Answer;
- 0.81 0r 81%
Step-by-step explanation;
-The frequency of the allele for this condition is 0.10
-It means q =0.1,
but; p+q = 1
p= 1-q
Homozygous dominant
= p²= (0.9)²
= 0.81 or 81%
thus, p =0.9
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