Mark-Release-Recapture (Practical)
A quadrat works for a plant that stays put, but how do you count fish in a lake, or beetles in a paddock, when they move and hide? You cannot. Instead, ecologists use a clever trick: catch some, mark them, let them mix back in, then catch a second batch and see what fraction carries a mark. That single proportion unlocks the whole population. This practical models the method with beans or beads and a simple, powerful equation.
Orient yourself
Make a prediction, scan the goals and the key terms, and see which misconception this lesson is built to fix.
Before You Read
warm-upYou catch and mark 20 snails in a garden, release them, and a week later catch 25 snails, 5 of which carry your mark.
Before reading: the marked snails are now 5 out of 25 (one-fifth) of your second catch. If marked snails are one-fifth of the population too, and you marked 20, how many snails do you estimate are in the garden? Explain your reasoning.
Learning Intentions
goalsKnow
- The steps of the mark-release-recapture method
- The Lincoln index formula for estimating population size
- Why this method is used for mobile organisms
Understand
- Why the proportion of marked individuals reveals the total
- The assumptions the method relies on
- How breaking an assumption biases the estimate
Can Do
- Model the method and calculate a population estimate
- Apply the Lincoln index correctly
- Identify limitations and how to reduce error
Scan these before reading
vocabMatch each term to its meaning: use the Key Terms card above.
- Mark-release-recapture
- Lincoln index
- M
- C
- R
- The total number of individuals caught in the second sample
- A method for estimating the size of a mobile-animal population without counting it
- The number in the second sample already carrying a mark
- The formula N = (M × C) ÷ R
- The number caught, marked and released in the first sample
Misconceptions To Fix
watch outThe Method & the Lincoln Index
Why a proportion gives the whole
Core Content
The Method & the Lincoln Index
+5 XPWhy a proportion gives the whole
If marked animals mix evenly back into the population, the fraction of marked animals you find in a second catch equals the fraction of the whole population that is marked.
Mark-release-recapture: (1) capture a first sample (M), mark them harmlessly and release; (2) allow time to mix back into the population; (3) capture a second sample (C) and count how many are marked (R). The population is estimated using the Lincoln index: N = (M × C) ÷ R.
The logic: in the second catch, the marked proportion (R ÷ C) should equal the marked proportion of the whole population (M ÷ N). Rearranging gives N = (M × C) ÷ R. So a low fraction of recaptures means a large population, and a high fraction means a small one. This is the only practical way to estimate the abundance of mobile organisms like fish, insects and small mammals.
Pause, copy the highlighted three steps and the formula N = (M × C) ÷ R into your book before moving on.
In the Lincoln index N = (M × C) ÷ R, the letter R stands for the number of individuals in the second catch that are _____.
Modelling It & Doing the Calculation
A bean-bag model and a worked example
Modelling It & Doing the Calculation
+5 XPA bean-bag model and a worked example
We just set out the formula. That raises a question: how do we model it safely in class and use the numbers? This card answers it → with a container of beans standing in for the animals.
A bag of identical beans models a population: you "mark" some by swapping them for a different colour, mix, then sample again.
Model method: fill a container with many dried beans (the population, of unknown number). Remove a handful, this is your first capture M; "mark" them by replacing each with a marked bean (or a different colour), and return them. Shake to mix. Remove a second handful: count the total (C) and how many are marked (R). Apply N = (M × C) ÷ R.
Text description
- Second catch: 50 animals (C = 50), of which 8 are marked (R = 8, shown red).
- M = 40 were marked in the first catch.
- N = (40 × 50) ÷ 8 = 250.
The same arithmetic, on real animals. Agile wallabies in Kakadu: 45 were caught, tagged and released, and a second catch of 60 contained 15 tagged animals. N = (45 × 60) ÷ 15 = 180 wallabies. Notice that the numbers are small and the answer is not: 15 recaptures is what the whole estimate rests on, which is why the assumptions in the next card matter so much more in the field than they do in your container of beans.
