A 26-year-old woman of Norwegian descent seeks genetic counseling. Her brother died at age eight of documented cystic fibrosis. Both of their parents are deceased. The woman undergoes DNA testing for 70 CF mutations which collectively detects approximately 90% of CF carriers of northern European descent. Testing reveals that she is negative for all 70 mutations. What is the probability that she is a heterozygous carrier of CF?
I recommend to read on Bayes theorem or maybe watch one of many videos on the subject.
I figured that in order for her to be heterozygous carrier, she has a 2/3 chance x by the 10% of catching a rare form of CF mutation that wasn't detected
Based on your comment I am guessing you have correctly identified, the correct (2/3) probability of her being a carrier prior to learning the test result. Also you have correctly identified the probability of her carrying rare CF mutation prior to learning the test result.
Now you can draw a table:
She is a carrier She is not a carrier c d test came out positive e f test came out negative
In this table c, d, e and f are probabilities prior to knowing the test result. So far you have calculated the e = 2/3*10% = 1/15. I will leave to you to calculate the rest of the fields in the table.
After learning the negative test result we can cross out the fields c and d. The list of all possible options has shrunk to e and f. The new probability of her being a carrier given the test came out negative is:
probability of carrying a rare mutation / (probability of carrying a rare mutation + probability of her not being a carrier) = e / (e + f)
There are 4 possibilities:
Family has rare allele, person is a carrier
family has rare allele, person is not a carrier
family has common allele, person is a carrier
family has common allele, person is not a carrier
Work out all these so that they add up to 100%. The testing allows you to eliminate one of those options, so then recalculate.