Population Size

Lessons

Senior 3 · Biology

Population Size

5 lessons

Lesson 2 of 5

Estimating Animal Populations: Capture–Recapture

25 min

How many beetles live under the leaves?

Ground beetles hunt at night among the dead leaves of a school garden. By day they hide, and when you lift a leaf they run. A quadrat is no use: you cannot count animals that move, hide and run away. Yet ecologists regularly estimate the number of fish in a pond, rats in a store or birds in a wetland. The trick is to catch some animals, mark them, let them go, and then see how many marked animals turn up in a second catch.

Definition

Capture–recapture (mark–release–recapture): a method of estimating the size of a population of mobile animals by catching and marking a first sample, releasing it, and later counting the marked animals in a second sample.

Lincoln index (Lincoln–Petersen index): the formula that gives the estimate: $N = \frac{M \times C}{R}$ where $N$ = estimated population size, $M$ = number caught and marked in the first sample, $C$ = total number caught in the second sample, and $R$ = number of marked animals recaptured in the second sample.

Why the formula works

If the marked animals have mixed evenly through the population, the fraction of marked animals in the second sample should equal the fraction of marked animals in the whole population:

\frac{R}{C} = \frac{M}{N}

Multiply both sides by $N$ and by $C$ , then divide by $R$ :

N = \frac{M \times C}{R}

Figure 1: Capture–recapture with beetles. Here M = 8, C = 10 and R = 2, so N = (8 × 10) ÷ 2 = 40.

Carrying out the method

Catch a sample, for example with pitfall traps (jars sunk level with the soil) for beetles, nets for fish, or soft mist nets for birds.
Count and mark every animal in a way that does not harm it or make it easier for predators to see: a small dot of non-toxic paint under a beetle's wing case, a numbered ring on a bird's leg, a clipped corner of a fish's tail fin.
Release the animals where they were caught, and leave time for them to mix back into the population, usually a day to a week.
Catch a second sample in the same way and at the same places; count all the animals and the marked ones.
Calculate $N$ with the Lincoln index.

Worked example 1: ground beetles

Pitfall traps caught 60 ground beetles, which were marked and released. Two days later the traps caught 50 beetles, of which 12 were marked. (Illustrative data.) Estimate the population.

Identify: $M = 60$ , $C = 50$ , $R = 12$ .
Substitute: $N = \frac{M \times C}{R} = \frac{60 \times 50}{12} = \frac{3000}{12}$
Calculate: $N = 250$ beetles.

Sense check: 12 of the 50 in the second catch were marked, about a quarter; 60 marked beetles would be about a quarter of 250. ✓

Worked example 2: tilapia in a fish pond

A farmer netted 45 tilapia from her fish pond, clipped a small corner of each fish's tail fin, and returned them. A week later she netted 60 tilapia, of which 9 had clipped fins. (Illustrative data.) (a) Estimate the number of tilapia in the pond. (b) The pond has a surface area of 200 m². Calculate the density.

(a) $M = 45,\ C = 60,\ R = 9$ : $N = \frac{45 \times 60}{9} = \frac{2700}{9} = 300 \text{ tilapia}$
(b) $\text{density} = \frac{300}{200\ \text{m}^2} = 1.5$ tilapia per m² of pond surface.

The assumptions behind the method

The Lincoln index gives a fair estimate only if:

the marks do not rub off, and do not harm the animals or make them easier for predators to see;
the marked animals have had time to mix evenly with the unmarked ones;
marked and unmarked animals are equally easy to catch (animals do not learn to avoid, or to seek, the traps);
there are no births, deaths, immigration or emigration between the two samples, or they are very few.

Worked example 3: evaluating a result

In Worked example 2, suppose 3 of the clipped fins had grown back so fully that the farmer could not see them. (a) What was the true number of marked fish in the second catch? (b) Recalculate $N$ . (c) Was the first estimate too high or too low, and why?

