Chapter 7: Estimates of Variance

William Beavis; Kendall Lamkey; Katherine Espinosa; and Anthony Assibi Mahama

Estimating heritability is a fundamental concept of quantitative genetics. One method for obtaining estimates of heritability is the use of variance and covariance of a known collection of relatives from various types of progeny.

Learning Objectives
  • Model components of genetic variances and covariances for purposes of estimating heritability, a fundamental concept of quantitative genetics.
  • Explain why estimates of components of genetic variability are limited to the population from which they are estimated.
  • Students will derive variance components and recognize the differences amongst components obtained from different progeny used for estimating heritability.
  • Students will write out the correct linear models for the correct mean squares and expected mean squares in a ANOVA table, and correctly interpret the ANOVA and algebraically extract the correct values for estimating heritability.
  • Leverage of the powerful algebraic equivalence of covariances within groups of relatives to variances among the same groups.

Covariance of Relatives

Recall that Cov(Yij,Yij’) = Var(Gi), for j ≠ j’. In the context of genotypic sampling of relatives, this general relationship has a profound and powerful impact on interpretation of ANOVA. It means that the covariance among a sample of relatives can be used to estimate components of genetic variance associated with the genotypic effect.

Table 1 A general ANOVA table for any type of related progeny.
n/a EMS
Source df MS Variances Covariances
Reps [latex]r-1[/latex] n/a n/a n/a
Progeny [latex]p-1[/latex] [latex]MS_3[/latex] [latex]\sigma ^2_w + k\sigma ^2 +rk \sigma ^2_p[/latex] [latex]\sigma^2_w + k\sigma^2+rk[\textrm{Cov(progeny)}][/latex]
Error [latex]\small(r-1)(p-1)[/latex] [latex]MS_2[/latex] [latex]\sigma^2_w + k \sigma^2[/latex] [latex]\sigma^2_w + k \sigma^2[/latex]
Total [latex]rp-1[/latex] n/a n/a n/a
Within Progeny [latex]rp(k-1)[/latex] [latex]MS_1[/latex] [latex]\sigma^2_w[/latex] [latex]\sigma^2_{we} + [\sigma^2_T - \textrm{Cov(progeny)}][/latex]

Note that there are p progeny grown in r reps. Cov(progeny) refers to the covariance of the progeny, where the progeny can be full-sibs, half-sibs, S1-progeny, S2-progeny, testcross progeny, etc. The key is to know the progeny type and take advantage of the general rule that the variance among progeny is equal to the covariance of the progenies.

Note the use of σ2T instead of σ2G in the within progeny line of the ANOVA table. This is because σ2G is usually equal to σ2A + σ2D the total variance in a non-inbred random mating population. If the population does not have a random mating structure, then the total variance will be something other than σ2A + σ2D. For example, the total genetic variance for an F3 population is as in Equation 1.

[latex]\sigma^2_F = {3 \over 2}\sigma^2_A + {3 \over 2}\sigma^2_D[/latex]

[latex]\textrm{Equation 1}[/latex] Formula for total genetic variance for and F3 population.

where:
[latex]\sigma^2_F[/latex] = total genetic variance for F3 population,
[latex]\sigma^2_A[/latex] = additive variance,
[latex]\sigma^2_D[/latex] = dominance variance.

Linear Models for Phenotypic Values

The covariance of relatives is simply that relatives tend to show more phenotypic similarities than with each other than with unrelated individuals. For example let Xij represent an individual from the mating of parent i and parent j:

Table 2 Descriptions of relationships between individuals Xij and Xi’j’.
Conditions Description
[latex]i = i', j = j'[/latex] Full-sibs
[latex]i = j', j = i'[/latex] Reciprocal Full-sibs
[latex]i = i', j \neq j'[/latex] Maternal half-sibs
[latex]i \neq i', j = j'[/latex] Paternal half-sibs

Specifying covariance of relatives in terms of genetic variances has the following assumptions:

  1. Regular diploid and solely Mendelian inheritance
  2. No environmental correlations among relatives
  3. No gametic disequilibrium
  4. The relatives are not inbred
  5. The relatives are considered to be random members of some non-inbred population

With these assumptions, we can specify the covariance of relatives as in Equation 2.

