Single Marker Association Analysis for Unrelated Samples

Abstract

Methods for single marker association analysis are presented for binary and quantitative traits. For a binary trait, we focus on the analysis of retrospective case–control data using Pearson's chi-squared test, the trend test, and a robust test. For a continuous trait, typical methods are based on a linear regression model or the analysis of variance. We illustrate how these tests can be applied using a public available R package “Rassoc” and some existing R functions. Guidelines for choosing these test statistics are provided.

1. Introduction

Statistical procedures for testing whether there is an association between a phenotype and a single nucleotide polymorphism (SNP) are described and illustrated. Usually, the phenotype of interest is either a binary or quantitative one. For a binary trait, we focus on a retrospective case–control study, in which cases and controls are randomly drawn from case and control populations, respectively. For a continuous trait, the data are obtained from a random sample of the general population. Although a large number of SNPs is available for testing association, single marker analysis is often employed. The significance level to test a single hypothesis is 0.05. When multiple SNPs are tested, the Bonferroni correction can be applied.

Denote the genotypes of an SNP as G₀, G₁, and G₂. For case–control data, denote the penetrances as f₀, f₁, and f₂ with respect to the three genotypes, respectively. Under the null hypothesis H₀, we have f₀ = f₁ = f₂ = Pr(case). A genetic model is recessive if f₁ = f₀, additive if f₁ = (f₀ + f₂)/2, or dominant if f₁ = f₂. For a single SNP, the observed case–control data consist of genotype counts (r₀, r₁, r₂) among r cases and (s₀, s₁, s₂) among s controls. Denote n_j = r_j + s_j (j = 0, 1, 2) and n = r + s. The Cochran-Armitage trend test (referred to as the trend test) is one of the two most commonly used statistics for the analysis of case–control data. It can be written as (1, 2)

$$T_1\left(x\right)=\frac{\sum _{j=0}^2{x}_j\{(1-\phi )r_j-\phi s_j\}}{{\left[n\phi (1-\phi )\left\{\sum _{j=0}^2{x}_j^2{n}_j∕n-{\left(\sum _{j=0}^2{x}_j{n}_j∕n\right)}^2\right\}\right]}^{1∕2}}$$ (1)

where (x₀, x₁, x₂) = (0, x, 1), x is determined by the genetic model and φ = r/n. Under H₀, given x, T₁(x) asymptotically follows a standard normal distribution N(0,1). When the genetic model is recessive and the risk allele is known, T₁(0) is used. When the genetic model is dominant and the risk allele is known, T₁(1) is used. When we only know that the genetic model is recessive (or dominant) but not the risk allele, T₁(0) (or T₁(1)) cannot be used alone. When the genetic model is additive regardless of the risk allele, T₁(1/2) is used. Pearson's chi-squared test (referred to as Pearson's test) is another commonly used test. It can be written as

$$T_2=\sum _{j=0}^2\frac{{(r_j-r{n}_j∕n)}^2}{(r{n}_j∕n)}+\sum _{j=0}^2\frac{{(s_j-s{n}_j∕n)}^2}{(s{n}_j∕n)}.$$

Under H₀, T₂ asymptotically follows a chi-squared distribution with two degrees of freedom (df), denoted as ${\chi }_2^2$. The robust test, MAX3, is given by

$$\mathrm{MAX}3=\text{max}(\mid T_1\left(0\right)\mid ,\mid T_1(1∕2)\mid ,T_1\left(1\right)\mid ),$$

where T₁(0), T₁(1/2), and T₁(1) are the trend tests given by Eq. 1 with different x values. Under H₀, the asymptotic distribution of MAX3 is far more complex than those of T₁(x) and T₂. A procedure for determining its asymptotic null distribution and P-value is discussed in Note 2. For other robust tests, see Note 3.

The power of each statistical test depends on the underlying genetic model. It will be seen that MAX3 is more robust than any single trend test or Pearson's chi-squared test when the genetic model is unknown, thus, it should be used in practice. Since MAX3 does not follow any chi-squared distribution, we discuss how to find its P-value using the R package Rassoc (3). This package is available from the Comprehensive R Archive Network at “http://CRAN.R-project.org/package=Rassoc.”

