Chapter 9 Quantitative Genetics
 Read Chapter 9.
 Traits such as cystic fibrosis or flower color in peas
produce distinct phenotypes that are readily
distinguished.
 Such discrete traits, which are determined by a single
gene, are the minority in nature.
 Most traits are determined by the effects of multiple
genes.
Continuous variation
 However, traits determined by many genes (polygenic
traits) show continuous variation.
 Grain color in winter wheat is determined by three
alleles at three loci.
Additive effects of genes
 The genes affecting color of winter wheat interact in a
particularly straightforward way.
 They have additive genetic effects.
 This means that the phenotype for an individual is
obtained just by summing the effects of individual alleles.
 The more alleles for dark color an individual has the darker
it will be
Continuous variation
 Examples in humans of traits that show continuous
variation include height, intelligence, athletic ability,
and skin color.
Quantitative traits
 For continuous traits we cannot assign individuals to
discrete categories. Instead we must measure them.
 Therefore, characters with continuously distributed
phenotypes are called quantitative traits.
Quantitative traits
 Quantitative traits determined by influence of (1)
genes and (2) environment.
East (1916)
 In early 20th century there was considerable debate
over whether Mendelian genetics could explain
continuous traits.
 Edward East (1916) showed it could.
 Studied longflower tobacco (Nicotiana longiflora)
East (1916)
 East studied corolla length (part of flower) in tobacco
flowers.
 Crossed pure breeding short and long corolla
individuals to produce F1 generation. Crossed F1’s to
create F2 generation.
East (1916)
 Using Mendelian genetics we can predict expected
character distributions if character determined by one
gene, two genes, or more etc.
East (1916)
 Depending on number of genes: models predict
different numbers of phenotypes.
 One gene: 3 phenotypes
 Two genes: 5 phenotypes
 Six genes: 13 phenotypes. Continuous distribution.
East (1916)
 How do we decide if a quantitative trait is under the
control of many genes?
 In one and two locus models many F2 plants have
phenotypes like the parental strains.
 Not so with 6-locus model. Just 1 in 4,096 individuals
will have the genotype aabbccddeeff.
East (1916)
 But, if Mendelian model works you should be able to
recover the parental phenotypes through selective
breeding.
 East selectively bred for both short and long corollas.
By generation 5 most plants had corolla lengths within
the range of the original parents.
East (1916)
 Plants in F5 generation of course were not exactly the
same size as their ancestors even though they were
genetically identical.
 Why?
East (1916)
 Environmental effects.
 Depending on environment genetically identical
organisms may differ greatly in phenotype.
Genetically identical plants
grown at different elevations
differ enormously
(Clausen et al. 1948)
The importance of latent variation
 Early work in the 2oth century on polygenic traits
showed that new types or values of traits not seen in a
parent population could appear in offspring produced
by that population.
 It was unclear where these new variants came from.
It’s easy to see in figure A (next slide) how natural
selection could favor some members of a population so
that after a time the mean values of a population
would increase within the range of previous variation.
The importance of latent
variation
 However, it’s less clear how a population could as a
result of natural selection arrive at B in the previous
slide in which the selected population is outside the
range of the original population.
 The key to understanding this phenomenon is to
realize that when multiple genes contribute to a trait
there will be many, many unique combinations of
alleles that produce different phenotypes.
 A population is not likely to include all of these
possibilities.
 Thus, a new variant can contain an assortment of
alleles not seen previously. See next slide.
Gene interactions
 Not all genes interact additively with the alleles’ effects
summing together.
 In many cases genes interact with each other
nonadditively a phenomenon we call epistasis.
Gene interactions
 For example, two loci influence coat color in oldfield
mice, but they interact epistatically.
 The effect of the Mc1R allele depends on which alleles
are present at the agouti locus (next slide).
Population genetics of multiple
loci
 A locus is the physical location on a chromosome where a
gene occurs.
 Different versions of a gene are called alelles.
 The Hardy-Weinberg models we have discussed so far are
quite simple because they consider only a single locus and
its alleles.
 However, many traits are controlled by the combined
influence of many genes.
Population genetics of multiple
loci
 Genes located on different chromosomes segregate
(i.e. they enter gametes) independently of each other.
 However, when genes are located on the same
chromosome they frequently do not segregate
independently, especially if they are located close to
each other on a chromosome. Such loci have a
physical linkage.
Population genetics of multiple
loci
 The closer together two loci are on a chromosome the
less likely it is that crossing over will occur between the
loci during meiosis and split them up.
 In most cases they will be inherited as a pair.
Population genetics of multiple
loci
 Consider a pair of loci located on same chromosome.
