6.891: Computational
Evolutionary Biology
R.C. Berwick & a cast of thousands
Today: the forces of evolution, III
The forces of evolution, III
Does selection maximize fitness?
Does sex make you fitter?
Avoiding the ratchet
The multivariate case: sickle cell anemia example
& stability
The weak force: mutation
The superstrong force: migration
The thermodynamic force: drift
Natural selection works by local gradient ascent
w
dw
dp
p1
" p(1 ! p)
#
$ w
frequency which makes p a bit less than 1 puts it in a region where p continues to decrease away
from 1. The equilibrium at p = 0.333 looks locally stable, but a casual glance is not enough to
determine its stability.
If we assume (as is true in our example) that f (p), and hence also ∆p, are continuous functions
of p, we can make a simple algebraic analysis of local stability. In the vicinity of an equilibrium let
us assume that the ∆p curve can be approximated by a straight line. If x is the distance between
p and pe , so that p = pe + x, then we will approximate ∆p by ax. The quantity a will be the slope
of the ∆p curve as it passes through p = p e . In the next generation, the deviation x ! from the
equilibrium will be
Stability Considerations
x! = p + ∆p − pe = pe + x + ∆p − pe
(II-102)
equilibrium within that region resulted in the gene frequency returning
to the equilibrium. Thus if
= x + ∆p
" x + ax
= x(1 + a)
the gene frequency will return to its equilibrium when changed by less than (say) 1%, we describe
When we are close to the equilibrium, the value of x is thus multiplied by 1 + a each generation.
the equilibrium
as locally stable. If a perturbation of (say) 20% would result in no return to the
After t generations, it will be (1 + a)t times its current value.
slope
curve
= is
axnot locally stable, we
equilibrium,
this (1
equilibrium
is not
globally
stable.
an!p
equilibrium
When then
a is positive
+ a)t is a positive
number
greater than
1, and If
itof
will
grow
with t. This
is the
near the equilibria p = 0 and p = 1, where the slope a of the ∆p curve is positive.
say that
itsituation
is unstable.
Any movement of p from p = 0!p
to a very small positive quantity will create a positive deviation x
Towhich
investigate
it is sufficient
consider
what
when the gene frequency
then grows local
until p stability,
leaves the immediate
vicinity of pto
= 0.
Near p = 1,
if p is happens
set just below
1,
this
is
a
negative
value
of
x
which
also
becomes
steadily
more
negative
until
p
departs
from
the
is moved an infinitesimal amount. If it always returns, it is necessarily locally stable, if not it
region near 1. Thus the algebra confirms our suspicions about the stability of p = 0 and p = 1.
is unstable.
the
point,0 ∆p
0. Figure
shows
plot of ∆p aginst p for an
When At
−1 <
a <equilibrium
0, 1 + a lies between
and 1.=Raising
1 + a to 2.4
the t-th
poweramakes
t
it approach case
zero without
ever becoming
negative.
This : is 1the
overdominant
in which
fitnesses
are 0.7
0.85.
There
equilibrium points, at
a :>case
0 in→which
(1
+p approaches
a)are
>three
1;the
x! grows
equilibrium smoothly without ever overshooting. Whatever the initial sign of the deviation x, it
p = 0,remains
p = 1,of and
p = 0.333. It seems that p = 0 and p = 1 must be unstable equilibria. When p
the same sign but goes to zero. When −2 < a < −1, 1 + a lies between -1 and 0.
is perturbed
just
p=
0, ∆p
is positive
region.
p will
continue to increase away
Multiplying
x byabove
1 + a will
change
its sign
but reducein
its that
magnitude.
ThatThus
corresponds
to the
case
where
there
is
overshooting
of
the
equilibrium,
but
the
overshoot
leaves
the
gene
frequency
p
from the equilibrium. By much the same reasoning p = 1 also seems unstable. Any change of gene
each time closer to the equilibrium than it was. The gene frequency oscillates, but with decreasing
frequency
which
aFigure
bit2.4:less
than
1 puts it in a region where p continues to decrease away
amplitude,
and makes
ultimatelypconverges
toThe
the
equilibrium.
change in gene frequency (∆p) plotted against the gene frequency in a
when a < −2, 1+a
<p
−1
thewhere
deviation
x AA
changes
sign
each
and growsglance is not enough to
of
overdominance
fitnesses locally
of
: Aa : aa
are 0.85
: 1generation
:but
0.7.
from 1. Finally,
The equilibrium
atcase
=so that
0.333
looks
stable,
a casual
in amplitude. The overshoot is so great as to leave the population farther from the equilibrium
determine
its stability.
each time.
It oscillates away from the equilibrium. Extrapolation of this behavior would lead to
an absurdity:
frequency
would the
ultimately
be a)greater
than and
1 or less
than
0. toThis
need
If
we
assume
(asgene
is true
in
our
f relevant
(p),
hence
also
∆p,
continuous
functions
not
us, sinceexample)
multiplier (1 +that
is
only
in a region
small
enough
allow
us are
gure 2.4: The the
change
introuble
gene
frequency
(∆p)
plotted
against
the
gene
frequency
in a
to approximate the ∆p curve as a straight line. Farther from the equilibrium the higher-order
of p, we can make a simple
algebraic
analysis
of
the
vicinity
of
an
equilibrium
let
derivatives
of ∆p become relevant,
inevitably
in local
such a waystability.
as to keep the geneIn
frequency
between
61 of AA : Aa : aa are 0.85 : 1 : 0.7.
se of overdominance where
fitnesses
0 and 1.
us assume that the ∆p curve
canforbe
by a straight line. If x is the distance between
Our criterion
local approximated
stability is this:
!
"
d(∆p)
p and pe , so that p = pe + x, then we will −2approximate
∆p by ax. The(II-103)
quantity a will be the slope
<
< 0,
dp
of the ∆p curve as it passes throughStability
p = p e . In Analysis
the next generation, the deviation x ! from the
the brackets indicating evaluation at p = p . There are two sorts of qualification of this picture
necessary. We have not investigated what will happen when a is exactly equal to 0, -1, or -2. In
equilibrium will be
p = pe
e
each case the exact behavior depends on the higher-order terms in ∆p. The results do not modify
(II-103) in any essential way. The second qualification is a more serious one. When generations
are continuous instead of discrete, oscillation is no longer a possibility. In that case the stability
e of dp/dt (the equantity analogous to ∆p). Ife it is positive below
is simply determined by the sign
the equilibrium and negative above it, the equilibrium is stable, and not otherwise. Overshooting
is impossible because the gene frequency would have to pass smoothly through p e in order to
overshoot, and once it reached pe it would not change further. There is an analogous damping of
x! = p + ∆p
− pa) is
= only
p + relevant
x + ∆p − pin a region small enough to allow u
ble us, since the multiplier
(1 +
(II-102)
oximate the ∆p curve as a straight line. Farther from the equilibrium the higher-ord
= x +inevitably
∆p
"in xsuch
+ axa way as to=keep
x(1 the
+ a)gene frequency betwee
ves of ∆p become relevant,
62
When we are close to the equilibrium, the value of x is thus multiplied by 1 + a each generation.
if: current value.
After
t generations,
will beLocally
(1is+this:
a)t stable
times its
criterion
for localitstability
When a is positive (1 + a)t is a positive number greater than 1, and it will grow with t. This
!
"
is the situation near the equilibria p = 0 and
p = 1, where the slope a of the ∆p curve is positive.
d(∆p)
< 0, will create a positive deviation(II-10
Any movement of p from p = 0 −2
to a <
very small
quantity
x
dp positive
p = pe
which then grows until p leaves the immediate vicinity of p = 0. Near p = 1, if p is set just below
1, this is a negative value of x which also becomes steadily more negative until p departs from the
ckets
indicating evaluation at p = p e . There are two sorts of qualification of this pictu
region near 1. Thus the algebra confirms our
suspicions about the stability of p = 0 and p = 1.
y. We
have
not
investigated
what
will
a is1 exactly
equal
0, -1,
or -2. I
When −1 < a < 0, 1 + a lies between happen
0 and 1.when
Raising
+ a to the
t-th to
power
makes
e the
exactzero
behavior
on the negative.
higher-order
in ∆p.
The results
do notthe
modi
it
approach
withoutdepends
ever becoming
This terms
is the case
in which
p approaches
in any essential
way.
Theever
second
qualification
is a the
more
serious
one.
When generation
equilibrium
smoothly
without
overshooting.
Whatever
initial
sign of
the deviation
x, it
remains instead
of the same
sign but goes
to zero. isWhen
−2 < aa possibility.
