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

Question

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.

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Eden 1 month 2021-09-17T13:15:33+00:00 1 Answer 0

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    2021-09-17T13:16:37+00:00

    The complete question is;

    Suppose a species of bird called the red-crested warer 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 and in a seperate population in South America. The trait is in hardy-Weinberg equilibrium in each population.

    An island nature preserve brings in 112 NA warblers and 452 SA warblers. Out of the 112 NA birds, 55 have long plumage. Out of the 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:

    the number of these offspring that are expected to have long plumage = 272

    Step-by-step explanation:

    We are going to have different allele frequencies for the two populations

    North American:

    112 birds total => 224 alleles

    112 – 55 birds = 57 short plume birds.

    q² = (2*17)/224 = 0.1518

    Frequency(short plume allele) = q = √0.1518 = 0.3896

    Frequency(long plume allele) = p = 1 – q = 0.6104

    From this North American population we get;

    0.3896 * 224 ≈ 87 long plume alleles

    0.6104 * 224 ≈ 137 short plume alleles

    South American:

    452 birds total => 904 alleles

    452 – 75 birds = 377 short plume birds.

    q² = (2 * 377)/904 = 0.834

    Freq(short plume allele) = q = √0.834 = 0.913

    Freq(long plume allele) = p = 1 – q = 0.087

    From this South American population we get;

    0.087 * 904 ≈ 79 long plume alleles

    0.913 * 904 ≈ 825 short plume alleles

    Blended population has;

    87 + 79 = 166 long plume alleles

    137 + 825 = 962 short plume alleles

    Total alleles = 166 + 962 = 1128

    p = 166/1128 = 0.147

    q = 962/1128 = 0.853

    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 p² + 2pq, and we will multiply that by the population size 1000 to get the final answer.

    Thus;

    population size * (p² + 2pq) =

    1000 * (0.147² + (2*0.147*0.853)) ≈ 272 birds

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