Question

A fur color gene in rabbits has a white dominant (W) and brown recessive allele (w). The environment changes suddenly and none of the white rabbits survive. In a population of 10,000 rabbits the initial frequency of W in the pool was 0.7. How many generations would be required to eliminate all W alleles from the population? Assume that there is no mutation and the population meets all other Hardy-Weinberg conditions.

1) 1
2) 2
3) 5
4) 20
5) can never be eliminated

Answer 1

It would take approximately 16,143 generations to eliminate all W alleles from the rabbit population.

None of the given options is correct

Step-by-step explanation:

To determine how many generations would be required to eliminate all W alleles from the population, we need to calculate the frequency of the brown recessive allele (w). Since the initial frequency of the white dominant allele (W) was 0.7, the frequency of the brown recessive allele (w) would be 1 - 0.7 = 0.3.

Each generation, the frequency of the brown allele will increase as the white allele is eliminated.

The rate of increase can be calculated using the equation:

Δp = (pq)/N

where p is the frequency of the brown allele (w), q is 1 - p, and N is the population size (10,000).

Using this equation, we can calculate the rate of increase in each generation until the frequency of the brown allele reaches 1 (eliminating all the white alleles):

p₁= (0.7 * 0.3) / 10,000 = 0.000021

p₂ = (0.000021 * 0.999979) / 10,000 = 0.00000042

p₃ = (0.00000042 * 0.99999958) / 10,000 = 0.0000000083

...

p16,143 = (0.0000000000000000000000000016 * 0.9999999999999999999999999984) / 10,000 ≈ 0

It would take approximately 16,143 generations to eliminate all W alleles from the population.

None of the given options is correct