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

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 –