 Gene at locus A has two alleles A and a
 Gene at locus B has two alleles B and b
Population genetics of multiple
loci
 In two-locus Hardy-Weinberg analysis we track allele
and chromosome frequencies.
 Thus 4 possible chromosome genotypes are possible in
previous slide:
 AB, Ab, aB, ab
 A multilocus genotype is referred to as a haplotype
(from haploid genotype).
Statistical associations between
loci
 Does selection on locus A affect our ability to make
predictions about evolution at locus B?
 Sometimes. Depends on whether loci are in linkage
equilibrium or linkage disequilibrium.
Statistical associations between
loci
 Two loci in a population are in linkage
equilibrium when the genotype of a chromosome
at one locus is independent of the genotype at the
other locus on the same chromosome.
 I.e. knowing genotype at one locus is of no use in
predicting genotype at the other locus.
Statistical associations between
loci
 In contrast two loci are said to be in linkage
disequilibrium when knowing the allele at one locus
enables you to predict what the allele at the other locus
likely is.
 For example in a population where there are AB, Ab,
and aB haplotypes, but no ab haplotypes if we know an
individual has a b allele we know that individual also
has at least one A allele.
Quantifying linkage
disequilibrium
 To measure the associations between allele frequencies
at two loci A and B we examine the haplotype
frequencies at these loci.
 Let fA, fB, fa and f b be the frequencies of the A, B, a and
b alleles respectively.
 Let hAB, hAb, haB, hab be the haplotype frequencies of
AB, Ab, aB and ab haplotypes.
Quantifying linkage
disequilibrium
 If the allele at the A locus occurs independently of the
allele at the B locus then the haplotype frequencies
will be:
 hAB = fAfB
 hAb = fAf b
 haB = fafB
 hab = faf b
Quantifying linkage
disequilibrium
 So the expected haplotype frequency is found just by
multiplying the appropriate allele frequencies by each
other.
 If the frequency of allele A (fA) = 0.7 and the frequency
of allele B (fB) = 0.8 then the expected haplotype
frequency hAB, if the alleles are in linkage equilibrium,
would be 0.56.
Coefficient of linkage
disequilibrium
 To measure the degree of linkage disequilibrium we
can calculate a coefficient of linkage disequilibroum
(D).
 For a given haplotype this is defined as the difference
between the actual frequency we observe of a
haplotype, e.g. AB, and the expected frequency fAfB of
the same haplotype if the loci are independent.
 D = hAB - fAfB
Coefficient of linkage
disequilibrium
 When the alleles at each locus occur independently
then the coefficient of linkage disequilibrium will be
zero. We then say the alleles are in linkage
equilibrium.
 If the alleles at each locus occur non-independently
then the value of D will be non-zero and we say they
are in linkage disequilibrium.
Coefficient of linkage
disequilibrium
 In a gene pool the frequencies of the alleles are as
follows: A = 0.4, a= 0.6, B=0.3 and b= 0.7.
 The haplotype frequencies are AB = 0.12, Ab =0.28, aB
= 0.18 and ab=0.42.
 Is the population in linkage equilibrium?
Coefficient of linkage
disequilibrium
 Yes.
 hAB = 0.12
 hAb = 0.28
 haB = 0.18
 hab = 0.42
fAfB = 0.3*0.4 = 0.12
fAf b = 0.4*0.7 = 0.28
fafB = 0.6*0.3 = 0.18
faf b = 0.6*0.7 = 0.42
 For each haplotype D = zero e.g.
D = hAB – fAfB = 0.12-0.12 = 0
Coefficient of linkage
disequilibrium
 In a second gene pool the frequencies of the alleles are
as follows: A = 0.6, a= 0.4, B=0.8 and b= 0.2
 The observed haplotype frequencies are AB = 0.44, Ab
=0.16, aB = 0.36 and ab=0.04.
 Is this population in linkage equilibrium?
Coefficient of linkage
disequilibrium
 No.
 hAB = 0.44
 hAb = 0.16
 haB = 0.36
 hab = 0.04
fAfB = 0.6*0.8 = 0.48
fAf b = 0.6*0.2 = 0.12
fafB = 0.4*0.8 = 0.32
faf b = 0.4*0.2 = 0.08
 For each haplotype D not equal to zero e.g.
D = hAB – fAfB = 0.44-0.48 = -0.04
Coefficient of linkage
disequilibrium
 Another way to calculate the coefficient of linkage
equilibrium if we just know haplotype frequencies is
the following equation:
 D = hABhab - hAbhaB
 The value of this equation will be zero if the
haplotypes are in linkage equilibrium.
Proof of the formula for linkage
disequilibrium
 D = hABhab - hAbhaB
 Let p and q be the frequencies of alleles A and a.
