Biology • Year 11 • Module 4 • Lesson 10
Ecological Sampling: Quadrats, Transects and Mark-Recapture
Apply sampling methods to real data sets — calculate population estimates, interpret distribution patterns, and identify sources of error in ecological sampling designs.
1. Interpret quadrat data — kangaroo grass survey
A student surveyed a 4 000 m² grassland using ten 1 m × 1 m quadrats placed randomly. The counts of Themeda triandra (kangaroo grass) were recorded below. 8 marks
| Quadrat | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Count | 11 | 6 | 18 | 8 | 14 | 5 | 20 | 9 | 13 | 6 |
1.1 Calculate the mean density of kangaroo grass per m². Show your working. 2 marks
1.2 Estimate the total kangaroo grass population in the 4 000 m² grassland. Show your working. 1 mark
1.3 The student notices that quadrats 3 and 7 (both placed near a creek) had the highest counts. Explain how this distribution pattern could affect the reliability of the population estimate, and suggest one improvement to the sampling design. 3 marks
1.4 Suggest one other source of error in this design (not the creek clustering). Explain how it would affect the estimate. 2 marks
2. Interpret transect data — NSW rocky shore
An ecologist conducted a belt transect from the high-tide mark to the low-tide mark on an NSW rocky shore. The percentage cover of three species was recorded at four zones. 7 marks
| Zone (distance from high tide) | Limpets (% cover) | Barnacles (% cover) | Mussels (% cover) |
|---|---|---|---|
| 0–2 m (high zone) | 38 | 5 | 0 |
| 2–5 m (upper-mid zone) | 18 | 54 | 2 |
| 5–10 m (lower-mid zone) | 4 | 42 | 22 |
| 10–15 m (low zone) | 1 | 12 | 61 |
2.1 Describe the distribution pattern shown by limpets across the transect. 2 marks
2.2 Identify the zone where barnacles are most abundant. Suggest one abiotic factor that explains why barnacle cover is highest in this zone rather than in the high zone. 2 marks
2.3 Explain why a line transect would be less useful than a belt transect for this investigation. 2 marks
2.4 The abiotic gradient driving the zonation in this data is desiccation (drying out). Predict what would happen to the distribution of mussels if sea levels rose by 0.5 m. 1 mark
3. Mark-recapture — eastern grey kangaroos
Researchers in a Victorian reserve captured and ear-tagged 38 eastern grey kangaroos. One week later they caught 52 kangaroos, of which 11 had ear tags. 8 marks
3.1 Identify M, C and R for this study. 1 mark
3.2 Calculate N using the Lincoln-Petersen formula. Show full working. 2 marks
3.3 Kangaroos give birth during spring. Explain how this could affect the validity of the population estimate if the study spanned a birthing season. 2 marks
3.4 The researchers used bright yellow ear tags. Suggest one reason why this choice of mark might reduce the accuracy of the estimate. 1 mark
3.5 State one other assumption of the Lincoln-Petersen method that this study relied on, and describe how it could be violated in this population. 2 marks
Q1.1 — Mean density
Total count = 11 + 6 + 18 + 8 + 14 + 5 + 20 + 9 + 13 + 6 = 110. Total quadrat area = 10 × 1 = 10 m². Mean density = 110 / 10 = 11 plants per m². Award 1 mark for correct total, 1 mark for correct density calculation.
Q1.2 — Population estimate
Population estimate = 11 × 4 000 = 44 000 plants. Accept any value based on a correct mean from 1.1 × 4 000.
Q1.3 — Clustering near creek
Kangaroo grass is non-randomly distributed — it is clumped near the creek where water and nutrients are higher. If the student randomly placed quadrats and happened to land two near the creek, this overweights the high-density zone in the mean, producing an overestimate of overall population density [1]. Conversely, if few quadrats land near the creek, the mean underestimates true density [1]. Improvement: use stratified random sampling — divide the grassland into zones (creek-adjacent and drier) and place quadrats randomly within each zone, then weight the estimates by zone area [1].
Q1.4 — Other error
Acceptable answers include: edge effects — plants on the quadrat boundary may be counted inconsistently, inflating or deflating the count; size mismatch — a 1 m × 1 m quadrat may be too large or too small for the plant spacing, increasing counting error; too few quadrats — with only 10 quadrats, natural variation is large and the mean is imprecise. Each answer needs the error name (1 mark) and direction of effect (1 mark).
Q2.1 — Limpet distribution
Limpets are most abundant in the high zone (38% cover) and decrease progressively towards the low zone (1% cover) [1]. Their distribution is strongly concentrated at the top of the intertidal, where they are exposed to air for the longest periods [1].
Q2.2 — Barnacles
Barnacles are most abundant in the upper-mid zone (2–5 m, 54% cover). One abiotic factor: desiccation tolerance — in the upper-mid zone, barnacles are submerged long enough to filter-feed during high tide but can also tolerate the exposure periods that exclude mussels and anemones. In the high zone, exposure time is too great for barnacles to filter-feed enough to sustain dense populations (accept: temperature extremes, wave action, or amount of submersion time with reasonable justification).
Q2.3 — Line vs belt transect
A line transect records only whether species touch the line — it produces presence/absence or relative abundance data, not actual cover values [1]. This investigation requires quantitative percentage-cover data at each zone to compare species abundances accurately; a belt transect provides this by surveying all individuals within a strip of defined width [1].
Q2.4 — Sea level rise prediction
If sea level rose by 0.5 m, the low zone would be submerged for longer. Mussels — which are already most abundant in the low zone — would likely extend their range upward into what is currently the lower-mid zone, as desiccation stress decreases. Accept also: mussel cover in the low zone may increase further as submersion time increases.
Q3.1 — Variables
M = 38 (tagged and released in first sample); C = 52 (total in second sample); R = 11 (marked individuals in second sample).
Q3.2 — Calculation
N = (M × C) / R = (38 × 52) / 11 = 1 976 / 11 ≈ 180 kangaroos (accept 179–180). Award 1 mark for correct substitution, 1 mark for correct answer.
Q3.3 — Birthing season
If joeys are born between the two captures, the population is no longer closed — new unmarked individuals enter the population [1]. This increases the total population (C includes unmarked joeys), but R remains the same proportion of tagged adults, so N is overestimated [1].
Q3.4 — Bright yellow tags
Bright yellow tags could attract predators or make tagged animals more visible to researchers during recapture (inflating R), or make tagged animals more conspicuous and stressed, altering behaviour (trap shyness or avoidance). Either could bias the recapture ratio and invalidate the estimate. Accept any one plausible reason with brief justification.
Q3.5 — Additional assumption
Acceptable assumptions: (a) Marks are retained — violated if ear tags fall out before the second sample, reducing R and inflating N. (b) Second sample is random — violated if traps are placed only near waterholes where kangaroos gather, overrepresenting those individuals. (c) Marks do not affect behaviour or survival — violated if tagged animals are conspicuous to predators, reducing their survival between captures. Any one assumption (1 mark) with a specific way it could be violated in this kangaroo population (1 mark).