Influence of population stratification on population-based marker-disease association analysis

Summary

Population-based genetic association analysis may suffer from the failure to control for confounders such as population stratification (PS). There has been extensive study on the influence of PS on candidate gene-disease association analysis, but much less attention has been paid to its influence on marker-disease association analysis. In this paper, we focus on the Pearson chi-square test and the trend test for marker-disease association analysis. The mean and variance of the test statistics are derived under presence of PS, so that the power and inflated type I error rate can be evaluated. It is shown that the bias and the variance distortion are not zero in the presence of both PS and penetrance heterogeneity (PH). Unlike the candidate gene-disease association analysis, when PS is present, the bias is not zero no matter whether PH is present or not. This work generalizes the results of Ewens and Spielman (1995), where only the fully recessive penetrance model is considered and only the bias is calculated. It is shown that candidate gene-disease association analysis can be treated as a special case of marker-disease association analysis. Consequently, our results extend previous study on the candidate gene-disease association analysis. A simulation study confirms the theoretical findings.

INTRODUCTION

Population-based gene-disease association analysis is the most commonly used statistical method for detecting genetical variants underlying human diseases (Risch and Merikangas, 1996; Risch, 2000). Such an approach makes use of the case-control design, which is easy to carry out and cost-effective. However, the case-control studies often suffer from a failure to account for confounders such as population stratification (PS), resulting in spurious associations (Knowler et al., 1988; Lander and Schork, 1994; Cardon and Palmer, 2003; Campbell et al., 2005). When parental genotypes of affected individuals are available, the transmission disequilibrium test (TDT) can be used to control for false positives due to PS. However, for diseases with a late age of onset, the parental genotypes are generally unavailable and, therefore, TDT is not applicable. There have been studies in recent years on the impacts of PS on gene-disease association analysis, particularly with respect to the bias and/or variance distortion of the test statistic (Ewens and Spielman, 1995; Gorroochurn et al., 2004; Heiman et al. 2004; Qin et al., 2006; Whittemore, 2006; Li et al., 2009; Zheng et al. 2009). Most of the existing studies focus on a candidate locus, where the null hypothesis states that the penetrance does not depend on genotype in any subpopulation. Markers are widely used in preliminary association analyses for detecting disease genes, especially in genome-wide association analyses. However, the impact of PS on marker-disease association has not been studied, with the exception of the work of Ewens and Spielman (1995), where the bias of the test statistic for marker-disease association was obtained by assuming a very special disease model, namely a fully recessive penetrance model, but the variance distortion and power function were not given.

In this paper, we extend the results of Ewens and Spielman (1995) to a more general class of models, without assuming any mode of inheritance. Besides the bias, we also derive the variance distortion and power function under both the null hypothesis and the alternative hypothesis. The null hypothesis in the marker locus case states that the linkage disequilibrium (LD) measures are zero in any subpopulation. It is shown that the bias and variance distortion under the null hypothesis are not zero in the presence of both PS and penetrance heterogeneity (PH). In addition, the bias is not zero when PS is present, even if PH is not, in contrast to the result for candidate gene-disease association analysis, where the bias is equal to zero when PH is absent. We demonstrate that candidate gene-disease association analysis can be treated as a special case of marker-disease association analysis, so that our results are extensions of the previous work on candidate gene-disease association analysis. Because the null hypothesis in the marker locus case is different from that in the candidate locus case, the existing results under the null hypothesis for a candidate locus cannot be transformed through simple reparameterization to yield our results.

Our contributions consist of the following: 1) we extend the existing results to the general case of marker-disease association analysis; 2) we find that the presence of PS can lead to bias of the marker-disease association test statistic even when the PH is absent, while in the candidate locus case the bias is always zero when PH is absent; 3) we derive the power functions for the Pearson chi-square test and the trend test so that one can study the impact of PS and PH on both the type I and type II errors of the two tests.

The rest of the paper is organized as follows. Some notation and definitions are given in the next section. The subsequent sections give the mean and variance of the Pearson chi-square test statistic and the trend test statistic and their power functions. A small-scale simulation study is conducted to verify the theoretical results. This is followed by some concluding remarks.

NOTATION

Suppose that in a case-control study n₁ cases and n₂ controls are sampled from their respective populations, where n = n₁+ n₂. A marker with alleles M and m is then genotyped, with the counts of genotypes and alleles given in Tables 1 and 2, respectively.

Table 1.

Genotype counts

	MM	Mm	mm	Sum
Cases	D2	D1	D0	n1
Controls	C2	C1	C0	n2
Sum	r2	r1	r0	n

Table 2.

Allele counts

	M	m	Sum
Cases	2D2+D1	2D0+D1	2n1
Controls	2C2+C1	2C0+C1	2n2
Sum	2r2+r1	2r0+r1	2n

Let the proportions of cases and controls with allele M be denoted by q̂_D = (2D₂+ D₁)/(2n₁) and q̂_C = (2C₂+ C₁)/(2n₂), respectively. In addition, let q̂ = (2D₂+ D₁+ 2C₂+C₁)/(2n) be the proportion of the pooled sample with allele M. The commonly used Pearson chi-square test statistic based on allele counts is the square of the following test statistic:

$$T=\frac{{\widehat{q}}_D-{\widehat{q}}_C}{V^{1/2}},$$ (1)

where

$$V=\widehat{q}(1-\widehat{q})(\frac{1}{2n_1}+\frac{1}{2n_2})$$ (2)

is an estimate of the variance of q̂_D−q̂_C. The estimate V is used when Hardy-Weinberg equilibrium (HWE) holds in the overall population. In the “variance adjustment” section, we will present a variance estimate that is valid even when HWE does not hold.

