Answer:
Frequency of individuals with AA genotype is ![0.81](https://tex.z-dn.net/?f=0.81)
Explanation:
As per Hardy-Weinberg equation, frequency of allele "a" is ![0.1](https://tex.z-dn.net/?f=0.1)
Which means that ![q=1](https://tex.z-dn.net/?f=q%3D1)
where "q" denotes frequency of allele "a"
As per I equation of Hardy-Weinberg -
![p+ q =1\\](https://tex.z-dn.net/?f=p%2B%20q%20%3D1%5C%5C)
where "p" denotes frequency of allele "A"
Substituting the value of "q" we get ,
![p+0.1=1\\p=1-0.1\\p=0.9](https://tex.z-dn.net/?f=p%2B0.1%3D1%5C%5Cp%3D1-0.1%5C%5Cp%3D0.9)
Frequency of individuals with AA genotype is represented as
which is equal to
![0.9^2\\= 0.81](https://tex.z-dn.net/?f=0.9%5E2%5C%5C%3D%200.81)
Frequency of individuals with AA genotype is ![0.81](https://tex.z-dn.net/?f=0.81)
B. Jim is revealing to Huck that most people are not what they seem.
I think, I am sorry if I am wrong ;/
what is the topic of essay? you have not mentioned the topic , then what is the purpose of giving all the other information.