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

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

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

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

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

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

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

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