Epistasis and covariance: how gene interaction translates into genomic relationship

Abstract

Key message

Models based on additive marker effects and on epistatic interactions can be translated into genomic relationship models. This equivalence allows to perform predictions based on complex gene interaction models and reduces computational effort significantly.

Abstract

In the theory of genome-assisted prediction, the equivalence of a linear model based on independent and identically normally distributed marker effects and a model based on multivariate Gaussian distributed breeding values with genomic relationship as covariance matrix is well known. In this work, we demonstrate equivalences of marker effect models incorporating epistatic interactions and corresponding mixed models based on relationship matrices and show how to exploit these equivalences computationally for genome-assisted prediction. In particular, we show how models with epistatic interactions of higher order (e.g., three-factor interactions) translate into linear models with certain covariance matrices and demonstrate how to construct epistatic relationship matrices for the linear mixed model, if we restrict the model to interactions defined a priori. We illustrate the practical relevance of our results with a publicly available data set on grain yield of wheat lines growing in four different environments. For this purpose, we select important interactions in one environment and use this knowledge on the network of interactions to increase predictive ability of grain yield under other environmental conditions. Our results provide a guide for building relationship matrices based on knowledge on the structure of trait-related gene networks.

Supporting information

Dominance in this marker effect model and in classical quantitative genetics

Different ways of incorporating dominance into effect models exist in literature (see, e.g., Zeng et al. 2005). The most famous approach uses a linear regression of the three genotypic values of a diploid organism and adds a dominance term $d_\mathrm{aa}, d_\mathrm{Aa}, d_\mathrm{AA}$ which describes the difference between the linear fit and the respective genotypic value (Falconer and Mackay 1996). Figure 4 compares this standard approach and the way dominance is modeled by the pair epistasis and dominance model (a fit by a polynomial of degree two). The scheme illustrates that the linear coefficient of the polynomial is not necessarily identical to the coefficient of the linear regression. Moreover, the term $h_{k,k}$ which is the second coefficient of the polynomial regression cannot be identified with $d_{\bullet \bullet }$. The parameterization of the two approaches is different. Analogously, the parameterization of the interactions in a two-locus model differs from the classical subdivision into additive by additive, additive by dominance and dominance by dominance effects. Regarding the effects of the pairs in a two-locus model, the framework used in this work does not give the freedom to choose any arbitrary effect for each allele combination of the two loci: The nine possible combinations of the states of two loci of a diploid organism (each with two alleles) are parameterized by less than nine parameters. Moreover, since the additive parameter of marker j is also present in all other two-loci effect models of j with any other locus k, the individual two-locus models are not independent but connected. For more information on two-locus models and their different parametrization, see the work of Hallgrímsdóttir and Yuster (Hallgrímsdóttir and Yuster 2008).

Derivation of the statistical equivalence of Eq. (3) and Eqs. (6, 7)

We consider the distribution of $\mathbf {y}$. The vector $\mathbf {1}\mu$ is nonrandom (fixed effect), the second summand of Eq. (3) translates into $\mathbf {g}$ of Eq. (6) which has a multivariate Gaussian distribution (with mean zero and covariance matrix $\sigma ^2_{\beta }\mathbf {MM}')$. The vector of errors is multivariate Gaussian, too. What has to be considered in more detail, is the third summand of Eq. (3)

$$\begin{aligned} \sum \limits _{k=1}^{p}\sum \limits _{j=k+1}^{p} M_{i,j} M_{i,k} h_{j,k}. \end{aligned}$$

We rewrite this equation for all n genotypes simultaneously in matrix notation

$$\begin{aligned} \underbrace{\begin{pmatrix} M_{1,1} M_{1,2} & M_{1,1} M_{1,3} & \cdots & M_{1,1} M_{1,p} & M_{1,2} M_{1,3} & \cdots & M_{1,p-1} M_{1,p} \\ M_{2,1} M_{2,2} & M_{2,1} M_{2,3} & \cdots & M_{2,1} M_{2,p} & M_{2,2} M_{2,3} & \cdots & M_{2,p-1} M_{2,p} \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ M_{n,1} M_{n,2} & M_{n,1} M_{n,3} & \cdots & M_{n,1} M_{n,p} & M_{n,2} M_{n,3} & \cdots & M_{n,p-1} M_{n,p}\\ \end{pmatrix}}_{=\mathbf {N}} \begin{pmatrix} h_{1,2} \\ h_{1,3} \\ \vdots \\ h_{1,p}\\ h_{2,3} \\ \vdots \\ h_{p-1,p} \\ \end{pmatrix}, \end{aligned}$$