< −1, 1 + a In
lies that
between
andstabilit
0.
inuous
of discrete,
oscillation
no longer
case-1the
x by
+ a sign
will change
its sign
reduce its
magnitude.
That corresponds
to thebelo
yMultiplying
determined
by1 the
of dp/dt
(thebut
quantity
analogous
to ∆p).
If it is positive
case where there is overshooting of the equilibrium, but the overshoot leaves the gene frequency
Stability Considerations
Cases:
(1) a > 0, (1 + a)t > 1, so x! = x(1 + a)t unbounded growth
(2) −1 < a < 0, 1 + a lies between 0 and 1, so (1 + a)t → 0
(3) −2 < a < −1, 1 + a lies between −1 and 0.
then multiplying x by 1+a changes its sign, reduces its magnitude, so overshoot,
but converges
(4) a < −2, 1 + a < −1, then deviation changes sign and grows
Summary conditions for local stability:
−2 < [
d(#p)
]p=pe < 0
dp
Human variation at genetic code level (genotype) to
variation in protein to variation in...
ATG GTG CAC CTG ACT CCT GAG GAG AAG TCT GCC GTT ACT
ATG GTG CAC CTG ACT CCT GTG GAG AAG TCT GCC GTT ACT
MVHLTPEEKSAVT (E is the single letter abbreviation for
glutamic acid)
MVHLTPVEKSAVT (V is the single letter abbreviation
for valine)
Glutamic acid is a hydrophilic amino acid. Valine is a
hydrophobic amino acid.
Analysis of Hb-a, HB-s, and HB-c data
(from Cavalli-Sforza, 1977)
AA
SS
CC
AS
AC
SC
Observed
25374
67
108
5482
1737
130
Expected
25616
307
75
4967
1769
165
Obs/Exp
0.99
0.22
1.45
1.10
0.98
0.79
Relative
fitness
0.89
0.20
1.31
1
0.89
0.70
Suppose just A, S alleles
p̂S =
w22 ! w12
(w11 ! w12 ) + (w22 ! w12 )
=
0.2 ! 1.0
0.89 ! 2.0 + 0.2
= 0.1209
pA =0.8791
Suppose just a few C
alleles introduced
w = 0.90
wC = pA wAC + pS wSC + pC wCC , when pc ! 0,
wC = pA wAC + pS wSC = 0.8670
C cannot invade when rare, even
though this yields global fitness!
= frequency of S (sickle)
= frequency of A (normal)
A
A
S
C
0.685
0.763
0.679
C
+
S
0.763
0.151
0.545
C
0.679
0.545
1.000
pairwise compare: C–A
0.685 : 0.679 : 1.0 → underdominance; C fixed?
0.9
0.8
0.7
0.6
pairwise compare: S–C
0.151 : 0.545 : 1.0 → C fixed?
0.5
0.692
0.4
0.691
0.3
0.2
*
!
!
+
S
0.69
0.693
A
pairwise compare: S–A
Figure 2.9: Contours of fitness plotted against allele frequencies in a three-hemoglobin
0.763 :of 0.151
→is proportional
overdominance;
S−
polymorphism. At any 0.685
point, the: frequency
each allele
to the attitude
to the side opposite the corner labeled with that allele’s symbol. Minima (-), Maxima
(+) and saddle-points (*) of the fitness surface are indicated.
A mix
where all three alleles are present. There is an equilibrium there, as the appropriate linear equations
show, but it is not stable. In fact, if there is a stable equilibrium with one fewer allele (including
stability to introduction of the missing allele), the interior equilibrium must be unstable. This fact
follows from the identification of peaks of w̄ with stable equilibria, and from the quadratic nature
of w̄, though it will not be proven here. Since there is a stable equilibrium with only alleles A and
S, there cannot be a stable equilibrium with all three alleles.
Note that the simple plotting of w̄ over the triangle, as done in Figure 2.9, immediately shows the
locations of the two peaks, and shows that there is an interior equilibrium which is unstable, as it is
a saddle of the fitness surface. Such a plot of w̄ will always convey the full picture in multiple-allele
cases, and is by far the easiest and most accessible way of analyzing these situations. Of course,
the plot must be done to sufficient accuracy: both the saddle point and the A-S polymorphic
equilibrium would have been missed if only the contours 0.1 apart (the coarsely dashed curves)
were plotted.
The small size of the peak for the A-S polymorphism would seem to indicate that West African
populations are proceeding to fixation for allele C. This is not necessarily so, because these popu-
Does sex make you fitter?
76
Why Sex Hurts
Mixing due to diploid mating actualy reduces fitness!
In other words: an asexually reproducing (haploid)
population could actually do better!
Let’s see how much – this is called segregational load
numerical example will
be useful
here.
Puzzle:
why does
sexSuppose
survive? that we have a diploid population with
es
AA Aa aa
0.4
1 0.8
d a gene frequency of 0.2 in the population at the beginning of a generation. After fertilization,
0.2; Hardy-Weinberg proportions 0.04 : 0.32 : 0.64, the mean
the genotype frequenciesp(A)
are in=their
p(AA) = 0.04; p(aa) = 0.64; p(Aa) = 2p(1 − p) = 0.32
will be
H − W ratios :0.04 : 0.32 : 0.04
0.04 × 0.4What
+ 0.32is×mean
1 + 0.8
× 0.64 = 0.848.
fitness?
atural selection (which in this example is most easily conceived of as differential viability) will
A genotype
numerical
exampleexample
will
usefuluseful
here. Suppose
that
we
havea adiploid
diploid
population
with
numerical
that
we we
have
population
with with
he
frequencies
to bewill
AAnumerical
example
will be
be usefulhere.
here. Suppose
Suppose
that
have a diploid
population
nesses fitnesses
fitnesses
AAAaAa aaaa
AA ×
0.04 × 0.4/0.848 : 0.32
1/0.848
0.8 × 0.64/0.848
AA
Aa : aa
0.4 0.4
1 1 10.80.80.8
0.4
and a gene frequency of 0.2 in the population
at the beginning of a generation. After fertilization,
and a gene
frequency
of 0.2 inofthe
population
at the
of aofgeneration.
After
fertilization,
and
a
gene
frequency
0.2
in
the
population
atbeginning
the beginning
a 0.04
generation.
Afterthe
fertilization,
when the genotype frequencies
aremean
in their Hardy-Weinberg
proportions
: 0.32 : 0.64,
mean
What
is
fitness
without
sex?
thewill
genotype
frequencies
are in Hardy-Weinberg
their
Hardy-Weinberg
proportions
0.04: :0.32
0.32: : 0.64,
0.64, the
the mean
hen the when
genotype
frequencies
are in0.0189
their
proportions
0.04
mean
: 0.3774
: 0.6038
fitness
be
ness willfitness
be will be
+ 0.32 × 1 + 0.8 × 0.64
= 0.848.
hese genotype frequencies are 0.04
not×in0.4Hardy-Weinberg
proportions:
there is an excess of het0.04
×
0.4
+
0.32
×
1
+
0.8
×
0.64
=
0.848.
× 0.4
+ 0.32 ×If1isthe
+most
0.8genetic
× 0.64 system
= 0.848.
gotes as a result
their0.04
high
viability.
asexual,
thesewill
would be
Naturalof
selection
(which
in
this example
easily conceived
ofwere
as differential
viability)
enotype frequencies
at frequencies
the
start
of the
nextis most
generation.
The mean
fitness would
then
alter
the genotype
Natural
selection
(which
into
this
example
easily conceived
of as differential
viability)
will be
Natural
selection
(which
in
this
example
is
most
easily
conceived
of
as
differential
viability)
will
After
selection
(viability,
roughly):
the genotype
frequencies
to= 0.862, an increase of 0.02 over the value before selection.
× 0.4 +alter
0.3774
× 1 + 0.6038
×
0.8
0.04 × 0.4/0.848 : 0.32 × 1/0.848 : 0.8 × 0.64/0.848
er the genotype frequencies to
ver meiosis intervenes and makes
the next: generation
start
Hardy-Weinberg proportions at
0.04 × 0.4/0.848
0.32 × 1/0.848
: 0.8 in
× 0.64/0.848
or
w gene frequency of 0.04
0.0189
+ 0.3774/2
=×
0.2076.
frequencies are then
× 0.4/0.848
: 0.32
1/0.848The
: 0.8genotype
× 0.64/0.848
or
0.0189 : 0.3774 : 0.6038
These genotype frequencies0.0431
are not0.0189
Hardy-Weinberg
proportions: there is an excess of het: 0.3774
:in0.3290
: 0.6279
Suppose just
apply
fitness
to: 0.6038
this
in next round
erozygotes as a result of their high
viability.
