Transcript Chapter 5

Chapter 5
Mendelian Selection in
Populations:
Selection and Mutation
Population Genetics
 Integrates Darwin’s theory of evolution by natural
selection with Mendelian genetics
Changes in relative abundance of traits in a
population can be tied to changes in relative
abundance of alleles that influence them
 Case Studies
CCR5-D32
Eugenics sterilization
Cystic fibrosis
Population Genetics

CCR5-D32 allele
Confers resistance to HIV-1 infection

Buck vs. Bell
U.S. Supreme Court upheld states’ mandatory sterilization
laws for feebleminded people
– Eugenics movement to remove feeblemindedness gene from
population

Cystic fibrosis
Genetic disease that affects Caucasians
Autosomal recessive
Causes chronic infections of Pseudomonas aeruginosa
– Severe lung damage
Hardy-Weinberg Equilibrium
Principle
 Must understand how alleles behave in
populations to understand case studies
 Develop a null model for behavior of genes in
populations
 Model specifies what will happen to
frequencies of alleles and genotypes
 Applies to all diploid sexual organisms
Hardy-Weinberg Equilibrium
Principle
Population = group of interbreeding individuals and
their offspring
 Life Cycle

Adults produce gametes
Gametes combine to make zygotes
Zygotes grow up to become next generation of adults

Track fate of Mendelian genes across generations in
a population
Find out if particular alleles become more or less
common over time
Hardy-Weinberg Equilibrium
Principle
 Imagine that mice have a particular locus A
with two alleles: A and a
Could also call them A1 and A2 or A and B
All equivalent
Track these alleles and follow them through one
complete turn of the cycle to see if frequencies
change
Hardy-Weinberg Equilibrium
Principle
 A numerical example
Assume adults choose their mates at random
– Matings are random within the gene pool
Diploid organisms, so each has two alleles for A
locus
Meiosis caused one allele to be in each gamete
for the A locus
Hardy-Weinberg
Hardy-Weinberg Equilibrium
Principle
 A numerical example
Imagine 60% of eggs and sperm received allele A
and 40% received allele a
Frequency of A allele in gene pool is 0.6, of a
allele is 0.4
When egg and sperm meet, what proportion of
genotypes will be AA?
60% egg will be A, 60% sperm will be A
– 0.6 X 0.6 = 0.36
– 36% of zygotes will have genotype AA
Hardy-Weinberg Equilibrium
Principle
 A numerical example
How many would be aa?
– 0.4 X 0.4 = 0.16
How many would be Aa?
– 0.6 X 0.4 X 2 = 0.48
Notice that 0.36 + 0.48 + 0.16 = 1
– All possible genotype frequencies will add up to one
Hardy-Weinberg
Hardy-Weinberg Equilibrium Principle

If zygotes grow up, what will the frequency of A in the
next generation be?
AA is 36%
– All gametes carry A
Aa is 48%
– Half will carry A, half will carry a
aa is 16%
– All gametes carry a
Frequency of A in next generation will be:
– 0.36 + (1/2)0.48 = 0.6
Frequency of a will be:
– 0.16 + (1/2)0.48 = 0.4
Hardy-Weinberg Equilibrium
Principle
 A numerical example
0.6 + 0.4 = 1
Allele frequencies are the same as in the first
generation
Allele frequencies are in equilibrium
The population does not evolve
If a population is in Hardy-Weinberg Equilibrium
it will NEVER EVOLVE!!!
– Regardless of starting frequencies
Hardy-Weinberg Equilibrium
Principle
 The General Case
Imaginary population
Single locus with A and B as the alleles
Three possible diploid genotypes
– AA, AB, BB
Frequency of allele A is called p
Frequency of allele B is called q
p+q=1
Hardy-Weinberg Equilibrium
Principle
 The General Case
Let gametes make zygotes
Four combinations
– A + A = AA
– A + B = AB
– B + A = BA
– B + B = BB
p X p = p2
p X q = pq
q X p = qp
q X q = q2
p2 + 2pq + q2 = 1
– Genotype frequencies in Hardy-Weinberg Equilibrium
General case
General Case
General case
p2 + 2pq + q2 = 1
Next generation (zygotes grow and reproduce)
What is the Allele freq of A?
p2 + (½)2pq
= p2 + pq
Subst. (1-p) for q
=p
Hardy-Weinberg Equilibrium
Principle
 Two fundamental conclusions
Conclusion 1: the allele frequencies in a
population will not change, generation
after generation
Conclusion 2: if the allele frequencies in a
population are given by p and q, the
genotype frequencies will be given by p2,
2pq, and q2
Hardy-Weinberg Equilibrium
Principle
 Why do we use Hardy-Weinberg Equilibrium
Principle?
Shows evolution does not happen (Huh?)
Gives specific set of testable assumptions
If an assumption is violated, the Conclusions do
not hold
It is a null model that we can use to test for
evidence of evolution occurring!!!
Hardy-Weinberg Equilibrium Principle
 Assumptions of Hardy-Weinberg
There is no selection
– All members contribute equally to gene pool
There is no mutation
– No new alleles are created
There is no migration
– All alleles stay in gene pool
There is a large population size
– No random events = genetic drift
Panmixia
– Mates are chosen randomly
How we can violate H-W shows us
how evolution can occur!
Hardy-Weinberg Equilibrium Principle
By having explicit assumptions, the violations of
assumptions can be known to determine which
forces are causing disequilibrium = evolution
 Example with CCR5-D32

