Suppose a species of bird called the red-crested warbler has a plumage length that is controlled by a single gene. the plm allele produces long plumage and is dominant over the plm allele. one population exists in north america (na) and a separate population exists in south america (sa). the trait is in hardy–weinberg equilibrium in each population. an island nature preserve brings in 72 na warblers and 252 sa warblers. out of the 72 na birds, 55 have long plumage. out of the 252 sa warblers, 75 have long plumage. after the birds from the combined populations mate randomly, the island population produces 1000 offspring. calculate the number of these offspring that are expected to have long plumage. round your answer to two significant figures.

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)

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