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Population Genetics
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Population genetics
• Concerned with changes in genetic variation
within a group of individuals over time
• Want to know:
– Extent of variation within population
– Why variation exists
– How variation changes over time
• Study the gene pool
– All alleles of every gene in a population
Population
• Group of individuals of
the same species that
occupy the same region
and can interbreed with
one another
• Local population (demes)
– Smaller groups within
population
– Separated by moderate
geographic barrier
– More likely to interbreed
Two local populations of
douglas fir
Figure 24.1
Populations
• Dynamic
– Size
• Feast or famine
– Geographic location
• Migrate to new site with new environment
– Consequences of changes in size and
location
• Changes in genetic composition
Polymorphism
• “many forms”
• Traits display variation
within population
• Ex. happy face spider
– Differ in alleles that affect
color and pattern
• At the DNA level:
– Multiple alleles for gene
– Involves changes in
DNA: deletion,
duplication, single
nucleotide change
• SNP – most common
– Monomorphism – single
allele within population
Figure 24.3
Polymorphism in humans
• Human b-globin gene
• HbA - normal gene
• HbS - differs by SNP
– Homozygous for HbS 
sickle cell anemia
• Deletion
– Loss of function allele
Two important frequency
calculations
• Allele frequency = (Number of copies of
an allele in population) / (Total number
of all alleles for that gene in the
population)
• Genotype frequency = (Number of
individuals with a particular genotype in
a population) / (Total number of
individuals in a population)
An example
• 100 pea plants
– 64 tall (TT)
– 32 tall (Tt)
– 4 dwarf (tt)
• Allele frequency = number of copies of allele / total
number of alleles
– To calculate the frequency of ‘t’
• (32 + 8)/200 = 0.2 or 20 %
• What would be the frequency of ‘T’?
• Genotype frequency = number of individuals with
particular genotype / total number of individuals
– To calculate the frequency of ‘tt’
• 4/100 = 0.04 = 4%
• Frequencies must always be less than or equal to 1
(100%)
Hardy Weinberg Equilibrium
• Allele and genotype frequencies in a
population are not changing over the
course of many generations
Hardy-Weinberg Equation
• Allows us to calculate allele and
genotype frequencies
• If gene is polymorphic and exists in two
alleles, then p + q = 1
• p2 + 2pq + q2 = 1
• p2 = genotype frequency of AA
• 2pq = genotype frequency of Aa
• q2 = genotype frequency of aa
Applying the HW equation
• Assume that gene exists as two
different alleles: G and g.
• If p = 0.8, then what is the genotype
frequency of gg?
Applying the HW equation
• Assume that gene exists as two
different alleles: G and g.
• If p = 0.8, then what is the genotype
frequency of gg?
• p = allele frequency of G
• p + q = 1 so q = 0.2
• q2 = genotype frequency of gg = (0.2)2 =
0.04 = 4%
Figure 24.4
Confirm with product rule
HW – predicting carriers
• Allows us to determine the frequency of
heterozygotes for recessive genetic
diseases
• Ex. Cystic fibrosis
• Frequency of affected individuals
(homozygous recessive) is 1 in 2500
• What is the frequency of heterozygous
carriers?
HW – predicting carriers
•
•
•
•
q2 = 1/2500
q = √(1/2500) = 0.02
p + q = 1 so p = 0.98
Frequency of heterozygous carriers =
2pq = 2(0.02)(0.98) = 0.0392 = 3.92%
Conditions for Hardy-Weinberg
with regard to gene of interest
• No new mutations: The gene of interest does not
incur any new mutations
• No genetic drift: The population is so large that allele
frequencies do not change due to random sampling
effects
• No migration: Individuals do not travel between
different populations
• No natural selection: All of the genotypes have the
same reproductive success
• Random mating: With respect to the gene of interest,
the members of the population mate with each other
without regard to their phenotypes and genotypes
Figure 24.5
Relationship between allele frequencies and
genotype frequencies according to Hardy-Weinberg
Use chi-square to determine if population
exhibits Hardy-Weinberg equilibrium for gene
• MN blood type
• 200 people
– 168 MM
– 30 MN
– 2 NN
• First, calculate allele frequency of M and
allele frequency of N
• Then find expected MM, MN, and NN and
solve chi-square equation
Chi square analysis
• Allele frequency of M = (2(168)+30)/400
= 0.915 = p
• Allele frequency of N = (2(2)+30)/400 =
0.085 = q
• Expected frequency of MM = p2 =
(0.915)2 = 0.837
– Expected number of MM individuals = (200
people)(0.837) = 167
• Expected frequency of MN = 2pq =
2(0.915)(0.085) = 0.007
– Expected number of MN individuals = (200
people)(0.007) = 1
• Expected frequency of NN = q2 =
(0.085)2 = 0.156
– Expected number of NN individuals = (200
people)(0.156) = 31
Chi square analysis
• Apply the ᵪ2 formula
• ∑((observed-expected)2/expected)
• = ((168-167)2/167) + ((30-31)2/31) +
((2-1)2/1) = 1.04
• Degrees of freedom = n-1
– Gene exists in 2 alleles so n=2 so DofF = 1
– See chart
– p > 0.05
– Do not reject the Null hypothesis
• Genes appear to be in HW equilibrium
• Genetic variation in
natural population
typically changes
over many
generations
– Changes in gene
pool
Mutations
• Random events that occur
spontaneously at a low rate or are
caused by mutagens at a higher rate
• Mutations can provide new alleles to a
population but do not substantially alter
allele frequencies
Mutations – an example
• Mutation rate – probability that gene will
be altered by a new mutation
– 10-5 to 10-6 per generation
• How much does mutation affect allele
frequency?
