Exercise 2 solution

import random
from collections import Counter

def test_cross_counts(rng, n):
    # AaBb x aabb : offspring show the gamete from the first parent
    c = Counter()
    for _ in range(n):
        c[rng.choice("Aa") + rng.choice("Bb")] += 1
    return c

rng = random.Random(13)
n = 40000
c = test_cross_counts(rng, n)
for k in ("AB", "Ab", "aB", "ab"):
    print(k, c[k], round(c[k] / n, 3))
print(max(c.values()) - min(c.values()))

Output:

AB 10020 0.251
Ab 9999 0.25
aB 9953 0.249
ab 10028 0.251
75

Reading it: the other parent only ever gives "ab", so each offspring shows exactly what the first parent's gamete carried. All four classes come out close to one quarter, the signature of independent assortment. The gap between the largest and smallest class is 75 out of 40,000, which is chance noise.