Partitioning Genetic Variance - University of Colorado …psych.colorado.edu/~carey/Courses/PSYC5102/handouts/genetic... · PSYC 5102 Genetic Variance: 1 Partitioning Genetic Variance

PSYC 5102 Genetic Variance: 1

Partitioning Genetic Variance

Introduction

Table 1 presents the notation that will be used here for deriving genetic variances for a

single locus on a quantitative phenotype. For simplicity’s sake, only two alleles for this locus

are used, but the same substantive results would have been obtained for more than two alleles

(Crow & Kimura, 197X). Let the two alleles be denoted as A and a, giving the three genotypes

aa, Aa, and AA. Algebraic quantities will be subscripted by the genotypes. Although this

notation looks quite cumbersome, it actually makes the equations easier to read in English.

Table 1. Notation for a two allele locus.

Genotypic Values Contrast Codes

Genotype Frequency RawDeviations

from the meanDeviations

from the midpoint Linear Quadraticaa faa

γαα µ + gaa m - α -1 -1Aa fAa γAa µ + gAa m + δα 0 2AA fAA γAA µ + gAA m + α 1 -1

To be general, the model does not assume Hardy-Weinberg equilibrium. Hence, the

genotypic frequencies are simply denoted by f in place of the more familiar notation of p and q.

There are three different ways to model the genotypic or genetic values of the three

genotypes. The genotypic or genetic value for a genotype is simply the mean phenotypic value

of that genotype. The first way of modeling this is to simply let the genotypic values equal an

algebraic quantity. In Table 1, γ is used to denote the genotypic values and is given in the column

labeled “Raw”.

The second method is to express the genotypic values as a deviation from the population

mean. Here, let µ denote the population phenotypic mean and let the quantity g denote the

deviation of the genotypic value from the population mean. For example, if the population mean

is 100 and the genotypic value for genotype aa is 97.3, then gaa is 97.3 - 100 = -2.7. The

genotypic value for genotype aa is then µ + gaa. The algebraic expressions for these genotypic

means are given in the column labeled “Deviations from the mean” in Table 1.


The third way is to express the genotypic values in terms of displacements from the

midpoint between the two homozygotes. For example, if the genetic value for aa is 97.3 and the

genetic value for AA is 104.9, then the midpoint is simply the average of these two quantities or

101.1. Let m denote this midpoint. Then the genetic value for genotype AA may be written as m

+ α and that for aa may be written as m - α, where α is the difference between the genotypic

value and the midpoint. The genotypic value for the heterozygote may be written as m + δα

where δ denotes a dominance parameter. When δ = 1, then allele A is completely dominant and

when δ = -1, allele a is completely dominant. When δ = 0, there is no dominance and when δ > 1

or δ < -1 there is overdominance1. These algebraic quantities are given in the column “Deviations

from the midpoint” in Table 1.

Because these three different parameterizations are effectively “saying the same thing but

with different words,” the three different quantities for a single genotype will be mathematically

identical. For example, for genotype aa,

γ aa = µ + gaa = m −α .

Phenotypic Mean

The phenotypic mean equals a weighted mean of the three genotypic means, the weight in

this case being the frequency of the genotypes. Thus, using the first parameterzation,

µ = faaγ aa + fAaγ Aa + fAAγ AA .

For the second parameterization,

µ = faa (µ + gaa ) + fAa (µ + gAa ) + fAa (µ + gAa )= ( faa + fAa + fAA )µ + faagaa + fAa gAa + fAA gAA

Now, the sum of the frequencies of the genotypes must be 1.0, so faa + f Aa + fAA = 1.0 .

Also, it is a mathematical necessity that the sum of the deviations from the mean must equal 0, so

faa gaa + fAag Aa + fAAg AA = 0 . Substituting these quantities into the equation gives the

identity µ = µ.

For the third parameterization,

1 This parameterization is not applicable for the extremely unlikely case where the genotypic values for bothhomozygotes are equal but different from the value of the heterozygote. For this case simply let δ be expressed inabolute units (instead of a fraction of α) so that the genotypic value for the heterozygote is m + δ.


µ = faa (m −α ) + fAa (m + δα) + f AA (m + α)which reduces to

µ = m +α( fAA − faa + fAaδ ).

Phenotypic Variance

The equation for the phenotypic value for the ith person with the jth genotype (or Pij)

may be written in a general form as

Pij = γ j + Rij .

where Rij denotes a residual deviation from the population mean. This residual value will include

all environmental factors as well as the influence of all loci other than the A locus. For example,

the phenotypic value for the ith person with genotype aa will be

Pi.aa = γ aa + Ri.aa = µ + gaa + Ri.aa = m −α + Ri.aa .

