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

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