For a continuous trait Υ, a typical model is ${\rm Y}=\mu +g+\in$, where μ is a fixed overall mean of the trait under H₀, g is the random genetic effect due to G, and ∈ is a random error. The genetic value of g is –a when G = G₀, d when G = G₁, and a when G = G₂. Under H₀, we have a = d = 0. A genetic model is recessive, additive, or dominant if d = a, d = 0, or d = a, respectively. The observed data consist of pairs $({{\rm Y}}_{ij},G_j)$ for i = 1,. . ., n_j and j = 0, 1, 2, where ${{\rm Y}}_{ij}$ is the trait value of the ith individual with genotype G_j. Denote n = n₀ + n₁ + n₂.

For the analysis of a quantitative trait, linear regression and the analysis of variance (ANOVA) are routinely used. Let (x₀, x₁, x₂) = (0, x, 1) be the values for the genotypes (G₀, G₁, G₂), where x = 0, 1/2, or 1 for the recessive, additive, or dominant models, respectively. Using the data $({{\rm Y}}_{ij},G_i),i=1,\dots ,n_j$ and j = 0, 1, 2, the F-test derived from a linear regression model is given by

$$F\left(x\right)=\frac{(n-2)\left\{\sum _j\sum _i(x_j-\stackrel{‒}{x}){({{\rm Y}}_{ij}-\stackrel{‒}{{\rm Y}})}^2\right\}}{\sum _j\sum _i{(x_j-\stackrel{‒}{x})}^2\sum _j\sum _i{({{\rm Y}}_{ij}-\stackrel{‒}{{\rm Y}})}^2-{\left\{\sum _j\sum _i(x_j-\stackrel{‒}{x})({{\rm Y}}_{ij}-\stackrel{‒}{{\rm Y}})\right\}}^2},$$ (2)

where $\stackrel{‒}{x}={\sum }_{j=0}^2{n}_j{x}_j∕n$ and $\stackrel{‒}{{\rm Y}}={\sum }_{j=0}^2{\sum }_{i=1}^{n_j}{{\rm Y}}_{ij}∕n$. Given x, F(x) has an asymptotic F-distribution with (1,n – 2) df under H₀.

Alternatively, denote ${\stackrel{‒}{{\rm Y}}}_{.j}={\sum }_{i=1}^{n_j}{{\rm Y}}_{ij}∕n_j$ for j = 0, 1, 2, $S{S}_b={\sum }_{j=0}^2{n}_j{({\stackrel{‒}{{\rm Y}}}_{.j}-\stackrel{‒}{{\rm Y}})}^2$ and $S{S}_e={\sum }_{j=0}^2{\sum }_{i=1}^{n_j}{({{\rm Y}}_{ij}-{\stackrel{‒}{{\rm Y}}}_{.j})}^2$. The F-test derived from the ANOVA is given by

$$F=\frac{S{S}_b∕2}{S{S}_b∕(n-3)},$$ (3)

which under H₀, has an asymptotic F-distribution with (2,n – 3) df. We illustrate how to use these F-tests using some existing R functions later.

The allele-based analysis, valid under Hardy–Weinberg equilibrium (HWE), has similar performance to the genotype-based analysis under the additive model. Therefore, we focus on the genotype-based analysis, which does not require HWE. In Note 4, we present some power comparisons of different F-tests for quantitative traits.

2. Methods

2.1. Analysis of Case–Control Data

Denote the genotypes as (G₀, G₁, G₂) = (AA, AB, BB). If we happen to know the risk allele, it is always denoted as B. The choice of test statistic depends on which of the following four situations holds: (1) the genetic model and the risk allele are known, (2) the genetic model is known but not the risk allele, (3) the risk allele is known but not the genetic model, and (4) neither the genetic model nor the risk allele is known. Common genetic models include recessive, additive, and dominant. It is important that one does not determine the genetic model and/or the risk allele from the same data that will be used in the subsequent association analysis. Of course, the genetic model and/or the risk allele may be known based on scientific knowledge or information from previous data. In our view, (3) and (4) are the most common situations in practice.

2.1.1. Which Test to Use?

Which test to choose depends on each of the four situations outlined above (see Note 1). Let ${\chi }_2^2(1-\alpha )$ be the upper 100(1 – α)th percentile of ${\chi }_2^2$ and z(1 – α) be the upper 100(1 – α)th percentile of N(0,1).