In English, this equation states that that an individual’s phenotypic value equals the genotypic

mean for his/hers genotype plus the effects of “all other genetic and environmental factors.” At

this point, we note that this model assumes no gene-gene interaction (aka epistasis) involving the

A locus and no gene-environment interaction involving the A locus. (There may indeed be

epistasis and/or gene-environment interaction involving other loci; this assumption applies only

to the A locus.) We also introduce a second assumption of no covariance between the genotypic

values at the A locus and any other genes and environments. Because of this assumption, the

mean value for R for each of the three genotypes will be 0.

Just as the phenotypic mean was the weighted sum of the genotypic means, the

phenotypic variance will equal the weighted squared deviations from the mean. To see how this

is so, we first write the phenotypic variance in summation notation,


VP =1N

(µ + gaa + Ri.aa − µ)2i= 1

Naa

∑ + (µ + gAa + Ri.Aa − µ)2i =1

NAa

∑ + (µ + gAA + Ri.AA − µ)2i=1

NAA

∑

=

1N

(gaa + Ri.aa )2

i=1

Naa

∑ + (gAa + Ri.Aa )2

i =1

NAa

∑ + (gAA + Ri.AA)2

i =1

NAA

∑

=

1N

(gaa2 + 2gaaRi.aa + Ri.aa

2 )i =1

Naa

∑ + (gAa2 + 2gAaRi.Aa + Ri.Aa

2 )i =1

NAa

∑ + (gAA2 + 2gAARi.AA + Ri.AA

2 )i =1

NAA

∑

Now expressions such as gAa2

i=1

NAa

∑ will equal NAa gAa2

because the constant g Aa2

is simply being

summed NAa times. Quantities such as 2gAaRi.Aai =1

NAa

∑ = 2gAa Ri.Aai =1

NAa

∑ = 0 . This occurs because

the mean of the Rs for each genotype equal 0, so the sum of the Rs, or Ri.Aai =1

NAa

∑ for example,

must also equal 0. Substituting these quantities into the above equation gives

VP =1N

Naagaa2 + NAagaa2 + NAAgAA2 + Ri.aa2 +i =1

Naa

∑ Ri.Aa2 +i= 1

NAa

∑ Ri.AA2

i =1

NAA

∑

=

Naa

Ngaa2 +

NAa

NgAa2 +

NAA

NgAA2 +

Ri.aa2 +

i =1

Naa

∑ Ri.Aa2 +

i=1

NAa

∑ Ri.AA2

i =1

NAA

∑

N=

faagaa2 + fAagAa

2 + fAAgAA2 +

Rij2

i=1

Nj

∑j =aa

AA∑

NThe quantity faa gaa

2 + fAag Aa2 + fAAg AA

2 can be shown to equal the variance of the

genotypic values, which will be denoted here as VG. Also, the quantity

Rij2

i=1

Nj

∑j =aa

AA∑

N equals the

variance of the residuals, or say VR. Thus, the phenotypic variance is the sum of two variances--

the variance of the genotypic values (or as it is most commonly called, the total genetic variance)

and the variance of the residuals,

VP = VG + VR .


This result is almost intuitive to those who have had quantitative genetics. The purpose

of this exercise was not to demonstrate the obvious, but to demonstrate the techniques whereby

many further equations may be derived.2

Additive genetic variance

For a single locus, the total genetic variance is partitioned into two types of variance, the

additive genetic variance and dominance variance. Here we give the derivation for additive genetic

variance. We begin by noting the orthogonal contrast codes for the three genotypes at the right

hand side of table 1. There are two of these, a linear contrast code of -1, 0, and 1 for genotypes

aa, Aa, and AA, respectively, and a quadratic contrast code of -1, 2, and -1. The variance

associated with the linear contrast code is the additive genetic variance. To find out the algebraic

formula for this variance, we use a simple linear regression to regress the phenotypic values on

the contrast codes. The general equation for this is

Pij = a + bXlj + Uij

where Pij, as before, is the phenotypic value for the ith person with the jth genotype, X l jis the

value of the contrast code for the jth genotype, Uij is a residual, and a and b are respectively the

intercept and slope for the regression line. The equations for the two regression parameters are

b =cov(P,Xl )

VXl

and

a = µ − bX l .

