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
Genetics
10/6/09
Quantitative Genetics is the statistic relationship between
genotype 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 evolution
can only act on genotypes that are associated with particular
phenotypes. Population parameters such as allele frequency
and mating factors can strongly effect these statistical
associations!
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
Relative Frequency in Population
Most Traits Are Influenced By Both Many Genes and
Environmental Variation: Frequently Results in a Normal
Distribution. E.g. Cholesterol in Framingham, MA
150
220
290
Total Serum Cholesterol in mg/dl
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:
n
x

i
i

1
x
=
x

x

x





x
/
n
=


1
2
3
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:


s

x
x

x
x





x
x
/(
n
1






2
2
2
1
2
2
n
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 for
The entire population:
ijPij/n
Where n is the number of individuals sampled.
Fisher’s Model
Pij = + gi + ej
The genotypic deviation for genotype i is the
Average phenotype of genotype i minus the
Average 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 deviation
Of an individual’s phenotype from the
Average Phenotype of his/her Genotype:
ej =Pij - jPij/ni = Pij-(gi+)=Pij--gi
Fisher’s Model
Pij = + gi + ej
Although called the “environmental” deviation,
ej is really all the aspects of an individual’s
Phenotype that is not explained by genotype in
This simple, additive genetic model.
Fisher’s Model
2

p=
Phenotypic Variance
2p = Average(Pij - 
2p = Average(gi + ej)2
Fisher’s Model
2p = Average(gi + ej)2
2
2
2
 p = Average(gi + 2giej + ej )
2
2
 p = Average(gi ) +
Average(2giej) +
Average(ej2)
Fisher’s Model
2p = Average(gi2) + Average(2giej) + Average(ej2)
Because the “environmental” deviation is
really 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(gi2) + Average(ej2)
2p = 2g + 2e
Phenotypic Variance
Fisher’s Model
2p = Average(gi2) + Average(ej2)
2p = 2g + 2e
Genetic Variance
Fisher’s Model
2p = Average(gi2) + Average(ej2)
2p = 2g + 2e
Environmental Variance
(Really, the variance not
Explained by the
Genetic model)
Fisher’s Model
2p =
2g
+
2e
Phenotypic Variance = Genetic Variance + Unexplained Variance
In this manner, Fisher partitioned the causes
Of phenotypic variation into a portion explained
By genetic factors and an unexplained portion.
Fisher’s Model
2p =
2g
+
2e
Phenotypic Variance = Genetic Variance + Unexplained Variance
This partitioning of causes of variation can only be
Performed at the level of a population.
An individual’s phenotype is an inseparable
Interaction of genotype and environment.
ApoE and Cholesterol in a Canadian Population
3/3
Relative Frequency
= 174.6
2p = 732.5
3/4
4/4
2/2
2/3
2/4
Total Serum Cholesterol (mg/dl)
2
0.078
3
0.770
4
0.152
Random Mating
Geno3/3
type
H-W
0.592
Freq.
Mean
173.8
Pheno.
3/2
3/4
2/2
2/4
4/4
0.121
0.234
0.006
0.024
0.023
161.4
183.5
136.0
178.1
180.3
Step 1: Calculate the Mean Phenotype of the Population
Geno3/3
type
H-W
0.592
Freq.
Mean
173.8
Pheno.
3/2
3/4
2/2
2/4
4/4
0.121
0.234
0.006
0.024
0.023
161.4
183.5
136.0
178.1
180.3
= (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
Step 2: Calculate the genotypic deviations
Geno3/3
type
Mean
173.8
Pheno.
gi
3/2
3/4
2/2
2/4
4/4
161.4
183.5
136.0
178.1
180.3
173.8-174.6 161.4-174.6 183.5-174.6 136.0-174.6 178.1-174.6 180.3-174.6
-0.8
-13.2
8.9
 = 174.6
-38.6
3.5
5.7
Step 3: Calculate the Genetic Variance
Geno3/3
type
H-W
0.592
Freq.
gi
-0.8
3/2
3/4
2/2
2/4
4/4
0.121
0.234
0.006
0.024
0.023
-13.2
8.9
-38.6
3.5
5.7
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
2g = 50.1
Step 4: Partition the Phenotypic Variance into
Genetic and “Environmental” Variance
2p = 732.5
 2g
50.1
 2e
682.4
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 genetic
Variation as a Contributor to Phenotypic
Variation Within a Generation
The more important (and difficult) question
Is how Phenotypic Variation is Passed on to
The Next Generation.
Environment
Deme
3/3
3/2
3/4
2/2
2/4
4/4
0.592 0.121 0.234 0.006 0.024 0.023
Gene Pool
2
0.078
Deme
Development
h2B
Meiosis
3
0.770
= 174.6
2p = 732.5
4
0.152
Random Mating Environment
3/3
3/2
3/4
2/2
2/4
4/4
0.592 0.121 0.234 0.006 0.024 0.023
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 Type
2. 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
1
ij
ii
2
ii
ij
i
j

