Different allele frequencies are expected in the two groups.
North American:
72 birds altogether => 144 total alleles
From 72, 55 birds have short plumes, hence 17 are short-plumed.
q^2 = (2*17) / 114 = 0.236
Frequency(short plume allele) = q = 0.486
Frequency(long plume allele) = p = 1 - q = 0.514
Thus, from the North American group:
0.486 * 114 = 55 long plume alleles
0.514 * 114 = 59 short plume alleles
South American:
252 birds totaling => 504 alleles
Out of 252 birds, 75 have long plumage, 177 are short-plumed.
q^2 = (2*177) / 504 = 0.702
Frequency(short plume allele) = q = 0.838
Frequency(long plume allele) = p = 1 - q = 0.162
Thus, from this South American group:
0.162 * 504 = 82 long plume alleles
0.838 * 504 = 422 short plume alleles
For the combined population:
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 reflects the p and q from this blended result. It consists of 1000 individuals. The share of long plumed birds will be the sum of homozygous long plumed and heterozygous long plumed, calculated as: p^2 + 2pq, which you multiply by the population count of 1000 for the outcome.
population size * (p^2 + 2pq) = 1000 * (0.222^2 + 2*0.222*0.778) = 395 birds (final result)
