EEMB
lec_4_cond_13

# Ccr5 ccr5 ccr5 δ ccr5 δ ccr5 δ ccr5 78 the data

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CCR5 / CCR5 CCR5 / Δ CCR5 Δ CCR5 / Δ CCR5 78 0 ? ? ? The Data 114 8 645 582 GENOTYPE antibody (-) FOR HIV Observed Expected Antibody (+) FOR HIV Observed Expected ? ? ? (uninfected) (infected) 18 We do not use the antibody(+) data since these individuals would already have had some natural selection operating on the CCR5/ Δ CCR5 alleles. Uninfected antibody(-) sample Calculate p & q Calculate HW expected P= p 2 H=2 p q Q= q 2 from null model with no selection Compare HW expected numbers: PN, HN, & QN to observed numbers in the infected antibody(+) sample Road map

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Genotype Δ CCR5 / Δ CCR5 CCR5 / Δ CCR5 Excellent fit q = 0.825 2 582 Expected Observed Number Number p = 0.0085 2 8 2 p q= 0.167 114 Assuming random mating and HW equilibrium & p = 0.092 Proportion N=704 2 p qN =117.62 q N = 580.42 2 CCR5 / CCR5 Assuming random mating and HW equilibrium & p = 0.092 We must use the p estimated from the unselected data (i.e., from the antibody(-) data) so as to not bias p upwards due to selection (which is what we are trying to measure). Same analysis but for antibody (+) data & using gene proportion data from antibody(-) data (unselected) Next: Genotype Expected Observed Number Number Δ CCR5 / Δ CCR5 CCR5 / Δ CCR5 CCR5 / CCR5 p = 0.0085 2 0 2 p q= 0.167 78 q = 0.825 2 645 Very Poor fit Far too few Δ CCR5 homozygotes and heterozygotes Strong evidence that Δ CCR5 gene is protective in natural population Proportion q N = 596.48 p N = 6.15 2 2 p qN =120.74 2 Assuming random mating and HW equilibrium & p = 0.092 antibody (+) data (INFECTED) antibody (-) data FOR HIV (uninfected) p = proportion of Δ CCR5 = { 2(PN) + HN } / 2N = { 2(8) + 1(114) } / 2(704) = 0.092 So q = ( 1 - p ) = 0.908 Now we can solve for the expected genotype frequencies assuming the Δ CCR5 gene has no benefit GENOTYPE CCR5 / CCR5 CCR5 / Δ CCR5 Δ CCR5 / Δ CCR5 Observed ? ? ? 582 = QN 114 = HN 8 = PN Expected antibody (-) FOR HIV 704 = N (uninfected) 19 p = .092 x .092 = 0.0085 2 q = .908 x .908 = 0.825 2 2 p q = 2 x .092 x .908 = 0167 (which will be our “A1” allele) p N = 5.96 2 Hardy-Weinberg expected values represent a null model of what nature would look like in the absence of any evolutionary forces operating So for the unexposed population, the data fit the null model of no selection very well
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• Winter '07
• Evan
• null model

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