Quantitative genetics: traits
controlled my many loci
Key questions: what controls the rate of
adaptation?
Example: will organisms adapt to increasing
temperatures or longer droughts fast enough
to avoid extinction?
What is the genetic basis of complex traits
with continuous variation?
Quantitative genetics vs. population
genetics
Population genetics
Quantitative genetics
Breeder’s equation
Breeder’s equation: R = h2S
Polygenic Inheritance Leads
to a Quantitative Trait
TRAIT
A
B
C
Z1
# of individuals
LOCUS
D
Z
E
F
S, the selection differential
R, the selection response,
Problems in predicting the
evolution of quantitative traits
-
Dominance
Epistasis
Environmental effects
Environment alters gene
expression - epigenetics
yellow: no difference;
red or green = difference
age 3
age 50
Overall: not all variation is
heritable
Major questions in quantitative
genetics
• How much phenotypic variation is due to
genes, and how much to the environment?
• How much of the genetic variation is due
to genes of large effect, and how much to
genes of small effect?
Measuring heritability
Variance:
_
Vp= Σi (Xi – X)2
--------------(N – 1)
Variance components
VP = total phenotypic variance
VP = VA + VD + VE + VGXE
VA = Additive genetic variance
VD = Dominance genetic variance (nonadditive – can include epistasis)
VE = Variance among individuals experiencing
different environments
VGXE = Variance due to environmental variation
that influences gene expression (not covered in
text)
Heritability = h2 = VA / VP
The proportion of phenotypic variance due to additive
genetic variance among individuals
h2 = VA / (VA + VD + VE + VGXE)
Heritability can be low due to:
Additive vs. dominance variance
Additive:
heterozygote is
intermediate
Dominance:
heterozygote is
closer to one
homozygote
(Difference
from line is due
to dominance)
Dominance and heritability
Dominance
Genotype
AA
AA’
A’A’
f(A) = p = 0.5
f(A’) = q = 0.5
Start in HWE
f(AA) = 0.25
f(AA’) = 0.50
f(A’A’) = 0.25
Toe len (cm)
0.5
1.0
1.0
Codominant (additive)
Genotype
Toe len (cm)
AA
0.5
AA’
0.75
A’A’
1.0
f(A) = p = 0.5
f(A’) = q = 0.5
Start in HWE
f(AA) = 0.25
f(AA’) = 0.50
f(A’A’) = 0.25
Dominance and heritability II
Dominance
Codominant (additive)
1
1
0. 9
0. 9
0. 8
0. 8
0. 7
0. 7
Starting
Genotype
frequencies
0. 6
0. 5
0. 4
0. 3
0. 2
0. 1
0. 6
0. 5
0. 4
0. 3
0. 2
0. 1
0
0
AA
AA'
A 'A '
AA
Genotype
1
1
0.9
0.9
0.8
0.8
0.7
0.6
0.5
Mean =
0.875 cm
0.7
0.6
Starting
Phenotype
frequencies
0.4
0.3
0.2
0.1
AA'
Genotype
A 'A '
Mean =
0.75 cm
0.5
0.4
0.3
0.2
0.1
0
0
0.5
0.75
1
Phenotype
(toe len – cm)
0.5
0.75
1
Phenotype
(toe len – cm)
Select toe length = 1 cm
Dominance
Additive
1
1
0. 9
0. 9
0. 8
0. 8
0. 7
0. 7
Starting
Genotype
frequencies
0. 6
0. 5
0. 4
0. 3
0. 2
0. 1
0. 6
0. 5
0. 4
0. 3
0. 2
0. 1
0
0
AA
AA'
A 'A '
AA
Genotype
1
1
0.9
0.9
0.8
0.8
0.7
0.6
0.5
Mean =
1.0 cm
0.7
0.6
Starting
Phenotype
frequencies
0.4
0.3
0.2
0.1
AA'
Genotype
A 'A '
Mean =
1.0 cm
0.5
0.4
S=
0.3
0.2
0.1
0
0
0.5
0.75
0.5
1
Phenotype
(toe len – cm)
S=
0.75
1
Phenotype