We assume that the total population consists of K subpopulations, with HWE holding within each subpopulation. Throughout this paper, we shall use S_i to denote the event that a randomly selected individual is from subpopulation i and w_i to denote the proportion of the total population that belongs to subpopulation i. Assume that only one locus, with alleles A and a, is responsible for the disease. For subpopulation i, let p_i and q_i denote the frequencies of alleles A and M, respectively. Thus, the frequencies of alleles A and M in the overall population are $p={\sum }_{i=1}^K{w}_i{p}_i$ and $q={\sum }_{i=1}^K{w}_i{q}_i$, respectively. Furthermore, let x_i1, x_i2, x_i3 and x_i4 denote the frequencies of gametes MA, Ma, mA and ma, respectively, and δ_i = x_i1x_i4−x_i2x_i3 the LD measure between the marker locus and the disease locus. Finally, denote by f_2i, f_1i and f_0i the penetrances of genotypes AA, Aa and aa, respectively. Under the HWE, the frequencies of genotypes AA, Aa and aa at the disease locus for subpopulation i are $p_{2i}=p_i^2$, p_1i = 2p_i(1−p_i) and p_0i = (1−p_i)², respectively. The null hypothesis of linkage equilibrium becomes

$$H_0:{\delta }_1=\dots ={\delta }_K.$$ (3)

Under the null hypothesis H₀, all LD measures δ_i, i = 1,…,K, are equal to 0, while under the alternative hypothesis, at least one LD measure is not equal to 0. It is clear that the null hypothesis implies that the marker is not associated with the disease.

Definition 1

PS is said to be present if the allele frequencies at the marker locus are heterogenous, i.e., the q_i vary with i.

EXPECTATION OF FREQUENCY DIFFERENCE

In this section, we calculate the expectations of q̂_D and q̂_C under both the null and the alternative hypotheses. We study the null expectation of q̂_D−q̂_C, which is termed bias. Hereafter, let Y = 1 denote the event that a randomly chosen individual is a case, and Y = 2 the event that a randomly chosen individual is a control.

By definition, the expectations of q̂_D and q̂_C are equal to

$$E({\widehat{q}}_D)=P(MM\mid Y=1)+\frac{1}{2}P(Mm\mid Y=1)$$ (4)

and

$$E({\widehat{q}}_C)=P(MM\mid Y=2)+\frac{1}{2}P(Mm\mid Y=2),$$ (5)

respectively. The disease prevalence, which we denote by B, satisfies $B={\sum }_{i=1}^K{w}_i{\sum }_{j=0}^2{f}_{ji}{p}_{ji}$ by the Law of Total Probability. In Appendix I, we show that,

$$P(MM\mid Y=1)=\frac{1}{B}\{\sum _{i=1}^K{w}_i{q}_i^2\sum _{j=0}^2{f}_{ji}{p}_{ji}+\sum _{i=1}^K{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i})+2\sum _{i=1}^K{w}_i{q}_i{\delta }_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})]\}$$ (6)

and

$$P(Mm\mid Y=1)=\frac{2}{B}\{\sum _{i=1}^K{w}_i{q}_i(1-q_i)\sum _{j=0}^2{f}_{ji}{p}_{ji}-\sum _{i=1}^K{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i})+\sum _{i=1}^K{w}_i(1-2q_i){\delta }_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})]\}.$$ (7)

Similarly, we have

$$P(MM\mid Y=2)=\frac{1}{1-B}\{\sum _{i=1}^K{w}_i{q}_i^2\sum _{j=0}^2(1-f_{ji})p_{ji}-\sum _{i=1}^K{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i})-2\sum _{i=1}^K{w}_i{q}_i{\delta }_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})]\}$$ (8)

and

$$P(Mm\mid Y=2)=\frac{2}{1-B}\{\sum _{i=1}^K{w}_i{q}_i(1-q_i)\sum _{j=0}^2(1-f_{ji})p_{ji}+\sum _{i=1}^K{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i})-\sum _{i=1}^K{w}_i(1-2q_i){\delta }_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})]\}.$$ (9)

Substituting the above four probabilities for those in (4) and (5) gives

$$\mathrm{\Delta }=E({\widehat{q}}_D-{\widehat{q}}_C)=\frac{A_1+A_2}{B(1-B)},$$ (10)

where

$$A_1=\sum _{j=0}^2[\sum _{i=1}^K{w}_i{f}_{ji}{p}_{ji}{q}_i-q\sum _{i=1}^K{w}_i{f}_{ji}{p}_{ji}]$$ (11)

and

$$A_2=\sum _{i=1}^K{w}_i{\delta }_i[(f_{2i}-f_{1i})p_i+(f_{1i}-f_{0i})(1-p_i)].$$ (12)

Since A₂ = 0 under the null hypothesis, the bias is A₁/[B(1−B)]. Define a random variable Z with probability function P(Z = i) = w_i, i = 1, ···, K. Then $A_1=\text{Cov}(q_Z,{\sum }_{j=0}^2{f}_{jZ}{p}_{jZ})$, where q_z, f_jZ and p_jZ are conditional probabilities that are equal to q_i, f_ji and p_ji, respectively, conditional on Z = i.