Everyone survives equally
No new D32 mutations arise
No one moves between populations
Populations are infinitely large
People choose mates at random
We can test for Hardy-Weinberg to see which
assumptions are broken to understand cause of
evolution and future of the D32 allele
Selection
 Individuals with particular phenotypes survive
to reproduce more than others
Differential reproductive success
 Phenotypes must be heritable
 Can selection change the frequencies of alleles
in the gene pool from one generation to the
next?
Can violation of the no-selection assumption violate
Conclusion 1 of Hardy-Weinberg Equilibrium?
The Hardy-Weinberg principle
Given certain assumptions, whatever the initial
genotype frequencies for two autosomal alleles may
be, after one generation of random mating, the
genotype frequencies will be p2:2pq:q2, and both
these genotype frequencies and the allele
frequencies will remain constant in succeeding
generations.
This means, as soon as Evolution stops occurring,
the population will go back into H-W equilibrium
within 1 generation!
The Math
N = total number of individuals
n1 = number of A1A1 individuals
n2 = number of A1A2 individuals
n3 = number of A2A2 individuals
p = frequency of the first allele
q = frequency of the second allele
The Math

p = frequency of the first
allele
q = frequency of the
second allele
n2
(n1 
)
2
 p
N
n2
(n3  )
2 q
N
p + q = 1 always in a two allele system
The Basic Symbols and their
Meaning
x = f(A1A1) = n1 / N
y = f(A1A2) = n2 / N
z = f(A2A2) = n3 / N
p = x + ½ y
q = z + ½ y
Back door method:
q = √q2 = √z
Only works if the population is in H-W equilibrium
Why is it so?
p + q = 1
(p + q)2 = 12
p2 + 2pq + q2 = 1
f(A)
f(a)
f(A)
f(a)
f(AA) = f(A) x f(A)
f(aA) = f(a) x f(A)
f(Aa) = f(A) x f(a)
f(aa) = f(a) x f(a)
• p2 = f(AA)
• pq + pq = 2pq = f(Aa)
• q2 = f(aa)
Three alleles
p = f(A1)
q = f(A2)
r = f(A3)
p + q + r = 1
(p + q + r)2 = 12
p2 +2pq + q2 + 2pr + 2qr + r2 = 1
f(A1A1) + f(A1A2) + f(A2A2) + f(A1A3) + f(A2A3) +
f(A3A3) = 1
What’s it good for?
Predicting genotype frequencies given allele
frequencies
Genotypes will approximate a binomial
distribution – (p + q)2 = 1 – after 1 generation of
random mating.
If we know the allele frequencies in generation 1,
we can predict the genotype frequencies in
generation 2.
Allele and genotype frequencies will not change
as long as the assumptions are met.
For example:
Generation 1:
f(AA) = 0.2, f(Aa) = 0.8, f(aa) = 0.0
What are the genotype frequencies in generation
2?
First, find p and q:
p = x + ½ y = 0.2 + (0.8 / 2) = 0.6
q = z + ½ y = 0.0 + (0.8 / 2) = 0.4
Generation 2:
F(AA) = p2 = 0.36
F(Aa) = 2pq = 0.48
F(aa) = q2 = 0.16
Another example:
Generation 1:
