A disease is caused by homozygosity for the g allele (G is the corresponding wild-type allele). 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?
A) 1/4
B) 3/4
C) 1/8
D) 3/16
E) 9/16
D) 3/16
Explanation: The disease is seen only in gg individuals. Because the disease is 75% penetrant, 75% of gg individuals will show disease symptoms. The chance that the child of two heterozygotes (Gg × Gg) is gg is 1/4. If the child is gg, there is a 75% (or 3/4) chance the child will show disease symptoms. Because both of those events must happen for the child to have the disease, the probabilities are multiplied (1/4 × 3/4) to equal 3/16.
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