Calculating Probability of Detinogenesis Imperfecta in Children

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In summary: The first question was correct. I think you have your probability equation for this problem correct.In summary, the probability of having two children with dentinogenesis imperfecta and four normal children, given that both parents are heterozygous dominant for the gene, is 6 * (3/4)^2 * (1/4)^4 = 0.00390625.
  • #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)?
 
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  • #2
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
Both parents are heterozygous dominant and have the disease for dentinogenesis imperfecta.
 
  • #4
Then your second equation is correct if you change to k=2.
 

1. What is detinogenesis imperfecta?

Detinogenesis imperfecta is a rare genetic disorder that affects the development of teeth. It can cause teeth to be discolored, weak, and prone to fractures.

2. How is the probability of detinogenesis imperfecta calculated?

The probability of detinogenesis imperfecta is calculated by looking at the genetic makeup of the parents, specifically if they carry a gene mutation that can cause the disorder. It is also important to consider any family history of the disorder.

3. Can detinogenesis imperfecta be prevented?

Currently, there is no known way to prevent detinogenesis imperfecta. However, early detection and treatment can help manage the symptoms and prevent further damage to the teeth.

4. What are the chances of passing on detinogenesis imperfecta to a child?

If both parents carry a gene mutation for detinogenesis imperfecta, there is a 25% chance that their child will inherit the disorder. If only one parent carries the gene mutation, the chances decrease to 50%.

5. Are there any other factors that can affect the probability of detinogenesis imperfecta in children?

Aside from genetic factors, other factors such as environmental influences and lifestyle choices can also play a role in the development of detinogenesis imperfecta. These factors may include poor dental hygiene, exposure to certain substances, and nutritional deficiencies.

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