Analysis of Heritability

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nalysis of HeritabilityBrent D. Burch a; Ian R. Harris b

a Department of Mathematics and Statistics, Northern Arizona University, Flagstaff, Arizona, U.S.A. b

Department of Statistical Science, Southern Methodist University, Dallas, Texas, U.S.A.

Online Publication Date: 23 April 2003

To cite this Section Burch, Brent D. and Harris, Ian R.(2003)'Analysis of Heritability',Encyclopedia of BiopharmaceuticalStatistics,1:1,36 — 41

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Analysis of Heritability

Brent D. BurchNorthern Arizona University, Flagstaff, Arizona, U.S.A.

Ian R. HarrisSouthern Methodist University, Dallas, Texas, U.S.A.

INTRODUCTION

Statistical methods applied to the field of quantitative

genetics provide insight concerning the role that nature

and nurture play in the development of life forms.

Physical traits of organisms depend on genetic as well

as environmental influences and quantifying these effectsare major concerns in the study of plants, animals, and

human beings. The heritability of a measurable char-

acteristic is the most commonly used statistic for ex-

pressing the degree to which genetic material is trans-

ferred from parent to progeny. Heritability is defined

in terms of the variation in the trait under study and

relies on the partitioning of this variation into compo-

nents that are attributable to genetic and environmental

sources. Variance components in mixed linear models

are used to measure how different sources contribute to

the variation of a characteristic.

Statistical models serve as the basis for analyzingheritability. The analysis of heritability is accomplished

by measuring the trait of interest, and other important

variables, on a collection of subjects. In addition, the re-

lationships among the subjects are recorded and used

in the statistical model. These relationships provide in-

formation about the pedigree structure or lineage of the

subjects. Point estimators, confidence intervals, and hy-

pothesis tests are some of the statistical inferential pro-

cedures employed to analyze heritability.

HERITABILITY

The phenotype of an individual is a quantitative cha-

racteristic or trait that is the result of combined influen-

ces of genetic and nongenetic effects. In equation form,

P ¼ G þ E ð1Þ

where P is the phenotypic value, G is the genotypic value

which represents the effect of the set of genes that

contribute to the phenotypic value, and E is the

environmental (nongenetic) effect on the phenotypic

value. The genotypic value (G) is further partitioned into

distinct genetic effects. Specifically,

G ¼ A þ D þ I ð2Þ

where A is the additive genetic effect, D is the do-

minance genetic effect, and I is the epistatic or in-

teraction genetic effect. The additive genetic effect ( A)

represents that part of G which is transferable from one

generation to the next. In some applications, A is referred

to as the breeding value of the individual because it

is the primary factor in determining the resemblance

between relatives. The dominance genetic effect ( D) is a

result of the dominance of alleles at a locus. In other

words, D represents the interaction of alleles within a

locus or within-locus interactions. The epistatic genetic

effect ( I ) accounts for genotypic values that depend on

more than one locus. Genes are said to be epistatic if the effects of the different loci are not independent. The

overall genotypic value for an individual is determined

by the sum of the effects of the individual loci.

Heritability is defined in terms of the variances

associated with the genetic influences on the phenotypic

value. If Var(G) and Var(P) are the variances of the

genotypic and phenotypic values, respectively, then

heritability in the broad sense is defined as

H 2 ¼ VarðGÞ

VarðPÞ ð3Þ

H 2 is the proportion of the total phenotypic variance

attributable to genetic effects. H 2 is sometimes referred to

as the coefficient of genetic determination. The more

commonly used form of heritability is known as he-

ritability in the narrow sense. If Var( A) is the variance

of the additive genetic effect, then heritability in the

narrow sense is defined as

h2 ¼ Varð AÞ

VarðPÞ ð4Þ

36 Encyclopedia of Biopharmaceutical Statistics

DOI: 10.1081/E-EBS 120007612

Copyright D 2003 by Marcel Dekker, Inc. All rights reserved.



and is the proportion of the total phenotypic variance

resulting from additive genetic effects. The coefficient

h2 measures the importance of transmissible genetic

material because it quantifies to what extent the varia-

tion in the phenotype of a progeny is determined by

the genes transmitted from its parents. See Refs. [1–4]for a thorough discussion of heritability and quantitat-

ive genetics.