[latex]\textrm{Cov} = ασ^2_A + δσ^2_D + α^2σ^2_{AA} + αδσ^2_{AD} + δ^2σ^2_D + α^3σ^2_{AAA} + ...[/latex]

[latex]\textrm{Equation 2}[/latex] Formula for covariance of relatives,

where:
[latex]α[/latex] = the coefficient of relative relationship,
[latex]σ^2_A[/latex] = additive genetic variance,
[latex]δ[/latex] = the dominance relationship coefficient,
[latex]σ^2_D[/latex] = the dominance variance,
[latex]σ^2_{AA}, σ^2_{AD}, σ^2_{AAA}[/latex] = the epistatic variances.

Common Types of Relatives

Using the result of Equation 1 for some common types of relatives, it can be shown that:

Covariance of half-sibs with one common parent is represented by Equation 3.

[latex]Cov(HS) = \frac{(1 + F_A)}{4}\sigma^2_A +(\frac{(1+F_A)}{4})^2\sigma^2_{AA} + ...[/latex]

[latex]\textrm{Equation 3}[/latex] Formula for calculating covariance of half-sibs,

where:
[latex]F_A[/latex] = inbreeding coefficient of parent A.

Covariance of full-sibs with parents A and B is estimated using Equation 4.

[latex]Cov(FS) = \frac{(2+F_A + F_B)}{4}\sigma^2_A + \frac{(1 + F_A)(1 + F_B)}{4}\sigma^2_D + (\frac{(2+F_A+F_B)}{4})^2\sigma^2_{AA} +\\ (\frac{(2+F_A+F_B)}{4})(\frac{(1+F_A)(1+F_B)}{4})\sigma^2_{AD} + (\frac{(1+F_A)(1-F_B)}{4})^2\sigma^2_{DD} + (\frac{(2+F_A+F_B)}{4})^3\sigma^2_{AAA} + ... \\[/latex]

[latex]\textrm{Equation 4}[/latex] Formula for calculating covariance of full-sibs,

where:
[latex]F_A[/latex] = inbreeding coefficient of parent A,
[latex]F_B[/latex] = inbreeding coefficient of parent B,
[latex]σ^2_{AA}, σ^2_{AD}, σ^2_{AAA}[/latex] = the epistatic variances.

F2 and F3 Progenies

Table 3 F3 progeny genotypes, frequencies, genotypic values and progeny mean values representation.
 

 

Genotype

 

 

Freq

 

 

GV

F3 Progeny  

 

F3 Progeny Mean

[latex]AA[/latex] [latex]Aa[/latex] [latex]aa[/latex]
[latex]AA[/latex] 1/4 a 1 0 0 a
[latex]Aa[/latex] 1/2 d 1/4 1/2 1/4 1/2d
[latex]aa[/latex] 1/4 -a 0 0 1 -a

F2 and F3 Variances

Total genetic variance among F2 individuals determined suing Eqaution 5:

[latex]\sigma^2_{F_2} = \sigma^2_A + \sigma^2_D = {1 \over 2}a^2 + {1 \over 4}d^2[/latex]

[latex]\textrm{Equation 5}[/latex] Formula for calculating total genetic variance among F2 individuals,

where:
[latex]a, d[/latex] = genotypic values of AA or aa and Aa genotypes, respectively.

Total F2 phenotypic variation:

[latex]\sigma^2_{F_2} = \sigma^2_{A} + \sigma^2_D + \sigma^2_{E1} = {1 \over 2}a^2 + {1 \over 4}d^2 +E1[/latex]

[latex]\textrm{Equation 6}[/latex] Formula for calculating total phenotypic variance among F2 individuals,

where:
[latex]\sigma^2_{F_2}[/latex] = total phenotypic variance among F2 individuals,
[latex]E1[/latex] = the non-genetic variation among F2 plants,
[latex]\textrm{other terms}[/latex] are as described previously.

