Editing Bayes' theorem (section)

=== Using genetic test results ===
Parental genetic testing can detect around 90% of known disease alleles in parents that can lead to carrier or affected status in their children. Cystic fibrosis is a heritable disease caused by an autosomal recessive mutation on the CFTR gene,<ref>"Types of CFTR Mutations". Cystic Fibrosis Foundation, www.cff.org/What-is-CF/Genetics/Types-of-CFTR-Mutations/.</ref> located on the q arm of chromosome 7.<ref>"CFTR Gene – Genetics Home Reference". U.S. National Library of Medicine, National Institutes of Health, ghr.nlm.nih.gov/gene/CFTR#location.</ref>

Here is a Bayesian analysis of a female patient with a family history of cystic fibrosis (CF) who has tested negative for CF, demonstrating how the method was used to determine her risk of having a child born with CF: because the patient is unaffected, she is either homozygous for the wild-type allele, or heterozygous. To establish prior probabilities, a Punnett square is used, based on the knowledge that neither parent was affected by the disease but both could have been carriers:
{| class="wikitable" style="text-align:center;"
! {{diagonal split header|<br /><br />Father|Mother}}
!W
Homozygous for the wild-<br />type allele (a non-carrier)
!M
Heterozygous<br />(a CF carrier)
|-
!W
Homozygous for the wild-<br />type allele (a non-carrier)
|WW
|MW
|-
!M
Heterozygous (a CF carrier)
|MW
|MM

(affected by cystic fibrosis)
|}
Given that the patient is unaffected, there are only three possibilities. Within these three, there are two scenarios in which the patient carries the mutant allele. Thus the prior probabilities are {{frac|2|3}} and {{frac|1|3}}.

Next, the patient undergoes genetic testing and tests negative for cystic fibrosis. This test has a 90% detection rate, so the conditional probabilities of a negative test are 1/10 and 1. Finally, the joint and posterior probabilities are calculated as before.
{| class="wikitable" style="text-align:center;"
!Hypothesis
!Hypothesis 1: Patient is a carrier
!Hypothesis 2: Patient is not a carrier
|-
!Prior Probability
|2/3
|1/3
|-
!Conditional Probability of a negative test
|1/10
|1
|-
!Joint Probability
|1/15
|1/3
|-
!Posterior Probability
|1/6
|5/6
|}
After carrying out the same analysis on the patient's male partner (with a negative test result), the chance that their child is affected is the product of the parents' respective posterior probabilities for being carriers times the chance that two carriers will produce an affected offspring ({{frac|1|4}}).