Jane and Bill are married. Each of them has a brother with cystic fibrosis, which is autosomal recessive. Neither Jane, Bill or any of their parents have the disease. Based on this information, what is the probability that if this couple has a child, the child will have cystic fibrosis?
Are both Jane and Bill carriers?
I got 1/4 (25%) but I'm not sure if this is the answer.
Wait I think the answer may be 1/9
It is possible that both Jane and Bill could be homozygous for the dominant trait and not have cystic fibrosis. Jane and Bill may not be carriers.
However
There is a 2/3 chance Bill could be a carrier.
There is a 2/3 chance Jane could be a carrier.
If Jane and Bill (both as carriers) has a child there is a 1/4 chance from them having a child with cystic fibrosis
Multiplying all the probabilities, 2/3 X 2/3 X 1/4 = 1/9
sS xsS>>ss sS Ss SS yes 1:4
thank you
To determine the probability of their child having cystic fibrosis, we first need to assess whether Jane and Bill are carriers of the disease.
Given that both Jane and Bill have a brother with cystic fibrosis, there is a reasonable assumption that they are carriers of the disease since it is autosomal recessive. This means that they have inherited one copy of the cystic fibrosis gene (recessive allele) and one copy of a normal gene (dominant allele). However, without explicit information about their genetic status, we cannot be certain whether they are carriers or not.
To calculate the probability of their child having cystic fibrosis, we need to consider the possible combinations of gene transmission from Jane and Bill. Since neither Jane nor Bill have cystic fibrosis, they can only be carriers if they inherit one copy of the cystic fibrosis gene from one of their parents.
Let's assume that both Jane and Bill are carriers:
Jane: Cc (Carrier)
Bill: Cc (Carrier)
The Punnett square can help us determine the potential outcomes:
C c
C CC Cc
c Cc cc
In this Punnett square, each column and row represent the potential alleles that each parent can contribute to their child. The "C" represents the dominant allele (normal gene), and the "c" represents the recessive allele (cystic fibrosis gene).
The results of the Punnett square show us there is a 25% chance (1 in 4) that their child will inherit two copies of the recessive allele (cc), which leads to cystic fibrosis.
So, based on the assumption that both Jane and Bill are carriers, the probability of their child having cystic fibrosis is indeed 1/4 or 25%.
However, please note that this probability is contingent on both Jane and Bill being carriers, which is an assumption based on the given information. If they are not carriers, the probability may vary. It is always best to consult with a medical geneticist or genetic counselor for a more accurate assessment of the probability in specific cases.