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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 - PowerPoint PPT Presentation
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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.