Exercise 2 solution

from itertools import combinations_with_replacement

def blood_group(genotype):
    alleles = set(genotype)
    if alleles == {"A", "B"}:
        return "AB"
    if "A" in alleles:
        return "A"
    if "B" in alleles:
        return "B"
    return "O"

def child_groups(p1, p2):
    return {blood_group(a + b) for a in p1 for b in p2}

genos = ["AA", "AO", "BB", "BO", "AB", "OO"]
pairs = list(combinations_with_replacement(genos, 2))
o_pairs = [p for p in pairs if "O" in child_groups(*p)]
print(len(pairs), len(o_pairs))
print([p for p in pairs if len(child_groups(*p)) == 4])
print([p for p in pairs if "AB" in child_groups(*p) and "O" in child_groups(*p)])

Output:

21 6
[('AO', 'BO')]
[('AO', 'BO')]

Reading it: there are 21 unordered pairs of parent genotypes, and only 6 of them can have a child of the first group O (each parent must carry an O allele). Just one pair, AO with BO, can have children in all four groups, so it is also the only pair that can have both an AB child and an O child.