Transcript Mendelian Genetics in Populations

Population Genetics
Hardy Weinberg
Equilibrium
6.1 Mendelian Genetics in
Populations:
The Hardy-Weinberg Equilibrium Principle
Population Genetics
• Population genetics is concerned with the
question of whether a particular allele or
genotype will become more common or less
common over time in a population, and Why.
• Example:
– Given that the CCR5-D32 allele confers immunity to
HIV, will it become more frequent in the human
population over time?
Predicting Allele Frequencies
Populations in Hardy-Weinberg
equilibrium
Yule vs. Hardy
• What are the characteristics of a population that
is in equilibrium or another words, not evolving.
• Yule thought that allele frequencies had to be
0.5 and 0.5. for a population to be in equilibrium.
• Hardy proved him wrong by developing the
Hardy-Weinburg equation.
Punnett square
• 60 % of the eggs
carry allele A and
40% carry allele a
• 60% of sperm carry
allele A and 40%
carry allele a.
Sample problem
• In a population of 100 people, we know that
36% are AA , 48% are Aa, and 16% are aa.
• Determine how many alleles in the gene
pool are A or a.
– Each individual makes two gametes....
– How many A alleles are in this population’s
gene pool? _____
120 (36*(2)+48)
– How many a alleles? _____
80 (16*(2) +48)
What percent of the alleles are A or
a?
120 / 200 = .6 or 60% A ; or
.6 = frequency of allele A
80 / 200 = .4 or 40% a ; or
.4 = frequency of allele a
• Creating the HardyWienburg equation is
a matter of combining
probabilities found in
the Punnett square.
Combining Probabilities
• The combined probability of two
independent events will occur together is
equal to the product of their individual
probabilities.
– What is the probability of tossing a nickel and
a penny at the same time and having them
both come up heads?
•½ x ½ = ¼
Combining Probabilities
• The combined probability that either of two
mutually exclusive events will occur is the
sum of their individual probabilities. When
rolling a die we can get a one or a two
(among other possibilities), but we cannot
get both at once. Thus, the probability of
getting either a one or a two is
• 1/6 + 1/6 = 1/3
Calculating Genotype Frequencies
• We can predict the genotype frequencies by
multiplying probabilities.
Hardy-Weinburg equation
Genotype Frequencies
Zygotes
Allelic frequency
Genotype frequency
AA
(p)(p)
p2
Aa
(p)(q)
2pq
aA
(q)(p)
aa
(q)(q)
q2
Genotype frequencies described by
p2+2pq+q2=1.0
The relationship between allele
and genotype frequency
• Let original A frequency be represented by
p and original a frequency be represented
by q
• Since there are only two alleles possible for
this gene locus, The frequencies of A and a
must equal 1.0
• Therefore, p + q =1.0
Sample: calculating genotype
frequencies from allele frequencies?
If a given population had the following allele frequencies:
allele frequency (p) for A of 0.8
allele frequency (q) for a of 0.2
Determine the genotype frequencies of this population?
AA
0.64
Aa
0.32
aa
0.04
AA = p2 ; Aa = 2pq ; and aa = q2 as follows…
We can also calculate the frequency of alleles from
the genotype frequencies.
When a population is in equilibrium the
genotype frequencies are represented as..
P2 + 2pq +q2
The allele frequency can therefore be calculated
as follows.
A = p2 + ½(2pq)
and
a = q2 + ½(2pq)
Examining our example again we see
that if we use the frequencies we
calculated for each genotype….
p2
2pq
q2
0.64 AA
.32 Aa
.04 aa
A = p2 + ½ (2pq)
A=.64 + ½ (.32)
A = 0.8
and since q = 1-p ; then a = 1-(0.8 ) a = 0.2
These rules hold as long as a
population is in equilibrium.
Hardy Weinberg Equilibrium
describes the conclusions and
assumptions that must be present
to consider a population in
equilibrium.
Hardy Weinberg Conclusions
1.
The allele frequencies in a population will not change
from generation to generation.
You would need at least 2 generations of data to
demonstrate this.
2.
If the allele frequencies in a population are given by p
and q then the genotype frequencies will be equal to
p2; 2pq ; q2.