In the analysis of h2, the primary source of variation

under study is that which results from additive genetic

effects. The other sources of variation (environmental,

dominance, epistatic) play a secondary role and are often

combined into a single term. In this manner

P ¼ A þ ð D þ I þ E Þ

¼ A þ Other ð5Þ

so that

VarðPÞ ¼ Varð AÞ þ VarðOtherÞ ð6Þ

with the understanding that additive genetic effects and

the other effects combined are treated as uncorrelated. In

most applications, h2 is written in terms of variance

components. Variance components correspond to the

sources that produce variation in the phenotypic value. If

Varð AÞ ¼ s21 ð7Þ

and

VarðOtherÞ ¼ s22 ð8Þ

it follows that heritability (in the narrow sense) is a

function of s1

2 and s2

2, say, h2 = f (s1

2,s2

2 ), where

h2 ¼ s2

1

s21 þ s2

2

ð9Þ

This form of h2 enables the investigator to utilize

statistical models to analyze heritability. See Refs. [5–7]

for further details concerning statistical models involving

genetic variances.

STATISTICAL MODELING OF HERITABILITY

The statistical models used to analyze heritability contain

the effects that influence the quantitative characteristic

under study. Effects have different values of interest

called levels. If all levels of an effect are included in a

model, or only those particular levels of an effect included

in the model are of interest, the effect is called a fixed

effect. Gender (Male, Female) and smoking status (Yes,

No) are examples of fixed effects. If the levels of an effect

included in the model may be considered as a random

sample of all possible levels, the effect is called a random

effect. The additive genetic effect is considered random if

the subjects in the study are a random sample from a

population of subjects. In this scenario, the focus of attention is not on the specific levels of the effect but on

the underlying distribution of the random effect. In

particular, the variance of the random effect, for example,

the variance of the additive genetic effect, is used to

measure the contribution of the effect on the overall

variation of the characteristic under study.

Mixed linear models, models that include a linear

combination of fixed and random effects, are used in the

analysis of heritability. Using matrix notation,

Y ¼ X þ Zu þ eee ð10Þ

where Y is the vector of measured characteristics, is thevector of parameters associated with the fixed effects, u is

the vector of unobservable additive genetic effects, and eee

is the vector of unobservable random variables corres-

ponding to ‘‘other’’ effects as given in Eq. 5. For a

particular individual, Y i is the phenotypic value, xi’ is the

mean phenotypic value for members of the population

associated with xi where xi is the ith row of X , ui is the

additive genetic effect, and eeei represents the combined

effect of sources of variation in the phenotypic value not

accounted for by the additive genetic effect.

For the purpose of statistical inference, it is assumed

that u and eee are independent and multivariate normallydistributed. Specifically, u N (0,s1

2 R) and eee N (0,s2

2 I)

where the matrix R is referred to as the relationship

matrix and I is the identity matrix. It follows that

Y N ( X ,V ) where

V ¼ s22I þ s2

1 ZR Z0 ð11Þ

The matrix R describes the degree to which the subjects in

the study are genetically related.

Consider a simple example of the modeling process.

Suppose the phenotypic values of six subjects have been

recorded. The gender (M,F) and the relationships amongthe six subjects are displayed in Fig. 1. For instance,

subject 1 is a male having a son (subject 3), a daughter

Fig. 1 Pedigree structure.

Analysis of Heritability 37

A



(subject 4), and a grandson (subject 6). Note that the

parents of subjects 1 and 2 have not been recorded, and

subjects 3, 4, 5, and 6 have only one identifiable parent

each. Based on Fig. 1, one can also see that subjects 1 and

2 are unrelated and thus the offspring of subject 1 will be

unrelated to the offspring of subject 2.Let Y i be the phenotypic value of the ith subject,

