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I have a gene say "X" of 3000 base pairs. 300 of those 3000 bp belong to ring domain. Rest 2700 do not belong to ring domain. In a particular tumor, the ring region has 4 alterations and the rest of the region has 4 alterations too. Clearly the frequency in ring region is higher, but is there a statistical test that will give me a p value that tells the probability of having a mutation in ring region is higher or not?

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  • $\begingroup$ Does yours ring domain mean Really Interesting New Gene finger domain or it refer to some kind of loop? Maybe you should strip the question of domain specific knowledge and edit it? Moreover, you should clearly state what model of mutation do you assume, something like Junkes & Cantor's, Kimura's or something else? $\endgroup$ – Adam Przedniczek Apr 5 '16 at 21:19
  • $\begingroup$ In Junkes & Cantor's each nucleotide could alternate to any other with the same probability. In Kimura's model the transitions (purine to purine or pyrimidine to pyrimidine) are more likeli than transversions (between purines and pyrimidines). Or maybe you should even use something more general. Have a look on those models and specify the question further en.wikipedia.org/wiki/Models_of_DNA_evolution $\endgroup$ – Adam Przedniczek Apr 5 '16 at 21:28
  • $\begingroup$ This question about RING is because in much more elaborated model we could make use of the very specific structure of RING finger domain: $Cys_3 His Cys_4$. Because we know their codons we probably could make some assuptions and put some bonds on nucleotide percentage content. $\endgroup$ – Adam Przedniczek Apr 5 '16 at 22:13
  • $\begingroup$ When i say ring domain i mean Really Interesting New Gene. I want to know if the occurrence of mutation in this region of the gene is significantly higher than the mutations occurring in the non RING part of the gene. $\endgroup$ – priboston Apr 6 '16 at 15:37
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As I wrote in comments, I don't know how complicated your question is. Maybe I'm exaggerating a bit with those models or you don't know anything else about the situation.

If that is question should be very simple (homework or something like that) you could have a look on simplest Welch's T test. With this test you could check wheter two non-overlapping populations have the same means.

Let's say, you have two populations $\mathfrak{F}$ (RING - foreground) and $\mathfrak{B}$ (the rest - background) represented by two arrays of lengths $L_F = 300$ and $L_B = 2700$, respectively. In the picture below, with blue balls are marked positions of bases which were altered.

enter image description here

For array $\mathfrak{F}$ we have independent binary random variables $F_1, \dots, F_{L_F}$, where $F_i = 1$ denotes at this position $i$ an alternation took place. I assume that any position in $\mathfrak{F}$ the probablity of alternation is the same(, but unknown) $p_f$ and each variable $F_i$ denotes single Bernoulli trial, so $F_i \sim Bernoulli(p_f)$. For array $\mathfrak{B}$ we have the same situation, but corresponding variables $B_1, \dots, B_{L_B}$ are distributed with maybe different probablity $p_b$, so $B_i \sim Bernoulli(p_b)$.

According to the Central Limit Theorem (we have more than 30 random variables with each array, so CLT may be used) the mean of $F_i$s and mean of $B_i$s have Normal distribution: $$\text{(mean of $F_{i}$)} ~~~~~~ \mu_F =\frac{\sum\limits_{i} F_i}{L_F} \sim Normal(p_f, \frac{p_f \cdot (1 - p_f)}{L_F} )$$

$$\text{(mean of $B_{i}$)} ~~~~~~ \mu_B = \frac{\sum\limits_{i} B_i}{L_B} \sim Normal(p_b, \frac{p_b \cdot (1 - p_b)}{L_B})$$

Because you don't know anything about variations, thus use Welch's T test https://en.wikipedia.org/wiki/Welch's_t_test (rather than Student's T test).

In this test you may compare the corresponding means with hypotheses: $$H_0 : \mu_F = \mu_B ~~ \text{against} ~~ H_A: \mu_F > \mu_B$$

In R you use method t.test{stats}:

t.test(arr_f, arr_b, alternative = "greater", paired = FALSE, var.equal = FALSE)

arr_f and arr_b are simply an 0-1 arrays where 1 points to alternated bases and 0 elsewhere.

The two trailing parameters are by default false, but I put them to make you aware of them. The last var.equal = FALSE indicates the Welch's T test instead of Student's T test with TRUE.

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  • $\begingroup$ That should explain how Welch's T test could work in your case. Remember only the assumption that the probabilities of alternation $p_f$ and $p_b$ are the same all over arrays. This simplest approach above would work, but maybe in your model you should give different probabilities for transversions and transitions. $\endgroup$ – Adam Przedniczek Apr 6 '16 at 16:46

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