Overview of QTL detection in plants and tests for synergistic epistatic interactions

Abstract

Improvements in the usefulness of QTL analysis arise from better statistical methods applied to the problem, ability to analyze more complex mating designs, and the fitting of less simplified genetic models. Here we review the advantages of different plant mating designs in QTL analysis and conclude that diallel designs have several favorable properties. We then turn to the detection of systematic genome-wide synergistic epistasis. This form of epistasis has important implications from evolutionary (maintenance of sexual reproduction and concealment of cryptic genetic variation) and practical perspectives (response to pyramided favorable alleles). We develop two methods for detecting systematic synergistic epistasis, one based on analyzing interactions between locus effects and predicted individual genotypic values and one based on analyzing pairwise locus interactions. Using the first method we detect synergistic epistasis in a barley and a wheat dataset but not in a maize dataset. We fail to detect synergistic epistasis with the second method. We discuss our results in the light of theoretical questions concerning the mechanisms of synergistic epistasis.

Appendices

Appendix A

In this Appendix, we show that the contrast between parental versus recombinant haplotype effects cannot be greater in magnitude than the magnitude of the smaller single-locus main effect. Consider the two-locus linear model $y_{i} = \mu + x_{i1} \beta _{1} + x_{i2} \beta _{2} + x_{i1} x_{i2} \varepsilon _{12} + e_{i} ,$ where β ₁ and β ₂ are single-locus main effects and ε ₁₂ is the contrast between parental versus recombinant haplotype effects and x _i1 and x _i2 are defined as in the text. Then $\hat{\beta }_{1} = \tfrac{1}{4}\left( {\bar{y}_{ + + } + \bar{y}_{ + - } - \bar{y}_{ - + } - \bar{y}_{ - - } } \right),$ $\hat{\beta }_{2} = \tfrac{1}{4}\left( {\bar{y}_{ + + } - \bar{y}_{ + - } + \bar{y}_{ - + } - \bar{y}_{ - - } } \right),$ $\hat{\varepsilon }_{12} = \tfrac{1}{4}\left( {\bar{y}_{ + + } - \bar{y}_{ + - } - \bar{y}_{ - + } + \bar{y}_{ - - } } \right).$Without loss of generality, constrain $\bar{y}_{ - - } = 0,$ $0 \le \bar{y}_{ + - } \le \bar{y}_{ + + } ,$ and $0 \le \bar{y}_{ - + } \le \bar{y}_{ + + }$ and suppose $\bar{y}_{ + - } > \bar{y}_{ - + }$ such that $\hat{\beta }_{1} > \hat{\beta }_{2} \ge 0.$ We want to show that $\left| {\hat{\varepsilon }_{12} } \right| \le \hat{\beta }_{2}.$ If $\hat{\varepsilon }_{12} > 0,$ then $\hat{\beta }_{2} - \left| {\hat{\varepsilon }_{12} } \right| = \tfrac{1}{2}\bar{y}_{ - + } \ge 0,$ such that the condition is met. If $\hat{\varepsilon }_{12} > 0,$ then $\hat{\beta }_{2} - \left| {\hat{\varepsilon }_{12} } \right| = \tfrac{1}{2}\left( {\bar{y}_{ + + } - \bar{y}_{ + - } } \right) \ge 0,$ such that the condition is again met.

Appendix B

In this Appendix, we show that the probability that a redundant locus carries the favorable allele conditional on the phenotypic value can be decomposed as

$$ p\left( {x_{ir} = 1\left| {y_{i} } \right.} \right) = p\left( {x_{ir} = 1\left| {x_{iq} = 1 \cup\,x_{ir} = 1} \right.} \right) \times p\left( {x_{iq} = 1 \cup\,x_{ir} = 1\left| {y_{i} } \right.} \right) $$

We show the complete development then describe its steps.

