quantitative genetics: traits controlled my many loci key questions: what controls the rate of...
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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
A
B
C
D
E
F
LOCUS TRAIT
Z1
# o
f in
div
idu
als
Z
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 (non- additive – 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
DominanceGenotype Toe len (cm)AA 0.5
AA’ 1.0A’A’ 1.0
f(A) = p = 0.5f(A’) = q = 0.5Start in HWEf(AA) = 0.25f(AA’) = 0.50f(A’A’) = 0.25
Codominant (additive)Genotype Toe len (cm)AA 0.5
AA’ 0.75A’A’ 1.0
f(A) = p = 0.5f(A’) = q = 0.5Start in HWEf(AA) = 0.25f(AA’) = 0.50f(A’A’) = 0.25
Starting Genotype
frequencies0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
AA AA' A'A'
0
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0.9
1
AA AA' A'A'
Codominant (additive)
Dominance and heritability II
Dominance
0
0.1
0.2
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0.4
0.5
0.6
0.7
0.8
0.9
1
0.5 0.75 10
0.1
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0.7
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0.9
1
0.5 0.75 1
Starting Phenotype frequencies
Genotype Genotype
Phenotype (toe len – cm)
Phenotype (toe len – cm)
Mean = 0.875 cm
Mean = 0.75 cm
0
0.1
0.2
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1
AA AA' A'A'
0
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0.5 0.75 1
0
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1
AA AA' A'A'
0
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1
0.5 0.75 1
Starting Genotype
frequencies
Additive
Select toe length = 1 cm
Dominance
Starting Phenotype frequencies
Genotype Genotype
Phenotype (toe len – cm)
Phenotype (toe len – cm)
Mean = 1.0 cm
Mean = 1.0 cm
S =
S =
0
0.1
0.2
0.3
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0.9
1
AA AA' A'A'
0
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0.9
1
AA AA' A'A'
Genotype freq after selection, before mating
Codominant (additive)
Effects of dominance: genotypes after random mating
Dominance
Genotype Genotype
Genotype freq after mating
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
AA AA' A'A'
Genotype
0
0.1
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1
AA AA' A'A'
Genotype
Effects of dominance: phenotypes after random mating
0
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0.5 0.75 1
0
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0.5 0.75 1
0
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0.5 0.75 1
Phenotype freq after selection, before mating
Codominant (additive)Dominance
Genotype
Phenotype freq after mating
PhenotypePhenotype
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
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1
0.5 0.75 1
mean = 0.954
mean = 1.0
R = 954 - .875 = 0.079
R=1 - .75
= 0.25
Effect of dominance on heritability
Dominant
S =
R =
R = h2S
h2 = R / S =
Codominant (additive)
S =
R =
R = h2S
h2 =
Second problem predicting outcome of selection: epistasis
Example: hair color in mammals
MC1-receptor
MC1 agouti Agouti is antagonist for MC1R.
If agouti binds, no dark pigment produced
Second problem predicting outcome of selection: epistasis
Epistasis
MC1 agouti
MC1-receptor
Normal receptor Mutant: never dark pigment (yellow labs)
Mutant: always dark pigment
Epistasis example
Genotype Phenotype EE / AA dark tips, light band ee / -- blond / gold / red Ed- / -- all dark EE / Ad- blond / gold / red Want dark fur: population ee / AA Ee / AdA Ee / AA
Epistasis example iiCross: Ee / AdA x Ee / AdA genotype phenotype freqEE / AdAd 1/16EE / AdA 1/8EE / AA 1/16 Ee / AdAd 1/8Ee / AdA 1/4Ee / AA 1/8 ee / AdAd 1/16ee / AdA 1/8ee / AA 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
Smith and Dhondt (1980)
Offspring vs. biological parent (h2); vs. foster parent (VE)
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 malemajor 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
trait, z
freq.
Gen. 0
Gen. 1
Mode of selection: directional
trait, z
freq.
trait, z
freq.
Mode of selection: stabilizing
trait, z
freq.
trait, z
freq.
Mode of selection: disruptive
trait, z
freq.
trait, z
freq.
Disruptive selection
Distribution of mandible widths in juveniles that died (shaded) and survived (black)
Complications: correlations
• Darwin’s finches: beak width is correlated with beak depth
Fitn
ess
Beak depth
Be
ak w
idth
Beak depth
Fitn
ess
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
LegendM. lewisii specific
M. cardinalis specific
non-specific
Mimulus cardinalisMimulus lewisii
Microsatellite
length 250 (M. l.) or 254 (M. c.)
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
M. cardinalisM. lewisii
F1
F2 plants
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. 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?
2. 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?
3. 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.