If
the
genetic
system were asexual, these would be
0.0189
:
0.3774
:
0.6038
These genotype
frequencies
are
notofinthe
Hardy-Weinberg
proportions:
there is an excess of het(asexual
reproduction),
just multiply
the genotype
frequencies
at the
start
nextsogeneration.
The mean fitness would then be
gives aerozygotes
mean
fitness
at
the
start
of
that
generation
of
0.0431×0.4+0.3290×1+0.6279×0.8
as
result
their
high
Ifanthe
genetic
system
asexual,
would
be =
by× of
without
meiosis
rejuggling:
0.0188 × 0.4
+a0.3774
1fitness
+are
0.6038
×in
0.8viability.
= 0.862,
increase
of 0.02
over were
thethere
value
before
selection.
These genotype
frequencies
not
Hardy-Weinberg
proportions:
is
anthese
excess
of het6. Thistheis genotype
considerably
loweratthan
0.868
theininitial
meanfitness
fitness
of 0.848.
The
frequencies
themakes
startthe
ofbut
thestill
nextabove
generation.
The mean
would
then
meiosis
and
next
generation
Hardy-Weinberg
proportions
at be
ozygotes However
as a result
of intervenes
their high
viability.
If the
geneticstart
system
were asexual,
these would
be
0.0188
× 0.4
+ 0.3774
× 1 due
+
×
0.8
= ×
0.862,
increase
of 0.02
over back
the
before
selection.
0.0188
×0.6038
0.4
+an
0.6038
0.8
=
0.862
>value
0.848
se of mean
fitness
by
0.02
to+natural
selection
has
been
rolled
bythen
meiosis
to a net
the new
gene
frequency
of
0.0189
+0.3774
0.3774/2
=10.2076.
The ×
genotype
frequencies
are
e genotype
frequencies
at the start
of thethenext
generation.
The
mean fitness would
then at
be
However
meiosis
intervenes
and makes
next on
generation
start
in Hardy-Weinberg
proportions
(by
0.02) effect
se
of
0.00056.
The
disruptive
of
meiosis
genotypic
combinations
is
apparent.
0188 × 0.4
0.3774
1 + 0.6038
0.8 =
0.862,: an
of 0.02
over frequencies
the value before
0.0431
0.3290
: 0.6279
the+new
gene×frequency
of×
0.0189
+ 0.3774/2
= increase
0.2076.
The
genotype
are thenselection.
the
population
lacked
meiosis,
how
much
higher
could
its
fitness
be?
This
question
was at
first
owever meiosis intervenes and makes the next generation start in Hardy-Weinberg proportions
which gives a mean fitness at the start of that generation of 0.0431×0.4+0.3290×1+0.6279×0.8 =
0.0431
: 0.3290The
: 0.6279
Morton,
Crow, and
Muller
who
and
calculated
the segregational
load. This
eby
new
gene
frequency
0.0189 +(1956),
0.3774/2
= defined
0.2076.
genotype
are
thenThe
0.84856.
This isof
considerably
lower
than 0.868
but still above
the initialfrequencies
mean fitness of
0.848.
defined which
asincrease
the
fractional
reduction
in thethat
fitness
of a has
population
a by
result
oftothe
existence
of amean
selection
been rolled as
back
meiosis
a net
gives
meanfitness
fitnessbyat0.02
the due
starttoofnatural
generation
of 0.0431×0.4+0.3290×1+0.6279×0.8
=
0.0431
:
0.3290
:
0.6279
increaseThis
of 0.00056.
The
disruptive
effect
of
meiosis
on
genotypic
combinations
is
apparent.
ndelian0.84856.
segregation.
In
the
presence
of
overdominance,
an
asexual
population
could
come
is considerably lower than 0.868 but still above the initial mean fitness of 0.848. The to
and a gene frequency of 0.2 in the population at the beginning of a generation. After fertilization,
when the genotype frequencies are in their Hardy-Weinberg proportions 0.04 : 0.32 : 0.64, the mean
A numerical example will be useful here. Suppose that we have a diploid population with
fitness will be
fitnesses
AA Aa aa
0.04 × 0.4 + 0.32 × 1 + 0.8 × 0.64 = 0.848.
0.4
1 0.8
and
a
gene
frequency
of
0.2
in
the
population
at
beginning
of a generation.
Afterviability)
fertilization,
Natural selection (which in this example is mostthe
easily
conceived
of as differential
will
What
is
mean
fitness
with
sex?
when
the
genotype
frequencies
are
in
their
Hardy-Weinberg
proportions
0.04
:
0.32
:
0.64,
the
mean
alter the genotype frequencies to
fitness will be
Before
meiosis (AA, Aa, aa frequencies not in
0.04 × 0.4/0.848 : 0.32 × 1/0.848 : 0.8 × 0.64/0.848
0.04 × 0.4 + 0.32 × 1 + 0.8 × 0.64 = 0.848.
H-W proportions):
or
Natural selection (which in this example
most easily
conceived of as differential viability) will
0.0189 is: 0.3774
: 0.6038
alter the genotype frequencies to
These genotype frequencies are not in Hardy-Weinberg proportions: there is an excess of hetH-W (i.e.,
sex)
spreads
some
(half)
of
the were
A’s asexual,
from Aa
to A,
erozygotes as a result of0.04
their
high viability.
If 1/0.848
the genetic
these would be
× 0.4/0.848
: 0.32 ×
: 0.8system
× 0.64/0.848
noindent atand
thenext
a’sgeneration.
from Aa to
the genotype frequencies
the some
start ofofthe
Theaa:
mean fitness would then be
or
0.0188 × 0.4 + 0.3774 × 1 + 0.6038 × 0.8 = 0.862, an increase of 0.02 over the value before selection.
: 0.3774
: 0.6038
However meiosis intervenes and makes0.0189
the next
generation
start in Hardy-Weinberg proportions at
the These
new gene
frequency
of 0.0189
+ not
0.3774/2
= 0.2076. The proportions:
genotype frequencies
areexcess
then of hetgenotype
frequencies
are
in Hardy-Weinberg
there is an
0.2076If the0.7924
erozygotes as a result of their high viability.
genetic system were asexual, these would be
0.0431 : 0.3290 : 0.6279
the genotype frequenciesSo
at after
the start
of theAA,
nextAa,
generation.
fitness
meiosis,
aa ratios The
are mean
p2 , 2pq,
q 2 , would
or: then be
0.0188
×
0.4
+
0.3774
×
1
+
0.6038
×
0.8
=
0.862,
an
increase
of
0.02
over
the
value
before
selection.
: 0.3290
: 0.6279
which gives a mean fitness0.0431
at the start
of that
generation of 0.0431×0.4+0.3290×1+0.6279×0.8
=
However
meiosis
intervenes
and
makes
the
next
generation
start
in
Hardy-Weinberg
proportions
at
0.84856. This is considerably lower than 0.868 but still above the initial mean fitness of 0.848. The
mean
fitness
= = 0.2076.
the new gene
frequency
+ 0.3774/2
frequencies
thento a net
increase
of mean
fitness of
by0.0189
0.02 due
to natural
selectionThe
hasgenotype
been rolled
back by are
meiosis
1 on
+ 0.6279×
= 0.84856
increase of 0.00056. The 0.0431×
disruptive0.4+0.3290×
effect of meiosis
genotypic 0.8
combinations
is apparent.
0.0431 : 0.3290 : 0.6279
If the population lacked meiosis, how much higher could its fitness be? This question was first
posed
Morton,
(1956),
defined and
calculated the segregational load. This
which by
gives
a meanCrow,
fitnessand
at Muller
the start
of thatwho
generation
of 0.0431×0.4+0.3290×1+0.6279×0.8
=
they
defined
as
the
fractional
reduction
in
the
fitness
of
a
population
as
a
result
of
the
existence
0.84856. This is considerably lower than 0.868 but still above the initial mean fitness of 0.848. The
of
Mendelian
segregation.
the due
presence
of overdominance,
asexual
could to
come
to
increase
of mean
fitness byIn0.02
to natural
selection has an
been
rolledpopulation
back by meiosis
a net
consist
of heterozygotes.
our of
parameterization,
it would
then haveisa apparent.
mean fitness of
increaseentirely
of 0.00056.
