Question

Huntington's disease is an inherited autosomal dominant disorder that can affect both men and women. Imagine a couple has had seven children, and later in life, the husband develops Huntington's disease. He is tested and it is discovered he is heterozygous for the disease allele, Hh. The wife is also genetically tested for the Huntington's disease allele, and her test results show she is unaffected, hh.
What is the percent probability that the first child of this couple will have Huntington's disease?
probability: %

What is the percent probability that four of the seven children will have Huntington's disease?
probability: %

Answer 1

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%

Genetical probabilities

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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