Question

DESeq2: Compare expression of one group to the mean of others

1

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johntlovell ▴ 10

@johntlovell-8263

Last seen 9.5 years ago

United States

Hi Folks, this is a conceptual question:

I have a RNA-seq dataset where I am trying to make comparisons among 3 levels of a factor - two parental genotypes and an F1. In addition to asking whether there is pairwise DE between genotypes (simple using contrast arguments in results), I would like to ask whether the F1 is intermediate between the two parents. This test is usually accomplished by comparing the mid-parent value (the estimate of the parental means) to the distribution of expression values of the F1. Is there a way to implement this test in DESeq2 - perhaps by setting a contrast between a value (mid parent mean) and the F1 count distribution?

Thanks,

John Lovell

Postdoctoral Fellow, UT Austin.

deseq2 heterosis contrast design and contrast matrix • 4.8k views

ADD COMMENT • link updated 9.5 years ago by Michael Love 43k • written 9.5 years ago by johntlovell ▴ 10

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Could you show us your sample table? Especially, do you have several related trios of father, mother, offspring as individuals (as in mice) or rather of whole collections (as in flies)?

ADD REPLY • link 9.5 years ago Simon Anders ★ 3.8k

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FYI... the way I am currently running the analysis follows:

extract varianceStabilizedTransformation data
calculate arithmetic mean of parents (mid parent vale)
conduct a one-sample t-test with mu = midParentValue, data= F1VstData
FDR adjust P-values.

I am not sure that this is appropriate, but I couldn't think of how to implement it within DESeq. I guess another option would be to extract the estimateSizeFactors normalized data, then run glm.nb.

ADD REPLY • link 9.5 years ago johntlovell ▴ 10

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johntlovell ▴ 10

@johntlovell-8263

Last seen 9.5 years ago

United States

Hi Simon, Thanks for your response.

I have many biological replicates (clones) derived from a single individual of each parent and the F1 for each genotype. So it is just one trio. A library was prepared for each clone, so there was no pooling of samples.

Here is the colData from the DESeq object. I am currently fitting a model where only genotype (halfil=f1, hal=parent1, fil=parent2) is the response variable, but I am open to fitting other models that could more accurately pick up the mid parent value.