The disruptiveIneffect
meiosis on genotypic
combinations
1. An
outcrossing
population
would
come
to
equilibrium
at
a
gene
frequency
of
p
+ t)first
for
e = t/(swas
If the population lacked meiosis, how much higher could its fitness be? This question
allele
would
result
a mean
fitness
of defined and calculated the segregational load. This
posed A,
by which
Morton,
Crow,
and in
Muller
(1956),
who
Sexin hurts
they defined as the fractional reduction
the fitness of a population as a result of the existence
w̄ = 1 − sp2e − t(1 − pe )2
of Mendelian segregation. In the presence of overdominance, an asexual population could come to
1 − parameterization,
st2 /(s + t)2 − ts2 /(s
t)2 then have a mean fitness of
consist entirely of heterozygotes. =
In our
it +
would
2
= come
1 − st(s
+ t)/(s + t)at
1. An outcrossing population would
to equilibrium
a gene frequency of p e = t/(s + t) for
mean
fitness
without
sex
=
allele A, which would result in a mean
= 1fitness
− st/(sof+ t)
(II-116)
0.01881× 0.4+0.3774× 1 + 0.6038× 0.8 = 0.862
2
= 1with
− by
sp2esex
− t(1
mean w̄
= − pe ) of segregation, relative to the fitness of
The fraction by which fitness
isfitness
reduced
the
presence
2
Aa, is st/(s + t). This is0.0431×
half the harmonic
oft)+
s2 −
and
= 1 − stmean
/(s +1
ts2t./(s + 0.8
t)2 = 0.84856
0.4+0.3290×
0.6279×
The segregational load calculation
misinterpreted
= 1is−often
st(s +
t)/(s + t)2 as meaning that a population segregating at an overdominant locus somehow suffers a loss in fitness. Keep in mind that we have been
= 1 − st/(s + t)
(II-116)
The fraction by which fitness is reduced by the71
presence of segregation, relative to the fitness of
Aa, is st/(s + t). This is half the harmonic mean of s and t.
The segregational load calculation is often misinterpreted as meaning that a population segregating at an overdominant
locus somehowwhy
suffersthen
a loss indoes
fitness.sex
Keepexist?
in mind that we have been
Question:
71
Sex helps!
A
A
B
A
B
D
Sex helps!
A
A
B
B
A
C
D
B
C
D
II.11
Frequency-Dependent Fitnesses
If the fitnesses of genotypes vary as a function of the gene frequency or the genotype frequencies,
various complex outcomes are possible, including oscillation of the gene frequency and chaotic
fluctuation. We are most interested in a simpler outcome, stable polymorphism. A natural condition
When does
fitness
maximization
to examine is frequency-dependent
selection
in which
the rare allele is fail?
at an advantage. There
are a number of biological
mechanisms
which
have
been
proposed
which
lead to frequency
The case of frequency dependent would
selection
dependent selection:
1. Specialization on different limiting resources. If two genotypes eat different foods, then an
individual of the rare genotype will have a more abundent source of food, by virtue of the
rareness of other individuals who eat that food. The same argument will hold for many other
limiting nonfood resources, such as breeding sites.
2. Different diseases or parasites for different genotypes. If each genotype has its own diseases
and parasites, then whichever type is rarer will be less likely to come into contact with carriers
of its own particular pests.
86
3. Specialization of different predators on different genotypes. When each genotype has its own
predators, then the genotype which is rare will presumably sustain a lower population density
of predators, and hence might suffer a lower mortaility rate from predation.
4. Predator search images: apostatic selection. Many intelligent visual predators form “search
images” of the desired appearance of their prey. They tend to reject potential prey which do
not fit this image. The search image depends on the last few prey eaten. Thus the predators
may tend to avoid taking the rare genotypes, which they have not encountered recently.
5. Rare male advantage. In some species, notably Drosophila melanogaster, males of a rare
genotype seem to have an advantage in mating simply because they are rare. This pattern of
female choice may be an adaptation to avoid inbreeding.
3. Specialization of different predators on different genotypes. When each genotype has its own
6. Social Interactions. In a social species, if the genotypes differ in their social behavior, the
predators, then the genotype which is rare will presumably sustain a lower population density
fitnessof of
a genotype may depend on the frequencies of the genotypes among the individuals
predators, and hence might suffer a lower mortaility rate from predation.
it encounters in the population.
4. Predator search images: apostatic selection. Many intelligent visual predators form “search
The first
four ofofthese
mechanisms
involve
ecological
interactions,
the last
two behavioral
images”
the desired
appearance
of their
prey. They
tend to reject
potential
prey which interdo
When
does
fitness
maximization
fail?
actions. Innot
many
of
these
cases
the
natural
selection
would
be
expected
to
be
density-dependent
fit this image. The search image depends on the last few prey eaten. Thus the predators
as well as frequency-dependent.
Fortheexample,
when population
density
low, the first
mechanism
may tend to avoid taking
rare genotypes,
which they have
not is
encountered
recently.
(different resources) will not operate, since individuals of both genotypes will find an abundance
of food. 5.When
density
high,
the fitnesses
will dependmelanogaster,
on the genotype
Rare population
male advantage.
In is
some
species,
notably Drosophila
males frequencies.
of a rare
To analyzegenotype
the outcome
of
these
kinds
of
frequency-dependent
selection
requires
a model
seem to have an advantage in mating simply because they are rare. This
patternofofthe
specific case,
including
the numbers
of inbreeding.
predators or parasites, or the amount of each
female
choice variables
may be anfor
adaptation
to avoid
kind of food resource available. The details of the model will be strongly dependent on the spe6. Social
Interactions.
a social
if the genotypes
differ in their selection
social behavior,
thethis
cific biology
involved.
In this In
section,
wespecies,
will examine
frequency-dependent
without
fitness of a genotype
may depend
on the frequencies
of the genotypes
among the
biological specificity.
We will allow
the fitnesses
to be arbitrarily
chosen functions
of individuals
the gene freit encounters
in thetypes
population.
quency, in order
to see what
of evolutionary outcome are possible, and what the implications
of frequency-dependence are for the mean population fitness.
The first four of these mechanisms involve ecological interactions, the last two behavioral inter-
actions. In many
of these
cases the natural
selection
be two
expected
to be density-dependent
ASEXUALS
AND
HAPLOIDS.
Suppose
that would
we have
genotypes,
A and a, with the
as fitness
well as frequency-dependent.
For genotype
example, when
population
densityp,isin
low,
the firstlinear
mechanism
relative
of A depending on the
(or gene)
frequency,
a simple
fashion.
(different
resources)
will
not
operate,
since
individuals
of
both
genotypes
will
find
an
abundance
Let
of food. When population density is high, the fitnesses will depend on the genotype frequencies.
To analyze the outcome of these kinds of frequency-dependent selection requires a model of the
wA = 1 + t − sp
specific case, including variables for the numbers of predators or parasites, or the amount of each
(II-137)
kind of food resource available. The details of the model will be strongly dependent on the spew
=
1.
a
cific biology involved. In this section, we will examine frequency-dependent selection without this
biological specificity. We will allow the fitnesses to be arbitrarily chosen functions of the gene freThe
equations for the evolution of genotype frequencies then become
quency, in order to see what types of evolutionary outcome are possible, and what the implications
of frequency-dependence are for thep!mean population fitness.
p
= (1 + t − sp)
(II-138)
ASEXUALS AND HAPLOIDS.
1 − p! Suppose that we1 have
− p two genotypes, A and a, with the
relative fitness of A depending on the genotype (or gene) frequency, p, in a simple linear fashion.
and Let
p! =
p(1 + t − sp)
.
(II-139)
cific biology involved. In this section, we will examine frequency-dependent selection without this
ogical
specificity.
We We
willwill
allow
thethe
fitnesses
functionsofofthe
the
gene
biological
specificity.
allow
fitnessestotobe
bearbitrarily
arbitrarily chosen
chosen functions
gene
fre-frequency,
in order
to what
see what
types
of evolutionaryoutcome
outcomeare
are possible,
possible, and
ncy,
in order
to see
types
of evolutionary
andwhat
whatthe
theimplications
implications
of frequency-dependence
mean
populationfitness.
fitness.
requency-dependence
are are
for for
thethe
mean
population
ASEXUALS
AND
HAPLOIDS.
thatwe
we have
have
two genotypes,
AAand
Simplest
case Suppose
ofSuppose
frequency
dependence:
A,theathe
ASEXUALS
AND
HAPLOIDS.
that
two
genotypes,haploid
anda, a,with
with
relative
fitness
A depending
genotype(or
(orgene)
gene) frequency,
frequency, p,
fashion.
tive
fitness
of Aofdepending
on on
thethe
genotype
p, in
in aasimple
simplelinear
linear
fashion.