Recall that the F2 is our reference population for interpretation of genetic results. To estimate the total genetic variation of an F2, we need the parents and the F1 (to estimate environmental effects)as well as the F2 generation.

F3 Variances

F3 population mean is equal to [latex]\frac{1}{4}d[/latex]

Variance among F3 progeny means is determined using Equation 7.

[latex]\sigma^2_{\bar{F_3}} = [{1 \over 4}a^2 + {1 \over 2}({1 \over 2}d)^2 + {1 \over 4}(-a^2)] - ({1 \over 4}d)^2 ={1 \over 2}a^2 + {1 \over 16}d^2 =\sigma^2_A + {1 \over 4}\sigma^2_D[/latex],

[latex]\textrm{Equation 7}[/latex] Formula for calculating variance among F3 progeny means,

where:
[latex]\sigma^2_{\bar{F_3}}[/latex] = variance among F3 progeny means,
[latex]\textrm{other terms}[/latex] are as described previously.

Variance within F3 progeny means is determined using Equation 8.

[latex]\bar{\sigma}^2_{F_3} = [[{1 \over 4}a^2 + {1 \over 2}d^2 + {1 \over 4}(-a^2)] - ({1 \over 2}d)^2] = {1 \over 4}a^2 + {1 \over 8}d^2 = {1 \over 2}\sigma^2_A + {1 \over 2}\sigma^2_D[/latex],

[latex]\textrm{Equation 8}[/latex] Formula for calculating total phenotypic variance within F3 progeny means,

where:
[latex]\bar{\sigma}^2_{F_3}[/latex] = variance within F3 progeny means,
[latex]\textrm{other components}[/latex] are as described previously.

Total variance among F3 individuals is then estimated from Equation 9:

[latex]\sigma^2_{F_3} = {3 \over 2}\sigma^2_A + {3 \over 4}\sigma^2_D[/latex]

[latex]\textrm{Equation 9}[/latex] Formula for calculating total variance among F3 individuals,

where:
[latex]\sigma^2_{F_3}[/latex] = total variance among F3 individuals,
[latex]\textrm{other terms}[/latex] are as described previously.

F3 progenies can be grown in replicated trials, so a set of equations like the following could be written to estimate the variance in different generations (Equation 10).

[latex]\sigma^2_{F_2} = \sigma^2_A + \sigma^2_D + \sigma^2_{E1}[/latex],

[latex]\sigma^2_{\bar{F_3}} = \sigma^2_A + {1 \over 4}\sigma^2_D + \sigma^2_{E2}[/latex],

[latex]\bar{\sigma}^{\ 2}_{F_3} = {1 \over 2}\sigma^2_A + {1 \over 2}\sigma^2_D + \sigma^2_{E1}[/latex],

[latex]and, \sigma^2_{E2} = \frac{\sigma^2}{r}[/latex],

[latex]\textrm{Equation 10}[/latex] Formulae for calculating variances for F2 and F3,

where:
[latex]E2[/latex] = non-genetic variation among mean variance of F3 progeny,
[latex]r[/latex] = number of replications,
[latex]\textrm{other terms}[/latex] are as described previously.

ANOVA for F3 Progenies

ANOVA for F3 progenies can be calculated from a replicated experiment.

