Question

you are studying a population of sexual, diploid insects. one particular trait, wing size, is controlled by a single locus with two alleles. The allele for narrow wing size is dominant (N) and the allele for wide wing size is recessive (n). The population you are studying is comprised of 83 NN individuals, 90 Nn individuals, and 83 nn individuals. What is the EXPECTED FREQUENCY OF THE DOMINANT ALLELE (N) AFTER SELECTION? Assume that w11=0.9, w12=1, and w22=0.7.

i dont know what information to add, as this is all the quiz gives me to solve the problem. i am not sure what you'd need, but whatever it is i assume i calculate it from the given data.

If it helps, i did find w (overall fitness of the population after selection) to be 0.739228 ?

Answer 1

From the Hardy-Weinberg equilibrium: p2 + q2+ 2pq = 1 (p = w11, q = w22, pq = w12)

So from this, the average fitness of the population (w- bar)= p2*w11 +2pq* w12 +q2*w22

Where p2 is the frequency of individuals who are homozygous dominant (NN), q2 is the frequency of individuals with the recessive genotype (nn),2pq is the frequency of heterozygous individuals in the population(Nn), w11 is the ability of individuals who are homozygous dominant to reproduce successfully and so on.

Genotype frequency = the number of individuals with a particular genotype/ total number of individuals.

Here, total number of individuals:

= 83 (NN) + 90 (Nn) + 83 (nn)

= 256

p2 = 83/256 = 0.324

q2 = 83/256 = 0.324

pq= 90/256 = 0.351

By substituting these values in the formula, we get

Average fitness of population (w- bar):

= 0.324 + 0.351 + 0.324

= 0.999 approximately equals to 1.

The expected frequency of dominant allele (N) will be NN + Nn/2 ÷ total frequency of all allele.

= 83 + 90/2 ÷ (83 + 90 + 83)

= 83 + 45 ÷ 256

= 0.5

Frequency of recessive allele : 45 +83/ 256

= 0.5.