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Homework Help: Genetic probability

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    What is the probability that two children have detinogenesis imperfecta and four are normal

    2. Relevant equations
    n= # of offsprings
    k= certain phenotype

    binomial coefficient:

    This probability is given by:
    Prob(Offspring has trait)^k*Prob(Offspring does not have trait)^(n-k)

    other valuable information:
    Dentinogenesis imperfecta is an autosomal dominant tooth disorder.

    3. The attempt at a solution
    where n=6 and k=1? (I'm unsure of k, would it be 2 because there's 2 out of 6 offsprings who exhibit the imperfecta, or just one because it's either this phenotype or the other - w/ or w/o imperfecta?)

    A. First, I use the binomial coefficient to get the orders of offspring:
    n!/k!(n-k)! = 6!/1!(6-1)! = 6 5 4 3 2 1/1 (5 4 3 2 1) = 6 / 1 = 6

    B. After so, I implement the probability equation
    Prob(Offspring has trait)^k*Prob(Offspring does not have trait)^(n-k)
    (3/4)^1*(1/4)^5 = .000732422

    Multiply the answers of part A & B to get the answer.
    6*.000732422= .00439??? Would this be correct?

    Also I have another question concerning X-linked genes. If a male inherits the gene, he's affected because he can only carry one X chromosome, but if a female inherits just one of them she would be a heterozygous carrier unless the disease was x-linked dominant (she would need to have 2 of the genes in order to be diseased correct)?
  2. jcsd
  3. Sep 8, 2009 #2


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    2017 Award

    First, the binomial coefficient you need to calculate is n = 6, k = 2. This basically tells you the numbers of ways of arranging two affected children and four normal children (alternatively, you could calculate the binomial coefficient for n = 6 and k = 4 and it would give you the same answer).

    Second, what is the genotype of the parents? Without that, I don't know whether your numbers for Prob(Offspring has trait) is correct.
  4. Sep 9, 2009 #3
    Both parents are heterozygous dominant and have the disease for dentinogenesis imperfecta.
  5. Sep 9, 2009 #4


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    2017 Award

    Then your second equation is correct if you change to k=2.
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