8. You have caught 100 flies, marked them, and released them back in a greenhouse of
area 100 m2. Assume the fly population doubles every 4 days; assume no flies die. After
12 days, you catch 400 flies and find that 10 out of them are marked. What is your
estimate of the number of flies in the greenhouse at the start of your experiment?
a. 500
b. 800
c. 1,000
d. 4,000
The correct answer is a. 500. This CSIR NET Life Sciences question tests the mark-recapture method (Lincoln Index) adjusted for population growth in a controlled greenhouse setting.
Problem Breakdown
You marked and released 100 flies (M = 100) into a 100 m² greenhouse at day 0, when the initial population was N₀. The population doubles every 4 days with no deaths, so after 12 days (3 doublings), the total population N = N₀ × 2³ = 8N₀.
Catching 400 flies (n = 400) with 10 marked (m = 10) allows the Lincoln Index estimate: N ≈ (M × n) / m = (100 × 400) / 10 = 4,000.
Thus, 8N₀ = 4,000, so N₀ = 4,000 / 8 = 500. This assumes random mixing, no mark loss, and equal catchability.
Option Analysis
| Option | Estimate | Why Incorrect/Correct |
|---|---|---|
| a. 500 | N₀ = 500 | Correct: Accounts for 3× doubling (N = 8N₀), Lincoln gives 4,000 total, so initial = 500. |
| b. 800 | N₀ = 800 | Wrong: Ignores growth; assumes static population of ~3,200 total (wrong Lincoln application). |
| c. 1,000 | N₀ = 1,000 | Wrong: Direct Lincoln without growth (N = 4,000), overestimates initial by 8×. |
| d. 4,000 | N₀ = 4,000 | Wrong: Treats 4,000 as initial, ignores doubling entirely. |
Introduction to Fly Mark Recapture Population Estimation
Fly mark recapture population estimation is a key ecology technique in CSIR NET Life Sciences, using the Lincoln Index to calculate animal populations in closed systems like greenhouses. This method shines in problems where fly populations double every 4 days, requiring growth adjustments over time (e.g., 12 days = 3 doublings).
Lincoln Index Formula with Growth
The core formula is N = (M × n) / m, where M = marked released, n = second sample caught, m = marked recaptured. For growing populations: adjust for multiplication factor (here, 2^(12/4) = 8). Greenhouse constraints (100 m², no deaths) validate closed population assumptions.
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Step 1: Compute total at recapture: (100 × 400) / 10 = 4,000.
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Step 2: Divide by growth: 4,000 / 8 = 500 initial flies.
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Common pitfalls: Forgetting doublings leads to option c or d errors.
CSIR NET Exam Relevance
This mirrors PYQs on mark recapture method in population ecology, testing integration of exponential growth (r = ln(2)/4 days) with sampling. Practice reveals why 800 or 4,000 distracts uninformed solvers.
