hi Zhu,

You don't need to split into separate datasets.

So, the easiest way to test, e.g. B/A = D/C is using interaction terms. First you need to code some new variables. Here I create a little toy dataset which doesn't have the 3 replicates per genotype x condition, but you'll see how to do that same to your colData. Create two variables which each divide A,B,C,D into two groups. Maybe you can give some more meaningful names here than geno2 and geno3 (for example, some description of whatever A and C have in common, and the same for whatever A and B have in common).

> d <- expand.grid(geno=letters[1:4], time=c("1","2"))
> d
  geno time
1    a    1
2    b    1
3    c    1
4    d    1
5    a    2
6    b    2
7    c    2
8    d    2
> d$geno2 <- ifelse(d$geno %in% c("a","c"), "1", "2")
> d
  geno time geno2
1    a    1     1
2    b    1     2
3    c    1     1
4    d    1     2
5    a    2     1
6    b    2     2
7    c    2     1
8    d    2     2
> d$geno3 <- ifelse(d$geno %in% c("a","b"), "1", "2")
> d
  geno time geno2 geno3
1    a    1     1     1
2    b    1     2     1
3    c    1     1     2
4    d    1     2     2
5    a    2     1     1
6    b    2     2     1
7    c    2     1     2
8    d    2     2     2

Now if you didn't have two time points, you would just use a design ~geno2 + geno3 + geno2:geno3 and run DESeq() and then test the interaction term using something like results(dds, name="geno22.geno32") ... you'll have to look at resultsNames(dds) to see what the name of the interaction term is for your dataset. The interaction term will have a low p-value if there is a small probability of seeing such an large difference in observed fold changes under the null hypothesis: B/A = D/C.

However, you can simply add time to the design in the following way to create more model terms to account for the two time points, including two interaction terms:

~ time + time:geno2 + time:geno3 + time:geno2:geno3

Now you can test name="time1.geno22.geno32" for B/A = D/C at time 1, and likewise for time 2. See the names in resultsNames(dds).

Note: I'm assuming that all of these terms are factors. You can make sure with: class(dds$time), etc.

1) yes the interaction tests if the ratio is equal. note that testing

B/A = D/C 

is identical to testing

B/D = A/C

Non-significance does not imply equality, for example, you could have an underpowered experiment. This is a general rule in frequentist statistics: p-value > alpha does not necessarily mean you have evidence that the null is true.

2) There is a single interaction term for the test of the ratios being equal. You can control which levels are in the denominator of the fold change by setting the reference level of the factors using relevel() (see vignette).

3) My code above for a toy dataset does not have replicates, but these would just be repeated rows of exactly what I have there. You would not define new variables.