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Genetic diseases probability help

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Some genetic diseases (e.g. hemophilia, muscular dystrophy) can be transmitted to the child only if the mother is a carrier; the father cannot transmit the disease to the child, even if he is a carrier. This type of disease affects only males, carrier females rarely
    exhibit any symptoms. In some situations, it can be determined that the probability that
    a woman is a carrier for the disease is 1/2 (for instance if the woman is the daughter of a
    known carrier of the disease). The probability that she will pass this disease to a son is also 1/2 (since the disease cannot be passed from father to son). The birth of a normal son adds evidence, but certainly not conclusive evidence, that the mother is not carrier of the disease gene. What is the probability that a women is a carrier, given that she has one normal son?

    2. Relevant equations

    Pr(A/B) = Pr(AnB)
    Pr(B)

    3. The attempt at a solution

    Let C - That a woman is a carrier for the disease.
    Therefore, Pr(C) = 0.5

    Let S - The probability that a woman passes the disease to her son
    Therefore, Pr(S) = 0.5

    So, the probability that the women is a carrier, given that she has one normal son(complement of S) would be:
    Pr(C/S^c) = Pr(CnS^c) ,
    Pr(S^c)
    Where Pr(S^c) = 1- Pr(S) = 1-0.5 = 0.5

    How is Pr(C n S^c) calculated?
     
  2. jcsd
  3. Sep 25, 2007 #2

    D H

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    Staff Emeritus
    Science Advisor

    First a few comments:
    This was true in the past because males used to die before they could father a child. The daughters of male hemophiliacs are carriers, and may be female hemophiliacs if the mother is a carrier. The status of male offspring of male hemophiliacs depends solely on their mother.

    You have the probability correct but not the reason. A female carrier has one normal and one defective X chromosome. The probability that she will pass the defective chromosome to any one of her offspring is 50%.

    Better stated as what is the probability that the daugher of a known carrier is also a carrier, given that ...

    This is a conditional probability. The probability that the woman passes the disease to her son is zero if she is not a carrier. Thus you should use Pr(S|C) rather than Pr(S).

    Do you know Bayes' Law?
    I will denote the birth of a normal son as ~S. Therefore you want to calculate the prior probability Pr(C|~S). Apply Bayes' Law.
     
  4. Sep 25, 2007 #3
    So, since the woman cannot pass on the disease if she doesn't have it:
    Pr(S/C^c) = 0, and Pr(~S/C^c) = 0

    If the woman has the disease,then the prob. that she has a son with the disease is:
    Pr(S/C) = 0.50
    And the prob. that she has a son without the disease is:
    Pr(~S/C) = 1- 0.50 = 0.50

    And if the probability that she passes it on is: Pr(C) = 0.50, then Pr(C^c) = 1- 0.5 = 0.50


    According to Bayes' Law:

    Pr(C/~S) = [Pr(Cn~S)] / [Pr(~S)]
    = [Pr(~S/C) P(C)] / [Pr(~S/C) Pr(C) + Pr(~S/C^c) Pr(C^c)]
    = [0.50 (0.50)] / [0.50(0.50) + 0(0.50)]
    = 0.25/0.25
    = 1
    This answer doesn't make sense to me, but I can't see where I went wrong.
     
  5. Sep 25, 2007 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Your mistake is in bold.

    The probability that a non-carrier mother will have a non-diseased son is not zero.
     
  6. Sep 25, 2007 #5
    Okay - it's 1 right?
    Thank you for all of your help!!!
     
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