4.10.7Additional lessons
Hardy-Weinberg Principle — IB Biology HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 90 of 101
Part 1 of 5Equations
The principle describes a population that is not evolving.
Allele equation
- p + q = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele.
Genotype equation
- p² + 2pq + q² = 1, where p² is homozygous dominant, 2pq is heterozygous and q² is homozygous recessive.
Equilibrium
- If the conditions are met, allele and genotype frequencies stay constant from one generation to the next.
Use
- The model gives a null hypothesis;
- a departure from the predicted values suggests that evolution is occurring.
| Condition | Reason it is needed |
|---|---|
| Very large population | Avoids chance changes from drift |
| Random mating | Genotypes combine in proportion to frequency |
| No mutation | No new alleles appear |
| No migration | No alleles enter or leave |
| No natural selection | All genotypes equally fit |
| Diploid sexual reproduction | Genotypes arise from random fusion of gametes |
Part 2 of 5Calculating From the Recessive Phenotype
1
Start with the frequency of the recessive phenotype, which equals q² because only aa individuals show it.
2
Take the square root of q² to find q, the recessive allele frequency.
3
Find p by subtraction, p = 1 − q, and then calculate 2pq for heterozygous carriers.
4
Multiply any frequency by the population size to estimate the number of individuals.
Part 3 of 5Limits of the Model
Limits of the Model
Idealised
- No real population meets every condition perfectly, so results are estimates.
Dominance
- The calculation assumes a simple two-allele gene with complete dominance.
Testing
- Observed genotype counts can be compared with expected counts using a chi-squared test.
Evolution
- Differences between generations indicate that at least one condition is not being met.
Part 4 of 5Worked Example
Question
- A recessive disorder affects 1 in 2 500 people (q² = 0.0004).
- Calculate q, p and the carrier frequency, and the number of carriers in a town of 10 000.
- q = √0.0004 = 0.02.
- p = 1 − 0.02 = 0.98.
- 2pq = 2 × 0.98 × 0.02 = 0.0392.
- Carriers = 0.0392 × 10 000 = 392.
Answer q = 0.02, p = 0.98, carrier frequency 0.0392, about 392 carriers.
Part 5 of 5Exam Checklist & Tips
4.10.7 – Key Points for the Exam
- Hardy-Weinberg equilibrium A state where allele and genotype frequencies do not change between generations.
- p Frequency of the dominant allele.
- q Frequency of the recessive allele.
- 2pq Frequency of heterozygotes.
- Null hypothesis A statement of no change used as a baseline for testing.
Common Mistakes
- Using q² as the frequency of the recessive allele, rather than of the recessive homozygotes.
- Forgetting to take the square root when moving from q² to q.
- Forgetting the factor 2 in 2pq.
- Assuming that dominant alleles must be the most common.
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