Package 'powerEQTL'

Calculation of Minimum Detectable Effect Size for EQTL Analysis Based on Un-Balanced One-Way ANOVA

Description

Calculation of minimum detectable effect size ($\delta/\sigma$) for eQTL analysis that tests if a SNP is associated to a gene probe by using un-balanced one-way ANOVA.

Usage

minEffectEQTL.ANOVA(MAF,
                typeI = 0.05,
                nTests = 2e+05,
                myntotal = 200,
                mypower = 0.8,
                verbose = TRUE)

Arguments

MAF	Minor allele frequency.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
myntotal	integer. Number of subjects.
mypower	Desired power for the eQTL analysis.
verbose	logic. indicating if intermediate results should be output.

Details

The assumption of the ANOVA approach is that the association of a SNP to a gene probe is tested by using un-balanced one-way ANOVA (e.g. Lonsdale et al. 2013). According to SAS online document https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_power_a0000000982.htm, the power calculation formula is

$power=Pr\left(\left.F\geq F_{1-\alpha}\left(k-1, N-k\right)\right| F\sim F_{k-1, N-k, \lambda}\right),$

where $k=3$ is the number of groups of subjects, $N$ is the total number of subjects, $F_{1-\alpha}\left(k-1, N-k\right)$ is the $100(1-\alpha)$-th percentile of central F distribution with degrees of freedoms $k-1$ and $N-k$, and $F_{k-1, N-k, \lambda}$ is the non-central F distribution with degrees of freedoms $k-1$ and $N-k$ and non-central parameter (ncp) $\lambda$. The ncp $\lambda$ is equal to

$\lambda=\frac{N}{\sigma^2}\sum_{i=1}^{k} w_i \left(\mu_i-\mu\right)^2,$

where $\mu_i$ is the mean gene expression level for the $i$-th group of subjects, $w_i$ is the weight for the $i$-th group of subjects, $\sigma^2$ is the variance of the random errors in ANOVA (assuming each group has equal variance), and $\mu$ is the weighted mean gene expression level

$\mu=\sum_{i=1}^{k}w_i \mu_i.$

The weights $w_i$ are the sample proportions for the 3 groups of subjects. Hence, $\sum_{i=1}^{3}w_i = 1$.

We assume that $\mu_2-\mu_1=\mu_3-\mu_2=\delta$, where $\mu_1$, $\mu_2$, and $\mu_3$ are the mean gene expression level for mutation homozygotes, heterozygotes, and wild-type homozygotes, respectively.

Denote $p$ as the minor allele frequency (MAF) of a SNP. Under Hardy-Weinberg equilibrium, we have genotype frequencies: $p_2=p^2$, $p_1=2 p q$, and $p_0=q^2$, where $p_2$, $p_1$, and $p_0$ are genotype for mutation homozygotes, heterozygotes, and wild-type homozygotes, respectively, $q=1-p$. Then ncp can be simplified as

$ncp=2pq N\left(\frac{\delta}{\sigma}\right)^2,$

Value

minimum detectable effect size $\delta/\sigma$.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Lonsdale J and Thomas J, et al. The Genotype-Tissue Expression (GTEx) project. Nature Genetics, 45:580-585, 2013.

See Also

powerEQTL.ANOVA, powerEQTL.ANOVA2, ssEQTL.ANOVA, ssEQTL.ANOVA2

Examples

minEffectEQTL.ANOVA(
          MAF = 0.1,
          typeI = 0.05,
          nTests = 200000,
          myntotal = 234,
          mypower = 0.8,
          verbose = TRUE)

---

Calculation of Minimum Detectable Minor Allele Frequency for EQTL Analysis Based on Un-Balanced One-Way ANOVA

Description

Calculation of minimum detectable minor allele frequency (MAF) for eQTL analysis that tests if a SNP is associated to a gene probe by using un-balanced one-way ANOVA.

Usage

minMAFeQTL.ANOVA(effsize, 
                 typeI = 0.05,
                 nTests = 200000,
                 myntotal = 200,
                 mypower = 0.8,
                 verbose = TRUE)

Arguments

effsize	Effect size $\delta/\sigma$.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
myntotal	integer. Number of subjects.
mypower	Desired power for the eQTL analysis.
verbose	logic. indicating if intermediate results should be output.

