Why do I obtain Identical Adj.P value when running edgeR ?
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Entering edit mode
Mohamed ▴ 30
@aa1ae679
Last seen 15 months ago
United Kingdom

I am running EdgeR pipeline on an RNA seq data comparing two condition (mock; 2 samples vs infected ; 10 samples) while adjusting for time points. I normalized using TMM method, and then fitted generalized linear model (glmQLFit) on the data and then run DE test using (glmQLFTest) When I checked the summary of the decideTests, no genes were sig DE at 0.05 or even 0.1 cutoff. I looked into toptag function to look at the adj.pvalue and found they are all identical and close to 1. So, this was the reason of no sig. genes retrieved. I have also tried all the P.adj methods, they gave the same results. I have calculated adj p-value manually using function (p.unadju), they gave the same results. May be because of the 2 biological replicates vs 10 replicates is reducing the sig. DE power ?

Any explanation of why do I get identical adj. p value ? although P-values are different ? Can I use P-values for sig DE genes at 0.05 ?

My code:

# Convert filtered raw counts to DGEList list 
L10_dgeObj <- DGEList(L10_counts.keep)
# TMM normalization 
L10_dgeObj <- calcNormFactors(L10_dgeObj)
# create design matrix to compare the two levels of Status ( Mock, infected) and adjust for time points
L10_design <- model.matrix(~Time_point+Status)
# dispersion estimation
L10_y <- estimateDisp(L10_dgeObj, L10_design)
L10_y$common.dispersion
L10_y$trended.dispersion
Tagwisedie<- L10_y$tagwise.dispersion
# instaling statmod package helps to robustness the DE model 
install.packages('statmod')
library('statmod')
# fitting the linear model.Robust = TRUE
    L10_fit<- glmQLFit(L10_y, L10_design, robust = TRUE)
# running the DE test 
    L10_tr <- glmQLFTest(L10_fit, coef = "StatusInfected")
# get summary of decideTest 
    summary(decideTests(L10_tr)) 
# Get list of top sig DE gene based on adj. pvalue 
    topTags(L10_tr, n = 15420, adjust.method = 'BH', sort.by = 'PValue', p.value = 0.05)

Session Info:

R version 4.2.3 (2023-03-15 ucrt)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 10 x64 (build 19045)

Matrix products: default

locale:
[1] LC_COLLATE=English_United Kingdom.utf8  LC_CTYPE=English_United Kingdom.utf8   
[3] LC_MONETARY=English_United Kingdom.utf8 LC_NUMERIC=C                           
[5] LC_TIME=English_United Kingdom.utf8    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
 [1] vegan_2.6-4          lattice_0.21-8       permute_0.9-7        xlsx_0.6.5          
 [5] statmod_1.5.0        readxl_1.4.3         BiocManager_1.30.22  edgeR_3.40.2        
 [9] metagenomeSeq_1.40.0 RColorBrewer_1.1-3   glmnet_4.1-8         Matrix_1.6-0        
[13] limma_3.54.2         Biobase_2.60.0       BiocGenerics_0.44.0 

