
The reproductive success of an individual
over its lifetime is known as its fitness.

When individuals differ in their fitness
selection takes place.

Fitness can be difficult to measure over
an organisms lifetime.

Instead other measures that correlate
well with lifetime fitness are used to
estimate fitness: e.g. survival to
reproductive age or reproductive
success in a single season.

The goal in studying selection is to relate
variation in fitness to variation in
phenotype.

E.g. we can try to compare variation in
fitness to an animal’s size or camouflage
color or some other phenotypic
measure.

Remember, fitness is a result of the
organisms entire phenotype (and
genotype).

Population genetics, however, looks at
the evolution of alleles at a single locus.

Population geneticists condense all the
components of fitness (survival, mating
success, etc.) into one value of fitness
called w.

Evolution depends on changes in the
gene pool so we need to consider how
alleles affect fitness rather than how
genotypes affect fitness.

The general selection model (next slide)
enables us to assess how individual
alleles contribute to fitness.
Genotype
Initial freq
Fitness
Abundance
In gen t+1
A1A1
p2
w11
A1A2
2pq
w12
A2A2
q2
w22
p2 X w11
2pq X w12 q2 X w22
Weighted
freq. gen t+1 (p2 X w11)/w (2pq X w12)/w (q2 X w22)/w
The term “Abundance in gen t+1” tells us for each
genotype its abundance relative to other
genotypes in the next generation
Abund. gen t+1
p2 X w11
2pq X w12
q2 X w22
To convert these to true frequencies we standardize
them by dividing them by the average fitness of
the population w.

w = p2 X w11 + 2pq X w12 + q2 X w22

Note that the formula is the sum of the
fitness values for each genotype
multiplied by (i.e. weighted by) the
genotype frequencies.
Normalized weighted freq. gen t+1
(p2 X w11)/w (2pq X w12)/w (q2 X w22)/w
These are the frequencies of each genotype
in generation t +1.

Using these weighted genotype
frequencies we can calculate the allele
frequencies in generation t+1.

Need to sum alleles across genotypes.

For the allele A1 it will be the frequency
of the A1A1 homozygotes plus half the
frequency of the A1A2 heterozygotes.

Frequency of allele A1 [p(t+1)]

p(t+1) = (p2 X w11 + pq X w12)/w

Frequency of allele A2 [q(t+1)]

q(t+1) = (q2 X w22 + pq X w12)/w
Starting allele frequencies: A1 = 0.8, A2 = 0.2
Fitness
w11
0.9
w12
1.0
w22
0.2
w = p2 X w11 + 2pq X w12 + q2 X w22
= (0.64 X 0.9) + (0.32 X 1) + (0.04 X 0.2)
= 0.576 + 0.32 + 0.008
= 0.904
P(t+1) = (p2 X w11 + pq X w12)/w
P(t+1) = 0.64 X 0.9 + 0.16 X 1)/0.904
= 0.576 + 0.16/0.904
= 0.814
Allele A1 has increased in abundance slightly.
In this example the success of the alleles A1
and A2 is very sensitive to the frequency of
A2.

In this example, heterozygotes have the
highest fitness, but if A2 becomes too
common A2A2 homozygotes begin to
appear and these have very low fitness.
At lower frequencies of A2 then A2A2
homozygotes will be rarer and the A2 allele
will increase.
 In next slide we lower frequency of A2 to 0.1.

Allele frequencies: A1 = 0.9, A2 = 0.1
Fitness
w11
w12
w22
0.9
1.0
0.2
w = p2 X w11 + 2pq X w12 + q2 X w22
= (0.81 X 0.9) + (0.18 X 1) + (0.01 X 0.2)
= 0.729 + 0.18 + 0.002
= 0.911
P(t+1) = (p2 X w11 + pq X w12)/w
P(t+1) = (0.81 X 0.9 + 0.09 X 1)/0.911
= (0.729 + 0.09)/0.911
= 0.899 (allele A1has declined very
slightly from frequency of 0.9 and
allele A2 has increased to a frequency
of 0.101

There are other ways of computing the
effects of selection on allele frequency.

One approach uses something called
the average excess of fitness.

A relatively simple formula allows us to
calculate the net fitness contribution of
an allele, which is called the average
excess of fitness.

This is the difference between the
average fitness of individuals with that
allele and the average fitness of the
entire population.
For example, for the allele A1 the
average excess of fitness is
 aA1= [p X (w11 – w)] + [q X (w12 – w)]





Where w11 – w is the difference in fitness between A1A1
individuals and the mean fitness of the population w.
W12 is fitness of A1A2 heterozygotes. W is mean fitness of
population
P and q are allele frequencies
See Box 6.5 in your text page 168 for derivation of this formula.

The average excess of fitness can be used
to calculate how much an allele frequency
will change between generations

Δp = p X (aA1/w)




Δp is change in allele frequency from one generation to the
next
p is the frequency of the A1 allele
aA1 is the average excess of fitness
Average fitness of the population is w

If the average excess of fitness is positive
then an allele will increase in frequency.

If average excess of fitness is negative
then the allele will decrease in
frequency.

Δp = p X (aA1/w)
The equation tells us that how fast an allele
increases or decreases depends on both
the strength of selection (value of aA1) AND
how common an allele is in the population
(p).
 For rare alleles even strong selection may
not result in a rapid increase in an allele’s
frequency.


Alleles can differ greatly in their fitness. E.g.
some alleles cause severe diseases and are
strongly selected against.
 Many
alleles however differ only
slightly in their average excess of
fitness, but because the effect of the
fitness difference compounds over
time (just like interest on money) even
small differences can result in big
changes.