Details

The assumption of the ANOVA approach is that the association of a SNP to a gene probe is tested by using un-balanced one-way ANOVA (e.g. Lonsdale et al. 2013). According to SAS online document https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_power_a0000000982.htm, the power calculation formula is

$power=Pr\left(\left.F\geq F_{1-\alpha}\left(k-1, N-k\right)\right| F\sim F_{k-1, N-k, \lambda}\right),$

where $k=3$ is the number of groups of subjects, $N$ is the total number of subjects, $F_{1-\alpha}\left(k-1, N-k\right)$ is the $100(1-\alpha)$-th percentile of central F distribution with degrees of freedoms $k-1$ and $N-k$, and $F_{k-1, N-k, \lambda}$ is the non-central F distribution with degrees of freedoms $k-1$ and $N-k$ and non-central parameter (ncp) $\lambda$. The ncp $\lambda$ is equal to

$\lambda=\frac{N}{\sigma^2}\sum_{i=1}^{k} w_i \left(\mu_i-\mu\right)^2,$

where $\mu_i$ is the mean gene expression level for the $i$-th group of subjects, $w_i$ is the weight for the $i$-th group of subjects, $\sigma^2$ is the variance of the random errors in ANOVA (assuming each group has equal variance), and $\mu$ is the weighted mean gene expression level

$\mu=\sum_{i=1}^{k}w_i \mu_i.$

The weights $w_i$ are the sample proportions for the 3 groups of subjects. Hence, $\sum_{i=1}^{3}w_i = 1$.

We assume that $\mu_2-\mu_1=\mu_3-\mu_2=\delta$, where $\mu_1$, $\mu_2$, and $\mu_3$ are the mean gene expression level for mutation homozygotes, heterozygotes, and wild-type homozygotes, respectively.

Denote $p$ as the minor allele frequency (MAF) of a SNP. Under Hardy-Weinberg equilibrium, we have genotype frequencies: $p_2=p^2$, $p_1=2 p q$, and $p_0=q^2$, where $p_2$, $p_1$, and $p_0$ are genotype for mutation homozygotes, heterozygotes, and wild-type homozygotes, respectively, $q=1-p$. Then ncp can be simplified as

$ncp=2pq N\left(\frac{\delta}{\sigma}\right)^2,$

Value

minimum detectable MAF.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Lonsdale J and Thomas J, et al. The Genotype-Tissue Expression (GTEx) project. Nature Genetics, 45:580-585, 2013.

See Also

powerEQTL.ANOVA, powerEQTL.ANOVA2, ssEQTL.ANOVA, ssEQTL.ANOVA2

Examples

minMAFeQTL.ANOVA(effsize = 1, 
                 typeI = 0.05,
                 nTests = 200000,
                 myntotal = 234,
                 mypower = 0.8,
                 verbose = TRUE)

---

Minimum Detectable Minor Allele Frequencey Calculation for EQTL Analysis Based on Simple Linear Regression

Description

Minimum detectable minor allele frequency (MAF) calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using simple linear regression.

Usage

minMAFeQTL.SLR(slope,
               typeI = 0.05,
               nTests = 200000,
               myntotal = 200,
               mypower = 0.8,
               mystddev = 0.13,
               verbose = TRUE)

Arguments

slope	Slope of the simple linear regression.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
myntotal	integer. Number of subjects.
mypower	Desired power for the eQTL analysis.
mystddev	Standard deviation of the random error term $\epsilon$ in simple linear regression.
verbose	logic. indicating if intermediate results should be output.

Details

To test if a SNP is associated with a gene probe, we use the simple linear regression

$y_i = \beta_0+\beta_1 x_i + \epsilon_i,$

where $y_i$ is the gene expression level of the $i$-th subject, $x_i$ is the genotype of the $i$-th subject, and $\epsilon_i$ is the random error term. Additive coding for genotype is used. To test if the SNP is associated with the gene probe, we test the null hypothesis $H_0: \beta_1=0$.

Denote $p$ as the minor allele frequency (MAF) of the SNP. Under Hardy-Weinberg equilibrium, we can calculate the variance of genotype of the SNP: $\sigma^2_x=2 p (1-p)$, where $\sigma^2_x$ is the variance of the predictor (i.e. the SNP) $x_i$.

We then can use Dupont and Plummer's (1998) power/sample size calculation formula to calculate the minimum detectable slope, adjusting for multiple testing.

