Estimating Population Size by Mark, Release and Recapture · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Counting animals that run away

Science · Scientific Inquiry · ages 15-16
Name ______________________   Date ____________

  1. 20 fish were marked. A second catch of 30 includes 6 marked fish. Assuming the method's conditions hold, what is the estimated population?

    Answer: ______________

  2. For which task is mark and recapture a suitable choice?

    • Estimating rooted plants with randomly placed quadrats
    • Estimating mobile fish using marked and unmarked catches
    • Comparing plant cover at intervals along a transect

  3. Why must marked and unmarked animals mix before the second catch?

    • To keep marked animals near the release point
    • To let the catch reflect the population's marked share
    • To make the marked fraction identical in each small group

  4. Which change violates the closed-population assumption of the two-catch method?

    • New animals are born and join the population between catches
    • Marked animals mix with unmarked animals before recapture
    • Marks stay readable throughout the gap between catches

  5. 15 beetles were marked. A second catch of 40 includes 10 marked beetles. Assuming the method's conditions hold, what is the estimated population?

    Answer: ______________

  6. Predators remove marked animals more often than unmarked animals. Other sampling conditions stay the same. Which bias is expected?

    • Too high: fewer marks recaught make the divisor smaller
    • Too low: fewer survivors mean a smaller calculated total
    • Unchanged: the second catch still includes unmarked animals

  7. Lou calculates 100 fish from two catches and says, "Exactly 100 fish live here." What should Lou change?

    • Repeat the arithmetic to make 100 an exact population count
    • Report 100 as a census because both catches were counted
    • Report an estimated 100, with sampling uncertainty

  8. Marked animals remain clustered at the release spot. A catch taken only there has a larger marked fraction than the whole population. How is the estimate biased?

    • Too high: a larger marked fraction scales the total upward
    • Too low: the marked share overstates its population value
    • Unchanged: the same first marked count is used
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Counting animals that run away W1-mt_SrISE2trf9-s1

  1. 100 · 20 × 30 ÷ 6 = 100, so the estimated population is 100 fish.
  2. Estimating mobile fish using marked and unmarked catches · Two catches can reveal the marked share of a mobile fish population. Quadrats and transects fit the other stated tasks.
  3. To let the catch reflect the population's marked share · Mixing supports a representative marked share. It does not give every small group an identical fraction.
  4. New animals are born and join the population between catches · A closed population has no births, deaths, arrivals or departures during the study. Births change its membership.
  5. 60 · 10/40 is 1/4. If 15 marked beetles represent one quarter of the population, the estimate is 4 × 15 = 60 beetles.
  6. Too high: fewer marks recaught make the divisor smaller · Extra loss of marked animals lowers their expected share in the second catch. The smaller marked-count divisor tends to inflate the estimate.
  7. Report an estimated 100, with sampling uncertainty · The calculation uses samples rather than a count of every fish. Chance variation remains even when its assumptions hold.
  8. Too low: the marked share overstates its population value · The total is estimated by dividing the first marked count by the observed marked fraction. An excessive fraction makes that total too small.
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