Fold change calculation

Question

[anm17]

Hello again ,

I'm still stucked in recalculating values from a series of GEO. This an example for one gene

group1.1 group1.2   group2.1    group2.2    loggroup1.mean  loggroup2.mean log.ratio     
34,1324  33,2059     164,307    84,3506     5,073218826     6,87928828     1,806069454

The values above are original data from the experiment itself, and now I wanted to calculate the fold change and did it in that way:

  2^(1,806069454)=3,496882825 
 fold change= 3,496882825/5,073218826(loggr1.mean)=0,689282869

But according to the paper the fold change is 4,5

I don't get how the assed this number. I hope someone can help me

[anm17]

First thank you for replying! I get how the ratio of both mean values was calculated but I still couldn't figure out how they assessed a fold change of 4,5 :/

[Istvan Albert]

(I moved this to a comment, because it was not a new answer. I can only move it as a comment to the top level.)

But then as for a better help, it is not clear to me what the right answer is.

[anm17]

I got an error when I tried to post it as a comment

Answer 1

[Istvan Albert]

Figuring out what someone means in a column should not be difficult but in practice it is. Becomes sort of a task of reverse engineering. Here is what I think:

a = (34 + 33) / 2
b = (164 + 84) / 2

then

log2(a) = 5.07
log2(b) = 6.95
log2(b/a) = 1.8

[Learner]

@Istvan Albert I just read your answer to this question which is interesting , so you average the biological replicate ? then transform them then calculate fold change ? I tried different ways and each gives different answer

for example one way

> log2(34)
[1] 5.087463
> log2(33)
[1] 5.044394
> 5.087463+5.044394/2
[1] 7.60966
> log2(164)
[1] 7.357552
> log2(84)
[1] 6.392317
> 7.357552+6.392317/2
[1] 10.55371
> log2(10.55371/7.60966)
[1] **0.4718463**

The second way

> a = (34 + 33) / 2
> b = (164 + 84) / 2
> log2(b/a)
[1] 1.888107

which one is really the right way to do when there is replicate or biological replicate

[Istvan Albert]

I was just reverse engineering how the numbers were obtained. I was not stating that this was indeed the most appropriate method.

[Learner]

@Istvan Albert Thanks I see , what is the most appropriate method? do you have any comment

[Istvan Albert]

We actively discourage asking new questions in answer or comment sections - it makes the site unfocused. You should post this as a new question.

[abhishekniroula7]

@Learner This is an old post and maybe most readers have/will notice that there are syntax errors in your calculations in your first way. Example a+b/2 will give the sum of a and half of b NOT half of (a+b). You should have (a+b)/2.

Answer 2

[Shyam]

Fold change is calculated using -

Test_read_count/Control_read_count

For logFC, either get the actual FC and convert to log or convert the means of red count to log first and calculate logFC using

log(test_mean) - log(control_mean)

In your case, the log FC should be 1.8 (the log ratio column). I am not sure why the paper reported 4.5 Even the untransformed FC is around 3.6

can you get any info on how they did the analysis in the paper.

[anm17]

This is their method description of their analysis:

Short version: They normalized data with the MAS 5.0 method, filtered the data based on background noise (75 RFU) calculated the mean value for two replicates for each sample and compared each mean of group with mean of control, then done filtering again : twofold up or down compared to control, identified in the new gene list common and unique genes

Long version(taken from the paper itself : http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3988454/)

After scanning, the microarrays were normalized using the Affymetrix Microarray Suite 5.0 (MAS 5.0). Signal values were scaled by global methods to a target value of 500. A pivot table was generated from MAS 5.0 and imported into GeneSpring GX (version 7.3; Agilent Technologies). The mean signal value for the two replicates for each of the samples was compared to the mean signal value of the two undosed control samples. For each of the pairwise comparisons, the 54,681 transcripts were first filtered to retain transcripts that had signal values above the background (> 150 RFU in one of two samples, either treated or control). The background/noise level is ~75 RFU for these arrays. Next, filtering was performed to keep transcripts that change at least twofold (up or down) compared to the control. The resulting gene lists, for each of the 10 samples, were then used to identify genes that were common to the 10 samples and to select genes that were unique to the 10 samples. The Affymetrix microarray data for this experiment have been submitted to the gene expression omnibus (GEO) repository and can be accessed under GEO record GSE18235.

[Shyam]

I am not sure why there is a difference in your and the paper result. And they didn't mention log transformation. I think the value they gave is just FC. How you get the values you gave in your original post. Are they extracted from their analysis or you calculated them by yourself. The only thing I can think of is the effect of normalization on the expression values (not sure).