• Consider a mutation that converts A to a
where the allele frequency of “A” = p
and the allele frequency of “a” = q
Mutation – an exmple
• Consider that Dq = mp (p decreases as
A is converted to a)
• Let’s assume: p = frequency of A =
80%; q = frequency of a = 20%
• Assume m (rate of conversion of A to a)
= 10-5
• To calculate change in allele frequency
– (1-m)t = (pt/p0)
– pt = allele frequency of A at time t; p0 =
allele frequency of A at time 0; t = time; m =
mutation rate
Mutations – an example
• What would be the allele frequency of p
after 1 generation?
– (1-10-5)1 = pt/0.8
– p1 = 0.799992
– q1 = 0.200008
• What would be the allele frequency of p
after 1000 generations?
– (1-10-5)1000 = pt/0.8
– p1000 = 0.792
– q1000 = 0.208
• Notice that the change in allele
frequency is very low.
Figure 24.6
Genetic Drift
• Changes in allele
frequencies due to
random fluctuations
• Small sample size
– Large fluctuations
between generations
– Allele eliminated or
fixed at 100% over
fewer generations
• Large sample size
– Random sampling
has smaller effect
New allele fixation / elimination
• Must first calculate the expected
number of new mutations
– Depends on mutation rate (μ) and
population size (N)
– 2Nμ
• Note: New mutation is more likely in a large
population
• Probability of fixation = 1/2N
• Probability of elimination = 1- (1/2N)
New allele fixation / elimination
• Probability of fixation = 1/2N
• Probability of elimination = 1- (1/2N)
• Large population
– Greater likelihood for mutation
– But, greater likelihood that new mutation
will be eliminated from the population
• Small population
– Decreased likelihood for mutation
– But, greater likelihood that new mutation
will be fixed in the population
How long will fixation take?
• ṫ = average # of generations to achieve
fixation
• ṫ = 4N
– Note: Allele fixation takes much longer in
larger populations
Genetic Drift – Take home
message
• Operates in random manner with regard
to allele frequency
– Regardless of the type of allele
• Deleterious, neutral, beneficial
• Eventually leads to allele fixation or
elimination
• Occurs more quickly in smaller
population
Changes in population size
and genetic drift
• Bottleneck effect
– Decrease in population size because of
events such as earthquake, drought,
destruction of habitat
• Elimination without regard to genetic
composition
• Founder effect
– Migration of small group of individuals to
new location
• Small group may not have same genetic
composition as parent population
Figure 24.7
Bottleneck effect
An example:
Low genetic variation thought to be due
to previous bottleneck
Founder effect
• An example:
– Old Order Amish in Lancaster County, PA
– 1770: 3 couples immigrated to US
– 1960: 800 individuals
• Very high prevalence of recessive form of
dwarfism
• Thought that one of original ancestors carried
recessive mutation or new mutation occurred
early on
Migration
• Between two different established
populations
– May alter allele frequencies
• New population (after migration) =
conglomerate
• Change in allele frequency in conglomerate
DpC = m(pD-pR)
m = proportion of migrants that make up conglomerate
pD = allele frequency in donor population
pR = allele frequency in original recipient population
Migration – an example
• A donor population has an “A” allele
frequency of 0.7. A recipient population
has an “A” allele frequency of 0.3. 20
people join the recipient population
which originally had 80 members. What
is the allele frequency in the
conglomerate?
Migration – an example
DpC = m(pD-pR)
• m = 20/(20+80)
• pD = 0.7
• pR = 0.3
DpC = 0.2(0.7-0.3) = 0.08
pC = allele frequency in conglomerate
pC = pR + DpC = 0.3 + 0.08 = 0.038
Migration
• Often bidirectional flow
• Tends to reduce differences in allele
frequencies between neighboring
populations
Natural selection
• Conditions found in nature result in the
selective survival and reproduction of
individuals whose characteristics make
them well adapted to their environment
• Surviving individuals are more likely to
reproduce and contribute offspring to
the next generation
• Must consider Darwinian fitness
– Relative likelihood that a phenotype will survive and
contribute to the gene pool of the next generation
– measure of reproductive superiority
– Not same as physical fitness
• Consider a gene with two alleles, A and a
– The three genotypic classes can be assigned fitness
values according to their reproductive potential
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24-45
• Suppose the average reproductive success is
– AA  5 offspring
– Aa  4 offspring
– aa  1 offspring
• By convention, the gene with the highest
reproductive ability is given fitness value of 1.0
– The fitness values of the other genotypes are assigned
relative to 1
• Fitness values are denoted by the variable W
– Fitness of AA: WAA = 1.0
– Fitness of Aa: WAa = 4/5 = 0.8
– Fitness of aa: Waa = 1/5 = 0.2
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24-46
• Differences in reproductive achievement could
be due to the
– 1. Fittest phenotype is more likely to survive
– 2. Fittest phenotype is more likely to mate
– 3. Fittest phenotype is more fertile
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24-47
• Natural selection acts on phenotypes (which are
derived from an individual’s genotype)
• With regard to quantitative traits, there are four
ways that natural selection may operate
– 1. Directional selection
• Favors the survival of one extreme phenotype that is better
adapted to an environmental condition
– 2. Stabilizing selection
• Favors the survival of individuals with intermediate phenotypes
– 3. Disruptive (or diversifying) selection
• Favors the survival of two (or more) different phenotypes
– 4. Balancing
• Favors the maintenance of two or more alleles
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24-48
• Directional Selection favor extreme phenotype
– A gene exist in two alleles A and a
– The three fitness values are
• WAA = 1.0
• WAa = 0.8
• Waa = 0.2
– In the next generation, the HW equilibrium will be modified
in the following way by directional selection:
• Frequency of AA: p2WAA
• Frequency of Aa: 2pqWAa
• Frequency of aa: q2Waa
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24-49
– These three terms may not add up to 1.0, as they would
in the HW equilibrium
– Instead, they sum to a value known as the mean fitness
of the population
• p2WAA + 2pqWAa + q2Waa = W
– Divide both sides of the equation by the mean fitness of
the population
p2WAA
W
+
2pqWAa
W
+
q2Waa
=1
W
– Now, calculate the expected genotype and allele
frequencies after one generation of natural selection
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24-50
– Frequency of AA genotype =
p2WAA
W
– Frequency of Aa genotype =
2pqWAa
W
– Frequency of aa genotype =
q2Waa
W
– Allele frequency of A: pA =
p2WAA
+
W
– Allele frequency of a: qa =
q2Waa
W
pqWAa
W
+
pqWAa
W
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24-51
• Example:
– Starting allele frequencies are A = 0.5 and a = 0.5
– Fitness values are 1.0, 0.8 and 0.2 for genotypes AA, Aa
and aa, respectively
• p2WAA + 2pqWAa + q2Waa = W
W = (0.5)2(1) + 2(0.5)(0.5)(0.8) + (0.5)2(0.2)
= 0.25 + 0.4 + 0.05
= 0.7
– After one generation of selection:
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24-52
– Frequency of AA genotype =
p2WAA
(0.5)2(1)
=
= 0.36
0.7
W
– Frequency of Aa genotype =
2pqWAa
W
– Frequency of aa genotype =
2(0.5)(0.5)(0.8)
=
= 0.57
0.7
(0.5)2(0.2)
=
= 0.07
0.7
q2Waa
W
– Allele frequency of A: pA =
p2WAA
+
W
pqWAa
W
= 0.36 + 0.57/2 = 0.64
– Allele frequency of a: qa =
q2Waa
W
+
pqWAa
W
= 0.07 + 0.57/2 = 0.36
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24-53
• After one generation
– f(A) has increased from 0.5 to 0.64
– f(a) has decreased from 0.5 to 0.36
– This is because the AA genotype has the highest fitness
• Natural selection raises the mean fitness of the
population
– Assuming the individual fitness values are constant
W = p2WAA + 2pqWAa + q2Waa
= (0.64)2(1) + 2(0.64)(0.36)(0.8) + (0.36)2(0.2)
= 0.80
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24-54
General trend
• increase A, decrease a, and increase the mean fitness
of the population
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24-55