The additive genetic variance is the variance associated with the slope of the regression line or

VA = b2VXl

=cov(P, Xl )2

VXl

.

To derive this quantity, it is necessary first to obtain expressions for the mean and

variance of variance Xl, the linear contrast codes. As before, we find the mean as a weighted sum

of the genotypic means, or

2 The summation notation will always give the correct result, but it is much more cumbersome than usingmathematical expectations. Students who wish to pursue this topic are urged to express the algebra in terms ofexpectaions.


X l = faa (−1) + fAa (0) + fAA (1) = f AA − faa1 .

The variance equals a weighted sum of the squared deviations of the contrast codes from the

contrast code mean,

VXl= faa (−1− X l )

2 + fAa (0 − X l )2 + fAA (1− X l )

2

which reduces to

VXl = faa + fAA − ( faa − fAA )2.

The final quantity is the covariance between the phenotypic values and the contrast

codes. First, write the phenotypic values according to the linear model given in Equation X.X.

The calculation of the covariance begins with multiplying the deviation of an individual’s

phenotypic value from the phenotypic mean (or Pij − µ ) by the deviation of the individual’s

contrast code from the contrast code mean (or Xl j− X l ). Once this has been done for all

individuals, these “cross products” are then summed over all individuals. Dividing by the total

number of individuals gives the covariance. The algebraic formula, expressed in summation

notation is

cov(P, Xl ) =1N

(µ + gaa + Ri.aa − µ)(−1 − X l )i=1

Naa

∑ + (µ + gAa + Ri.Aa − µ)(0 − X l )i =1

N Aa

∑

+ (µ + gAA + Ri.AA − µ)(1− X l )i=1

N AA

∑

which reduces (mercifully) to

cov(P, Xl ) = fAAgAA − faa gaa .

Substituting this expression into that for the slope gives

b =cov(P,Xl )

VXl

=fAAgAA − faagaa

faa + fAA − ( faa − fAA )2.

Hence, the additive genetic variance equals

VA = b2VXl

=( fAA gAA − faa gaa )2

faa + fAA − ( faa − fAA )2.

The numerator for this expression is important. In English, it equals the square of the

weighted difference between the two heterozygote means. Hence, even if the two homozygotes


had identical genetic values (admittedly, an implausible case), there could still be additive genetic

variance. A more reasonable situation where additive genetic variance is small is when gAA and gaa

are of the same sign and their respective frequencies are such that fAAgAA almost equals

faa gaa . This is the classic situation of overdominance where the heterozygote genotypic value

is much greater than (or much less than) the average value of the two homozygotes.

The final, and indeed most important case of small additive genetic variance occurs with a

rare recessive gene. For the sake of exposition, assume that aa is the recessive genotype.

Because aa is very rare, the quantity faa will be very small, so the term faa gaa will be quite

small. On the other hand, the population mean will be very close to the genotypic value of AA,

making the difference between the population mean and the genetic value of AA--i.e., the quantity

gAA--will be very small. This makes the expression fAAgAA small. Consequently, the numerator

in Equation X.X will be tiny and there will be little additive genetic variance.

Dominance Variance

The term “dominance” variance is unfortunate because it is often misinterpreted as

dominant transmission of a trait. We shall see that a rare dominant allele actually has very little

dominance variance. A better term would be something akin to “nonadditive main effect

variance,” but the usage of dominance variance is so widespread that custom dictates its use here.

Dominance variance is literally the difference between the total genetic variance and the

additive genetic variance. In terms of a regression model, one would estimate dominance variance

as the explanatory variance gained after entering the quadratic term into the model. That is, one

would perform two regressions. In the first, one would enter only the linear contrast. In the

second, one would enter both the linear and the quadratic contrasts. The R2 (i.e., multiple

correlation squared) from the first model is the additive genetic heritability for the locus. The R2

from the second model is the total heritability. The dominance heritability is simply the

difference between the two R2s.

Tedious algebra shows that the dominance variance will equal

faa fAa fAA (2gAa − gaa − g AA )2

faa + fAA − ( fAA − faa )2.


The numerator for this expression reveals what dominance variance is. The important term is

(2g Aa − gaa − gAA )2

. Because this is squared (and because the expression involving all the fs

must be positive), the result must always be greater than or equal to 0. There will be no

dominance variance when gaa + gAA = 2gAa. This will occur only when the genetic value for the

heterozygote is exactly midway between the genetic values of the two homozygotes.