Let
wA = 1 + t − sp
wA = 1 + t − sp
wa
wa
(II-137)
(II-137)
= 1.
= 1.
The equations for the evolution of genotype frequencies then become
2.10: ∆p
as a function
of p forthen
a case
of frequency-dependent selection. The
The equations forFigure
the evolution
of genotype
frequencies
become
!
p
p
relative fitness of genotype
1.6sp)
− p.
= (1A+ist −
(II-138)
!
!
1
−
p
1 p− p
p
= (1 + t − sp)
(II-138)
1 − p!
1−p
and
p(1 + t − sp)
.10: ∆p as a function of p forp!a =case
of frequency-dependent
selection.
.
(II-139)The
+
(tt −
−frequency;
sp)p
When we compute the change p(1
of1 gene
it is, from (II-139)
+
sp)
!
fitness of genotype A is 1.6 −pp.
=
.
(II-139)
Figure 2.10:
a1 function
of −p[1for
case of +frequency-dependent
s
+
(t
87 −
∆p = ∆p
p! − as
p =
(p(1
+sp)p
t − sp)
+ (t a− sp)p])/(1
(t − sp)p)
relative fitness of genotype A is 1.6 − p.
(II-140
=
87
p(1 − p)(t − sp)/(1 + (t − sp)p).
The equilibria of the genotype frequency are the values of p at which ∆p = 0. For this to
pute the change
of gene
frequency;ofit(II-140)
is, from
(II-139)
occur, either
the denominator
must
be infinite, which is not possible, or the numerato
must be zero. The equilibria then occur at p = 0, p = 1, and p = t/s. This last equilibrium wil
= When
p! −liep we
(p(1
t the
− sp)
+of
−ifsp)p])/(1
(t −
in=
the
[0, 1] +
interval
ifchange
s−
>[1
t>
0(tor
0 >
t > s.+Otherwise
always be greater than
A will(II-139)
compute
gene
frequency;
it sp)p)
is, w
from
(less than) wa , and although selection will be frequency-dependent, it will nevertheless
always b
(II-140)
!
directional
selection
which
the
substitution
of one
another. + (t − sp)p
= p(1
p−
− p)(t
pleads=
(p(1
+
− [1genotype
+ (t −for
sp)p])/(1
= ∆p
−tosp)/(1
+ t(t−−sp)
sp)p).
Figure 2.10 shows ∆p plotted for particular values of s and t. These values lead the relative
ofEquilibrium
A to frequency
be higherp when
is rare
and
is rare. ∆p
It would
on intuitive
ground
bria of thefitness
genotype
areit !p
the
of when
p at awhich
= seem
0. For
this to
when
=values
0 lower
=
p(1
−
p)(t
−
sp)/(1
+
(t
−
sp)p).
that thisi.e.,
lead must
to a=0,
stable
polymorphism.
shows that
positive below th
numerator
p = 0 orwhich
p = 1The
pgraph
=possible,
t/s
he denominator
ofshould
(II-140)
besoinfinite,
isornot
or ∆p
theisnumerator
equilibrium point and negative above it. This shows that p = 0 and p = 1 are both unstabl
Last occur
in [0,of1]at
if ps =
>
t0,>p0 =
or1,iffrequency
0 > tp>=s t/s.
(o.w.,
wAvalues
>
wa ) of p at which
The equilibria
then
This
last
will
The
equilibria
the
genotype
areprovided
the
∆p
equilibria.
The equilibrium
p=
t/s will
beand
a stable one
that equilibrium
(by the stability
criterion
interval
if
seither
> t above)
>the
0 or
if 0 > t > of
s. (II-140)
Otherwise
w A be
willinfinite,
always which
be greater
developed
occur,
denominator
is notthan
possibl
! at must
"
Fitness
is not maximized
this
stable
point!
d(∆p)
andmust
although
selection
will
be
frequency-dependent,
it
will
nevertheless
always
be
−2 <occur at p = 0,
< 0.
(II-141l
be zero. The equilibria then
p
=
1,
and
p
=
t/s.
This
dp p = t/s
ctionlie
which
leads
to
the
substitution
of
one
genotype
for
another.
= 1.0;
pe =Otherwise
t/s = 0.6 w A will alw
in the [0, 1] interval if s > t > 0 ors if
0 >t =t 0.6;
> s.
After differentiating (II-140), substituting in p = t/s, and doing some tedious collection
of term
0 shows ∆p plotted for particular values of s and t. These values lead the relative
(less we
than)
wa , and although !selection
will be #
frequency-dependent,
it will nev
find that
" rare. It would
$ on intuitive grounds
be higher when
it is rare and lower when
a
is
seem
d(∆p)
t
directional selection which leads
to the substitution
= −t 1 − of. one genotype for anothe
(II-142
ld lead to a stable polymorphism. Thedpgraph
shows that ∆p
s is positive below the
p = t/s
Figure 2.10 shows ∆p plotted for particular values of s and t. These valu
int and negative above it. This shows that p = 0 and p = 1 are both unstable
fitness ofpA=tot/s
bewill
higher
it is
rare
and
88 lower when a is rare. It would seem
e equilibrium
be awhen
stable
one
provided
that (by the stability criterion
polymorphism. The graph shows that ∆p is
ve) that this should lead !to a stable
"
equilibrium point and d(∆p)
negative above it. This shows that p = 0 and p = 1
Figure 2.10:
selection. The
−2
< ∆p as a function of p for a case of<frequency-dependent
0.
(II-141)
equilibria. Therelative
equilibrium
=
fitness of genotype
− p. t/s will be a stable one provided that (by th
dp A isp1.6
p = t/s
developed above)
! some tedious
"
entiating (II-140), substituting in p = t/s, and doing
collection of terms
When we compute the change of gene frequency; it is, from (II-139)
!
∆p
" p −p
=
!
=
d(∆p)
−2 <$ + (t − sp)p)
(p(1 + t − sp) − #
[1 + (t − sp)p])/(1
< 0.
food.
When population
is high, the
fitnesses
depend
onstable
the
genotype
frequencies.
equilibrium
is unstable.
Thisfrequency
correponds
to
frequency-dependent
fitnesses
in which
wa when
stable
equilibrium
atpossible
thedensity
gene
at
which
wwill
= t/s
wa , iswith
equilibria
at w
pA
=<
0t and
A if
quite
that
there
will
be
oscillations
or
chaos
the=particular
The
equilibrium
is only
a relevant
one
between
zero
and
one.
ispnegative,
the
w̄
=
pw
+frequency-dependent
(1
−this
p)w
=
+ cases,
(1 − p)
1.
Ain
a If
o
analyze
the
outcome
of
these
kinds
of
frequency-dependent
selection
requires
a
model
of
the
A
is
rare,
and
the
opposite
when
A
is
common.
It
should
be
obvious
that
will
lead
to an
= 1. When
t
is
positive,
the
quantity
(II-142)
will
be
negative
(if
t/s
<
1,
which
we
assume).
linear
dependence
of
w
on
p
which
we
have
used
here
has
ruled
this
out.
equilibrium
is
unstable.
This
correponds
to
frequency-dependent
fitnesses
in
which
w
<
w
when
A
A
a
pecific
including
variables
for
numbers
of
orstable
the fitnesses,
amount
ofsoeach
equilibrium
at theon
gene
frequency
atpredators
which
w Aor=
w
equilibria
atwhen
p = t0 and
a , with
ereunstable
is case,
oneAfurther
restriction
tthe
and
s.
It makes
no sense
toparasites,
have
negative
is
rare,
and
the
opposite
when
A
is
common.
It
should
be
obvious
that
this will
lead
to mean
an
Does
frequency-dependent
selection
necessarily
maximize
some
measure
of the
fitness?
The
maximum
value
of
w̄
can
be
found
by
writing
nd
of
food
resource
available.
The
details
of
the
model
will
be
strongly
dependent
on
the
spep
=
1.
When
t
is
positive,
the
quantity
(II-142)
will
be
negative
(if
t/s
<
1,
which
we
assume).
d s are positive
weequilibrium
must have 1at+the
t −gene
s > 0,
so that sat−which
t < 1. wConsideration
of
the right-hand
unstable
frequency
=
w
,
with
stable
equilibria
at
p
=
0
and
A
a
This
is
investigated
in
the
present
case. to
Athave
theselection
polymorphic
equilibrium
fic There
In easily
this
section,
wetwill
frequency-dependent
without
thisso when tp = t/s, w A = 1 =
isinvolved.
one(II-142)
further
restriction
on
andexamine
s.−
Itt),makes
noitsense
negative
fitnesses,
e ofbiology
equation
shows
it to the
be
−(t/s)(s
so
that
will
never
below
-1
inthe
biologically
whichever
has
higher
fitness.
