1. The percent probability that the first child of the couple will have Huntington's disease is 50%
2. The percent probability that four of the seven children will have Huntington's disease is 6.25%
<h3>Genetical probabilities</h3>
The disease is autosomal dominant. Only one dominant allele is needed for the disease to be manifested.
The husband is Hh and the wife is hh.
Hh x hh
Hh Hh hh hh
Hh (affected) = 1/2
hh (unaffected) = 1/2
1. Probability that the first child of this couple will have Huntington's disease is the same as the probability of having the disease. Thus, it is 1/2 (50%)
2. Probability that four of the seven children will have Huntington's disease is the same as the probability of having the disease in 4 places.
1/2 x 1/2 x 1/2 x 1/2 = 1/16 or 6.25%
More on genetic probabilities can be found here: brainly.com/question/13764507
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