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I was wondering if there was a way to analyse my RNA seq dataset in a way that I could work out which day the specific gene I have is most significantly expressed?

I have a RNA seq dataset with 6 days and their repeats and then many genes I tried a MANOVA but this didn't work.

I have included a picture of my dataset.

I would quite like a test that would give me a result like this so I can compare the days?

:GeneID : :S1<->S2: S1<->S3 S2<->S3
at1g01040.2 0.027832572 0.04020203 0.13481563
at1g01050.1 0.852379466 0.31471871 0.36326955
at1g01070.1 0.003200692 0.00113536 0.02236621
at1g01080.2 0.086426813 0.03092924 0.45999438
at1g01090.1 0.090387087 0.04638872 0.04978092
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  • $\begingroup$ Can you give a hint what significantly "expressed" is in your opinion? by the way by what value do you decide this. on all values? as in: if gene x is significantly higher on 6 features out of 10 and thus all other genes are only higher on 5 genes? I believe we need more information here. $\endgroup$
    – Patrick Bormann
    Mar 15, 2021 at 9:03
  • $\begingroup$ im trying to figure out, which day out of the 6 days looked at each, of the genes are more expressed on. I think the day with the highest expression of that genes with a p=0.05 in comparison to the other days. this is kind of what I want :GeneID S1->S2 S1->S3 S2->S1 S2->S3 S3->S1 S3->S2 1 at1g01040.2 1.6713881 2.0806706 0.5983051 1.2448758 0.4806143 0.8032930 2 at1g01050.1 1.0273222 1.2709185 0.9734045 1.2371177 0.7868325 0.8083305 $\endgroup$
    – Chloe Reid
    Mar 15, 2021 at 9:28
  • $\begingroup$ so that I can see in comparison to each day the pvalue of expression on that specific day to see which days that gene is most significantly expressed $\endgroup$
    – Chloe Reid
    Mar 15, 2021 at 9:32
  • $\begingroup$ I still have problems to understand your data. What is your level of data aggregation? A gene is measured 24 times on day and I have 6 days would result in 144 data points for one gene. Im not quite sure what we have here. Because the aggregation level also determines the sort of the method not only the respective outcome. ATM it looks like you want results as in an ANOVA but I have the feeling that we have cross sectional data with time points: mixture of these: differencebetween.com/… due to the repeated measures $\endgroup$
    – Patrick Bormann
    Mar 15, 2021 at 11:18
  • $\begingroup$ So a gene is measured 23 times. There are 6 days each with a number of repeats this is the 23 times. So for each gene there is 23 data points $\endgroup$
    – Chloe Reid
    Mar 15, 2021 at 11:52

1 Answer 1

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I believe you need a Mixed Model Analysis as you want to have ANOVA like results but on time-dependent observations,

Some advantages of the MM:

  • in an ANOVA you probably had to average the observations of your daily genes. not in an MM.

  • You dont need a balanced ata set in terms of observations (e.g. the test which is a small brother of ANOVA needs fairly distributed groups)

  • Missing values in a MM are allowed

try this link : https://m-clark.github.io/mixed-models-with-R/introduction.html

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