Population Genetics:
Selection and mutation as
mechanisms of evolution

Population genetics: study of Mendelian
genetics at the level of the whole
population.
Hardy-Weinberg
Equilibrium

To understand the conditions under
which evolution can occur, it is
necessary to understand the
population genetic conditions under
which it will not occur.
Hardy-Weinberg
Equilibrium

The Hardy-Weinberg Equilibrium
Principle allows us do this. It enables
us to predict allele and genotype
frequencies from one generation to
the next in the absence of evolution.
Hardy-Weinberg
Equilibrium


Assume a gene with two alleles A and
a with known frequencies (e.g. A =
0.6, a = 0.4.)
There are only two alleles in the
population so their frequencies must
add up to 1.
Hardy-Weinberg
Equilibrium


Using allele frequencies we can predict
the expected frequencies of genotypes
in the next generation.
With two alleles only three genotypes
are possible: AA, Aa and aa
Hardy-Weinberg
Equilibrium


Assume alleles A and a are present in
eggs and sperm in proportion to their
frequency in population (i.e. 0.6 and
0.4)
Also assume that sperm and eggs
meet at random (one big gene pool).
Hardy-Weinberg
Equilibrium



Then we can calculate expected
genotype frequencies.
AA: To produce an AA individual, egg
and sperm must each contain an A
allele.
This probability is 0.6 x 0.6 = 0.36
(probability sperm contains A times
probability egg contains A).
Hardy-Weinberg
Equilibrium


Similarly, we can calculate frequency
of aa.
0.4 x 04 = 0.16.
Hardy-Weinberg
Equilibrium

Probability of Aa is given by probability
sperm contains A (0.6) times
probability egg contains a (0.4). 0.6 x
04 = 0.24.
Hardy-Weinberg
Equilibrium


But, there’s a second way to produce
an Aa individual (egg contains A and
sperm contains a). Same probability
as before: 0.6 x 0.4= 0.24.
Hence the overall probability of Aa =
0.24 + 0.24 = 0.48.
Hardy-Weinberg
Equilibrium





Genotypes in next generation:
AA = 0.36
Aa = 0.48
Aa= 0.16
These frequencies add up to one.
Hardy-Weinberg
Equilibrium

General formula for Hardy-Weinberg.
Let p= frequency of allele A and q =
frequency of allele a.

p2 + 2pq + q2 = 1.

Hardy Weinberg
Equilibrium with more
than 2 alleles

If three alleles with frequencies P1, P2
and P3 such that P1 + P2 + P3 = 1
Then genotype frequencies given by:
 P12 + P22 + P32 + 2P1P2 + 2P1 P3 +
2P2P3

Conclusions from HardyWeinberg Equilibrium


Allele frequencies in a population will not
change from one generation to the next just
as a result of assortment of alleles and
zygote formation.
Assortment of alleles simply means what
occurs during meiosis when only one copy
of each pair of alleles enters any given
gamete (remember each gamete only
contains half the DNA of a body cell).
Conclusions from HardyWeinberg Equilibrium

If the allele frequencies in a gene pool
with two alleles are given by p and q,
the genotype frequencies will be given
by p2, 2pq, and q2.
Working with the H-W
equation


You need to be able to work with the HardyWeinberg equation.
For example, if 9 of 100 individuals in a
population suffer from a homozygous
recessive disorder can you calculate the
frequency of the disease-causing allele? Can
you calculate how many heterozygotes are
in the population?
Working with the H-W
equation



p2 + 2pq + q2 = 1. The terms in the equation
represent the frequencies of individual genotypes.
[A genotype is possessed by an individual organism
so there are two alleles present in each case.]
P and q are allele frequencies. Allele frequencies
are estimates of how common alleles are in the
whole population.
It is vital that you understand the difference
between allele and genotye frequencies.
Working with the H-W
equation

9 of 100 (frequency = 0.09) of
individuals are homozygous for the
recessive allele. What term in the H-W
equation is that equal to?
Working with the H-W
equation



It’s q2.
If q2 = 0.09, what’s q? Get square root of
q2, which is 0.3, which is the frequency of
the allele a.
If q=0.3 then p=0.7. Now plug p and q into
equation to calculate frequencies of other
genotypes.
Working with the H-W
equation



p2 = (0.7)(0.7) = 0.49 -- frequency of AA
2pq = 2 (0.3)(0.7) = 0.42 – frequency of
Aa.
To calculate the actual number of
heterozygotes simply multiply 0.42 by the
population size = (0.42)(100) = 42.
Working with the H-W
equation: 3 alleles


