Announcements

-Pick up Problem Set 3 from the front

-Ave.- 41.7 SD- 4.1

Group work

-Amy is lecturing on Thursday on Methods in Quantitative Genetics

-Test Oct. 22

Introduction to Quantitative Genetics10/6/09

Quantitative Genetics is the statistic relationship betweengenotype and phenotype. It does not concern itself with the mechanistic theory of how genotype (interacting with environment) produces phenotype (developmental genetics).Only statistical relationships are important, because evolutioncan only act on genotypes that are associated with particularphenotypes. Population parameters such as allele frequencyand mating factors can strongly effect these statisticalassociations!

Physical Basis of Evolution• DNA can replicate

• DNA can mutate and recombine

• DNA encodes information that interacts with the environment to influence phenotype

Goal of Today

• Derive a measure to describe the phenotype that each individual gamete contributes on to the next generation relative to the average phenotype.

• Two of these measures (Average Excess and Average Effect)

Mendelism After 1908• Hardy-Weinberg Helped Establish That

Many Traits Were Mendelian• Still, the Majority of Quantitative Traits

Could Not Be Put Into A Mendelian Framework

• Most Traits Were Quantitative• Therefore, Many Believed That An

Alternative and More Important Mechanism of Heredity Existed in Addition to Mendelism

Ronald A. Fisher

• By Age 22 (1912) was laying the basis for much of modern statistics, but could not get an academic position

• In 1913 Became Convinced of Mendelism

• In 1916 submitted a paper that explained the inheritance of quantitative traits through Mendelian factors

Fisher’s 1916 Paper: The Correlation Between Relatives On The Supposition of

Mendelian Inheritance• Was rejected by several journals• Fisher personally paid to have it published by the Royal

Society of Edinburgh in 1918• When published, it ended all serious opposition to

Mendelism• It established the genetic basis for natural selection• It laid the foundation for modern plant and animal

breeding, and genetic epidemiology• It presented new statistical techniques (e.g., ANOVA --

ANalysis Of VAriance) soon used by virtually all empirical sciences.

Two Basic, Non-Mutually Exclusive Ways of Having Discrete Genotypes Yield Continuous Phenotypes

1. Polygenes

2. Environmental Variation

Polygenes

Fewer Loci More Loci

Environmental Variation

Most Traits Are Influenced By Both Many Genes and Environmental Variation: Frequently Results in a Normal

Distribution. E.g. Cholesterol in Framingham, MAR

elat

ive

Fre

quen

cy in

Pop

ulat

ion

Total Serum Cholesterol in mg/dl

150 220 290

The Normal Distribution Can Be Completely Described by Just 2 Numbers:

The Mean () and Variance ()

Let x be an observed trait value• The mean () is the average or expected

value of x.

• The mean measures where the distribution is centered

• If you have a sample of n observations, x1, x2, …, xn, Then is estimated by:

x = x1 x2 x3 xn /n =

xi

i1

n

n

• The variance () is the average or expected value of the squared deviation of x from the mean; that is, (x)2

• The variance measures the amount of dispersion in the distribution (how “fat” the distribution is)

• If you have a sample of n observations, x1, x2, …, xn, Then given is estimated by:

s2 = [(x1- )2 + (x2-)2 + … + (xn- )2]/n• If you do not know then is estimated by:

s2 x1 x 2 x2 x 2 xn x 2 /(n -1)

By 1916, Fisher Realized

1. Could Examine Causes of Variation, but not cause and effect of quantitative phenotypes.

2. Therefore, what is important about an individual’s phenotype is not its value, but how much it deviates from the average of the population; That is, focus is on variation.

3. Quantitative inheritance could not be studied in individuals, but only among individuals in a population.

Fisher’s Model

Pij = + gi + ej

The mean (average) phenotype forThe entire population:

ijPij/nWhere n is the number of individuals sampled.

Fisher’s Model

Pij = + gi + ej

The genotypic deviation for genotype i is theAverage phenotype of genotype i minus theAverage phenotype of the entire population:

gi = jPij/ni - Where ni is the number of individuals with genotype i.

Fisher’s Model

Pij = + gi + ej

The environmental deviation is the deviationOf an individual’s phenotype from the

Average Phenotype of his/her Genotype:ej =Pij - jPij/ni = Pij-(gi+)=Pij--gi

Pij = + gi + ej

Fisher’s Model

Although called the “environmental” deviation,ej is really all the aspects of an individual’s

Phenotype that is not explained by genotype inThis simple, additive genetic model.

