Genomic selection for marker-assisted improvement in line crosses

Abstract

Efficiency of genomic selection with low-cost genotyping in a composite line from a cross between inbred lines was evaluated for a trait with heritability 0.10 or 0.25 using a low-density marker map. With genomic selection, selection was on the sum of estimates of effects of all marker intervals across the genome, fitted either as fixed (fixed GS) or random (random GS) effects. Reponses to selection over 10 generations, starting from the F₂, were compared with standard BLUP selection. Estimates of variance for each interval were assumed independent and equal. Both GS strategies outperformed BLUP selection, especially in initial generations. Random GS outperformed fixed GS in early generations and performed slightly better than fixed GS in later generations. Random GS gave higher genetic gain when the number of marker intervals was greater (180 or 10 cM intervals), whereas fixed GS gave higher genetic gain when the number of marker intervals was low (90 or 20 cM). Including interactions between generation and marker scores in the model resulted in lower genetic gains than models without interactions. When phenotypes were available only in the F₂ for GS, treating marker scores as fixed effects led to considerably lower genetic gain than random GS. Benefits of GS over standard BLUP were lower with high heritability. Genomic selection resulted in greater response than MAS based on only significant marker intervals (standard MAS) by increasing the frequency of QTL with both large and small effects. The efficiency of genomic selection over standard MAS depends on stringency of the threshold used for QTL detection. In conclusion, genomic selection can be effective in composite lines using a sparse marker map.

Appendix

The complete derivation is given below.

Using

$$ V(A) = {\mathop E\limits_B }[V(A|B)] + V[{\mathop E\limits_B }(A|B)]. $$

Let

$$ A = \beta _{i} \times {\text{MS}}_{{ijk}} \quad {\text{and}}\quad B = {\text{MS}}_{{ijk}} . $$

It was assumed that marker intervals were independent and each marker interval contributed equal variance.

Then

$$ V(\beta _{i} \times {\text{MS}}_{{ijk}} ) = {V_{{\text{G}}} } \mathord{\left/ {\vphantom {{V_{{\text{G}}} } N}} \right. \kern-\nulldelimiterspace} N, $$

where N = number of marker intervals (90 or 180).

$$ V(A) = {\mathop E\limits_{{\text{MS}}} }[V(\beta _{i} \times {\text{MS}}_{{ijk}} |{\text{MS}}_{{ijk}} = i)] + V{\mathop {[E}\limits_{{\text{MS}}} }(\beta _{i} \times {\text{MS}}_{{ijk}} |{\text{MS}}_{{ijk}} = i)], $$

where

$$ \begin{aligned}{} & {\mathop E\limits_{{\text{MS}}} }[V(\beta _{i} \times {\text{MS}}_{{ijk}} |{\text{MS}}_{{ijk}} = i)] \\ & \quad = {\sum\limits_{i = 0}^4 {P({\text{MS}}_{{ijk}} = i) \times V(\beta _{i} \times {\text{MS}}_{{ijk}} |{\text{MS}}_{{ijk}} = i)} } \\ & \quad = {\sum\limits_{i = 0}^4 {P({\text{MS}}_{{ijk}} = i) \times i^{2} \times V(\beta _{i} )} } \\ \end{aligned} $$

and

$$\begin{aligned}{} & V[{\mathop E\limits_{{\text{MS}}} }(\beta _{i} \times {\text{MS}}_{{ijk}} |{\text{MS}}_{{ijk}} = i)] \\ & \quad = {\sum\limits_{i = 0}^4 {P({\text{MS}}_{{ijk}} = i) \times [E(\beta _{i} \times {\text{MS}}_{{ijk}} |{\text{MS}}_{{ijk}} = i)]^{2} - \left[{\sum\limits_{i = 0}^4 {P({\text{MS}}_{{ijk}} = i) \times E(\beta _{i} \times {\text{MS}}_{{ijk}} |{\text{MS}}_{{ijk}} = i)}}\right]^{2}}} \\ & \quad = {\sum\limits_{i = 0}^4 {P({\text{MS}}_{{ijk}} = i) \times [i \times E(\beta _{i} )]^{2} - \left[{\sum\limits_{i = 0}^4 {P({\text{MS}}_{{ijk}} = i) \times i \times E(\beta _{i} )}}\right]^{2}}} \\ & \quad = \,0\,\quad {\text{because}}\,\quad E(\beta _{i} ) =\frac{{\mu _{{\text{Phen}}\_{\text{Line1}}} - \mu _{{\text{Phen}}\_{\text{Line2}}}}} {N}, \\ \end{aligned} $$

which in our case is equal to zero because the expected means of the two lines are equal. This would be true for the 50/50 case but not for the 75/25 case.

Hence,

$$ {V_{{\text{G}}} } \mathord{\left/ {\vphantom {{V_{{\text{G}}} } N}} \right. \kern-\nulldelimiterspace} N = {\sum\limits_{i = 0}^4 {P({\text{MS}}_{{ijk}} = i) \times i^{2} \times V(\beta _{i} )} }\quad \,{\text{and}}\quad \,V(\beta _{i} ) = V_{{\text{G}}} /[N \times i^{2} \times {\sum\limits_{i = 0}^4 {P({\text{MS}}_{{ijk}} = i)} }] $$