Solved – Meta-analysis in R with multiple SNPs

geneticsmeta-analysisr

I'm conducting a meta-analysis on two genome wide association studies (GWAS), each consisting of 150 SNPs, for which I computed summary statistics for association.
I'm using the R package meta and the function metagen.
It seems to me that this function is able to combine studies but on single measurement effect (I have 150), one for each study.

For example, I may have 3 SNPs on each study

STUDY    SNP    OR    SE
GWAS_A  rs694739    0.6691  0.07588
GWAS_A  rs9858968   0.1091  0.01588
GWAS_A  rs1529267   0.9291  0.02588
GWAS_B  rs694739    0.6128  0.37344
GWAS_B  rs9858968   0.0332  0.27344
GWAS_B  rs1529267   0.3481  0.81284

so I would like an output in a table format (e.g., as in PLINK), with 3 rows and the corresponding pooling statistics.
I know I could manage it with for loops or by grouping with the by function on the summary output of metagen,
but I wonder whether there is another R package/function to easily accomplish this.

Best Answer

The MetABEL part of GenABEL does this. For 150 SNPs, you might find coding the loop yourself quicker than ensuring it's doing exactly what you want. (Neither should take very long)

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Solved – How to calculate confidence intervals for pooled odd ratios in meta-analysis

In most meta-analysis of odds ratios, the standard errors $se_i$ are based on the log odds ratios $log(OR_i)$. So, do you happen to know how your $se_i$ have been estimated (and what metric they reflect? $OR$ or $log(OR)$)? Given that the $se_i$ are based on $log(OR_i)$, then the pooled standard error (under a fixed effect model) can be easily computed. First, let's compute the weights for each effect size: $w_i = \frac{1}{se_i^2}$. Second, the pooled standard error is $se_{FEM} = \sqrt{\frac{1}{\sum w}}$. Furthermore, let $log(OR_{FEM})$ be the common effect (fixed effect model). Then, the ("pooled") 95% confidence interval is $log(OR_{FEM}) \pm 1.96 \cdot se_{FEM}$.

Update

Since BIBB kindly provided the data, I am able to run the 'full' meta-analysis in R.

library(meta)
or <- c(0.75, 0.85)
se <- c(0.0937, 0.1029)
logor <- log(or)
(or.fem <- metagen(logor, se, sm = "OR"))

> (or.fem <- metagen(logor, se, sm = "OR"))
    OR            95%-CI %W(fixed) %W(random)
1 0.75  [0.6242; 0.9012]     54.67      54.67
2 0.85  [0.6948; 1.0399]     45.33      45.33

Number of trials combined: 2 

                         OR           95%-CI       z  p.value
Fixed effect model   0.7938  [0.693; 0.9092] -3.3335   0.0009
Random effects model 0.7938  [0.693; 0.9092] -3.3335   0.0009

Quantifying heterogeneity:
tau^2 < 0.0001; H = 1; I^2 = 0%

Test of heterogeneity:
    Q d.f.  p.value
 0.81    1   0.3685

Method: Inverse variance method

References

See, e.g., Lipsey/Wilson (2001: 114)

Solved – How to interpret false positive from QQ-plot in genome-wide association studies

Under the null hypothesis, your p-values should adhere to a uniform distribution (I'm ignoring some issues with dependencies here, as they are not really relevant to the discussion). This includes the false positives: just by coincidence, you should get a few extreme p-values, but since this is exactly what is in the expected distribution, this will not distort the QQ-plot: false positives are already accounted for in this type of QQ-plot.

On the other hand, if some SNPs were tested where the null hypothesis is not true, typically this should show up as a lower p-value than expected, and these will indeed distort the image.

What the author of the article you link seems to mean is: if you have wicked strong distortion, there are unexpectedly (as per the uniform distribution) many extreme p-values, so something is likely to be wrong.

So, in short: a uniform QQ-plot of p-values (like this) should show relatively little digression from the straight line, and only in the extremely low p-values (showing you have more low p-values than if the null hypothesis were true everywhere).

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