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Bayes Theorem in Genetic Counseling

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The picture shows a pedigree (as good as the dot software allows) of a family with a recessive x-linked disorder. The mutated X-chromosome is symbolized by the small letter x while the normal X has a capital letter. Each female (symbolized by a circle) has two X-chromosomes and is healthy even if she carries a mutated x.

graphviz image

The question posed to Bayes analysis is what is the risk that the two females—the patients sister (3)and his nice (5)—in this pedigree are carriers of the mutated allele. Modified from Bayesian Analysis and Risk Assessment in Genetic Counseling and Testing[1].

The Bayes theorem states:

P(A|B) = \frac{P(A)P(B|A)}{P(B)}

A priori probability

We call P(A) the a priori probability. In our case it is the probability that the patient's sister (3) is a carrier. This probability is exactly 1/2 or 50%.

Conditional probability

Next, we call P(B|A) the conditional probability, also called likelihood[2], of of event B (three healthy sons) if the is a carrier. This probability is 1/8 or 12.5% as there exist exactly 8 possibilities to inherit the x chomosomes to her three sons and in only one all three of them are healthy. The event B|A (B given A) is a subset with only one element from the set of possible events C.

C = \{xxx,Xxx,xXx,xxX,XXx,xXX,XxX,XXX\};

B|A = \{XXX\};

B|A \in E

By use of the cardinality symbol \#E we can write

P(B|A) = \frac{\#B|A}{\#E} = 1/8

Joint probability

The term "joint" probability is a little bit misleading in general Bayesian analysis, so some other publications prefer the term normalizing factor [2], but it works fine with this example as it is the probability of the mother (3) having three healthy sons regardless her own carrier status. Two datasets can be distinguished one (E) as above if she is a carrier and one (N) if she is non-carrier. For better distinguishing her two X chromosomes we use indexes.

C = \{x_{c1}x_{c1}x_{c1},X_{c2}x_{c1}x_{c1},x_{c1}X_{c2}x_{c1},x_{c1}x_{c1}X_{c2},X_{c2}X_{c2}x_{c1},x_{c1}X_{c2}X_{c2},X_{c2}x_{c1}X_{c2},X_{c2}X_{c2}X_{c2}\};

N = \{X_{n1}X_{n1}X_{n1},X_{n2}X_{n1}X_{n1},X_{n1}X_{n2}X_{n1},X_{n1}X_{n1}X_{n2},X_{n2}X_{n2}X_{n1},X_{n1}X_{n2}X_{n2},X_{n2}X_{n1}X_{n2},X_{n2}X_{n2}X_{n2}\};

As we do nor know her carrier status, we have to consider all these possibilities, the union set J of sets C and N. (Here is where the term "joint" makes sense.)

J = C \cup N

The set B of three healthy sons contains the whole set N and one element of set C.

B = \{X_{n1}X_{n1}X_{n1},X_{n2}X_{n1}X_{n1},X_{n1}X_{n2}X_{n1},X_{n1}X_{n1}X_{n2},X_{n2}X_{n2}X_{n1},X_{n1}X_{n2}X_{n2},X_{n2}X_{n1}X_{n2},X_{n2}X_{n2}X_{n2},X_{c2}X_{c2}X_{c2}\};

Again using cardinality we receive.

P(B) = \frac{\#B}{\#J} = 9/16

Posterior probability

The posterior probability is the probability that we receive if putting all this into Bayes formula.

P(A|B) = \frac{P(A)P(B|A)}{P(B)} = \frac{1/2 \cdot 1/8}{9/16} = \frac{16}{2 \cdot 8 \cdot 9} = 1/9

As to now we solved the question what is the probability that the patient's sister (3) is a carrier given she has three healthy sons. It is 1/9 \approx 11\%.

Now we have to calculate her daughter's (6) probability being a carrier. Of course we may use the entire Bayesian process again, but it is easier to calculate that if her mother is a carrier she has a 50% chance to be a carrier too. That is her probability being a carrier is 1/18 \approx 5.6\%.

Besides we receive the same result with Bayes formula taking into account that her prior probability is 1/4 (half her mothers), and all the other terms remain the same.


Tags: Genetics Statistics


Categories: Mathematics

 
   

(c) Mato Nagel, Weißwasser 2004-2013, Disclaimer