The following are some scenarios that occur in practice.

Scenario 1

If PS is absent, then random variable q_Z degenerates to a constant. In this scenario, A₁ = 0 and the bias is zero.

Scenario 2

If PS is present but PH is absent (i.e., f_ji = f_j1, j = 0,1,2, i = 1,···,K), then the random variables f_jZ, j = 0,1,2, degenerate to constants, but A₁ is not zero. Hence the bias is not zero in general since q_Z and p_jZ, j = 0,1,2, are not necessarily constant.

Scenario 3

If both PS and PH are present, then the bias is not zero in general.

Remark 1

When f_2i = 1, f_0i = f_1i = 0, i = 1, ···,K, the model degenerates to the so-called fully recessive penetrance model and the expectation becomes

$$\frac{{\displaystyle \sum _{i=1}^K}{w}_i{q}_i{p}_i^2-q{\displaystyle \sum _{i=1}^K}{w}_i{p}_i^2+{\displaystyle \sum _{i=1}^K}{w}_i{\delta }_i{p}_i}{B(1-B)}.$$ (13)

The above expression is almost identical to expression (5) in Ewens and Spielman (1995).

Remark 2

If the marker locus and the disease locus coincide, so that p_i = q_i and the LD measures are ${\delta }_i=p_i-p_i^2$, then the marker locus becomes a candidate locus. In this case, (6)–(9) become

$$\begin{array}{c}P(MM\mid Y=1)=(\sum _{i=1}^K{w}_i{f}_{2i}{p}_{2i})/(\sum _{i=1}^K{w}_i\sum _{j=0}^2{f}_{ji}{p}_{ji}),\\ P(Mm\mid Y=1)=(\sum _{i=1}^K{w}_i{f}_{1i}{p}_{1i})/(\sum _{i=1}^K{w}_i\sum _{j=0}^2{f}_{ji}{p}_{ji}),\\ P(MM\mid Y=2)=(\sum _{i=1}^K{w}_i(1-f_{2i})p_{2i})/(\sum _{i=1}^K{w}_i\sum _{j=0}^2(1-f_{ji})p_{ji})\end{array}$$

and

$$P(Mm\mid Y=2)=(\sum _{i=1}^K{w}_i(1-f_{1i})p_{1i})/(\sum _{i=1}^K{w}_i\sum _{j=0}^2(1-f_{ji})p_{ji}).$$

The resulting expectation corresponds to the candidate locus case studied by Li et al. (2009). In the candidate locus case, the null hypothesis is f_0i = f_1i = f_2i for i = 1,…, K, and the bias is equal to zero if either PS or PH is absent. In the marker locus case that we study in the current paper, however, the bias is generally not zero if PH is absent but PS is present.

VARIANCE OF THE FREQUENCY DIFFERENCE

In Appendix II, we derive the following variance formula for q̂_D:

$$\text{Var}({\widehat{q}}_D)=\frac{1}{4n_1}[4P(MM\mid Y=1)(1-P(MM\mid Y=1))+P(Mm\mid Y=1)(1-P(Mm\mid Y=1))-4P(MM\mid Y=1)P(Mm\mid Y=1)].$$ (14)

Under the null hypothesis H₀, the conditional probabilities P(MM|Y = 1) and P(Mm|Y = 1) given by (6) and (7) are equal to

$$P_{H_0}(MM\mid Y=1)=\frac{1}{B}\sum _{i=1}^K{w}_i{q}_i^2\sum _{j=0}^2{f}_{ji}{p}_{ji}=\sum _{i=1}^K{q}_i^2{\alpha }_i={\overline{q}}_D^2+{\sigma }_D^2$$ (15)

and

$$P_{H_0}(Mm\mid Y=1)=\frac{2}{B}\sum _{i=1}^K{q}_i(1-q_i)\sum _{j=0}^2{f}_{ji}{p}_{ji}=\sum _{i=1}^{K}2q_i(1-q_i){\alpha }_i=2{\overline{q}}_D(1-{\overline{q}}_D)-2{\sigma }_D^2,$$

respectively, where

$${\overline{q}}_D=\sum _{i=1}^K{\alpha }_i{q}_i,{\sigma }_D^2=\sum _{i=1}^K{\alpha }_i{(q_i-{\overline{q}}_D)}^2\text{and}{\alpha }_i=\frac{w_i}{B}\sum _{j=0}^2{f}_{ji}{p}_{ji}.$$

It follows from (14), (15) and (16) that the null variance of q̂_D is

$${\text{Var}}_{H_0}({\widehat{q}}_D)=\frac{{\overline{q}}_D(1-{\overline{q}}_D)+{\sigma }_D^2}{2n_1}.$$