Table 4 ANOVA for F3 Progenies.
Source df MS EMS
Reps [latex]r-1[/latex] n/a n/a
Progeny [latex]p-1[/latex] M3 [latex]\sigma^2_e+r\sigma^2_{F_2}[/latex]
Error [latex](r-1)(p-1)[/latex] M2 [latex]\sigma^2[/latex]
Total [latex]rp-1[/latex] n/a n/a
Within Progeny [latex]rp(k-1)[/latex] M1 [latex]\sigma^2_{F_a}={1\over2}\sigma^2_A + {1\over2}\sigma^2_D+\sigma^2_{E1}[/latex]

Then using Equation 11,

[latex]\sigma^2_{\bar{F_3}} = \frac{M3 - M2}{r} = \sigma^2_A + {1 \over 4}\sigma^2_D[/latex]

[latex]\sigma^2_{E2} = \frac{M2}{r} + \frac{\sigma^2}{r}[/latex]

[latex]\textrm{Equation 11}[/latex] Formulae for calculating variances using MS and EMS,

[latex]\sigma^2_{E1}[/latex] is estimated as environmental variance within P1 or P2 plots (the inbred lines).

Note that the phenotypic variance among F3 families is determined with Equation 12:

[latex]\hat{\sigma}^2_p = \frac{M3}{rk} = \frac{\sigma^2_w}{rk} + \frac{\sigma^2}{r} + \sigma^2_c[/latex],

[latex]\textrm{Equation 12}[/latex] Formula for calculating total phenotypic variance among F3 families,

where:
[latex]\hat{\sigma}^2_p[/latex] = estimate of total phenotypic variance,
[latex]\sigma^2_w[/latex] = within progeny variance,
[latex]\sigma^2_c[/latex] = phenotypic variance among F3 families = the genotypic variance,
[latex]r[/latex] = number of replications,
[latex]k[/latex] = individuals withing progeny type.

Estimate of Heritability

A type of heritability estimates on a progeny mean basis can be calculated as shown in Equation 13:

[latex]h^2 = \frac{\sigma^2_c}{\sigma^2_p} = \frac{{1 \over 2}\sigma^2_A + {1 \over 4}\sigma^2_D}{\frac{\sigma^2_w}{rk}+{\sigma^2 \over r} + \sigma^2_c}[/latex],

[latex]\textrm{Equation 13}[/latex] Formulae for calculating heritability on progeny entry mean basis,

where:
Terms are as described previously.

Note that this estimate of heritability contains both additive and dominance variance. Recall that this is an estimate of intra-class correlation, thus it is a type of broad-sense heritability.

Limitations of this method (often referred to as Mather’s methods)

  1. Estimates apply only to specific parents.
  2. Estimates for σ2E1 may vary among generations.
  3. Estimates for a particular set of F2 plants can be obtained in only one environment.
  4. Linkage will bias estimates.
  5. Epistasis is assumed to be absent.

Bi-Parental Progenies

Bi-parental progenies are just crosses between individual plants; thus, genetically, they are full-sibs. For example, in a random mating maize population, you could cross two individual plants reciprocally and bulk the seed from the two ears. This would produce enough seed to plant FS progeny in 10-20 replications. We could then think about n plants and making n / 2 full-sib families. The covariance then be computed using Equation 14.

Table 5 ANOVA Table for bi-parental progenies
Source df MS EMS
Reps [latex]r-1[/latex] n/a n/a
Among families [latex]{n \over 2} - 1[/latex] [latex]\textrm{M3} =\sigma^2_w + k\sigma^2 + rk\sigma^2_c[/latex] [latex]\sigma^2_w + k\sigma^2 + rk[Cov(FS)][/latex]
Error [latex](r-1)({n \over 2} -1)[/latex] [latex]\textrm{M2} =\sigma^2_w + k\sigma^2[/latex] [latex]\sigma^2_w + k\sigma^2[/latex]
Total [latex]r{n \over 2} - 1[/latex] n/a n/a
Within families [latex]r{n \over 2}(k-1)[/latex] [latex]\textrm{M1} =\ \sigma^2_w[/latex] [latex]\sigma^2_{w} + [\sigma^2_G - Cov(FS)][/latex]

[latex]\sigma^2_c = Cov(FS) = {1 \over 2}\sigma^2_A + {1 \over 4}\sigma^2_D; \sigma^2_c = \frac{M3 - M2}{rk}; \sigma^2_w = \sigma^2_{w} + {1 \over 2}\sigma^2_A + {3 \over 4}\sigma^2_D; \hat{\sigma}^2_w = M1[/latex]

[latex]\textrm{Equation 14}[/latex] Extracting different variance components.