There are three alleles in a population
A1, A2 and A3 whose frequencies
respectively are 0.2, 0.2 and 0.6 and
there are 100 individuals in the
population.
How many A1A2 heterozygotes will
there be in the population?
Working with the H-W
equation: 3 alleles
Just use the formulae P1 + P2 + P3 = 1
and P12 + P22 + P32 + 2P1P2 + 2P1 P3 +
2P2P3 = 1
Then substitute in the appropriate
values for the appropriate term
2P1P2 = 2(0.2)(0.2) = 0.08 or 8 people
out of 100.

Other examples of working with
HW equilibrium: is a population in
HW equilibrium?





In a population there are 100 birds
with the following genotypes:
44 AA
32 Aa
24 aa
How would you demonstrate that this
population is not in Hardy Weinberg
equilibrium
Three steps



Step 1: calculate the allele
frequencies.
Step 2: Calculate expected numbers of
each geneotype (i.e. figure out how
many homozygotes and heterozygotes
you would expect.)
Step 3: compare your expected and
observed data.
Step 1 allele frequencies




Step 1. How many “A” alleles are there
in total?
44 AA individuals = 88 “A” alleles
(because each individual has two
copies of the “A” allele)
32 Aa alleles = 32 “A” alleles
Total “A” alleles is 88+32 =120.
Step 1 allele frequencies



Total number of “a” alleles is similarly
calculated as 2*24 + 32 = 80
What are allele frequencies?
Total number of alleles in population is
120 + 80 = 200 (or you could
calculate it by multiplying the number
of individuals in the population by two
100*2 =200)
Step 1 allele frequencies



Allele frequencies are:
A = 120/200= 0.6. Let p = 0.6
a = 80/200 = 0.4. Let q = 0.4
Step 2 Calculate expected
number of each genotype
Use the Hardy_Weinberg equation
p2 + 2pq + q2 = 1 to calculate what expected
genotypes we should have given these observed
frequencies of “A” and “a”


Expected frequency of AA = p2 = 0.6 * 0.6 = 0.36

Expected frequency of aa = q2 = 0.4*0 .4 =0.16

Expected frequency of Aa = 2pq = 2*.6*.4 = 0.48
Step 2 Calculate expected
number of each genotype




Convert genotype frequencies to
actual numbers by multiplying by
population size of 100
AA = 0.36*100 = 36
aa = 0.16*100 = 16
Aa = 0.48*100 = 48
Step 3 Compare Observed
and Expected values
Observed population is:
44 AA 32 Aa 24 aa
Expected population is:
36AA 48Aa 16aa
These numbers are not the same so the
population is not in Hardy-Weinberg
equilibrium. An assumption of the Hardy
Weinberg equilibrium is being violated.
What are those assumptions?
Assumptions of HardyWeinberg

1. No selection.
– When individuals with certain genotypes
survive better than others, allele
frequencies may change from one
generation to the next.
Assumptions of HardyWeinberg

2. No mutation
– If new alleles are produced by mutation
or alleles mutate at different rates, allele
frequencies may change from one
generation to the next.
Assumptions of HardyWeinberg

3. No migration
– Movement of individuals in or out of a
population will alter allele and genotype
frequencies.
Assumptions of HardyWeinberg

4. No chance events.
– Luck plays no role. Eggs and sperm collide at
same frequencies as the actual frequencies of p
and q.
– When assumption is violated, and by chance
some individuals contribute more alleles than
others to next generation, allele frequencies may
change. This mechanism of allele frequency
change is called Genetic Drift.
Assumptions of HardyWeinberg

5. Individuals select mates at
random.
– If this assumption is violated allele
frequencies will not change, but genotype
frequencies may.
Hardy-Weinberg
Equilibrium


Hardy Weinberg equilibrium principle
identifies the forces that can cause
evolution.
If a population is not in H-W
equilibrium then one or more of the
five assumptions is being violated.
5.10
Can selection change
allele frequencies?