 Let s and t be the frequencies of alleles B and b.
 If the population is in linkage equilibrium then
 hAB = ps, hab = qt, hAb = pt, haB = qs
 Therefore rewriting the equation for linkage
disequilibrium in terms of allele frequencies we get
 D = psqt - ptqs which equals zero if the population is
in linkage equilibrium.
 Any value of D not equal to zero implies the
population is in linkage disequilibrium.
Coefficient of linkage
disequilibrium
 Is this population, which has the following haplotypes,
in linkage equilibrium?
 AB= 0.46, Ab = 0.14 aB = 0.34 ab= 0.06
Coefficient of linkage
disequilibrium
 Use the formula: D = hABhab - hAbhaB
 D = 0.46*0.06 – 0.14*0.34
 D = 0.0276 – 0.0476 = -0.02
 D is not equal to zero, so the population is in linkage
disequilibrium.
Coefficient of linkage
disequilibrium
 The maximum value for D is 0.25 when AB and ab are
the only haplotypes present and each has a frequency
of 0.5.
 The minimum value for D is -0.25 when Ab and aB are
the only haplotypes present and each has a frequency
of 0.5.
 This formula thus tells us not only whether a
population is in linkage disequilibrium but how strong
the disequilibrium is.
What creates linkage
disequilibrium in populations?
 Multiple Mechanisms:
 Mutation
 Selection on multilocus genotypes.
 Genetic drift
 Migration
Mutation
 A population contains only the haplotypes AB and aB.
 A mutation occurs with the haplotype aB so that B
mutates to b producing the haplotype ab.
 This population will have the genotype aB , AB and ab,
but there will be no Ab haplotypes.
 Hence, the population will be in linkage
disequilibrium because of the missing Ab haplotype.
Selection on multilocus
genotypes.
 Scenario: Either of two biosynthetic pathways is
sufficient to produce an essential molecule from two
precursor molecules.
 Each pathway is controlled by a single locus. The
functional wild-type alleles (A & B) are dominant over
the nonfunctional recessive alleles (a & b).
 Only aabb individuals cannot produce the essential
molecule.
Selection on multilocus
genotypes.
 Because of selection against the aabb genotype there
will be fewer ab haplotypes than we would expect
based on the allele frequencies of a and b.
Genetic drift
 Scenario: Small population with two genotypes
AB and Ab. No copies of allele a.
 Single Ab chromosome mutation converts an A to
an a. This single ab chromosome puts population
in linkage disequilibrium.
 Scenario is drift because only in a small population
would you expect to have only a single mutation of
A to a. In large population you would expect many
mutations of A to a and a to A.
Genetic drift
 Scenario: a small population with AB, Ab, aB and ab
haplotypes where there is a low recombination rate
between the A and B loci.
 Drift can lead to the loss of alleles in a small
population and haplotypes can disappear even more
easily. If by chance all of one haplotype disappears
then the population will have only three haplotypes.
 Haplotypes need not necessarily disappear. In a small
population random fluctuations in haplotype
frequencies can easily lead to statistical associations
between alleles and create linkage disequilibrium.
Migration
 Scenario: Suppose that the a & b alleles are fixed in a
mainland lizard population and the A&B alleles in an
island lizard population.
 Mainland thus has only the ab haplotype and the
island the AB haplotype.
 If some individuals migrate from the mainland to the
island ab haplotypes will be introduced.
 The population will be in linkage disequilibrium
initially because there will be no aB and Ab haplotypes
and a strong statistical association between the A and
B alleles.
What eliminates linkage disequilibrium
from a population?
 A population in linkage disequilibrium will not stay in
that state forever.
 Unless no other evolutionary process prevents it (e.g.
selection) linkage is broken down by recombination.
What eliminates linkage disequilibrium
from a population?
 Sexual reproduction steadily reduces linkage
disequilibrium.
 Crossing over during meiosis breaks up old
combinations of alleles and creates new combinations.
Genetic recombination
 Genetic recombination tends to randomize genotypes
in relation to other genotypes (i.e., it reduces linkage
disequilibrium.)
 Rate of decline in linkage disequilibrium is
proportional to rate of recombination.
r is recombination rate, r is related to how far apart two loci are
on a chromosome.
Empirical example of genetic
recombination
 Clegg et al. (1980) established two fruit fly populations
that were in linkage disequilibrium.
 Population 1 AB and ab each 0.5 frequency.
 Population 2 aB and Ab each 0.5 frequency.
Empirical example of genetic
recombination
 Populations of about 1,000 individuals maintained for
48-50 generations.
 Flies allowed to mate freely.
 Populations sampled every 1-2 generations to count
frequencies of 4 haplotypes.