It is instructive to view the relationship between additive and dominance variance with

dominant and recessive alleles. Let allele A be the dominant allele and let the genotypic values for

aa, Aa, and AA be respectively 0, 1 and 1. Then the phenotypic mean is simply fAa + fAA, and the

genotypic values expressed as deviations from the mean become faa - 1, faa and f aa . A bit of

algebra reveals the numerator for VA as faa2 ( fAa + 2 f AA )

2 and the numerator for VD as

faa fAa fAA . Hence, the ratio of additive to dominance variance is

VAVD

=faa2 ( fAa + 2 fAA )2

faa fAa fAA.

At this point, it will be convenient to express the genotypic frequencies in terms of Hardy-

Weinberg frequencies. Let p denote the frequency of the dominant allele A and q = 1 - p, the

frequency of allele a. Thus, faa = q2, fAa = 2pq, and fAA = p2. Then substitution and algebraic

reduction of the equation gives

VAVD

=qp .

This equation shows that when there is complete dominance or recessivity, then the ratio

of additive to dominance variance depends only on the allele frequencies! When there is a rare

dominant, p is very small, q must be large, so the ratio is very large. Hence, a rare dominant gives

a large amount of additive genetic variance. When the locus is a rare recessive which, of course, is

the same as a common dominant, then p is large, q is small, and the ratio is very small. Hence, a

rare recessive will have little additive genetic variance but large dominance variance.

Relationship between additive and dominance variance

The relationship between additive and dominance variance is depicted in Figure 1. The

solid squares give the genetic values for the three genotypes. The straight line represents the line


of best fit when regressing these genotypic values upon the linear contrast codes. The variance

associated with this straight line is the additive genetic variance.

The deviations of the actual genotypic values from their values predicted on the values of

the regression line are depicted by the double headed arrows. These are prediction errors from

the simple additive model. The variance associated with these prediction errors are the

dominance variance. Literally, computation of dominance variance would begin by measuring the

length of a double headed arrow. Then square that length and then multiply this squared length

by the frequency of the genotype. Summing these “weighted square lengths” over the three

genotypes gives the dominance variance.

Figure 1: Additivity and Dominance

aa Aa AA

Genotype

Genotypic

Value

Epistasis

Epistasis occurs when genes and/or gene products interact and epistatic variance is the

statistical interaction variance. It is important to emphasis the term statistical in this definition.

It is entirely possible for biochemical products of loci to physically interact but this does not

necessarily lead to a statistical interaction3. Classic examples of epistasis for behavior can be

seen in many rare genetic disorders. A Tay-Sachs genotype, for example, interacts with those

loci that contribute to individual differences in normal cognitive development during infancy. If

an infant has Tay-Sachs disease, the expression of these normal loci is inhibited. For those

without Tay-Sachs disease, the other loci will be expressed. However, epistatic variance for the

Tay-Sachs locus will be very small because the disorder is very rare.

3 The same must be said of the gene-environment interaction. In casual discourse, this term often implies that bothgenes and environment are important for behavior. In quantitative genetics, however, the term implies a statisticalinteraction. Hence, when heritability is less than 1.0, there is always gene-environment interaction in the loosesense, but there may be no gene-environment interaction in the strict sense.


Statistical epistasis has the same meaning as the statistical interaction in ANOVA or

regression. We can think of the linear and the dominance terms as the main effects of an

ANOVA. With two loci or more loci, epistatic variance is equivalent to the interaction terms in

ANOVA. Visually, epistatic variance occurs when the regression lines for one locus are not

parallel when they are plotted as a function of another locus. Figures 2 and 3 illustrate cases of,

respectively, no epistatic variance and epistatic variance. In Figure 2, the regression lines for

genotypes bb, Bb, and BB are parallel; hence there is no interaction. In Figure 3, however, the A

locus has no influence on individual differences in the presence of genotype bb. With genotype

Bb, aa is recessive but with genotype BB, the A locus is totally additive. The fact that the three

lines for genotypes at the B locus are not parallel indicates a statistical interaction between the A

and the B locus.

Figure 2: No epistasis

aa Aa AA

Genotype: A locus

Gen

otyp

ic V

alue bb

BbBB

Figure 2: Epistasis

aa Aa AA

Genotype: A locus

Gen

otyp

ic V

alue

bbBbBB

The total epistatic genetic variance is subdivided into several statistical components. To

examine the statistical components, let us return to the contrast codes for additive and dominance

effects as they would apply to two loci, the W and the B locus. These are given in the left hand

columns of Table 2 under the label “Main Effects.” In actually performing an analysis of this

type, one would regress the phenotype on the contrast codes. The first regression would enter

the additive codes for both locus W and locus B at the same time, i.e. a multiple regression with

Add.W and Add.B as the independent variables. The R2 for this regression gives the additive

genetic variance for the trait. The second regression would add the two dominance contrast

codes, Dom.W and Dom.B. The R2 from this less the initial R2 would give the dominance

variance.


