Bio 321 F13 lecture 5

# aa 121 aa 211 aa 85 total inds417 p aaaa2total

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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 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 Was your intuition correct? Use of HW in conserva7on biology: an example http://www.arkive.org/giant-kangaroo-rat/dipodomys-ingens/imageG6826.html Expected % heterozygotes Observed % heterozygotes * * * * * * * * 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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## This note was uploaded on 10/01/2013 for the course BIO 321 taught by Professor John during the Spring '12 term at SUNY Stony Brook.

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