Value

The estimated minimum detectable MAF.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Dupont, W.D. and Plummer, W.D.. Power and Sample Size Calculations for Studies Involving Linear Regression. Controlled Clinical Trials. 1998;19:589-601.

See Also

powerEQTL.SLR, ssEQTL.SLR

Examples

minMAFeQTL.SLR(slope = 0.1299513,
               typeI = 0.05,
               nTests = 200000,
               myntotal = 176,
               mypower = 0.8,
               mystddev = 0.13,
               verbose = TRUE)

---

Minimum Detectable Slope Calculation for EQTL Analysis Based on Simple Linear Regression

Description

Minimum detectable slope calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using simple linear regression.

Usage

minSlopeEQTL.SLR(
  MAF,
  typeI = 0.05,
  nTests = 2e+05,
  myntotal = 200,
  mypower = 0.8,
  mystddev = 0.13,
  verbose = TRUE)

Arguments

MAF	Minor allele frequency.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
myntotal	integer. Number of subjects.
mypower	Desired power for the eQTL analysis.
mystddev	Standard deviation of the random error term $\epsilon$ in simple linear regression.
verbose	logic. indicating if intermediate results should be output.

Details

To test if a SNP is associated with a gene probe, we use the simple linear regression

$y_i = \beta_0+\beta_1 x_i + \epsilon_i,$

where $y_i$ is the gene expression level of the $i$-th subject, $x_i$ is the genotype of the $i$-th subject, and $\epsilon_i$ is the random error term. Additive coding for genotype is used. To test if the SNP is associated with the gene probe, we test the null hypothesis $H_0: \beta_1=0$.

Denote $p$ as the minor allele frequency (MAF) of the SNP. Under Hardy-Weinberg equilibrium, we can calculate the variance of genotype of the SNP: $\sigma^2_x=2 p (1-p)$, where $\sigma^2_x$ is the variance of the predictor (i.e. the SNP) $x_i$.

We then can use Dupont and Plummer's (1998) power/sample size calculation formula to calculate the minimum detectable slope, adjusting for multiple testing.

Value

The estimated minimum detectable slope.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Dupont, W.D. and Plummer, W.D.. Power and Sample Size Calculations for Studies Involving Linear Regression. Controlled Clinical Trials. 1998;19:589-601.

See Also

powerEQTL.SLR, ssEQTL.SLR

Examples

minSlopeEQTL.SLR(
  MAF = 0.1,
  typeI = 0.05,
  nTests = 2e+05,
  myntotal = 176,
  mypower = 0.8,
  mystddev = 0.13,
  verbose = TRUE)

---

Power Calculation for EQTL Analysis Based on Un-Balanced One-Way ANOVA

Description

Power calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using un-balanced one-way ANOVA.

Usage

powerEQTL.ANOVA(MAF,
                typeI = 0.05,
                nTests = 2e+05,
                myntotal = 200,
                mystddev = 0.13,
                deltaVec = c(0.13, 0.13),
                verbose = TRUE)

Arguments

MAF	Minor allele frequency.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
myntotal	integer. Number of subjects.
mystddev	Standard deviation of gene expression levels in one group of subjects. Assume all 3 groups of subjects (mutation homozygote, heterozygote, wild-type homozygote) have the same standard deviation of gene expression levels.
deltaVec	A vector having 2 elements. The first element is equal to $\mu_2-\mu_1$ and the second elementis equalt to $\mu_3-\mu_2$, where $\mu_1$ is the mean gene expression level for the mutation homozygotes, $\mu_2$ is the mean gene expression level for the heterozygotes, and $\mu_3$ is the mean gene expression level for the wild-type gene expression level.
verbose	logic. indicating if intermediate results should be output.

