The impact of linkage disequilibrium and epistasis in the studies of inheritance based on Hayman’s diallel and generation mean analysis

Abstract

This simulation-based study assessed the impact of linkage disequilibrium (LD) and epistasis on Hayman’s diallel and generation mean analysis, assuming hundreds of genes, variable degree of dominance, and seven types of digenic epistasis. The diallel parents were 15 doubled haploids from a high LD population. The generation mean analysis was based on seven generations, assuming association. Under low LD and no epistasis, the diallel analysis provided confident results about the inheritance of the quantitative trait and high correlation between number of recessive genes and W_r + V_r, but biased estimates of the dominance components and genetic parameters. The additional consequences of high LD under no epistasis were rejection of the additive-dominance model assuming high heritability and lower correlation. Assuming 100% of epistatic genes, for four epistasis types there was evidence of inadequacy of the additive-dominance model. Assuming 30% of epistatic genes, there was a tendency for accepting the additive-dominance model for low heritability traits and for rejecting for high heritability traits. Linkage and epistasis affect the estimates of the genetic components of the generation means. Even assuming 100% of interacting genes, for most epistasis types there was no statistical evidence of epistasis. Assuming positive partial dominance, the signs of the epistatic components do not allow discriminate complementary, recessive, dominant and recessive, duplicate genes with cumulative effects, and non-epistatic genic interaction. Negative epistatic components evidence dominant epistasis. When the additive × additive and dominance × dominance components are positive and the additive × dominance component is negative, there is duplicate epistasis.

Appendix

The expectation of the epistatic effects in the F_n+2 generation is $E\left( I \right)^{\left( n \right)} = E\left( I \right)^{\left( 0 \right)} + p_{22}^{\left( n \right)} I_{22} + p_{21}^{\left( n \right)} I_{21} + p_{20}^{\left( n \right)} I_{20} + p_{12}^{\left( n \right)} I_{12} + p_{11}^{\left( n \right)} I_{11} + p_{10}^{\left( n \right)} I_{10} + p_{02}^{\left( n \right)} I_{02} + p_{01}^{\left( n \right)} I_{01} + p_{00}^{\left( n \right)} I_{00}$, where n is the number of selfing generations and

$$p_{22}^{\left( n \right)} = \left( {F/2} \right)\left( {f_{21}^{\left( 0 \right)} + f_{12}^{\left( 0 \right)} } \right) + P_{1}^{\left( n \right)}$$

$$p_{21}^{\left( n \right)} = \left( {1 - F} \right)\left[ {f_{21}^{\left( 0 \right)} + \left( {1 - c^{n} } \right)f_{11}^{\left( 0 \right)} /2} \right] - f_{21}^{\left( 0 \right)}$$

$$p_{20}^{\left( n \right)} = \left( {F/2} \right)\left( {f_{21}^{\left( 0 \right)} + f_{10}^{\left( 0 \right)} } \right) + P_{2}^{\left( n \right)}$$

$$p_{12}^{\left( n \right)} = \left( {1 - F} \right)\left[ {f_{12}^{\left( 0 \right)} + \left( {1 - c^{n} } \right)f_{11}^{\left( 0 \right)} /2} \right] - f_{12}^{\left( 0 \right)}$$

$$p_{11}^{\left( n \right)} = \left[ {\left( {1 - F} \right)c^{n} - 1} \right]f_{11}^{\left( 0 \right)}$$

$$p_{10}^{\left( n \right)} = \left( {1 - F} \right)\left[ {f_{10}^{\left( 0 \right)} + \left( {1 - c^{n} } \right)f_{11}^{\left( 0 \right)} /2} \right] - f_{10}^{\left( 0 \right)}$$

$$p_{02}^{\left( n \right)} = \left( {F/2} \right)\left( {f_{01}^{\left( 0 \right)} + f_{12}^{\left( 0 \right)} } \right) + P_{2}^{\left( n \right)}$$

$$p_{01}^{\left( n \right)} = \left( {1 - F} \right)\left[ {f_{01}^{\left( 0 \right)} + \left( {1 - c^{n} } \right)f_{11}^{\left( 0 \right)} /2} \right] - f_{01}^{\left( 0 \right)}$$

$$p_{00}^{\left( n \right)} = \left( {F/2} \right)\left( {f_{01}^{\left( 0 \right)} + f_{10}^{\left( 0 \right)} } \right) + P_{1}^{\left( n \right)}$$

where $F = 1 - \left( {1/2} \right)^{n}$ is the inbreeding coefficient, $f_{ij}^{\left( 0 \right)}$ is a genotype probability in F₂ (i and j = 2, 1, or 0), $c = 1 - 2r\left( {1 - r} \right)$, $P_{1}^{\left( n \right)} = \left( {1/4} \right)\left\{ {\left[ {F - \left( {1 - F} \right)\left( {1 - c^{n} } \right)} \right]f_{11}^{\left( 0 \right)} + c_{1}^{\left( n \right)} \left( {1 - 2r} \right)\Delta } \right\}$, $P_{2}^{\left( n \right)} = \left( {1/4} \right)\left\{ {\left[ {F - \left( {1 - F} \right)\left( {1 - c^{n} } \right)} \right]f_{11}^{\left( 0 \right)} - c_{1}^{\left( n \right)} \left( {1 - 2r} \right)\Delta } \right\}$, $c_{1}^{\left( n \right)} = \left\{ {2\left[ {1 - \left( {\left( {1 - 2r} \right)/2} \right)^{n} } \right]/\left( {1 + 2r} \right)} \right\}$, and $\left| \Delta \right| = \left( {1 - 2r} \right)/4$ (positive for coupling and negative for repulsion). Defining $P_{11}^{\left( 0 \right)}$, $P_{10}^{\left( 0 \right)}$, $P_{01}^{\left( 0 \right)}$, and $P_{00}^{\left( 0 \right)}$ as the gamete probabilities of the F₁, where, for example, $P_{11}^{\left( 0 \right)} = P_{00}^{\left( 0 \right)} = \left( {1 - r} \right)/2$ and $P_{10}^{\left( 0 \right)} = P_{01}^{\left( 0 \right)} = r/2$ for coupling, the probability of the genotype AABb in F₂ is $f_{21}^{\left( 0 \right)} = 2P_{11}^{\left( 0 \right)} P_{10}^{\left( 0 \right)}$.