# Genetic probability

1. Sep 8, 2009

### Anthem26

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. Sep 8, 2009

### Ygggdrasil

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.

3. Sep 9, 2009

### Anthem26

Both parents are heterozygous dominant and have the disease for dentinogenesis imperfecta.

4. Sep 9, 2009

### Ygggdrasil

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