Details

The assumption of the ANOVA approach is that the association of a SNP to a gene probe is tested by using un-balanced one-way ANOVA (e.g. Lonsdale et al. 2013). According to SAS online document https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_power_a0000000982.htm, the power calculation formula is

$power=Pr\left(\left.F\geq F_{1-\alpha}\left(k-1, N-k\right)\right| F\sim F_{k-1, N-k, \lambda}\right),$

where $k=3$ is the number of groups of subjects, $N$ is the total number of subjects, $F_{1-\alpha}\left(k-1, N-k\right)$ is the $100(1-\alpha)$-th percentile of central F distribution with degrees of freedoms $k-1$ and $N-k$, and $F_{k-1, N-k, \lambda}$ is the non-central F distribution with degrees of freedoms $k-1$ and $N-k$ and non-central parameter (ncp) $\lambda$. The ncp $\lambda$ is equal to

$\lambda=\frac{N}{\sigma^2}\sum_{i=1}^{k} w_i \left(\mu_i-\mu\right)^2,$

where $\mu_i$ is the mean gene expression level for the $i$-th group of subjects, $w_i$ is the weight for the $i$-th group of subjects, $\sigma^2$ is the variance of the random errors in ANOVA (assuming each group has equal variance), and $\mu$ is the weighted mean gene expression level

$\mu=\sum_{i=1}^{k}w_i \mu_i.$

The weights $w_i$ are the sample proportions for the 3 groups of subjects. Hence, $\sum_{i=1}^{3}w_i = 1$.

Value

power of the test after Bonferroni correction for multiple testing.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Lonsdale J and Thomas J, et al. The Genotype-Tissue Expression (GTEx) project. Nature Genetics, 45:580-585, 2013.

See Also

minEffectEQTL.ANOVA, powerEQTL.ANOVA2, ssEQTL.ANOVA, ssEQTL.ANOVA2

Examples

powerEQTL.ANOVA(
          MAF = 0.1,
          typeI = 0.05,
          nTests = 200000,
          myntotal = 234,
          mystddev = 0.13,
          deltaVec = c(0.13, 0.13))

---

Power Calculation for EQTL Analysis Based on Un-Balanced One-Way ANOVA

Description

Power calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using un-balanced one-way ANOVA (assuming Hardy-Weinberg equilibrium).

Usage

powerEQTL.ANOVA2(effsize,
                MAF,
                typeI = 0.05,
                nTests = 2e+05,
                myntotal = 200,
                verbose = TRUE)

Arguments

effsize	effect size $\delta/\sigma$, where $\delta=\mu_2-\mu_1=\mu_3-\mu_2$, $\mu_1$, $\mu_2$, $\mu_3$ are the mean gene expression level of mutation homozygotes, heterozygotes, and wild-type homozygotes, and $\sigma$ is the standard deviation of gene expression levels (assuming each genotype group has the same variance).
MAF	Minor allele frequency.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
myntotal	integer. Number of subjects.
verbose	logic. indicating if intermediate results should be output.

Details

The assumption of the ANOVA approach is that the association of a SNP to a gene probe is tested by using un-balanced one-way ANOVA (e.g. Lonsdale et al. 2013). According to SAS online document https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_power_a0000000982.htm, the power calculation formula is

$power=Pr\left(\left.F\geq F_{1-\alpha}\left(k-1, N-k\right)\right| F\sim F_{k-1, N-k, \lambda}\right),$

where $k=3$ is the number of groups of subjects, $N$ is the total number of subjects, $F_{1-\alpha}\left(k-1, N-k\right)$ is the $100(1-\alpha)$-th percentile of central F distribution with degrees of freedoms $k-1$ and $N-k$, and $F_{k-1, N-k, \lambda}$ is the non-central F distribution with degrees of freedoms $k-1$ and $N-k$ and non-central parameter (ncp) $\lambda$. The ncp $\lambda$ is equal to

$\lambda=\frac{N}{\sigma^2}\sum_{i=1}^{k} w_i \left(\mu_i-\mu\right)^2,$

where $\mu_i$ is the mean gene expression level for the $i$-th group of subjects, $w_i$ is the weight for the $i$-th group of subjects, $\sigma^2$ is the variance of the random errors in ANOVA (assuming each group has equal variance), and $\mu$ is the weighted mean gene expression level

$\mu=\sum_{i=1}^{k}w_i \mu_i.$

The weights $w_i$ are the sample proportions for the 3 groups of subjects. Hence, $\sum_{i=1}^{3}w_i = 1$.

We assume that $\mu_2-\mu_1=\mu_3-\mu_2=\delta$, where $\mu_1$, $\mu_2$, and $\mu_3$ are the mean gene expression level for mutation homozygotes, heterozygotes, and wild-type homozygotes, respectively.

