# Methods for Degenerated Codons

## Problem

Sometimes statistics on nucleotide sequences are limited to X-fold degenerated codons. The following code provides some functions to solve this problem, by generating a subsequence.

## Solution

``````from Bio.Data.CodonTable import unambiguous_dna_by_id

def altcodons(codon, table):
"""List codons that code for the same aminonacid / are also stop.

@param codon
@table code table id
@return list of codons

"""
tab = unambiguous_dna_by_id[table]

if codon in tab.stop_codons:
return tab.stop_codons

try:
aa = tab.forward_table[codon]
except:
return []

return [k for (k, v) in tab.forward_table.iteritems()
if v == aa and k[0] == codon[0] and k[1] == codon[1]]

def degeneration(codon, table):
"""Determine how many codons code for the same amino acid / are also stop

@param codon the codon
@param table code table id
@param the number of codons also coding for the amino acid codon codes for

"""
return len(altcodons(codon, table))

def is_x_degenerated(x, codon, table):
"""Determine if codon is x-fold degenerated.

@param codon the codon
@param table code table id
@param true if x <= the degeneration of the codon

"""
return (x <= len(altcodons(codon, table)))

def degenerated_subseq(seq, x, table):
"""Get a subsequence consisting of the x-fold degenerated codons only."""
data = ""
for i in range(0, len(seq), 3):
codon = seq[i:i + 3].tostring()
if isXdegenerated(x, codon, table):
data += codon
return data

``````