$$ \begin{aligned} p\left( {x_{ir} = 1\left| {y_{i} } \right.} \right) & = \frac{{p\left( {x_{ir} = 1,y_{i} } \right)}}{{p\left( {y_{i} } \right)}} \\ & = \frac{{p\left( {x_{ir} = 1,x_{iq} = 1 \cup x_{ir} = 1,y_{i} } \right)}}{{p\left( {y_{i} } \right)}} \\ & = \frac{{p\left( {x_{ir} = 1,x_{iq} = 1 \cup x_{ir} = 1,y_{i} } \right)}}{{p\left( {x_{iq} = 1 \cup x_{ir} = 1,y_{i} } \right)}} \times \frac{{p\left( {x_{iq} = 1 \cup x_{ir} = 1,y_{i} } \right)}}{{p\left( {y_{i} } \right)}} \\ & = \frac{{p\left( {x_{ir} = 1,x_{iq} = 1 \cup x_{ir} = 1} \right)}}{{p\left( {x_{iq} = 1 \cup x_{ir} = 1} \right)}} \times \frac{{p\left( {x_{iq} = 1 \cup x_{ir} = 1,y_{i} } \right)}}{{p\left( {y_{i} } \right)}} \\ = p\left( {x_{ir} = 1\left| {x_{iq} = 1 \cup x_{ir} = 1} \right.} \right)p\left( {x_{iq} = 1 \cup x_{ir} = 1\left| {y_{i} } \right.} \right) \\ \end{aligned} $$

The transition from line one to two follows because P(A, A ∪ B) = P(A). The transition from line two to three follows by simple algebraic manipulation. The transition from line three to four uses the fact that x _ir  = 1 is independent of y _i , conditional on x _iq  = 1 ∪x _ir  = 1. That is, $p\left( {x_{ir} = 1\left| {x_{iq} = 1 \cup\,x_{ir} = 1,y_{i} } \right.} \right) = p\left( {x_{ir} = 1,\left| {x_{iq} = 1 \cup x_{ir} = 1} \right.} \right).$ Finally, the transition from line four to five follows from the definition of conditional probability.

To see the independence of x _ir  = 1 from y _i conditional on x _iq  = 1 ∪ x _ir  = 1, note that, because of the redundancy relationship between loci q and r,

$$ p\left( {y_{i} \left| {x_{ir} = 1} \right.} \right) = p\left( {y_{i} \left| {x_{iq} = 1 \cup\,x_{ir} = 1} \right.} \right) $$

Consequently,

$$ p\left( {y_{i} ,x_{iq} = 1 \cup\,x_{ir} = 1} \right) = \frac{{p\left( {x_{iq} = 1 \cup\,x_{ir} = 1} \right)p\left( {y_{i} ,x_{ir} = 1} \right)}}{{p\left( {x_{ir} = 1} \right)}}. $$

Therefore,

$$ \begin{aligned} p\left( {x_{ir} = 1\left| {x_{iq} = 1 \cup\,x_{ir} = 1,y_{i} } \right.} \right) & = \frac{{p\left( {x_{ir} = 1,x_{iq} = 1 \cup\,x_{ir} = 1,y_{i} } \right)}}{{p\left( {x_{iq} = 1 \cup\,x_{ir} = 1,y_{i} } \right)}} \\ & = \frac{{p\left( {x_{ir} = 1,x_{iq} = 1 \cup\,x_{ir} = 1,y_{i} } \right)p\left( {x_{ir} = 1} \right)}}{{p\left( {x_{iq} = 1 \cup\,x_{ir} = 1} \right)p\left( {x_{ir} = 1,y_{i} } \right)}} \\ & = \frac{{p\left( {x_{ir} = 1} \right)}}{{p\left( {x_{iq} = 1 \cup\,x_{ir} = 1} \right)}} \\ & = p\left( {x_{ir} = 1\left| {x_{iq} = 1 \cup\,x_{ir} = 1} \right.} \right) \\ \end{aligned}. $$

This equality defines conditional independence.