Huntington's disease is an autosomal dominant trait. Given the pedigree below, if individual IV-4 has three children with a normal woman, what is probability that they would have at least one child with the disorder

Respuesta :

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=[tex]\frac{1}{2}[/tex]

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

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

                                                                      = [tex]1-(1/2)^{3}[/tex]

                                                                     = [tex]\frac{7}{8}[/tex]  

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

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