Answer:
Frequency of individuals with AA genotype is [tex]0.81[/tex]
Explanation:
As per Hardy-Weinberg equation, frequency of allele "a" is [tex]0.1[/tex]
Which means that [tex]q=1[/tex]
where "q" denotes frequency of allele "a"
As per I equation of Hardy-Weinberg -
[tex]p+ q =1\\[/tex]
where "p" denotes frequency of allele "A"
Substituting the value of "q" we get ,
[tex]p+0.1=1\\p=1-0.1\\p=0.9[/tex]
Frequency of individuals with AA genotype is represented as [tex]p^{2}[/tex] which is equal to
[tex]0.9^2\\= 0.81[/tex]
Frequency of individuals with AA genotype is [tex]0.81[/tex]