i = 1,. . .,6. Then Y ’ = [Y 1,Y 2,Y 3,Y 4,Y 5,Y 6] is the vector

of responses for the subjects. The mixed linear model for

this example may be expressed as

Y 1Y 2Y 3Y 4Y 5Y 6

2666664

3777775 ¼

1 00 11 00 11 01 0

2666664

3777775

b1

b2

þ

1 0 0 0 0 01 0 0 0 0

1 0 0 01 0 0

1 01

26666643777775

u1u2

u3

u4

u5

u6

266664377775þ

12

3

4

5

6

266664377775 ð12Þ

The influence of the fixed effect gender on the

responses is taken into account by the 2 1 vector ,

where the first element of corresponds to males and the

second element of corresponds to females. The

influence of additive genetic effects on the responses is

taken into account by the vector 6 1 vector u. The

additive genetic effect is a random effect in this example.From Fig. 1, subjects 1, 3, 4, and 6 are related and

subjects 2 and 5 are related. The relationship matrix R for

the example is

R ¼

1 0 1=2 1=2 0 1=4

1 0 0 1=2 0

1 1=4 0 1=2

1 0 1=81 0

1

2666664

3777775

ð13Þ

Because subject 3 receives one-half of its genetic materialfrom subject 1 (the other half coming from an unidentified

source), the genetic covariance between subjects 1 and 3,

denoted by Cov(u1,u3), is 1/2s1

2. Similarly, because

subjects 3 and 4 are half-sibs, that is, they have only

one parent in common, Cov(u3,u4) = 1 / 4s1

2. Note, for

instance, that Cov(u1,u5) = 0 as subjects 1 and 5 do not

share any common genetic material. A general proce-

dure used to construct the relationship matrix is given in

Ref. [8].

Heritability of the characteristic under study is based

on the variance components in the statistical model. It

is assumed that s1

2 0 and s2

2 > 0. This setup allows

for the possibility that the genetic influence on the varia-

bility in the phenotypic value may be negligible. It fol-

lows, from Eq. 9, that 0 h2 < 1. In the following three

sections, the analysis of heritability is considered by

applying the statistical inferential procedures of pointestimation, interval estimation, and hypothesis testing.

POINT ESTIMATION

The point estimation techniques used for h2 are essentially

the same as those used to estimate the individual variance

components. That is, if bs21 and bs2

2 are the estimates of s1

2

and s2

2, respectively, then bh2 ¼ f ð bs21; bs2

2Þ, where

bh2 ¼ bs2

1 bs21 þ bs2

2

ð14Þ

is the estimate of h2. In this section, we discuss estimators

based on analysis of variance (ANOVA), maximum

likelihood (ML), and restricted maximum likelihood

(REML) procedures. Because ML estimators are trans-

lation invariant, the properties of the ML estimators of h2

carry over from the properties of the ML estimators of s1

2

and s2

2. In addition to the estimators considered here,

alternative estimators of h2 may be of interest.

Analysis of Variance Estimation

The ANOVA estimator of h2 may be obtained from the

ANOVA table associated with the model. The procedure

involves equating the sums of squares to their expected

values and solving for the estimates of the resulting

variance components. In general, if Bi is a symmetric

matrix having the same dimension as the length of the Y

vector, then the expected value of Y ’BiY is

E ðY 0BiY Þ ¼ trðBiV Þ þ 0 X 0Bi X ð15Þ

tr(BiV ) is the trace of the matrix BiV and is a linear

combination of the variance components. The expectation

will be free of the fixed effects if ’ X ’Bi X is zero.

There exists a B i such that

E ðY 0BiY Þ ¼ r is22 þ r ilis

21 ð16Þ

where 0 l1 < < ld are the distinct eigenvalues,

each having mulitiplicity r i, of a matrix related to the Bi’s.

See Ref. [9] for more details. Under normal theory the

quadratic forms, Y ’BiY , more simply written as Qi, are

38 Analysis of Heritability



independently distributed as (s2

2 +lis1

2)w2 variates with

degrees of freedom r i.

In many applications l1 = 0, which implies that re-

plications exist in the experiment. The ANOVA table as-

sociated with this scenario is given in Table 1.

For those applications in which l1 > 0, there is not a

simple estimator of s2

2. In either case, the ANOVA es-

timator of h2 takes the form of

bh2 ¼

Xk

i ¼ 1

r iXd

i ¼ k þ1

Qi Xd

i ¼ k þ1

r iXk

i ¼ 1

Qi

Xd

i ¼ k þ1

r iðli 1ÞXk

i ¼ 1

Qi Xk

i ¼ 1

r iðli 1ÞXd

i ¼ k þ1

Qi

ð17Þ

where the choice of k represents the partitioning of

Q1,. . .,Qd into two distinct groups and lies at the

discretion of the investigator. Only for the case when

d = 2 is the ANOVA estimator of h2 unique. This occurs

when the model is based on balanced data. Note that

0 h2 < 1 whereas bh2 may be negative.