Model with beans: take a handful (M), replace with marked beans, mix, take a second handful, count total (C) and marked (R), then N = (M × C) ÷ R. Example: M = 40, C = 50, R = 8 → N = (40 × 50) ÷ 8 = 250.
Pause, copy the model method and the worked calculation into your book.
Why the model uses beans and not animals. The dot point asks you to model the method, and that wording is deliberate. Capturing, handling and marking live animals causes stress, can injure them and requires training, ethics approval and, in New South Wales, a scientific licence under the Biodiversity Conservation Act 2016. None of that is available in a school laboratory, so the population is modelled with beans or beads, which reproduces the mathematics exactly while removing the welfare problem entirely. The trade-off is worth naming: beans cannot die, breed, emigrate, learn to avoid the scoop or lose their mark, so the model satisfies every assumption of the method perfectly and therefore cannot show you how badly a real survey can go wrong. That is what the next card is for.
Risk assessment for the model investigation. The hazards are small but real and must be identified before you start.
| Hazard | Risk | Control |
|---|---|---|
| Dried beans or beads | Choking if placed in the mouth; a serious hazard if a younger student is nearby | Nothing goes near the mouth; count on the bench, not in the air; keep containers closed when not sampling |
| Spilled beans on the floor | Slipping | Sample over a tray; sweep spills immediately rather than at the end |
| Legume dust and food allergies | Allergic reaction in a susceptible student | Check for legume and nut allergies before choosing the material; substitute plastic beads; wash hands afterwards |
| Marking pens or paint | Skin and eye irritation, fumes in a closed room | Use pre-coloured beads rather than wet marker where possible; ventilate; do not mark beads that will be handled while wet |
If a version of this investigation is run on live invertebrates such as slaters or garden snails, the ethical requirements are not optional extras. Animals must be collected and returned to the exact place they came from within the lesson, held in humid shaded containers away from heat, marked only with a small dot of non-toxic, quick-drying, water-based paint on a hard surface such as a shell, never on soft tissue and never with correction fluid or solvent-based marker, and handled as little and as gently as possible. The teacher, not the student, decides whether the species and the site make this acceptable.
30 animals were marked. A second catch of 40 contained 6 marked. Using N = (M × C) ÷ R, the estimated population is:
Apply it: Run the Numbers
Work the activity in your own book before you read on. This is the first point where you are on your own.
Activity 1
Run the Numbers
Pattern B, Work with data
In a model, 60 beans were marked and returned. Three second samples gave: (C = 50, R = 12), (C = 48, R = 10), (C = 52, R = 14). (a) Estimate the population from each sample using N = (M × C) ÷ R. (b) Average the three estimates. (c) Explain why averaging several samples gives a more reliable result than relying on one.
Assumptions & Sources of Error
When the estimate goes wrong, and why
Assumptions & Sources of Error
+5 XPWhen the estimate goes wrong, and why
We just calculated a population. That raises a question: when can we trust the number, and what makes it wrong? This card answers it → the method relies on several assumptions that must hold.
The Lincoln index is only valid if the marked animals truly behave like an unbiased random sample of the population the second time around.
The method assumes: (1) marked individuals mix randomly and fully back into the population before recapture; (2) the mark does not harm the animal, fall off, or make it easier/harder to catch; (3) there is no significant birth, death, immigration or emigration between samples (a "closed" population); and (4) marked and unmarked individuals are equally likely to be caught.
If an assumption breaks, the estimate is biased. For example, if marked animals are more easily caught (or learn to avoid the trap), R changes and N is wrong; if animals die or emigrate, fewer marks are recaptured, over-estimating N. To improve reliability and accuracy: leave enough time for mixing but not so long that births/deaths occur, use a harmless and durable mark, take a large enough sample, and repeat to average. Acknowledge these assumptions whenever you report a result.
Assumptions: marks mix randomly; mark is harmless, stays on, and does not change catchability; closed population (no birth/death/migration between samples); marked = unmarked catchability. Breaking any of these biases N. Improve by allowing mixing time (not too long), harmless durable marks, large samples, and repeats.