Summary

Table 6 Data from Cockerham, 1983.
Progeny Type Cov(progeny) Total Variance, [latex]\sigma^2_T[/latex]
Half-sib [latex]\small{{1 \over 4}\sigma^2_A}[/latex] [latex]\small{\sigma^2_A + \sigma^2_D}[/latex]
Full-sib [latex]\small{{1 \over 2}\sigma^2_A + {1 \over 4}\sigma^2_D}[/latex] [latex]\small{\sigma^2_A + \sigma^2_D}[/latex]
S1(F2:3) [latex]\small{\sigma^2_A + {1 \over 4}\sigma^2_D + D_1 + {1 \over 8}D_2}[/latex] [latex]\small{{3 \over 2}\sigma^2_A + {1 \over 2}\sigma^2_D + 2D_1 + {1 \over 2}D_2 + {1 \over 4}H^*}[/latex]
S2(F3:4) [latex]\small{{3 \over 2}\sigma_A^2 + {1 \over 8}\sigma^2_D + 2.5D_1 + {9 \over 16} D^2 + {1 \over 16}H^*}[/latex] [latex]\small{{7 \over 4}\sigma^2_A + {1 \over 4}\sigma^2_D + 3D_1 + {3 \over 4}D_2 + {3 \over 16}H^*}[/latex]
Sn(F4:5) [latex]\small{{7 \over 4}\sigma^2_A + {1 \over 16}\sigma^2_D + 3.25D_1 + {25 \over 32}D_2 + {3 \over 64}H^*}[/latex] [latex]\small{{15 \over 8}\sigma^2_A + {1 \over 8}\sigma^2_D + 3.5D_1 + {7 \over 8}D_2 + {7 \over 2\sigma^2_A + 4D_1 + D_264}H^*}[/latex]
S∞ [latex]\small{2\sigma^2_A + 4D_1 + D_2}[/latex] [latex]\small{2\sigma^2_A + 4D_1 + D_2}[/latex]

Expected Mean Squares

The AOV tables cannot be interpreted without understanding the expected sources of variability represented by the Mean Squares. In the case of balanced field plot designs with only a few sources of variation, the expected mean squares are easily determined. If a particular design involves many sources of random and fixed factors, students have found the approach of Lorenzen and Anderson (1993, Design of Experiments: A No-Name Approach. p 71-72) to be useful.

  1. Write the terms of the model with associated subscripts down the left side of the page. Across the top, write the single letter subscripts (i,j,k, etc.). Above each subscript, place either F or R if the factor associated with that transcript is fixed or random. Above that, place the number of levels associated with that subscript (I, J, K, etc.).
  2. Enter a 1 in every slot where the subscript at the top is contained within brackets in the term at the left.
  3. Enter a 0 in every slot where the subscript at the top is fixed and also contained in the term as the left. Enter a 1 in every slot where the subscript at the top is random and also contained in the terms at the left.
  4. Fill in the remaining slots with the number of levels at the top of each column.
  5. To compute the Expected Mean Squares (EMS) for a given term having df > 0, start at the bottom and work up. Only consider terms whose indices include all the indices in the term whose EMS you are deriving. Compute the coefficient of this term by covering the columns corresponding to the indices in the term whose EMS you are deriving and multiplying the values in the remaining columns. If there is a 0 column that is not covered, this term need not be written in the EMS. A factor is considered fixed and denoted with a Φ only if all of its indices are fixed. Otherwise, it is considered random and denoted by the appropriate σ2 term.

