The BH p-adjustment algorithm is as follows:

BH = {
        i <- lp:1L
        o <- order(p, decreasing = TRUE)
        ro <- order(o)
        pmin(1, cummin(n/i * p[o]))[ro]
}

n/i is the total number of tests divided by the index of which test we are looking at. For the smallest p-value, this will adjust using Bonferroni (n * p), and for each successive p-value, the penalty is reduced, so for the second smallest p-value, it's half the Bonferroni cutoff, then a third, then a quarter, etc.

If you have four contrasts, one of which contains a p-value that will be Bonferroni significant if you combine all the p-values, and the other three containing p-values that get really close to the Bonferroni cutoff (for that contrast), then when you combine all the p-values, all those 'close' p-values will get relaxed penalties and will get under the cutoff. As an example, consider this.

## for reproducibility
> set.seed(0xabeef)
## we expect no significant p-values from this
> z <- runif(1000)
> head(sort(p.adjust(z, "BH")))
[1] 0.2806372 0.2806372 0.2806372
[4] 0.2806372 0.2806372 0.3232420

## so no significant p-values. Let's add one that is right at Bonferonni

> z <- c(z, 0.05/1001)
> head(sort(p.adjust(z, "BH")))
[1] 0.0500000 0.2340982 0.2340982
[4] 0.2340982 0.2340982 0.2340982

## now we have one significant gene, and the FDR for some of the others is smaller
## let's add more genes, none of which meet Bonferroni

> z <- c(z[1:1000], 0.05/seq(1000, 990, -1))
## none are significant at Bonferroni cutoff
> head(sort(p.adjust(z, "bonferroni")))
[1] 0.05055000 0.05060060 0.05065130
[4] 0.05070211 0.05075301 0.05080402

## but now all ten p-values we added are significant using BH, and the ones that were 
## originally at a BH of 0.28 are now at 0.082!
> head(sort(p.adjust(z, "BH")), 16)
 [1] 0.004641873 0.004641873
 [3] 0.004641873 0.004641873
 [5] 0.004641873 0.004641873
 [7] 0.004641873 0.004641873
 [9] 0.004641873 0.004641873
[11] 0.004641873 0.053937065
[13] 0.081557728 0.081557728
[15] 0.081557728 0.088663830

And to reiterate. There aren't multiple copies of the same gene. There are multiple p-values that were generated using data from the same gene, using different sets of flies. Those are not the same thing, and there is no need to adjust, other than for the fact that you have made multiple comparisons. It's no different from a situation where one might have three groups. One control, and two treated with two different compounds. Would you expect that you have to make some adjustment for the fact that the comparison of compound A and control includes all the same genes that are compared in compound B and control? If so, why?