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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Arclight Arclight · Answer 1 · 2020-02-25T10:39:39+01:00

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,

[tex]p^2+q^2 + 2pq = 1\\[/tex]

where 2 pq is the frequency of heterozygour genotype "Hh"

[tex]p^2[/tex] is the frequency of dominant homozygous genotype "HH"

and [tex]q^2[/tex] is the frequency of recessive homozygous genotype "hh"

p is the frequency of allele "H" and q is the frequency of allele "h"

[tex]p^ 2 = 0.4^2 \\p^2 = 0.16[/tex]

and

[tex]q^2 = 0.6^2\\q^2 = 0.36[/tex]

Substituting the values in above equation, we get -

[tex]0.16 + 0.36 + 2pq = 1\\2pq = 1 - 0.16 -0.36 \\2pq = 0.48[/tex]