Hardy-Weinberg Equilibrium
Genotype frequencies in a non-evolving population.
Formula
Variables
Example
q=0.1 ⇒ 2pq=0.18
How it works
The Hardy-Weinberg equation predicts how often each genotype should appear in a population if the allele frequencies are p and q and the population is not being changed by selection, mutation, migration, genetic drift, or non-random mating. The terms come from simple probability: p^2 is the chance of getting two dominant alleles, q^2 is the chance of getting two recessive alleles, and 2pq is the chance of getting one of each. Because there are only two alleles for a gene, their frequencies must add to 1, so all genotype possibilities together also add to 1.
When to use it
Use this formula in population genetics problems when you know an allele frequency and want the expected genotype frequencies, or when you want to test whether a population is roughly in genetic equilibrium. It is also useful in biology exams for estimating carrier frequencies and comparing observed vs. expected genotype counts.
Worked example, step by step
Suppose the recessive allele frequency is q = 0.3. Step 1: Find the dominant allele frequency: p = 1 - q = 1 - 0.3 = 0.7 Step 2: Calculate homozygous dominant frequency: p^2 = (0.7)^2 = 0.49 Step 3: Calculate heterozygous frequency: 2pq = 2(0.7)(0.3) = 0.42 Step 4: Calculate homozygous recessive frequency: q^2 = (0.3)^2 = 0.09 Step 5: Check the total: 0.49 + 0.42 + 0.09 = 1.00 Final answer: expected genotype frequencies are 49% dominant homozygotes, 42% heterozygotes, and 9% recessive homozygotes.
Common mistakes to avoid
- Using p and q as genotype frequencies instead of allele frequencies
- Forgetting that p + q must equal 1 before calculating genotype frequencies
- Leaving out the factor of 2 in the heterozygote term 2pq
Did You Know?
The Hardy-Weinberg principle gives the baseline of a non-evolving population, so deviations reveal evolution at work.
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