Hardy–Weinberg Equilibrium on the MCAT

Hardy–Weinberg describes what allele and genotype frequencies look like in a population that is not evolving. Two equations carry it: p + q = 1 for allele frequencies, and p² + 2pq + q² = 1 for genotype frequencies.

In those equations, p is the frequency of one allele and q the frequency of the other. Then p² is the fraction of the population homozygous for the first allele, q² the fraction homozygous for the second, and 2pq the fraction of heterozygotes.

The MCAT's favourite version of this is a recessive disease. You are told the incidence of affected individuals, which is q². Take the square root to get q, subtract from one to get p, and compute 2pq for the carrier frequency. Most questions on this topic are that calculation with different numbers attached.

A population is in Hardy–Weinberg equilibrium only if five conditions hold, and questions frequently ask which one has been violated.

No mutation, so allele frequencies are not being altered at the source. No natural selection, so no genotype confers a survival or reproductive advantage. No gene flow — no migration into or out of the population. Random mating, with no preference by genotype. And a population large enough that genetic drift is negligible, usually stated as infinitely large.

The point of the model is that when all five hold, allele frequencies do not change from generation to generation. That is why Hardy–Weinberg is described as a null hypothesis for evolution: deviation from the predicted genotype frequencies is evidence that something is acting on the population.

A passage will often describe a scenario — a founder event, assortative mating, heterozygote advantage in a malarial region — and ask which assumption fails. Founder effects and bottlenecks are drift. Heterozygote advantage is selection. Mate choice by phenotype is non-random mating.

Suppose a recessive condition affects 1 in 10,000 individuals in a population at equilibrium. What fraction are carriers?

Affected individuals are homozygous recessive, so q² = 1/10,000 = 0.0001. Taking the square root, q = 0.01. Then p = 1 − q = 0.99.

Carriers are heterozygotes, so their frequency is 2pq = 2 × 0.99 × 0.01 = 0.0198, or roughly 2 percent — about 1 in 50.

Notice the scale of the result. A disease affecting one in ten thousand people has carriers at one in fifty, two hundred times more common. That gap is not an accident of the numbers; it is a general feature of rare recessive alleles, because when q is small, q² is very small while 2pq is approximately 2q. Recognising that approximation lets you estimate carrier frequency in your head, which is worth having when the arithmetic is under time pressure.

For an X-linked recessive trait, males have only one X chromosome, so a single copy of the allele produces the phenotype. The frequency of affected males is therefore q itself, not q².

Females need two copies, so the frequency of affected females is q², and the frequency of female carriers is 2pq — the usual autosomal expressions.

This produces the familiar pattern that X-linked recessive conditions are far more common in males, and the ratio is dramatic for rare alleles. If q = 0.01, one in a hundred males is affected while only one in ten thousand females is.

The MCAT tests this by giving you the frequency of affected males and asking for something about females, or the reverse. The step students miss is the first one: for X-linked traits, read the male frequency directly as q rather than square-rooting it.

The most common error is confusing allele frequency with genotype frequency. If a question says the recessive allele has a frequency of 0.2, that is q, and the affected fraction is q² = 0.04. If it says 4 percent of the population is affected, that is q², and q is 0.2. Reading which quantity you were given is half the problem.

The second is answering "carrier frequency" with q. Carriers are heterozygotes, so the answer is 2pq. The allele frequency q counts copies of the allele across the whole gene pool, including both copies in affected individuals; it is not the fraction of people carrying one copy.

The third is forgetting that a dominant phenotype includes two genotypes. The frequency of individuals showing the dominant phenotype is p² + 2pq, not p². Questions phrase this as "what proportion display the trait," and the extra term is easy to drop.

Finally, if a passage reports observed genotype frequencies that do not match the predicted ones, do not assume an arithmetic error. That mismatch is usually the point — it is the evidence that one of the five assumptions is being violated, and the question is asking which.

Some questions give you genotype counts directly and ask for allele frequencies. This does not require the equations at all, only counting.

Each homozygous individual contributes two copies of one allele; each heterozygote contributes one of each. So the frequency of an allele is the number of copies of that allele divided by twice the number of individuals.

For example, in a population of 100 with 36 homozygous dominant, 48 heterozygous, and 16 homozygous recessive individuals, the recessive allele appears 48 + 32 = 80 times out of 200 total copies, so q = 0.4 and p = 0.6. Checking against the equations, p² = 0.36 and q² = 0.16 and 2pq = 0.48 — exactly the observed counts, so this population is in equilibrium.

That check is worth knowing as a technique in its own right. Comparing observed genotype counts against the frequencies predicted from the allele frequencies is how you determine whether a population is at equilibrium, which is a question the MCAT does ask directly.

Verbloom's MCAT biology practice includes genetics questions in passage format with distractor explanations, which matters here because the wrong answers are usually one identifiable substitution — q for 2pq, or p² for p² + 2pq.

Seeing which substitution you keep making is more useful than repeating the calculation correctly a dozen times.