A zebra’s diet consists of grasses, shrubs, and leaves. A zebra doesn’t eat other animals to survive. What kind of an animal is
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To first understand your question, here are some vocabulary terms I will be using throughout this problem.
- Homozygous dominant: (BB) <em>This </em><u><em>is not</em></u><em> having sickle cell disease.</em>
- Homozygous recessive: (bb) <em>This </em><u><em>is</em></u> <em>having sickle cell disease.</em>
- Heterozygous: (Bb) <em>This is being a </em><u><em>carrier</em></u><em> for sickle cell disease.</em>
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 <em>both</em> 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).
<u>Now we can multiply the probability of both of Karen and Steve being a carrier for sickle cell, which is </u><u>2/3 each</u><u>, with the probability of their children having it, which is </u><u>1/4</u><u>.</u>
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 <u>1/9</u>.