Therefore if
AA can not be predicted by p2
Aa cannot be predicted by 2pq and
aa cannot be predicted by q 2
then the population is not in equilibrium
There are 5 assumptions which must be
met in order to have a population in
equilibrium
1. There is no selection. In other words there
is no survival for one genotype over another
2. There is no mutation. This means that none
of the alleles in a population will change over
time. No alleles get converted into other forms
already existing and no new alleles are
formed
3. There is no migration (gene flow)New
individuals may not enter or leave the
population. If movement into or out of the
population occurred in a way that certain
allele frequencies were changed then the
equilibrium would be lost
Exceptions to Hardy Weinberg cont.
4. There are no chance events (genetic drift)
This can only occur if the population is
sufficiently large to ensure that the chance of
an offspring getting one allele or the other is
purely random. When populations are small
the principle of genetic drift enters and the
equilibrium is not established or will be lost as
population size dwindles due to the effects of
some outside influence
5. There is no sexual selection or mate
choice Who mates with whom must be
totally random with no preferential selection
involved.
Problem #6 on page 219
• Go to your text page 219 and answer
question number 6.
• In humans, the COL1A1 locus codes for a
certain collagen protein found in bone. The
normal allele at the locus is denoted with S. A
recessive allele s is associated with reduced
bone mineral density and increased risk of
fractures in both Ss and ss women. A recent
study of 1,778 women showed that 1,194 were
SS, 526 were Ss, and 58 were ss.
• Are these two alleles in Hardy-Weinberg
equilibrium in this population?
• What information would you need to determine
whether the alleles will be in Hardy-Weinberg
equilibrium in the next generation?
Problem approach
• Check that conclusion #2 holds
– First figure genotype frequencies from the data
(percentages)
– Then from the data, count the actual A alleles in the
population and the actual a alleles in the population.
What are their frequencies?
• Then calculate the predicted genotype
frequencies of Hardy Weinberg and compare to
actual numbers.
The genotype frequencies are:
SS =1194/1778 = .67 Ss= 526/1778 = .30 ss= 58/1778 = .03
Calculate allele frequencies from genotype frequencies
2914/ 3556 S alleles = 0.82 S frequency or .67+1/2(.30) = 0.82
642 / 3556 s alleles = 0.18 s frequency or .03 + ½ (.30)=0.18
If the population is in Hardy-Weinberg equilibrium, the allele
frequencies should predict the genotype frequencies.
SS genotype frequency would be (0.82)2 , = 0.67;
Ss frequency would be 2 (.82) (.18), = 0.30;
ss frequency would be (0.18)2 , = 0.03.
These numbers almost exactly match the measured
genotype frequencies - so this population may be in HardyWeinberg equilibrium
However, what also must we
know to be sure?
• We would need to check in future
generations to make sure that the allele
frequencies are not changing.
• So here we confirmed conclusion #2 but
have not yet verified conclusion#1.
Example
Initial frequencies
15 B1B1
50 B1B2
15 B2B2
80
total
Example
Initial frequencies
15 B1B1
50 B1B2
15 B2B2
Calculate genotype
frequencies
15/80 = .1875
50/80 = .625
15/80 = .188
80
total
Example
Initial frequencies
15 B1B1
50 B1B2
15 B2B2
Calculate genotype
frequencies
15/80 = .1875
50/80 = .625
15/80 = .188
Calculate allele
frequencies in the
population
B1=
15+1/2(50)/80
= .5
B2=
15+1/2(50)/80
= .5
80
total
Example
Initial frequencies
15 B1B1
50 B1B2
15 B2B2
Calculate genotype
frequencies
15/80 = .1875
50/80 = .625
15/80 = .188
Calculate allele
frequencies in the
population
B1=
15+1/2(50)/80
= .5
Can we predict the
genotype frequency
from the allele
frequency?
B2=
15+1/2(50)/80
= .5
(Frequency of B1)2
2(B1 B2)
(Frequency of B1)2
(0.5) 2 = .25
2(.5)(.5) = .5
(0.5) 2 = .25
80
total
Example
Initial frequencies
15 B1B1
50 B1B2
15 B2B2
Calculate genotype
frequencies
15/80 = .1875
50/80 = .625
15/80 = .188
Calculate allele
frequencies in the
population
B1=
15+1/2(50)/80
= .5
Can we predict the
genotype frequency
from the allele
frequency?
