We are going to have different allele frequencies for the two populations
North American:
72 birds total => 144 alleles
72 - 55 birds = 17 short plume birds.
q^2 = (2*17) / 114 = 0.236
Freq(short plume allele) = q = 0.486
Freq(long plume allele) = p = 1 - q = 0.514
From this North American population we get
0.486 * 114 = 55 long plume alleles
0.514 * 114 = 59 short plume alleles
South American:
252 birds total => 504 alleles
252 - 75 birds = 177 short plume birds.
q^2 = (2*177) / 504 = 0.702
Freq(short plume allele) = q = 0.838
Freq(long plume allele) = p = 1 - q = 0.162
From this South American population we get
0.162 * 504 = 82 long plume alleles
0.838 * 504 = 422 short plume alleles
Blended population has
55 + 82 = 137 long plume alleles
59 + 422 = 481 short plume alleles
137 + 481 = 618 total alleles
p = 137/618 = 0.222
q = 481/618 = 0.778
The new population has the p and q from the blended population. The new population has 1000 individuals. The portion of long plumed will be homozygote long plumage + heterozygote, which is p^2 + 2pq, and you multiply that by the population size 1000 to get the final answer.
population size * (p^2 + 2pq) = 1000 * (0.222^2 + 2*0.222*0.778) = 395 birds (answer)
