Probability and statistics in genetics

Genetic predictions are statements of probability. A cross between two heterozygotes does not guarantee that one offspring in four will be recessive; it predicts that, on average, one in four will be. To use genetics in practice, whether in a breeding program, a research laboratory or a genetic counselling session, we need to calculate probabilities for families and samples and to decide whether observed results fit a hypothesis. This lesson develops the product and sum rules, conditional probability, the binomial distribution and the chi-square goodness-of-fit test, the main statistical tools of transmission genetics.

The product and sum rules revisited

Lesson 1 introduced two rules. The product rule gives the probability that independent events all occur: multiply their probabilities. The sum rule gives the probability that any one of several mutually exclusive events occurs: add their probabilities. Most genetics problems combine the two. The trick is to identify which events are independent (different children, different genes that assort independently, egg and sperm) and which outcomes are mutually exclusive (the possible genotypes of a single child).

A frequent error is to forget that the order of births creates separate outcomes. If a question asks for one affected child among two, the two orders must both be counted. When many children are involved, counting orders by hand becomes tedious, and the binomial distribution does it for us.

The binomial distribution

When each of n independent trials has two outcomes, with probability p for one outcome and q = 1 − p for the other, the probability of exactly k occurrences of the first outcome is given by the binomial formula:

The coefficient \(\binom{n}{k}\), read “n choose k”, counts the number of different orders in which k successes can occur among n trials. In the previous example, n = 2, k = 1, p = 1/4, and \(\binom{2}{1} = 2\), giving \(2 \times \tfrac{1}{4} \times \tfrac{3}{4} = \tfrac{3}{8}\), as before.

The “at least one” calculation uses the complement: it is usually easier to calculate the probability that something never happens and subtract from 1 than to add up all the ways it could happen. The code below lists the whole distribution for a family of five.

from math import comb
from fractions import Fraction

p = Fraction(1, 4)          # probability a child is affected (Aa x Aa)
n = 5
print("affected  probability")
for k in range(n + 1):
    pk = comb(n, k) * p**k * (1 - p)**(n - k)
    print(f"{k:5d}     {str(pk):>10s} = {float(pk):.4f}")
print("P(at least one) =", float(1 - (1 - p)**n).__round__(4))

affected  probability
    0       243/1024 = 0.2373
    1       405/1024 = 0.3955
    2        135/512 = 0.2637
    3         45/512 = 0.0879
    4        15/1024 = 0.0146
    5         1/1024 = 0.0010
P(at least one) = 0.7627

Conditional probability

Often we know something that changes the odds. Consider a healthy brother of a child with cystic fibrosis. Both parents must be carriers. Before we know anything about the brother, his chance of being a carrier is 1/2. But we know he is unaffected, which rules out the aa genotype. Among the remaining possibilities, AA (probability 1/4) and Aa (probability 1/2), the carrier genotype is twice as likely. His conditional probability of being a carrier is therefore (1/2) ÷ (3/4) = 2/3, not 1/2.

This 2/3 figure appears constantly in genetic counselling. It is the starting point for calculating risks in later generations, and it illustrates a general principle: probabilities must always be calculated from the information actually available.

Testing hypotheses: observed versus expected

Real data never fit expected ratios exactly. Suppose a cross expected to give a 3 : 1 ratio yields 84 dominant and 36 recessive offspring out of 120. The expected numbers are 90 and 30. Is the difference just chance, or does the hypothesis need revising? Statistics cannot prove a hypothesis true, but it can tell us how often deviations this large would occur by chance if the hypothesis were correct.

The chi-square goodness-of-fit test compares observed (O) and expected (E) counts across all categories:

Each term measures how far a category is from expectation, scaled by the expected size so that a difference of 6 matters more in a small class than in a large one. The larger χ², the worse the fit. To interpret χ² we need the degrees of freedom (df), which for a goodness-of-fit test is the number of categories minus one when the expected ratio is specified in advance. We then find the P value: the probability of obtaining a χ² at least this large if the hypothesis is true.