Add the four assumptions and how to reduce error to your notes before the check below.
Work out the direction of the error, not just that there is one. Because N = (M × C) ÷ R, R sits on the bottom of the fraction, so anything that reduces the number of marks you recapture inflates the estimate, and anything that increases it deflates the estimate. Reason it through with the formula in front of you rather than guessing.
| What goes wrong | Effect on R | Effect on the estimate N |
|---|---|---|
| Marks rub off or fade before the second catch | Falls, marked animals are counted as unmarked | Over-estimate |
| Marked animals die, or emigrate from the study area | Falls | Over-estimate |
| Too little time to mix, so marked animals are still near the release point and the second sample is taken elsewhere | Falls | Over-estimate |
| The mark makes an animal conspicuous or slower, so it is caught more easily | Rises | Under-estimate |
| Animals become trap-happy, returning to a trap that fed them | Rises | Under-estimate |
| Animals become trap-shy after a frightening first capture | Falls | Over-estimate |
| Unmarked young are born between the samples | Unchanged, but C now includes animals that were never at risk of being marked | Over-estimate |
Two of those, trap-happy and trap-shy, push in opposite directions, which is exactly why a well-run survey uses a trap and a handling routine that does neither, and why the assumption is stated as "marked and unmarked individuals are equally likely to be caught" rather than simply "the mark is harmless".
The Lincoln index assumes marked individuals mix randomly back into the population before recapture.
The estimated population is simply the total number of animals caught across both samples.
If the mark makes an animal harder to catch the second time, the population estimate will be biased.
See where it lands in the real world
Why this matters: counting wildlife and fish stocks
Why this matters: counting wildlife and fish stocks
Mark-release-recapture is used worldwide to estimate populations that cannot be counted directly, from tagging fish to set sustainable catch limits, to banding birds and microchipping threatened mammals like quolls. Fisheries managers depend on these estimates to prevent over-fishing, and conservation scientists use them to tell whether an endangered population is recovering. The simple proportion you modelled with beans is, scaled up, one of the most important tools for managing living resources responsibly.
Odd one out: three of these follow from the card above about real wildlife and fisheries estimates. Which one does not?
Drive It: Estimate a Population You Cannot Count
+XP for exploringN = (M × C) ÷ R works only while its assumptions hold
Use the simulation. You mark M = 50 animals, and a second catch of C = 40 contains R = 10 marked animals. What is the estimate?
Use the simulation. True or false: if marks rub off before the second catch, the estimate of N is too low.
In a farm dam, 120 fish were caught, marked and released. A week later a second catch of 80 fish contained 16 that were marked. Apply the Lincoln index to estimate the population, and state the effect on your estimate if some marks had been lost in the intervening week.
Apply asks you to use the method in a new situation. The second clause asks for the direction of an error, which is reasoning about the formula rather than a further calculation.
5 marks. Identify each symbol against the scenario, substitute, calculate, check the answer is reasonable, then reason about the effect of lost marks.
In the Lincoln index N equals M multiplied by C, divided by R. Here M is the first capture, the 120 fish marked and released. C is the second capture, the whole of the second catch, which is 80 fish. R is the number in that second catch that carried a mark, which is 16. Note that C is the entire second catch and not the unmarked part of it.
Substituting gives N equals 120 multiplied by 80, divided by 16. The numerator is 120 multiplied by 80, which is 9600. Dividing by 16 gives 9600 divided by 16, which is 600. The estimated population of the dam is therefore about 600 fish.
Check the estimate against the logic of the method rather than only the arithmetic. Marked fish were 16 out of 80 in the second catch, which is one-fifth, so if the second catch is representative then the 120 marked fish are about one-fifth of the population, giving 120 multiplied by 5, or 600. The two routes agree. The estimate must also always be at least as large as M and as C, since both are samples from the population, and 600 exceeds both. An estimate smaller than the number marked, or one of several hundred thousand from a farm dam, would indicate an error rather than a finding.