f(AA) = 0.5, f(Aa) = 0.2, f(aa) = 0.3
What are the genotype frequencies in generation
2?
First, find p and q:
p = x + ½ y = 0.5 + (0.2 / 2) = 0.6
q = z + ½ y = 0.3 + (0.2 / 2) = 0.4
Generation 2:
F(AA) = p2 = 0.36
F(Aa) = 2pq = 0.48
F(aa) = q2 = 0.16
Did you notice?
We started with different genotype
frequencies in each example, but the
same allele frequencies (p & q).
We ended (generation 2) with the same
genotype frequencies in both examples.
If all of the assumptions are met, what
will be the genotype frequencies in
generation 10?
What else is it good for?
Null hypothesis:
H0: the population is in H-W equilibrium
Ha: the population is not in H-W
equilibrium
Use a χ2 test to determine if the population
is significantly different from H-W
expectations
If it is, one (or more) of the assumptions
must be violated!
Assumptions of H-W
Equilibrium
1.
2.
3.
4.
5.
6.
Random mating – panmixia
Infinitely large population
No gene flow – genes are not added from outside
the population
No mutation – genes do not change from one
allelic state to another
No natural selection – all individuals have equal
probabilities of survival and reproduction
Discrete generations – offspring do not mate with
previous generations
What happens if the
assumptions are violated?
Evolution!
Evolution (in terms of H-W) is change in allele
frequency over time.
Allele frequencies do not change in a
population in H-W equilibrium, therefore:
A population in H-W equilibrium is not
evolving.
Violations of H-W
assumptions
Violations due to sampling error
Drift
Inbreeding
Systematic changes
Mutation
Selection
Gene flow / migration
Violations of H-W
assumptions
Factors that increase variation
Mutation
Gene flow
Factors that decrease variation
Selection
Drift
Inbreeding
Selection
 Example with mice
B locus affects probability of survival
Frequency of B1 = 0.6
Frequency of B2 = 0.4
After random mating, genotype frequencies are
0.36, 0.48, and 0.16
Will use a population of 1000 individuals
– B1B1 = 360 individuals
– B1B2 = 480
– B2B2 = 160
Selection
 Example with mice
Incorporate selection
All B1B1 individuals survive
75% B1B2 individuals survive
50% B2B2 individuals survive
Now 800 adults left
– 360 B1B1 (360)
– 360 B1B2 (480)
– 80 B2B2 (160)
Selection
 Example with mice
Frequencies of genotypes are:
– B1B1 360/800 = 0.45
– B1B2 360/800 = 0.45
– B2B2 80/800 = 0.1
When they produce gametes
– B1 = 0.45 + (1/2)0.45 = 0.675
– B2 = (1/2)0.45 + 0.1 = 0.325
– Allele frequencies
Frequency of B1 rose by 7.5%
Frequency of B2 fell by 7.5%
Selection
Population evolved in response to selection
 Rarely is selection so strong
 Usually requires many generations to change
allele frequencies much

