Economics of genomic selection: the role of prediction accuracy and relative genotyping costs

Abstract

Studies that incorporate economic aspects of plant breeding into the evaluation of selection strategies tend to focus on a specific example. This makes it difficult to generalize the conditions under which one strategy is more cost-effective than the other. We provide a general, average cost framework for quantifying the effects of genomic selection prediction accuracy and varying cost ratios of phenotyping to genotyping on the economic performance of genomic selection relative to traditional phenotypic selection. We assess prediction accuracy as a stochastic function of trait heritability, population-specific effective number of chromosome segments underlying a trait, and training population size. In addition, we set up an analytical method for determining the economically optimal size of the training population under varying cost scenarios for traits that differ with respect to heritability. The results provide quantitative estimates of the economic performance of genomic selection under a wide range of scenarios. The benefits of increasing the training population size beyond the typical size of 400 lines tend to be higher for traits with heritability below 0.1 and for populations with the effective number of chromosome segments above 100. Genomic selection may offer promising economic advantages, but only for traits with heritability roughly below 0.25, unless the per-line cost of phenotyping is higher than the genotyping cost, and the effective number of chromosome segments is smaller than about 100. The model can be used for preliminary economic assessment of genomic selection, and it can be expanded to account for subjective risk preferences of plant breeders.

Appendices

Appendix 1

First and second partial derivatives of the squared prediction accuracy function ($r_{{jN_{j} }}^{2}$) with respect to the size of the training population, N _j heritability, $h_{j}^{2}$, and the effective numbers of chromosome segments, M.

Since we are interested only in the signs of the derivatives, for simplicity, we use $r_{{jN_{j} }}^{2}$ rather than $r_{{jN_{j} }}$ to derive the first and second derivatives. $r_{{jN_{j} }}^{2}$ is a monotonically increasing transformation of $r_{{jN_{j} }}$, so the signs of the first derivatives of the two function are the same. If the second derivative of $r_{{jN_{j} }}^{2}$ is negative, the second derivative of $r_{{jN_{j} }}$ must also be negative. Only in the case of a positive second derivative of $r_{{jN_{j} }}^{2}$ the sign of the second derivative of $r_{{jN_{j} }}$ could not be determined using this method, but none of the second derivatives of $r_{{jN_{j} }}^{2}$ are positive.  

$$\frac{{\partial r_{{jN_{j} }}^{2} }}{{\partial N_{j} }} = \frac{{h_{j}^{2} \left( {N_{j} h_{j}^{2} + M} \right) - N_{j} h_{j}^{4} }}{{\left( {N_{j} h_{j}^{2} + M} \right)^{2} }} = \frac{{h_{j}^{2} M}}{{\left( {N_{j} h_{j}^{2} + M} \right)^{2} }} > 0.$$

$$\frac{{\partial r_{{jN_{j} }}^{2} }}{{\partial^{2} N_{j} }} = \frac{{2M\left( {N_{j} h_{j}^{2} + M} \right)}}{{\left( {N_{j} h_{j}^{2} + M} \right)^{3} }} < 0$$

$$\frac{{\partial r_{{jN_{j} }}^{2} }}{{\partial h_{j}^{2} }} = \frac{{N_{j} \left( {N_{j} h_{j}^{2} + M} \right) - N_{j}^{2} h_{j}^{2} }}{{\left( {N_{j} h_{j}^{2} + M} \right)^{2} }} = \frac{{N_{j} M}}{{\left( {N_{j} h_{j}^{2} + M} \right)^{2} }} > 0$$

$$\frac{{\partial r_{{jN_{j} }}^{2} }}{{\partial^{2} h_{j}^{2} }} = \frac{{2M\left( {N_{j} h_{j}^{2} + M} \right)}}{{\left( {N_{j} h_{j}^{2} + M} \right)^{3} }} < 0$$

$$\frac{{\partial r_{{jN_{j} }}^{2} }}{\partial M} = \frac{1}{{\left( {N_{j} h_{j}^{2} + M} \right)^{2} }} < 0$$

$$\frac{{\partial r_{{jN_{j} }}^{2} }}{{\partial^{2} M}} = \frac{1}{{2\left( {N_{j} h_{j}^{2} + M} \right)^{3} }} > 0$$

Prediction accuracy increases with $N_{j}$, but at a decreasing rate. This further implies diminishing returns (in terms of genetic gain) to the size of the training population. Similar to the size of the training population, the first and second derivatives of $r_{{jN_{j} }}^{2}$ with respect to $h_{j}^{2}$ suggest that the genetic gain response function increases at a decreasing rate with trait heritability. On the other hand, genetic gain falls at a decreasing rate as $M$ increases.