1. The genetic model and the risk allele are known. T₁(x) is optimal and should be used. Since the risk allele is known, a one-sided H₁ is used and z(0.95) = 1.645. For the recessive (additive, or dominant) model, reject H₀ if T₁(0) > z(0.95) (T₁(1/2) > z(0.95), or T₁(1) > z(0.95)). In each case, the P-value equals the probability of Z > T₁(x), where Z ~ N(0,1).

2. The genetic model is known but not the risk allele. When the model is additive, use T₁(1/2) and reject H₀ if |T₁(1/2)| > z (0.975) = 1.96. The P-value equals two times the probability of Z > |T₁(1/2)|. When the model is recessive or dominant, use MAX3 (see (3) next).

3. The genetic model is unknown (regardless of the risk allele). Use MAX3. Three approaches are available to calculate the P-value of MAX3 using the R package Rassoc. But we recommend using the one based on the asymptotic null distribution of MAX3.

4. The same as (3). Thus, we only discuss (3) in the following.

Note that we do not recommend T₂, because it is always less powerful than MAX3 (see Note 1). If T₂ is used, reject H₀ if $T_2>{\chi }_2^2\left(0.95\right)=5.9915$. The P-value equals the probability of T > T₂, where $T\sim {\chi }_2^2$.

2.1.2. Examples Using R

The R package Rassoc can be loaded from

> library(Rassoc);

There are two functions CATT(data,x) and MAX3(data, method,m) in the package for computing the trend tests and MAX3 and their P-values. Pearson's test and its P-value can be obtained using an existing R function. In both functions, the “data” comprises a 2 × 3 contingency table, i.e., genotype counts (r₀, r₁, r₂) for cases and (s₀, s₁, s₂) for controls. In the first function, “x” is 0, 0.5, or 1 for the recessive, additive, or dominant models, respectively. In the second function, the “method” refers to the procedure to calculate the P-value of MAX3. Three methods are available: “boot” for the bootstrap procedure, “bvn” for the bivariate normal procedure, or “asy” for the asymptotic procedure.

The first two procedures are simulation-based and the last one is based on the asymptotic distribution of MAX3 (see Note 2). The “m” in the second function refers to the number of replicates when “boot” or “bvn” is used. When “asy” is used, “m” can be any positive integer.

For illustration, we use a SNP (rs420259) reported by the WTCCC (4), which was the only SNP showing strong association with bipolar disorder in a genome-wide association study (GWAS) with 500,000 SNPs (the actual number of SNPs tested after quality control steps is less than 500,000). The genome-wide significance level used by (4) was 5 × 10^–7 for strong association. The genotype counts are (r₀, r₁, r₂) = (83, 755, 1020) and (s₀, s₁, s₂) = (260, 1134, 1537). The data can be entered as follows.

> = c(83,755,1020);
> s = c(260,1134,1537);
> a = matrix(rbind(r,s), nrow = 2, byrow = FALSE);

To check that the data are correctly entered, just type a.

> a;
83 755 1020
260 1134 1537

We may not have sufficient scientific knowledge to claim a priori which allele (A or B) is the risk one and what the true genetic model is. For illustration purpose, let us say we know a priori the true model is dominant and that B is the risk allele. The analysis is carried out based on the three situations outlined before.

1. If we know the genetic model is dominant and the risk allele is B, apply T₁(1) using the R function CATT as follows.

> CATT(a,1);
The Cochran-Armitage trend test
data: a
statistic = 5.7587 p-valve = 8.478e-09

The output shows that |T₁(1)| = 5.7587 and its P-value is 8.478 × 10^–9. This is a two-sided test. We use a one-sided test because the risk allele is known. Thus, the actual P-value is half of the reported one, that is, 4.239 × 10^–9, which is less than the significance level 5 × 10^–7. Hence we reject H₀.

• 2If we know the genetic model is dominant but not the risk allele, use MAX3 not T₁(1). Suppose we happen to enter the data as b and apply T₁(1) as before.

