Effects of selection on allele frequencies
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. For example, 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. As a result, population geneticists condense
all the components of fitness (survival, mating success, etc.) into one value of fitness called w.
Converting genotype fitness to allele fitness
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, shown below, enables us to assess how individual alleles contribute to
fitness.
Genotype
A1A1
A1A2
A2A2
Initial freq
p2
2pq
q2
Fitness
w11
w12
w22
Abundance In gen t+1
p2 X w11
2pq X w12
q2 X w22
Weighted freq. gen t+1
(p2 X w11)/wbar (2pq X w12)/ wbar
(q2 X w22)/ wbar
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 wbar. [NOTE: I’m using the term wbar here refer to the average fitness of a population.
Usually wbar would be indicated by a w with a line or bar over the w, but I cannot produce that in this
word processing software, hence my use of wbar]
Formula for wbar (average fitness of population) for two alleles A1 and A2
wbar = 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.
General selection model for diploid organisms
Normalized weighted freq. gen t+1
(p2 X w11)/ wbar
(2pq X w12)/ wbar
(q2 X w22)/ wbar
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.
To do this we need to sum alleles across genotypes.
For the allele A1 this will be the frequency of the A1A1 homozygotes plus half the frequency of the A1A2
heterozygotes (since the A1 allele gets half the credit for the reproductive success of A1A2 hterozygotes)
Frequency of allele A1 [p(t+1)]
p(t+1) = (p2 X w11 + pq X w12)/ wbar
Frequency of allele A2 [q(t+1)]
q(t+1) = (q2 X w22 + pq X w12)/ wbar
An example of allele change under selection
We have a situation where we have two alleles A1 and A2 and the different genotypes of these two
alleles have different fitness levels.
Starting allele frequencies: A1 = 0.8, A2 = 0.2.
Fitness
w11
w12
w22
0.9
1.0
0.2
In this example the heterozygote has the highest fitness, the A1A1 homozygote slightly lower fitness and
the A2A2 homozygote has very low relative fitness.
First we must calculate the average fitness of the population wbar
wbar = 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
Now to calculate the frequency of the allele A1 in the next generation we plug our numbers into the
formula below to calculate the allele frequency after selection in the next generation.
P(t+1) = (p2 X w11 + pq X w12)/wbar
[remember in the formula above that you use pq and not 2pq because the allele A1 gets only half the
“credit” for the fitness of the heterozygotes].
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 (from 0.8 to 0.814). Δp the change in frequency of the
allele A1 from one generation to the next is given by
Δp = P(t+1) – P(t)
Δp = 0.814 -0.8 = 0.014
In this example, the success of the alleles A1 and A2 is very sensitive to the frequency of A2.
In the previous 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 the next example we increase the frequency of A2 to 0.6
Allele frequencies: A1 = 0.4, A2 = 0.6
Fitness
w11
w12
w22
0.9
1.0
0.2
First calculate wbar the average fitness of the population
wbar = p2 X w11 + 2pq X w12 + q2 X w22
= (0.16 X 0.9) + (0.48 X 1) + (0.36 X 0.2)
= 0.144 + 0.48 + 0.072
= 0.696
Now we calculate the frequency of allele A1 after one generation of selection.
P(t+1) = (p2 X w11 + pq X w12)/wbar
P(t+1) = (0.16 X 0.9 + 0.24 X 1)/0.696
= (0.144 + 0.384)/0.696
= 0.552 (allele A1has increased substantially from a frequency of 0.4 and allele A2 has
declined to a frequency of 0.448 from 0.6)
Δp (change in the frequency of the allele A1 from generation t to generation t+1)
Δp = P(t+1) – P(t). All you have to do to calculate Δp is to subtract the frequency in one generation from the
frequency in the next.
Δp = 0.552-0.4 = +0.152
Some more examples of allele frequency change under selection.
Allele frequencies
A1 = 0.7
A2 = 0.3
In this example allele A1 is a dominant allele so it determines the phenotype (and fitness) of the A1A1
homozygote and the A1A2 heterozygote and so they have the same fitness.
Fitness
W11
w12
w22
2
2
4
wbar = p2 X w11 + 2pq X w12 + q2 X w22
wbar = 0.49x2 + 0.42 x 2 + 0.09 x 4
wbar = 0.98 + 0.84 + 0.36 = 2.18
Frequency of allele A1 in the next generation is given by
P(t+1)
= (p2 X w11 + pq X w12)/wbar
= 0.49 X 2 + 0.21 X 2 / 2.18
= 0.98 + 0.42 /2.18
= 1.4/2.18 = 0.642
Similarly,
q(t+1)
= (q2 X w22 + pq X w12)/wbar
= 0.09X4 + 0.21X2/ 2.18
= 0.36 + 0.42 /2.18
=0.84/2.18
= 0.358
So in one generation the recessive allele A2 which has higher fitnesshas increased from 0.3 to 0.358 and
the dominant, but lower fitness allele A1 has declined from0.7 to 0.642.
Average excess of fitness
Another formula is also used to calculate allele frequency changes over time. It is a relatively simple
formula that allows us to calculate the net fitness contribution of an allele, which is called the average
excess of fitness.
Average excess of fitness is the difference between the average fitness of individuals with that allele
and the average fitness of the entire population.
Equation for average excess of fitness for allele A1 (aA1)
For example, for the allele A1 the average excess of fitness is
aA1= [p X (w11 – wbar)] + [q X (w12 – wbar)]
Where w11 – wbar is the difference in fitness between A1A1 individuals and the mean fitness of the
population wbar. W12 is fitness of A1A2 heterozygotes. Wbar 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 easily calculate how much an allele frequency will change
between generations
Δp = p X (aA1/wbar)
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 wbar
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.
Allele frequency change between generations
Δp = p X (aA1/wbar)
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.
Natural selection is more powerful in large populations
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.
Directional Selection
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.
Effects of gene interactions on selection
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.
Relationships among alleles at a locus
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.
Directional selection at one locus with two alleles. (A) In directional selection, one allele A1 is favored
over another, A2. This can occur in different ways: A1 can be dominant (red), A1 and A2 can be
codominant (blue), or A1 can be recessive (orange). (B) The trajectories of p, the frequency of the A1
allele, are illustrated from a starting value of p = 0.005.
Empirical examples of allele frequency change under selection
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.
Empirical examples of allele frequency change under selection: Jaeken syndrome
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 Hardy-Weinberg
equilibrium to do so.
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
Expected frequencies
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.
Test of theory
Dawson (1970). Flour beetles. Two alleles at locus: + and l. +/+ and +/l genotypes are phenotypically
normal, but the l/l genotype is 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.
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)
0.9025AA
0.095Aa
aa (0.05)2
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++
w+l
wll
1.0
1.0
0.0
Allele frequencies + = 0.5, l = 0.5
Genotype frequencies in initial generation
++ = 0.25 (p2)
+l = 0.5 (2pq)
ll = 0.25 (q2)
wbar = 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++ – wbar)] + [q X (w+l – wbar)]
a+ = [0.5 (1 - 0.75 ) + [0.5 X (1 - 0.75)]
= 0.25
Δp = p (a+ / wbar)
= 0.5 (0.25/0.75) = 0.167
P t+1 = P + Δp = 0.5 + 0.167 = 0.667
Graph shows allele frequency was exactly as predicted in beetle population.