Jump to content

Calculating genotype frequency from allele frequencies


horsebox

Recommended Posts

As an example, I'll use the rs33 SNP. For Europeans, A has a frequency of 0.220 and G = 0.780. Then for the genotypes: AA = 0.051, AG=0.339, GG=0.610. I would assume the genotype frequencies are a matter of statistics. Maths isn't my strong point, I can't figure out how to calculate them from the allele frequencies. Also I've seen some weird things on dbsnp like two allele frequencies A=0.50,C=0.50 adding up to AC=1.00 which makes me wonder if its even possible to calculate them with a simple mathematical formula. But if thats the case, how can they determine the allele frequencies if they're not even related to the genotype frequencies?

 

EDIT: I checked out one of those online calculators, and they give the expected results (for AA and GG, the value is what you get when you square the allele frequencies), they don't work on these real life examples.

Edited by horsebox
Link to comment
Share on other sites

I'm not sure I follow 100%.

 

If [math]p + q = 1[/math] is true but [math]p^2 + 2pq + q^2 = 1[/math] is false, then one of the Hardy-Weinberg assumptions has been violated.

 

ideal - - actual

0.0484 - - 0.051

0.3432 - - 0.339

0.6084 - - 0.610

 

They're approximately the same. I wouldn't expect them to ever be exact. Population size is always finite, thus so are number of mate pairings and number of deaths/selections, leaving some room for chance deviation.

Edited by MonDie
Link to comment
Share on other sites

You gotta multiply the frequencies.

The frequencies are basically percentages(0.780=78%)

So to calculate AG you multiply 0.78 by 0.22 and get something around 0.18.

Note that numbers become smaller after multiplication because you multiply numbers which are smaller then one.

Link to comment
Share on other sites

What an embarrassing mistake!

 

If [math]p + q = 1[/math] is true but [math]p^2 + 2pq + q^2 = 1[/math] is false, then one of the Hardy-Weinberg assumptions has been violated.

I meant:

If [math]p^2, 2pq,[/math] and [math]q^2[/math] do not reflect the observed genotype frequencies!

 

You gotta multiply the frequencies. The frequencies are basically percentages(0.780=78%) So to calculate AG you multiply 0.78 by 0.22 and get something around 0.18. Note that numbers become smaller after multiplication because you multiply numbers which are smaller then one.

Multiply by two for the heterozygous genotype!

Edited by MonDie
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.