Homework Answers
C). Given that both the parents are heterozygous. Cross between them will result in the offspring with the following genotypes.
Aa* Aa -----> A_ (3/4, normal), aa (1/4, affected)
The probability of having all 4 four children with normal phenotypes is, = (3/4)^4 = 81/256
d). The probability of having at least one affected children is = 1 - probability of having no affected children = 1-81/256 = 175/256.
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