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I have used a python script to identify target sequences in a DNA sequence file.

There are two classes of sequence: coding and non-coding. I have identified $728$ sequences of interest. $597$ of these fall into the coding regions and $131$ of these fall into the non-coding regions. This is the equivalent of $18%,$ non-coding, but the total non-coding region in the sequence file is $13% $.

Is there a statistical tool to demonstrate the python script identified target sequences in a non-random fashion way?

If the script identified sequences that were randomly distributed then $13% $ of them would have been found in the non-coding region, from a total of $728$ sequences. This seems like it should be reliable.

I hope my question is clear.

Your null hypothesis is $H_0: p = 0.13$ against the alternative
$H_a: p ne 0.13,$ where $p = P(text{Non Coding}).$
You observe $X =131$ non-coding sequences among $n = 728$ observed,
which gives you $hat p = 0.1812$ as the observed frequency.
Because the observed frequency is substantially different from $p = 0.13$
you wonder whether this might have been an 'unlucky' draw, or whether
you have statistically significant evidence that the method of sampling is unfair.

This is called a "one-sample binomial test". Often this test is done by
using a normal approximation to the binomial distribution. You can find
that method in elementary statistics textbooks. The output below from
Minitab statistical software uses the binomial distribution to give an
exact P-value. [It seems that that SciPy also implements a version of this test, but I have not tried it.]

If the P-value is less than 5%, one says that the null
hypothesis is rejected at the 5% level of significance. Here the P-value
is printed as 0.000 which means that the P-value is smaller than 0.0005.
So it is extremely unlikely that an unbiased draw would give an observed
proportion of non-coding sequences so far from $p = 0.13.$

Test and CI for One Proportion 



Test of p = 0.13 vs p ≠ 0.13



                                                    Exact

Sample    X    N  Sample p         95% CI         P-Value

1       131  723  0.181189  (0.153769, 0.211239)    0.000

Another way to interpret the output is that a 95% confidence interval
for $p$ is $(0.154, 0.211),$ which is centered at $hat p = 0.1812,$ but
does not contain $p = 0.13.$ Thus it is difficult to believe that
the sampling procedure would have given close to the true value $p = 0.13.$

Note: Yet another approach is to note that quantiles .025 and .975 of
the 'null distribution' $mathsf{Binom}(n = 723, p = 0.13)$ are 77 and 112, respectively. Thus the observed value $X = 131$ falls considerably
above the upper 'critical value' of the null distribution for a two-sided test at the 5% level. (Computation in R.)

 qbinom(c(.025,.975), 723, .13)

 [1]  77 112