Question

Karen and Steve each have a sibling with sickle-cell disease (recessive disorder). Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.

Answer 1

To first understand your question, here are some vocabulary terms I will be using throughout this problem.

• Homozygous dominant: (BB) This is not having sickle cell disease.
• Homozygous recessive: (bb) This is having sickle cell disease.
• Heterozygous: (Bb) This is being a carrier for sickle cell disease.

We know that having sickle cell disease is a recessive disorder so it will be homozygous recessive (bb).

We also know that Karen and Steve each have a sibling with sickle cell disease, but they don't. This means that both of their parents have to be heterozygous (Bb), or a carrier for the disease.

If their parents are both heterozygous (Bb), there is a 1/4 chance that Karen and Steve are completely healthy (BB) and a 2/4 chance that they are carriers (Bb). We know for a fact that they aren't sick so that is out of the question.

This means that the probability of Karen and Steve being a carrier for this disease is 2/3.

Now, we can make a Punnett square for the probability of their children, which I have attached below. Since being heterozygous is most likely, I made both Karen and Steve heterozygous in the Punnett square.

This means that their children are 1/4 likely to be completely healthy (BB), 2/4 likely to be a carrier (Bb), and 1/4 likely to have sickle cell disease (bb).

Now we can multiply the probability of both of Karen and Steve being a carrier for sickle cell, which is 2/3 each, with the probability of their children having it, which is 1/4.

2/3 × 2/3 = 4/9

4/9 × 1/4 = 1/9

The final probability of Karen and Steve having a child with sickle cell disease is 1/9.