(a) She counted 9, but 3 more were marked: $R = 9 + 3 = 12$ .
(b) $N = \frac{45 \times 60}{12} = \frac{2700}{12} = 225$ .
(c) 300 was too high. Losing marks makes $R$ too small, and a smaller number on the bottom of the fraction makes $N$ too large.

Rule to remember: anything that lowers $R$ (lost marks, marked animals dying or avoiding traps) overestimates the population; anything that raises $R$ (marked animals that are easier to catch) underestimates it.

Warning

Common mistakes. (1) Using the number of unmarked animals as $R$ : $R$ is the number of marked animals in the second catch. (2) Putting $C$ on the bottom of the fraction. (3) Marking with something that harms the animal or makes it easy to see. Handle animals gently, release them where they were caught, and never mark protected species without permission.

Sampling Animal Populations: Mark-release-recapture Technique (Cognito). Watch the worked calculation and the list of assumptions; the video uses the same formula with different letters.

Tip

Memory hook: "first times second over recaptured". $N = \frac{\text{first catch} \times \text{second catch}}{\text{marked in second catch}}$ . If your answer is smaller than either catch, you have made a mistake: the population can never be smaller than the number of animals you caught.

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Practice: capture–recapture

Why is capture–recapture used for animals instead of quadrats?
40 snails were marked and released. Later, 30 snails were caught, of which 8 were marked. Estimate the population.
In a wetland, 120 birds were ringed. Later, 150 birds were caught, of which 36 had rings. Estimate the population.
Rearrange the Lincoln index to make $R$ the subject.
A population is thought to be 500. If 50 animals are marked and 100 are caught in the second sample, how many marked animals would you expect to recapture?
State three assumptions of the capture–recapture method.
Beetles were marked with bright white paint on their backs, and many were eaten by birds before the second sample. Explain the effect on the estimate.
Explain why the second sample should not be taken ten minutes after the first.

Show answersHide answers

Animals move and hide, so they cannot be counted in a fixed frame; capture–recapture uses marked animals to estimate a population that cannot be seen all at once.
$N = \frac{40 \times 30}{8} = \frac{1200}{8} = 150$ snails.
$N = \frac{120 \times 150}{36} = \frac{18\,000}{36} = 500$ birds.
$R = \frac{M \times C}{N}$ .
$R = \frac{50 \times 100}{500} = 10$ marked animals.
For example: marks do not come off or harm the animals; marked animals mix evenly with the population; all animals are equally likely to be caught; no births, deaths or migration between samples.
Fewer marked beetles survive, so $R$ is too small and the population is overestimated. The bright paint broke the assumption that marks do not affect survival.
The marked animals would not have had time to mix with the rest of the population; many would still be near the traps, so $R$ would be too large and the population underestimated.

Key points

Lincoln index: $N = \frac{M \times C}{R}$ , based on $\frac{R}{C} = \frac{M}{N}$ .
The estimate is fair only if marks last and do no harm, animals mix, all are equally catchable, and the population does not change between samples.
Too few recaptures (a small $R$ ) gives an overestimate; too many gives an underestimate.

Sources and credits

19.1: Population Demographics and Dynamics, from Concepts of Biology by Samantha Fowler, Rebecca Roush and James Wise (OpenStax), via LibreTexts Biology, licensed under CC BY 4.0. Adapted by Ganzaa. (section on mark and recapture)
Photo: Aphelocoma coerulescens -Florida, USA -leg rings-8, Lauren, CC BY 2.0, via Wikimedia Commons.
Video: A-Level Biology - Sampling Animal Populations: Mark-release-recapture Technique, Cognito, embedded from YouTube.

Written by Ganzaa: Figure 1, the derivation as set out here, the worked examples (illustrative data) and the practice questions.

1How many beetles live under the leaves?

2Why the formula works

3Carrying out the method

4The assumptions behind the method

How many beetles live under the leaves?

Why the formula works

Carrying out the method

The assumptions behind the method