Gray bodies (Eb^+) are dominant over ebony bodies (Eb) in fruit flies. A researc

Question

Gray bodies (Eb^+) are dominant over ebony bodies (Eb) in fruit flies. A research lab has two populations of flies Population 1 is true-breeding for gray bodies and has 1000 members. Population 2 has 700 members, 50 of which have ebony bodies. A careless freshman research assistant leaves the door open, and 70 members from population 2 sneak into population 1. What are the allele frequencies of the conglomerate? Assuming Hardy-Weinberg equilibrium is reached after one generation, what will the genotypic frequencies of the conglomerate be after one generation? Equations: p^2 + 2pq + q^2 = 1 Delta p_c = m(pD - pR) Delta q = mu p (1 - mu)^t = p_t/p_0 (Initial allele #)/2N t = 4N W = p^2 W_AA + 2pq W_Aa + q^2 W_aa

Explanation / Answer

In population 1, previously out of 700 , 650 were true Eb+ and 50 are EB-

So the allele frequencies of population 2, where p for Eb+ and q is for Eb-

P = 650/700 = 0.93, q =1-p = 1-0.93 = q= 0.07

Now 70 were flied to population 1 so new population has 1070 individuals now.

Now the frequency in conglomerate population is calculated by Pc = m(Pd – Pr)

Where Pc is the change of conglomerate frequency, Pd = frequency of donor population and Pr = frequency of donor population, m is the migrated population frequency in new population.

In population 1 there is no ebony bodies so Pr = 0, donor frequency Pd =0.07 and m =70/1070 = 0.065

Now the Pc = 0.065(0.07-0) = 0.00455

Now total frequency of population one is =0+0.00455 = 0.00455

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