(19)

and name the left hand matrix $\mathbf {N}$. This presentation shows that the third summand of the covariance matrix analog of Eq. (3) is also multivariate Gaussian distributed with mean zero and covariance matrix $\mathbf {N N}' \sigma ^2_h$ (according to a definition of the multivariate Gaussian distribution, since the interactions are assumed to be i.i.d. Gaussian distributed random variables). To see which structure the covariance has, we compare $\mathbf {NN}'$ to $\mathbf {MM}'$ by regarding the entries $(\mathbf{NN'})_{i,l}$ and $(\mathbf{MM'})_{i,l}$.

$$\begin{aligned} (\mathbf{NN'})_{i,l}=\sum \limits _{k=1,j>k}^p M_{i,k}M_{i,j}M_{l,k}M_{l,j}=\sum \limits _{k=1}^p \left( M_{i,k} M_{l,k} \sum \limits _{j>k}^p M_{i,j} M_{l,j} \right) \end{aligned}$$

(20)

and

$$\begin{aligned} (\mathbf{MM'})_{i,l}=\sum \limits _{k=1}^p M_{i,k}M_{l,k}. \end{aligned}$$

(21)

Eqs. (20) and (21) show that the structure of the additional covariance matrix $\mathbf {NN}'$ of the epistasis model resembles $\mathbf {G}$, but the summands of each entry are weighted by products of entries of matrix $\mathbf {M}$.

Derivation of the statistical equivalence of Eq. (4) and Eqs. (6, 8)

Analogously, to the procedure we applied to the first model, we regard the third term

$$\begin{aligned} \sum \limits _{j=1}^{p}\sum \limits _{k=1}^{p} M_{i,j}M_{i,k} h_{j,k}, \end{aligned}$$

and write

$$\begin{aligned} { \underbrace{\begin{pmatrix} M_{1,1}^2 & M_{1,1} M_{1,2} & \cdots & M_{1,1} M_{1,p} & M_{1,2} M_{1,1} & M_{1,2}^2 & \cdots & M_{1,p}^2 \\ M_{2,1}^2 & M_{2,1} M_{2,2} & \cdots & M_{2,1} M_{2,p} & M_{2,2} M_{2,1} & M_{2,2}^2 & \cdots & M_{2,p}^2 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ M_{n,1}^2 & M_{n,1} M_{n,2} & \cdots & M_{n,1} M_{n,p} & M_{n,2} M_{n,1} & M_{n,2}^2 & \cdots & M_{n,p}^2 \\ \end{pmatrix}}_{=\mathbf {Q}} \begin{pmatrix} h_{1,1} \\ h_{1,2} \\ \vdots \\ h_{1,p}\\ h_{2,1} \\ h_{2,2} \\ \vdots \\ h_{p,p} \\ \end{pmatrix}} \end{aligned}$$

(22)

where the letter $\mathbf {Q}$ is defined to be the left hand matrix. Thus, we are interested in $(\mathbf{QQ'})_{i,l}$ which is given by

$$\begin{aligned} (\mathbf{QQ'})_{i,l}= \begin{pmatrix} M_{i,1}^2&M_{i,1} M_{i,2}&\cdots&M_{i,1} M_{i,p}&M_{i,2} M_{i,1}&M_{i,2}^2&\cdots&M_{i,p}^2 \end{pmatrix} \begin{pmatrix} M_{l,1}^2 \\ \\ M_{l,1} M_{l,2} \\ \\ \vdots \\ \\ M_{l,1} M_{l,p} \\ \\ M_{l,2} M_{l,1} \\ \\ M_{l,2}^2 \\ \\ \vdots \\ \\ M_{l,p}^2 \end{pmatrix} =\sum \limits _{k,j=1}^p M_{i,k}M_{i,j}M_{l,k}M_{l,j} = \left( \sum \limits _{k=1}^p M_{i,k} M_{l,k} \right) ^2 \end{aligned}$$

(23)

which means that $\mathbf {QQ}'$ represents the Hadamard product $\mathbf {MM}' \circ \mathbf {MM}'$.