Two alleles B1 and B2
Frequency of B1= 0.6 and frequency of
B2 = 0.4.
Random mating gives genotype
frequencies 0.36 B1B1 0.48B1B2
0.16B2B2
Can selection change
allele frequencies?



Assume 100 individuals
36 B1B1 48 B1B2 16 B2B2
Incorporate selection. Assume all B1B1
survive, 75% of B1B2 survive and 50%
of B2B2 survive.
Can selection change allele
frequencies?





Population now 80 individuals: 36 B1B1
36 B1B2 8 B2B2
Allele frequencies now:
B1 = 72 + 36/160 = 0.675
B2 = 36+16/160 = 0.325
Selection resulted in allele frequency
change.
FIG 5.11
Can selection change allele
frequencies?



Selection in previous example very
strong.
What patterns expected with weaker
selection.
Initial frequencies B1 = 0.01, B2 =
0.99.
Fig 5.12
Note line colors yellow and red have been flipped in the table.
Can selection change allele
frequencies?

Rate of change of B1is rapid when
selection pressure is strong, but much
slower, although still steady, under
weak selection.
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
Empirical examples of allele
frequency change under selection


Two Drosophila populations
maintained: one fed food spiked with
ethanol, control fed unspiked food.
Populations maintained for multiple
generations.
Empirical examples of allele
frequency change under selection



Experimental population showed
consistent long-term increase in
frequency of ADHF
Flies with ADHF allele have higher
fitness when ethanol is present.
ADHF enzyme breaks down ethanol
twice as fast as ADHS enzyme.
Fig 5.13
Empirical examples of allele
frequency change under selection:
Jaeken syndrome

Jaeken syndrome: patients severely
disabled with skeletal deformities and
inadequate liver function.
Jaeken syndrome



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.
Jaeken syndrome


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.
Jaeken syndrome


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.
Jaeken syndrome


Researchers studied 54 patients and
identified most common mutation as
R141H.
Dividing population into R141H and
other alleles [Other]. Allele
frequencies are:
Other: 0.6 and R141H: 0.4.
Jaeken syndrome




If disease alleles are in H-W
equilibrium then we predict genotype
frequencies of
Other/other: 0.36
Other/R141H: 0.48
R141H/R141H: 0.16
Jaeken syndrome

Observed frequencies are:
Other/Other: 0.2
Other/R141H: 0.8
R141H/R141H : 0
Clearly population not in H-W
equilibrium.
Jaeken syndrome


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.
Using H-W to predict
potential spread of CCR532 allele


Will AIDS epidemic cause CCR5-32
(delta32) allele to spread? Offers
protection against HIV infection.
In principle it could, but models
suggest it probably will not in any real
population.
Spread of CCR5-32 allele


Scenario 1. Initial allele frequency
20%. 25% of heterozygotes and
those homozygous for normal allele
die of AIDS. Homozygous 32
individuals do not die of AIDS.
Over 40 generations allele increases to
almost 100%.
FIG 5.15a
Spread of CCR5-32 allele

In human population with high HIV
infection rate and high frequencies of
32 allele, 32 could spread rapidly.
Spread of CCR5-32 allele

Scenario 2. In areas with low HIV
infection rates (under 1%), but high
levels of 32 (as found in Europe),
selection too weak to raise delta 32
frequency much.
FIG 5.15b
Spread of CCR5-32 allele


Scenario 3. In sub-Saharan Africa HIV
infection rate about 25%. However,
32 allele almost absent
Under these conditions 32 frequency
will hardly change because most
copies of allele are in heterozygotes,
which are not protected from HIV.
FIG 5.15c
Testing predictions of
population genetics theory


Theory predicts that if an individual carrying
an allele has higher than average fitness
then the frequency of that allele will
increase from one generation to the next.
Obviously, the converse should be true and
a deleterious allele should decrease in
frequency if its bearers have lower fitness.
Testing predictions of
population genetics theory


Mathematical treatment of effect of
selection on gene frequencies is given in
Box 6.3 (page 186) of your text.
Main point is that if the average fitness of
an allele A when paired at random with
other alleles in the population is higher than
the average fitness of the population, then it
will increase in frequency.
Tests of theory

Dawson (1970). Flour beetles. Two
alleles at locus: + and l.