Similarly,

$${\text{Var}}_{H_0}({\widehat{q}}_C)=\frac{{\overline{q}}_C(1-{\overline{q}}_C)+{\sigma }_C^2}{2n_2},$$

where

$${\overline{q}}_C=\sum _{i=1}^K{\gamma }_i{q}_i,{\sigma }_C^2=\sum _{i=1}^K{\gamma }_i{(q_i-{\overline{q}}_C)}^2\text{and}{\gamma }_i=\frac{w_i}{1-B}\sum _{j=0}^2(1-f_{ji})p_{ji}.$$

Under the alternative hypothesis, the variance of q̂_D can be expressed as

$$\text{Var}({\widehat{q}}_D)=\frac{{\overline{q}}_D(1-{\overline{q}}_D)+{\sigma }_D^2}{2n_1}+\frac{1}{2n_{1}B}\sum _{i=1}^K{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i})-\frac{1}{n_1{B}^2}{\{\sum _{i=1}^K{w}_i{\delta }_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})]\}}^2+\frac{1}{2n_{1}B}\sum _{i=1}^K{\delta }_i{w}_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})](1+2q_i-4{\overline{q}}_D).$$ (16)

We refer to Appendix III for its detailed derivation. Similarly, the variance of q̂_C is equal to

$$\text{Var}({\widehat{q}}_C)=\frac{{\overline{q}}_C(1-{\overline{q}}_C)+{\sigma }_C^2}{2n_2}-\frac{1}{2n_2(1-B)}\sum _i{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i})-\frac{1}{n_2{(1-B)}^2}{\{\sum _i{w}_i{\delta }_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})]\}}^2-\frac{1}{2n_2(1-B)}\sum _i{w}_i{\delta }_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})](1+2q_i-4{\overline{q}}_C).$$ (17)

By virtue of the independence between the cases and controls, the variance of q̂_D−q̂_C is

$${\sigma }^2\equiv \text{Var}({\widehat{q}}_D)+\text{Var}({\widehat{q}}_C).$$ (18)

In particular, the null variance of q̂_D−q̂_C is

$${\sigma }_0^2\equiv \frac{{\overline{q}}_D(1-{\overline{q}}_D)+{\sigma }_D^2}{2n_1}+\frac{{\overline{q}}_C(1-{\overline{q}}_C)+{\sigma }_C^2}{2n_2}.$$ (19)

Definition 2

We say variance distortion exists if under the null hypothesis, the variance estimator V for q̂_D−q̂_C as given by (2), is not asymptotically equivalent to the true variance, that is, if $V/{\sigma }_0^2$ does not converge to 1 with probability 1 under the null hypothesis.

By the Law of Large Numbers, q̂_D → q̄_D and q̂_C → q̄_C, which imply that q̂ converges to c₁q̄_D + c₂q̄_C with probability 1, where c_j = lim_n→∞n_j/n, j = 1,2. It follows that under the null hypothesis V is asymptotically equivalent to

$${\overline{\sigma }}^2\equiv [(c_1{\overline{q}}_D+c_2{\overline{q}}_C)(1-c_1{\overline{q}}_D-c_2{\overline{q}}_C)](\frac{1}{2n_1}+\frac{1}{2n_2}).$$ (20)

Remark 3

If PS is absent, then under the null hypothesis q̄_D = q̄_C = q and ${\sigma }_D^2={\sigma }_C^2=0$. Hence, ${\sigma }_0^2={\overline{\sigma }}^2$ and the variance distortion vanishes under the null hypothesis and HWE. Otherwise, the variance distortion is present in general.

VARIANCE ADJUSTMENT

In this section, we derive the power function of the test statistic T, which is given by (1). By the Central Limit Theorem, T_A = (q̂_D−q̂_C−Δ)/σ converges in distribution to the standard normal distribution, since the mean and variance of q̂_D−q̂_C are Δ and σ². The two-sided T test at level of significance α is determined by rejection region {|T|> u_α/2}, where u_α/2 is the upper α/2 -quantile of the standard normal distribution. The corresponding power function is therefore approximated by

$$1-\mathrm{\Phi }(\frac{u_{\alpha /2}\overline{\sigma }-\mathrm{\Delta }}{\sigma })+\mathrm{\Phi }(\frac{-u_{\alpha /2}\overline{\sigma }-\mathrm{\Delta }}{\sigma }),$$ (21)

where Φ is the standard normal distribution function and σ̄² is defined by (20).

As we mentioned earlier, variance distortion exists in presence of PS. Therefore, it is necessary to use a consistent estimate of the variance σ². Notice that under the null hypothesis, σ² becomes ${\sigma }_0^2=({\overline{q}}_D(1-{\overline{q}}_D)+{\sigma }_D^2)/(2n_1)+({\overline{q}}_C(1-{\overline{q}}_C)+{\sigma }_C^2)/(2n_2)$. We can estimate it with

$$V^{\ast }=\frac{{\widehat{q}}_D(1-{\widehat{q}}_D)+{\widehat{\sigma }}_D^2}{2n_1}+\frac{{\widehat{q}}_C(1-{\widehat{q}}_C)+{\widehat{\sigma }}_C^2}{2n_2},$$ (22)

where ${\widehat{\sigma }}_D^2=D_2/n_1-{\widehat{q}}_D^2$ and ${\widehat{\sigma }}_C^2=C_2/n_2-{\widehat{q}}_C^2$ are consistent estimates of ${\sigma }_D^2$ and ${\sigma }_C^2$, respectively. The estimator V^* was used by Li et al. (2009) for the candidate locus. In the marker locus case, we can show that V^* is asymptotically equivalent to σ² under both the null hypothesis and the alternative hypothesis. Actually, V^* is a special estimate of the trend test statistic that will be studied in the next section, and it will be shown that V^* is asymptotically equivalent to σ² even when HWE does not hold.