Proof of Eq. (9)

From Eq. (20) we know that

$$\begin{aligned} 2 (\mathbf{NN'})_{i,l}=2 \sum \limits _{k=1}^{p} \left( M_{i,k} M_{l,k} \sum \limits _{j > k}^p M_{i,j} M_{l,j}\right) \mathop {=}\limits ^{*}\sum \limits _{k=1}^{p} \left( M_{i,k} M_{l,k} \sum \limits _{j \ne k}^p M_{i,j} M_{l,j}\right) = \left(\mathbf{ MM'} \circ \mathbf{MM'} \right) _{i,l} -\sum \limits _{k=1}^p {M_{i,k}M_{l,k}M_{i,k}M_{l,k}}=\left(\mathbf{ MM'} \circ \mathbf{MM'} \right) _{i,l} -\left( (\mathbf{M}\circ\mathbf{ M}) (\mathbf{M \circ M})'\right) _{i,l} \end{aligned}$$

To see equality $*$, consider the left side as sum over products defined by all tuples $\{(k,j)|j>k\}$ with fixed i, l. Since j and k can be exchanged with each other, this is equal to the sum over all products defined by $\{(k,j)|j<k\}$. Thus, multiplying the sum defined by the tuples $\{(k,j)|j>k\}$ by 2 equals adding the sum defined by $\{(k,j)|j<k\}$.

Derivation of Eq. (5)

We maximize

$$\begin{aligned} F(\mu ,\mathbf {g},\mathbf {y}_\mathrm{train}):{=}- \left( \left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \mathbf {y}_\mathrm{test} \end{pmatrix}- \ \mathbf {1} \mu -\mathbf {g}\right) ' \mathbf {I}_n \frac{1}{\sigma ^2_\epsilon } \left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \mathbf {y}_\mathrm{test} \end{pmatrix}- \ \mathbf {1} \mu -\mathbf {g} \right) + \mathbf {g}' \frac{1}{\sigma _\beta ^2} \mathbf {G}^{-1} \mathbf {g} \right) \end{aligned}$$

with respect to $\mu$, $\mathbf {g}$ and $\mathbf {y}_\mathrm{test}$ by calculating the partial derivatives and the corresponding zeros. Let m denote the number of genotypes in the test set. Thus, we have to solve

$$\begin{aligned} (i) \qquad \frac{\partial F}{\partial \mu } \text{ gives: } \begin{pmatrix} \mathbf {1} _{n-m} \\ \mathbf {1} _{m} \end{pmatrix} '\left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \hat{\mathbf {y}}_\mathrm{test} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{n-m} \\ \mathbf {1} _{m} \end{pmatrix} \hat{\mu } - \begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix}\right) =\mathbf{0} \end{aligned}$$

$$\begin{aligned} (ii) \qquad \frac{\partial F}{\partial \mathbf {g}} \text{ gives: } \left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \hat{\mathbf {y}}_\mathrm{test} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{n-m} \\ \mathbf {1} _{m} \end{pmatrix} \hat{\mu } - \begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix} \right) - \sigma _\epsilon ^2 \left( \frac{1}{\sigma ^2_\beta } \mathbf {G}^{-1}\right) \begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix}=\mathbf{0} \end{aligned}$$

$$\begin{aligned} (iii) \qquad \frac{\partial F}{\partial \mathbf {y}_\mathrm{test}} \text{ gives: } \begin{pmatrix} \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf {I}_m \end{pmatrix}\left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \hat{\mathbf {y}}_\mathrm{test} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{n-m} \\ \mathbf {1} _{m} \end{pmatrix} \hat{\mu } - \begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix} \right) =\mathbf{0} \end{aligned}$$

$\mathbf {1}_i$ here denotes the vector of length i with every entry equal to 1 and $\mathbf {I}_{m}$ the m-dimensional identity matrix. Eq. (iii) can be rewritten to $\hat{\mathbf {y}}_\mathrm{test}=\hat{\mathbf {g}}_\mathrm{test} + \mathbf {1}_{m} \hat{\mu }$. Plugging this into Eq. (i) gives $\hat{\mu }= \mathbf {1}_{n-m}'(\mathbf {y}_\mathrm{train}-\hat{\mathbf {g}}_\mathrm{train})(n-m)^{-1}$. Using the rewritten version of (iii) in (ii) gives

$$\begin{aligned} (iii) \text{ in } (ii): \qquad &\left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} + \mathbf {1}_m\hat{\mu } \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{n-m} \\ \mathbf {1} _{m} \end{pmatrix} \hat{\mu } - \begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix} \right) - \sigma _\epsilon ^2 \left( \frac{1}{\sigma ^2_\beta } \mathbf {G}^{-1}\right) \begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix}=\mathbf{0} \\ &\Longleftrightarrow \\ &\left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \mathbf{0} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{n-m} \\ \mathbf{0} \end{pmatrix} \hat{\mu } - \begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \mathbf{0} \end{pmatrix} \right) - \sigma _\epsilon ^2 \left( \frac{1}{\sigma ^2_\beta } \mathbf {G}^{-1}\right) \begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix}=\mathbf{0} \\ &\Longleftrightarrow \\ &\left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \mathbf{0} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{n-m} \\ \mathbf{0} \end{pmatrix} \hat{\mu } \right) - \left( \underbrace{ \mathbf {I}_n - \begin{pmatrix} \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf {I}_m \end{pmatrix}}_{=\mathbf {T}_\mathrm{train}}+ \sigma _\epsilon ^2 \left( \frac{1}{\sigma ^2_\beta } \mathbf {G}^{-1}\right) \right) \begin{pmatrix}\hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix}=\mathbf{0} \end{aligned}$$

