Question

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

Answer 1

The answer to the given question is =7/8

Step-by-step 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=
`(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)`

=
`(7)/(8)`

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