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

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.  

Aa

aaAaa

aaAaa

Number of children affected=

Number child affected out of three =

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

                                                                      =

                                                                     =  

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

i think the answer is b.