(toe len – cm)
Effects of dominance: genotypes
after random mating
Dominance
Codominant (additive)
1
1
0. 9
0. 9
0. 8
0. 8
0. 7
0. 7
0. 6
Genotype freq
after selection,
before mating
0. 5
0. 4
0. 3
0. 2
0. 1
0. 6
0. 5
0. 4
0. 3
0. 2
0. 1
0
AA
AA'
0
A 'A '
AA
Genotype
AA'
Genotype
A 'A '
1
1
0. 9
0. 9
0. 8
0. 8
0. 7
0. 7
0. 6
0. 6
Genotype freq
after mating
0. 5
0. 4
0. 3
0. 2
0. 5
0. 4
0. 3
0. 2
0. 1
0. 1
0
0
AA
AA'
A 'A '
AA
Genotype
AA'
Genotype
A 'A '
Effects of dominance: phenotypes
after random mating
Dominance
Codominant (additive)
1
1
0.9
0.9
0.8
0.8
0.7
Phenotype freq
after selection,
before mating
0.6
0.5
0.4
0.3
0.2
0.1
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.5
0.75
Genotype
0
1
0.5
1
0.7
0.6
1
0.9
0.9
0.8
0.75
1
0.8
mean =
0.954
0.7
0.6
0.5
0.5
Phenotype freq
after mating
0.4
0.3
0.2
mean =
1.0
0.4
0.3
R=1 - .75
0.2
0.1
0.1
0
0
0.5
0.75
Phenotype
1
0.5
R = 954 - .875 =
0.079
0.75
Phenotype
1
= 0.25
Effect of dominance on heritability
Dominant
S=
R=
R = h 2S
h2 = R / S =
Codominant (additive)
S=
R=
R = h 2S
h2 =
Second problem predicting
outcome of selection: epistasis
Example: hair color in mammals
MC1
agouti
MC1receptor
Agouti is
antagonist for
MC1R.
If agouti binds, no
dark pigment
produced
Second problem predicting
outcome of selection: epistasis
Epistasis
MC1
agouti
MC1receptor
Normal receptor
Mutant: never Mutant: always
dark pigment dark pigment
(yellow labs)
Epistasis example
Genotype
Phenotype
EE / AA
dark tips, light band
ee / -blond / gold / red
Ed- / -all dark
EE / Adblond / gold / red
Want dark fur: population
ee / AA
Ee / AdA
Ee / AA
Epistasis example ii
Cross: Ee / AdA
genotype
EE / AdAd
EE / AdA
EE / AA
x Ee / AdA
phenotype
freq
1/16
1/8
1/16
Ee / AdAd
Ee / AdA
Ee / AA
1/8
1/4
1/8
ee / AdAd
ee / AdA
ee / AA
1/16
1/8
1/16
Measuring h2: Parent-offspring
regression
Estimating h2
Analysis of related individuals
Measuring the response of a population, in
the next generation, to selection
Heritability is estimated as the
slope of the least-squares
regression line
h2 data: Darwin’s finches
h2 example: Darwin’s finches
mean before selection: 9.4
mean after selection: 10.1
S = 10.1 - 9.4 = 0.7
h2 example: Darwin’s finches II
mean before selection: 9.4
mean of offspring after
selection: 9.7
Response to selection:
9.7 – 9.4 = 0.3
R = h2S; 0.3 = h2 * 0.7
h2 = R/S = 0.3 / 0.7 = 0.43
Controlling for environmental
effects on beak size
• song sparrows: cross fostering
Offspring vs.
biological
parent (h2);
vs. foster
parent (VE)
Smith and Dhondt (1980)
Cross fostering in song sparrows
Testing for environmental effects
How can we determine the effect of the
environment on the phenotype?
Two genotypes in two environments:
possible effects on phenotype
Example of environmental effects: locusts
Environmental effects: carpenter
ant castes
queen
male
major
worker
minor
worker
Effect of GxE: predicting outcome
of selection
7 yarrow (Achillea
millefolium) genotypes
2 gardens
Clausen, Keck, and Heisey (1948)
What happens to h2 when selection
occurs?
