In a large population of ruby-throated hummingbirds, it was observed that the frequency of allele H was 0.4 and the frequency of allele h wa

Question

In a large population of ruby-throated hummingbirds, it was observed that the frequency of allele H was 0.4 and the frequency of allele h was 0.6. Calculate the expected genotype frequencies if the population is in Hardy-Weinberg equilibrium. Show your work, i.e., show how you got your anser.

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Josie 6 days 2021-11-21T15:50:52+00:00 1 Answer 0 views 0

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    2021-11-21T15:52:32+00:00

    Answer:

    Frequency of genotype “HH” is 0.16

    Frequency of genotype “hh” is 0.36

    Frequency of genotype “Hh” is 0.48

    Step-by-step explanation:

    It is given that the population of ruby-throated hummingbirds is in Hardy Weinberg’s equilibrium.

    As per second Hardy Weinberg’s equilibrium equation,

    p^2+q^2 + 2pq = 1\\

    where 2 pq is the frequency of heterozygour genotype “Hh”

    p^2 is the frequency of dominant homozygous genotype “HH”

    and q^2 is the frequency of recessive homozygous genotype “hh”

    p is the frequency of allele “H” and q is the frequency of allele “h”

    p^ 2 = 0.4^2 \\p^2 = 0.16

    and

    q^2 = 0.6^2\\q^2 = 0.36

    Substituting the values in above equation, we get –

    0.16 + 0.36 + 2pq = 1\\2pq = 1 - 0.16 -0.36 \\2pq = 0.48

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