The latter equivalence uses the equality $\begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \mathbf{0} \end{pmatrix}=\mathbf {T}_\mathrm{train}\begin{pmatrix}\hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix}$. The left summand can be rewritten to

$$\begin{aligned} \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \mathbf{0} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{s} \\ \mathbf{0} \end{pmatrix} \hat{\mu } = \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \mathbf{0} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{s} \\ \mathbf{0} \end{pmatrix} \mathbf {1}_{s}'(\mathbf {y}_\mathrm{train}-\hat{\mathbf {g}}_\mathrm{train})s^{-1} = \begin{pmatrix}\mathbf {y}_\mathrm{train} \\ \mathbf{0} \end{pmatrix} - \begin{pmatrix} \mathbf {J} _{s \times s} \\ \mathbf{0} \end{pmatrix} (\mathbf {y}_\mathrm{train}-\hat{\mathbf {g}}_\mathrm{train})s^{-1}=\begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \mathbf{0} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{s} \bar{y}_\mathrm{train}\\ \mathbf{0} \end{pmatrix} +\begin{pmatrix} \mathbf {J} _{s \times s} \\ \mathbf{0} \end{pmatrix} \hat{\mathbf {g}}_\mathrm{train}s^{-1}. \end{aligned}$$

by plugging in the rewritten version of (i) into (ii), using $s=n-m$ for the number of genotypes of the training set, and defining the empirical mean $\bar{y}_\mathrm{train}=s^{-1}\sum \mathbf {y}_\mathrm{train}$. Moreover, writing $\hat{\mathbf {g}}_\mathrm{train} = (\mathbf {I}_{s},\mathbf{0}) \begin{pmatrix} \hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix}$, gives

$$\begin{aligned} \left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \mathbf{0} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{s} \bar{y}_\mathrm{train}\\ \mathbf{0} \end{pmatrix} \right) - \left( \mathbf {T}_\mathrm{train} -s^{-1} \begin{pmatrix} \mathbf {J}_{s \times s } & \mathbf{0} \\ \mathbf{0}& \mathbf{0} \end{pmatrix} + \sigma _\epsilon ^2 \left( \frac{1}{\sigma ^2_\beta } \mathbf {G}^{-1}\right) \right) \begin{pmatrix}\hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix}=\mathbf{0} \end{aligned}$$

and thus

$$\begin{aligned} \begin{pmatrix}\hat{\mathbf {g}}_\mathrm{train} \\ \hat{\mathbf {g}}_\mathrm{test} \end{pmatrix}= \left( \mathbf {T}_\mathrm{train} - s^{-1} \begin{pmatrix} \mathbf {J}_{s \times s } & \mathbf{0} \\ \mathbf{0}& \mathbf{0} \end{pmatrix} + \sigma _\epsilon ^2 \left( \frac{1}{\sigma ^2_\beta } \mathbf {G}^{-1}\right) \right) ^{-1} \times \left( \begin{pmatrix} \mathbf {y}_\mathrm{train} \\ \mathbf{0} \end{pmatrix} - \begin{pmatrix} \mathbf {1} _{s} \bar{y}_\mathrm{train}\\ \mathbf{0} \end{pmatrix} \right) \end{aligned}$$

which represents Eq. (5).

Proof of the statement of “Equivalent linear models for higher order polynomial functions of the markers”

We know that the statement is true for $D=2$. What has to be shown is that if it is true for $D=j-1$ then it is also true for $D=j$ (mathematical induction). Let j be the degree. Analogously to Eq. (22), we consider the matrix $\mathbf {Q}^{(j)}$ which has all products of j factors in the respective row of matrix $\mathbf {M}$ as entries. The i-th row $\mathbf {Q}^{(j)}_{i,\bullet }$ of matrix $\mathbf {Q}^{(j)}$ can be written as

$$\begin{aligned} \mathbf {Q}^{(j)}_{i,\bullet }= \mathbf {M}_{i,\bullet } \otimes \mathbf {Q}^{(j-1)}_{i,\bullet } \end{aligned}$$