Allele frequency does
not predict genotype
frequency.
B2=
15+1/2(50)/80
= .5
(Frequency of B1)2
2(B1 B2)
(Frequency of B1)2
(0.5) 2 = .25
2(.5)(.5) = .5
(0.5) 2 = .25
Population is not in Hardy-Weinberg equilibrium
because it violates conclusion 2
80
total
Using the Hardy-Weinberg
equilibrium with more than two
alleles
• In a population of mice, coat color is determined
by 1 locus with 4 alleles: A, B, C, and D. The
possession of an A allele confers black coat
color with another A allele, or a D allele. If a B
allele is present with an A allele then coat color
is brown, and if C is present with and A allele,
coat color is grey. All other phenotypes are light
tan. Given that the frequencie of the A, B, and C
alleles are .05, .4, and .3, respectivelly, what are
the phenotpic frequencies of black, brown, grey
and light tan mice when the population is in
Hardy-Weinberg equilibrium?
Adding Selection to the HardyWeinberg Analysis
• How do you know if a population is responding
to selection.
1. Some phenotypes allow greater survival to
reproductive age.
-or2. Equal numbers of individuals from each
genotype reach reproductive age but some
genotypes are able to produce more
viable (reproductively successful) offspring.
If these differences are heritable then
evolution may occur over time.
Caution
• most phenotypes are not strictly the
result of their genotypes.
• Environmental plasticity and
• interaction with other genes may also
be involved.
• not as simple as we are making it here.
We will look at two possible effects of
natural selection on the gene pool
1. Selection may alter allele frequencies or
violate conclusion #1
2. Selection may upset the relationship
between allele frequencies and genotype
frequencies. Conclusion #1 is not
violated but conclusion #2 is violated.
An example of what we might
see happen to allele
frequencies when natural
selection is at work
Let B1 and B2 = the allele frequencies of the
initial population
with frequencies of B1 = .6 and B2 = .4
• After random mating which produces 1000
zygotes we get:
Selection Example
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
1000 total
Selection Example
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to
reduced numbers of
some genotypes
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
1000 total
Selection Example
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to
reduced numbers of
some genotypes
number surviving
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
360
360
80
1000 total
800 total
Selection Example
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype
frequencies of mating
individuals which
survive is
.45
.45
.10
differential survival of
offspring leads to
reduced numbers of
some genotypes
1000 total
800 total
Selection Example
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype
frequencies of mating
individuals which
survive is
.45
.45
.10
B1 =
B2 =
differential survival of
offspring leads to
reduced numbers of
some genotypes
Using the genotype
frequencies, calculate
the allelic frequencies in
the new population
1000 total
800 total
Selection Example
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype
frequencies of mating
individuals which
survive is
.45
.45
.10
differential survival of
offspring leads to
reduced numbers of
some genotypes
Using the genotype
frequencies, calculate
the allelic frequencies in
the new population
B1 =
.45+1/2(.45)
= 0.675
1000 total
800 total
Selection Example
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype
frequencies of mating
individuals which
survive is
.45
.45
.10
B1 =
.45+1/2(.45)
= 0.675
B2 =
1/2(.45)+0.10
= 0.325
differential survival of
offspring leads to
reduced numbers of
some genotypes
Using the genotype
frequencies, calculate
the allelic frequencies in
the new population
1000 total
800 total
Selection Example
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype
frequencies of mating
individuals which
survive is
.45
.45
.10
B1 =
.45+1/2(.45)
= 0.675
B2 =
1/2(.45)+0.10
= 0.325
an increase of
.075
a decrease of
.075
differential survival of
offspring leads to
reduced numbers of
some genotypes
Using the genotype
frequencies, calculate
the allelic frequencies in
the new population
1000 total
800 total
Selection Example
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype
frequencies of mating
individuals which
survive is
.45
.45
.10
B1 =
.45+1/2(.45)
= 0.675
B2 =
1/2(.45)+0.10
= 0.325
an increase of
.075
a decrease of
.075
differential survival of
offspring leads to
reduced numbers of
some genotypes
Using the genotype
frequencies, calculate
the allelic frequencies in
the new population
1000 total
800 total
Thus, conclusion #1 is violated because the allele frequencies
are changed; we are not in equilibrium. The population is
evolving!