Now, a modification of T takes the form

$$T^{\ast }=\frac{{\widehat{q}}_D-{\widehat{q}}_C}{{(V^{\ast })}^{1/2}}.$$ (23)

The T^* test with rejection region {|T^*|>u_α/2} has an approximate power function

$$1-\mathrm{\Phi }(u_{\alpha /2}-\mathrm{\Delta }/\sigma )+\mathrm{\Phi }(-u_{\alpha /2}-\mathrm{\Delta }/\sigma ).$$ (24)

EXTENSION TO TREND TEST

The trend test statistic is defined as

$$T_x=\frac{(D_2/n_1-C_2/n_2)+x(D_1/n_1-C_1/n_2)}{V_x^{1/2}},$$

where x is a given real number between 0 and 1 and V_x is an estimator of the variance of the numerator. From (6)–(9), it follows that the expectation of (D₂/n₁−C₂/n₂)+ x(D₁/n₁−C₁/n₂) is

$${\mathrm{\Delta }}_x=\frac{1}{B(1-B)}\{\sum _{i=1}^K{w}_i[q_i^2+2{xq}_i(1-q_i)]\sum _{j=0}^2{f}_{ji}{p}_{ji}-\sum _{i=1}^K{w}_i\sum _{j=0}^2{f}_{ji}{p}_{ji}\sum _{i=1}^K[q_i^2+2{xq}_i(1-q_i)]+(1-2x)\sum _{i=1}^K{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i})+2\sum _{i=1}^K{w}_i{\delta }_i[q_i+x(1-2q_i)][p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})].$$

Under the null hypothesis, this expectation is equal to $\text{Cov}({\sum }_{j=0}^2{p}_{jZ}{f}_{jZ},q_Z^2+2{xq}_Z(1-q_Z))$, where Z is the random variable defined below (12). In the absence of PS, the random variable Z becomes non-random, making the null expectation 0. Otherwise, the expectation is nonzero in general. Furthermore, under the assumptions in Remark 2, the expression Δ_x reduces to that given by Zheng et al. (2009).

For notational simplicity, we use g₂₁ = P(MM|Y = 1), g₁₁ = P(Mm|Y = 1), g₂₂ = P(MM|Y = 2) and g₁₂ = P(Mm|Y = 2) for the expressions given by (6)–(9). Using the facts that (D₂, D₁, D₀) and (C₂,C₁,C₀) follow trinomial distributions, we obtain the following formula

$$\begin{array}{l}{\sigma }_x^2=\text{Var}[(\frac{D_2}{n_1}-\frac{C_2}{n_2})+x(\frac{D_1}{n_1}-\frac{C_1}{n_2})]\\ =\frac{1}{n_1}[g_{21}(1-g_{21})+x^2{g}_{11}(1-g_{11})-2{xg}_{11}{g}_{21}]+\frac{1}{n_2}[g_{22}(1-g_{22})+x^2{g}_{12}(1-g_{12})-2{xg}_{12}{g}_{22}].\end{array}$$

Replacing the g_ij by their consistent estimators, we get the following estimate of the variance of (D₂/n₁−C₂/n₂)+ x(D₁/n₁−C₁/n₂):

$$V_x=\frac{1}{n_1}[\frac{D_2}{n_1}(1-\frac{D_2}{n_1})+x^2\frac{D_1}{n_1}(1-\frac{D_1}{n_1})-2x\frac{D_1}{n_1}\frac{D_2}{n_1}]+\frac{1}{n_2}[\frac{C_2}{n_2}(1-\frac{C_2}{n_2})+x^2\frac{C_1}{n_2}(1-\frac{C_1}{n_2})-2x\frac{C_1}{n_2}\frac{C_2}{n_2}].$$ (25)

By the Law of Large Numbers, V_x is a consistent estimate under both the null hypothesis and the alternative hypothesis, even if HWE does not hold in any subpopulation. When x = 0.5, we have that V_0.5 = V^* and T_0.5 = T^*. This shows that V^* is a consistent estimate of the variance of q̂_D−q̂_C.

The asymptotic power function of the T_x test with rejection region {|T_x|> u_α/2} is

$$1-\mathrm{\Phi }(u_{\alpha /2}-{\mathrm{\Delta }}_x/{\sigma }_x)+\mathrm{\Phi }(-u_{\alpha /2}-{\mathrm{\Delta }}_x/{\sigma }_x).$$ (26)

A SIMULATION STUDY

To study the finite sample performance of the mentioned tests, we conducted some simulations. We studied the impact of PS on the powers and type I error rates of the T test (defined in (1)) and the T^* test (defined in (23)). In the simulations, we assumed the study population consisted of 2 subpopulations of equal sizes.

