Question

A couple of normal ancestry has two normal children and an infant affected with sickle cell anemia. The sister of the husband wants to marry the brother of the wife. In such a mating, what would be the probability of their first child having sickle cell anemia?

Answer 1

Sickle cell anemia is an autosomal recessive disorder. The blood group of person having this disease will be Hb^SHb^S.

So, if a child has this blood group, it means that both of his parents are carriers, with the blood group Hb^SHb^A.

So, the cross was Hb^SHb^A X Hb^SHb^A

The normal children will have blood group Hb^AHb^A. Note that chances of having normal: affected children are 3:1.

This further implies that one of the parents of the wife and husband are also carriers of the disease.

Parents of the wife and husband have the genotype: Hb^SHb^A X Hb^AHb^A i.e. one parent is normal and the other is a carrier. The result of the above cross will be:

1(Hb^AHb^A normal) : 2 (Hb^SHb^A, carriers) : 1 (Hb^SHb^S, affected). Note that chances of having a carrier child are 50% or ½, and chances of having an affected child from the carrier are ¼ or 25%.

So, if the sister of the husband marries the brother of the wife, the probability of their first child having sickle cell anemia = (chances of being a carrier * chances of having an affected child from the carriers) = ½ * ¼ = 1/8 (12.5%).