On Wed, 2 Feb 2000 23:53:39 +1100, "James" <[EMAIL PROTECTED]> wrote: >Given a daughter has breast cancer, how to calculate the probability that >her mother has breast cancer ? Suppose the probability that the mother has >breast cancer is p1 if she carries a certain gene a, and p0 if she does not >carry the gene. The allele frequency of gene a is assumed to be q. Given that most of your readers are not biologists, it might be worth telling us: (a) what "allele frequency of gene a" means, (b) the probability of inheriting the gene from the parents, and (c) the impact of one or two copies of the gene. Since breast cancer may be age related, (d) can you also say the probability that the daughter would have got breast cancer was p1 if she carries the gene a, and p0 was she does not carry the gene and that the gene has no impact of the liklihood of producing daughters? You really need all this before applying Bayes theorem. If (e) the relevant chromosone has gene a with probability q, (f) given the mother has two copies then the daughter will have two with probability q and one with probability 1-q given the mother has one copy then the daughter will have two with probability q/2, one with probability 1/2 and none with probability (1-q)/2 given the mother has no copies then the daughter will have one with probability q and none with probability 1-q (g) with two copies the probability is p2, with one p1 and with none p0 (h) and for all women the probabilities refer to the chance at birth of suffering sometime in their lives, without affecting the probability of producing a given number of daughters. then we can say: (i) among women a proportion q^2 have two copies 2.q.(1-q) have one and (1-q)^2 have no copies (j) given the daughter has two copies then the mother will have two with probability q and one with probability 1-q given the daughter has one copy then the mother will have two with probability q/2, one with probability 1/2 and none with probability (1-q)/2 given the daughter has no copies then the mother will have one with probability q and none with probability 1-q (k) the probability the daughter has two copies given she has breast cancer is r2=q^2.p2/(q^2.p2+2.q.(1-q).p1+(1-q)^2.p0) the probability the daughter has one copy given she has breast cancer is r1=2.q.(1-q).p1/(q^2.p2+2.q.(1-q).p1+(1-q)^2.p0) the probability the daughter has one copy given she has breast cancer is r0=(1-q)^2.p0/(q^2.p2+2.q.(1-q).p1+(1-q)^2.p0) (l) the probability the mother has two copies given the daughter has breast cancer is s2=r2.q+r1.q/2 the probability the mother has one copy given the daughter has breast cancer is s1=r2.(1-q)+r1/2+r0.q the probability the mother has no copies given the daughter has breast cancer is s0=r1.(1-q)/2+r0.(1-q) (m) so the probability at birth of the mother developing breast cancer, given that the daughter has done so is p2.s2+p1.s1+p0.s0 NOTE: this result can be simplified, it may contain wrong assuptions, it may also contain errors, and the fact that the mother has presumably not previously had breast cancer diagnosed should affect the result. =========================================================================== This list is open to everyone. Occasionally, people lacking respect for other members of the list send messages that are inappropriate or unrelated to the list's discussion topics. Please just delete the offensive email. For information concerning the list, please see the following web page: http://jse.stat.ncsu.edu/ ===========================================================================
