how to caculate enrichment score
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4.3 years ago
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Hi there, i am processing a genome-wide methylation data, and i wanna to find out which chromosome were enriched with differentially methylated probes(dmp). I noticed a research compared the distribution of dmps located on each chromosome.

for example, 8.87% of CpG probes on Chromosome 13 were identified as Bonferroni-significant dDMPs (relative enrichment = 1.26, P = 3.68 × 10−10), whereas only 4.09% of CpG probes on Chromosome 19 reach this criterion (relative enrichment = 0.55, P = 7.01 × 10−79).

(reference: Spiers H, Hannon E, Schalkwyk LC, et al. Methylomic trajectories across human fetal brain development. Genome Research. 2015;25(3):338-352. doi:10.1101/gr.180273.114.)

The METHOD part mentioned that the enrichment analysis was done by fisher's exact test, so i try to use fisher.test in R to replicate the result. However, i get a same P value but i have no idea how to caculate the enrichment score. The imput data as following:

Chromosome  TotalProbes  SignificantProbes (%)  Enrichment(95% CI)  p value
1   40079   3109 (7.76) 1.09 (1.04 -1.13)   3.56E-05
Total   399364  28718 (7.19)    NA  NA


The METHOD part mentioned that the enrichment analysis was done by fisher's exact test, so i try to use fisher.test in R to replicate the result. However, i get a same P value but i have no clue how to caculate the enrichment score. Any idea? Any suggestion will be appreciated!

    > x<-c(40079-3109,3109,399364-28718,28718)
> dim(x)<-c(2,2)
> fisher.test(x)

Fisher's Exact Test for Count Data

data:  x
p-value = 3.558e-05
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
0.8864884 0.9578336
sample estimates:
odds ratio
0.9213249

R fisher • 2.6k views
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odds ratio gives you the enrichment.

res <- fisher.test(x)
odds <- res$estimate conf.int <- res$conf.int

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Thanks for your reply. But my OR is 0.92 (95%CI: 0.886-0.958) differs from the enrichment score in paper: enrichment score=1.09 (1.04 -1.13). I used the same input with the research, and our p values are consistent. Do i miss any steps?

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It's just flipped: 1/(0.92)=1.0869 which is basically what is reported in the paper