Selection
 Empirical example with Drosophila
Cavener and Clegg
Two alleles for alcohol dehydrogenase locus
– AdhF and AdhS
– Break down alcohol at different rates
Maintained two populations of flies spiked with
alcohol and two controls without alcohol
At each generation took random sample of flies and
determined their genotypes
Selection
 Empirical example with Drosophila
Control populations appeared to be in HardyWeinberg Equilibrium
– Allele frequencies did not change
Populations under selection pressure (alcohol)
showed a decline in AdhS allele
Hardy-Weinberg Conclusion 1 did not hold in
experimental populations
The populations evolved because of selection for
better ability to break down alcohol
Selection
 How does selection affect Conclusion 2?
Can we still calculate genotype frequencies by
multiplying allele frequencies
 Sometimes Conclusion 2 is violated
Allele frequencies do not change but genotype
frequencies cannot be calculated by HardyWeinberg equation
Can use Chi Square test to determine if
genotype frequencies vary significantly from
Conclusion 2’s expectations
Selection
 CCR5-D32
Will the frequency of the allele increase in
response to HIV epidemic?
Three potential models
– Model 1
 Frequency of allele 20%
 1/4 of people with genotype +/+ or +/D32 die
before reproducing
 All D32/D32 individuals survive
 After 40 generations (1000 years) the D32 allele is
nearly 100%
Selection
 CCR5-D32
Model 2
– Frequency of allele 20%
– HIV infection rate less than 1%
– All D32/D32 individuals survive
– After 40 generations (1000 years) the D32
allele is still at 20%
– Selection is too weak to cause a large change
in allele frequencies
Selection
 CCR5-D32
Model 3
– Frequency of allele 1%
– 1/4 of people with genotype +/+ or +/D32 die
before reproducing
– All D32/D32 individuals survive
– After 40 generations (1000 years) the D32
allele is still at 1%
– Most copies of D32 would be heterozygotes
and hidden from selection
Patterns of Selection
 Dominant and Recessive alleles (Directional
selection)
 Flour beetles with l locus
Two alleles: + and l
– Individuals with genotype +/+ or +/l are normal
– Individuals with genotype l/l do not survive
 Recessive lethal allele
Dawson started colonies with heterozygotes
– Allele frequencies 0.5 for each
Patterns of Selection
 Flour beetles with l locus
Because l/l have lower fitness, expect
population to evolve to lower l frequencies
Measured allele frequency over 12 generations
Frequency of l allele dropped to 0.25 but was not
eliminated
Dominance and allele frequency interact
– If recessive is common, evolution is rapid
– When recessive is rare, evolution is very slow
– When rare, recessive allele is usually hidden from
selection
Patterns of Selection
 Selection coefficient
w = fitness of an allele
s = strength of selection on an allele
w++, w+l, wll
– w++ = 1, w+l = 1, wll = 1 + s
– s gives strength of selection on homozygous
recessive phenotype
 Positive s is selection in favor of allele
 Negative s is selection against allele
Patterns of Selection
 Selection coefficient
Selection on dominant allele
w++ = 1 + s, w+l = 1 + s, wll = 1
– Positive s is selection in favor of allele
– Negative s is selection against allele
Adaptive landscapes of fitness
Patterns of Selection
 Selection on heterozygotes and
homozygotes (Balancing selection)
 When one allele is dominant and one is
recessive, heterozygote fitness is equal to
that of one kind of homozygote
 Other scenarios possible
Often fitness is intermediate to two homozygotes
– Changes rate of evolution
– Eventually one allele may become fixed and the other
lost
Patterns of Selection



Sometimes fitness of heterozygote is superior or inferior to
either homozygote
Different evolutionary outcomes produced
Example with Drosophila melanogaster
Single locus
– Homozygotes for one allele viable
– Homozygotes for other allele not viable
Used heterozygotes to start two populations
– Initial allele frequencies 0.5 each
Frequency of viable allele increased rapidly at first
Rate slowed and viable allele reached equilibrium at 0.79
Patterns of Selection
 Example with Drosophila melanogaster
Started new populations with viable allele at 0.975
Frequency of viable allele dropped to 0.79
Patterns of Selection
 Example with Drosophila melanogaster
How did selection maintain a lethal allele at
0.21?
Heterozygote superiority = overdominance
Heterozygotes have higher fitness than
either homozygote
Overdominance maintains genetic diversity
Patterns of Selection

Second example with Drosophila
Heterozygotes may have lower fitness than either
homozygote
Fruit flies with compound chromosomes
– Homologous chromosomes that have swapped entire arms
– During meiosis, compound chromosomes may not segregate
– Four types of gametes can be produced