Creating an equation that allows for
selection
•
First, we analyze the population on the basis of
the fitness of the offspring.
•
Fitness (w) is defined as the survival rates, or
percentage of individuals which survive to
reproduce.
•
We can use the fitness (w) of each genotype to
calculate the average fitness of the population
Fitness formulas
MEAN FITNESS
• If :
w11 = fitness of allele #1 homozygote
w12 = fitness of the heterozygote
w22 = fitness of allele #2 homozygote
mean fitness of the population will be described by the
formula:
ŵ = p2w11 + 2pqw12 + q2w22
For our previous example
• B1= 0.6 and B2 = 0.4 and
fitness of B1B1 = 1.0 (100% survived)
fitness of B1B2 = .75 ( 75% survived)
fitness of B2B2 = .50 (50% survived)
• Figure the mean fitness now.
For our previous example
• B1= 0.6 and B2 = 0.4 and
fitness of B1B1 = 1.0 (100% survived)
fitness of B1B2 = .75 ( 75% survived)
fitness of B2B2 = .50 (50% survived)
• Figure the mean fitness now.
• ŵ= (.6)2(1)+
For our previous example
• B1= 0.6 and B2 = 0.4 and
fitness of B1B1 = 1.0 (100% survived)
fitness of B1B2 = .75 ( 75% survived)
fitness of B2B2 = .50 (50% survived)
• Figure the mean fitness now.
• ŵ= (.6)2(1)+(2(.6)(.4)(.75)) +
For our previous example
• B1= 0.6 and B2 = 0.4 and
fitness of B1B1 = 1.0 (100% survived)
fitness of B1B2 = .75 ( 75% survived)
fitness of B2B2 = .50 (50% survived)
• Figure the mean fitness now.
ŵ= (.6)2(1)+(2(.6)(.4)(.75)) + (.4)2 (.5) = .80
We can also use the fitness to
calculate the expected frequency of
each genotype in the next
generation. We can do it the long
way, if we know actual numbers
OR…….
B1B1 = P2w11
ŵ
B1B2 = 2pqw12
ŵ
B2B2 = q2w22
ŵ
We can also use the fitness to calculate
the expected frequency of each allele
in the next generation.
B1 = p2w11+pqw12
ŵ
B2 = pqw12+q2w22
ŵ
Finally, we can calculate the change (D) in
the frequency of B1 or B2 directly as follows:
Δ B1 = Δp =
p (pw11+qw12 – ŵ)
ŵ
Δ B2 = Δq = q (pw12+qw22 – ŵ)
ŵ
In class problem
• Go back to problem # 6 on page 219.
Taking this current population that you
have already analyzed, figure out what the
new genotype and allele frequencies will
be if the fitness of these individuals is
actually as follows:
• SS individuals 0.7 ; Ss individuals 1.0 and
the ss individuals 0.8.
We calculated S = .82 and s = .18
and the fitnesses are w11(SS)=.7; w12(Ss)=1; w22(ss)=.8
We know
this to start
ŵ = p2w11 + 2pqw12 + q2w22
ŵ= (.82) 2 (.7) + 2(.82)(.18)(1.0) + (.18)2 (.8)
ŵ = .470 + .295 + .026 = .791
B1B1 = P2w11
ŵ
; SS = (.82)2(.7) / .791 = .595
B1B2 = 2pqw12
ŵ
;Ss = 2(.82)(.18)(1.0) / .791 = .373
B2B2 = q2w22
ŵ
;ss = (.18)2(.8) / .791 = .032
B1 =
B2 =
p2w11+pqw12
ŵ
pqw12+q2w22
ŵ
S = (.82)2(.7) + (.82)(.18)(1.0) = .78
.791
s = (.82)(.18)(1.0) + (.18)2(.8) = .22
.791
So…… B1B1 = .595
B1B2 = .373
B2B2 = .032
and
B1 =
.78
B2 = .22
Is this population in equilibrium?
Have the allele frequencies changed?
Can we predict the genotype frequencies from the
allelic frequencies?
Experimental confirmation of
loss of Hardy Weinberg
equilibrium
Fruit fly experiments of
Cavener and Clegg
• Worked with fruit flies having two
versions of the ADH (alcohol
dehydrogenase) enzyme, F and S. (for
fast and slow moving through an
electrophoresis gel)
• Grew two experimental populations on
food spiked with ethanol and two control
populations on normal, non-spiked food.