Insince
doing
sobeit
alters
fitness
of assume).
A for the worse.
p=w
1.agenotype
When
t isallow
positive,
the
quantity
(II-142)
will
be negative
(if
t/s gene
< 1,frewhich we
ological
specificity.
We
will
the
fitnesses
to
be
arbitrarily
chosen
functions
of
the
,
so
that
the
mean
relative
fitness
is
1,
and
s are positive
we must
have
+t−
> 0, equilibrium
so that
− twith
< so
1.noConsideration
of
the
right-hand
whichever
the 1have
higher
fitness.
In sdoing
itovershooting.
alters
the
fitness
evant
cases.
In genotype
this case
wehas
always
a sstable
While
itof
is Aa for the worse.
w̄
=
pw
+
(1
−
p)w
A
There
one
further
restriction
on t outcome
and
s. Itare
makes
nocurrent
sense
to fitness
have
negative
fitnesses,
so to
when
t
Natural
selection
will
maximize
mean
fitness
if
is-1apoint!
good
guide
future
fitness.
uency,
inoforder
toissee
what
types
ofFitness
evolutionary
possible,
and
what
the
implications
equation
(II-142)
shows
it
to bemean
−(t/s)(s
−in
t),frequency-dependent
soonly
that
it will
be
below
in
biologically
ischaos
not
maximized
atnever
this
stable
iteside
possible
that
there
will
be
oscillations
or
cases,
the
particular
Natural
selection
will
maximize
fitness
only
if
current
fitness
is
a
good
guide
to
future
fitness.
and s are positive
must have
1w̄+ =
t −fitness.
s>
that
s a− t=< p1.
Consideration
of the
right-hand
frequency-dependence
arecase
forwe
the
population
pw
+ so
(1 −
p)w
(1p(1
− p)
(II-143)
A 0,
=+overshooting.
+ t=− 1.
sp)
+it(1
relevant cases. In this
we mean
always
have
a stable
equilibrium
with no
While
is− p)
ear dependence of wA on p which we have used here has ruled this out.
side of equation
(II-142)
shows it to beor−(t/s)(s
− t), so that it will never
be below
-1 in biologically
quitefrequency-dependent
possible AND
that there
will
be oscillations
chaos
in frequency-dependent
ASEXUALS
HAPLOIDS.
Suppose
thatmaximize
we have
two
genotypes,
a,mean
withthe
theparticular
Does
selection
necessarily
some
measureAofand
thecases,
fitness?
DIPLOIDS.
All
ofpthe
the
phenomena
which
we
see
in
haploids
and
cases
relevant The
cases.
In on
this
case
we
have
afound
stable
equilibrium
no
overshooting.
itofisfrequencyvalue
ofhave
w̄(orfitness.
can
be
by
writing
=with
1and
+t/s,
p(t
−asexual
sp).=
whichever
genotype
the
higher
In
doing
so
alters
the
for While
the
dependence
ofmaximum
w
which
wealways
used
here
has
this
out.
DIPLOIDS.
All
of
phenomena
which
we
inin1it
haploids
asexual
of worse.
frequencylative
fitness
of A depending
onhas
the
genotype
frequency,
p,
a=
simple
fashion.
is linear
is easily
investigated
inAAt
the
present
case.
At the
polymorphic
equilibrium
= fitness
w Aof=A1cases
equilibrium,
pgene)
=
t/s,
wsee
=
w
Aruled
a plinear
quite
possible
that
there
will
be
oscillations
or
chaos
in
frequency-dependent
cases,
the
particular
also
occur
in
diploids.
Ifwefitness
we
consider
a diploid
population
with
multiplicative
fitnesses,
Natural
selection
will
maximize
mean
only
current
is a of
good
to
future fitness.
Does
selection
necessarily
maximize
somefitness
measure
theguide
mean
fitness?
et
dependence
also
occur
inmean
diploids.
If
consider
aif diploid
population
with
multiplicative
fitnesses,
, sodependence
that
thefrequency-dependent
mean
relative
fitness
is 1,relative
since
so
fitness
is
1:
linear
dependence
of
w
on
p
which
we
have
used
here
has
ruled
this
out.
A
w̄
=
pw
+
(1
−
p)w
little
is
changed.
the
fitnesses
This
is are
a are
quadratic
which
can
A of p
a
This
easily
investigated
infitnesses
the present
case. At thefunction
polymorphic
equilibrium
p =be
t/s,maximized
w A = 1 = by equating its
littleis is
changed.
IfIfthe
Does
frequency-dependent
selection
necessarily
maximize
some measure of
the mean fitness?
w
=
1
+
t
−
sp
w̄
=
pw
+
(1
−
p)w
=
p
+
(1
−
p)
=
1.
(II-143)
A
wa , so thatDIPLOIDS.
the mean
relative
fitness
is
1,
since
a
(II-144)
All of Atheinphenomena
wep(1
see+
int haploids
and
cases
of frequency− sp) + (1
− asexual
p)
This is easily investigated
the present which
case.=At
the
polymorphic
equilibrium
p = t/s,
wA = 1 =
(II-137)
d
w̄
dependence
also mean
occur relative
in diploids.
If we
a diploid population with
= tmultiplicative
− 2sp (II-143)
= 0,fitnesses,
wa , so that
1,consider
The maximum
valuethe
of w̄ can
by(1writing
waAfitness
1.
w̄ be
= found
pw
+=
−isp)w
==p +1 (1
1.
+−
p(tp)−=sp).
asince
dp
little is changed. If the fitnesses are
AA
AA
AaAa aa aa
The The
equations
for thevalue
evolution
of genotype
frequencies
then
become
2
w̄
=
pw
+
(1
−
p)w
=
p +s(p)
(1 − p)1 = 1.
(II-143)
2
maximum
of w̄w̄can
be
found
by
writing
A
a
= pw + (1
[1−+p)w
s(p)] 1 +
der
A [1 +ofs(p)]
a
This is aso
quadratic
p which
be maximized
1 can
+ s(p)
1
that thefunction
maximum
or
minimum
occurs
atby equating its derivative to zero:
!
p writing
(II-144)
p(1
++t t−
sp)
+ (1
− p)
The maximum value pof=w̄
w̄ =
can
AA
aa
(II-138)
=(1be
pwfound
(1 by
−d
p)w
w̄ a Aa
A−+sp)
!
t
1−p
−2 p = t − 2sp = 0,
(II-145)
s(p)]
1+
s(p)
1 thisthis
function
of1of
gene
frequency,
willwill
have
exactly
the same
gene
with the
the selection
selection coefficient
coefficient
function
gene
frequency,
have
exactly
the same
gene
=s sa1 a+
p(t [1
−+
sp).
p
=
dp
(II-144)
= w̄
p(1 =
+ t pw
− sp)++(1(1−−p)w
p)
2s
A
a
frequency behavior
which
fitnesses
areare
nd frequency
behaviorasasthe
thehaploid
haploidcase
caseinin
which
fitnesses
=
1
+
p(t
−
sp).
This is awith
quadratic
function
of
p
which
can
be
maximized
by
equating
its derivative
zero: the same
p(1
+
t
−
sp)
so
that
the
maximum
or
minimum
occurs
at
(II-144)
! s a function
= p(1
+ t −frequency,
sp) + (1 −this
p)
the selection coefficient
of gene
will have to
exactly
gene
where
p =
(II-139)
.