+/+ and +/l phenotypically normal.

l/l lethal.
Dawson’s flour beetles


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.
Fig 5.16a
Dawson’s flour beetles


Predicted and observed allele
frequencies matched very closely.
l allele declined rapidly at first, but
rate of decline slowed.
Dawson’s flour beetles


Dawson’s results show that when a
recessive allele is common, evolution
by natural selection is rapid, but slows
as recessive allele becomes rarer.
Hardy-Weinberg explains why.
Dawson’s flour beetles




When recessive allele (a) common e.g. freq
a=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 we expect
rapid population change.
Dawson’s flour beetles





When recessive allele (a) rare [e.g. freq
a=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 (>97%) are
hidden from selection as heterozygotes.
Expect only slow change in frequency of a.
Maintaining multiple
alleles in gene pool


Dawson’s beetle work shows that
deleterious rare alleles may be very
hard to eliminate from a gene pool
because they remain hidden from
selection as heterozygotes.
(this only applies if the allele is not
dominant.)
Maintaining multiple
alleles in gene pool

There are several different ways in
which multiple alleles may be
maintained in populations even, as we
saw, if one allele is deleterious.
– 1. Deleterious recessive allele can hide in
heterozygote
– 2. There may be heterozygote advantage
– 3. Frequency-dependent selection
Maintaining multiple alleles in gene
pool-2. Heterozygote advantage


Another way in which multiple alleles
may be maintained in a population is
through heterozygote advantage.
Classic example is sickle cell allele.
Sickle Cell Anemia
Sickle cell anemia is a condition
common among West Africans (and
African Americans of West African
ancestry).
In sickle cell anemia red blood cells are
sickle shaped.
Usually fatal by about age 10.
About 1% of West Africans have sickle
cell anemia.
A single mutation that causes a valine
amino acid to replace a glutamine
in an alpha chain of the hemoglobin
molecule.
Mutation causes molecules to stick
together.
Why isn’t mutant sickle cell gene
eliminated by natural selection?
Only individuals homozygous for
sickle cell gene get sickle cell anemia.
Individuals with one copy of sickle cell gene
(heterozygotes) get sickle cell trait
(mild form of disease).
Individuals with sickle cell allele (one or two
copies) don’t get malaria.
Heterozygotes have higher survival than
either homozygote. Heterozygote advantage.
Sickle cell homozygotes die of sickle
cell anemia.
“Normal” homozygotes more likely
to die of malaria.
Stabilizing selection for sickle cell allele.
Maintaining multiple alleles in gene
pool- 3. Frequency-dependent
selection


Another way in which multiple alleles
are maintained is frequencydependent selection.
Frequency-dependent selection occurs
when rare alleles have a selective
advantage because they are rare.
Frequency-dependent
selection



Color polymorphism in Elderflower
Orchid
Two flower colors: yellow and purple.
Offer no food reward to bees. Bees
alternate visits to colors.
How are two colors maintained in the
population?
Frequency-dependent
selection

Gigord et al. hypothesis: Bees tend to
visit equal numbers of each flower
color. As a result the rarer color will
have an advantage (because each
individual rarer color flower will get
more visits from pollinators).
Frequency-dependent
selection


Experiment: provided five arrays of
potted orchids with different
frequencies of yellow orchids in each.
Monitored orchids for fruit set and
removal of pollinaria (pollen bearing
structures)
Frequency-dependent
selection

As predicted, reproductive success of
yellow varied with frequency.
5.21 a
Mutation as an
evolutionary force

It is obvious that selection is a very
powerful evolutionary force but how
strong is mutation alone as an
evolutionary force?

To check: Two alleles A and a.

Frequency of A = 0.9, a = 0.1.
Mutation as an
evolutionary force


Assume A mutates to a at rate of 1
copy per 10,000 per generation (high
rate, but within observed range) and
all mutations occur in gametes.
How much does this change gene pool
in next generation?
Mutation as an
evolutionary force



Hardy Weinberg genotypes in current
generation:
0.81 AA, 0.18 Aa, 0.01 aa
With no mutation allele frequency in
gene pool 0.9 A, 0.1 a
Mutation as an
evolutionary force




But mutation reduces frequency of A
and increases frequency of a
A
a
0.9 - (0.0001)(0.9) 0.1 +
(0.0001)(0.9)
0.89991A
0.10009a
5.23
Mutation as an
evolutionary force


Not a big change.
After 1000 generations frequency of A
= 0.81.
5.24
Mutation as an
evolutionary force



Mutation alone clearly not a powerful
evolutionary force.
But mutation AND selection together make
a very powerful evolutionary force.
Remember mutation provides the raw
material for selection to work with.
Lenski’s E. coli work

Lenski et al. studied mutation and selection
together in an E. coli strain that did not
exchange DNA (hence mutation only source
of new variation).

