Lecture 6: Introduction to
Quantitative genetics
Bruce Walsh lecture notesLiege May 2011 courseversion 25 May 2011
Quantitative GeneticsThe analysis of traits whose
variation is determined by botha number of genes andenvironmental factors
Phenotype is highly uninformative as tounderlying genotype
Complex (or Quantitative) trait• No (apparent) simple Mendelian basis for variation in the
trait
• May be a single gene strongly influenced by environmentalfactors
• May be the result of a number of genes of equal (ordiffering) effect
• Most likely, a combination of both multiple genes andenvironmental factors
• Example: Blood pressure, cholesterol levels– Known genetic and environmental risk factors
• Molecular traits can also be quantitative traits
– mRNA level on a microarray analysis
– Protein spot volume on a 2-D gel
Consider a specific locus influencing the trait
For this locus, mean phenotype = 0.15, whileoverall mean phenotype = 0
Goals of Quantitative Genetics
• Partition total trait variation into genetic (nature)vs. environmental (nurture) components
• Predict resemblance between relatives– If a sib has a disease/trait, what are your odds?
• Find the underlying loci contributing to geneticvariation– QTL -- quantitative trait loci
• Deduce molecular basis for genetic trait variation
• eQTLs -- expression QTLs, loci with a quantitativeinfluence on gene expression– e.g., QTLs influencing mRNA abundance on a microarray
Dichotomous (binary) traits
Presence/absence traits (such as a disease) can (and usually do) have a complex genetic basis
Consider a disease susceptibility (DS) locus underlying a disease, with alleles D and d, where allele D significantly increases your disease risk
In particular, Pr(disease | DD) = 0.5, so that thepenetrance of genotype DD is 50%
Suppose Pr(disease | Dd ) = 0.2, Pr(disease | dd) = 0.05
dd individuals can rarely display the disease, largelybecause of exposure to adverse environmental conditions
If freq(d) = 0.9, what is Prob (DD | show disease) ?
freq(disease) = 0.12*0.5 + 2*0.1*0.9*0.2 + 0.92*0.05 = 0.0815
From Bayes’ theorem, Pr(DD | disease) = Pr(disease |DD)*Pr(DD)/Prob(disease) = 0.12*0.5 / 0.0815 = 0.06 (6 %)
dd individuals can give rise to phenocopies 5% of the time,showing the disease but not as a result of carrying therisk allele
Pr(Dd | disease) = 0.442, Pr(dd | disease) = 0.497
Thus about 50% of the diseased individuals are phenocopies
Basic model of Quantitative Genetics
Basic model: P = G + E
Phenotypic value -- we will occasionallyalso use z for this value
Genotypic value
Environmental value
G = average phenotypic value for that genotypeif we are able to replicate it over the universeof environmental values, G = E[P]
Basic model of Quantitative Genetics
Basic model: P = G + E
G = average phenotypic value for that genotypeif we are able to replicate it over the universeof environmental values, G = E[P]
G x E interaction --- G values are differentacross environments. Basic model nowbecomes P = G + E + GE
Q1Q1 Q2Q1 Q2Q2
C C + a(1+k) C + 2aC C + a + d C + 2a
C -a C + d C + a
2a = G(Q2Q2) - G(Q1Q1)
d = ak =G(Q1Q2 ) - [G(Q2Q2) + G(Q1Q1) ]/2
d measures dominance, with d = 0 if the heterozygoteis exactly intermediate to the two homozygotes
k = d/a is a scaled measure of the dominance
Contribution of a locus to a trait
Example: Apolipoprotein E &Alzheimer’s
84.375.568.4Average age of onset
EEEeeeGenotype
2a = G(EE) - G(ee) = 84.3 - 68.4 --> a = 7.95
ak =d = G(Ee) - [ G(EE)+G(ee)]/2 = -0.85
k = d/a = 0.10 Only small amount of dominance
Example: Booroola (B) gene
2.662.171.48Average Litter size
BBBbbbGenotype
2a = G(BB) - G(bb) = 2.66 -1.46 --> a = 0.59
ak =d = G(Bb) - [ G(BB)+G(bb)]/2 = 0.10
k = d/a = 0.17
Fisher’s (1918) Decomposition of GOne of Fisher’s key insights was that the genotypic valueconsists of a fraction that can be passed from parent tooffspring and a fraction that cannot.
µG =!
