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Punnet Square


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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 ?

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?

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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.

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I thought it would but there seem to be some difference

I 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)

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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

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