Appendix 2: Derivation of the optimal training population size ($N_{{Opt_{j} }}$)

Simplifying notation:

$$N_{j} = N$$

$$C_{{p_{j} }} = a$$

$$C_{{p_{j} }} N_{T} = b$$

$$S_{{b_{j} }} = c$$

$$h_{j}^{4} = d$$

$$h_{j}^{2} = e$$

$$M = g$$

Direct link to the derivation: http://www.derivative-calculator.net/#expr = sqrt%283%2F2%29%2A%28a%2An%2Bb%29%2F%28c%2Asqrt%28n%2Ad%2F%28e%2An%2Bg%29%29%29&diffvar = n&simplify = 1&showsteps = 1

$$\frac{\partial }{\partial n}f\left( {a,b,c,d,e,g,N} \right) = \frac{\partial }{\partial N}\left( {\frac{{\sqrt 3 \left( {aN + b} \right)}}{{\sqrt 2 c\sqrt {\frac{dN}{eN + g}} }}} \right) = \frac{{\sqrt 3 \frac{\partial }{\partial N}\left( {\frac{aN + b}{{\sqrt {\frac{dN}{eN + g}} }}} \right)}}{\sqrt 2 c}$$

$$= \frac{{\sqrt 3 \frac{{\sqrt {\frac{dN}{eN + g}} \frac{\partial }{\partial N}\left( {aN + b} \right) - \frac{\partial }{\partial N}\left( {\sqrt {\frac{dN}{eN + g}} } \right)\left( {aN + b} \right)}}{{\frac{dN}{eN + g}}}}}{\sqrt 2 c}$$

$$= \frac{{\sqrt 3 \left( {eN + g} \right)\left( {a\sqrt {\frac{dN}{eN + g}} - \frac{1}{{2\sqrt {\frac{dN}{eN + g}} }}\frac{\partial }{\partial N}\left( {\frac{dN}{eN + g}} \right)\left( {aN + b} \right)} \right)}}{\sqrt 2 cdN}$$

$$= \frac{{\sqrt 3 \left( {eN + g} \right)\left( {a\sqrt {\frac{dN}{eN + g}} - \frac{{d\frac{\partial }{\partial N}\left( {\frac{N}{eN + g}} \right)\left( {aN + b} \right)}}{{2\sqrt {\frac{dN}{eN + g}} }}} \right)}}{\sqrt 2 cdN}$$

$$= \frac{{\sqrt 3 \left( {eN + g} \right)\left( {a\sqrt {\frac{dN}{eN + g}} - \frac{{d\frac{{\left( {eN + g} \right)\frac{\partial }{\partial N}\left( N \right) - \frac{\partial }{\partial N}\left( {eN + g} \right)N}}{{\left( {eN + g} \right)^{2} }}\left( {aN + b} \right)}}{{2\sqrt {\frac{dN}{eN + g}} }}} \right)}}{\sqrt 2 cdN}$$

$$= \frac{{\sqrt 3 \left( {eN + g} \right)\left( {a\sqrt {\frac{dN}{eN + g}} - \frac{{d\left( {aN + b} \right)\left( {1\left( {eN + g} \right) - eN} \right)}}{{2\left( {eN + g} \right)^{2} \sqrt {\frac{dN}{eN + g}} }}} \right)}}{\sqrt 2 cdN}$$

$$= \frac{{\sqrt 3 \left( {eN + g} \right)\left( {a\sqrt {\frac{dN}{eN + g}} - \frac{{dg\left( {aN + b} \right)}}{{2\left( {eN + g} \right)^{2} \sqrt {\frac{dN}{eN + g}} }}} \right)}}{\sqrt 2 cdN}$$

$$= \frac{{\sqrt 3 d\left( {2eaN^{2} + agN - bg} \right)}}{{2^{{\frac{3}{2}}} c(eN + g)^{2} \left( {\frac{dN}{eN + g}} \right)^{{\frac{3}{2}}} }}$$

The optimal N is subject to

$$\frac{\partial }{\partial N}f\left( {a,b,c,d,e,g,N} \right) = 0 .$$

or

$$\,2aeN^{2} + agN - bg = 0.$$

which implies solutions for the quadratic equation

$$\, N_{1,2} = \frac{{ - ag \pm \sqrt {a^{2} g^{2} - 8aebg} }}{4ae}.$$

or

$$N_{{Opt_{j} }} \frac{{ - C_{{p_{j} }} M + \sqrt {C_{{p_{j} }}^{2} M^{2} + 8C_{{p_{j} }} C_{{g_{j} }} h_{j}^{2} MN_{T} } }}{{4C_{{p_{j} }} h_{j}^{2} }}$$