Mark–Release–Recapture Virtual Lab: The Lincoln Index

Q: Can the class link open on the mark–release–recapture screen?

Yes. In the link's starting values, set Screen to Mark–release–recapture (animals) and pin First sample (M) and Second sample (n) at 100. Students still tick the Break an assumption boxes themselves.

Updated 2026-10-07

This mark–release–recapture virtual lab covers the capture–recapture method for animals that move, as taught in A-level, IB and NSW HSC Biology, and as a GCSE or IGCSE ecology extension. Students catch woodlice in pitfall traps, mark and release them, wait a day, take a second sample and apply the Lincoln index N = M × n ÷ m. The simulation knows the true population, so students can test the method: repeat trials, change the sample sizes, then break one assumption at a time and see which way the estimate moves. For plants, use the quadrat sampling virtual lab on the other tab of the same simulation.

Quadrat Sampling – Estimating Population Size

Curriculum links

AQA A-level Biology 7402, section 3.7.4 (populations in ecosystems): estimating the size of a population of motile organisms by mark–release–recapture, and the assumptions the method makes.
IB Biology (first assessment 2025), C4.1.4: capture–mark–release–recapture and the Lincoln index for motile organisms.
NSW HSC Biology, Module 4 (Ecosystem Dynamics): estimating populations in fieldwork.

Simulic is not affiliated with or endorsed by AQA, the International Baccalaureate or NESA.

Before the lab (5 min)

Ask students to commit to a prediction, on paper or as question 1 of the class link:

"Some woodlice molt between the two samples and lose their paint marks. Will the Lincoln index estimate be too high, too low or still about right?"

Many say "too low". Leave it open.

Method in the simulation

Click the Mark–release–recapture (animals) tab. Keep Animal on Woodlouse, True population at 300 and Time between samples at 1 day.
Run one trial by hand (M = n = 50): press Capture and mark, Release, Wait 1 day and Recapture and count. Calculate N from M, n and m.
Set First sample: M = 100 and Second sample: n = 100. Press Clear results, then Run 10 trials. Record the mean and range from the summary line.
Press Clear results, tick Marks lost (moulting, paint wears off) and press Run 10 trials. Record the mean, then untick the box.
Repeat step 4 for Births, deaths, migration, Release all under one log, and Marked animals are trap-shy (avoid traps), then trap-happy (seek traps). Break only one assumption at a time.

Assumption broken	Mean of 10 estimates	Range	Too high, too low or about right?
None
Marks lost
Births, deaths, migration
Release all under one log
Trap-shy
Trap-happy

a freshly opened simulation on the Mark–release–recapture (animals) tab with Woodlouse, True population 300, M = 100, n = 100 and Time between samples 1 day, after one press of Run 10 trials: the results table with ten rows and the graph of ten estimates scattered around the dashed green true-size line at 300

Expected results

Captures are random, so every run differs. These values come from the browser and from 5,000 simulated runs of 10 trials.

M = n = 50: m is about 8. Single estimates spread from 208 to 500 in 90% of trials; two browser runs of 10 gave means of 324 and 308.
M = n = 100: m is about 33. Browser means were 296, 297, 300 and 308; 98% of simulated runs fell between 279 and 334.
One assumption broken (M = n = 100), typical means:

Assumption broken	Typical mean	Estimate is
None	304	about right
Marks lost	414	too high
Births, deaths, migration	373	too high
Release all under one log	354	too high
Trap-shy	504	too high
Trap-happy	210	too low

With marks lost, browser means were 412, 430, 434 and 449: 37–50% too high.

Questions for students

(Prediction, asked again after the lab) If marked woodlice lose their marks, is the estimate too high, too low or about right?
When you compare "no assumption broken" with "marks lost", which quantities must stay the same?
With M = n = 100 and no assumption broken, what is the mean of 10 estimates?
With marks lost, by what percentage does the mean exceed the true population?
Explain why trap-shy marked animals make the estimate too high.

Answers for teachers: (1) Too high. (2) The animal, the true population, M, n and the time between samples. (3) Accept 260–350; the true size is 300. (4) Accept 15–70%; most runs give 35–50%. (5) Marked animals avoid the traps, so the second sample holds fewer marked animals than their true share. m/n is smaller than M/N, so N = M × n ÷ m comes out too large.

Common misconceptions

"More marked animals recaptured means a bigger population." It means a smaller one: N falls as m rises.
"One trial is enough." With M = n = 50, single estimates ran from about 200 to 500. Use the mean of repeats.
"Small samples are fine if you repeat them." With M = n = 20, about a quarter of trials recapture no marked animals, so N cannot be calculated.

Extension

Chapman's formula: with M = n = 50, small m pushes the Lincoln mean about 9% above 300 (326 in simulated runs). Recalculating each trial with N = (M + 1)(n + 1)/(m + 1) − 1 brings it to 297.
Design the sample: find the smallest M and n that keep all 10 estimates within 20% of the true size.

FAQ

Can the class link open on the mark–release–recapture screen?

Yes. In the link's starting values, set Screen to Mark–release–recapture (animals) and pin First sample (M) and Second sample (n) at 100. Students still tick the Break an assumption boxes themselves.

Why do my students get different answers?

Each capture is random, like real pitfall traps, so questions 3 and 4 accept ranges. Pool the class means and compare their spread with the true size.

Does it cover fieldwork ethics?

Yes. The note under the table gives the welfare rules: handle animals gently, use non-toxic paint, release them where caught and check traps at least daily.