Breeders for each generation were
picked at random.
• Took random samples of flies every few
generations and calculated the allele
frequencies for AdhF and AdhS
Figure 5.13 pg 158
Can we identify which equilibrium
assumption is being violated?
• only difference is ethanol in food
migration?
random mating?
drift?
mutations?
• If we eliminate the other factors...
Must be selection for the fast form of gene.
• Indeed studies show that AdhF form breaks down
alcohol at twice the rate as the AdhS form.
• Therefore offspring carrying this allele are more fit and
leave more offspring and the make-up of the gene
pool changes.
Cavener and Clegg
demonstrated that selection
pressure can lead to changes
in allele frequencies in just a
few generations
A second selection scenario
• Selection may upset the relationship
between allele frequencies and
genotype frequencies.
• Conclusion #1 ( allele frequencies do
not change) is not violated but
conclusion #2 (that we can predict
genotype frequencies from allele
frequencies) is violated.
Genetic variation for resistance to
kuru in a Fore population
• Kuru is a fatal neurological disorder that is
caused by the misfolding of a prion protein in
neurological tissue.
• Humans contract the disease by eating
contaminated tissue.
• Symptoms start with shivering and trembling and
lead to staggering and trouble talking and
swallowing to coma and death.
Genetic variation for resistance to
kuru in a Fore population
• The greatest outbreak of kuru occurred in the
1950’s in the Fore people. They practiced ritual
funereal cannibalism.
• Researchers wanted to know whether certain
genotypes were more susceptible than others.
• They looked at 2 different alleles for the
encoded protein that was responsible for the
disease.
– One allele contained Met at position 129 while
the other contained a Val at the same
position.
Genotypes of the survivors
• Researchers surveyed a population of 30
females who had eaten dead relatives and
yet survived the kuru epidemic without
getting sick.
• The numbers of individuals with their
genotypes were as follows:
– Met/Met
4
Met/Val
23
Val/Val
3
Do these numbers deviate significantly
from H.W. equallibrium?
– Met/Met
4
Met/Val
23
Val/Val
3
1. Calculate the allele frequencies.
Met
(8+23) =0.52
60
Val
(23+6) =0.48
60
2. Calculate the genotype frequencies
expected under the Hardy-Weinberg
principle.
– Met/Met
(.52)2
Met/Val
2(.52)(.48)
= .27
=.5
Val/Val
(.48)2
= .23
3. Calculate the expected number of
individuals of each genotype under
Hardy-Weinberg equilibrium.
Met/Met
(.27)(30)
=8
Met/Val
(.5)(30)
= 15
Val/Val
(.23)(30)
=7
These numbers are different from the ones that
were actually observed in the population.
=4
= 23
=3
Is the difference statistically
significant?
• Is it plausible that the difference between
the expected and observed values arose
by chance?
• Our null hypothesis is that the difference is
simply due to chance.
Chi-square
4. Calculate the test statistic using Chi-square.
2
2
(observed
–
expected)
c =
(expected)
c2 = (4-8)2 + (23-15)2 + (3-7)2 = 8.55
8
15
7
Is the value 8.55 statistically significant or could it
reasonably have occurred by chance?
5. Determine whether the value of the test
statistic is significant.
•
Look up the value of 8.55 in the table of “Critical values of
the chi-square distribution”.
–
To use this table we need to calculate the degrees of freedom (df)
for the test statistic.
df = number of classes – number of independent values.
There are three classes: the number of genotypes.
We calculated two values in determining the expected
values: the total number of individuals, and the
frequency of the Val allele
df = 1
• The critical value for most research studies is P
= 0.05.
– This means that there is a 5% chance that are null
hypothesis is correct.
– Any P value less than 0.05 means that we reject our
null hypothesis.
• Our c2 value of 8.55 has a P value of .0034.
• Therfore we reject our null hypothesis that the
difference between the expected and the
observed is simple due to chance events.
What pattern of allele frequency
changes might be caused by selection
• If selection is acting, does the rate of
evolution of a particular allele depend
on whether it is….
dominant or recessive?
heterozygote or homozygote?