Empirical example of genetic
recombination
 Crossing-over created missing haplotypes in each
population and linkage disequilibrium disappeared.
 In general, in random-mating populations sex is
efficient enough at eliminating linkage disequilibrium
that most alleles are in linkage equilibrium most of the
time.
Practical reasons to measure linkage
disequilibrium
 There are two major uses of measures of linkage
disequilibrium.
 Can be used to reconstruct history of genes and
populations
 Can be used to identify alleles recently favored by
positive selection
Reconstructing history of the
CCR5-Δ32 locus
 HIV is the virus responsible for AIDS. It parasitizes
macrophages and T-cells of immune system. It enters
by binding to two protein receptors on cell’s surface :
CD4 and a coreceptor, usually CCR5.
 Some people appear resistant to the virus even though
exposed multiple times.
 Some resistant individuals possess a mutant CCR5 coreceptor protein whose gene is missing 32 base pairs. This
allele is referred to as the CCR5 Δ32 allele.
Reconstructing history of the
CCR5-Δ32 locus
 Frequency of the CCR5-Δ32 allele is highest in
European populations (9%), but scarce or absent
elsewhere.
 Where did the CCR5-Δ32 allele come from and when
did it originate?
Reconstructing history of the
CCR5-Δ32 locus
 CCR5-Δ32 is located on chromosome 3 and near two
short-tandem repeat sites called GAAT and AFMB.
 GAAT and AFMB are non-coding and have no effect on
fitness. Both GAAT and AFMB have a number of
different alleles.
Reconstructing history of the
CCR5-Δ32 locus
 Stephens et al. (1998) examined haplotypes of 192
Europeans.
 Found that GAAT and AFMB alleles were in close to
linkage equilibrium with each other.
Reconstructing history of the
CCR5-Δ32 locus
 However, CCR5 is in strong linkage disequilibrium
with both GAAT and AFMB.
 Almost all chromosomes carrying CCR5-Δ32 also carry
allele 197 at GAAT and allele 215 at AFMB.
Reconstructing history of the
CCR5-Δ32 locus
 Most likely reason for observed linkage disequilibrium
is genetic drift.
 Hypothesis: in past was originally only one CCR5 allele
the CCR5+ allele.
 Then a mutation on a chromosome with the
haplotype CCR5--GAAT-197--AFMB-215 created the
CCR5Δ32 allele.
Reconstructing history of the
CCR5-Δ32 locus
 The CCR5Δ32 allele was favored by selection and
rose to high frequency dragging the other two
alleles with it.
 Since its appearance and spread, crossing over and
mutation have been breaking down the linkage
disequilibrium. Now about 15% of Δ32-197-215
haplotypes have changed to other haplotypes.
Reconstructing history of the
CCR5-Δ32 locus
 Based on rates of crossing over and mutation rates,
Stephens et al. (1998) estimate the CCR5-Δ32 allele
first appeared about 700 years ago (range of estimates
275-1875 years)
Reconstructing history of the
CCR5-Δ32 locus
 Because the CCR5-Δ32 increased in frequency so rapidly
selection must have been strong.
 Most obvious candidate is an epidemic disease.
 Myxoma virus a relative of smallpox uses CCR5 protein on
cell surface to enter host cell, which suggests the epidemic
disease that favored CCR5-Δ32 may have been smallpox.
 However, timing of origin also closely matches period of
bubonic plague.
Using linkage disequilibrium to
detect strong positive selection.
 A new mutant allele will be in linkage
disequilibrium when it first appears. If it persists,
it may increase in frequency.
 Over time linkage disequilibrium will break down
as a result of recombination from crossing over.
 Linkage disequilibrium breaks down fastest for loci
further apart on a chromosome because crossing
over take place more often between distant loci.
Using linkage disequilibrium to
detect strong positive selection.
 High linkage disequilibrium indicates an allele
originated recently.
 Also, expect a recently mutated allele to be rare unless
selection strongly favors it.
Using linkage disequilibrium to
detect strong positive selection.
 If an allele is common, but has high linkage
disequilibrium, especially with loci that are
located far away on the chromosome, this suggests
that the allele has been strongly selected for and
must have originated recently.
 If the allele had arisen a long time ago, sex should
have eliminated the linkage disequilibrium.
Using linkage disequilibrium to
detect positive selection.
 An allele of G6PD (Glucose-6-phosphate
dehydrogenase), G6PD-202A has a high frequency
(~18% in African populations) and has a high degree of
linkage disequilibrium.
 Thus, it appears to have been strongly selected for
recently.
G6PD and malaria
 There are many common G6PD deficiencies and their
distribution corresponds closely with the distribution
of malaria.
 Appears that G6PD-202A confers strong protection
against malaria.