If some marks were lost during the week, a fish that was genuinely marked would be recorded as unmarked, so R would fall below its true value while M and C stayed the same. In N equals M times C divided by R, R is the divisor, so a smaller R makes N larger. The estimate would therefore be too high, and the population would be over-estimated.
The effect can be substantial. If 6 of the 16 marks had been lost, R would be 10 and the estimate would be 120 multiplied by 80 divided by 10, which is 960 rather than 600, an over-estimate of 60 per cent from losing six marks. This is a systematic bias with a known direction rather than random scatter, so taking more recapture samples and averaging would not remove it: every sample would be inflated the same way. The remedy is a durable, harmless mark and a short interval between samples.
In the Lincoln index N equals M multiplied by C divided by R, where M is the first capture, the 120 fish marked and released, C is the second capture, meaning the whole of the second catch of 80 fish and not merely its unmarked part, and R is the number in the second catch carrying a mark, which is 16. Substituting gives N equals 120 multiplied by 80 divided by 16; the numerator is 9600, and 9600 divided by 16 is 600, so the estimated population is about 600 fish. Checking against the logic of the method, marked fish were 16 out of 80 in the second catch, which is one-fifth, so if that catch is representative the 120 marked fish are about one-fifth of the population, giving 120 multiplied by 5, or 600, and the two routes agree; the estimate must also be at least as large as both M and C, since both are samples from the population, and 600 exceeds both, whereas an estimate smaller than the number marked, or one of several hundred thousand from a farm dam, would indicate an error rather than a finding. If some marks were lost during the week, a genuinely marked fish would be recorded as unmarked, so R would fall below its true value while M and C stayed the same, and since R is the divisor a smaller R makes N larger, so the population would be over-estimated. The effect can be substantial: had 6 of the 16 marks been lost, R would be 10 and the estimate would be 120 multiplied by 80 divided by 10, which is 960 rather than 600, an over-estimate of 60 per cent from six lost marks. This is a systematic bias with a known direction rather than random scatter, so repeating the recapture and averaging would not remove it, every sample being inflated the same way, and the remedy is instead a durable, harmless mark and a short interval between samples.
Explain why the mark-release-recapture method requires a closed population, and explain the direction of the error introduced if the population is not closed.
Explain means giving reasons. There are two here: why the assumption is needed at all, and which way the estimate moves when it fails, so treat them as separate tasks.
5 marks. State the assumption the method rests on, show why a closed population is needed for it to hold, take each way the population can open in turn, then state the practical remedy.
Mark-release-recapture works by proportion, not by counting. It assumes the proportion of marked individuals in the second sample equals the proportion of marked individuals in the whole population, so that R divided by C equals M divided by N, which rearranges to N equals M times C divided by R. Everything the method claims depends on that one equality.
A closed population is one in which no individuals are added or removed between the two samples, so there is no significant birth, death, immigration or emigration. It is required because the equality above compares a proportion measured at the time of the second sample with a population defined at the time of marking. If the population changes in between, the two sides of the equality refer to different populations and the estimate has no clear meaning.
If marked individuals die or emigrate between the samples, fewer marked individuals remain available to be recaptured, so R falls while M and C are unchanged. Since R is the divisor in N equals M times C divided by R, a smaller R makes N larger, and the population is over-estimated.
If unmarked individuals are born into the population or immigrate between the samples, the second catch is drawn from a population containing a higher proportion of unmarked individuals than existed at marking. R therefore falls as a proportion of C, the marked fraction of the second sample is diluted, and the estimate rises again. Both ways of opening the population push the estimate in the same direction, upward, which is why the method characteristically over-estimates when the interval is too long.