Natural Selection is most potent as an
evolutionary force when selection acts on
recessive alleles which are common and the
dominant form is relatively rare.
• Dawson’s Flour beetle example
• Studied a gene locus that had a wild
type (+) allele and a lethal allele.
• +/+ or +/L are normal L/L is lethal.
• Two experimental populations
composed of all heterozygotes +/L; (+ =
0.5 and L =0.5).
• Expected populations to evolve toward
lower frequency of the L allele.
Results showed that the
recessive lethal did
drop rapidly at first but
slowed down over
successive generations.
WHY?
In each succeeding generation all LL
are lost and ++ makes up a greater
proportion of the survivors.
With each new generation there are
less and less homozygous lethals for
selection to act on and the lethal allele
hides in the heterozygotes
Summary
• Dawson showed that dominance and allele frequency interact to
determine the rate of evolution when acted on by selection
• when a recessive allele is common and there is a great difference in
fitness between the phenotypes then evolution is rapid because
both the recessive and dominant phenotype are well represented for
selection to act on.
• Example if A= .05 a= .95
–
AA
(0.05)2
=
.0025
Aa
2(0.05)(0.95)
=.095
aa
(0.95)2
= .9025
Almost 10 % in the population have the dominant phenotype, while 90 % have
the recessive phenotype
Thus if the two phenotypes differ in fitness there will be a change in
allele frequency
Summary
• If recessive allele is rare and the dominant allele is
common, evolution is slow.
• Example if A= .95 a= .05
–
AA
(0.95)2
=
.9025
Aa
2(0.95)(0.05)
=.095
aa
(0.05)2
= .0025
• Approximately 100% of the population has the dominant phenotype.
• Even if the phenotypes differ greatly in fitness, there are so few of the
minority phenotype that there will be little change in allele frequencies
in the next generation.
Selection on Heterozygotes and
Homozygotes
Selection on Heterozygotes and
Homozygotes
• Normally in a recessive/ dominant gene,
the fitness of the heterozygote will be
equal to that of the dominant homozygotes
• Also, it is possible for the heterozygotes to
have a fitness intermediate to the two
homozygotes. (incomplete dominance)
• Thirdly we may find Heterozygote
Superiority or Inferiority
Scenario #1 from Mukai and Burdick
Fruit fly experiment
• Studied a gene in which
 Homozygotes for one allele are
viable (VV)
 Homozygotes for the other allele are
not viable and lethal. (LL)
• Started with all heterozygotes to establish
a new population (each allele =.5)
• Predict the frequency of V after 15
generations.
The experiment
-After several generations
equilibrium was reached at
.79 frequency for the viable
allele
-This means that the lethal
allele was at a frequency of
0.21!
How could this be?
To further test these results, Mukai and
Burdick did a second experiment...
-This time the population
began with a frequency
of .975 of V allele.
- Expected the population to
eliminate all lethal
alleles and fix the viable
allele at 1.0.
-However, Same equilibrium
around a frequency of
.79 was reached for
viable allele!
This is a case of Heterozygote superiority
or overdominance (also called heterosis)
• There is some advantage to the
heterozygote condition and the
heterozygote actually has a superior
fitness to either homozygote.
• Example in humans is sickle cell anemia,
• Leads to the maintenance of genetic
diversity = balanced polymorphism
Can also have Heterozygote inferiority
or underdominance
• Where the heterozygote condition is
inferior to either of the homozygotes
• What do you predict would happen
here?
• Leads to fixation of one allele in the
population, while the other is lost.
• Either allele may be fixed depending on
conditions and beginning frequencies of
each allele in the gene pool.
Example from G.G. Foster’s work with
Fruit flies
• Looked at
chromosome
differences where
different chromosome
forms behave like
single alleles.
• In meiosis compound
chromosomes may or
may not segregate.
Only certain chromosomal;
combinations will lead to viable zygotes
+
A
B
+
C
+
D
+
Which of these combinations
would give viable offspring?
What is the evolutionary
impact?
• Leads to a loss of genetic diversity
• Although if different alleles are fixed in
different populations can help maintain
genetic diversity among populations
SUMMARY
• When one allele is consistently favored it
will be driven to fixation
• When heterozygote is favored both alleles
are maintained and at a stable equilibrium
(balanced polymorphism) even though
one of the alleles may be lethal in the
homozygous state.
