In 2017, CSIRO scientists published an estimate: Australia is home to between 2.1 and 6.3 million feral cats. But cats hide. They are nocturnal, territorial, and avoid humans. No one counted every cat. Instead, scientists used cameras, statistical models, and mark-recapture methods to estimate what cannot be directly observed. This lesson teaches you the same techniques ecologists use to measure the invisible.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Before you read, commit to a prediction. You will revisit these at the end.
Q1. A national park covers 5,000 hectares. Rangers need to know how many brushtail possums live there. The possums are nocturnal, tree-dwelling, and hide in hollows during the day. Describe two methods you could use to estimate the population, and explain why you cannot simply walk through the park and count every individual.
Q2. Scientists place a 1 m×1 m quadrat randomly in a grassland and count 8 kangaroo grass plants inside it. There are 200 such quadrats that could fit across the entire field. Predict whether simply multiplying 8 × 200 gives an accurate population estimate. What could go wrong with this approach?
Counting every plant in a forest is impossible. Ecologists instead count organisms in small, representative areas called quadrats, then use those counts to estimate abundance across the whole habitat.
1. Define the habitat boundary. Mark the total area to be studied (e.g., a 2-hectare grassland).
2. Place quadrats randomly. Use a random number grid or random coordinate generator to eliminate observer bias. Never place quadrats where they look convenient.
3. Record data in each quadrat. Options include:
4. Calculate mean density.
Mean density = Total individuals counted / Total quadrat area sampled
5. Extrapolate to the whole habitat.
Population estimate = Mean density × Total habitat area
Worked example — Kangaroo grass in a paddock:
A researcher places ten 1 m×1 m quadrats randomly across a 5,000 m² paddock. The counts are: 7, 12, 9, 15, 8, 11, 6, 13, 10, 9.
Total individuals = 7 + 12 + 9 + 15 + 8 + 11 + 6 + 13 + 10 + 9 = 100
Total quadrat area = 10 × 1 = 10 m²
Mean density = 100 / 10 = 10 plants per m²
Population estimate = 10 × 5,000 = 50,000 plants
Quadrats tell you how many organisms are in an area. Transects tell you where they are — revealing how distribution changes along an environmental gradient such as tide height, altitude, or distance from a disturbance.
A straight line is laid across the habitat. The researcher records every species that touches the line, noting the distance along the line where contact occurs.
Best for: Quick assessment of presence/absence and relative abundance along a gradient. Shows zonation patterns clearly.
Limitation: Does not quantify actual density — a species that barely touches the line is recorded the same as one covering 50% of the area.
A strip of defined width (e.g., 1 m or 2 m) is surveyed on either side of the line. All individuals within the belt are counted or percentage cover is estimated.
Best for: Quantitative data — actual counts, density, and biomass estimates along the gradient. More time-consuming but scientifically more valuable.
Advantage: Produces data comparable to quadrat sampling, but arranged spatially.
Australian example — Rocky shore zonation:
On the New South Wales coast, ecologists lay belt transects from the high-tide mark to the low-tide mark. The pattern is remarkably consistent:
This zonation pattern is driven primarily by abiotic tolerance (desiccation, temperature, wave action) rather than biotic competition, though competition does sharpen the boundaries between zones.
For animals that move too fast to count in quadrats, ecologists use a clever statistical trick: catch some, mark them, release them, then see what fraction of marked animals appear in a second catch. The mathematics does the rest.
Step 1: Capture
Catch a sample of animals. Count and record: M (marked in first sample).
Step 2: Mark
Attach a non-harmful mark: ear tag, leg band, dye, microchip, or photographic ID. Release and allow mixing.
Step 3: Recapture
After mixing, capture a second sample. Count total caught: C. Count marked in this sample: R (recaptured).
Step 4: Calculate
Use the Lincoln-Petersen formula to estimate total population N.
N = (M × C) / R
Where N = total population, M = marked and released, C = total in second sample, R = marked recaptures
Worked example — Agile wallabies:
Researchers in Kakadu National Park capture, tag, and release 45 agile wallabies (M = 45). Two weeks later, they capture 60 wallabies (C = 60) and find that 15 carry tags (R = 15).
N = (45 × 60) / 15 = 2,700 / 15 = 180 wallabies
The estimate suggests approximately 180 agile wallabies in the study area.
The Lincoln-Petersen estimate is only reliable if these assumptions hold:
1. Closed population
No births, deaths, immigration, or emigration between the two sampling events. Violated if the study spans breeding season or migration.
2. Marks do not affect survival or behaviour
Marked animals must not be more likely to die, be preyed upon, or alter their behaviour (e.g., become trap-shy or trap-happy).
3. Marks are retained
Tags must not fall off, fade, or be overlooked during the second capture. Double-marking can reduce this risk.
4. Second sample is random
Every individual, marked or unmarked, must have an equal chance of being captured. Violated if traps are clustered or if marked animals learn to avoid them.
Every sampling method introduces uncertainty. Understanding these errors helps you design better studies and interpret published data critically.
Apply the sampling methods you have learned to real ecological data. Show your working for full marks.
Researchers studying eastern grey kangaroos in a Victorian reserve caught and ear-tagged 38 individuals. One week later, they caught 52 kangaroos, of which 11 had ear tags.
A student surveyed a 2,000 m² meadow using 1 m×1 m quadrats. The counts of white clover plants were:
| Quadrat | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Count | 14 | 9 | 22 | 7 | 16 | 11 | 19 | 10 |
Definition
Quadrat sampling: A method for estimating the abundance of sessile or slow-moving organisms by counting individuals within randomly placed squares of known area, then extrapolating to the whole habitat.
Definition
Mark-recapture (Lincoln-Petersen): A method for estimating mobile animal populations by marking a sample, releasing it, and using the proportion of marked individuals in a second sample to calculate total population: N = (M × C) / R.
Key assumption
Mark-recapture assumes a closed population, random mixing of marked and unmarked individuals, marks that do not affect survival or behaviour, and retention of marks until recapture.
Error checklist
Quadrat errors: observer bias, non-random placement, edge effects, size mismatch. Mark-recapture errors: open population, trap shyness/happiness, mark loss, heterogeneous capture probability.
Australian context
CSIRO scientists estimated 2.1–6.3 million feral cats in Australia using remote camera mark-recapture combined with spatial modelling — demonstrating how sampling methods scale from local studies to national estimates.
Syllabus link
ACSBL049, ACSBL050, ACSBL060: Explain why sampling is necessary; investigate procedures to measure distribution and abundance; analyse models to justify factors affecting distribution and abundance.
Now that you have completed the lesson, review your initial answers. What did you get right? What surprised you?
Q1. A national park covers 5,000 hectares. Rangers need to know how many brushtail possums live there. The possums are nocturnal, tree-dwelling, and hide in hollows during the day. Describe two methods you could use to estimate the population, and explain why you cannot simply walk through the park and count every individual.
Q2. Scientists place a 1 m×1 m quadrat randomly in a grassland and count 8 kangaroo grass plants inside it. There are 200 such quadrats that could fit across the entire field. Predict whether simply multiplying 8 × 200 gives an accurate population estimate. What could go wrong with this approach?
In this lesson you learned: