Population Genetics
Allele Frequency – proportion of a specific allele relative to all other alleles in a
population.
Genotypic Frequency – proportion of a specific genotype relative to all other
genotypes in a population.
Phenotypic Frequency – proportion of a specific phenotype relative to all other
phenotypes in a population.
Calculating Allele Frequencies: Two alleles
 For two alleles A and a with relative proportions p and q, respectively, the
sum of their proportions equals one:
o p+q=1
o q=p–1
 The frequency of an allele = frequency of homozygotes for that allele + ½
frequency of heterozygotes carrying that allele
o If p = fA in a population, then p = fAA + ½(fAa)
o If q = fa in a population, then q = faa + ½(fAa)
Hardy-Weinberg Equilibrium –used to explain the persistence of recessive traits
and the maintenance of deleterious genes (which should be selected against) in
populations.
 Assumed conditions for this model:
o Random mating
o Infinitely large population
o No mutation
o No selection
o No migration (in or out)
 Under these conditions, two alleles of an autosomal gene with allelic
frequencies fA = p and fa = q will have genotypic frequencies of:
o fAA = p2
o fAa = 2pq
o faa = q2
 These genotypic frequencies will persist (i.e. remain in equilibrium) as long
as the five conditions of the model are met.
Factors That Can Disturb H-W Equilibrium
 Non-random mating – e.g. assortive mating (like with like—sharing traits)
or consanguineous mating.
o Leads to increased homozygosity.
 Small Population Size – leads to fixation of genes (the allele in the
population goes to 100%). All members of a population become homozygous
for a particular allele. Also causes genetic drift.
o Founder effect – one person of a small group with an uncommon
allele moves to an isolated area; the allele becomes prominent.
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o Bottle-Neck effect – an abrupt decrease in population size (often due
to an environmental cause) results in an increase in the prevalence
(frequency) of alleles carried by the survivors.
Mutation – introduction of new alleles to the population.
Selection – due to reproductive or genetic fitness (refers to the tendency for
alleles to increase in the population, often due to some advantage conferred).
o Heterozygote advantage – causes mutations that result in disease
(when homozygous) to increase in prevalence within a population,
because those individuals who are heterozygous are given a survival
advantage (e.g. sickle cell anemia and malaria).
Gene Flow –flux of genes into or out of a population due to migration.
Proof of H-W Equilibrium
 The model is based upon equivalence of random “mating” of genotypes (i.e.
mating partners are selected irrespective of genotype) with random
combination of gametes at fertilization.
 Therefore, if chance determines the genotypes of mating individuals, then
the allele (A or a) carried by the egg or sperm is equally subject to chance:
o (p + q)2 = (1)2 = p2 + 2pq + q2
Sperm
Eggs
p (A)
q (a)
p (A)
p2 (AA)
pq (Aa)
q (a)
pq (Aa)
q2 (aa)
Using H-W Equilibrium
Given the frequency (q = 0.3) of a deleterious, recessive allele (i.e. allele a) in a
population, and knowing that the population is in H-W equilibrium, it is possible to
determine 1) the frequency of phenotypically normal individuals in the population,
2) the frequency of carriers of allele a in the population, and 3) the frequency of
affected individuals in the population.
1. Frequency of Normal Phenotypes:
a. Given that q = 0.3, we know p = 0.7
b. Amount of normal phenotypes = p2 + 2pq = (0.7)2 + 2(0.7)(0.3) = 0.91
2. Frequency of Carriers:
a. Frequency of carriers = frequency of heterozygotes = 2pq
b. 2pq = 2(0.7)(0.3) = 0.42
3. Frequency of Affected Individuals:
a. Affected = aa genotype = q2
b. q2 = 0.09
Example 2 – given a disease frequency of 1/10,000 for a recessive, monogenic trait
in a population that is in H-W equilibrium, what is the frequency of carriers in the
population?
 q2 (faa) = 1/10,000 , therefore q = 1/100 = fa
 p = 1 – q = 99/100
 f carrier = fAa = 2pq = 2 (0.99)(0.01) = ~0.02
Special Case – X-linked traits
 Males and females have different equilibrium frequencies due to gene
dosage differences and sex chromosome differences.
 Because females carry two X chromosomes, allele frequencies of p + q = 1
result in genotype frequencies of p2 +2pq + q2 = 1
 Males only have one X chromosome, so both allele and genotype frequencies
are represented by p + q = 1
 Problem:
o Given a disease frequency of 1/10,000 for an X-linked recessive trait
among males in a population in H-W equilibrium, 1) what is the
expected frequency of carriers for the trait in this population? 2) what
is the expected frequency of affected females?
1. Calculate the frequency of the two alleles (fXA = p and fXa = q).
a. Most of the affected will be male, and because X
chromosomes are inherited directly from the mother and Y
chromosomes directly from the father, individuals will
either be XAY (normal) found at frequency p, or XaY
(affected) found at frequency q.
b. The frequency of affected males is determined by the
frequency of the Xa allele among females.
c. q = 1/10,000 = fXa ~ frequency of affected males.
d. p = 1 – q = 9,999/10,000 = fXA ~ frequency of unaffected
males.
e. Carriers of an X-linked recessive trait must be female, and
in a population that is in H-W equilibrium, an X-linked trait
behaves in females as if it were an autosomal recessive
trait.
f. Thus, frequency of carriers = frequency of heterozygous
females = 2pq = 2 (9,999/10,000)(1/10,000) = ~0.0002
2. Because the frequency of affected females is <<<< frequency of
affected males, p approaches 1.0 for rare, X-linked traits.
Frequency of affected females = 0.00000001 (much < than
observed frequency in males) and can be ignored.
Mutation-Selection Equilibrium
 In populations, new alleles arise by mutation and are maintained or
removed by selection.
 Fitness (f)– measure of the number of offspring of affected persons who
survive to reproductive age, compared with an appropriate control group.
o Fitness range is between 0-1.
 Coefficient of Selection (s) = 1 – f
o Acts as a measure of loss of fitness.
 Ex: Achondroplastic dwarves will have ~20% of the number of offspring
compared to those of normal stature.
o Fitness = 0.2 and therefore the coefficient of selection = 0.8
Warfarin Resistance in Rodents – a model for how deleterious, recessive alleles can
be maintained (Evolution in real time).
 Warfarin rodenticide introduced in the 1940s-50s. Mechanism was to block
recycling of vitamin K (clotting cofactor), and rodents that ingested it would
hemorrhage following injury.
 This model exhibits selection for a single-gene, vitamin K epoxide reductase
complex subunit 1 (Vkorc1) – an allele that confers resistance to warfarin.
 When warfarin was used, it created selective pressure on rat populations
for resistance to warfarin-induced bleeding. There was selection for the
Vkorc1 resistance (R) allele, versus the susceptible allele (S).
o Studies found relative fitness of:
 0.37 for SS
 1.00 for SR
 0.68 for RR
o Frequency of R allele did not approach 1, because RR homozygotes
had a vitamin K deficiency.
Fitness and Autosomal Dominant Disorders
 Autosomal dominant disorders are often associated with reduced fitness.
 In the case of autosomal dominant disorders, fitness is inversely associated
with the proportion of patients with that disorder that have new mutant
genes (as in a somatic mutation that was not inherited).
 In disorders with f = 0:
o Patients with such disorders never reproduce, and all observed
cases are caused by newly arising mutations.
 In disorders with f = 1:
o Patients with such disorders have essentially normal reproductive
fitness, and observed cases are most likely due to inheritance of a
mutant allele, very rarely from newly arising mutations.
Mutation-Selection Equilibrium and H-W Equilibrium
 Autosomal Dominant Traits:
o For a mutant allele to remain in H-W equilibrium, the mutation rate /
generation (μ) must be sufficient to balance the proportion of
mutant alleles (q) lost in each generation from selection (s). For
autosomal dominant traits:
 μ = sq = (1-f)q
 Autosomal Recessive Traits:
o Selection against deleterious recessive mutations is much less
influential on allele frequencies, because the frequency of affected
individuals (faa) upon which selection acts represents a much
smaller fraction of the population (This is the fallacy of eugenics).
 X-linked Recessive Traits:
o If the phenotype is benign (i.e. affected males live to reproduce), then
1/3 of all mutant alleles will be in males (because they only have one
X chromosome), while 2/3 of all mutant alleles will be in females.
o Thus, the mutation rate / generation (μ) must equal the selection
coefficient (s) multiplied by q/3 (because selection only acts against
hemizygous males, not carrier females).
 μ = s(q/3) = (1-f) (q/3)
o In the case of X-linked genetic lethality, where f~0 and s~1, each
generation will lose 1/3 of all copies of the mutant allele.
 To maintain the observed disease incidence, mutant alleles lost
by selection must be replaced by recurrent production of
newly-arising mutations.
Inbreeding
Consanguineous mating: A mating between close relatives. Offspring from such a
mating are considered inbred.
Coefficient of Relationship – a measure of the proportion of genes shared by two
related individuals.
 Often calculated for closely-related people who form a mating pair.
 In such a relationship, both individuals will have a certain proportion of
alleles that are identical by descent (have a common history).
 Formula for the coefficient of relationship:
o (1/2)n , where n = number of matings that connect related
individuals.
 Example 1 – first cousins.
o Number of mating pairs connecting = 3 (grandparents, parents of
male cousin and parents of female cousin). (1/2)3 = 1/8
 Example 2 – parent-child relationship
o (1/2)1 = ½
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The proportion of genes shared = the probability that a particular allele
(such as a mutant allele causing a recessive trait) is shared.
Coefficient of Inbreeding (F)– A measure of the probability that a homozygote has
received both alleles from a common, recent ancestor. Also a measure of the
proportion of loci for which a person is homozygous (i.e. both alleles are identical by
descent).
 F = ½ coefficient of relationship
o This formula represents P(inheriting a common allele) x P(other
parent also passing on that allele)
 **This relationship only applies when there are low levels of inbreeding, and
not to livestock breeding or inbred laboratory strains (requires more
complex formula).
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Inbred Strains – produced by 20 or more generations of brother-sister
matings.
o After 20 generations of brother sister matings, 99% of loci will be
homozygous.
o Ex: albino and pigmented rats.
Risk Assessment
Risk Calculation
 Events in probability calculations are either:
o mutually exclusive (“or” event, meaning addition of probabilities)
o or independent (“and” event, meaning multiplication of
probabilities).
 **Note: The sex of an offspring is an independent event:
o P(male) ~ P(female) ~ 0.5
o Extremely important in calculating the risk of inheriting a
deleterious phenotype.
Bayesian Analysis: Used in genetics to assess risk. Calculation of risk requires a
combination of several probability values.
 The prior probability distribution of an uncertain quantity “p” represents
the probability of an uncertain event before any data is collected.
o Ex: When constructing a pedigree for a trait that exhibits autosomal
dominance, any affected individual has a prior probability of ½ for
being a heterozygote (we are unsure if this person is nonheterozygous or heterozygous and affected by the dominant trait).
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o Provided that you have adequate family history, complete penetrance
and can identify heterozygotes; it is a rule that in pedigrees exhibiting
an autosomal dominant trait there is a 50% chance of passing on the
mutant allele.
Conditional probability is the probability that an event will occur, when
another event is known to occur or to have occurred. In genetics, conditional
probability is used to represent the probability that one event (that has
occurred) accounts for another event.
o Ex: If an individual is affected by an autosomal dominant trait, the
prior probability of that individual being a heterozygote is ½.
Conditionally, since we know that the individual is affected, he/she
must be either a heterozygote or a homozygote for the mutant allele
(each possibility with a probability of ½).
Joint probability = prior probability x conditional probability.
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o Following the examples above, an affected individual has 0.5 chance of
being a heterozygote, and conditionally there is a 0.5 chance of a
heterozygous genotype being the CAUSE of the disease (the other 0.5
chance would be a homozygous genotype).
o Therefore in this example, the joint probability = ½ * ½ = ¼
Posterior probability of a random event is the conditional probability that
is assigned after relevant evidence is taken into account (i.e. evidence from a
survey or study).
o To calculate posterior probability, you must find the joint
probabilities of two potential outcomes. The joint probability of each
outcome is then divided by the sum of the joint probabilities from
both outcomes:
o Example – Pedigree For Huntington Disease (Autosomal Dominant)
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I-2 has Huntington
Disease.
It has been observed that
50% of heterozygotes
manifest disease by age
50.
In this example, we want to determine the probability that III-1 is
heterozygous for Huntington Disease. To determine this, we must know the
probability that II-1 carries the mutant allele.
Since we do not know whether I-2 is heterozygous or homozygous, we assign
a prior probability of ½ to the chance that she is heterozygous or nonheterozygous.
In assessing the probability that II-1 is carrying the mutant allele, there is a
conditional probability of 1 (certainty) that, if her mother is nonheterozygous, she is carrying the allele.
o In other words, if her mother is not a heterozygote, she must be
homozygous for the mutant allele (because she was affected),
therefore it is certain that II-1 is a carrier.
In assessing the probability that II-1 is carrying the mutant allele, there is a
conditional probability of ½ that IF her mother (I-2) was a heterozygote,
then II-2 is also a heterozygote (there is a ½ chance that she is homozygous
recessive).
The joint probabilities of these independent events is calculated by
multiplying prior and conditional probabilities, and the posterior or
relative possibilities are found by dividing each joint probability by the sum
of both joint probabilities.
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Because II-1 has a P(1/3) for being heterozygous, and there is a P(1/2) of her
passing on the mutant allele, the probability that III-1 is heterozygous is:
o P = (1/3)(1/2) = 1/6