Derivation
of
the General Selection Equation
______________________________________________________________
Genotype
AA
AB
BB
______________________________________________________________
(1) Frequency p2
+
2pq
+
q2
=
1
before selection
(2) Fitness W0
W1
W2
(3) Relative
p2W0
+
2pqW1
+ q2W2
=
Contribution
(4) Frequency p2W0/
+ 2pqW1/
+ q2W2/
= /
= 1
after selection
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(1) Genotype distributions before selection follow Hardy-Weinberg
expectations.
(2) Each genotype AA, AB, and BB has a
distinct
phenotype: W0, W1, & W2,
respectively.
W is the expectation
that an individual with a particular genotype will survive &
reproduce
(3) Each genotypic class makes a relative
contribution
to the next generation,
which is proportional to its
initial frequency, weighted by its fitness.
[e.g., if the AA genotype has a
frequency
of 0.25 and 80% survive to reproduce,
the
relative contribution of AA to the next generation
is
(0.25)(0.8)=0.20]
The sum of the relative contributions of all
three genotypes is
(read as, "W bar") = mean
population fitness
In
this simple model,< 1,
because
not everyone on Line (1) survives.
(4) Because < 1,
the relative
contributions have to be "normalized":
Dividing the contribution of each genotype by returns
the sum to unity,
and the final values are the relative genotype
frequencies
after selection.
To derive the allele
frequencies after selection, take Line (4) above and
recall q = f(BB) + (1/2) f(AB)
so q' = q2W2/
+ (1/2) 2pqW1/
=
q(qW2 + pW1)/
then q
=
q' - q = qafter
- qbefore
=
q(qW2 + pW1)/
- q/
=
[(q)(qW2 + pW1) - (q)(p2W0
+ 2pqW1 + q2W2)]
/
=
[(q)(qW2 + pW1 - p2W0
- 2pqW1 - q2W2)]
/
=
[(q)(pqW2 + W1p(1-2q) - pW0p)]
/
[Note
1]
=
[(pq)(qW2 + W1(1-2q) - W0p)] /
=
[(pq)(W2q + W1(p-q) - W0p)] /
[Note
2]
=
[(pq)(W2q + W1p - W1q - W0p)/
q
=
[pq] [(q)(W2 - W1) + (p)(W1 -
W0)] / []
________________________________________________
Notes: [1] (q - q2) =
q(1
- q) = pq [2]
(1
- 2q) = (1 - q) - q = p - q
Optional Homework
Exercise: repeat this
derivation
for p
=
p'
- p
Text
material
© 2005 by Steven
M. Carr