Exercise 2 solution

B = "TCAG"
AA = "FFLLSSSSYY**CC*WLLLLPPPPHHQQRRRRIIIMTTTTNNKKSSRRVVVVAAAADDEEGGGG"
CODON = {a + b + c: AA[16 * i + 4 * j + k]
         for i, a in enumerate(B) for j, b in enumerate(B) for k, c in enumerate(B)}

def translate(dna):
    protein = []
    for i in range(0, len(dna) - 2, 3):
        aa = CODON[dna[i:i + 3]]
        if aa == "*":
            break
        protein.append(aa)
    return "".join(protein)

def classify(old, new):
    if len(old) != len(new):
        return "frameshift" if (len(new) - len(old)) % 3 else "in-frame indel"
    a, b = translate(old), translate(new)
    if a == b:
        return "silent"
    return "nonsense" if len(b) < len(a) else "missense"

gene = "ATGGCCTTGAAATGGTAA"
counts = {"silent": 0, "missense": 0, "nonsense": 0}
for pos in range(15):
    for n in "ACGT":
        if n != gene[pos]:
            counts[classify(gene, gene[:pos] + n + gene[pos + 1:])] += 1
print(counts, sum(counts.values()))

Output:

{'silent': 6, 'missense': 35, 'nonsense': 4} 45

Reading it: Every substitution is classified by translating both versions. The classifier counts changes; it does not judge how serious each one is.