loaded via a namespace (and not attached):
  [1] Rtsne_0.16             colorspace_2.1-0       ellipsis_0.3.2         XVector_0.38.0        
  [5] fs_1.6.3               rstudioapi_0.15.0      remotes_2.4.2.1        fansi_1.0.4           
  [9] codetools_0.2-19       splines_4.2.3          cachem_1.0.8           knitr_1.44            
 [13] pkgload_1.3.2.1        ade4_1.7-22            jsonlite_1.8.7         phyloseq_1.44.0       
 [17] rJava_1.0-6            cluster_2.1.4          shiny_1.7.5            compiler_4.2.3        
 [21] fastmap_1.1.1          cli_3.6.1              later_1.3.1            htmltools_0.5.5       
 [25] prettyunits_1.1.1      tools_4.2.3            igraph_1.5.1           gtable_0.3.4          
 [29] glue_1.6.2             GenomeInfoDbData_1.2.9 reshape2_1.4.4         dplyr_1.1.2           
 [33] Rcpp_1.0.11            cellranger_1.1.0       vctrs_0.6.3            Biostrings_2.66.0     
 [37] rhdf5filters_1.10.1    multtest_2.54.0        ape_5.7-1              nlme_3.1-162          
 [41] iterators_1.0.14       xfun_0.40              stringr_1.5.0          xlsxjars_0.6.1        
 [45] ps_1.7.5               mime_0.12              miniUI_0.1.1.1         lifecycle_1.0.3       
 [49] gtools_3.9.4           devtools_2.4.5         zlibbioc_1.44.0        MASS_7.3-60           
 [53] scales_1.2.1           promises_1.2.1         parallel_4.2.3         biomformat_1.26.0     
 [57] rhdf5_2.42.1           yaml_2.3.7             memoise_2.0.1          gridExtra_2.3         
 [61] ggplot2_3.4.3          stringi_1.7.12         S4Vectors_0.36.2       foreach_1.5.2         
 [65] caTools_1.18.2         pkgbuild_1.4.2         shape_1.4.6            GenomeInfoDb_1.34.9   
 [69] rlang_1.1.1            pkgconfig_2.0.3        matrixStats_1.0.0      bitops_1.0-7          
 [73] Wrench_1.16.0          evaluate_0.21          purrr_1.0.2            Rhdf5lib_1.20.0       
 [77] htmlwidgets_1.6.2      tidyselect_1.2.0       processx_3.8.2         plyr_1.8.8            
 [81] magrittr_2.0.3         R6_2.5.1               IRanges_2.32.0         gplots_3.1.3          
 [85] generics_0.1.3         profvis_0.3.8          pillar_1.9.0           mgcv_1.9-0            
 [89] survival_3.5-5         RCurl_1.98-1.12        tibble_3.2.1           crayon_1.5.2          
 [93] KernSmooth_2.23-22     utf8_1.2.3             microbiome_1.20.0      rmarkdown_2.24        
 [97] urlchecker_1.0.1       usethis_2.2.2          locfit_1.5-9.8         grid_4.2.3            
[101] data.table_1.14.8      callr_3.7.3            digest_0.6.33          xtable_1.8-4          
[105] tidyr_1.3.0            httpuv_1.6.11          stats4_4.2.3           munsell_0.5.0         
[109] sessioninfo_1.2.2
edgeR • 965 views
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Regarding identical values of adj.P.Val, you may want to see this post and link: Limma adj.p.value the same for many different p.values

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I went through this, but what I got is that this is normal. Does this now means I do not have any sig DE genes at 0.05 based on adj.P value ?

Thanks

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@gordon-smyth
Last seen 35 minutes ago
WEHI, Melbourne, Australia

I looked into toptag function to look at the adj.pvalue and found they are all identical and close to 1

That means there is no evidence for any differential expression in your dataset. If there is no DE, then the FDR will be close to 1 even for the top candidate gene.

May be because of the 2 biological replicates vs 10 replicates is reducing the sig. DE power?

More replicates generally mean more power but the small sample size is unlikely to be the main reason for getting no results here. If power was the only problem then you would still get some smaller FDR vales. With FDR=1, the reason is more likely that there is just no separation between the two groups.

You should always check the quality of your data by doing MDS plots etc. You haven't reported doing any data exploration or checking.

Any explanation of why do I get identical adj. p value ? although P-values are different ?

That's how the FDR algorithm works. If none of the p-values are small then the FDR will be 1.

Can I use P-values for sig DE genes at 0.05 ?

No, not unless you want to publish non-reproducible results.

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Thanks Gordon, I run nMDs on the data and here it looks like, supporting the idea of no differential expression. Actually this nMDS is run on some samples excluding the outliers... Looks separated but I can not believe that small sample size is not affecting that ...

Thanks

enter image description here

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I run nMDs on the data

No you haven't! The data shown in the nMDS plot is obviously different from the data that you ran the DE analysis on. The data shown in the plot has more mock samples and has been curated differently including removal of outliers.

There is no need for a non-metric MDS here and its meaning of the x and y axes is unclear. You should simply run edgeR::plotMDS(L10_dgeObj).

supporting the idea of no differential expression

On the contrary, the plot shows a complete separation of the two groups. If edgeR was run on the data shown in the plot then it would return heaps of significant DE.

If your data has outliers samples then you should remove them before running the edgeR DE analysis. Outliers will have a dramatic detrimental effect on the analysis.

Low samples sizes could also be an issue because you say that you are adjusting for time points. Adjusting for time point will be very difficult if you only have two mock samples. You don't show the design of your study so it is impossible to tell ...

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