Sur.*Fec.*K = 1.8 offspring/lifetime
W*
CCR5/
CCR5
=
Sur.*Fec.*K = 1.4 offspring/lifetime
Gene
pool
p
=freq(A1)
q
=freq(A2)
Genotype
pool
P
= prop(A1A1)
H = prop(A1A2)
Q = prop(A2A2)
Gen=0
Gene
pool
p
!
=freq(A1)
q
!
=freq(A2)
Gen=1
Natural selection
is on phenotypes
produced by the
genotypes
W
1
, W
2
, W
Gene Pool
Gen=0
Selection
Gen=
1
W
1
freq (A
2
)
=
q
W
2
qW
2
W
W
1
Gene Pool
W
total
ʼ
=
W
W
=
1
Gene Pool
q
ʼ
=
qW
2
W
p
W
1
= 1-
p
ʼ
p
ʼ
=
p
W
1
freq (A
1
) =
p
(Total) =
p
+q
=
1
Next, let
p
= proportion of
!
CCR5
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W
=
p
(W
1
) + q (W
2
) =
.1 (.91) + .9 (.72) = 0.739
W
2
= q (W
22
) +
p
(W
12
) =
.9 (.7) + .1 (.9) = 0.72
W
1
=
p
(W
11
) + q (W
12
) =
.1 (1) + .9 (.9) = 0.91
Because humans have such a long generation
time, substantial evolutionary adaptation to the
HIV/AIDS will take at least about 2500 years (a
hundred generations)
But eventually it would occur
Let,
Δ
CCR5 = A1
CCR5 = A2
p
= 0.1
q = 0.9
(rounding 9.1% to 10%)
Then,
to make math easier
p
’ =
p
in next generation
=
p
W
1
W
= 0.1
.91
.739
= .123
=
p
’ -
p
=
.123
- .1 = 0.023
So the change is SUBSTANTIAL but SLOW
Δ
p
= Change in
p
If evolution is so slow, then how did
Δ
CCR5
rise to a frequency of 9% in Northern Europe?
Recent evidence indicates that another virus also uses
the CCR5 receptor to enter cells
the virus that causes smallpox,
so 100’s of years of past adaptation to smallpox
may have lead to the high frequency of
Δ
CCR5
in N. Europeans.
We do not yet know for certain, but
Very Important to do study
questions from this lecture
material
p
= 0.1
p
’ = .123
24
0.05
0.15
0.25
0.1
0.2
0.3
Generation
0
1
2
3
4
5
W is the average fitness of an allele in the gene pool and it also is
the average fitness of an individual in the population = average
population fitness
Frequency of
Δ
CCR5 (
p
)
4 years with
2-week life cycle
But only
W
CCR5/
Δ
CCR5
=
W
12
= W*
12
/ W*
MAX
= 1.8/2 = 0.9
W
Δ
CCR5/
Δ
CCR5
=
W
11
= W*
11
/ W*
MAX
= 2/2 = 1.0
W
CCR5/CCR5
=
W
22
= W*
22
/ W*
MAX
= 1.4/2 = 0.7
0.1
0.5
1.0
100
0
(2500 years)
population
one
individual
W = individual fitness
= Darwinian fitness
average W
for all A
1
A
1
average W
for all A
2
A
2
average W
for all A
1
A
2
W
*
11
W
*
12
W
*
22
divide
by
largest
W
*
ij
W
11
W
12
W
22
relative
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- Winter '07
- Evan
- Evolution
-
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