By convention in biology, if P is greater than 0.05 we do not reject the hypothesis: the deviation is consistent with chance. If P is 0.05 or less, the deviation is considered statistically significant and we reject the hypothesis, recognizing that about 1 time in 20 we will reject a correct hypothesis by bad luck. The 0.05 threshold is a convention, not a law of nature, and modern practice favours reporting the P value itself.

Testing a dihybrid ratio

With four phenotype classes, as in a dihybrid cross, there are 3 degrees of freedom. The code below tests Mendel's own dihybrid data against 9 : 3 : 3 : 1 and then tests an invented data set against the same ratio. In the second case the fit is poor, and the pattern of deviation (too many parental types, too few recombinant types) is a clue that the two genes may be linked, a topic for lesson 8.

from scipy.stats import chisquare

def test(observed, ratio, label):
    n = sum(observed)
    expected = [n * r / sum(ratio) for r in ratio]
    stat, p = chisquare(observed, expected)
    print(label)
    print("  observed:", observed)
    print("  expected:", [round(e, 1) for e in expected])
    print(f"  chi-square = {stat:.2f}, df = {len(observed) - 1}, P = {p:.4f}")

test([315, 108, 101, 32], [9, 3, 3, 1], "Mendel's round/yellow dihybrid F2")
test([620, 112, 105, 163], [9, 3, 3, 1], "Invented F2 for two other genes")

Mendel's round/yellow dihybrid F2
  observed: [315, 108, 101, 32]
  expected: [312.8, 104.2, 104.2, 34.8]
  chi-square = 0.47, df = 3, P = 0.9254
Invented F2 for two other genes
  observed: [620, 112, 105, 163]
  expected: [562.5, 187.5, 187.5, 62.5]
  chi-square = 234.18, df = 3, P = 0.0000

Mendel's data fit extraordinarily well, with a χ² of less than 0.5. In 1936 the statistician Ronald Fisher pointed out that, across all of Mendel's experiments, the fit was better than chance alone would usually produce, and suggested the data might have been adjusted, perhaps unconsciously or by an assistant. Later analyses have argued about this conclusion for decades without settling it. Whatever happened, the ratios Mendel reported have been confirmed many times by others.

Using chi-square correctly

The chi-square test has rules that are easy to forget. It must be applied to counts, never to percentages or proportions, because the size of the sample matters: 30 : 10 and 300 : 100 have the same proportions but very different reliability. Expected counts should generally be at least 5 in every category; with smaller samples, an exact test such as the binomial test is better. And a non-significant result does not prove the hypothesis is right; it means only that the data do not give strong evidence against it. Small samples often fail to reject quite wrong hypotheses.

Sample size and power

How many offspring should a breeder count to be confident of distinguishing a 3 : 1 ratio from, say, a 1 : 1 ratio? The ability of a test to detect a real difference is called its power. Power rises with sample size. The simulation below repeatedly samples families from a true 1 : 1 ratio and tests them against 3 : 1, recording how often the false 3 : 1 hypothesis is correctly rejected.

import numpy as np
from scipy.stats import chisquare
rng = np.random.default_rng(201)

def power(n, true_p=0.5, null_p=0.75, reps=2000):
    rejected = 0
    for _ in range(reps):
        dom = rng.binomial(n, true_p)
        stat, p = chisquare([dom, n - dom], [n * null_p, n * (1 - null_p)])
        rejected += p <= 0.05
    return rejected / reps

for n in (8, 16, 32, 64, 128):
    print(f"n = {n:3d}: false 3:1 hypothesis rejected in {power(n):.1%} of samples")

n =   8: false 3:1 hypothesis rejected in 36.8% of samples
n =  16: false 3:1 hypothesis rejected in 59.4% of samples
n =  32: false 3:1 hypothesis rejected in 90.0% of samples
n =  64: false 3:1 hypothesis rejected in 99.1% of samples
n = 128: false 3:1 hypothesis rejected in 100.0% of samples

With 8 offspring the wrong hypothesis is rejected less than half the time; with 64 it is rejected almost always. This is a practical reason why geneticists prefer organisms with large broods, and why conclusions from small human families must be pooled across many families before they are trustworthy.