-2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Welcome to Biology.SE! Please take the tour and then go through the help pages starting with How to Ask questions effectively on this site. In general, we expect you to do some research on your own and then, informed by what you have learned, ask any questions you still have (ideally with references to reliable sources). For "homework" questions, you are required to show your attempt to answer the question and to use the "homework" tag. (Note that "homework" can apply to questions even if they are not assigned as homework.) Thanks! 😊 $\endgroup$ – tyersome Oct 27 at 3:18
  • 2
    $\begingroup$ Hi, So I have tried solving this by myself. 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. I got a 1/15 chance, however the correct answer was 1/6, so I'm not sure what I'm missing here $\endgroup$ – ZAINAB A Oct 27 at 17:51
  • $\begingroup$ Please edit that information into your post and include your reasoning for each step (e.g. where the 2/3 came from) — comments are ephemeral and often ignored so they can not be used to convey essential information. Please also check out the links from my first comment. $\endgroup$ – tyersome Oct 28 at 4:52
0
$\begingroup$

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)

| improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.