Using the Algorithm

Notice that this algorithm can be used to compute EMS for all terms in the model, including those that have zero df. A term that has zero df has no expected mean squares. For this reason, we will not compute EMS for terms having zero df even though such terms are in the algorithm to make the EMS of the other terms come out right. Note that this simple algorithm for determining the EMS in an AOV assumes that the data are balanced, i.e., each of the sources of variability (model parameters) have data for all levels, i, j, and k.

Step 1

The phenotype Y for this typical field trial will be something like in Equation 15:

[latex]Y_{ijk} = \mu + E_i +B{(E)}_{(i)k} + G_j + GE_{ij} + \Large{\varepsilon}\small{_{(ij)k}}[/latex],

[latex]\textrm{Equation 15}[/latex] Linear model for phenotype,

where:
[latex]Y_{ijk}[/latex] = phenotypic measure of trait for the jth genotype in the kth block nested within the ith environment,
[latex]\mu[/latex] = overall mean,
[latex]E_i[/latex] = ith environment,
[latex]B{(E)}_{(i)k}[/latex] = represents the kth block nested within the ith environment,
[latex]G_j[/latex] = the jth genotype,
[latex]GE_{ij}[/latex] = interaction effect between the jth genotype and the ith environment,
[latex]\large{\varepsilon}\small{_{(ij)k}}[/latex] = the residual for genotype j in the kth block nested within the ith environment.

Notice that this algorithm can be used to compute EMS for all terms in the model, including those that have zero df. A term that has zero df has no expected mean squares. For this reason, we will not compute EMS for terms having zero df even though such terms are in the algorithm to make the EMS of the other terms come out right. Note that this simple algorithm for determining the EMS in an AOV assumes that the data are balanced, i.e., each of the sources of variability (model parameters) have data for all levels, i, j, and k.

To illustrate, let us consider a slightly more complex but typical RCBD design used by plant breeders to evaluate many genotypes grown in replicates at several environments for purposes of identifying and discarding poor-performing genotypes in a cultivar development project. Equation 15 will appear for each of the steps below.

Write the terms of the model with associated subscripts down the left side of the page. Across the top, write the single letter subscripts (i,j,k, etc.). Above each subscript, place either F or R if the factor associated with that transcript is fixed or random. Above that, place the number of levels associated with that subscript (I, J, K, etc.).

Factors:

  • Factor E – Fixed
  • Factor G – Random
  • Blocks – Random
Source E G R EMS
F R R
i j k
[latex]E_i[/latex] n/a n/a n/a n/a
[latex]B(E)_{(i)k}[/latex] n/a n/a n/a n/a
[latex]G_j[/latex] n/a n/a n/a n/a
[latex]GE_{ij}[/latex] n/a n/a n/a n/a
[latex]\Large{\varepsilon\small_{(ij)k}}[/latex] n/a n/a n/a n/a

[latex]Y_{ijk} = \mu + E_i +B{(E)}_{(i)k} + G_j + GE_{ij} + \Large{\varepsilon}\small{_{(ij)k}}[/latex],

Step 2

The phenotype Y for this typical field trial will be something like:

[latex]Y_{ijk} = \mu + E_i +B{(E)}_{(i)k} + G_j + GE_{ij} + \Large{\varepsilon}\small{_{(ij)k}}[/latex],

Enter a 1 in every slot where the subscript at the top is contained within brackets in the term at the left.

Factors:

  • Factor E – Fixed
  • Factor G – Random
  • Blocks – Random
Source E G R EMS
F R R
i j k
[latex]E_i[/latex]n/a  n/a n/a n/a n/a
[latex]B(E)_{(i)k}[/latex] 1 n/a n/a /n/a
[latex]G_j[/latex] 1 n/a n/a n/a
[latex]GE_{ij}[/latex] 1 1 n/a n/a
[latex]\Large{\varepsilon\small_{(ij)k}}[/latex] 1 1 1 n/a

Step 3

The phenotype Y for this typical field trial will be something like:

[latex]Y_{ijk} = \mu + E_i +B{(E)}_{(i)k} + G_j + GE_{ij} + \Large{\varepsilon}\small{_{(ij)k}}[/latex],

Enter a 0 in every slot where the subscript at the top is fixed and also contained in the term as the left. Enter a 1 in every slot where the subscript at the top is random and also contained in the terms at the left.