Maximum Likelihood and Restricted

Maximum Likelihood Estimation

The ML and REML procedures estimate variance com-

ponents, and hence heritability, by determining the values

of the variance components that maximize a likelihood

function based on the response vector Y . ML methodsrequire the investigator to specify the distribution of the

responses and, in many applications, it is assumed that Y

is multivariate normally distributed. The ML estimation

procedure uses a likelihood function that is based on the

original responses and produces estimates of in addi-

tion to estimates of s1

2 and s2

2. The logarithm of the like-

lihood function, which is the function that is usually

maximized to obtain the ML estimators, is often denoted

by L ( ,s1

2,s2

2|Y ). The ML estimate of heritability may

be obtained by first estimating the variance components

and then using Eq. 14.

The REML estimation procedure uses a likelihood

function that is based on a transformation of the res-

ponses. The distribution of the transformed responses,

also referred to as error contrasts, does not depend on the

fixed effects part of the model in Eq. 10. The logarithm of

the likelihood function for REML estimation can bewritten as L (s1

2,s2

2|Q) where Q is the vector (Q1,. . .,Qd ).

The elements of Q represent the squares of the error

contrasts. It follows that the REML procedure is used

to estimate the variance components only. By applying

Eq. 14, one obtains the REML estimate of heritability.

See the Mixed Effects Model entry for more information

concerning ML procedures in mixed linear models.

By construction, both the ML methods result in

estimates of heritability that are in the parameter space.

That is, 0 bh2ML < 1 and 0 bh2

REML < 1. Whereas

the REML estimation procedure accounts for the degrees

of freedom that are associated with estimating fixed

effects, the ML estimation procedure does not. For all but

the simplest of models, closed-form expressions for the

ML and REML estimators do not exist and iterative

procedures are required to compute the estimates. It can

be shown that the REML estimator of h2 is

bh2REML ¼

Xd

i ¼ 1

aiðh2; li; r iÞQi

Xd

i ¼ 1

biðh2; li; r iÞQi

ð18Þ

where ai and bi are functions of h2, li, and r i thatserve as coefficients of Qi. In essence, the REML

estimator of h2 is a ratio of linear combinations of the

quadratic forms Q1,. . .,Qd whose coefficients depend in

part on the unknown parameter. The value of bh2REML

must be obtained iteratively by selecting a starting

value and relying on the convergence of the procedure.

See Refs. [7,10,11] for general information about ML

estimation procedures.

INTERVAL ESTIMATION

Confidence intervals for variance components and for h2

in particular are not typically symmetric. That is, the point

estimate of h2 does not necessarily lie at the center of the

interval. Intervals are obtained by inverting a quantity that

is a function of the responses and h2 whose distribution

is independent of h2. Specifically, if Y is multivariate

normally distributed, then the elements of Q are chi-

squared distributed and are independent as discussed in

the subsection ‘‘Analysis of Variance Estimation.’’

Because the ratio of independently chi-squared variates

Table 1 ANOVA table

Source df SS E(SS)

Model (free of )

Xd

i¼2

r i

Xd

i¼2

Qi s22

Xd

i¼2

r i þ s21

Xd

i¼2

r ili

Error r 1 Q1 r 1s2

2


A



give rise to F -distributed quantities, it follows that a

100(1 a)% confidence interval for h2 can be computed

by recognizing that

P F a1 X

d

i ¼ k þ1

Qi

1 þ h2ðli 1Þ= X

d

i ¼ k þ1

r i

Xk

i ¼ 1

Qi

1 þ h2ðli 1Þ=Xk

i ¼ 1

r i

F a2

0BBBB@ 1CCCCA¼ 1 a

ð19Þ

where a1 +a2 = a and F a1, F

a2 represent the a1 and a2

percentiles of the F -distribution having numerator and

denominator degrees of freedom equal toP

i = k + 1

d r iand

Pi = 1

k r i , respectively. By rewriting the joint in-

equality in the probability statement in Eq. 19 so that

h2

alone appears in the center, one can obtain a con-fidence interval for h2. If d > 2, this can only be ac-

complished using numerical techniques as no closed-

form expression for the endpoints of the confidence

interval exists. As previous discussions suggested, the

choice of k is up to the investigator. Additional details

are presented in Ref. [12].