Table 2. Contrast codes for two loci.

Main Effects Interactions

Additive DominanceAdditive*Additive

Additive*Domiance

Dominance*Dominance

Geno-type

Add.W

Add.B

DomW

DomB

Add.W*Add.B

Add.W*Dom.B

Dom.W*Add.B

Dom.W*Dom.B

WWBB 1 1 -1 -1 1 -1 -1 1WWBb 1 0 -1 2 0 2 0 -2WWbb 1 -1 -1 -1 -1 -1 1 1WwBB 0 1 2 -1 0 0 2 -2WwBb 0 0 2 2 0 0 0 4Wwbb 0 -1 2 -1 0 0 -2 -2wwBB -1 1 -1 -1 -1 1 -1 1wwBb -1 0 -1 2 0 -2 0 -2wwbb -1 -1 -1 -1 1 1 1 1

The epistatic or interaction variance is literally the results of multiplying the additive and

dominance contrast codes for the W locus with those for the B locus. Multiplying the additive

code for W (Add.W) with that for B (Add.B) gives the contrast code for the first epistatic

component, additive by additive epistasis. The variance associated with this contrast code is

called additive by additive epistatic variance and is usually abbreviated as VAA. On would

estimate this component by entering the contrast code into the regression equation. There are

now five independent variables (Add.W, Add.B, Dom.W, Dom.B, and Add.W*Add.B). The R2

for this less the R2 for the model containing dominance gives the estimate of VAA.

The second epistatic component involves multiplying the additive contrast codes for one

locus and the dominance codes for the other locus. There are two ways of doing this. The first

way is to multiply Add.W with Dom.B, and the second is to multiply Dom.W with Add.B.

Entering both of these contrast codes into the last regression equation and subtracting the R2 from

the previous R2 gives what is called the additive by dominance epistatic variance, or VAD.

The final epistatic component is VDD or dominance by dominance epistasis. This is

estimated by multiplying the two dominance contrast codes together, entering the resulting

contrast code into the last regression equation, and subtracting the R2s.


Epistatic components for additional loci will be formed in a similar way, but there will be

more epistatic components. For example, let us add the C locus to the above problem. There

would now be three additive by additive contrast codes, Add.W*Add.B, Add.W*Add.C, and

Add.B*Add.C. Entering these three contrast codes simultaneously into the regression and

subtracting the R2s would now give the estimate of VAA, the additive by additive epistatic

variance. Following this logic would give 6 contrast codes for estimating VAD (or all the two way

interactions between the additive and dominance contrast codes), and 3 for VD (or all the two way

interactions among the dominance contrast codes.

With the three locus case, however, there is also the possibility of three way interactions

among the loci. Just as the two way interactions were subdivided into components, so are the

three way epistatic interactions subdivided into individual components that reflect the products

of the additive and dominance contrast codes. The first of these would be VAAA or the additive by

additive by additive epistatic variance. The contrast code for this is simply

Add.W*Add.B*Add.C, or the product of the three additive contrast codes. The second

component would be VAAD (additive by additive by dominance epistatic variance), the next

component would be VADD (additive by dominance by dominance epistatic variance), and the final

three way interaction would be VDDD (the dominance by dominance by dominance epistatic

variance). Once again the contrast codes may be found by multiply all the relevant additive and

dominance main effects contrast codes, and the variance components would be estimated by

hierarchical multiple regression.

The total epistatic variance for the three locus case is simply the addition of all the

individual components of variance. Let VI denote the total epistatic variance. Then,

VI = VAA + VAD + VDD + VAAA +VAAD + VADD + VDDD .

Additional loci may be accommodated using identical logic. With n loci, the variance

component VAA is simply the sum of all two way additive by additive interactions among all n

loci, VAAD is the sum of all possible additive by additive by dominance interactions among n loci,

and so on.

Partitioning Genetic Variance - University of Colorado …psych.colorado.edu/~carey/Courses/PSYC5102/handouts/genetic... · PSYC 5102 Genetic Variance: 1 Partitioning Genetic Variance

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