Consider an autosomal recessive disease with an incidence of 1/10,000 in the general population of 100,000. Your best friend comes to you very upset because he has just taken a screening test for this disease and gotten a positive result. He is convinced he is a carrier, despite having no family history of the disease. You try to reassure him, but he says, "Don't bother. The Clinic said this test has 98% sensitivity and 90% specificity. With that level of sensitivity, it must be correct!" What is the chance your friend is NOT a carrier?

I have attempted it in this way: The reason I used Hardy-Weinberg is because there isn't any other way to determine whether he is a carrier or not.

Case 1: He is AA (double dominant). This has a 0.99^2 = 0.9801 chance. The chance that his result was positive is 1 - 0.90 = 0.1 since that is the chance that the specificity failed since specificity measures the chance of true negatives. Multiplying this gives us a 0.09801 probability of this case.

Case 2: He is Aa (heterozygous). This has a 2*0.99*0.01 = 0.0198 chance. Again his chance of a positive result should be 1 - 0.9 = 0.1. Multiplying this gives us 0.00198.

Case 3: He is aa (double recessive). This has a 0.01^2 = 0.0001 chance. The chance of getting a positive result is 0.98. Multiplying we get 0.000098.

Adding these up, I have 0.100088. We want the probability he is not a carrier, so 0.000098+0.09801 (cases 1 and 3) divided by 0.100088 (the total). This yields around 0.9802 or 98% chance, which isn't one of the choices. What am I doing wrong?

  • 3
    $\begingroup$ Welcome To Biology.SE. Homework questions are off-topic on Biology unless you have shown your attempt at an answer. For more information see our homework policy. Hint1: You have to use conditional probabilities. Hint2: The question has nothing to do with Hardy-Weinberg equilibrium (unlike you seem to think given the tags used), it is a pure mathematical probability question. $\endgroup$ – Remi.b Mar 24 '17 at 3:56
  • $\begingroup$ Hint 3: Seriously and truly, this question has nothing to do with Hardy-Weinberg. Hint 2 is a solid hint. Try rethinking your approach. What if your friend really has the disease? What if your friend doesn't? What is going to happen with the test in each case? What do specificity and sensitivity mean? $\endgroup$ – Bryan Krause Mar 24 '17 at 19:58
  • $\begingroup$ Does my logic make sense if I do it this way? 1-0.9 (failing of specificity to catch someone without the disease) = 0.1. 0.1*2 (since AA and Aa are two cases that are both cases without the disease) = 0.2 is the chance that he is either AA or Aa. The third probability is 0.98, the sensitivity to aa having the disease. Adding these we have 1.18, and AA + aa = 1.08. 1.08/1.18 giving 0.915. $\endgroup$ – Derrick Liang Mar 25 '17 at 4:02

Your Answer

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

Browse other questions tagged or ask your own question.