if the $\beta _{\varvec{\kappa },j}$ is ordered appropriately ($\otimes$ denotes the Kronecker product). Then the (i, l)-th entry of $\mathbf {Q}^{(j)} {\mathbf {Q}^{(j)}}^\prime$ is the matrix product

$$\begin{aligned} \left( \mathbf {Q}^{(j)} {\mathbf {Q}^{(j)}}^\prime \right) _{i,l}= \left( \mathbf {M}_{i,\bullet } \otimes \mathbf {Q}^{(j-1)}_{i,\bullet } \right) \left( \mathbf {M}_{l,\bullet } \otimes \mathbf {Q}^{(j-1)}_{l,\bullet } \right) ^\prime \end{aligned}$$

which is equal to

$$\begin{aligned} \left( \mathbf {M}_{i,\bullet } \otimes \mathbf {Q}^{(j-1)}_{i,\bullet } \right) \left( \left( \mathbf {M}_{l,\bullet }\right) ^{\prime } \otimes {\left( \mathbf {Q}^{(j-1)}_{l,\bullet }\right) }^{\prime } \right) = \underbrace{\left( \mathbf {M}_{i,\bullet } \left( \mathbf {M}_{l,\bullet } \right) ^{\prime } \right) }_{=G_{i,l}} \otimes \underbrace{\left( \mathbf {Q}^{(j-1)}_{i,\bullet } {\left( \mathbf {Q}^{(j-1)}_{l,\bullet } \right) }^{\prime } \right) }_{=G_{i,l}^{j-1}}=G_{i,l}^j, \end{aligned}$$

according to the calculation rules for the Kronecker product and matrix multiplication and the induction hypothesis.

Proof of Eq. (17)

Assuming one interaction between marker positions k and j leads to the equation

$$\begin{aligned} \mathbf{y}= \mathbf {1}\mu + \mathbf{M}\varvec{\beta} + \underbrace{\begin{pmatrix} M_{1,k}M_{1,j} \\ M_{2,k}M_{2,j} \\ \vdots \\ M_{n,k}M_{n,j} \end{pmatrix}}_{=\mathbf {R}} h_{k,j} + \varvec{\epsilon} , \end{aligned}$$

where we call the vector $\mathbf {R}$. According to the calculation rules for multivariate normal distributions, the covariance matrix of the corresponding interaction effect $\mathbf {g_2}$ is $\sigma _h^2 \mathbf {RR}^\prime$. We have to show that this expression is equal to Eq. (17). As before, we consider the entry (l, i) of $\sigma _h^2 \mathbf {RR}^\prime$ which is given by $M_{l,k}M_{l,j}M_{i,k}M_{i,j} \sigma _h^2$. Moreover the (l, i)-th entry of $\left( \mathbf {M}_{\bullet , k} \mathbf {M}^\prime _{\bullet , k} \right)$ is $M_{l,k} M_{i,k}$ and the (l, i)-th entry of $\left( \mathbf {M}_{\bullet , j} \mathbf {M}_{\bullet , j}^\prime \right)$ is $M_{l,j} M_{i,j}$ which proves the statement.

Fig. 1

Illustration of the differences between the pair epistasis model (left scheme) and the pair epistasis and dominance model (middle scheme). In the latter, the interactions are modeled as the sum of two variables and dominance is included by an interaction of each locus with itself. The right scheme illustrates a subnetwork which might fit the underlying biology of the trait best

Fig. 2

The $0.01~\%$ most important (highest absolute values) pairwise interactions of the wheat data set, calculated with Eq. (15) for the four different environments (positions of the markers according to Crossa et al. 2010)

Fig. 3

Prediction accuracy (here Fisher’s z-transformed correlation: arctanh(r)) in the different environments of the wheat data set with relationship matrices determined by variable selection in the other environments (0–90 $\%$ of the variables were removed in 5 $\%$ steps). Black, red, green and blue dots reflect the respective environment 1, 2, 3, or 4, which was used to determine the relationship matrices. The initial “jump” at zero represents the difference between GBLUP and the full epistatic model

Fig. 4

Comparison of the classical way to incorporate dominance effects (Falconer and Mackay 1996) and the fitting by a polynomial of degree two. Black points genotypic values of the genotypes aa, Aa, AA of a diploid organism with the two alleles a and A present in the population. Black line linear regression which defines the additive effect in the classical model. Blue lines The dominance terms (one dominance term for each genotype) which are given by the difference of the genotypic values and the regression line. Red curve Fit by a polynomial of degree two

Table 1 Average correlation between predicted and measured phenotypes obtained by 200 independent draws of a test set consisting of 60 lines in the respective environment