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)?