Frequency-Dependent Selection
• Evolution can be effected by the frequency
of a particular phenotype in the population.
Frequency-dependent
selection
• The Elderflower orchid
example in book
• Population’s allele
frequencies remain at or
near an equilibrium but it is
due to the direction of
selection fluctuating
• First one allele is favored
and then the other
• Population fluctuates
around an equilibrium point
The favored allele fluctuates
because
• Bumblebees visit yellow and purple flowers alternately
• The least frequent phenotype is visited more often and
receives more pollination events.
• In subsequent generations this color becomes more and
more frequent until it becomes the dominant color.
• Once this happens then the same color becomes less
frequently visited and the other color becomes favored.
• Oscillation between the two colors continues and the
favored allele alternates over time around some mean
equilibrium value.
End of Selection Effects
Effects of Mutation
• Mutation is the source of all new alleles
• Mutation provides the raw material on
which selection can act
Hardy Weinberg and Mutation
• Mutation alone is a weak or nonexistent
evolutionary force
• If all mutations that happened, occurred in
gametes so that they would be immediately
passed on to their offspring and ….
• the rate of mutation were high, say Aa at a
rate of 1 in 10,000 per generation.
• then the rates are very slow as shown in
figure 5.22
Figure 5.23 pg. 183
Mutation and Selection
In concert with selection, mutation
becomes a potent evolutionary
force.
Richard Lenski and colleagues
working with E. coli
• Used a strain of E. coli that cannot exchange
DNA (conjugation) so the only possible source of
genetic variation is mutation.
• Showed steady increases in fitness and size
over 10,000 generations in response to a
demanding environment. (little over 4 years)
• However, increases in fitness occurred in jumps
when a beneficial mutation occurred and then
spread rapidly through the population
Figure 5.25 pg 185
Mutation –Selection Balance
• Most mutations are deleterious
• Selection acts to eliminate them
• Deleterious Mutations persist because
they are created anew over and over
again
• When the rate at which they are formed
exactly equals the rate at which they are
eliminated by selection the allele is in
equilibrium. = mutation-selection
balance
Intuition tells us that ...
• If the mutation is only mildly deleterious and
therefore selection against it is weak; and
• Mutation rate is high then
• The equilibrium frequency of the mutated allele
will be relatively high in the population.
• If, on the other hand, there is strong selection
against a mutation (the mutation is highly
deleterious) and the mutation rate is low then
• Equilibrium ratio of the mutated allele will be
low
Example
• Spinal muscular atrophy, second most common
lethal autosomal recessive disease. Selection
coefficient is .9 against the disease mutations.
• However, 1 in 100 carry a disease causing allele
in Caucasians
• Research shows that the mutation rate for this
disease is quite high
• Mutation selection balance is proposed
explanation for persistence of mutant alleles.
Cystic Fibrosis
• Cystic fibrosis is the most common lethal
autosomal recessive disease in Caucasians
• Mutation-selection balance alone cannot
account for the high frequency of the allele = .02
• Appears to also be some heterozygote
superiority involved
• Heterozygotes are resistant to typhoid fever
bacteria and have superior fitness during typhoid
fever epidemic.
Cystic Fibrosis
• Pier and his colleagues have found, in 11 European countries, an
association between the severity of typhoid outbreaks and the
frequency of the delta-F508 allele (the most common loss-offunction mutation) a generation later.
• Salmonella typhi bacteria manipulate their host cells, causing them
to express more CFTR protein on their membranes.
• Pier et al. engineered cells homozygous for functional CFTR alleles,
homozygous for a common loss-of-function allele, and heterozygous
for the two. The loss-of-function homozygotes were virtually
impervious to invasion by typhus-causing bacteria; heterozygotes
were more vulnerable, but accumulated 86% fewer bacteria than did
the dominant homozygotes.
Quiz 1.
Multiply Choice question: Mark all the choices that are correct.
1. Which of the follow describes a population that is at equilibrium yet
has an allele that is lethal. The selective advantage enjoyed by the
lethal allele when it is in heterozygotes exactly balances the obvious
disadvantage it suffers when it is in homozygotes.
A homogenization across populations
B haplodiploidy
C heterozygote superiority
D overdominance