A certain disease is caused by a recessive mutant allele. (The wild-type allele is dominant.) However,...
A certain disease is caused by a recessive mutant allele. (The wild-type allele is dominant.) However, the penetrance of the disease is 75%. Two individuals known to be heterozygotes have a child. What is the probability that the child exhibits the disease?
Homework Answers
Normal = AA, Aa
Affected = aa
Aa × Aa
| Gametes | A | a |
| A | AA normal | Aa normal |
| a | Aa normal | aa affected |
Probability of affected child = 1/4
But the disease is only 75% penetrant. So, the probability will be 75% of 1/4 = 75/100 × 1/4 = 3/4 × 1/4 = 3/16.
Answer is 3/16.
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