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buffzyzz buffzyzz
wrote...
12 years ago
Is it enough to show that p^2+2pq+q^2=1 ?
Thanks.
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wrote...
12 years ago
if the allele frequency of q when it is squared is the same as the percentage of recessive members of the population, then it is at hardy-weinberg equilibrium
wrote...
12 years ago
If you have the genotype frequencies you add the dominant homozygot to half the heterozygot then divide by the total; this gives you "p".  Then add the recessive homozygot to half the hetrozygot then divide by the total; this is "q".  You then plug these values into the p^2, 2pq, and q^2 equations and compare your results to the original genotipic frequencies.  you can use chi squared analisys to verify the goodness of fit.
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