Deseq2 - disturbance in fold change value
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0
Entering edit mode
@madhusudhanajanga-13266
Last seen 7.6 years ago

Hi,

We are trying to understand DEseq2 fold change values.

Here I am giving example of some genes with their normalized read count values and calculated log fold change.

Normalized Read count values

Feature

A

B

C

D

E

F

G

Gene1

43.06

0.00

0.00

0.00

0.00

0.00

0.00

Gene2

679.19

0.00

0.00

0.00

0.00

0.00

0.00

Gene3

91.82

0.00

0.00

0.00

0.00

0.00

0.00

Gene4

1331.35

64.61

41.58

10.96

4.93

0.79

0.00

fold change

Feature

logFC - A vs B

logFC - A vs C

logFC - A vs D

logFC - A vs E

logFC - A vs F

logFC - A vs G

Gene1

8.02

7.89

7.78

7.51

7.57

7.78

Gene2

5.54

5.53

5.53

5.52

5.52

5.53

Gene3

9.15

8.87

8.59

7.99

8.13

8.64

Gene4

4.36

4.99

6.84

8.01

10.05

11.93

Here for gene2 we expect high fold change compare to gene1 and gene2, but we see disturbance in the fold change values.

I understand DEseq2 has complicated statistics to achieve this fold change values.

Please help us to understand why gene2 FC < gene1 FC.

again gene1 and gene 2 FC are comparable based on their read count values?.

we used following code:

library(DESeq2)

>de=DESeq(ddsHTSeq, betaPrior = TRUE)

    estimating size factors
    estimating dispersions
    gene-wise dispersion estimates
    mean-dispersion relationship
    final dispersion estimates
    fitting model and testing

resultsNames(de)

res_AvsB <- results(de, contrast = c("condition", "A", "B"))

Thank you for your help in this regard.

Madhu

deseq2 • 1.0k views
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Hi Michael,

Thank you for your quick reply. Sorry I did not mention details, for each condition i have three replicates. 7 tissues * 3 replicates.

read counts are averaged normalized read counts from three replicates for each tissue.

Thanks

madhu

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@mikelove
Last seen 11 days ago
United States

hi Madhu,

I'n not sure exactly, but you have to remember it's a very interdependent system that relies on accurate estimates of dispersion. Without biological replicates, you throw out the ability to do a good job estimating dispersion (because true differences look like noise). So a big difference can possibly be moderated more than a small difference. See ?DESeq for the discussion for experiments without replicates.

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Entering edit mode

Hi Michael,

Thank you for your quick reply. Sorry I did not mention details, for each condition i have three replicates. 7 tissues * 3 replicates.

read counts are averaged normalized read counts from three replicates for each tissue.

Thanks

madhu

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Entering edit mode

In that case, remember that moderation is influenced by dispersion, and so a higher mean count doesn't lead to a higher LFC if there is low statistical information to support it (the mean can be driven by high variance alone).

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