MATSEC IM Biology study materials
MATSEC IM Biology study resources

Two Mark–Recapture Estimates and One Unhappy Funder

MATSEC IM Biology · Put your knowledge to work in context.

All case studies

Two Mark–Recapture Estimates and One Unhappy Funder

A conservation group estimates a population of freshwater turtles in a reservoir. In week 1 they capture 200 turtles, mark each with a small notch in the shell and release them at the capture site. In week 2 they capture 150 turtles, of which 15 are marked. In week 3 they capture another 150, of which 30 are marked. Between weeks 1 and 2 heavy rain caused the reservoir to overflow into a neighbouring stream; turtles may have moved either way. One marked turtle was found dead, apparently predated. The group reports the week-2 estimate of 2,000 turtles to a funder, then the week-3 estimate of 1,000. The funder asks which figure is correct and whether the project should be funded at the higher level.

1. Calculate the week-2 and week-3 Lincoln–Petersen estimates and a pooled Schnabel estimate. Explain why the figures differ.

Read the worked answer

Lincoln–Petersen: N = (M × C)/R. Week 2: (200 × 150)/15 = 2,000. Week 3: (200 × 150)/30 = 1,000. Both use the same 200 marked animals, so the difference comes entirely from the recapture proportion, 10% versus 20%. A pooled Schnabel estimate uses all marked animals known to be at large: week 2, M = 200, C = 150, R = 15; week 3, M = 200 + (150 − 15) = 335, C = 150, R = 30. N = (150 × 200 + 150 × 335)/(15 + 30) = (30,000 + 50,250)/45 = 80,250/45 ≈ 1,783. The three estimates, 2,000, 1,000 and 1,783, show that sampling error and violated assumptions are large. A single figure is not defensible; the data are compatible with a population of roughly 1,000–2,000, and the direction of bias must be assessed before choosing a figure.

2. Identify the assumptions of mark–recapture that may have been violated and state the likely direction of bias for each.

Read the worked answer

Assumptions include: no births, deaths, immigration or emigration; all animals equally likely to be caught; marks not lost; marked animals mix randomly; and marking does not affect survival or catchability. The overflow between weeks 1 and 2 violates closure, because turtles could enter or leave the reservoir. Release at the capture site may mean marked animals are clustered and not yet mixed in week 2, which could lower the recapture proportion and inflate the estimate. Trap-shyness after handling would also lower recapture and inflate the estimate. Predation of a marked turtle removes marked animals and inflates the estimate further. If, by week 3, marked animals had mixed and become less trap-shy, the recapture proportion would rise and the estimate would fall, which is consistent with 1,000. Immigration of unmarked animals would dilute the marked fraction and inflate the estimate, whereas emigration of unmarked animals would raise the marked fraction and deflate it. The net direction is uncertain, so neither figure can be treated as exact.

3. Recommend how the group should estimate the population and report it to the funder. Justify each recommendation.

Read the worked answer

Use a robust multi-sample design: at least four or five capture occasions, individual permanent tags, and recording of capture histories so that detection probability and trap response can be modelled. Check tag retention and record any deaths. Allow a mixing period after release and spread release sites. If closure is doubtful, use an open-population model or restrict the study to a period without overflow. Present the funder with a range and confidence interval, not a single number, and state the assumptions. Fund at a level that supports the monitoring needed to resolve the uncertainty; the higher figure alone is not justified. A power analysis should determine how many recaptures are needed to distinguish 1,000 from 2,000. Consider complementary methods such as eDNA or standardised visual surveys to cross-check the estimate.