1 +case
(t −in
sp)p
2
frequency behavior as the haploid
t
= be1which
+ p(tfitnesses
− sp).by are
This is a quadratic function ofdpw̄which
can
tot zero:
= t−
2sp =maximized
0,
(II-145)
AA
p a=aequating itsw̄derivative
=
1
+
.
dp
2s
87
4s
+
s(p)
1. 1. by equating its derivative to zero:
w̄ p which 1can
This is a quadratic function dof
be
maximized
1A+
s(p)
=
t
−
2sp
=
0,
a
s is at
positive,
which will be the case when we (II-145)
have a stable polymorphic
that the maximum
whereor minimumIfoccurs
dp
2 2
1
+
s(p)
1.
t
d
w̄
Equilibria
will
atatp p==0,
any
of of
poffor
which
s(p)
= stationary
0.=If0.s(p)
is point
positive
quadratic
has
negative
coefficient
so which
that
the
p = t/(2s)
tand
w̄ value
==
10,+
.pp for
(II-146)
=attat
−
2sp
(II-145)
Equilibria
willoccur
occur
0,ppp=aat
=1,
1,
and
any
value
s(p)
If
s(p)
is positive
so that
the maximum
or
minimum
occurs
=
dp
4s
below a polymorphic
equilibrium
above
it, ofthe
equilibrium
could
but only and the
thus
no
correspondence
between
the
equilibrium
will There
occur
atisp =
0,and
pand
= negative
1,2s
and at any
value
p the
for
which
s(p)polymorphic
=
0.could
If be
s(p)stable,
is positive
below a Equilibria
polymorphic
equilibrium
negative
above
it,
equilibrium
be
stable, but only
t
If
s
is
positive,
which
will
be
the
case
when
we
have
a
stable
polymorphic
equilibrium,
the
if wild
from
overshooting
can
berelative
ruled
out.
writing
expression
for ∆p
containing
sooscillations
thata the
maximum
or minimum
occurs
at
below
polymorphic
equilibrium
negative
above
it, By
theBy
equilibrium
could
be stable,
but
only
maximizes
theand
mean
fitness.
Inwriting
fact,anmaximum
occurs
at
half
the equilibriu
pcan
=
ereif wild
oscillations
from
overshooting
be
ruled
out.
an
expression
for
∆p
containing
2
2s
2 we
has
a and
negative
coefficient
of
pbeso
the
stationary
point
=∆p
t/(2s)
is the
maximum.
s(p), ifdifferentiating
this
requiring
that
attthat
aBy
point
where
s(p)
=
0, pfor
we
find
that
forabove,
local it will h
tbe
wildquadratic
oscillations
from
overshooting
can
ruled
out.
writing
an expression
containing
in
this
case.
If 1the
population
approaches
thes(p)
equilibrium
from
s(p), differentiating
this
and
requiring
that
we
be
at
a
point
where
=
0,
we
find
that
for
local
w̄
=
+
.
(II-146)
There
is thus this
no correspondence
between
the
polymorphic
equilibrium
and
the
p which
p be
= at
s(p),of
differentiating
and requiring
that
we
at
a point
where s(p)
= 0, we
find
that
forvalue
local0ofand
stability
the polymorphism,
the
slope
of
s(p)
the
equilibrium
point
must
be
between
where
4s
2s
increasing
w̄.
But
if
instead
it
approaches
from
below,
w̄
at
first
increases,
then d
2
stability
of
the
polymorphism,
the fitness.
slopeofof
at the
equilibrium
point
must
be 0between
0 and
ts(p)
maximizes
thebe
mean
In
fact,
maximum
occurs
at half
thebeequilibrium
gene
stability
polymorphism,
slope
s(p)
the
equilibrium
point
must
between
and frequency
)].the will
If −2/[p
s is positive,
the relative
case the
when
we
have
a. at
stable
polymorphic
equilibrium,
the
e (1 − pwhich
eof
w̄
=
1
+
(II-146)
particular
example
in
the
tp =
0.6 and
s =maximum.
1, so itthat
the
equilibrium
lies at
−2/[p
−
)].
where
4s Figure,
in
this
If theof population
approaches
the
equilibrium
from
will
have
a continually
e (1
ethree
−2/[p
(1p
−
pecase.
)].
adratic
has
a negative
coefficient
p 2have
so that
the
stationary
point
=wrecks
t/(2s)functions
is
the above,
eall
When
genotypes
fitnesses
which
are
arbitrary
of
the
gene
frequency,
2
Frequency
dependent
fitness
method
of
t stable
The
mean
fitness
there
is
maximum
mean
fitness
isfrequency,
achieved
at
s isno
positive,
which
will
be
thehave
casepolymorphic
when
have
aThe
polymorphic
equilibrium,
the
increasing
w̄.
But
ifthe
instead
itfitnesses
approaches
below,
atmodel.
first
increases,
decreases.
Forp max
the = 0
three
genotypes
have
which
are
functions
gene
When
all all
three
genotypes
fitnesses
which
are
arbitrary
functions
ofthen
the
gene
frequency,
erethere
is Ifthus
correspondence
between
the
equilibrium
and
the
value
ofofThe
pthe
which
w̄we
=
11.from
+
.arbitrary
(II-146)
is When
hardly
any
limit
to
complexity
of
the
behavior
of w̄
the
equations
of change
2
4s
using
slope
of
w
to
direct
local
search
quadratic
has
a
negative
coefficient
of
p
so
that
the
stationary
point
p
=
t/(2s)
is
the
maximum.
there
is hardly
to
the
complexity
of0.6
theand
of
The
equations
of
particular
example
infact,
the complexity
Figure, t =
sbehavior
=the
1,equilibrium
sothe
that
thegene
equilibrium
lies
at change
p e = 0.6/1
= 0.6.
w̄limit
= to
1.09.
aximizes
the
mean
relative
fitness.
In
maximum
occurs
atbehavior
half
frequency
there
isgene
hardly
anyany
limit
the
of
the
ofmodel.
the
model.
The
equations
change
of the
frequency
are
the
usual
ones,
but
now
with
fitnesses
being
functions
ofthe
p. ofThe
Ifthe
sgene
is
positive,
which
will
be
the
case
when
wemean
havethe
athe
stable
polymorphic
equilibrium,
There
is Ifthe
thus
no
correspondence
between
the
polymorphic
equilibrium
and
the
value
of
p
which
of
frequency
are
the
usual
ones,
but
now
with
fitnesses
being
functions
of
p.
The
this
case.
population
approaches
the
equilibrium
from
above,
it
will
have
a
continually
The
mean
fitness
there
is
1.
The
maximum
fitness
is
achieved
at
p
=
0.6/2
=
0.3,
max
Why
is usual
w̄points
not
maximized?
Since
wpoint
is pa=function
offunctions
p, the current
fitness o
of
thequadratic
gene
frequency
are
the
ones,
but
now
with the
being
of p. where
The
A fitnesses
2p so
equilibria
of
the
model
are
at
the
=
0,
p
=
1,
and
has
a
negative
coefficient
of
p
that
the
stationary
t/(2s)
is
the
maximum.
maximizes
the
mean
relative
fitness.
In
fact,
maximum
occurs
at
half
the
equilibrium
gene
frequency
equilibria
of1.09.
the model
are at the
points
p=
= 1,increases,
and
reasing w̄.
But
if=
instead
it approaches
from
below,
w̄ 0,
at pfirst
then decreases. For the
w̄
not
necessarily
a good
its
future
fitness.
Natural
selection increases
equilibria
model
areapproaches
at the points
pthe
= polymorphic
0,guide
p=
1,toabove,
and
There
isinthe
thus
no is
correspondence
equilibrium
and
the
value
of p which
in
thisexample
case. of
If
the
from
itat
will
have
a continually
rticular
the population
Figure,
t =maximized?
0.6 and sbetween
= the
1,Since
soequilibrium
that
the
equilibrium
lies
p
=
0.6/1
=
0.6.
e
Why
is w̄ not
w
is
a
function
of
p,
the
current
fitness
of
the
A genotype
A
w
(p
waaaa
Aa
e ) halfthen
wmaximum
(p
))−−
w
(p(p
maximizes
mean
relative
fitness.
Infitness
fact,
occurs
the equilibrium
genethe
frequency
Aa
e )at
increasing
w̄.there
Butthe
ifis instead
it
approaches
from
below,
w̄ eeat
first
increases,
decreases.
For
e mean
fitness
1.
The
maximum
mean
is
achieved
at
p
=
0.6/2
=
0.3,
where
p
=
(II-147)
max
p
=
(II-147)
e
is not necessarily ae good
guide
its w(p
future
Natural
selection
increases the frequency of
)fitness.
−
w
(p
)AA
Aa
aa
ewabove,
(p(pees))to
−
w
)](p
+e[w
[w
)−
w
)]89
[w
−1,
wthe
+
(p(p
(pat
)]itpeewill
case. inIfthe
theFigure,
population
approaches
equilibrium
from
have =
a 0.6.
continually (II-147)
Aa
aa
Aa
AA
example
0.6
and
=
so(p
that
the
equilibrium
lies
= 0.6/1
Aa
aa
ee)]
Aa
e )e−
e(p
= particular
1.09. in this
pe t==[w
increasing
w̄.
Butis if1.
instead
itisapproaches
from
w̄ [w
at Aa
first
decreases.
[w
(p
wp,aabelow,
(pise )]
+
) of−=
w0.6/2
(p
The ismean
fitness
there
Thewmaximum
mean
fitness
achieved
at(ppincreases,
=
0.3, where For the
e) −
emax
AA
e )]
Why
w̄
not
maximized?
