Calculating Probability of Detinogenesis Imperfecta in Children

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Discussion Overview

The discussion revolves around calculating the probability of two children having dentinogenesis imperfecta while four are normal. It includes aspects of probability theory, specifically the binomial distribution, and touches on genetic inheritance patterns related to autosomal dominant traits.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a formula for calculating the probability using the binomial coefficient and expresses uncertainty about the correct value of k, questioning whether it should be 1 or 2.
  • Another participant clarifies that the binomial coefficient should use n = 6 and k = 2, indicating this represents the arrangement of two affected children and four normal children.
  • A third participant confirms that both parents are heterozygous dominant for dentinogenesis imperfecta, which is relevant for determining the probability of offspring traits.
  • There is a discussion about the implications of X-linked genes, noting that males affected by such genes would express the trait while females could be carriers unless the condition is X-linked dominant.

Areas of Agreement / Disagreement

Participants generally agree on the use of the binomial coefficient but have not reached a consensus on the implications of parental genotypes for determining the probabilities. There is also ongoing discussion regarding the inheritance patterns of X-linked genes.

Contextual Notes

The discussion lacks clarity on the exact probabilities associated with the trait, as the genotype of the parents influences these calculations. Additionally, the assumptions regarding the inheritance of X-linked traits are not fully resolved.

Who May Find This Useful

Students studying genetics, probability theory, or those interested in the inheritance patterns of genetic disorders may find this discussion relevant.

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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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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