First we considered an additive mode of inheritance in each subpopulation. The underlying models were specified as follows. The allele frequencies of M at a disease locus were 0.2 and 0.2 (PS is absent) or 0.1 and 0.3 (PS is present) for the two subpopulations. The allele frequencies of M at a marker locus were 0.3 and 0.3 (PS is absent) or 0.4 and 0.2 (PS is present) for the two subpopulations. HWE was assume to hold in the 2 subpopulations at both the marker and the disease loci. The penetrances of genotypes aa, Aa and AA in subpopulation 1 were 0.1, 0.2 and 0.3, respectively, and they were either 0.2, 0.3, 0.4 for subpopulation 2 (PH is present) or the same as those in subpopulation 1 (PH is absent). The LD measures in the two subpopulations were the same, that is, either 0 (null hypothesis) or 0.05 (alternative hypothesis).

We randomly generated the genotypes of 1000 cases and 1000 controls. The empirical type I error rates/powers at a 0.05 level of significance were estimated based on 5,000,000 replications. The asymptotic type I error rates/powers of the T^* test were calculated using formula (24). The resulting powers are presented in Table 3.

Table 3.

Type I error rates/powers for marker locus under additive mode of inheritance

			T test	T* test
Hypothesis 1	PS 2	PH 3	Empirical	Asymptotic	Empirical
Null	Absent	Absent	0.050	0.050	0.050
Null	Absent	Present	0.050	0.050	0.050
Null	Present	Absent	0.217	0.203	0.204
Null	Present	Present	0.885	0.876	0.876
Alternative	Absent	Absent	0.808	0.809	0.809
Alternative	Absent	Present	0.604	0.604	0.605
Alternative	Present	Absent	0.402	0.385	0.386
Alternative	Present	Present	0.153	0.142	0.143

¹“Null”: δ₁ =δ₂ = 0; “Alternative”: δ₁ =δ₂ = 0.05.

²“Absent”: p₁ = p₂ = 0.2 and m₁ = m₂ = 0.3; “Present”: p₁ = 0.1, p₂ = 0.3 and m₁ = 0.4, m₂ = 0.2.

³“Absent”: the penetrances of genotypes aa, Aa, AA are 0.1,0.2 and 0.3, respectively, in both of the subpopulations; “Present”: the penetrances are 0.1, 0.2 and 0.3 in subpopulation 1 and 0.2, 0.3 and 0.4 in subpopulation 2.

For the T^* test, it is seen that the asymptotic type I error rates/powers and the empirical type I error rates/powers are very close to each other, with differences of no more than 0.001, showing an accurate approximation of the power function. As expected, when PS is absent, the type I errors are virtually equal to the nominal level 0.05; when PS is present, the type I error could be inflated a great deal, especially when PH is also present (0.876). The power is also influenced by the presence of PS and PH. For example, when both PS and PH are present, the power is only 0.142, compared with 0.809 for the case where neither PS nor PH is present.

The T test has type I error rates/powers close to those of the T^* test in the absence of PS, with differences of no more than 0.001. In the presence of PS, there are minor differences that vary from 0.009 to 0.017.

The above simulations assumed that HWE held in any subpopulation. Our further simulations without an assumption of HWE (results not shown) showed that the T^* test had type I error rates close to the nominal levels in the presence of PS, but the T test could distort the type I error rate, with its magnitude depending upon the strength of Hardy-Weinberg disequilibrium.

Second we considered a fully penetrance recessive model in each subpopulation, with the penetrance being 1 for AA and 0 for Aa and aa. The other parameters are the same as those in Table 3, except that the LD measures under the null hypothesis are 0.01. The simulation results are reported in Table 4. For this mode of inheritance, the impact of PS on the type I error rates and the powers has a trend the same as that for the additive mode of inheritance.

Table 4.

Type I error rates/powers for marker locus under fully recessive mode of inheritance¹

		T test	T* test
Hypothesis2	PS3	Empirical	Asymptotic	Empirical
Null	Absent	0.050	0.050	0.050
Null	Present	1.000	1.000	1.000
Alternative	Absent	0.941	0.941	0.941
Alternative	Present	0.838	0.828	0.828

¹In both of the subpopulation, the penetrances of genotypes aa, Aa and AA are 0, 0 and 1, respectively.

²“Null”: δ₁ =δ₂ = 0; “Alternative”: δ₁ =δ₂ = 0.01.

³“Absent”: p₁ = p₂ = 0.2 and m₁ = m₂ = 0.3; “Present”: p₁ = 0.1, p₂ = 0.3 and m₁ = 0.4, m₂ = 0.2.

Third we considered a special case, where the the marker locus and the disease locus coincide, with a common allele frequency p_i for the i th subpopulation, i = 1,2, and where the LD measures are ${\delta }_i=p_i-p_i^2$. The other parameters are the same as those in Table 3 except that the null hypothesis (in each subpopulation the penetrances are independent of the genotypes) is different and the mode of inheritance under the alternative hypothesis is recessive. The detailed parameter settings are described and the simulation results are presented in Table 5. As expected, the T^* test has type I error rates controlled at the nominal level when either PS or PH is absent (this is different from the marker locus case), but the type I error rate is inflated when both PS and PH are present. Furthermore, the presence of PS and/or PH also has an impact on the powers of the T^* test, with the trend similar to that for the marker locus case. There are only minor differences between the T test and the T^* test, except under the null hypothesis with the absence of PS.