Both homologous chromosomes
Just one member of pair
Just other member of pair
Neither member of pair
Patterns of Selection
 Second example with Drosophila
When two compounds mate, 1/4 of zygotes
have correct chromosome dose and are
viable
– Other 3/4 have too many or too few pieces and
are not viable
When compound flies mate with normal flies,
no zygotes are viable
Patterns of Selection
 Second example with Drosophila
Established populations with some founders with
compound chromosomes, C(2), and some with
compound chromosomes, C(3)
– Treat chromosomes as alleles
– C(2)C(2) and C(3)C(3) genotypes
Fitnesses of mixed populations
– wC(2)C(2) = 0.25
– wC(2)C(3) = 0
– wC(3)C(3) = 0.25
Genotypes exhibit underdominance
Patterns of Selection
 Algebraic treatment predicts that genetic
equilibrium will be achieved if both allele are
at 0.5.
 The equilibrium is unstable if either allele
freq. goes above or below 0.5
 It will quickly rise to 1 if above 0.5 and go to 0
if below 0.5
Patterns of Selection
Patterns of Selection
 Second example with Drosophila
Established populations with some founders with
compound chromosomes, C(2), and some with
normal chromosomes, N(2)
– Treat chromosomes as alleles
– C(2)C(2) and N(2)N(2) genotypes
Fitnesses of mixed populations
– wC(2)C(2) = 0.25
– wC(2)N(2) = 0
– wN(2)N(2) = 1
Genotypes exhibit underdominance
Patterns of Selection

Second example with Drosophila
Set up mixed 13 populations with initial C(2) frequencies
ranging from 0.71 to 0.96
Populations with higher initial C(2) frequency, C(2) rose to
fixation
Populations with lower C(2) frequency, C(2) was lost
Unstable equilibrium reached at 0.9 C(2)
Heterozygote inferiority reduces genetic diversity within
populations by pushing alleles to fixation
It can help to maintain genetic diversity among
populations
Frequency-Dependent Selection
So far we have assumed selection is constant over
time
 Direction of selection may fluctuate

One allele favored and then the other

Elderflower orchids
Persist in two colors, yellow and purple
Flowers attract bumble bees but they get no reward
Alternate between colors
Frequency-Dependent
Selection
 Bee visits one color 1st, no reward,
alternates to other color, no reward.
 This keeps going until the bee leaves
 Bees tend to visit equal numbers of
flowers
 Thus, more rare color is visited more often
per plant
Frequency-Dependent
Selection
 If pollinator visits equal reproductive
success, then rare-color has advantage
Selection by bees favors yellow until it
becomes too common, then switch to purple.
Frequency-Dependent Selection
 Experiment, 10 experimental arrays of 50
plants each
 Freq. of yellow flowers varied among arrays
with 2 arrays at 0.1,0.3,0.5,0.7 and 0.9
 Monitored orchids for removal of pollinaria,
deposition of pollinaria and fruit set
Frequency-Dependent Selection
Freq of yellow morph
Frequency-Dependent Selection
 Fitnesses change each generation with freq.
of allele (yellow flower)
 Move toward equilibrium freq.
 Frequency-Dependant selection can maintain
genetic diversity
Compulsory Sterilization
Evolutionary consequences of eugenics
sterilization
 Reduce fitness of feeblemindedness genotype to
zero to reduce frequency of alleles responsible for
feeblemindedness
 Defined as: “One who is capable of earning his
living under favorable circumstances, but is
incapable from mental defect existing from birth or
from an early age. . .”
 Feeblemindedness thought to behave as a simple
Mendelian recessive trait

Compulsory Sterilization

Remember that recessive genetic diseases are not
easily eliminated from populations
Rare recessives decline slowly

Scientists thought feeblemindedness was very
common and increasing
At least 1% of population: ff

If q is frequency of recessive allele f:
q = √0.01 = 0.1
wff = 0 if all are sterilized
 Over 250 generations frequency declines from 0.01
to 0.0025

Compulsory Sterilization
Is this decrease good?
 Most copies of allele are in heterozygotes and
hidden from selection

Compulsory Sterilization
 Problems with the practice
Feeblemindedness is not a recessive allele
Caused by many heritable and nonheritable factors
People were classified based on hearsay
Eugenics movement not based on sound
population genetics understanding
Mutation

How do deleterious alleles like cystic fibrosis
remain at high frequencies in population?
Heterozygote superiority?
Introduced anew by mutation?
Mutation introduces new alleles into a population
 How effective is mutation as a force of evolution?
 Can mutation violate Conclusions of HardyWeinberg Equilibrium Principle?

