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Sagot :
We have that the the frequency of A1A2 individuals in the offspring is mathematically given as
2pq = 0.65
Frequency
Generally the equation for the Hardy-Weinberg equilibrium is mathematically given as
[tex](P+q)^2=p^2+2pq+q^2\\\\(P+q)^2=1\\\\[/tex]
Therefore
q2= the frequency of individual Al Al
[tex]q2= (72/10+72) \\\\q2= 72/82[/tex]
thus,
P = [10/82]^(1/2)
q= [72/82]^(1/2)
Hence
2pq = 0.65
For more information on frequency visit
https://brainly.com/question/3004869
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