Fisher’s Model

2p = Phenotypic Variance

2p = Average(Pij -

2p = Average(gi + ej)2

Fisher’s Model

2p = Average(gi + ej)2

2p = Average(gi

2 + 2giej + ej2)

2p = Average(gi

2) + Average(2giej) + Average(ej

2)

Fisher’s Model

2p = Average(gi

2) + Average(2giej) + Average(ej2)

Because the “environmental” deviation isreally all the aspects of an individual’s

Phenotype that is not explained by genotype,This cross-product by definition has an average

Value of 0.

Fisher’s Model

2p = Average(gi

2) + Average(ej2)

2p = 2

g + 2e

Phenotypic Variance

Fisher’s Model

2p = Average(gi

2) + Average(ej2)

2p = 2

g + 2e

Genetic Variance

Fisher’s Model

2p = Average(gi

2) + Average(ej2)

2p = 2

g + 2e

Environmental Variance(Really, the variance not

Explained by theGenetic model)

Fisher’s Model

2p = 2

g + 2e

Phenotypic Variance = Genetic Variance + Unexplained Variance

In this manner, Fisher partitioned the causesOf phenotypic variation into a portion explainedBy genetic factors and an unexplained portion.

Fisher’s Model

2p = 2

g + 2e

Phenotypic Variance = Genetic Variance + Unexplained Variance

This partitioning of causes of variation can only bePerformed at the level of a population.

An individual’s phenotype is an inseparableInteraction of genotype and environment.

ApoE and Cholesterol in a Canadian Population

Total Serum Cholesterol (mg/dl)

3/4

3/3

2/3

4/4

2/2

2/4

Rel

ativ

e F

requ

ency

= 174.62

p = 732.5

2

0.078

3

0.770

4

0.152

Random Mating

Geno-

type3/3 3/2 3/4 2/2 2/4 4/4

H-W

Freq.0.592 0.121 0.234 0.006 0.024 0.023

Mean

Pheno.173.8 161.4 183.5 136.0 178.1 180.3

Geno-

type3/3 3/2 3/4 2/2 2/4 4/4

H-W

Freq.0.592 0.121 0.234 0.006 0.024 0.023

Mean

Pheno.173.8 161.4 183.5 136.0 178.1 180.3

Step 1: Calculate the Mean Phenotype of the Population

= (0.592)(173.8)+(0.121)(161.4)+(0.234)(183.5)+(0.006)(136.0)+(0.024)(178.1)+(0.023)(180.3)

= 174.6

Geno-

type3/3 3/2 3/4 2/2 2/4 4/4

Mean

Pheno.173.8 161.4 183.5 136.0 178.1 180.3

gi

173.8-174.6

-0.8161.4-174.6

-13.2183.5-174.6

8.9136.0-174.6

-38.6178.1-174.6

3.5180.3-174.6

5.7

Step 2: Calculate the genotypic deviations

= 174.6

Geno-

type3/3 3/2 3/4 2/2 2/4 4/4

H-W

Freq.0.592 0.121 0.234 0.006 0.024 0.023

gi -0.8 -13.2 8.9 -38.6 3.5 5.7

Step 3: Calculate the Genetic Variance

2g = 50.1

2g= (0.592)(-0.8)2 +(0.121)(-13.2)2 +(0.234)(8.9)2 +(0.006)(-38.6)2 +(0.024)(3.5)2 +(0.023)(5.7)2

Step 4: Partition the Phenotypic Variance intoGenetic and “Environmental” Variance

2g

50.12

e

682.4

2p = 732.5

Broad-Sense Heritability

h2B is the proportion of the

phenotypic variation that can be explained by the modeled genetic

variation among individuals.

Broad-Sense Heritability

For example, in the Canadian Population for Cholesterol Level

h2B = 50.1/732.5 = 0.07

That is, 7% of the variation in cholesterol levels in this population is explained by

genetic variation at the ApoE locus.

Broad-Sense Heritability

Genetic Variation at the ApoE locus is therefore a cause of variation in cholesterol

levels in this population.

ApoE does not “cause” an individual’s cholesterol level.

An individual’s phenotype cannot be partitioned into genetic and unexplained factors.

Broad-Sense Heritability

Measures the importance of geneticVariation as a Contributor to Phenotypic

Variation Within a Generation

The more important (and difficult) questionIs how Phenotypic Variation is Passed on to

The Next Generation.

