Originally posted by whataBamBam
View Post
Seqanswers Leaderboard Ad
Collapse
Announcement
Collapse
No announcement yet.
X
-
-
Originally posted by rskr View PostIf the hypothesis test was significant, this would indicate that there isn't a problem with power, since a lower powered test should make it more difficult to get significant results, if the test was actually answering the question you were asking. Though in theory this does seem a little confusing, because getting a hypothesis to fit thousands of sample should be harder than just a few samples, however with thousands of samples the means in the population can be known very accurately, so even trivial differences like color between two placebos can be significant.
http://en.wikipedia.org/wiki/Lindley's_paradox
The next part I'm less sure about.. but I think this paradox, Lindleys paradox would only apply if there were a very large number of replicates? Which we aren't ever likely to see.
Comment
-
Originally posted by whataBamBam View PostYeah the first part is what I meant - well kind of. Yes a lower power test makes it more difficult to observe significant results - but we observe them. So the test had enough power to detect the differences it detected but there could be other differences it did not detect because it did not have enough power. This is what I mean by saying it's conservative.
The next part I'm less sure about.. but I think this paradox, Lindleys paradox would only apply if there were a very large number of replicates? Which we aren't ever likely to see.
Comment
-
Hi
I want to use the DESEQ package between a control (3 biological replicates) and treatment (1 biological replicate).
IN DESeq I herefore used the following code, and got 266 genes with padj < 0.05:
table <- read.delim("test.txt")
row.names(table) <- table$Feature_ID
count_table <- table[, -1]
conds <- c("ctrl", "ctrl", "ctrl", "treatment")
cds <- newCountDataSet(count_table, conds)
cds <- estimateSizeFactors(cds)
cds <- estimateDispersions(cds, method="blind", sharingMode="fit-only")
results <- nbinomTest(cds, "ctrl", "treatment")
In DESeq2 I used the follwing command, but got > 10000 genes with padj < 0.05:
table <- read.delim("test.txt")
row.names(table) <- table$Feature_ID
count_table <- table[, -1]
colData <- DataFrame(condition=factor(c("ctrl", "ctrl", "ctrl", "treatment")))
dds <- DESeqDataSetFromMatrix(count_table, colData, formula(~ condition))
results <- DESeq(dds, minReplicatesForReplace=Inf)
So probably I need to add extra parameters to the DESEQ2 analysis but for now I can't figure out how?
Thank you for helping
WannesLast edited by tompoes; 12-02-2015, 12:31 PM.
Comment
-
Originally posted by Simon Anders View PostTo be honest, we couldn't yet be bothered to explain how to analyse such data in DESeq2. It's tricky to write up because too many people will misinterpret whatever I write as if it were actually possible to conduct a meaningful statistical analysis when comparing just two samples.
So, if you promise to not use any such comparisons for actual science, here is how you do it:
Start as above:
Code:library(DESeq2) library(pasilla) data("pasillaGenes") countData <- counts(pasillaGenes) countData<-countData[,c("treated1fb","untreated1fb")] colData <- pData(pasillaGenes)[c("treated1fb","untreated1fb"),c("condition","type")] dds <- DESeqDataSetFromMatrix( countData = countData, colData = colData, design = ~ condition)
Code:rld <- rlogTransformation( dds )
Code:res <- data.frame( assay(rld), avgLogExpr = ( assay(rld)[,2] + assay(rld)[,1] ) / 2, rLogFC = assay(rld)[,2] - assay(rld)[,1] )
Code:> head( res[ order(res$rLogFC), ] ) treated1fb untreated1fb avgLogExpr rLogFC FBgn0011260 7.830359 6.627326 7.228842 -1.203033 FBgn0001226 10.128636 8.929985 9.529311 -1.198652 FBgn0034718 8.503006 7.315640 7.909323 -1.187366 FBgn0003501 7.927864 6.743974 7.335919 -1.183889 FBgn0033635 11.126300 9.973979 10.550139 -1.152321 FBgn0033367 13.411814 12.269436 12.840625 -1.142378
The advantage of this procedure is that it does not produce any p values (which would be misleading anyway).
just a silly question... why is a (-) and not a division (/) to calculate the rlogFC;
res <- data.frame(
assay(rld),
avgLogExpr = ( assay(rld)[,2] + assay(rld)[,1] ) / 2,
rLogFC = assay(rld)[,2] / assay(rld)[,1] )
Thank you!!
Comment
-
Originally posted by FJwlf View PostHi everyone!
just a silly question... why is a (-) and not a division (/) to calculate the rlogFC;
res <- data.frame(
assay(rld),
avgLogExpr = ( assay(rld)[,2] + assay(rld)[,1] ) / 2,
rLogFC = assay(rld)[,2] / assay(rld)[,1] )
Thank you!!
Comment
Latest Articles
Collapse
-
by seqadmin
The field of epigenetics has traditionally concentrated more on DNA and how changes like methylation and phosphorylation of histones impact gene expression and regulation. However, our increased understanding of RNA modifications and their importance in cellular processes has led to a rise in epitranscriptomics research. “Epitranscriptomics brings together the concepts of epigenetics and gene expression,” explained Adrien Leger, PhD, Principal Research Scientist...-
Channel: Articles
04-22-2024, 07:01 AM -
-
by seqadmin
Proteins are often described as the workhorses of the cell, and identifying their sequences is key to understanding their role in biological processes and disease. Currently, the most common technique used to determine protein sequences is mass spectrometry. While still a valuable tool, mass spectrometry faces several limitations and requires a highly experienced scientist familiar with the equipment to operate it. Additionally, other proteomic methods, like affinity assays, are constrained...-
Channel: Articles
04-04-2024, 04:25 PM -
ad_right_rmr
Collapse
News
Collapse
Topics | Statistics | Last Post | ||
---|---|---|---|---|
Started by seqadmin, Yesterday, 08:47 AM
|
0 responses
16 views
0 likes
|
Last Post
by seqadmin
Yesterday, 08:47 AM
|
||
Started by seqadmin, 04-11-2024, 12:08 PM
|
0 responses
60 views
0 likes
|
Last Post
by seqadmin
04-11-2024, 12:08 PM
|
||
Started by seqadmin, 04-10-2024, 10:19 PM
|
0 responses
60 views
0 likes
|
Last Post
by seqadmin
04-10-2024, 10:19 PM
|
||
Started by seqadmin, 04-10-2024, 09:21 AM
|
0 responses
54 views
0 likes
|
Last Post
by seqadmin
04-10-2024, 09:21 AM
|
Comment