Thanks, John

jgi.id generation genotype
1 F1_G948_376 F1 halfil
2 F1_G969_327 F1 halfil
3 F1_H005_732 F1 halfil
4 F1_H016.1_219 F1 halfil
5 F1_H021_969 F1 halfil
6 F1_H040_457 F1 halfil
7 F1_H042.1_688 F1 halfil
8 F1_H064_616 F1 halfil
9 F1_H107_918 F1 halfil
10 F1_H128_677 F1 halfil
11 F1_H131_966 F1 halfil
12 F1_H140_842 F1 halfil
13 F1_H148_386 F1 halfil
14 F1_H151_243 F1 halfil
15 F1_H185_520 F1 halfil
16 F1_H220_365 F1 halfil
17 F1_H229_417 F1 halfil
18 F1_H231_250 F1 halfil
19 F1_H239_8 F1 halfil
20 F1_H249_22 F1 halfil
21 F1_H253_725 F1 halfil
22 F1_H270_964 F1 halfil
23 F1_H273_254 F1 halfil
24 F1_H317_925 F1 halfil
25 F1_H343_981 F1 halfil
26 F1_H347.1_410 F1 halfil
27 F1_H350_879 F1 halfil
28 F1_H352_759 F1 halfil
29 F1_H355_38 F1 halfil
30 FIL2_G956_490 F0 fil
31 FIL2_G957_7 F0 fil
32 FIL2_G964_479 F0 fil
33 FIL2_G968_100 F0 fil
34 FIL2_G976_42 F0 fil
35 FIL2_G979_865 F0 fil
36 FIL2_H033_594 F0 fil
37 FIL2_H037_176 F0 fil
38 FIL2_H058_486 F0 fil
39 FIL2_H068_368 F0 fil
40 FIL2_H092_19 F0 fil
41 FIL2_H097_895 F0 fil
42 FIL2_H098_248 F0 fil
43 FIL2_H114_549 F0 fil
44 FIL2_H119_563 F0 fil
45 FIL2_H120_2 F0 fil
46 FIL2_H144_888 F0 fil
47 FIL2_H170_247 F0 fil
48 FIL2_H172_475 F0 fil
49 FIL2_H182_534 F0 fil
50 FIL2_H192_228 F0 fil
51 FIL2_H207_610 F0 fil
52 FIL2_H208_103 F0 fil
53 FIL2_H211_367 F0 fil
54 FIL2_H213_58 F0 fil
55 FIL2_H228_485 F0 fil
56 FIL2_H235_289 F0 fil
57 FIL2_H267_130 F0 fil
58 FIL2_H275_502 F0 fil
59 FIL2_H293_55 F0 fil
60 FIL2_H297_504 F0 fil
61 FIL2_H326_12 F0 fil
62 FIL2_H338_904 F0 fil
63 FIL2_H345_323 F0 fil
64 HAL2_G950_542 F0 hal
65 HAL2_G959.1_220 F0 hal
66 HAL2_G988_482 F0 hal
67 HAL2_G996_353 F0 hal
68 HAL2_H008_403 F0 hal
69 HAL2_H020_447 F0 hal
70 HAL2_H057_24 F0 hal
71 HAL2_H074_147 F0 hal
72 HAL2_H090_76 F0 hal
73 HAL2_H101_237 F0 hal
74 HAL2_H126_821 F0 hal
75 HAL2_H152_332 F0 hal
76 HAL2_H166_959 F0 hal
77 HAL2_H188_341 F0 hal
78 HAL2_H193_752 F0 hal
79 HAL2_H195.1_306 F0 hal
80 HAL2_H236_73 F0 hal
81 HAL2_H246_762 F0 hal
82 HAL2_H261_344 F0 hal
83 HAL2_H266_370 F0 hal
84 HAL2_H305_910 F0 hal
85 HAL2_H309_469 F0 hal
86 HAL2_H312_538 F0 hal
87 HAL2_H319_128 F0 hal
88 HAL2_H351_435 F0 hal

ADD COMMENT • link 9.5 years ago johntlovell ▴ 10

score 4 · Accepted Answer · 2015-06-25

4

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Michael Love 43k

@mikelove

Last seen 4 days ago

United States

Hi John,

I think that you can perform this test with a numeric contrast. If you have a design ~ genotype, and the reference level of genotype is the F1 generation (you can set this using relevel(), see the vignette), then the resultsNames(dds) should be something like:

Intercept, genotypehalfil, genotypehal, genotypefil

You want a numeric contrast of halfil against 1/2 (hal + fil):

results(dds, contrast=c(0, 1, -1/2, -1/2))

Or equivalently, using the list style of contrast:

results(dds, contrast = list("genotypehalfil", c("genotypehal","genotypefil")), 
        listValues=c(1, -1/2))

This gives you the log fold change of halfil over the midparent. So small p-values for genes in which halfil is above or below the midparent value.

ADD COMMENT • link 9.5 years ago Michael Love 43k

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Dear Michael,

In the latest version of DESeq2, resultsNames(dds) do not show something like:

Intercept, genotypehalfil, genotypehal, genotypefil

but like this:

>resultsNames(dds)
[1] "Intercept"        "condition_b_vs_a" "condition_c_vs_a"

if I want to do the above comparation, I should use DESeq(dds, betaPrior = TRUE), then I find something confusing.