i i
j
t
t
a

g

g

t
(
ij
|i
)
g


i
ij
p
p
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:

t
Prob(
iigiven
i
)

t
(
ii
|i
)
ii
p
i
1
2ij
t
Prob(
ijgiven
i
)

t
(
ij
|i
)

p
i
when
j
i
The Average Excess
Note, under random mating tii = pi2 and tij = 2pipj, so:
t
ii
Prob(
iigiven
i
)

t
(
ii
|i
)


p
i
p
i

t
Prob(
ijgiven
i
)

t
(
ij
|i
)
 
p
j
p
i
1
2ij
a
p

i
jg
ij
j
whe
j
i
Average Excess of An Allele
Gene Pool
2
0.078
3
0.770
4
0.152
Random Mating
Deme
3/3
3/2
3/4
2/2
2/4
4/4
0.592 0.121 0.234 0.006 0.024 0.023
Average Excess of An Allele
Gene Pool
2
0.078
What Genotypes Will an 2
allele find itself in after
random mating?
Average Excess of An Allele
Gene Pool
What are the probabilities of
these Genotypes after random
mating given an 2 allele?
2
0.078
Random Mating
Deme
3/2
2/2
2/4
Average Excess of An Allele
Gene Pool
2
0.078
3
0.770
4
0.152
Random Mating
Deme
3/2
2/2
2/4
0.770
0.078
0.152
These are the Conditional Probabilities of the genotypes
Given random mating and a gamete with the 2 allele.
Average Excess of An Allele
Gene Pool
2
0.078
3
0.770
4
0.152
Random Mating
Deme
3/2
0.770
Environment
h2B
Development
Genotypic
Deviations
2/2
2/4
0.078 0.152
-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
Average Excess of Allele 3
Gene Pool
2
0.078
3
0.770
4
0.152
Random Mating
Deme
3/4
3/3
3/2
0.770 0.078 0.152
h2B
Development
Genotypic
Deviations
-0.8
-13.2
Environment
8.9
Average Excess of 3 =
(0.770)(-0.8)+(0.078)(-13.2)+(0.152)(8.9) = -0.3
Average Excess of Allele 4
Gene Pool
2
0.078
3
0.770
4
0.152
Random Mating
Deme
Development
Genotypic
Deviations
3/4
0.770
h2B
8.9
2/4
4/4
0.078 0.152
Environment
3.5
5.7
Average Excess of 4 =
(0.770)(8.9)+(0.078)(3.5)+(0.152)(5.7) = 8.0
Gene Pool
Alleles
Frequencies
“Phenotype”
(Average
Excess)
2
0.078
-12.6
3
0.770
-0.3
4
0.152
8.0
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 Effect
Derived in Quant. Gen. terms with RM assumption
Jim’s DefinitionThe mean phenotypic deviation of a
gamete from the population mean
when paired with another allele at
random.
AND
The additive contribution of a gamete
to the next generation
The Average
Effect
Templeton (1987)
showed:
ai
i 
1f
Fisher’s Model
The 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
Geno3/3
type
H-W
0.592
Freq.
gi
-0.8
Alleles
Frequencies
Average Excess
=Effect (rm)
gai
3/2
3/4
2/2
2/4
4/4
0.121
0.234
0.006
0.024
0.023
-13.2
8.9
-38.6
3.5
5.7
2
0.078
-12.6
3
0.770
-0.3
4
0.152
8.0
-0.3+(-0.3)
-0.3+(-12.6)
-0.3+8.0
-12.6 -12.6
-12.6+8.0
8.0 + 8.0
-0.6
-12.9
7.7
-25.4
-4.6
16.0
The Additive Genetic Variance
Geno3/3
type
H-W
0.592
Freq.
3/2
3/4
2/2
2/4
4/4
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
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
2a = 44.7
The Additive Genetic Variance
Note that 2g = 50.1 > 2a = 44.7
It is always true that 2g > 2a
Have now subdivided the genetic
variance into a component that is
transmissible to the next generation and
a component that is not:
2g = 2a + 2d
The Additive Genetic Variance
2g = 2a + 2d
The non-additive variance, 2d, is called the
“Dominance Variance” in 1-locus models.
Mendelian dominance is necessary but not
sufficient for 2d > 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 2a = 44.7
Since 2g = 2a + 2d
50.1 = 44.7 +2d
2d = 50.1 - 44.7 = 5.4
Partition the Phenotypic Variance into
Additive Genetic, non-Additive Genetic
and “Environmental” Variance
2p = 732.5
 2g
50.1
 2a
44.7
 2d
5.4
 2e
682.4
 2e
682.4
The Additive Genetic Variance
2g = 2a + 2d + 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 2i > 0.
2i depends upon epistasis, genotype
frequencies, allele frequencies and system of
mating.
The Partitioning of Variance
2p = 2a + 2d + 2i + 2e
As more loci are added to the model, 2e goes
down relative to 2g such that hB2 = 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) Heritability
For 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.