1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    variables:
    n= # of offsprings
    k= certain phenotype

    binomial coefficient:
    n!/k!(n-k)!

    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

    Ygggdrasil

    User Avatar
    Science Advisor

    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

    Ygggdrasil

    User Avatar
    Science Advisor

    Then your second equation is correct if you change to k=2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Genetic probability
  1. Genetic probability (Replies: 1)

Loading...