Mutation
Mutation alone is not a potent evolutionary force
 Model mice population

Locus A
Frequency of allele A = 0.9
Frequency of allele a = 0.1
a is recessive loss of function mutation
Copies of A are converted to a at a rate of 1 copy per
10,000 generations
– Very high mutation rate
Mutation

Model mice population
Back mutations to A are negligible
Assume all mutations happen in gametes in gene pool
Adult genotype frequencies
–
–
–
–
AA = 0.81
Aa = 0.18
aa = 0.01
In Hardy-Weinberg proportions
Alleles in gametes are still 0.9 and 0.1
Mutation

Model mice population
Now 1 of 10,000 A alleles mutates to a
New frequency of A (p) is old frequency minus fraction
lost to mutation
– p = 0.9 - (0.0001)(0.9)
New frequency of a (q) is old frequency plus fraction
gained by mutation
– q = 0.1 + (0.0001)(0.9)
When gametes make zygotes
– AA = 0.80984
– Aa = 0.18014
– aa = 0.01002
Mutation

Model mice population
New allele and genotype frequencies
are almost identical to old frequencies
Mutation had virtually no effect
Over many generations, mutation could change allele
frequencies
After 1000 generations, frequency of A would be 0.81
Mutation can cause evolution,
but it is usually does so slowly
Mutation and Selection
Mutation alone cannot cause great changes in allele
frequencies but it is still important in evolution
 In combination with selection, mutation can be a
potent evolutionary force
 Lenski’s E. coli study

Studied a strain incapable of conjugation
Mutation is only form of genetic variation
Grew 12 colonies in minimal salts medium
Mutation and Selection

Lenski’s E. coli study
After growing billions of cells, they removed
approximately 5 million cells from each and transferred
them to new medium
Daily transfers for 1500 days
– 10,000 generations
At intervals, researchers froze samples of transferred
cells
– Can freeze E. coli and revive it later
Could directly measure differential fitness of descendant
populations in competition at the same time
Also monitored cell size
Mutation and Selection
 Lenski’s E. coli study
During study fitness and cell size increased in
response to natural selection
Fitness increases occurred in jumps
Beneficial mutations swept through population
to fixation
Mutations caused bacteria to divide faster and
increase in size
Mutation is ultimate source of genetic variation
Mutation-Selection Balance
 Most mutations are deleterious
 Selection eliminates these mutations
 Mutations are created anew
 When rate of deleterious alleles being
eliminated by selection equals rate of
creation by mutation: Mutation-Selection
Balance
Mutation-Selection Balance

Deleterious recessive allele is at equilibrium when:
^ =
/m
q
√ s
m = mutation rate
s = selection coefficient
– Between 0 and 1 expressing strength of selection
If selection is small and mutation is high, equilibrium
frequency of allele will be high
If selection is high and mutation is low, equilibrium
frequency will be low
Mutation-Selection Balance

Is cystic fibrosis maintained by mutation-selection
balance?
Recessive loss-of-function allele on locus on
chromosome 7
Cystic fibrosis transmembrane conductance regulator
(CFTR)
CFTR normally destroys Pseudomonas aeruginosa
– Cause chronic lung infections
Few affected people survive to reproductive age or are
infertile
Mutation-Selection Balance

Cystic fibrosis
Frequency q = 0.02 in people with European ancestry
s=1
q = 0.02
Therefore, m = 4 X 10-4
Actual mutation rate is 6.7 X 10-7
Mutation rate is not causing high frequency of the
recessive allele
Mutation-Selection Balance
 Cystic fibrosis
Researchers discovered that the high frequency
is maintained by heterozygote superiority
Heterozygotes were partially resistant to typhoid
fever infection
Cystic fibrosis maintained by mutation and
heterozygote superiority
Hardy-Weinberg Equilibrium
 Have examined how selection and mutation
violate the Conclusions of the HardyWeinberg Equilibrium Principle
 Will next examine the other three forces:
Migration
Genetic Drift
Nonrandom mating