Bacteria grown in challenging environment
(low salts and low glucose medium) so
selection would be strong.
Lenski’s E. coli work

12 replicate populations tracked over
about 10,000 generations.

Fitness and cell size of populations
increased over time.

Pattern of change interesting: steplike.
Why is it steplike?

5.25
Lenski’s E. coli work

Step-like pattern results when a new
mutation occurs and sweeps through
the population as mutant bacteria outreproduce competitors.

Remember, without mutation evolution
would eventually cease. Mutation is
the ultimate source of genetic
variation.
Mutation-selection
balance

Most mutations are deleterious and
natural selection acts to remove them
from population.

Deleterious alleles persist, however,
because mutation continually produces
them.
Mutation-selection
balance

When rate at which deleterious alleles
are being eliminated is equal to their
rate of production by mutation we have
mutation-selection balance.
Mutation-selection
balance

Equilibrium frequency of deleterious
allele q = square root of µ/s where µ is
mutation rate and s is the selection
coefficient (measure of strength of
selection against allele; ranges from 0
to 1). See Box 6.10 page 215 for
derivation of equation.
Mutation-selection
balance

Equation makes intuitive sense.

If s is small (mutation only mildly
deleterious) and µ (mutation rate) is high
than q (allele frequency) will also be
relatively high.

If s is large and µ is low, than q will be low
too.
Mutation-selection
balance



Spinal muscular atrophy is a generally
lethal condition caused by a mutation
on chromosome 5.
Selection coefficient estimated at 0.9.
Deleterious allele frequency about
0.01 in Caucasians.
Inserting above numbers into equation
and solving for µ get estimated
mutation rate of 0.9 X 10-4
Mutation-selection
balance


Observed mutation rate is about 1.1
X10-4, very close agreement in
estimates.
High frequency of allele is accounted
for by observed mutation rate.
Is frequency of Cystic fibrosis
maintained by mutation
selection balance?

Cystic fibrosis is caused by a loss of function
mutation at locus on chromosome 7 that
codes for CFTR protein (cell surface protein
in lungs and intestines).

Major function of protein is to destroy
Pseudomonas aeruginosa bacteria.
Bacterium causes severe lung infections in CF
patients.
Cystic fibrosis


Very strong selection against CF
alleles, but CF frequency about 0.02 in
Europeans.
Can mutation rate account for high
frequency?
Cystic fibrosis

Assume selection coefficient (s) of 1 and q
= 0.02.

Estimate mutation rate µ is 4.0 X 10-4

But actual mutation rate is only 6.7 X 10-7

Is there an alternative explanation?
Cystic fibrosis



May be heterozygote advantage.
Pier et al. (1998) hypothesized CF
heterozygotes may be resistant to typhoid
fever.
Typhoid fever caused by Salmonella typhi
bacteria. Bacteria infiltrate gut by crossing
epithelial cells.
Cystic fibrosis


Hypothesized that S. typhi bacteria
may use CFTR protein to enter cells.
If so, CF-heterozygotes should be less
vulnerable to S. typhi because their
gut epithilial cells have fewer CFTR
proteins on cell surface.
Cystic fibrosis





Experimental test.
Produced mouse cells with three
different CFTR genotypes
CFTR homozygote (wild type)
CFTR/F508 heterozygote (F508
most common CF mutant allele)
F508/F508 homozygote
Cystic fibrosis



Exposed cells to S. typhi bacteria.
Measured number of bacteria that
entered cells.
Clear results
Fig 5.27a
Cystic fibrosis



F508/F508 homozygote almost
totally resistant to S. typhi.
Wild type homozygote highly
vulnerable
Heterozygote contained 86% fewer
bacteria than wild type.
Cystic fibrosis

Further support for idea F508
provides resistance to typhoid
provided by positive relationship
between F508 allele frequency in
generation after typhoid outbreak and
severity of the outbreak.
Fig 5.27b
Data from 11 European countries