Table 5.

Type I error rates/powers for candidate locus ¹

			T test	T* test
Hypothesis 2	PS 3	PH 4	Empirical	Asymptotic	Empirical
Null	Absent	Absent	0.050	0.050	0.050
Null	Absent	Present	0.050	0.050	0.050
Null	Present	Absent	0.057	0.050	0.050
Null	Present	Present	0.850	0.838	0.838
Alternative	Absent	Absent	0.749	0.730	0.731
Alternative	Absent	Present	0.482	0.467	0.467
Alternative	Present	Absent	0.891	0.867	0.868
Alternative	Present	Present	0.108	0.089	0.090

¹The marker locus and the disease locus coincide, and the LD measure δ_i in the i th subpopulation is $p_i-p_i^2$ with p_i being the allele frequency of both marker and disease loci in the i th subpopulation.

²“Null”: the penetrances of the genotypes aa, Aa and AA are the same in each subpopulation; “Alternative”: the penetrances are different for the genotypes aa, Aa and AA in each subpopulation.

³“Absent”: p₁ = p₂ = 0.2; “Present”: p₁ = 0.3, p₂ = 0.1.

⁴“Absent”: under the null hypothesis, the penetrances of genotypes aa, Aa and AA are 0.2 in each subpopulation, and under the alternative hypothesis, the penetrances of genotypes aa, Aa and AA are 0.1, 0.1 and 0.2, respectively, in both of the two subpopulations; “Present”: under the null hypothesis, the penetrances of genotypes aa, Aa and AA are 0.2 in subpopulation 1 and 0.1 in subpopulation 2, and under the alternative hypothesis, the penetrances of genotypes aa, Aa and AA are 0.1, 0.1 and 0.2 in subpopulation 1 and 0.2, 0.2 and 0.3 in subpopulation 2.

DISCUSSION

Population-based marker-disease association analysis is a powerful tool but may suffer from PS. Our work provides closed forms for the expectation and variance of two commonly used test statistics, which enable us to study the type I error rate and power under various scenarios. We extend the work of Ewens and Spielman (1995) by relaxing the assumption of fully recessive penetrance and studying bias and variance distortion simultaneously. Our simulation results are in agreement with those from asymptotic approximations, confirming that the theoretical findings are correct. Both analysis and simulation results show that the presence of PS can inflate the type I error rate and decrease the power dramatically in the marker-disease association analysis. Therefore, it is necessary to modify the test statistics to accommodate PS. Methods have been proposed in the literature for correcting bias and/or variance distortion in candidate gene-disease association analysis, including genomic control (Devlin and Roeder, 1999; Devlin et al., 2001), structured association (Pritchard et al., 2000; Satten et al., 2001; Pritchard and Donnelly, 2001), the delta centralization (Gorroochurn et al., 2006). Whittermore (2006) suggested sensitivity analysis. However, the performance qualify for these methods for marker-disease association analysis is unclear and needs further investigations.

APPENDIX I. Proof of (6) and (7)

By the definition of the linkage equilibrium measures δ_i, the probabilities of gametes MA, Ma, mA and ma are

$$P(MA\mid S_i)=p_i{q}_i+{\delta }_i$$ (27)

$$P(Ma\mid S_i)=(1-p_i)q_i-{\delta }_i,$$ (28)

$$P(mA\mid S_i)=(1-q_i)p_i-{\delta }_i$$ (29)

and

$$P(ma\mid S_i)=(1-q_i)(1-p_i)+{\delta }_i,$$ (30)

respectively. Random mating gives

$$P((MM,AA)\mid S_i)={[P(MA\mid S_i)]}^2={[p_i{q}_i+{\delta }_i]}^2,$$ (31)

$$P((MM,Aa)\mid S_i)=2[p_i{q}_i+{\delta }_i][(1-p_i)q_i-{\delta }_i],$$ (32)

$$P((MM,aa)\mid S_i)={[(1-p_i)q_i-{\delta }_i]}^2,$$ (33)

$$P((Mm,AA)\mid S_i)=2[p_i{q}_i+{\delta }_i][p_i(1-q_i)-{\delta }_i],$$ (34)

$$P((Mm,Aa)\mid S_i)=2[p_i{q}_i+{\delta }_i][(1-p_i)(1-q_i)+{\delta }_i]+2[(1-p_i)q_i-{\delta }_i][p_i(1-q_i)-{\delta }_i],$$ (35)

$$P((Mm,aa)\mid S_i)=2[(1-p_i)q_i-{\delta }_i][(1-p_i)(1-q_i)+{\delta }_i].$$ (36)

