Question

Whip-poor-wills. (Camprimulgus vociferus! The population has found in two adult phenotypes: rufous(red) and gray. An ornithologist investigates the genetics of this color variation and discovers that rufous individuals are one of two genotypes (RR or Rr), while gray individuals are rr. The frequency of these genotypes at Generation 1 in northern New Jersey are: 60% RR, 20%Rr, and 20% rr.

What will the phenotype frequency be in the next generation (Generation+1)?

Answer 1

there will be no change in the allelle frequencies. in the n generation we have p and q, in the n+1 generation we also have p and q

Frequency of (A1A1) in the generation n+1=( p^2​)​2+1/2(2pq.p2)+1/2(2pq.p2)+1/4(2pq)2

=p⁴​+p³q+p³​q+p²q^​2=p²(p²+2pq+q²)=p²

^​​the frequency of A₁A₁ does not change between generation n and n+1 (same demonstration for A₂A₂​ and A₁A₂ genotypes). the genotype structure ​structure undergoes no further changes once the population reaches Hardy-Weinberg equilibrium. So the phenotypic frequencies remains same 80 % rufous and 20 % gray

or
​ furthermore here in the question frequency of dominant allelle A is P²​+pq, sine p+q=1 or q=1-p
​then above equation becomes p²​+p (1-p)= p²+p-p²​=p, ie the p stays the same from one generation to the next and same will be applicable for q also