Novel Sib Pair Selection Strategy Increases Power in Quantitative Association Analysis

Abstract

Quantitative-trait association studies have been widely used in search for genetic loci for complex traits in recent years. Yet, fiscal constraints still prohibit many on-going research projects from recruiting a large number of individuals for genotyping to reach a desired level of statistical power. Accordingly, in this article, we describe a novel sib pair sampling strategy for genotyping in QTL association studies. With the use of phenotypic scores (and IBD allele-sharing probabilities if available), the genetic effect of a biallelic additive trait locus can be properly modelled within the maximum-likelihood variance components framework proposed by Fulker et al. (Am J Hum Genet 64(1):259–267, 1999) and sib pairs can be rank-ordered by use of informativeness indices. The performance of our method was investigated using simulation. The power of our approach was shown to be higher when compared with other phenotypic selection schemes. An R-script implementing all the selection approaches (including the traditional phenotype-based ones) used in the simulation is available at http://statgen.hku.hk/jshkwan.

Appendix

For an additive locus with a trait-increasing allele, A ₁, of frequency p and the other allele, A ₂, of frequency 1 − p, the additive effects (a) of genotypes A ₁ A ₁, A ₁ A ₂ and A ₂ A ₂, can be coded as 1, 0.5 and 0, respectively. The expected additive effect of the locus in a random mating and large population, E(a), is therefore, 1 × p ² + 0.5 × 2p(1 − p) + 0 × (1 − p)² = p, and the variance,

$$ {\text{Var}}(a) = E(a^{2} )-\left[ {E(a)} \right]^{2} = [1^{2} \times p^{2} + (0.5)^{2} \times 2p(1-p) + 0^{2} \times (1-p)^{2} ]-p^{2} = p(1- p)/2 $$

Let Q = α + βa, where α and β are constants such that E(Q) = 0 and Var(Q) = 1, so

$$ \left\{ {\begin{array}{*{20}c} {E(Q) = \alpha + \beta \cdot E(a) = \alpha + \beta p = 0} \hfill \\ {{\text{Var}}(Q) = {\text{Var}}(\beta a) = \beta^{2} {\text{Var}}(a) = \beta p(1 - p)/2 = 1} \hfill \\ \end{array} } \right. $$

By solving the equations, one gets $\alpha = { - p/\sqrt {p(1 - p)/2} }$ and $\beta = {1/\sqrt {p(1 - p)/2} }$, respectively.