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Ann [662]
3 years ago
5

Huntington's disease is an autosomal dominant trait. Given the pedigree below, if individual IV-4 has three children with a norm

al woman, what is probability that they would have at least one child with the disorder
Biology
1 answer:
Bas_tet [7]3 years ago
8 0

Answer:

The answer to the given question is =7/8

Explanation:

In autosomal dominant traits, one copy of the affected gene is enough to cause the disease. Let ’A’ be the affected gene, ‘a’ be the non affected gene. Since IV-4’s parents are a couple of affected and non-affected. So, he has the genes Aa, i.e. a single affected gene.  

A a

a aA aa

a aA aa

Number of children affected=\frac{1}{2}

Number child affected out of three = (\frac {1}{2}) ^{3}

At least one child affected out of three = 1-P (number of children affected out of 3 )

                                                                      = 1-(1/2)^{3}

                                                                     = \frac{7}{8}  

So, the answer to the given question is 7/8  

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