>resultsNames(dds)
[1] "Intercept"  "conditiona" "conditionb" "conditionc"(conditiona is the hyb condition, and the other two are P1 and P2, I've set the conditionia  to be reference)
>mid<-results(dds, contrast=c(0, 1, -1/2, -1/2))
>mid[rownames(mid)=="geneA",]
log2 fold change (MAP): 0,+1,-0.5,-0.5 
Wald test p-value: 0,+1,-0.5,-0.5 
DataFrame with 1 row and 6 columns
                         baseMean   log2FoldChange             lfcSE
                        <numeric>        <numeric>         <numeric>
geneA 72.3609376218373 1.15447089513261 0.228187646879045
                             stat               pvalue                 padj
                        <numeric>            <numeric>            <numeric>
geneA 5.05930496642775 4.20787419598304e-07 6.10070739443769e-05
>mid_norm<-counts(dds, normalized=TRUE)
>mid_norm[rownames(mid_norm)=="geneA",]
    sample_hyb_1     sample_hyb_2     sample_hyb_3     sample_P1_1     sample_P1_2     sample_P1_3     sample_P2_1 
 75.965725  62.666961  69.122855   3.389493   3.060267   4.232354 143.554684 
    sample_P2_2     sample_P2_3 
148.588003 140.668097

So for geneA, the mean of parents(P1,P2) is similar to hyb. (3.38+3.06+4.23+143.55+148.58+140.66)/6=73.91,(75.96+62.66+69.12)/3=69.25, they are similar.

How could the log2FoldChange be as large as 1.154? that means a foldchange of 2^1.154=2.23. Could you please give some explanation?

or are there any other better methods for the latest version if I want to compare genotypehalfil to the mean of genotypehal and genotypefil? Looking forward to your reply. Thank you very much！

Best,

Aifu.

ADD REPLY • link 6.2 years ago Aifu • 0

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Entering edit mode

When you average the coefficients, it's not equivalent to averaging the data and then computing a LFC.

It's closer to this:

log2(70/70) - .5*log2(3/70) - .5*log2(140/70)

which gives an LFC above 1.

ADD REPLY • link 6.2 years ago Michael Love 43k

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Thank you very much for your explanation, Michael. But I still have some confusion. Your explanation seems reasonable for geneA above. I find another gene, geneB, seems different.

> bb<-counts(dds,normalized=T)
> b<-bb[rownames(bb)=="geneB",]
> b
   R_SY_1    R_SY_2    R_SY_3    R_MH_1    R_MH_2    R_MH_3    R_ZS_1    R_ZS_2 
 58.13703  70.72414  46.96809  80.50047  75.48658 110.04121  28.71094  26.10330 
   R_ZS_3 
 22.81104
> n<-mean(b[1:3])
> m<-mean(b[4:6])
> z<-mean(b[7:9])
> log2(n) - .5*log2(m) - .5*log2(z)
[1] 0.2910863
> res[rownames(res)=="geneB",]
log2 fold change (MAP): 0,+1,-0.5,-0.5 
Wald test p-value: 0,+1,-0.5,-0.5 
DataFrame with 1 row and 6 columns
                         baseMean   log2FoldChange             lfcSE
                        <numeric>        <numeric>         <numeric>
geneB 57.7203113936858 1.06110078378387 0.240486475709396
                             stat               pvalue                 padj
                        <numeric>            <numeric>            <numeric>
geneB 4.41230959310205 1.02273726663086e-05 0.000258779640785735

The log2FoldChange is 1.06110078378387, quite different from 0.29. Could you please explain this to me? Thank you for your attention.

ADD REPLY • link 6.2 years ago Aifu • 0

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There is shrinkage applied. The output is simply the linear combination of the coefficients. You can look at the coefficients yourself with coef(dds). You will find that multiplying 1, -.5, -.5 times the coefficients gives the LFC.

I unfortunately don’t have much spare time at the moment so I may not reply to further gene examples.

ADD REPLY • link 6.2 years ago Michael Love 43k

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Yes, that's true! Thank you very much!

Aifu.

ADD REPLY • link 6.2 years ago Aifu • 0