The capture recapture formula (mark-release-recapture)

The capture recapture formula estimates the size of an animal population: N = (M × C) ÷ R, where M is the number of animals caught, marked and released, C is the number caught in a second sample, and R is how many of that second sample were already marked. It is often called the Lincoln index. For example, if 40 snails are marked, and a later sample of 50 contains 8 marked ones, the population is about 40 × 50 ÷ 8 = 250.

The mark-release-recapture equation

Some books use other letters, such as N = (n₁ × n₂) ÷ m₂, but the idea is identical: first sample × second sample ÷ marked in the second sample.

Where the formula comes from

The method rests on one idea: once the marked animals have mixed back in, the fraction of the second sample that is marked should match the fraction of the whole population that is marked.

If 8 out of 50 in the second sample are marked, that is 16%. So the 40 marked animals should be about 16% of the population, which makes the population 40 ÷ 0.16 = 250. Rearranging R ÷ C = M ÷ N for N gives N = MC ÷ R.

How mark-release-recapture works

1. Catch a sample of the animals, using a method suited to them, such as pitfall traps for beetles or nets for fish. Count them: this is M.
2. Mark each one in a way that does not harm it or make it easier for predators to spot, such as a small dot of paint on a snail's shell.
3. Release them where they were caught, and leave enough time for them to mix back into the population.
4. Catch a second sample in the same way. Count the total, C, and how many are marked, R.
5. Work out N = MC ÷ R.

AQA's A-level Biology specification says that population size can be estimated using quadrats for slow-moving or non-motile organisms, and the mark-release-recapture method for motile organisms. It also expects students to know the assumptions made when using the method.

Capture recapture worked examples

Example 1: snails in a garden

40 snails are marked and released. A week later, 50 snails are collected, and 8 of them are marked. Estimate the population.

Example 2: fish in a pond

120 fish are netted, tagged and returned. A few days later, 90 fish are netted, of which 15 are tagged.

The pond holds about 720 fish.

Example 3: working backwards

An ecologist expects a population of about 600 voles. They will catch 75 in the second sample and want about 10 of them to be marked. How many should they mark first?

Rearranging N = MC ÷ R gives M = NR ÷ C:

Example 4: population density

The 250 snails from Example 1 live in a garden of 400 m². Their density is 250 ÷ 400 = 0.625 snails per m². Dividing by the area lets you compare habitats of different sizes.

Here is a quick check before the assumptions.

Why a small R makes the estimate unreliable

The graph shows the estimate N = 2000 ÷ R, for M = 40 and C = 50, as the number recaptured changes. Here x is the number of marked animals recaptured and y is the estimated population.

When R is small, the curve is steep: recapturing 4 marked animals instead of 5 changes the estimate from 400 to 500. A single lucky or unlucky catch makes a big difference. Larger samples, which give a bigger R, make the estimate far more reliable.

Assumptions of mark-release-recapture

The fraction R ÷ C only matches M ÷ N if the marked animals behave exactly like the rest. For that to be fair:

• the population is closed: no births, deaths, immigration or emigration between the two samples;
• the marked animals have had time to mix fully back into the population;
• marks are not lost, rubbed off or washed away before the second sample;
• marking does not harm the animals or change their chance of survival;
• marked animals are no more or less likely to be caught the second time.

OpenStax's Biology textbook gives real examples of the last two going wrong. Some animals learn to avoid traps after being caught, which inflates the estimate. Others come back for more if a trap has food in it, which leads to an underestimate. And some marking techniques harm the animals, reducing their survival.

Sources of error and which way they push the estimate

The pattern to remember: anything that makes R smaller makes N bigger, because R is on the bottom of the fraction.

A refinement: the Chapman estimate

The Lincoln index cannot be used if no marked animals are recaptured, because it would mean dividing by zero. Ecologists often use Chapman's version instead:

For Example 1 this gives (41 × 51) ÷ 9 − 1 = 231, a little lower than 250. The Ecology Calculator works out both the Lincoln index and the Chapman estimate, with a range for the Chapman estimate, and shows the working.

Try it

Ecology Calculator

Estimate populations by capture recapture and Simpson's diversity index

Open the calculatorGo to its page

Common mistakes

• Using the wrong number on the bottom. Divide by R, the marked animals in the second sample, not by C.
• Adding the samples together. N is M × C ÷ R, not M + C.
• Giving a decimal number of animals. Round to a whole number, and treat it as an estimate.
• Saying an error makes the estimate "inaccurate" without a direction. Say whether it would be too high or too low, and why.
• Forgetting the time gap. Too short and the animals have not mixed; too long and births, deaths and migration break the closed population assumption.

Practise ecology in a short timed paper

A-level Biology: Year 2 and University Biology: Year 1 each give you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so population sampling turns up regularly but not in every paper.

Read next

Biology · Ecology and evolutionSimpson's diversity index calculator and mark-recaptureFree ecology calculator: find Simpson's diversity index from species counts, or estimate a population by mark-release-recapture with a 95% interval.