Question

# (Hardy Weinberg Equation) In a population of rabbits, there are 423 A1A1, 1484 A2A2, and 1661...

(Hardy Weinberg Equation)

In a population of rabbits, there are 423 A1A1, 1484 A2A2, and 1661 A1A2 individuals. If the enviornment changes so that the homozygous recessive genotype suffers a reduction of fitness where it's fitness is now 0.59 , but the other genotypes are unaffected, what will be the frequency of the dominant allele in the NEXT generation?

Note: Assuming A1 dominant to A2.

Given, relative fitness of A1A1 = A1A2 = 1, relative fitness of A2A2 = 0.59, number of A1A1 = 423, number of A2A2 = 1484, number of A1A2 = 1661

So, total number of rabbits = 423 + 1484 + 1661 = 3568

So, frequency of A1A1 = Number of A1A1 / Total population = 423/3568 = 0.11855 (Up to 5 decimals)

So, frequency of A1A2 = Number of A1A2 / Total population = 1661/3568 = 0.46553 (Up to 5 decimals)

So, frequency of A2A2 = Number of A2A2 / Total population = 1484/3568 = 0.41592 (Up to 5 decimals)

Now, mean fitness = (Frequency of A1A1 x Fitness of A1A1) + (Frequency of A1A2 x Fitness of A1A2) + (Frequency of A2A2 x Fitness of A2A2) = (0.11855 x 1) + (0.46553 x 1) + (0.41592 x 0.59) = 0.11855 + 0.46553 + 0.24539 = 0.82947

So, after selection frequency of A1A1 = (Frequency of A1A1 x Fitness of A1A1) / mean fitness = 0.11855 / 0.82947 = 0.14292 (Up to 5 decimals)

So, after selection frequency of A1A2 = (Frequency of A1A2 x Fitness of A1A2) / mean fitness = 0.46553 / 0.82947 = 0.56124 (Up to 5 decimals)

So, frequency of A1 (Dominant allele) in the next generation = 1/2 of frequency of A1A2 + Frequency of A1A1 = 1/2 x 0.56124 + 0.14292 = 0.42354

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