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.

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,

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

is the frequency of dominant homozygous genotype "HH"

and is the frequency of recessive homozygous genotype "hh"

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

and

Substituting the values in above equation, we get -

c. (0.6,-1.4)

step-by-step explanation:

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