F13 lecture 5

# 294 0496 0209 expected 1226 2068 872

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Unformatted text preview: cted genotype numbers under H-W? given 417 total individuals #AA = p2(#inds) = (0.543)2(417) = 0.294(417) = 122.6 #Aa = 2pq(#inds) = 2(.543)(.457) (417) = .496(417) = 206.8 #aa = q2(#inds) = (0.457)2(417) = 0.209(417) = 87.2 Now let’s use a chi-square test to check our observed numbers to see if the null hypothesis of H-W equilibrium at this locus is supported by the data. chi- square special case •  a H- W goodness of ﬁt is a special case in terms of degrees of freedom –  df = # genotypes - #alleles –  for two alleles –  df = 3- 2=1 •  1 degree of freedom AA Aa aa total observed # 121 211 85 417 expected frequency 0.294 0.496 0.209 - expected # 122.6 206.8 87.2 - (O- E) - 1.6 4.2 - 2.2 - (O- E)^2 2.55 17.37 4.64 - [(O- E)^2]/E 0.02 0.08 0.05 0.16 total chi square P = 0.924036198 =chidist(0.16,1) Population is very close to H-W prediction. Data are consistent with the population being in H-W equilibrium at this locus W...
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## This document was uploaded on 02/27/2014.

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