HYPOTHESIS TESTS

Hypothesis tests for h2 are useful if the investigator

wants to determine if the sample information is con-sistent with a specific value of heritability under study.

The hypothesis test developed in Ref. [9] is equivalent to

the confidence interval given in the section ‘‘Interval

Estimation.’’ To conduct the hypothesis test H 0: h2 =

ho

2 vs. the two-sided hypothesis H 1: h2 6¼ ho

2, define the

test statistic to be

F * ¼

Xd

i ¼ k þ1

Qi

1 þ h2oðli 1Þ

=Xd

i ¼ k þ1

r i

Xk

i ¼ 1

Qi

1 þ h2

oðli 1Þ

=

Xk

i ¼ 1

r i

ð20Þ

Under H 0, F * is F -distributed with numerator and

denominator degrees of freedom equal toP

i = k + 1

d r i

andP

i = 1

k r i, respectively. For a prespecified value

of a, the type I error probability, reject H 0 if

F * < F a1 or F * > F

a2 and do not reject H 0 if F a1

F * F a2. One-sided tests may also be conducted using

a similar procedure.

The locally most powerful (LMP) test has the property

of being more powerful than any other test for values of

h2 that are close to h0

2. By definition, the LMP test is a

test whose power function has maximum slope at

h2 = h0

2. The test statistic for the LMP test is

T LMP ¼

ð1 h2oÞ

Xd

i ¼ 1

liQi

ð1 þ h2

oðl

i 1ÞÞ2

Xd

i ¼ 1

Qi

1 þ h2oðli 1Þ

ð21Þ

Recall that Qi (s2

2 +lis1

2)w2(r i) are independent for

i = 1,. . .,d . Under H 0, the distribution of the test sta-

tistic is

ð1 h2oÞXd

i ¼ 1

liw2ðr iÞ

1 þ h2oðli 1Þ

Xd

i ¼ 1 w

2

ðr iÞ

ð22Þ

where the chi-squared random variables are independ-

ent. This testing procedure is discussed in some detail

in Ref. [13]. Neyman– Pearson tests and alternative

testing procedures are given in Ref. [14].

DISCUSSION

In the preceding analyses, the investigator may be faced

with the task of splitting the information contained in thequadratic forms Q1,. . .,Qd into two separate pieces,

namely Q1,. . .,Qk and Qk + 1,. . .,Qd . This becomes appar-

ent by examining the ANOVA estimator of h2 in Eq. 17,

the confidence interval for h2 as determined by Eq. 19,

and the test statistic for h2 in Eq. 20. The choice of k

is simplified if l1 = 0. In this case, it is recommended

that k = 1 so that Q1 is separated from the remaining

quadratic forms. See Table 1 for an example of this ap-

plication. The choice of k = 1 corresponds to the Wald

method in Ref. [15] for the unbalanced one-way random

effects model. See Ref. [12] for guidance on how to select

the value of k when l1 > 0.The ML procedures are popular methods used to

estimate variance components. Many statistical software

packages will compute ML estimates of s1

2, s2

2 and

hence, h2, with very little effort required on the part of

the investigator. These methods are based on iterative

procedures that require initial estimates and employ well-

designed algorithms to arrive at the solutions. While these

procedures perform admirably in a wide variety of appli-

cations, there is no guarantee that the methods will al-

ways converge to a solution. Closed-form approximations

to the REML estimator of h2 are discussed in Ref. [16].

40 Analysis of Heritability



CONCLUSION

Other statistical procedures may be used to analyze

heritability. Bootstrap and jackknife approaches may be

useful. An overview of bootstrap estimates and bootstrap

confidence intervals for variance components may befound in Ref. [17]. The analysis of heritability using

Bayesian procedures may be also be performed. In

Ref. [7], the authors note that, because the REML es-

timation of variance components is based on a specific

marginal likelihood function, it is equivalent to Bayesian

estimation using noninformative priors.

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