Calculating Probability of Detinogenesis Imperfecta in Children

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The discussion focuses on calculating the probability of two children having dentinogenesis imperfecta while four are normal, using the binomial coefficient and probability equations. The participants clarify that for n=6 (total offspring) and k=2 (affected children), the binomial coefficient is calculated as 6, representing the arrangements of affected and normal children. The probability equation is applied, resulting in a probability of approximately 0.00439 for this specific genetic outcome. Additionally, there is a discussion about X-linked genes, noting that males are affected if they inherit the gene, while females can be carriers unless the condition is X-linked dominant. Understanding the parents' genotypes is crucial for accurate probability calculations.
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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)?
 
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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.
 
Both parents are heterozygous dominant and have the disease for dentinogenesis imperfecta.
 
Then your second equation is correct if you change to k=2.
 
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