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Unformatted text preview: n 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 chisquare test to check our observed
numbers to see if the null hypothesis of HW 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 HW prediction.
Data are consistent with the population being in HW equilibrium at this locus
Was your intuition correct? Use of HW in conserva7on biology: an example http://www.arkive.org/giantkangaroorat/dipodomysingens/imageG6826.html Expected % heterozygotes Observed % heterozygotes *
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* exercise ① Rocio hand out a deck of cards to everyone and then she will give each student two cards from our deck (don’t mix these up with your deck) • either two reds “a red homozygote” • or two blacks “a black homozygote” ② Take a look to see which one you have but then turn them back over ③ Shuﬄe your deck thoroughly and pick the top card • this will be used to see who “mates with” who ④ exchange cards as instructed ⑤ we will calculate Red Red, Red Black, and Black Black frequencies and see if they conform to H W...
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 Spring '12
 john

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