​A researcher captured 60 bivalves from a habitat on day 1 and tagged all of them. On day 2, she caught 40 bivalves out of which 20 were already tagged. She then estimated the population size of bivalves in the lake using this information. Which one of the options represents the percentage of the bivalve population that were marked on day 1?​

Correct option is B

The correct option is (b)

This problem involves the capture-recapture method commonly used in ecology to estimate animal populations.

On Day 1, the researcher caught and tagged 60 bivalves (let's call this M = number tagged initially).

On Day 2, she caught 40 bivalves (let's call this C = size of second sample).

Among the 40 caught on day 2, 20 were already tagged (let's call this R = Marked  individuals recaptured).

The basic assumption of the capture-recapture method is that the ratio of tagged individuals in the second catch is proportional to the ratio of tagged individuals in the entire population.
Using this, the estimated total population (N) is:

$$N = \frac{M \times C}{R}$$

Substituting values:

                                                                        $$N = \frac{60 \times 40}{20} = \frac{2400}{20} = 120$$

So, the estimated total population of bivalves is 120.

Since 60 were marked initially and the total population is 120, the percentage marked is:

                                                                           $$\text{Percentage marked} = \frac{60}{120} \times 100 = 50\%$$​​​