Sickle Cell Mendelian Probability

[biohzd79]

This question has been killing me (Sickle Cell is Autosomal Recessive):

The risk of an African American woman whose sister has Sickle cell Anemia, but who does not have any clinical features of Sickle Cell Disease asks you to determine the risk that she and her husband, who has no familoy history of Sickle Cell Disease will have a child with Sickle Cell Disease. He is also African American and the incidence of Sickle Cell Anemia in the African American population is approximately 1/625. You determine that their risk for having an affected child to be:

Correct answer is 1/75

How do you get the correct answer?????

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[Slack3r]

Bayes theorem maybe? I failed my first Genetics test though, so who knows.

 

[xanthomondo]

This question has been killing me (Sickle Cell is Autosomal Recessive):

The risk of an African American woman whose sister has Sickle cell Anemia, but who does not have any clinical features of Sickle Cell Disease asks you to determine the risk that she and her husband, who has no familoy history of Sickle Cell Disease will have a child with Sickle Cell Disease. He is also African American and the incidence of Sickle Cell Anemia in the African American population is approximately 1/625. You determine that their risk for having an affected child to be:

Correct answer is 1/75

How do you get the correct answer?????

EDIT: the superscripts appear when you type but not in the forum. Anytime there's a number following the variable (p or q) it should be an exponent.

You need to use Hardy-Weinberg equilibrium:
p=frequency of dominant allele, q=frequency of recessive allele
p+q=1, p2 + 2pq + q2 = 1
p2 represents homozygous dom, 2pq = carrier, q2 means homozygous recessive (affected)

SCD requires two recessive alleles
if the incidence of SCD is 1/625, that means q2 = 1/625 (so q = 1/25 and p = 24/25 which you can round to 1)

the chances of the husband being a carrier is 2pq = 2(1/25)(1) = 2/25

Now forget HW equilibrium. The chance of the mother being a carrier is 2/3, do the Punnett square but cross out qq, because shes not affected. 3 possible scenarios exist, pp, pq, qp.

So now we have dads chance of being carrier = 2/25, moms = 2/3. They both have a 1/2 chance of passing it down to the baby - which is the only way it will be affected with SCD.

(2/25 x 1/2 )* (2/3 x 1/2) = 1/75

 

[biohzd79]

Excellent, thank you so much for the answer. It was driving me nuts!