Denote $p$ as the minor allele frequency (MAF) of a SNP. Under Hardy-Weinberg equilibrium, we have genotype frequencies: $p_2=p^2$, $p_1=2 p q$, and $p_0=q^2$, where $p_2$, $p_1$, and $p_0$ are genotype for mutation homozygotes, heterozygotes, and wild-type homozygotes, respectively, $q=1-p$. Then ncp can be simplified as

$ncp=2pq N\left(\frac{\delta}{\sigma}\right)^2,$

Value

power of the test after Bonferroni correction for multiple testing.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Lonsdale J and Thomas J, et al. The Genotype-Tissue Expression (GTEx) project. Nature Genetics, 45:580-585, 2013.

See Also

minEffectEQTL.ANOVA, powerEQTL.ANOVA, ssEQTL.ANOVA, ssEQTL.ANOVA2

Examples

powerEQTL.ANOVA2(effsize = 1,
                MAF = 0.1,
                typeI = 0.05,
                nTests = 2e+05,
                myntotal = 234,
                verbose = TRUE)

---

Power Calculation for EQTL Analysis Based on Simple Linear Regression

Description

Power calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using simple linear regression.

Usage

powerEQTL.SLR(
  MAF,
  typeI = 0.05,
  nTests = 2e+05,
  slope = 0.13,
  myntotal = 200,
  mystddev = 0.13,
  verbose = TRUE)

Arguments

MAF	Minor allele frequency.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
slope	Slope $\beta_1$ of the simple linear regression $y_i = \beta_0+\beta_1 x_i + \epsilon_i,$ where $y_i$ is the gene expression level of the $i$-th subject, $x_i$ is the genotype of the $i$-th subject, and $\epsilon_i$ is the random error term. Additive coding for genotype is used.
myntotal	integer. Number of subjects.
mystddev	Standard deviation of the random error term $\epsilon$ in simple linear regression.
verbose	logic. indicating if intermediate results should be output.

Details

To test if a SNP is associated with a gene probe, we use the simple linear regression

$y_i = \beta_0+\beta_1 x_i + \epsilon_i,$

where $y_i$ is the gene expression level of the $i$-th subject, $x_i$ is the genotype of the $i$-th subject, and $\epsilon_i$ is the random error term. Additive coding for genotype is used. To test if the SNP is associated with the gene probe, we test the null hypothesis $H_0: \beta_1=0$.

Denote $p$ as the minor allele frequency (MAF) of the SNP. Under Hardy-Weinberg equilibrium, we can calculate the variance of genotype of the SNP: $\sigma^2_x=2 p (1-p)$, where $\sigma^2_x$ is the variance of the predictor (i.e. the SNP) $x_i$.

We then can use Dupont and Plummer's (1998) power/sample size calculation formula to calculate the minimum detectable slope, adjusting for multiple testing.

Value

power of the test after Bonferroni correction for multiple testing.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Dupont, W.D. and Plummer, W.D.. Power and Sample Size Calculations for Studies Involving Linear Regression. Controlled Clinical Trials. 1998;19:589-601.

See Also

ssEQTL.SLR, minSlopeEQTL.SLR

Examples

powerEQTL.SLR(
  MAF = 0.1,
  typeI = 0.05,
  nTests = 2e+05,
  slope = 0.13,
  myntotal = 176,
  mystddev = 0.13,
  verbose = TRUE)

---

Sample Size Calculation for EQTL Analysis Based on Un-Balanced One-Way ANOVA

Description

Sample size calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using un-balanced one-way ANOVA.

Usage

ssEQTL.ANOVA(
  MAF,
  typeI = 0.05,
  nTests = 2e+05,
  mypower = 0.8,
  mystddev = 0.13,
  deltaVec = c(0.13, 0.13))

Arguments

MAF	Minor allele frequency.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
mypower	Desired power for the eQTL analysis.
mystddev	Standard deviation of gene expression levels in one group of subjects. Assume all 3 groups of subjects (mutation homozygote, heterozygote, wild-type homozygote) have the same standard deviation of gene expression levels.
deltaVec	A vector having 2 elements. The first element is equal to $\mu_2-\mu_1$ and the second elementis equalt to $\mu_3-\mu_2$, where $\mu_1$ is the mean gene expression level for the mutation homozygotes, $\mu_2$ is the mean gene expression level for the heterozygotes, and $\mu_3$ is the mean gene expression level for the wild-type gene expression level.