The remedy is to keep the interval between samples short enough that births, deaths and movement are negligible, and to sample the same defined area both times so that movement across the boundary is limited. This conflicts with the other requirement of the method, that marked individuals must have time to mix randomly back into the population before the second sample: too short an interval breaks the mixing assumption and too long an interval breaks the closed-population assumption. The interval must be long enough to mix and short enough to stay closed, and choosing it is a judgement about the particular species rather than a fixed rule.
Mark-release-recapture works by proportion rather than by counting: it assumes the proportion of marked individuals in the second sample equals the proportion in the whole population, so that R divided by C equals M divided by N, which rearranges to N equals M times C divided by R, and everything the method claims rests on that equality. A closed population is one in which no individuals are added or removed between the two samples, with no significant birth, death, immigration or emigration, and it is required because the equality compares a proportion measured at the time of the second sample against a population defined at the time of marking; if the population changes in between, the two sides refer to different populations and the estimate has no clear meaning. If marked individuals die or emigrate between the samples, fewer remain available for recapture, so R falls while M and C are unchanged, and since R is the divisor a smaller R makes N larger and the population is over-estimated. If unmarked individuals are born or immigrate, the second catch is drawn from a population with a higher proportion of unmarked individuals than existed at marking, so R falls as a proportion of C, the marked fraction of the second sample is diluted, and the estimate rises again. Both ways of opening the population therefore push the estimate upward, which is why the method characteristically over-estimates when the interval is too long, and why the error cannot be averaged away. The remedy is to keep the interval short enough that births, deaths and movement are negligible and to sample the same defined area both times so movement across the boundary is limited, but this conflicts with the method's other requirement, that marked individuals mix randomly back into the population before the second sample: too short an interval breaks the mixing assumption and too long an interval breaks the closed-population assumption, so the interval must be long enough to mix and short enough to stay closed, which is a judgement about the particular species rather than a fixed rule.
Practise independently
Answer every question in your own words first. The model answers are for checking against, not for copying.
Multiple Choice
+5 XPA fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
Short Answer, 14 marks
+5 XPApplyBand 5(6 marks) 1. An ecologist marked 80 frogs and released them. Two weeks later they caught 100 frogs, 20 of which were marked. Estimate the population using the Lincoln index, and explain two assumptions of the method and how breaking one would affect the estimate.
2 marks: correct Lincoln-index calculation with working · 2 marks: two valid assumptions stated · 2 marks: explains how breaking one assumption biases the estimate
ApplyBand 4(4 marks) 2. In a model, 60 beans were marked and returned to the container. Three second samples gave (C = 50, R = 12), (C = 48, R = 10) and (C = 52, R = 14). (a) Estimate the population from each sample. (b) Give the mean estimate. (c) Explain what the spread of the three estimates tells you that a single sample could not.
1 mark: all three estimates calculated correctly (250, 288, 223 to the nearest whole bean) · 1 mark: mean stated as about 254 · 1 mark: identifies the spread, roughly 223 to 288, as a measure of how much a single estimate can vary by chance · 1 mark: concludes that one sample gives no indication of its own uncertainty, so repeats are what make the estimate defensible
EvaluateBand 5-6(4 marks) 3. A student estimates a snail population in a garden bed. They mark 30 snails with a dab of correction fluid, release them, and return four weeks later to collect the second sample after heavy rain. Evaluate the reliability and accuracy of any estimate produced this way.