Factors:

  • Factor E – Fixed
  • Factor G – Random
  • Blocks – Random
Source E G R EMS
F R R
i j k
[latex]E_i[/latex] n/a n/a n/a
[latex]B(E)_{(i)k}[/latex] 0 1 n/a n/a
[latex]G_j[/latex] 1 n/a n/a n/a
[latex]GE_{ij}[/latex] 1 1 n/a n/a
[latex]\Large{\varepsilon\small_{(ij)k}}[/latex] 1 1 1 n/a

Step 4

The phenotype Y for this typical field trial will be something like this:

[latex]Y_{ijk} = \mu + E_i +B{(E)}_{(i)k} + G_j + GE_{ij} + \Large{\varepsilon}\small{_{(ij)k}}[/latex],

Fill in the remaining slots with the number of levels at the top of each column.

Factors:

  • Factor E – Fixed
  • Factor G – Random
  • Blocks – Random
Source E G R EMS
F R R
i j k
[latex]E_i[/latex] 0 G R n/a
[latex]B(E)_{(i)k}[/latex] 0 G 1 n/a
[latex]G_j[/latex] E 1 R n/a
[latex]GE_{ij}[/latex] 1 1 R n/a
[latex]\Large{\varepsilon\small_{(ij)k}}[/latex] 1 1 1 n/a

Step 5

The phenotype Y for this typical field trial will be something like:

[latex]Y_{ijk} = \mu + E_i +B{(E)}_{(i)k} + G_j + GE_{ij} + \Large{\varepsilon}\small{_{(ij)k}}[/latex],

To compute the EMS for a given term having df > 0, start at the bottom and work up. Only consider terms whose indices include all the indices in the term whose EMS you are deriving. Compute the coefficient of this term by covering the columns corresponding to the indices in the term whose EMS you are deriving and multiplying the values in the remaining columns.

If there is a 0 column that is not covered, this term need not be written in the EMS. A factor is considered fixed and denoted with a Φ only if all of its indices are fixed. Otherwise it is considered random and denoted by the appropriate σ2 term.

Factors:

  • Factor E – Fixed
  • Factor G – Random
  • Blocks – Random
Source E G R EMS
F R R
i j k
[latex]E_i[/latex] 0 G R [latex]\sigma + G\sigma^2_{B(E)} + \sigma^2_{E}[/latex]
[latex]B(E)_{(i)k}[/latex] 0 G 1 [latex]\sigma^2 + G\sigma^2_{B(E)}[/latex]
[latex]G_j[/latex] E 1 R [latex]\sigma^2 + R\sigma^2_{GE} + RE\sigma^2_G[/latex]
[latex]GE_{ij}[/latex] 1 1 R [latex]\sigma^2 + R\sigma^2_{GE}[/latex]
[latex]\Large{\varepsilon\small_{(ij)k}}[/latex] 1 1 1 [latex]\sigma^2[/latex]

References

Cockerham, C.C. 1983. Covariances of relatives from self-fertilization. Crop Sci. 23: 1177-1180.

Lorenzen, T., and  V. Anderson. 1993. Design of Experiments: A No-Name Approach. Routledge & CRC Press.

 

How to cite this chapter: Beavis, W., K. Lamkey, K. Espinosa, and A. A. Mahama. 2023. Estimates of Variance. In W. P. Suza, & K. R. Lamkey (Eds.), Quantitative Genetics for Plant Breeding. Iowa State University Digital Press.

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Quantitative Genetics for Plant Breeding Copyright © 2023 by William Beavis; Kendall Lamkey; Katherine Espinosa; and Anthony Assibi Mahama is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.