The compounding effect of natural
selection is more effective in large
populations than small ones.

In small populations drift can easily
eliminate beneficial mutations. In larger
populations drift has less of an effect.
Effects of drift strong in small populations but
weaker in large populations
 Small advantages in fitness can lead to large
changes over the long term in large populations.

Under directional selection one allele is
consistently favored over the other allele so
selection drives allele frequencies in only
one direction towards a higher frequency
of the favored allele.
 Eventually a favored allele may replace
other alleles and become fixed.


Whether an allele is dominant, recessive
or has additive effects (is codominant)
will have a strong influence on how fast it
spreads in a population.
Additive: allele yields twice the
phenotypic effect when two copies
present
 Dominance: dominant allele masks
presence of recessive in heterozygote
 Recessive: two copies of recessive allele
need to be present for alleles effect to
be felt.


Clavener and Clegg’s work on
Drosophila.

Two alleles for ADH (alcohol
dehydrogenase breaks down ethanol)
ADHF and ADHS

Two Drosophila populations maintained:
one fed food spiked with ethanol,
control fed unspiked food.

Populations maintained for multiple
generations.

Experimental population showed
consistent long-term increase in
frequency of ADHF
Flies with ADHF allele have higher fitness
when ethanol present.
 ADHF enzyme breaks down ethanol
twice as fast as ADHS enzyme.

Fig 5.13

Jaeken syndrome: patients severely
disabled with skeletal deformities and
inadequate liver function.

Autosomal recessive condition caused by
loss-of-function mutation of gene PMM2
codes for enzyme phosphomannomutase.

Patients unable to join carbohydrates and
proteins to make glycoproteins at a high
enough rate.

Glycoproteins involved in movement of
substances across cell membranes.

Many different loss-of-function mutations
can cause Jaeken Syndrome.

Team of researchers led by Jaak Jaeken
investigated whether different mutations
differed in their severity. Used HardyWeinberg equilibrium to do so.

People with Jaeken syndrome are
homozygous for the disease, but may be
either homozygous or heterozygous for a
given disease allele.

Different disease alleles should be in
Hardy-Weinberg equilibrium.

Researchers studied 54 patients and
identified most common mutation as
R141H.

Dividing population into R141H and
“other” alleles. Allele frequencies are:
Other: 0.6 and R141H: 0.4.
If disease alleles are in H-W equilibrium
then we would predict genotype
frequencies of
 Other/other: 0.36
 Other/R141H: 0.48
 R141H/R141H: 0.16


Observed frequencies are:
Other/Other: 0.2
Other/R141H: 0.8
R141H/R141H : 0
Clearly population not in H-W equilibrium.
Researchers concluded that R141H is an
especially severe mutation and
homozygotes die before or just after
birth.
 Thus, there is selection so H-W assumption
is violated.


If an allele has a positive average excess of
fitness then the frequency of that allele
should increase from one generation to the
next.

Obviously, the converse should be true and
an allele with a negative average excess of
fitness should decrease in frequency.

Dawson (1970). Flour beetles. Two
alleles at locus: + and l.

+/+ and +/l phenotypically normal.

l/l lethal.

Dawson founded two populations with
heterozygotes (frequency of + and l
alleles thus 0.5).

Expected + allele to increase in
frequency and l allele to decline over
time.

Predicted frequencies based on
average excess of fitness estimates and
observed allele frequencies matched
very closely.

l allele declined rapidly at first, but rate
of decline slowed.
Fig 5.16a

Dawson’s results show that when the
recessive allele is common, evolution by
natural selection is rapid, but slows as the
recessive allele becomes rarer.

Hardy-Weinberg explains why.
When recessive allele (a) common e.g. 0.95
genotype frequencies are:
 AA (0.05)2
Aa (2 (0.05)(0.95) aa (0.95)2
 0.0025AA
0.095Aa
0.9025aa
 With more than 90% of phenotypes being
recessive, if aa is selected against expect
rapid population change.

When recessive allele (a) rare [e.g. 0.05]
genotype frequencies are:
 AA (0.95)2
Aa 2(0.95)(0.05)
aa
(0.05)2
 0.9025AA
0.095Aa
0.0025aa
 Fewer than 0.25% of phenotypes are aa
recessive. Most a alleles are hidden from
selection as heterozygotes. Expect only
slow change in frequency of a.


What is the predicted allele frequency after
one generation for the + allele in Dawson’s
beetle experiment?

We can calculate the average excess of
fitness and use our formula for Δp (change
in p) to find out.
Fitness
w++
1.0
w+l
1.0
Allele frequencies + = 0.5, l = 0.5
Genotype frequencies in initial generation
++ = 0.25 (p2)
+l = 0.5 (2pq)
ll = 0.25 (q2)
wll
0.0

w = p2 X w++ + 2pq X w+l + q2 X wll

= (0.25 X 1) + (0.5 X1) + (0.25 X 0)

= 0.75
For the + allele the average excess of
fitness is
 a+= [p X (w++ – w)] + [q X (w+l – w)]
 a+ = [0.5 (1 - 0.75 ) + [0.5 X (1 - 0.75)]

= 0.25
 Δp = p (a+ / w)

= 0.5 (0.25/0.75) = 0.167
 P t+1 = P + Δp = 0.5 + 0.167 = 0.667

Fig 5.16a

Graph shows allele frequency was
exactly as predicted in beetle
population.