Since
a Aa
function
of
the
current
fitness
the
Athen
genotype
Amany
This
equation
may
have
roots,
depending
on
the
way
in in
which
the
w’s
depend
on =
p. on
Wep. We
89
This
equation
may
have
many
roots,
depending
on
the
way
which
the
w’s
particular
example
in
the
Figure,
t
=
0.6
and
s
=
1,
so
that
the
equilibrium
lies
at
p e of
=depend
0.6/1
0.6.
= 1.09. a good guide to its future fitness. Natural selection increases the frequency
notw̄ necessarily
can
divide
w’s
by
w
(p)
and
write
fitnesses
as fitnessonis the
Aahave
canWhy
divide
by
w
(p)
and
write
fitnesses
as
This
may
roots,
depending
way
inpof
which
the
w’s
depend
The
mean
there
isSince
1.many
The
mean
achieved
at
= 0.3,
where on p. We
Aa
is equation
w̄w’s
not fitness
maximized?
wAmaximum
is
a function
of p, the current
fitness
the=A0.6/2
genotype
max
This
equation
has
many
roots,
depending
on
w
wrt
p
w̄
=
1.09.
iscan
notdivide
necessarily
to itswrite
future
fitness. Natural
selection increases the frequency of
w’s abygood
wAaguide
(p) and
fitnesses
as
89
we divideSince
w’s wby
w
(p)
to
get
fitnesses:
Why is w̄ not maximized?
is
a
function
of
p, the
current fitness of the A genotype
Aa
A
AA
Aa
aa
AA
Aa
aa
is not necessarily a good guide to its1 −
future
fitness.
Natural
selection
increases the frequency of
89 e ) 1 1 − t(pe )
s(p
1 − s(p
AAe )
1 Aa1 − t(paa
e)
in which case the equilibria are at 1p −
= s(p
0, p =
)891, 1and 1 − t(pe )
in which case the equilibria are at p = 0, pe= 1, and
t(pe )
.
(II-148)
in which case the equilibria are at ppe ==0,s(p
p e=
)t(p
+1,et(p
)and
e)
pe =
.
(II-148)
t(pthe
e) +
e ) gene frequency changes according to
The principle at work here is that in any s(p
generation
t(p
)
e
pe = equilibrium can. only occur if the fitnesses at the (II-148)
the momentary fitnesses, so that a polymorphic
s(pe ) + t(p
The principle at work here is that in any generation
the
e ) gene frequency changes according to
value of p yield an equilibrium at that value of p. However it is now neither necessary nor sufficient
the momentary fitnesses, so that a polymorphic equilibrium can only occur if the fitnesses at the
The
work here
that
in any
generation
gene
frequency
changes
according to
value
of pprinciple
yield an at
equilibrium
at is
that
value
of p.90
However it the
is now
neither
necessary
nor sufficient
the momentary fitnesses, so that a polymorphic equilibrium can only occur if the fitnesses at the
Why this causes problems:
Current fitness is no longer a good guide
to future fitness!
to the pool
of Ai gametes
will be proportional
the fitness
Ai A(II-35),
. Thethe 1
ij , wij being
1 ofand
2 pi pj w
EFFECT
OF SELECTION.
We can to
directly
generalize
equations
(II-31)
pijpjgene,
wij + 2
pj pi w
total basic
frequency
of
A
copies
in
the
gene
pool,
those
A
copies
that
come
from
the
left-hand
2
i
i
equations for gene frequency change with two alleles. The extension is straightforward.
The
j=1
j=1
1
! also
is thegenotype
sum overfrequency
j of 2 pi pofj wthe
Of course,genotype
an Ai inAthe
gene pool could
have comeis from
ij . (ordered)
after
fertilization
p i pj . an
This
i Aj immediately p
i =
1 !!
individual
of
genotype
A
A
,
from
which
it
is
a
copy
of
the
right-hand
gene,
so
there
is
a
similar
j
i
2
×
pi pj wij
holds for all i and all j, including the case where i = j. The contribution from these individuals
2
1
1
sum of
p
p
w
.
The
total
number
of
genes
in
the
gamete
pool
is
the
sum
of
these
expressions
over
j ipool
ij of Ai gametes will be proportional to pi pj wij , wij being the fitness of Ai Aj . The
to 2the
!
2
all i and
so the gene
frequency
of A in the gene pool is:
totalj,frequency
of A
pj wij )gene,
i copies in thei gene pool, those A i copies that come frompthe
i ( left-hand
1
is the sum over j of 2 pi pj wij . Of course, an Ai in the gene pool could also havej come from an
,
n
nof the right-hand =
individual of genotype Aj Ai , from 1which
it is a copy
gene, so there is a similar
!
1 !
w̄
1
p
p
w
+
p
p
w
i jinijthe gamete
j pool
i ij is the sum of these expressions over
2
sum of 2 pj pi wij . The total number 2ofj=1
genes
j=1
!
!
!
p
=
!
!
all i and j, so the gene frequency
is:
i
1 gene poolw
where
w̄ of=Ai 2in×the
mean fitness
fitness of the population. I
ijij , the
= mean
i 2 j ppiippjj w
Multiple allele case
!
straightforward
quantity:!
pi ( p!
n the mean fitness of the organisms in
jnwij )
1
j1
p
p
w
+
i j ij
2
= by 2the
,
(II-118) We call it w̄ i
weighted
ofpj pAi wi ijalleles they contain.
j=1 numbers
j=1
!
w̄
pi =
1 !!
2 × w̄
pi pj ww̄
! !
! used in the two-allele
to the quantities
ij a which we
2 A and
where w̄ =
the population.
In this
equation j pj wij is a quite
i
j pi pj wij , the mean fitness of!
as pofi ( thepj worganisms
straightforward quantity: rewritten
the mean fitness
in which A i alleles find themselves,
ij )
j
pi w̄i
=
mean
fitness of organisms in
which
weighted by the numbers of Ai alleles they
= contain. We ,call it w̄
(II-118)
i . As such these are direct! parallels
p
=
w̄
i
Ai alleles
find themselves
to the quantities !
w̄A!and w̄a which we used in the two-allele
argument.
Then (II-118)
can be w̄
!
where
w̄
=
p
p
w
,
the
mean
fitness
of
the
population.
In
this
equation
p
w
rewritten as
weighted by the number
ofa quite
i
j i j ij
j j ij is
p
w̄
i
i
which
also
leads
us
directly
to
!
straightforward quantity: the mean fitness
p =of the organisms in which A i alleles find themselves,
(II-119)
they contain
i
w̄ We callAiti w̄alleles
weighted by the numbers of Ai alleles they contain.
i . As such these are direct parallels
w̄i
whichtoalso
us directly
to w̄a which we used in the two-allele argument. Then (II-118)
theleads
quantities
w̄A and
can be
!
∆p
=
p
−
p
=
p
i
i
i
i
rewritten as
w̄i
w̄
∆pi = p!i −pp! i == ppiiw̄i − pi
(II-119)
i
w̄
w̄
(w̄i − w̄)
which also leads us directly to
(w̄i − w̄)
=
p
.
i
= pi
.
(II-120)
w̄i
w̄
!w̄
∆pi = pi − pi = pi
− pi
w̄
73
73
(w̄i − w̄)
= pi
.
(II-120)
w̄
73
Equllibrium conditions
At equillibrium, for allele i, either
pi =0 or w̄=0
− pi
The shape of things to come
!p =
p(1 " p) dw
w
dp
! G
!p = "w
w
w=!
i
!p p w
i
j
ij
j
!w
= 2" p j wij = 2wi
!pi
j
pi ' = pi
wi
w
pi ' =
pwi 2 !w
=
w
w !pi
"pi =
p j !w
pi !w
w
# pi substitute for w = $
2w !pi
w
j 2 !p j
"pi =
pi % !w
!w (
#
p
$ j !p *)
2w '& !pi
j
j
The multiple allele jet-fuel formula
" p1 %
" p1 (1 ( p1 )
1 $
!
$
'
!pi = ! $ p2 ' =
( p2 p1
2w $
$# p3 '&
$# ( p3 p1
" )w %
$
'
)p
( p1 p2
( p1 p3 % $ 1 '
$ )w '
p2 (1 ( p2 )
( p2 p3 '' $
'
)p2 '
$
'
( p3 p2
p3 (1 ( p3 ) &
$ )w '
$
'
variance
# )p3 &
! G
!p = "w
w
The shape of things to come...
grad mean fitness
wrt pi