Transcript Evolution 1/e


Read Chapter 7 of text
We saw in chapter 6 that a cross
between two individuals heterozygous
for a dominant allele produces a 3:1
ratio of individuals expressing the
dominant phenotype: to those
expressing the recessive phenotype.
 For example brachydachtyly (shortening
of the digits) displays this pattern of
inheritance.

In the early 1900’s when Mendel’s work
was rediscovered there was confusion
about how these simple patterns of
inheritance affected populations.
 Why, for example, was not 3 of every 4
people a person with brachdactyly?
 Why did not dominant alleles replace
recessive alleles?


The confusion stemmed from confusing
what was happening at the level of the
individual with what occurs at the
population level.

Individual-level thinking enables us to
figure out the result of particular crosses.

Population level thinking however is
needed to figure out how the genetic
characteristics of populations change
over time.

It enables us to figure out quantitatively
what is happening in a population as a
result of evolution. Remember, evolution
occurs when genotype frequencies
change over time.

Null models provide us with a baseline.
They tell us what we expect to be the
case if certain forces are not operating.

The Hardy-Weinberg equilibrium tells us
what we expect to happen to genotype
frequencies when forces such as natural
selection are not operating on a
population.

The Hardy-Weinberg model enables us
to determine what allele and genotype
frequencies we would expect to in a
population if all that is happening is
alleles are being randomly assigned to
gametes and those gametes meet up at
random.
The Hardy-Weinberg model examines a
situation in which there is one gene with
two alleles A1 and A2.
 There are three possible genotypes A1A1,
A2 A2,and A1 A2

Hardy and Weinberg used their model to
predict what would happen to allele
frequencies and genotype frequencies
in the absence of any evolutionary
forces.
 Their model produced three important
conclusions

The three conclusions of the H-W model.
In the absence of evolutionary processes
acting on them:
 1. The frequencies of the alleles A1 and
A2 do not change over time.
 2. If we know the allele frequencies in a
population we can predict the
equilibrium genotype frequencies
(frequencies of A1A1, A2 A2,and A1 A2).


3. A gene not initially at H-W equilibrium
will reach H-W equilibrium in one
generation.

1. No selection.
› If individuals with certain genotypes survived
better than others, allele frequencies would
change from one generation to the next.

2. No mutation
› If new alleles were produced by mutation or
alleles mutated at different rates, allele
frequencies would change from one
generation to the next.

3. No migration
› Movement of individuals in or out of a
population would alter allele and genotype
frequencies.

4. Large population size.
› Population is large enough that chance plays no
role. Eggs and sperm collide at same
frequencies as the actual frequencies of p and
q.
› If assumption was violated and by chance some
individuals contributed more alleles than others
to next generation allele frequencies might
change. This mechanism of allele frequency
change is called Genetic Drift.

5. Individuals select mates at random.
› Individuals do not prefer to mate with
individuals of a certain genotype. If this
assumption is violated allele frequencies will
not change, but genotype frequencies
might.

Assume two alleles A1 and A2 with known
frequencies (e.g. A1 = 0.6, A2 = 0.4.)

Only two alleles in population so their
allele frequencies add up to 1.

Can predict frequencies of genotypes in
next generation using allele frequencies.

Possible genotypes are: A1A1 , A1A2 and
A2A2

Assume alleles A1 and A2 enter eggs and
sperm in proportion to their frequency in
population (i.e. 0.6 and 0.4)

Assume sperm and eggs meet at
random (one big gene pool).
Then we can calculate expected
genotype frequencies.
 A1A1 : To produce an A1A1 individual,
egg and sperm must each contain an A1
allele.
 This probability is 0.6 x 0.6 or 0.36
(probability sperm contains A1 times
probability egg contains A1).


Similarly, we can calculate frequency of
A2A2.

0.4 x 04 = 0.16.

Probability of A1A2 is given by probability
sperm contains A1 (0.6) times probability
egg contains A2 (0.4). 0.6 x 04 = 0.24.

But, there’s a second way to produce an
A1A2 individual (egg contains A1 and
sperm contains A2). Same probability as
before: 0.6 x 0.4= 0.24.

Overall probability of A1A2 = 0.24 + 0.24 =
0.48.
Genotypes in next generation:
 A1A1 = 0.36
 A1A2 = 0.48
 A2 A2= 0.16
 Adds up to one.

General formula for Hardy-Weinberg.
 Let p= frequency of allele A1 and q =
frequency of allele A2.


p2 + 2pq + q2 = 1.

If there are 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


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.