Modes of selection
Gen. 2
Gen. 1
Gen. 0
freq.
trait, z
Mode of selection: directional
freq.
freq.
trait, z
trait, z
Mode of selection: stabilizing
freq.
freq.
trait, z
trait, z
Mode of selection: disruptive
freq.
freq.
trait, z
trait, z
Disruptive selection
Distribution of mandible widths
in juveniles that died (shaded)
and survived (black)
Fitness
Beak depth
Beak depth
Fitness
• Darwin’s finches: beak width
is correlated with beak depth
Beak width
Complications: correlations
Beak width
Detecting loci
affecting
quantitative traits
(QTL)
QTLs and genes of major effects
How important are genes of major
effect in adaptation?
QTL analysis: Quantitative Trait
Loci – where are the genes
contributing to quantitative traits?
• Approach
– two lineages consistently differing for trait of
interest (preferably inbred for homozygosity)
– Identify genetic markers specific to each
lineage (eg microsatellite markers)
– make crosses to form F1
– generate F2s and measure trait of interest
– test for association between markers and trait
– Estimate the effect on the phenotype of each
marker
Example: Mimulus cardinalis and
Mimulus lewisii
Mimulus cardinalis
Mimulus lewisii
Locate lineage-specific markers
Mimulus lewisii
Mimulus cardinalis
Microsatellite
length 250 (M. l.)
or 254 (M. c.)
M. lewisii specific
Legend
M. cardinalis specific
non-specific
Cross lines 
QTL Mapping: crosses
F1
F2
recombinants
intermediate
phenotype
scrambled
phenotypes
QTL analysis: trait associations
Homozygotes at
marker 2 are closer to
one parent
Heterozygotes at
marker 2 are
intermediate in trait
values
Trait analysis
For each marker,
ask whether
changing
genotype affects
phenotype
QTL probabilities
Estimate the
probability of
location of QTL
based on
association of
markers and
recombination
probabilities
(eg CD34B rarely
associated with
pH, CD34A nearly
always, TG63
rarely)
Markers
Example study: basis of floral traits in two
Mimulus species (Schemske and Bradshaw,
PNAS 1999)
Pollination
syndrome
changes during
the evolution of
the Mimulus
group:
hummingbird or
bee
F2 plants showed variation for most
floral traits
F1
M. lewisii
F2
plants
M. cardinalis
Most traits had
multiple QTL but
one explaining >
25% of variation
Few genes of
large effect
important in this
case
Case study: domestication traits in
sunflowers
• Most traits showed many genes of small
effect (< 10%)
• Problems: gene map resolution is low
– 35,000 genes per plant genome
– 100 markers on genetic map: 350 genes per
marker
• A “weak” QTL can be due to
– nearby gene of weak effect
– more distant gene of strong effect (looks weak
due to recombination between marker and
locus)
Recap: quantitative genetics
• Are traits heritable?
– Usually find heritability
– However, environmental effects can be large
• Are genes controlling quantitative traits of
large effect or small effect?
– Some important genes for adaptation of large
effect.
– Overall pattern still unclear
Questions
1.
2.
3.
Human height is highly heritable: among university students in the
US, the heritability is 0.84. Yet, during the 1980s, when
Guatemalan refugees fled the civil war to the United States, 12
year old Mayan children were four inches taller in the United
States than in Guatemala. How is this possible?
Consider the following scenario: In October 2006 you banded and
weighed all of the adult burrowing owls near Osoyoos. The
average owl weighed 207.8 gm. The winter of 2006-2007 was
especially cold, and many owls died before they bred in the spring.
When you weighed those owls that survived to breed in April of
2007, their average weight was 211.8 gm. If the heritability of owl
body weight is 0.25, what is the expected mean body weight in
the NEXT generation of adults, assuming none die before they
become adults?
Data demonstrating stabilizing selection for human head size at
birth were collected in 1951 in the United States. If you were to
collect the same data now, would you expect to find the same
pattern? Why or why not? Would it matter where in the world you
did your study?
Also, see posted quantitative genetics practice questions for further
practice.