Just to be clear, this is not homework. Suppose a biological family has five children, two of whom have a certain recessive genetic disorder that is not sex-linked. The parents themselves do not have this disorder. I want to estimate the probability that the parents are carriers. How can I do that, short of getting this probability directly from a lab study?

This problem appears to be where Bayesian statistics can help. All I need to know is: 1) the probability of the parents being carriers, given they don't have the disorder, but 2/5 of the children do. 2) The probability of the parents being carriers and 3) The probability of the children having the disorder.

3 I can probably find without too much effort. #2, I'm guessing, is a Hardy-Weinberg problem - I just need to find the relevant statistics for people of the parents' backgrounds. But I have no clue how to start with #1.


closed as off-topic by rg255, AliceD, James, fileunderwater, March Ho May 27 '16 at 10:22

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    $\begingroup$ You should read the homework policy. You will understand why I have voted to close as homework. $\endgroup$ – rg255 May 21 '16 at 6:17


  1. The parents are the real biological parents (no cheating allowed!)
  2. There is a purely dominance/recessivity relationship between the alleles and no individual can express the disorder without being homozygous for the disease causes allele
  3. De novo mutations for the disorder are rare enough to be ignored in the calculation of this probability
  4. Trisomy and monosomy is not involved

, then the two parents must necessarily be carrier (both of them) as long as at least one of their children express the disease.

Note, by the way that some of the above assumptions are too strict and could be partially released.

This question does not require any complicated bayesian statistics (or any statistics at all) or any use of Hardy-Weinberg rule. If a children expresses a purely recessive disease, then (s)he is necessarily aa, where a is the recessive allele and A is the dominant allele(s). It means that both parents must have at least one a. Because, no parents express the disease, both parents are necessarily Aa (or aA)

  • $\begingroup$ Suppose assumption #2 is untrue for this particular disorder. Then, what statistic would I want to look at? $\endgroup$ – moonman239 May 21 '16 at 0:19
  • $\begingroup$ We can reduce assumption #2 to "if no individual can express the disease without being aa". If this assumption hold it is fine, you are sure that both parents were carrier. I edited assumption #2. Other assumption can eventually be further reduced or specified. $\endgroup$ – Remi.b May 21 '16 at 0:21
  • $\begingroup$ I still don't understand. Mutations cause the alleles, and a recessive allele can only express itself if a dominant allele is not present. Thus, if we're talking Mendelian genetics, then it is biologically impossible for assumption #2 to be false. I suppose I'm getting something wrong, though, aren't I? $\endgroup$ – moonman239 May 21 '16 at 0:32
  • $\begingroup$ I don't really understand what you misunderstand! Assumption #2 says that if an individual is sick (express the disease), then this individual must be aa. We can imagine a case where there are some Aa individuals that would express the disease if they smoke a lot for example. All the assumptions are probably assumptions you wanted to make to start with. $\endgroup$ – Remi.b May 21 '16 at 0:35
  • $\begingroup$ Why dont hemophilia follow these steps? The father has no diagnosis of hemophilia and the mother can only be a carrier yet show no symptoms of a hemophilac. The child could only be a male with no x chromesome from the father because he recieved two y's from the mother, even without any family history of hemophilia the mutation is said to be spontaneous or inherited and the new adult version where it comes out of dormancy in males of 40 years of age for the first time. $\endgroup$ – user5434678 May 21 '16 at 6:56

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