1 mark: identifies correction fluid as a mark that may harm the snail or wash off, and links a lost mark to a lower R · 1 mark: identifies four weeks as long enough for births, deaths and migration, breaking the closed-population assumption · 1 mark: uses N = (M x C) / R to argue the direction, a reduced R inflates N, so the population is over-estimated · 1 mark: reaches a judgement and proposes specific fixes, a small dot of non-toxic water-based paint on the shell, a shorter interval of a few days, and repeated samples averaged
Show all answers
Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Short Answer Model Answer
Q1 (6 marks): Using the Lincoln index, N = (M × C) ÷ R = (80 × 100) ÷ 20 = 400 frogs. The method relies on several assumptions. Assumption 1: the marked frogs mix randomly and fully back into the population before the second capture, so the marked proportion in the second sample reflects the marked proportion of the whole population. Assumption 2: the population is "closed", there is no significant birth, death, immigration or emigration between the two samples, and the mark is harmless, stays on, and does not change how likely a frog is to be caught. If an assumption is broken, the estimate is biased. For example, if some marked frogs died or moved away during the two weeks, fewer marks would be recaptured (R falls); since N = (M × C) ÷ R, a smaller R makes N larger, so the population would be over-estimated. Conversely, if the mark made frogs easier to catch, R would rise and N would be under-estimated. Reporting the estimate should therefore state these assumptions, and reliability is improved by allowing time to mix (but not so long that births and deaths occur), using a harmless durable mark, taking large samples and repeating to average.
Q2 (4 marks): (a) Applying N = (M × C) ÷ R with M = 60 each time: sample 1 gives (60 × 50) ÷ 12 = 3000 ÷ 12 = 250; sample 2 gives (60 × 48) ÷ 10 = 2880 ÷ 10 = 288; sample 3 gives (60 × 52) ÷ 14 = 3120 ÷ 14 = 223 to the nearest whole bean [1]. (b) The mean of the three is (250 + 288 + 223) ÷ 3 = 761 ÷ 3 = about 254 beans [1]. (c) The three estimates come from the same unchanged container, so the differences between them are not real changes in the population, they are the variation produced by which beans happened to end up in each handful. The spread of roughly 223 to 288, about 25 % of the mean, is therefore a measurement of how much chance alone moves the answer [1]. A single sample gives a number with no indication of its own uncertainty attached: report 288 on its own and nothing in the result reveals that a repeat would plausibly have given 223 instead. Repeating and averaging both improves the estimate and, more importantly, shows how much confidence the estimate deserves [1].
Q3 (4 marks): The mark is the first problem. Correction fluid is solvent-based, it can harm the snail's shell and tissue, and it flakes; four weeks of weather including heavy rain make it likely that some marks are gone by the second sample, so genuinely marked snails are recorded as unmarked and R falls below its true value [1]. The interval is the second problem. Four weeks is long enough for snails to hatch, die, and move in and out of a garden bed, so the population is not closed: marked snails are lost from the area and unmarked snails that were never at risk of being marked have joined it, and both effects reduce the marked fraction of the second catch [1]. Both problems push the answer the same way. Since N = (M × C) ÷ R and R sits on the bottom of the fraction, an R that is too small makes N too large, so this design systematically over-estimates the snail population, and the error is a bias rather than random scatter, which means averaging more samples would not remove it [1]. My judgement is that the estimate is neither reliable nor accurate and should not be reported as a population figure. It could be made defensible by marking with a small dot of non-toxic, quick-drying, water-based paint on the hard shell rather than the body, shortening the interval to a few days, long enough to mix but too short for much birth or death, sampling the same defined area both times, and taking several recapture samples and averaging them so that the remaining uncertainty is at least visible [1].
Retrieve and reflect
Do the spaced-review set from memory first, then look back at the prediction you made before you read anything.
Check what actually stuck
A short spaced-review set drawn from this lesson and the lessons before it. Answer from memory first, that is what makes it stick.
Lock it in
bossYou can now run the mark-release-recapture method, apply the Lincoln index, and explain its assumptions and limitations. Next lesson turns to the living interactions between organisms, predation, competition, mutualism and more.
↩ Back to focus-area hubHow did your thinking change?
You estimated the snail population. Marked snails were 5 out of 25 (one-fifth) of the second catch, so the 20 snails you marked are about one-fifth of the whole population, giving an estimate of 20 × 5 = 100 snails. This is exactly the Lincoln index: N = (M × C) ÷ R = (20 × 25) ÷ 5 = 100. The key idea is that a proportion, not a head count, reveals the total.
If you started by trying to add up the catches, the method instead uses the marked fraction: a small recaptured fraction means a large population, and a large fraction means a small one.