> r = c(1020,755,82);
> s = c(1537,1134,260);
> b = matrix(rbind(r,s), nrow = 2, byrow = FALSE);
> CATT(b,1);
The Cochran-Armitage trend test
data: b
statistic = 1.6618 p-valve = 0.09656

The two-sided P-value is 0.09656, which is not significant. This example shows that knowing the risk allele is necessary for using T₁(0) or T₁(1). The use of MAX3 is illustrated in (3) later. We first show how to use the following R function to obtain Pearson's test T₂ = 33.165 and its P-value 6.285 × 10^–8. This P-value is also significant but larger than that of T₁(1) in case (1), because T₁(1) is optimal for the dominant model. If we apply T₂ to the dataset b, we would obtain the same results.

> chisq.test(a);
Pearson's Chi-squared test
data: a
X-squared = 33.165, df = 2, p-value = 6.285e-08

• 3If we do not know the genetic model, we apply MAX3 and calculate its P-value using the “asy” procedure. The reported statistic is MAX3 = 5.7587 with P-value 2.347 × 10^–8. Thus, we reject H₀. Note that this P-value is smaller than that of T₂ but larger than that of T₁(1) in case (1).

> MAX3(a, “asy”, 1);
The MAX3 test using the asy method
data: a
statistic = 5.7587 p-value = 2.347e-08

If x = 0.5 is used in T₁(x) regardless of the true genetic model and the risk allele, the following results show that the P-value of T₁(1/2) is not significant.

> CATT(a,0.5);
The Cochran-Armitage trend test
data: a
statistic = 3.6966 p-value = 0.0002185

2.2. Quantitative Trait

2.2.1. Which Test to Use?

For a continuous trait, the four situations outlined before also apply. Let F_u,v(1 – α) be the upper 100(1 – α)th percentile of an F-distribution with (u,v) df.

1. When the genetic model and the risk allele are known, the statistic F(x) given in Eq. 2 is used. Since the risk allele is known, a one-sided H₁ is used. For the recessive (additive, dominant) model, reject H₀ if F(0) > F_{1,n – 2}(0.95) (F(1/2) > F_{1,n – 2}(0.95), F(1) > F_{1,n – 2}(0.95)), where F(0) (F(1/2), F(1)) is the observed statistic. In each case, the P-value equals half of the probability of f_{1,n – 2} > F(x), where f_{1,n – 2} follows an F-distribution with (1,n – 2) df.

2. The genetic model is known but not the risk allele. When the genetic model is additive, F(x) given in Eq. 2 is used (with x = 1/2). Reject H₀ if F(1/2) > F_{1,n – 2}(0.95), where F(1/2) is the observed statistic. The P-value equals the probability of f_{1,n – 2} > F(1/2). When the genetic model is not additive (either recessive or dominant), F given in Eq. 3 is used. Reject H₀ if F > F_{2,n – 3}(0.95), where F is also the observed statistic.

The P-value equals the probability of f_{2,n – 3} > F, where f_{2,n – 3} follows an F-distribution with (2,n – 3) df.

• 3When the genetic model is unknown, F given in Eq. 3 is used. The rejection rule and P-value are similar to those in case (2) when F is used.
• 4The same as (3). Thus, we focus on (3) only.

2.2.2. Examples Using R

For illustration, we simulated a dataset called “QTLex.txt,” which contains $({\rm Y},G)$ for n = 100 individuals. In the simulation, the true model was dominant and the risk allele was B with population frequency 0.3. HWE was assumed in the population. The heritability was set to 0.1. The trait Υ was simulated from a normal distribution using the model given in Subheading 1 where μ = 0, E(∈) = 0, and Var(∈) = 1. The genotype G is AA, AB, or BB. The data can be read as follows.

> c = read.table(“QTLex.txt”,header=T)

If we know the true genetic model (dominant) and the risk allele a priori, we use F(x) given in Eq. 2 with x = 1 as follows.