Here (MM, AA) is the joint genotype at the marker locus (MM) and the disease locus (AA), so are the other 5 pairs. It follows from (31)–(36) that

$$P(MM\mid Y=1)=\frac{{\displaystyle \sum _{i=1}^K}{w}_i[P((MM,AA)\mid S_i)f_{2i}+P((MM,Aa)\mid S_i)f_{1i}+P((MM,aa)\mid S_i)f_{0i}]}{{\displaystyle \sum _{i=1}^K}{w}_i{\displaystyle \sum _{j=0}^2}{f}_{ji}{p}_{ji}}=\frac{1}{B}\{\sum _{i=1}^K{w}_i{q}_i^2\sum _{j=0}^2{f}_{ji}{p}_{ji}+\sum _{i=1}^K{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i})+2\sum _{i=1}^K{w}_i{q}_i{\delta }_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})]\}$$

and

$$P(Mm\mid Y=1)=\frac{{\displaystyle \sum _{i=1}^K}{w}_i[P((Mm,AA)\mid S_i)f_{2i}+P((Mm,Aa)\mid S_i)f_{1i}+P((Mm,aa)\mid S_i)f_{0i}]}{{\displaystyle \sum _{i=1}^K}{w}_i{\displaystyle \sum _{j=0}^2}{f}_{ji}{p}_{ji}}=\frac{2}{B}\{\sum _{i=1}^K{q}_i(1-q_i)\sum _{j=0}^2{f}_{ji}{p}_{ji}-\sum _{i=1}^K{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i})+\sum _{i=1}^K{w}_i(1-2q_i){\delta }_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})]\}.$$

APPENDIX II. Proof of (14)

Define two indicator functions

$$\begin{array}{l}{I}_{MM}=\{\begin{array}{l}1,\text{if}\text{randomly}\text{selected}\text{marker}\text{genotype}\text{of}\text{a}\text{case}\text{is}MM;\hfill \\ 0,\text{otherwise}.\hfill \end{array}\\ I_{Mm}=\{\begin{array}{l}1,\text{if}\text{randomly}\text{selected}\text{marker}\text{genotype}\text{of}\text{a}\text{case}\text{is}Mm;\hfill \\ 0,\text{otherwise}.\hfill \end{array}\end{array}$$

Then,

$${\widehat{q}}_D=\frac{\sum (2I_{MM}+I_{Mm})}{2n_1},$$

where the summation is taken over all cases. Since

$$\begin{array}{l}\text{Var}(I_{MM})=P(MM\mid Y=1)(1-P(MM\mid Y=1)),\\ \text{Var}(I_{Mm})=P(Mm\mid Y=1)(1-P(Mm\mid Y=1)),\\ \text{Cov}(I_{MM},I_{Mm})=-P(MM\mid Y=1)P(Mm\mid Y=1),\end{array}$$

the variance of q̂_D is

$$\begin{array}{l}\text{Var}({\widehat{q}}_D)=\text{Var}(\frac{\sum (2I_{MM}+I_{Mm})}{2n_1})\\ =\frac{1}{{(2n_1)}^2}[4n_1\text{Var}(I_{MM})+n_1\text{Var}(I_{Mm})+4n_1\text{Cov}(I_{MM},I_{Mm})]\\ =\frac{1}{4n_1}[4P(MM\mid Y=1)(1-P(MM\mid Y=1))+P(Mm\mid Y=1)(1-P(Mm\mid Y=1))-4P(MM\mid Y=1)P(Mm\mid Y=1)].\end{array}$$

APPENDIX III. Proof of (16)

Define $c_1\equiv {\sum }_i{q}_i^2{\alpha }_i$, c₂ ≡ Σ_i2q_i(1−q_i)α_i, x₁ ≡ P(MM|Y = 1)−c₁, x₂≡ P(MM|Y = 1)−c₂. Substituting c₁, c₂, x₁ and x₂ into (14), we have

$$\text{Var}({\widehat{q}}_D)=\frac{{\widehat{q}}_D(1-{\widehat{q}}_D)+{\sigma }_D^2}{2n_1}-\frac{{(2x_1+x_2)}^2}{4n_1}+\frac{4(1-2c_1-c_2)x_1+(1-2c_2-4c_1)x_2}{4n_1}$$ (37)

By the definitions,

$$2x_1+x_2=\frac{2}{B}\sum _i{\delta }_i{w}_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})],$$ (38)

$$2c_1+c_2=\sum _i[2q_i^2{\alpha }_i+2q_i(1-q_i){\alpha }_i]=2{\overline{q}}_D,$$ (39)

$$4x_1+x_2=\frac{2}{B}\sum _i{\delta }_i{w}_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})](1+2q_i)+\frac{2}{B}\sum _i{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i}).$$ (40)

It follows from (39) and (40) that

$$\frac{4(1-2c_1-c_2)x_1+(1-2c_2-4c_1)x_2}{4n_1}=\frac{4(1-2{\overline{q}}_D)x_1+(1-4{\overline{q}}_D)x_2}{4n_1}=\frac{4x_1+x_2-4(2x_1+x_2){\overline{q}}_D}{4n_1}=\frac{1}{2n_{1}B}\sum _i{\delta }_i{w}_i[p_i(f_{2i}-f_{1i})+(1-p_i)(f_{1i}-f_{0i})](1+2q_i-4{\overline{q}}_D)+\frac{1}{2n_{1}B}\sum _i{w}_i{\delta }_i^2(f_{2i}+f_{0i}-2f_{1i}).$$ (41)

Equation (16) follows from (37), (38) and (41).