genetics problem about sickle cell disease and probability?

Question

Karen and Steve each have a sibling with sickle cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to reveal the 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.

The answer they give in the back of my book is 1/9 but my question is WHY???

Answer 1

i know that if both partners have AS as there genotype it is either one of the children that have it or all of them. i don't understand the remainig part of the question

Answer 2

Okay here it is! Because both Karen and Steve are unaffected but have affected siblings, all four parents must be carriers. This would mean an Ss genotype. Given this a punnet square tells you that both Karen and Steve have a 2/4 chance of being a carrier. But since we know that they are not affected we can eliminate one of the possibilities. This means that the chance of them being carriers is actually 2/3. We also know that the only way they can have an affected child is if they are both carriers. Again a punnet square tells us that the probability of them having an affected child is 1/4. So... The answer comes from 2/3 x 2/3 x 1/4. That is the probability that He is a carrier, times the probability that she is a carrier, times the probability that if they are both carriers they will have an affected child. Yay!

Answer 3

It didn't sound correct, but I just figured it out! There are three possible combinations of K. and S,'s gene make-up. They are both carriers, only one is a carrier, or neither is a carrier. Only the first scenario could produce a child with the disease. So we are down to 1/3. There are three genetic poss., child with disease, child as carrier, or normal child. So, 1/3 times 1/3 equals 1/9.

Answer 4

Karen and Steve have the same histories, and therefore the same possible genotypes.

SS = sickle cell disease
SN = sickle cell trait
NN = normal hemoglobin

Each has a sibling with SS, so their parents must be SN (parents don't have the disease, but each parent must have one S ... SN is the only genotype possible for the parents).

Neither Karen nor Steve have sickle cell disease, so they are not SS. Parents SN x SN can produce offspring SS, SN, SN, NN which are the predicted genotypes in the Punnett square. Karen and Steve aren't SS, but they could be any one of the other three.

Karen: SN, SN, or NN
Steve: SN, SN, or NN
I'm naming the SN twice, because two of the boxes from the Punnett square are heterozygous.

The rest of the solution involves working the Punnett squares for all of Karen's and Steve's possible genotypes.

Karen's first SN x Steve's first SN -------> 1 SS out of 4
Karen's first SN x Steve's second SN --> 1 SS out of 4
Karen's first SN x Steve's NN -------------> 0 SS out of 4
Karen's second SN x Steve's first SN -------> 1 SS out of 4
Karen's second SN x Steve's second SN --> 1 SS out of 4
Karen's second SN x Steve's NN -------------> 0 SS out of 4
Karen's NN x Steve's first SN -------> 0 SS out of 4
Karen's NN x Steve's second SN --> 0 SS out of 4
Karen's NN x Steve's NN -------------> 0 SS out of 4

Total up the results from all these boxes in the 9 Punnett squares. All together, only 4 of the boxes are SS out of the 36 boxes in the nine squares. 4/36 = 1/9

Whew.
And, sadly enough, that was fun.