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
HardyWeinberg 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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