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.
The frequencies of the different
genotypes are a function of the
frequencies of the underlying alleles.
 The closer the allele frequencies are to
0.5 the greater the frequency of
heterozygotes.


You need to be able to work with the
Hardy-Weinberg 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 you would expect in the
population?
p2 + 2pq + q2 = 1. The terms in the
equation represent the frequencies of
individual genotypes.
 P and q are allele frequencies. It is vital
that you understand this difference.


9 of 100 (frequency = 0.09) of individuals
are homozygotes. What term in the H-W
equation is that equal to?

It’s q2.

If q2 = 0.09, what’s q? Get square root of
q2, which is 0.3.

If q=0.3 then p=0.7. Now plug p and q
into equation to calculate frequencies of
other genotypes.



p2 = (0.7)(0.7) = 0.49
2pq = 2 (0.3)(0.7) = 0.42
Number of heterozygotes = 0.42 times
population size = (0.42)(100) = 42.

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?
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.


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.

If we relax the H-W assumption of no
selection how does that affect allele
frequencies?

To quantify the strength of selection
against a recessive allele we can use a
parameter (s) called the selection
coefficient to describe the reduction in
fitness of one phenotype vs the other.
For example pocket mice coat color is
affected by a gene with two alleles D
and d. D allele is dominant.
 DD: dark phenotype
 Dd: dark phenotype
 Dd: light phenotype
 On dark backgrounds light phenotype
will be selected against.


The higher the value of s the more
strongly natural selection will act.

The mouse coat color example is an
example of frequency-independent
selection. The fitness of a trait is not
associated with how common the trait is.
The commonest form of frequencyindependent selection is 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 favored allele may replace
other alleles and become fixed.


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.


Theory predicts that if an average individual
carrying an allele has higher than average
fitness that 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.

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.

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 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.


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. A dominant allele is
expressed both as a heterozygote and a
homozygote and so is always visible to
selection.

One way in which multiple alleles may
be maintained in a population is through
heterozygote advantage (also called
overdominance).

Classic example is sickle cell allele.

Sickle cell anemia is a condition
common among West Africans and
those of West African descent.

Under low oxygen conditions the red
blood corpuscles are sickle shaped.

Untreated the condition usually causes
death in childhood.

About 1% of West Africans have sickle
cell anemia.

A single mutation causes a valine amino
acid to replace a glutamine in the alpha
chain of hemoglobin

The mutation causes hemoglobin
molecules to stick together.
Only individuals homozygous for the
allele get sickle cell anemia.
 Individuals with only one copy of the
allele (heterozygotes) get sickle cell trait
(a mild form of the disease)
 Individuals with the 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, many “normal” homozygotes
die of malaria.
 Stabilizing selection thus favors sickle cell
allele.


A heterozygote advantage (or
overdominance) results in a balanced
polymorphism in a population.

Both alleles are maintained in the
population as the heterozygote is the
best combination of alleles and a purely
heterozygous population is not possible.

Underdominance is when the
heterozygote has lower fitness than
either homozygote.

This situation is In this case one or other
allele will go to fixation, but which
depends on the starting allele
frequencies

In some cases the costs and benefits of a
trait depend on how common it is in a
population.

In this case the commoner a phenotype
is the more successful it is.

If two phenotypes are determined by
single alleles one allele will go to fixation
and the other be lost, but which one
depends on the starting frequencies.

In “flat” snails individuals mate face to
face and physical constraints mean only
individuals whose shells coil in the same
direction can mate successfully.

Higher frequencies of one coil direction
leads to more mating for that phenotype
and eventually it replaces the other
types.

Under negative frequency-dependent
selection a trait is increasingly favored
the rarer it becomes.

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?

Gigord et al. hypothesis: Bees tend to
visit equal numbers of each flower color
so rarer color will have advantage (will
get more visits from pollinators).

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)

As predicted, reproductive success of
yellow varied with frequency.
5.21 a

Another example of negative frequencydependent selection involves a scaleeating cichlid fish in Lake Tanganyika.

The fish come in left- and right-mouthed
morphs. They attack their victims from
behind.

Because each morph always attacks
the same side of its victims when the
frequency of a morph increases the
victims become good at guarding
against attacks from that side.

The common morph then suffers
reduced feeding success and declines in
abundance.

As a result the morphs fluctuate in
frequency over time.

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.

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?
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


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


Not a big change.

After 1000 generations frequency of A =
0.81.

Mutation alone clearly not a powerful
evolutionary force.

But mutation AND selection make a very
powerful evolutionary force.

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.

