are unreliable/meaningless.

1.5

Section 1 · Short answer · 2 marks · Band 3

Sample response. Any two from: (1) The population is closed — no births, deaths, immigration or emigration occur between the first and second captures. (2) Marks do not affect the survival or behaviour of the animal — tagged animals must not be more likely to die, be avoided by conspecifics, or become trap-shy or trap-happy. (3) Marks are retained until recapture — tags must not fall off or fade. (4) The second sample is random — every individual, marked or unmarked, has an equal chance of being captured in the second sample.

Marking notes. 1 mark per correctly stated assumption (max 2). The assumption must be stated clearly enough to distinguish it from the others. A vague answer such as "no animals die" without identifying it as the closed-population assumption scores 0.

2.1

Section 2 · Data response · 8 marks · Band 4–5

Sample response (a) — calculations.

Trial 1: N = (45 × 60) / 15 = 2 700 / 15 = 180 wallabies. [Full working required for marks: 1 mark substitution + 1 mark answer.]

Trial 2: N = (45 × 60) / 9 = 2 700 / 9 = 300 wallabies. [1 mark]

Trial 3: N = (45 × 60) / 25 = 2 700 / 25 = 108 wallabies. [1 mark]

Sample response (b). The variable that changed is R — the number of marked individuals recaptured in the second sample (15 in Trial 1, 9 in Trial 2, 25 in Trial 3). R appears in the denominator of the formula, so a smaller R produces a larger N (as in Trial 2, N = 300) and a larger R produces a smaller N (as in Trial 3, N = 108). R reflects the proportion of marked animals in the population: if few marked animals are recaptured, the formula infers that marked animals are rare relative to the total population, implying a larger total.

Marking notes (b). 1 mark for correctly identifying R as the variable that changed. 1 mark for explaining the inverse relationship: larger R → smaller N, smaller R → larger N (with reasoning or formula reference).

Sample response (c). The Trial 2 estimate (N = 300) is likely an overestimate, and the study's validity is compromised. The flood caused many wallabies to emigrate from the study area between the first and second captures, meaning the population was not closed — a critical assumption of the Lincoln-Petersen method. With emigration, the effective number of marked animals still in the study area is lower than M = 45 (some have left). Therefore, the true proportion of marked animals available for recapture is smaller than it appears, and R = 9 is lower not because the population is large but because many marked animals have left. The formula treats this low R as evidence of a large population, inflating N. In reality, the population in the study area at the time of Trial 2 was smaller than 300.

Marking notes (c). 1 mark for stating the direction of error (overestimate). 1 mark for identifying the closed-population assumption by name or clear description. 1 mark for explaining the mechanism: emigration removes marked animals from the study area, reducing R below its expected value, which inflates N.

3.1

Section 3 · Extended response · 7 marks · Band 4–6

Sample response. To estimate the kangaroo grass population in a 10-hectare paddock (= 100 000 m²) using quadrat sampling, I would proceed as follows.

Random placement: I would overlay a numbered grid on a map of the paddock and use a random number generator to select coordinate pairs for each quadrat location. This prevents observer bias — the researcher cannot subconsciously choose areas with more or less grass. [1 mark]

Number of quadrats: I would use at least 20–30 quadrats (each 1 m × 1 m). Fewer quadrats produce an imprecise mean because natural spatial variability in plant distribution is large; more quadrats reduce this variation and give a more reliable estimate. The exact number depends on the variability seen in pilot sampling. [1 mark]

Measurements: Within each quadrat, I would count the number of individual kangaroo grass plants (including tillers as separate plants, or rosettes if the species is more mat-forming — the counting rule must be defined and applied consistently). I would record the count for each quadrat. [1 mark]

Calculation: Mean density = total individuals counted across all quadrats / total quadrat area sampled (e.g. 25 quadrats × 1 m² = 25 m²). Population estimate = mean density × total paddock area = mean density × 100 000 m². [1 mark for formula; 1 mark for correct substitution with units]

Source of error and minimisation: Non-random distribution (clumping) — kangaroo grass may be denser near water points or shelter. If quadrats happen to land disproportionately in dense or sparse patches, the mean density will be biased. Minimisation: use stratified random sampling — divide the paddock into identifiable zones (e.g. near water vs away from water) and place quadrats randomly within each zone, then weight the zone estimates by area when calculating the total. This reduces the risk of missing or overweighting a particular density zone. [1 mark for named error + 1 mark for correct minimisation strategy]

Marking criteria.

  • 1 mark — Describes a method for achieving random placement (random number grid / coordinate generator) and states why randomness is required (prevents observer bias).
  • 1 mark — Justifies the number of quadrats in terms of reducing variability / improving precision of the mean; gives a reasonable minimum number (≥ 10).
  • 1 mark — Describes what is measured in each quadrat (count of individuals OR percentage cover) with a note about consistency of the counting rule.
  • 1 mark — States the mean density formula correctly: total count / total quadrat area.
  • 1 mark — States the population estimate formula correctly: mean density × total habitat area, including correct unit handling (m² consistently).
  • 1 mark — Identifies a specific, named source of error (observer bias, non-random plant distribution / clumping, edge effects, size mismatch).
  • 1 mark — Provides a specific, practical minimisation strategy matched to the error identified (e.g. stratified random sampling for clumping; consistent counting rule for edge effects).