- #1
Anthem26
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Homework Statement
What is the probability that two children have detinogenesis imperfecta and four are normal
Homework 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.
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)?