3/3

0.592

3/2

0.121

3/4

0.234

2/2

0.006

2/4

0.024

4/4

0.023

Deme

Development

Environment= 174.6

2p = 732.5

h2B

40.152

3

0.770

2

0.078

4/4

0.023

2/4

0.024

2/2

0.006

3/4

0.234

3/2

0.121

3/3

0.592

Random Mating

MeiosisGene Pool

DemeEnvironment

?Development

Fisher’s Model

1. Assume that the distribution of environmental deviations (ej’s) is the same every generation

2. Assign a “phenotype” to a gamete

Phenotypes of Gametes

1. Average Excess of a Gamete Type2. Average Effect of a Gamete Type

3. These two measures are identical in a random mating population, so we will consider only the average excess for now.

The Average Excess

The Average Excess of Allele i Is The Average Genotypic Deviation Caused By A Gamete Bearing Allele i After Fertilization With A Second Gamete Drawn From the

Gene Pool According To The Deme’s System of Mating.

The Average Excess

ai tii

pi

gii 12 tij

pi

gijji

t(ij | i)gijj

Where gij is the genotypic deviation of genotype ij, tij is the frequency of ij in the population (not necessarily HW), pi is the frequency of allele i, and:

Prob(ii given i) t( ii | i) tii

pi

Prob(ij given i) t( ij | i) 12 tij

pi

when j i

The Average Excess

Note, under random mating tii = pi2 and tij = 2pipj, so:

Prob(ii given i) t(ii | i) tii

pi

pi

Prob(ij given i) t(ij | i) 12 tij

pi

p j when j i

ai pjgijj

2

0.078

3

0.7704

0.152

Random Mating

Gene Pool

Deme3/3

0.592

3/2

0.121

3/4

0.234

2/2

0.006

2/4

0.024

4/4

0.023

Average Excess of An Allele

2

0.078Gene Pool

Average Excess of An Allele

What Genotypes Will an 2 allele find itself in after

random mating?

2

0.078

Random Mating

Gene Pool

Deme3/2 2/2 2/4

Average Excess of An Allele

What are the probabilities of these Genotypes after random

mating given an 2 allele?

2

0.078

3

0.770

4

0.152

Random Mating

Gene Pool

Deme3/2

0.770

2/20.078

2/40.152

Average Excess of An Allele

These are the Conditional Probabilities of the genotypesGiven random mating and a gamete with the 2 allele.

2

0.078

3

0.770

4

0.152

Random Mating

Gene Pool

Deme3/2

0.770

2/2

0.078

2/4

0.152

Average Excess of An Allele

Environmenth2BDevelopment

GenotypicDeviations

-13.2 -38.6 3.5

Average Genotypic Deviation of a 2 bearing gamete =(0.770)(-13.2)+(0.078)(-38.6)+(0.152)(3.5) = -12.6

2

0.078

3

0.770

4

0.152

Random Mating

Gene Pool

Deme3/3

0.7703/2

0.0783/4

0.152

Average Excess of Allele 3

Environmenth2

BDevelopment

GenotypicDeviations

-0.8 -13.2 8.9

Average Excess of 3 =(0.770)(-0.8)+(0.078)(-13.2)+(0.152)(8.9) = -0.3

2

0.078

3

0.770

4

0.152

Random Mating

Gene Pool

Deme3/4

0.770

2/4

0.078

4/4

0.152

Average Excess of Allele 4

Environmenth2

BDevelopment

GenotypicDeviations

8.9 3.5 5.7

Average Excess of 4 =(0.770)(8.9)+(0.078)(3.5)+(0.152)(5.7) = 8.0

2

0.078

-12.6

3

0.770

-0.3

4

0.152

8.0

Gene Pool

AllelesFrequencies

“Phenotype”(AverageExcess)

The critical breakthrough in Fisher’s paper was assigning a “phenotype” to a gamete, the

physical basis of the transmission of phenotypes from one generation to the next.

Average Excess of 4 =(0.770)(8.9)+(0.078)(3.5)+(0.152)(5.7) = 8.0

The Average Excess Depends Upon the Genotypic Deviations, which in turn Depend

Upon the Average Phenotypes of the Genotypes and And the Average Phenotype of the Deme, which in turn Depends Upon

The Genotype Frequencies.

Average Excess of 4 =(0.770)(8.9)+(0.078)(3.5)+(0.152)(5.7) = 8.0

The Average Excess Depends Upon the Gamete Frequencies in the Gene Pool and

Upon the System of Mating.