Details

The assumption of the ANOVA approach is that the association of a SNP to a gene probe is tested by using un-balanced one-way ANOVA (e.g. Lonsdale et al. 2013). According to SAS online document https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_power_a0000000982.htm, the power calculation formula is

$power=Pr\left(\left.F\geq F_{1-\alpha}\left(k-1, N-k\right)\right| F\sim F_{k-1, N-k, \lambda}\right),$

where $k=3$ is the number of groups of subjects, $N$ is the total number of subjects, $F_{1-\alpha}\left(k-1, N-k\right)$ is the $100(1-\alpha)$-th percentile of central F distribution with degrees of freedoms $k-1$ and $N-k$, and $F_{k-1, N-k, \lambda}$ is the non-central F distribution with degrees of freedoms $k-1$ and $N-k$ and non-central parameter (ncp) $\lambda$. The ncp $\lambda$ is equal to

$\lambda=\frac{N}{\sigma^2}\sum_{i=1}^{k} w_i \left(\mu_i-\mu\right)^2,$

where $\mu_i$ is the mean gene expression level for the $i$-th group of subjects, $w_i$ is the weight for the $i$-th group of subjects, $\sigma^2$ is the variance of the random errors in ANOVA (assuming each group has equal variance), and $\mu$ is the weighted mean gene expression level

$\mu=\sum_{i=1}^{k}w_i \mu_i.$

The weights $w_i$ are the sample proportions for the 3 groups of subjects. Hence, $\sum_{i=1}^{3}w_i = 1$.

Value

sample size required for the eQTL analysis to achieve the desired power.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Lonsdale J and Thomas J, et al. The Genotype-Tissue Expression (GTEx) project. Nature Genetics, 45:580-585, 2013.

See Also

minEffectEQTL.ANOVA, powerEQTL.ANOVA, powerEQTL.ANOVA2, ssEQTL.ANOVA2

Examples

ssEQTL.ANOVA(MAF = 0.1,
       typeI = 0.05,
       nTests = 200000,
       mypower = 0.8,
       mystddev = 0.13,
       deltaVec = c(0.13, 0.13))

---

Sample Size Calculation for EQTL Analysis Based on Un-Balanced One-Way ANOVA

Description

Sample size calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using un-balanced one-way ANOVA.

Usage

ssEQTL.ANOVA2(
  effsize,
  MAF,
  typeI = 0.05,
  nTests = 2e+05,
  mypower = 0.8
)

Arguments

effsize	effect size $\delta/\sigma$, where $\delta=\mu_2-\mu_1=\mu_3-\mu_2$, $\mu_1$, $\mu_2$, $\mu_3$ are the mean gene expression level of mutation homozygotes, heterozygotes, and wild-type homozygotes, and $\sigma$ is the standard deviation of gene expression levels (assuming each genotype group has the same variance).
MAF	Minor allele frequency.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
mypower	Desired power for the eQTL analysis.

Details

The assumption of the ANOVA approach is that the association of a SNP to a gene probe is tested by using un-balanced one-way ANOVA (e.g. Lonsdale et al. 2013). According to SAS online document https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_power_a0000000982.htm, the power calculation formula is

$power=Pr\left(\left.F\geq F_{1-\alpha}\left(k-1, N-k\right)\right| F\sim F_{k-1, N-k, \lambda}\right),$

where $k=3$ is the number of groups of subjects, $N$ is the total number of subjects, $F_{1-\alpha}\left(k-1, N-k\right)$ is the $100(1-\alpha)$-th percentile of central F distribution with degrees of freedoms $k-1$ and $N-k$, and $F_{k-1, N-k, \lambda}$ is the non-central F distribution with degrees of freedoms $k-1$ and $N-k$ and non-central parameter (ncp) $\lambda$. The ncp $\lambda$ is equal to

$\lambda=\frac{N}{\sigma^2}\sum_{i=1}^{k} w_i \left(\mu_i-\mu\right)^2,$

where $\mu_i$ is the mean gene expression level for the $i$-th group of subjects, $w_i$ is the weight for the $i$-th group of subjects, $\sigma^2$ is the variance of the random errors in ANOVA (assuming each group has equal variance), and $\mu$ is the weighted mean gene expression level

$\mu=\sum_{i=1}^{k}w_i \mu_i.$

The weights $w_i$ are the sample proportions for the 3 groups of subjects. Hence, $\sum_{i=1}^{3}w_i = 1$.

