Robust ranks of true associations in genome-wide case-control association studies

Notation and model

Consider a SNP with alleles D and d and frequencies p and q = 1 - p, respectively. In a case-control design, r cases and s controls are independently sampled from a population. The genotype counts of three genotypes G₀ = dd, G₁ = Dd, and G₂ = DD are denoted as (r₀, r₁, r₂) in cases and (s₀, s₁, s₂) in controls, which follow multinomial distributions mul(r: p₀, p₁, p₂) and mul(s: q₀, q₁, q₂), respectively. Denote the disease prevalence as k and penetrances as f _i = P(case|G _i ) for i = 0, 1, 2. By the Bayes Theorem, p _i = g _i f _i /k and q _i = g _i (1 - f _i )/(1 - k), where g _i = P(G _i ). Without loss of generality, assume that D has high risk. Then the null hypothesis of no association can be stated as H₀: f₀ = f₁ = f₂ = k. The alternative hypothesis is H₁: f₀ ≤ f₁ ≤ f₂ with at least one inequality. The genotype relative risks (GRRs) are defined as λ₁ = f₁/f₀ and λ₂ = f₂/f₀. The recessive, additive, and dominant models are referred to as λ₁ = 1, λ₁ = (1 + λ₂)/2, and λ₁ = λ₂, respectively [2–4].

Trend tests and robust tests

where (x₀, x₁, x₂) = (0, x, 1) and 0 ≤ x ≤ 1. Given x, Z _x follows asymptotically N(0,1). The choice of x is 0, 1/2, and 1 for the recessive, additive/multiplicative, and dominant models, respectively [5]. In practice, however, the true genetic model is unknown. Hence the robust tests, maximin efficiency robust test (MERT) and maximum test (MAX), can be applied, which are given by MERT = (Z₀ + Z₁)/{2(1 + ρ)}^1/2 and MAX = max(|Z₀|, |Z_1/2|, |Z₁|), where ρ = [n₀n₂/{(n₀ + n₁)(n₁ + n₂)}]^1/2 [4]. Note that Pearson's association test can also be used. However, Zheng et al. [6] showed that the MAX is often more powerful than the Pearson chi-squared test for a case-control design. Comparison of MERT and MAX can be found in Freidlin et al. [7]. The MAX and MERT have also been applied to other designs for GAW14 [8, 9].

Ranking markers with multiple statistics

When the genetic model is unknown, the three CATTs (Z₀, Z_1/2, Z₂) are calculated for each of M SNPs. Then the p-values of MERT and MAX can be obtained for ranking. However, computing the p-value of MAX needs extensive simulation. Thus, alternatively, the minimum of the p-values (min p) of the three CATTs can be used for ranking. Rather than ranking M SNPs based on any single CATT, we propose ranking the SNPs by the MERT and the minimum of the p-values. We expect that ranking SNPs based on this approach would be more robust compared to ranking by a single CATT when the ranks by the three CATTs are quite different.

Application to GAW15

As an application, we consider the first simulated data set of Problem 3 from GAW15. A simulated data set was considered, as we knew that there were eight candidate genes. One of them at chromosome 6 with physical location 32,484,648 bp was simulated based on the DRB1 locus of the HLA gene. We selected four SNPs closest in physical distance to the eight known candidate genes as candidate SNPs. We examined the ranks of the 32 candidate SNPs among all 9187 SNPs. All 2000 unrelated controls were used. For the affected sib-pair (ASP) data, we selected an affected sib (case) with the first individual ID from each family. A total of 1500 unrelated cases were used. In the simulated data set, genotypes of all 9187 SNPs from 22 chromosomes were generated (no missing genotypes and no genotyping errors). All SNPs had minor allele frequency (MAF) greater than 1% and there were no monomorphisms. Because we considered the CATTs, Hardy-Weinberg equilibrium in the population was not required [2]. If any genotype count in cases or controls was 0, 0.5 was added to all genotype counts in cases and controls.

An independent simulation study