caseclosed Posted November 22, 2005 Share Posted November 22, 2005 when finding the genotype ratio of Aa * Aa we could use binomial expansion (A+a)*(A+a) to get 1:2:1 ratio but how do we find AaBbCc * AaBbCc ? Link to comment Share on other sites More sharing options...
cosine Posted November 22, 2005 Share Posted November 22, 2005 when finding the genotype ratio of Aa * Aawe could use binomial expansion (A+a)*(A+a) to get 1:2:1 ratio but how do we find AaBbCc * AaBbCc ? Hm interesting observation. I expect it would be (A+a+B+b+C+c)*(A+a+B+b+C+c) unless there is something I misunderstand about the question? Link to comment Share on other sites More sharing options...
Dave Posted November 22, 2005 Share Posted November 22, 2005 I think you may want to use something called the multinomial expansion. Basically, the multinomial theorem says: [math](x_1 + \cdots + x_k)^n = \sum_{\stackrel{n_1, \dots, n_k \geq 0}{n_1 + \cdots + n_k = n}} \frac{n!}{n_1! \, n_2! \cdots n_k!} \, a_1^{n_1} \cdots a_k^{n_k}[/math] You can find out more about it at MathWorld. Link to comment Share on other sites More sharing options...
caseclosed Posted November 22, 2005 Author Share Posted November 22, 2005 did it wrong. damn it. I found this online calculator, just gonna use that.... Link to comment Share on other sites More sharing options...
cosine Posted November 22, 2005 Share Posted November 22, 2005 I thought it would but there seem to be some differenceI mean here is example for Aa * Aa' date=' on the punnet square we write, A a (on top) * A a (on side) for AaBbCc * AaBbCc, we write ABC ABc AbC Abc (on top) * ABC ABc AbC Abc (on side) notice it is 3 letters at a time instead of 1 letter.[/quote'] Oh yes, very different I see... I really have not seen a solution for this, maybe Dave's link will work. Are you saying that there are 4 different tops or 4 different columns? If the latter I expect the solution to still be represented by (ABC+ABc+AbC+Abc)*(ABC+ABc+AbC+Abc) Link to comment Share on other sites More sharing options...
caseclosed Posted November 22, 2005 Author Share Posted November 22, 2005 fixed it, I forgot to add the other variations. it would be this below ABC ABc AbC Abc aBC aBc abC abc (on top) * ABC ABc AbC Abc aBC aBc abC abc (on side) Link to comment Share on other sites More sharing options...
Mokele Posted November 23, 2005 Share Posted November 23, 2005 Because the alleles assort independently, you just "layer it" (or whatever the actual math term is). In essence, to find the probability of, say AABbCC, you say there's a 1/4th probability os AA, a 1/2 probability of Bb, and a 1/4th probability of CC, so the total probability is 1/32. Repeat ad nauseam. Of course, it can get lots more complicated, with linkage disequilibrium (where alleles *don't* assort independently), but that's a whole different mess. Mokele Link to comment Share on other sites More sharing options...
caseclosed Posted November 23, 2005 Author Share Posted November 23, 2005 found this online calculator, gonna use that. Link to comment Share on other sites More sharing options...
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