We assume that $\mu_2-\mu_1=\mu_3-\mu_2=\delta$, where $\mu_1$, $\mu_2$, and $\mu_3$ are the mean gene expression level for mutation homozygotes, heterozygotes, and wild-type homozygotes, respectively.

Denote $p$ as the minor allele frequency (MAF) of a SNP. Under Hardy-Weinberg equilibrium, we have genotype frequencies: $p_2=p^2$, $p_1=2 p q$, and $p_0=q^2$, where $p_2$, $p_1$, and $p_0$ are genotype for mutation homozygotes, heterozygotes, and wild-type homozygotes, respectively, $q=1-p$. Then ncp can be simplified as

$ncp=2pq N\left(\frac{\delta}{\sigma}\right)^2,$

Value

sample size required for the eQTL analysis to achieve the desired power.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Lonsdale J and Thomas J, et al. The Genotype-Tissue Expression (GTEx) project. Nature Genetics, 45:580-585, 2013.

See Also

minEffectEQTL.ANOVA, powerEQTL.ANOVA, powerEQTL.ANOVA2, ssEQTL.ANOVA

Examples

ssEQTL.ANOVA2(
  effsize = 1,
  MAF = 0.1,
  typeI = 0.05,
  nTests = 2e+05,
  mypower = 0.8
)

---

Sample Size Calculation for EQTL Analysis Based on Simple Linear Regression

Description

Sample size calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using simple linear regression.

Usage

ssEQTL.SLR(
  MAF,
  typeI = 0.05,
  nTests = 2e+05,
  slope = 0.13,
  mypower = 0.8,
  mystddev = 0.13,
  n.lower = 2.01,
  n.upper = 1e+30,
  verbose = TRUE)

Arguments

MAF	Minor allele frequency.
typeI	Type I error rate for testing if a SNP is associated to a gene probe.
nTests	integer. Number of tests in eQTL analysis.
slope	Slope $\beta_1$ of the simple linear regression $y_i = \beta_0+\beta_1 x_i + \epsilon_i,$ where $y_i$ is the gene expression level of the $i$-th subject, $x_i$ is the genotype of the $i$-th subject, and $\epsilon_i$ is the random error term. Additive coding for genotype is used.
mypower	Desired power for the eQTL analysis.
mystddev	Standard deviation of the random error term $\epsilon$.
n.lower	integer. Lower bound of the total number of subjects.
n.upper	integer. Upper bound of the total number of subjects.
verbose	logic. indicating if intermediate results should be output.

Details

To test if a SNP is associated with a gene probe, we use the simple linear regression

$y_i = \beta_0+\beta_1 x_i + \epsilon_i,$

where $y_i$ is the gene expression level of the $i$-th subject, $x_i$ is the genotype of the $i$-th subject, and $\epsilon_i$ is the random error term. Additive coding for genotype is used. To test if the SNP is associated with the gene probe, we test the null hypothesis $H_0: \beta_1=0$.

Denote $p$ as the minor allele frequency (MAF) of the SNP. Under Hardy-Weinberg equilibrium, we can calculate the variance of genotype of the SNP: $\sigma^2_x=2 p (1-p)$, where $\sigma^2_x$ is the variance of the predictor (i.e. the SNP) $x_i$.

We then can use Dupont and Plummer's (1998) power/sample size calculation formula to calculate the minimum detectable slope, adjusting for multiple testing.

Value

sample size required for the eQTL analysis to achieve the desired power.

Author(s)

Xianjun Dong <[email protected]>, Tzuu-Wang Chang <[email protected]>, Scott T. Weiss <[email protected]>, Weiliang Qiu <[email protected]>

References

Dupont, W.D. and Plummer, W.D.. Power and Sample Size Calculations for Studies Involving Linear Regression. Controlled Clinical Trials. 1998;19:589-601.

See Also

powerEQTL.SLR, minSlopeEQTL.SLR

Examples

ssEQTL.SLR(
  MAF = 0.1,
  typeI = 0.05,
  nTests = 2e+05,
  slope = 0.13,
  mypower = 0.8,
  mystddev = 0.13,
  n.lower = 2.01,
  n.upper = 1e+30,
  verbose = TRUE)

---