Quantitative genetics: traits controlled my many loci Key questions: what controls the rate of adaptation? Example: will organisms adapt to increasing.

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

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

AA AA' A'A'

Codominant (additive)

Dominance and heritability II

Dominance

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.5 0.75 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

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

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

AA AA' A'A'

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.5 0.75 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

AA AA' A'A'

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

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

0.4

0.5

0.6

0.7

0.8

0.9

1

AA AA' A'A'

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 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

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

AA AA' A'A'

Genotype

Effects of dominance: phenotypes after random mating

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.5 0.75 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.5 0.75 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

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

0.9

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.