Abstract
The power of testing for a population-wide association between a biallelic quantitative trait locus and a linked biallelic marker locus is predicted both empirically and deterministically for several tests. The tests were based on the analysis of variance (ANOVA) and on a number of transmission disequilibrium tests (TDT). Deterministic power predictions made use of family information, and were functions of population parameters including linkage disequilibrium, allele frequencies, and recombination rate. Deterministic power predictions were very close to the empirical power from simulations in all scenarios considered in this study. The different TDTs had very similar power, intermediate between one-way and nested ANOVAs. One-way ANOVA was the only test that was not robust against spurious disequilibrium. Our general framework for predicting power deterministically can be used to predict power in other association tests. Deterministic power calculations are a powerful tool for researchers to plan and evaluate experiments and obviate the need for elaborate simulation studies.
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Acknowledgements
We are grateful to Ian White and Dr O Southwood for helpful comments on earlier versions of this manuscript. This work has been supported by Sygen International, and by the Biotechnology and Biological Sciences Research Council of UK.
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Appendices
Appendix A
The NCP (λ) of two-way ANOVA can be expressed as25
Let σe2 be unity. Let B′ be the vector [μ, f1, f2, f3, g1, g2, g3] of parameters in the model, where μ is the sample mean, fi is the mean of the ith parental type, and gj the mean of the jth marker genotype across all parental types. Let K be a matrix of parameter contrasts reflecting the H0 being tested; for example, if H0: g1=g2 and g2=g3, then
The matrix X′X is
where nij is the number of records in the ith family and jth marker genotype class. X′X is a matrix of order 7 and rank 5; hence, there are seven unknowns and only five df. An appropriate generalisation of X′X is obtained deleting the first row and column, hence setting μ=0, and the last row and column, hence setting g3=0.25 Let G be the reduced X′X matrix. This G matrix can be partitioned as follows:
Then, if C=G−1, K*′ is the matrix K′ with the first and last columns deleted, and B* is the vector B with the first and last elements deleted, then
where C22 = K*CK* = (G22 - G21 G11- 1 G12)- 1 and (K*′ B*)′ = [g1 - g2, g2] When testing the QTL, Eq. (A2) gives
where the first part of (A3) corresponds to the sum of squares due to genotype, and the second part of (A3) corresponds to the sum of squares due to parental type.
However, it is a linked marker, rather than the QTL, what usually is being tested. Thus, Eq. (A3) needs to accommodate this fact. Using Tables 2 and 3 in the Materials and Methods section, the new λ can be written as
where bi is the expected marker genotype effect among progeny in the ith trio class, ni the number of trios in class i, Ii(j) an indicator variable equal to 1 if the trio is informative and 0 otherwise. Table A1 shows Fj, the mean value of the jth parental type, and fj, the number of these parental types.
It is also possible to use all trios, thus setting Ii(j)=1 for all i (j), without increasing the type-I error rate. By doing so, power increases slightly, through augmenting the residual df, and ascertainment of informative families becomes unnecessary.
This method of obtaining λ can be applied to derive the NCP for nested ANOVA; however, the algebra becomes more tedious. Finally, the NCP λO can be derived through Eq. (3), although a simpler method was described in the Materials and Methods section.
Appendix B
Let us consider two fixed effects, α and β, where α could represent the factor parental type, and β could represent the genotype of the progeny. Thus, the model can be written as yij=μ+αi+βi+eij, which corresponds to a two-way ANOVA model without interaction. We will now show that the original statistic F2,n−5 for 14 is equivalent to the F-ratio for testing the effects of β after having corrected for the effects due to μ and α, using the previous model.
For a constant k = 2/n - 5, we can see that
where SSμα and SSμαβ are the sum of squares explained by a model that fits μ and α, and by a model that fits μ, α, and β, respectively; SST is the total sum of squares; and Rβ∣μ,α2 and Re2 are the proportions of the total variance explained by β, after taking into account the effects of μ and α, and the proportion of unexplained variance, respectively. The null hypothesis of interest is whether factor β explains a significant amount of phenotypic variance over and above the amount explained by μ and α jointly. The F-ratio that appropriately reflects this null hypothesis is given in Eq. (B1).
Appendix C
Let assume T is a random variable following a t-distribution, and let σT be the standard deviation of T. A first-order Taylor's approximation for λ is λ = E(T/σT) ≈ E[T]/E[σT].32 In order to derive E[T] and E[σT], we used the probabilities of the 10 different types of trios and the expected effects of marker genotypes in the progeny contained in Tables 2 and 3. Hence, conditional on pM, pQ, c, and D′, E[T] = E[∑in (yi - ȳ)wi], and because all family trios are independent (ie unrelated) E[T] = NE[(y - ȳ)w], where y, the phenotype, and w, a weighting factor, are expectations for a single trio (Table 3). Thus, the expected value of the numerator of TDTR is approximately E[T] = NpM pm [pM2 (b2 - b3) + pM pm (b5 - b7) + pm2 (b8 - b9)]. When analysing the QTL, and assuming no dominance, the previous equation simplifies to E[T] = NpQ pqa.
The expected variance of T, E[σT2], is the same regardless of whether the locus being tested is the QTL or a marker. Equation (A1.23a) in Reference32 is 
, which reduces to 
if the second term can be ignored. Hence, E[σT2] = E[1/4 ∑in (yi - ȳ)2 Hi] = 1/4E[(y - ȳ)2 H] and, as the expectation of a random variable X given another random variable Y is E[X] = E[E[X∣Y]], E[(y - ȳ)2 H] = ∑H = 02 HPHE(y - ȳ)2 = pM pm σe2 + σQTL2 Finally, dividing E[T] by 
we obtain 
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Hernández-Sánchez, J., Haley, C. & Visscher, P. Power of QTL detection using association tests with family controls. Eur J Hum Genet 11, 819–827 (2003). https://doi.org/10.1038/sj.ejhg.5201042
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DOI: https://doi.org/10.1038/sj.ejhg.5201042
