Question

Suppose a species of bird called the red‑crested warbler has a plumage length that is controlled by a single gene. The P????mPlm allele produces long plumage and is dominant over the p????mplm short plumage 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 112 NA warblers112 NA warblers and 452 SA warblers.452 SA warblers. Out of the 112 NA birds,112 NA birds, 55 have long plumage. Out of the 452 SA warblers,452 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 the number to two significant figures.

Answer 1

Check the explanation

Step-by-step explanation:

North American:

112 birds total => 224 alleles

112 - 55 birds = 57 short plume birds.

`q^2`= (2*57) / 224 = 0.5089

Freq(short plume allele) = q = 0.7134

Freq(long plume allele) = p = 1 - q = 0.2866

From this North American population we get

0.2866 * 224 = 64 long plume alleles

0.7134 * 224 = 160 short plume alleles

South American:

452 birds total => 904 alleles

452 - 75 birds = 377 short plume birds.

`q^2` = (2*377) / 904 = 0.8341

Freq(short plume allele) = q = 0.9133

Freq(long plume allele) = p = 1 - q = 0.0867

From this South American population we get

0.0867 * 904 = 78 long plume alleles

0.9133 * 904 = 826 short plume alleles

Blended population has

64 + 78 = 142 long plume alleles

160 + 826 = 986 short plume alleles

142 + 986 = 1128 total alleles

p = 142/1128 = 0.1259

q = 986/1128 = 0.8741

All right. 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 * (0.1259^2 + 2*0.1259*0.8741) = 236 birds