> objF1=aov(Y ~ (as.integer(G)==1),data=c);
> summary(objF1);
Df Sum Sq Mean Sq F value Pr($>$F)
as.integer(G) == 1 1 30.230 30.2295 31.317 1.993e-07 ***
Residuals Df 98 94.599 0.9653
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

The function “as.integer(G)” assigns 1 for AA, 2 for AB, and 3 for BB. Thus, “objF1 = aov(Y ~ (as.integer(G)==1),data = c)” conducts the ANOVA comparing the mean trait values between the two genotype groups: AA and AB + BB. In the first line of the output, “F value” is the statistic F(1) and “Pr(>F)” is the P-value. These values are reported in the second line. In this example, F(1) = 31.317 and the P-value is 1.993 × 10^–7. The strength of association is indicated by “***” near the P-value, and the interpretation of this significance code is given in the last line of the output. Since the risk allele is known, a one-sided test should be used. Thus, the actual P-value is half of that reported one, i.e., 9.965 × 10^–8. This P-value is very significant compared to the 0.05 significance level.

If we did not know the genetic model, the statistic F given in Eq. 3 should be used. See the following output. In this case, we would not assign scores (1, 2, 3) to the three genotypes. The output given below shows F = 15.575 with P-value 1.361 × 10^–6, which is also significant, but larger than the P-value obtained from F(1), which is the most powerful test when the true model is dominant.

> objF=aov(Y ~ G,data=c);
> summary(objF);
Df Sum Sq Mean Sq F value Pr(>F)
G 2 30.343 15.1715 15.575 1.361e-06 ***
Residuals 97 94.485 0.9741
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

For illustration, we also calculate F(1/2) and F(0) and their P-values. F(1/2) can be obtained by

> objF05=aov(Y~ (as.integer(G),data=c);
> summary(objF05);

In this case, “as.integer(G)” is equivalent to using scores 1, 2, 3 for the three genotypes (under the additive model). The reported P-value is 1.011 × 10^–6, which is the P-value if we do not know the risk allele. If we know the risk allele, the P-value is 1.011 × 10^–6/2 = 5.055 × 10^–7. Both one-sided and two-sided P-values are significant. Interestingly, if we apply F(1/2) even when the risk allele is unknown, the two-sided P-value (1.011 × 10^–6) is smaller than that of F. The test F(0), which is optimal for arecessive model, is obtained as follows.

> objF0=aov(Y ~ (as.integer(G)==3),data=c);
> summary(objF0);

In this case, the ANOVA is applied with two genotype groups: AA + AB and BB. The reported P-value is 0.1398. If we know the risk allele, the P-value is 0.1398/2 = 0.0699, which is not significant at the 0.05 level. This illustrates loss of power can occur if the test used is not appropriate for the underlying genetic model.

Table 1.

Empirical power (%) and robustness of different tests for the analysis of case–control data

P	x	GRR	T1(0)	T1(1/2)	T1(1)	T 2	MAX3
0.10	0.0	3.15	80.40	30.30	11.56	70.58	72.42
	0.5	1.88	21.23	80.30	79.20	73.09	76.89
	1.0	1.47	8.94	77.60	79.85	72.78	75.07
	Min		8.94	30.30	11.56	70.58	73.42
	Max of min						72.42
0.30	0.0	1.65	80.37	52.07	15.52	71.18	72.46
	0.5	1.60	44.78	81.71	76.59	73.01	76.96
	1.0	1.38	12.91	71.00	80.73	70.92	72.54
	Min		12.91	52.07	15.52	70.92	72.54
	Max of min						72.54
0.45	0.0	1.45	79.69	61.13	16.46	70.25	72.27
	0.5	1.57	56.52	80.01	67.91	71.52	76.07
	1.0	1.44	14.63	64.92	80.81	71.72	73.74
	Min		14.63	61.13	16.46	70.25	73.74
	Max of min						73.74

Table 2.

Empirical power (%) for the analysis of a quantitative trait using different test statistics given p, h = 0.1, and n = 100

Model (x)	P	F(0)	F(1/2)	F(1)	F
0.0	0.10	62.29	31.54	12.88	59.70
	0.30	87.65	56.16	15.60	80.49
	0.45	89.86	70.22	17.00	82.87
0.5	0.10	53.80	89.05	87.65	82.89
	0.30	57.15	90.57	83.96	83.50
	0.45	70.88	90.49	77.42	84.03
1.0	0.10	40.03	87.82	89.75	84.44
	0.30	15.58	84.14	90.15	83.39
	0.45	17.76	77.45	89.62	82.85