Exercise 1 solution

Code:

from itertools import product
from collections import Counter

def cross(p1, p2):
    kids = Counter()
    for a, b in product(p1, p2):
        kids["".join(sorted(a + b))] += 1
    return kids

def dom_count(counts, allele="A"):
    return sum(n for g, n in counts.items() if allele in g)

for p1, p2 in (("Aa", "aa"), ("AA", "aa"), ("AA", "Aa")):
    c = cross(p1, p2)
    d = dom_count(c)
    print(p1, "x", p2, dict(c), "dominant:", d, "recessive:", sum(c.values()) - d)

Output:

Aa x aa {'Aa': 2, 'aa': 2} dominant: 2 recessive: 2
AA x aa {'Aa': 4} dominant: 4 recessive: 0
AA x Aa {'AA': 2, 'Aa': 2} dominant: 4 recessive: 0

Reading it:
- Aa x aa gives half Aa and half aa, so a 1 : 1 ratio of dominant to recessive appearance. This is the classic "test cross" pattern.
- AA x aa gives only Aa, so every offspring shows the dominant trait.
- AA x Aa gives half AA and half Aa; all four squares contain A, so all show the dominant trait even though the genotypes differ.