A population of finches lives on an island some distance from the mainland, wher

Question

A population of finches lives on an island some distance from the mainland, where another population of finches of the same species resides. In both populations, the size of the birds' beaks is influenced by two alleles at the beak locus, B and b. On the island, the frequency of the B allele is 0.8, and the frequency of the b allele is 0.2. On the mainland, both alleles are equally frequent (0.5).

(a) If the per-generation mutation rate from B to b (and from b to B) is 1x10-4, what is change in the frequency of the B allele in one generation due to mutation?

(The change in allele frequency resulting from mutation can be described as follows: if the mutation rate from A to a is u and the rate from a to A is v, then the frequency of A in the next generation is p' = p - u*p + v*q)

(b) If the migration rate from the mainland to the island is 0.1 (meaning that 10% of individuals in the island population migrate in from the mainland every generation), what is the change in the frequency of the B allele in one generation due to migration? Assume there is no migration from the island back to the mainland.

(c) If the genotypes at the beak locus have the following fitnesses: wBB = 1, wBb = 0.95, wbb = 0.9, what is the change in the frequency of the B allele in one generation due to natural selection?

Explanation / Answer

Ans. a) To calculate the frequency of B in next generation the formula is given below -

p' = p – u X p + v X q

Now put the given values in the formula,

=0.8 - (1 x 10-4 X 0.8) + (1 x 10-4 x 0.2)

= 0.79994

Therefore, the frequency of B in next generation is 0.79994.

b.) Change in the frequency of the B allele in one generation is calculated by following formula –

Change in frequency of B allele in one generation = (B x Migration rate) /100

Where B =. 0.5 (calculated by above mentioned formula)

= (0.5 x 0.1)/ 100

= 0.49995

Hence, change in the frequency of the B allele is 0.49995.

c.) As it is given, wBB = 1, wBb = 0.95 and wbb = 0.9

Frequency of B after one gen. of selection

wBB = p2 = 1 x (0.9)2 = 0.81

wBb = 2pq = 2 x 0.9 x 0.1 x 0.95 = 0.171

wbb = q2 = 1 x (0.9)2 = 0.095

Now, the the frequency of the B allele

=D+H/2

=0.81+0.171/2

=0.895

Hence, change in the frequency of the B allele

=0.91 - 0.895

=0.015

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