The Average Excess

The Portion of Phenotypic Variation That Is Transmissible Through a Gamete Via

Conditional Expectations

Purposes

• Population Genetics vs. Quantitative Genetics

Populations Genetics-Concerned pop gen forces-Average Excess- understand NS

Quantitative Genetics-Concerned with Breeding Values-Measurable values before genetics-Average Effect (of the gene)= Average Effect (Alan)-Average Effect (of allelic substitution)

The Average Effect

The Portion of Phenotypic Variation That Is Transmissible Through a Gamete Measured

At the Level of a Deme and Its Associated Gene Pool via Least-Squares Regression.

The Average EffectDerived in Quant. Gen. terms with RM assumption

Jim’s Definition-The mean phenotypic deviation of a gamete from the population meanwhen paired with another allele at random.

AND

The additive contribution of a gameteto the next generation

The Average Effect

Templeton (1987) showed:

i ai

1 f

Fisher’s ModelThe Next Step Is To Assign a “Phenotypic”

Value To a Diploid Individual That Measures Those Aspects of Phenotypic Variation That Can be Transmitted Through the Individual’s Gametes.

Breeding Value or Additive Genotypic Deviation Is The Sum of the Average Effects (=Average Excesses Under Random Mating) of Both Gametes Borne By An Individual.

Additive Genotypic Deviation

Let k and l be two alleles (possibly the same) at a locus of interest. Let k be the Average Effect of allele k, and l the Average Effect of allele l. Let gakl be the additive genotypic

deviation of genotype k/l. Then:gakl = k + l

Geno-

type3/3 3/2 3/4 2/2 2/4 4/4

H-W

Freq.0.592 0.121 0.234 0.006 0.024 0.023

gi -0.8 -13.2 8.9 -38.6 3.5 5.7

2

0.078

-12.6

3

0.770

-0.3

40.152

8.0

Alleles

FrequenciesAverage Excess

=Effect (rm)

gai-0.3+(-0.3)

-0.6-0.3+(-12.6)

-12.9-0.3+8.0

7.7-12.6 -12.6

-25.4-12.6+8.0

-4.68.0 + 8.0

16.0

Geno-

type3/3 3/2 3/4 2/2 2/4 4/4

H-W

Freq.0.592 0.121 0.234 0.006 0.024 0.023

gi -0.8 -13.2 8.9 -38.6 3.5 5.7

gai -0.6 -12.9 7.7 -25.4 -4.6 16.0

The Additive Genetic Variance

2a = 44.7

2a=(0.592)(-0.6)2+(0.121)(-12.9)2+(0.234)(7.7)2+(0.006)(-25.4)2+(0.024)(-4.6)2+(0.023)(16.0)2

The Additive Genetic Variance

Note that 2g = 50.1 > 2

a = 44.7

It is always true that 2g > 2

a

Have now subdivided the genetic variance into a component that is

transmissible to the next generation and a component that is not:

2g = 2

a + 2d

The Additive Genetic Variance2

g = 2a + 2

d

The non-additive variance, 2d, is called the

“Dominance Variance” in 1-locus models.

Mendelian dominance is necessary but not sufficient for 2

d > 0.

2d depends upon dominance, genotype

frequencies, allele frequencies and system of mating.

The Additive Genetic Variance

For the Canadian Population,

2g = 50.1 and 2

a = 44.7

Since 2g = 2

a + 2d

50.1 = 44.7 +2d

2d = 50.1 - 44.7 = 5.4

Partition the Phenotypic Variance intoAdditive Genetic, non-Additive Genetic

and “Environmental” Variance

2g

50.1

2p = 732.5

2e

682.42

a

44.72

e

682.42d

5.4

The Additive Genetic Variance2

g = 2a + 2

d + 2i

In multi-locus models, the non-additive variance is divided into the Dominance Variance and the

Interaction (Epistatic) Variance, 2i.

Mendelian epistasis is necessary but not sufficient for 2

i > 0.

2i depends upon epistasis, genotype

frequencies, allele frequencies and system of mating.

The Partitioning of Variance

2p = 2

a + 2d + 2

i + 2e

As more loci are added to the model, 2e goes

down relative to 2g such that hB

2 = 0.65 for the phenotype of total serum cholesterol in this

population. Hence, ApoE explains about 10% of the heritability of cholesterol levels, making it

the largest single locus contributor.

(Narrow-Sense) Heritability

h2 is the proportion of the phenotypic variance that can be explained by the additive genetic

variance among individuals.

(Narrow-Sense) HeritabilityFor example, in the Canadian

Population for Cholesterol Level

h2 = 44.7/732.5 = 0.06

That is, 6% of the variation in cholesterol levels in this population is transmissible through gametes to the next generation

from genetic variation at the ApoE locus.