In Genetics: A Woman Whose Brother Had Cystic Fibrosis Marries a Man Who Had a Child With Cystic Fibros. . .?
A question asked by ihaveques: In Genetics: A woman whose brother had cystic fibrosis marries a man who had a child with cystic fibros. . .?
In Genetics: A woman whose brother had cystic fibrosis marries a man who had a child with cystic fibrosis from a previous marriage. They plan to have 3 children. What is the probabilty that only one of the 3 will have cystic fibrosis?
If you know how to solve this, please help! Please explain how answer is derived also thanks.
Chosen answer:
Answer by CNTB
This requires a long answer that would better be answered in person but I will try.
Check back for edits just in case:
Cystic fibrosis (here on called CF) is an autosomal recessively inherited disease (I assume you know this). I can’t guarantee this is right but bare with me. First we need to make some assumptions. What we know for sure is that in order for a child to have the disease, it must inherit two alleles that are recessive. Since the question only tells us that the child of the man that the woman marries has CF (and that the man himself is not infected, nor does the question tell us the parents of the woman has CF), we can assume that:
1) The man is heterozygous for CF (he has to carry a recessive allele to give it to his child)
2) The woman is either a carrier of CF (heterozygous) or “normal” (no CF copies)- in order to be in a family where her brother had the disease, you assume that both parents were carriers of the disease )
So if we assume ANY family that has either one sibling or one child with CF and assuming the parents don’t have CF, then both parents were carriers (heterozygous). So if you cross two heterozygotes together, there is a 1/4 chance of getting CF, a 1/2 chance of being a carrier and 1/4 chance of being dominant (wild-type/”normal”)
With this information you have two possible pathways (depending on whether the woman is homozygous dominant or heterozygous):
a) If she’s dominant and mates with a heterozygous male, you’d expect that all children would not have CF(do one punnet square). There is a 1/3 chance she is dominant and normal-she can’t have CF so the demoninator of the fraction is reduced from 4 to 3.
b) If she’s a carrier (2/3 chance) you’d expect: 2/8ths of a chance that the child has CF (1/4 chance and another 1/4 chance because there is two times the probability she will be a carrier than wild-type)
BUT we aren’t sure what the genotype of the woman is (dominant homozygous or heterozygous) so we combine all the possibilities together. There is 12 total (3 punnet squares worth, one for each type the mother could be) and so there is a 2/12 or 1/6 chance However, the question asks what are the chances of having one of three children with the disease. So here we go. The chances of getting one child with CF is 1/6 and the chance of not getting one with CF is 5/6. So we need to figure out the probability for each time the couple has a child, and there are several combinations (i.e. order) in which the couple can have their children. The probabilties for each combo is the same.
CF, noCF, noCF (1/6*5/6*5/6)= (25/216)
noCF,noCF,CF = 25/216(same as above, only in a diff order)
noCF,CF,noCF=25/216
Note that there are three possible combos and you must consider them all so multiply 25/216 by 3 (3 combos) and you get: 75/648 or 0.11574 or 11.574% chance
Answer: 11.574% chance of getting 1 of 3 children with CF.
Whether you agree or disagree, why not leave your own thoughts below.
Tagged with: Brother • child • cystic • fibros. • fibrosis • GENETICS • marries • whose • Woman
Filed under: Cystic Fibrosis
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