Hello folks,
I have the excel file generated from Cuffdiff output for genes with the following columns
Gene, locus, sample_1, sample_2, status, value_1, value_2, log2(fold_change), test_stat, p_value, q_value, significant
Casp1, chr9:5298516-5307281, MUT, WT, OK, 123.019, 0.671358, -7.51758, 6.17607, 6.57E-10, 6.36E-07, yes
As per the excel file, sample_1 is Mutant and sample_2 is Wildtype. Log2(fold_change) is calculated as log2(sample_2/sample_1) --> log2(0.671358/123.019) is -7.51758 .
I thought it should be log2(final/initial), isn't it?
What is the difference between log2(Mutant/Wildtype) or log2(Wildtype/Mutant)?
How to show the fold change (like 4-fold, 5-fold change) in heatmap?
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In cuffdiff output, value_1 and value_2 is for control and experimental conditions. Cuffdiff calculates fold change as log2(value_2/value_1), meaning how much gene expression changes in experimental condition over control.
Further, it depends on which condition you have given as control and experimental while running cuffdiff. You need to share the command that you used for running cuffdiff.
What is the difference between log2(Mutant/Wildtype) or log2(Wildtype/Mutant)?
In log2(Mutant/Wildtype) , wildtype is value_2 and mutant is value_1. It will calculate expression changes in wildtype over mutant. Reverse is valid for log2(Wildtype/Mutant)
How to show the fold change (like 4-fold, 5-fold change) in heatmap?
You can use the color intensity/gradient scale to show the proportion of fold change in the heatmap. See heatmap.2 package in R.
Thanks, Renesh for the explanation.
I just have the excel file with me. Cuffdiff analysis was done by someone else. I believe he/she might have mistakenly given mutant as "CONTROL" and wildtype as "EXPERIMENTAL".
Hi genomax, that person moved out of the lab long time back. Only excel file has been provided by that person. I don't have the proper log of the analysis. Thanks for cautioning me about it.
Hi Genomax,
Does the order of the samples entered in Cuffdiff command impact the results of the differential gene expression?
b'cos Cuffdiff considers Label1 as sample1 and Label2 as sample2. So is it a hidden norm to mention the control always as LABEL1 and experimental/mutant as LABEL2?
Hi Renesh,
This is the code I have for another analysis for Lymph and spleen together from the person who did the below analysis as well,
cuffdiff -o diff_out_Lymphnodes_Spleen -b mouse_mm10.fa -p 12 -L Mutant_Lymph_Spleen,WT_Lymph_Spleen -u merged_asm/merged.gtf Mutant_Lymph/accepted_hits.bam,Mutant_Spleen/accepted_hits.bam Wildtype_Lymph/accepted_hits.bam,Wildtype_Spleen/accepted_hits.bam
I am currently working on the analysis of Lymph nodes, based on the above labeling I suspect that
cuffdiff -o diff_out_Lymphnodes -b mouse_mm10.fa -p 12 -L Mutant_Lymph,WT_Lymph -u merged_asm/merged.gtf Mutant_Lymph/accepted_hits.bam Wildtype_Lymph/accepted_hits.bam
If, for GeneX, Sample1's expression is 20 and Sample2's expression is 5, then:
log2(Sample1/Sample2) = 2
We can make the following statement: Sample1 has higher expression than Sample2 for GeneX
log2(Sample2/Sample1) = -2
We can make the following statement: Sample2 has lesser expression than Sample1 for GeneX
Both statements are implying the same thing. You can see, however, that the choice of nominator and denominator is important.
This should not be important for the heatmap. If a gene has higher expression than another in a particular sample, then the heatmap will have a shading representative of the higher level (e.g. it will be red if your colour scheme goes from green-to-black-to-red for low-to-normal-to-high expression). The heatmap function in R will usually scale your data to the Z scale for the purposes of heatmap colour-shading, thus, we are then referring to standard deviations from the mean, as opposed to fold-changes. You can switch off this function of the heatmap and transform the data in your own way using the following commands:
myBreaks <- seq(-3, 3, length.out=101)
heat <- t(scale(t(MyDataMatrix)))
heatmap.2(..., breaks=myBreaks, scale="none")
Thanks, Kevin. In general, won't they consider (final/initial) in wet-lab.
As you mentioned, inorder to get the Z-scale for the heatmap, I log transformed the FPKM values.
> FPKM <-read.table("dataset.csv",sep=",", header=TRUE, row.names=1)
> nrow(FPKM) [1] 21
> FPKM_log10 <- log10(FPKM+1)
> head(FPKM)
MUT_Lymph WT_Lymph
Actn1 29.42360 92.09680
Ccr7 61.08610 177.92300
Ctla4 53.45150 10.96750
Dapl1 43.16620 140.23300
> head(FPKM_log10)
MUT_Lymph WT_Lymph
Actn1 1.4832106 1.9689348
Ccr7 1.7929944 2.2526662
Ctla4 1.7360098 1.0780034
Dapl1 1.6450900 2.1499362
>Log_data_matrix <- data.matrix(FPKM_log10)
> heatmap.2(Log_data_matrix,scale="row", col=greenred, trace="none",margins = c(5,5),cexRow=0.7,cexCol=0.7,dendrogram='both',Rowv=TRUE,Colv=TRUE,reorderfun=function(d,w) reorder(d, w, agglo.FUN=mean), distfun=function(x) as.dist(1-cor(t(x))), hclustfun=function(x) hclust(x, method="complete"))
Inorder to just get the fold-change of mutant and wildtype in the heatmap, I believe that I should use the FPKM instead of log-transformed FPKM.
Yes, I would just use the FPKM values for the heatmap. The heatmap function will scale these itself and they will be transformed into Z scores.
Also take a look at my colleague's answer below.
Just use your FPKM data-matrix for the heatmap. Do not use the log-transformed one
Thanks, Kevin. Sure, I will use FPKM values alone.
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