## Answers

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.

.