Inferring mRNA from Protein .ipynb

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   "source": [
    "# Inferring mRNA from Protein\n",
    "\n",
    "Given a protein string, return the number of different RNA strings from which the protein could have been translated, answer in modulo 1,000,000"
   ]
  },
  {
   "cell_type": "code",
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   "source": [
    "# codon dictionary from previous problem\n",
    "codon_dict = {\n",
    "    'UUU': 'F',     'CUU': 'L',     'AUU': 'I',     'GUU': 'V',\n",
    "    'UUC': 'F',     'CUC': 'L',     'AUC': 'I',     'GUC': 'V',\n",
    "    'UUA': 'L',     'CUA': 'L',     'AUA': 'I',     'GUA': 'V',\n",
    "    'UUG': 'L',     'CUG': 'L',     'AUG': 'M',     'GUG': 'V',\n",
    "    'UCU': 'S',     'CCU': 'P',     'ACU': 'T',     'GCU': 'A',\n",
    "    'UCC': 'S',     'CCC': 'P',     'ACC': 'T',     'GCC': 'A',\n",
    "    'UCA': 'S',     'CCA': 'P',     'ACA': 'T',     'GCA': 'A',\n",
    "    'UCG': 'S',     'CCG': 'P',     'ACG': 'T',     'GCG': 'A',\n",
    "    'UAU': 'Y',     'CAU': 'H',     'AAU': 'N',     'GAU': 'D',\n",
    "    'UAC': 'Y',     'CAC': 'H',     'AAC': 'N',     'GAC': 'D',\n",
    "    'UAA': 'Stop',  'CAA': 'Q',     'AAA': 'K',     'GAA': 'E',\n",
    "    'UAG': 'Stop',  'CAG': 'Q',     'AAG': 'K',     'GAG': 'E',\n",
    "    'UGU': 'C',     'CGU': 'R',     'AGU': 'S',     'GGU': 'G',\n",
    "    'UGC': 'C',     'CGC': 'R',     'AGC': 'S',     'GGC': 'G',\n",
    "    'UGA': 'Stop',  'CGA': 'R',     'AGA': 'R',     'GGA': 'G',\n",
    "    'UGG': 'W',     'CGG': 'R',     'AGG': 'R',     'GGG': 'G'\n",
    "}"
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     "text": [
      "915648\n"
     ]
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   "source": [
    "def codon_freq():\n",
    "    tally = {}\n",
    "    for key, value in codon_dict.iteritems():\n",
    "        if not tally.has_key(value):\n",
    "            tally[value] = 0\n",
    "        tally[value] += 1\n",
    "    return tally\n",
    "\n",
    "\n",
    "def get_all_possible(file_path):\n",
    "    x = open(file_path).read().strip()\n",
    "    freq = codon_freq()\n",
    "    out = freq['Stop']\n",
    "    for i in x:\n",
    "        out *= freq[i]\n",
    "    return out % 1000000\n",
    "\n",
    "print get_all_possible(\"/home/scott/Dropbox/rosalind/rosalind_mrna.txt\")"
   ]
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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "------------------------------------\n",
    "\n",
    "\n",
    "\n",
    "## Explanation\n",
    "\n",
    "- Ignore Rosalind, it's 2016:\n",
    "    * Don't actually have to worry about storing large integers\n",
    "    * Easier to just modulo the final answer\n",
    "\n",
    "`codon_freq()` returns another dictionary from `codon_dict` with a total number of possible codons for each amino acid"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
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   "outputs": [
    {
     "data": {
      "text/plain": [
       "{'A': 4,\n",
       " 'C': 2,\n",
       " 'D': 2,\n",
       " 'E': 2,\n",
       " 'F': 2,\n",
       " 'G': 4,\n",
       " 'H': 2,\n",
       " 'I': 3,\n",
       " 'K': 2,\n",
       " 'L': 6,\n",
       " 'M': 1,\n",
       " 'N': 2,\n",
       " 'P': 4,\n",
       " 'Q': 2,\n",
       " 'R': 6,\n",
       " 'S': 6,\n",
       " 'Stop': 3,\n",
       " 'T': 4,\n",
       " 'V': 4,\n",
       " 'W': 1,\n",
       " 'Y': 2}"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "codon_freq()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "`get_all_possible()` loops through the amino acids, looks up the value in `codon_freq` and keeps a cumulative product.\n",
    "\n",
    "Have to add in a look-up for the `Stop` codon as we don't have that in our input from Rosalind."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def get_all_possible(x):\n",
    "    \n",
    "    freq = codon_freq()\n",
    "    \n",
    "    # as stop isn't in our input\n",
    "    out = freq['Stop'] \n",
    "    \n",
    "    # loop through aa sequence\n",
    "    for i in x:\n",
    "        out *= freq[i] # cumulative product\n",
    "        \n",
    "    return out % 1000000 # return total modulo 1M\n",
    "        "
   ]
  },
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   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
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   "outputs": [
    {
     "data": {
      "text/plain": [
       "12"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "test = \"MA\"\n",
    "get_all_possible(test)\n"
   ]
  }
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