Capture–Recapture in a Pond with Tag Loss and Trap-Shy Fish
A student estimated the number of fish in a pond. On Monday, 200 fish were caught, marked with a small fin clip, and released. On Tuesday, 150 fish were caught; 30 had clips. The student calculated 200 × 150 ÷ 30 = 1000 fish. A teacher noted three problems: the clip could be lost; marked fish might avoid the traps after handling; and Tuesday's catch was taken only from the shallow edge. The pond also contains a heron that eats fish. The student wants to report that 'the pond contains about 1000 fish' in a school poster. No fish were born or died from disease during the study, but the heron was seen hunting between the two samples. The student did not record the water depth at which each fish was caught.
1. State the assumption of the Lincoln index and calculate the estimate. Explain why the estimate is valid only if marked and unmarked fish are equally likely to be recaptured.
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The Lincoln index is N = (M × C) / R, where M is the number marked and released, C is the number caught in the second sample, and R is the number of marked fish in that sample. Here N = (200 × 150) / 30 = 1000. The estimate assumes that marked and unmarked fish are equally likely to be caught in the second sample. If marked fish are recaptured more often, R is too high and N is underestimated; if they are recaptured less often, R is too low and N is overestimated. The method also assumes a closed population with no births, deaths, immigration or emigration between samples, no mark loss, and random mixing of marked fish with the rest of the population.
2. For each of the three issues, predict whether the estimate is too high or too low and explain. Include the heron in your answer.
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Mark loss makes R too low, so N is too high. Trap avoidance by marked fish also makes R too low, so N is too high. Sampling only the shallow edge can bias the estimate in either direction: if marked fish move to deeper water, R is too low and N is too high; if unmarked fish concentrate in the shallows, R is also too low; if marked fish stay in the shallows, R is too high and N is too low. The heron is a mortality source between samples, which violates the closed-population assumption. If it eats marked and unmarked fish equally, the effective number of marked fish available to be recaptured is lower, so R falls and N is overestimated. The direction of bias from the heron depends on whether it selects marked fish.
3. Describe two improvements to the sampling design and explain how each reduces bias.
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Take several recapture samples over a longer period and use multiple marking occasions, for example the Schnabel method. This reduces the effect of one unusual sample and allows fish to mix more thoroughly. Sample at randomly chosen depths and times, and record depth for each fish, so that the recapture sample is representative of the whole pond. Use a permanent tag that is less likely to be lost, and check for tag loss by keeping a few marked fish in a pen. Use two independent trapping methods to reduce trap-shyness. These changes do not remove all bias, so the result should be reported as an estimate with a range, not as an exact count.