Lecture 3 Introduction on Quantitative Genetics: I …

1 Lecture 3 Introduction on Quantitative Genetics: IFisher s Variance DecompositionBruce Walsh. Aug 2004. Royal Veterinary and Agricultural University, DenmarkContribution of a Locus to the Phenotypic Value of a TraitThe basic model for Quantitative genetics is that thephenotypic valuePof a trait is the sum of agenetic valueGplus anenvironmental valueE,P=G+E( )The genetic valueGrepresents the average phenotypic value for that particular genotype if we wereable to replicate it over the distribution (oruniverse) of environmental values that the populationis expected to experience.

2 While it is often assumed that the genetic and environmental values areuncorrelated, this not need be the case. For example, a genetically higher-yield dairy cow mayalso be feed more, creating a positive correlation betweenGandE. In such cases the basic modelbecomesP=G+E+Cov(G;E)( )The genotypic valueGis usually the result of a number of loci that influences the trait. However,we will start by first considering the contribution of a single locus, whose alleles are allelesQ1andQ2. We need a parameterization to assign genotypic values to each of the three genotypes, and thereare three slightly different notations used in the literature:GenotypesQ1Q1Q1Q2Q2Q2CC+a(1 +k)C+2aAverage Trait Value:CC+a+dC+2aC aC+dC+aHereCis some background value, which we usually set equal to zero.

3 What matters here is thedifference2abetween the two homozygotes,a= [G(Q2Q2)-G(Q1Q1)]/2, and the relative positionof the heterozgotes compared to the average of the homozygotes. If it is exactly intermediate,d=k=0and the alleles are said to be additive. Ifd=a(or equivalentlyk=1)), then alleleQ2iscompletely dominant toQ1( ,Q1is completely recessive). Conversely, ifd= a(k= 1) thenQ1is dominant toQ2. Finally ifd>a(k>1) the locus showsoverdominancewith the heterozygotehaving a larger value than either homozygote.

4 Thusd(and equivalentlyk) measure the amount ofdominance at this locus. Note thatdandkare related byak=d;ork=da( )The reason for using bothdandkis that some expressions are simpler using one parameterizationover 3, pg. 1 Example: the Booroola (B) geneThe Booroola (B) gene influences fecundity in the Merino sheep of Australia. The mean litter sizesfor thebb,Bb, andBBgenotypes based on685total records are1:48,2:17, and2:66; these to be estimates of the genotypic values (Gbb;GBb;andGBB);the homozygous effectof theBallele is estimated bya=(2:66 1:48)=2=0:59.

5 The dominance coefficient is estimatedby taking the difference betweenbbandBbgenotypes,a(1 +k)=0:69, substitutinga=0:59, andrearranging to obtaink=0:17. This suggests slight dominance of the Booroola gene. Using thealternativednotation, from Equation 4,d=ak=0:59 0:17 = 0:10 Fisher s Decomposition of the Genotypic ValueQuantitative genetics as a field dates back to R. A. Fisher s brilliant (and essentially unreadable)1918 paper, in which he not only laid out the field of Quantitative genetics, but also introduced theterm variance and developed the analysis of variance (ANOVA).

6 Not surprisingly, his paper wasinitially had two fundamental insights. First, that parents do not pass on their entire genotypicvalue to their offspring, but rather pass along only one of the two possible alleles at each locus. Hence,only part ofGis passed on and thus we decomposeGinto component that can be passed alongand those that cannot. This insight is more fully developed below. Fisher s second great insightwas that phenotypic correlations among known relatives can be used to estimate the variances ofthe components ofG.

7 We develop this point in the next suggested that the genotypic valueGijassociated with an individual carrying aQiQjgenotype can be written in terms of theaverage effects for each allele and adominance deviation giving the deviation of the actual value for this genotype from the value predicted by the averagecontribution of each of the single alleles,Gij= G+ i+ j+ ij( )The predicted genotypic value isbGij= G+ i+ j, where Gis simply the average genotypicvalue, G=XGij freq(QiQj)Note that since we assumed the environmental values have mean zero, G= P, the mean phe-notypic value.

8 LikewiseGij bGij= ij, so that is the residual error, the difference between theactual value and that predicted from the regression. Since and represent deviations from theoverall mean, they have expected values of might notice that Equation looks like a regression. Indeed it is. Suppose we have onlytwo alleles,Q1andQ2. Notice that we can re-express Equation asGij= G+2 1+( 2 1)N+ ij( )whereNis the number of copies of alleleQ2, so that2 1+( 2 1)N=8> <>:2 1forN=0; ,Q1Q1 1+ 2forN=1; ,Q1Q22 2forN=2; ,Q2Q2( )Thus we have a regression, whereN(the number of copies of alleleQ2) is the dependentvariable, the genotypic valueGthe dependent variable,( 2 1)is the regression slope, and the ijare the residuals of the actual values from the predicted values.

9 Recall from the standard theory ofleast-squares regression that the correlation between the predicted value of a regression ( G+ i+ j)and the residual error ( ij) is zero, so that ( i; j)= ( k; j)= 3, pg. 2To obtain the , Gand values, we use the notation ofGenotypes:Q1Q1Q1Q2Q2Q2 Average Trait Value:0a(1 +k)2afrequency (HW):p212p1p2p22A little algebra gives G=2p1p2a(1 +k)+2p22a=2p2a(1 +p1k)( )Recall that the slope of a regression is simply the covariance divided by the variance of the predictorvariable, giving 2 1= (G.)

10 N2) 2(N2)=a[1+k(p1 p2) ]( )See Lynch and Walsh, Chapter 4 for the algebraic details leading to Equation Since we havechosen the to have mean value zero, it follows thatpi 1 p2 2=0 When coupled with Equation this implies (again, seeL&WChapter 4) 2=p1a[1+k(p1 p2)](3:6c) 1= p2a[1+k(p1 p2)](3:6d)Finally, the dominance deviations follow since ij=Gij G i j( )Note the important point that both and are functions of allele frequency and hence change as theallele (and/or genotype) frequencies change. While theGijvalues remain constant, their weights arefunctions of the genotype (and hence allele) frequencies.