Note that the equation for the probability of a dictionary (introduced in “Compute the Size of a Spectral Dictionary”¹),

$$\mathrm{Pr}(\textit{Dictionary}) = \displaystyle\sum_{\text{each peptide }\textit{Peptide}\text{ in }\textit{Dictionary}} \dfrac{1}{20^{|\textit{Peptide}|}},$$

is similar to an equation for the size of a dictionary,

$$|\textit{Dictionary}|= \displaystyle\sum_{\text{each peptide }\textit{Peptide}\text{ in }\textit{Dictionary}}1.$$

This similarity suggests that we can derive a recurrence for the probability of a dictionary using arguments similar to those used to find the size of a dictionary.

So define Pr(i, t) as the sum of probabilities of all peptides with mass i for which Score(Peptide, Spectrum'_i ) is equal to t. The set of peptides contributing to Pr(i, t) can be split into 20 subsets depending on their final amino acid. Each peptide Peptide ending in a specific amino acid a results in a shorter peptide Peptide_a if we remove a; Peptide_a has mass i − |a| and score t − s_i. Since the probability of Peptide is 20 times smaller than the probability of Peptide_a, the contribution of Peptide to Pr(i, t) is 20 times smaller than contribution of Peptide_a to Pr(i − |a|, t − s_i ). Therefore, Pr(i, t) can be computed as

$$\mathrm{Pr}(i, t) = \displaystyle\sum_{\text{all amino acids }a} \frac{1}{20} \cdot \mathrm{Pr}(i - |a|, t - s_i),$$

which differs from the recurrence for computing Size(i, t) only in the presence of the factor 1/20. We can now compute the probability of a spectral dictionary as

$$\mathrm{Pr}(\textit{Dictionary}_{\textit{threshold}}({\textit{Spectrum'}})) = \displaystyle\sum_{t \geq \textit{threshold}} \mathrm{Pr}(m, t)$$.

Assignment

Given a file with a spectral vector Spectrum', an integer threshold, and an integer max_score.

Return the probability of the dictionary Dictionary_threshold(Spectrum').

Example

>>> spectral_dictionary_probability('data02.txt'2, 19, 39)
0.005125

>>> spectral_dictionary_probability('data03.txt'3, 19, 43)
0.000375

>>> spectral_dictionary_probability('data04.txt'4, 11, 32)
0.0012348468750000001