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

Step-like pattern results when a new
mutation occurs and sweeps through the
population as mutant bacteria out-reproduce
competitors.

Remember, without mutation evolution would
eventually cease. Mutation is ultimate source
of genetic variation.

Most mutations are deleterious and natural
selection acts to remove them from
population.

Deleterious alleles persist, however,
because mutation continually produces
them.

When rate at which deleterious alleles being
eliminated is equal to their rate of production
by mutation we have 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 7.8 for derivation of equation.

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.
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

Observed mutation rate is about 1.1 X10-4,
very close agreement in estimates.
 High frequency of allele accounted for by
observed mutation rate.


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.

Very strong selection against CF alleles,
but CF frequency about 0.02 in
Europeans.

Can mutation rate account for high
frequency?

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?

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.

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.
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


Exposed cells to S. typhi bacteria.

Measured number of bacteria that
entered cells.

Clear results
Fig 5.27a

F508/F508 homozygote almost totally
resistant to S. typhi.

Wild type homozygote highly vulnerable

Heterozygote contained 86% fewer
bacteria than wild type.

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

Another assumption of Hardy-Weinberg
is that random mating takes place.

The most common form of non-random
mating is inbreeding which occurs when
close relatives mate with each other.

Most extreme form of inbreeding is self
fertilization.

In a population of self fertilizing organisms all
homozygotes will produce only
homozygous offspring. Heterozygotes will
produce offspring 50% of which will be
homozygous and 50% heterozygous.

How will this affect the frequency of
heterozygotes each generation?

In each generation the proportion of
heterozygous individuals in the
population will decline.

Because inbreeding produces an excess
of homozygotes in a population
deviations from Hardy-Weinberg
expectations can be used to detect
such inbreeding in wild populations.

Sea otters once abundant along the west
coast of the U.S were almost wiped out by
fur hunters in the 18th and 19th centuries.

California population reached a low of 50
individuals (now over 1,500). As a result of
this bottleneck the population has less
genetic diversity than it once had.

Population still at a low density and
Lidicker and McCollum (1997)
investigated whether this resulted in
inbreeding.

Determined genotypes of 33 otters for
PAP locus, which has two alleles S (slow)
and F (fast)

The genotypes of the 33 otters were:
› SS 16
› SF 7
› FF 10

This gives approximate allele frequencies
of S= 0.6 and F = 0.4

If otter population in H-W equilibrium,
genotype frequencies should be
› SS = 0.6* 0.6 = 0.36
› SF =2*0.6*0.4 = 0.48
› FF = 0.4*0.4 = 0.16

However actual frequencies were:
› SS= 0.485, SF= 0.212, FF =0.303

There are more homozygotes and fewer
heterozygotes than expected for a random
mating population.

Having considered alternative explanations
for deficit of heterozygotes Lidicker and
McCollum (1997) concluded that sea otter
populations show evidence of inbreedng.

Self-fertilization and sibling mating most
extreme forms of inbreeding, but matings
between more distant relatives (e.g.
cousins) has same effect on frequency
of homozygotes, but rate is slower.

F = Coefficient of inbreeding: probability
that two alleles in an individual are
identical by descent (both alleles are
copies of a particular ancestor’s allele in
some previous generation).

F increases as relatedness increases.

If we compare heterozygosity of inbred
population Hf with that of a random mating
population Ho relationship is

Hf = Ho (1-F)

Anytime F>0 frequency of heterozygotes is
reduced and frequency of homozygotes
naturally increases.

Calculating F. Need to use pedigree
diagrams.

Example: Female is daughter of two halfsiblings.

Two ways female could receive alleles
that are identical by descent.
Male
Female
Female
Male
Male
Fig 6.27a
Half-sibling mating
Fig 6.27b

Total probability of scenario is 1/16 + 1/16
= 1/8.

Inbreeding increases frequency of
homozygotes and thus the probability
that deleterious alleles are visible to
selection.

In humans, children of first cousins have
higher mortality rates than children of
unrelated individuals.
Each dot on
graph
represents
mortality
rates for a
human
population.
Fig 6.28
Mortality rate
for children
of cousins
consistently
about 4%
higher than
rate for
children of
non-relatives.

In a study of 2760 individuals from 25
Croatian islands Rudan et al. found a
strong positive relationship between high
blood pressure and the inbreeding
coefficent.

Inbreeding depression also documented
in studies of wild animals.

E.g. Great Tit. Two studies show that
survival of inbred nestlings is lower than
that of outbred individuals and that
hatching success of inbred eggs is lower
than that of outbred eggs.
Fig. 6.30