Gij · freq(QiQj )Mean value, with
Average contribution to genotypic value for allele i
Gij = µG + !i +!j + "ij
Consider the genotypic value Gij resulting from an AiAj individual
In particular, under sexual reproduction, parents onlypass along SINGLE ALLELES to their offspring
Since parents pass along single alleles to theiroffspring, the !i (the average effect of allele i)represent these contributions
Gij = µG + !i +!j + "ij
"Gij = µG +!i + !j
The genotypic value predicted from the individualallelic effects is thus
The average effect for an allele is POPULATION-SPECIFIC, as it depends on the types and frequencies of alleles that it pairs with
Gij = µG + !i +!j + "ij
"Gij !Gij = "ij
Dominance deviations --- the difference (for genotypeAiAj) between the genotypic value predicted from thetwo single alleles and the actual genotypic value,
"Gij = µG +!i + !j
The genotypic value predicted from the individualallelic effects is thus
Gij = µG + 2!1 + (!2 ! !1)N + "ij
Gij = µG + !i +!j + "ij
Fisher’s decomposition is a Regression
Predicted valueResidual error
A notational change clearly shows this is a regression,
Independent (predictor) variable N = # of Q2 alleles
Gij = µG + 2!1 + (!2 ! !1)N + "ij
2!1 + (!2 !!1)N =
#$%$&
2!1 forN = 0, e.g, Q1Q1
!1 + !2 forN = 1, e.g, Q1Q2
2!2 forN = 2, e.g, Q2Q2
Regression slopeIntercept
2aa(1+k)0Genotypic
value
Q2Q2Q2Q1Q1Q1Genotype
Consider a diallelic locus, where p1 = freq(Q1)
µG = 2p2 a(1 + p1k)Mean
Allelic effects
!2 = p1a [ 1 +k (p1 � p2 ) ]!1 = � p2a [ 1 + k (p1 � p2 ) ]
Dominance deviations "ij = Gij ! µG! !i ! !j
Average effects and Additive Genetic Values
A (Gij ) = !i + !j
A =n!
k=1
'!(k)
i + !(k)k
(
The ! values are the average effects of an allele
A key concept is the Additive Genetic Value (A) ofan individual
A is called the Breeding value or the Additive geneticvalue
A =n!
k=1
'!(k)
i + !(k)k
(
Why all the fuss over A?
Suppose father has A = 10 and mother has A = -2for (say) blood pressure
Expected blood pressure in their offspring is (10-2)/2 = 4 units above the population mean. Offspring A =average of parental A’s
KEY: parents only pass single alleles to their offspring.Hence, they only pass along the A part of their genotypicvalue G
Genetic Variances
Gij = µg + (!i + !j ) + "ij
#2(G) =n!
k=1
#2(!(k)i + !(k)
j ) +n!
k=1
#2("(k)ij )
#2(G) = #2(µg +(!i + !j ) + "ij) = #2(!i + !j ) + #2("ij)
As Cov(!,") = 0
Genetic Variances
#2(G) =n!
k=1
#2(!(k)i + !(k)
j ) +n!
k=1
#2("(k)ij )
#2G = #2
A + #2D
Additive Genetic Variance(or simply Additive Variance)
Dominance Genetic Variance(or simply dominance variance)
Key concepts (so far)• !i = average effect of allele i
– Property of a single allele in a particular population (depends ongenetic background)
• A = Additive Genetic Value (A)
– A = sum (over all loci) of average effects
– Fraction of G that parents pass along to their offspring
– Property of an Individual in a particular population
• Var(A) = additive genetic variance– Variance in additive genetic values
– Property of a population
• Can estimate A or Var(A) without knowing any of theunderlying genetical detail (forthcoming)
#2A = 2p1 p2 a2[ 1+ k (p1 p2 ) ]2One locus, 2 alleles:
Q1Q1 Q1Q2 Q2Q2
0 a(1+k) 2a
Dominance alters additive variance
When dominance present, Additive variance is anasymmetric function of allele frequencies
#2A = 2E[!2 ] = 2
m!
i=1
!2i pi
Since E[!] = 0, Var(!) = E[(! -µa)2] = E[!2]
#2D = E["2 ] =
m!
i=1
m!
j=1
"2ij pi pj
#2D = (2p1 p2 ak)2One locus, 2 alleles:
Q1Q1 Q1Q2 Q2Q2
0 a(1+k) 2a
Equals zero if k = 0
This is a symmetric function ofallele frequencies
Dominance variance
Can also be expressed in terms of d = ak
Epistasis
Gijkl = µG + (!i + !j + !k +!l) + ("ij + "kj)+ (!!ik +!!il + !!jk + !!jl)+ (!"ikl + !"jkl + !"kij + !"lij)+ (""ijkl)
= µG + A+ D + AA + AD + DD
These components are defined to be uncorrelated,(or orthogonal), so that
#2G = #2
A + #2D + #2
AA + #2AD +#2
DD
Gijkl = µG + (!i + !j + !k +!l) + ("ij + "kj)+ (!!ik +!!il + !!jk + !!jl)+ (!"ikl + !"jkl + !"kij + !"lij)+ (""ijkl)
= µG + A + D + AA + AD + DD
Additive x Additive interactions -- !!, AA
interactions between a single alleleat one locus with a single allele at another
Additive x Dominance interactions -- !", AD
interactions between an allele at onelocus with the genotype at another, e.g.allele Ai and genotype Bkj
Dominance x dominance interaction --- "", DD
the interaction between the dominancedeviation at one locus with the dominancedeviation at another.