The shape of the lesson
By the end of the lesson, learners can
- Two equations, one method: Start with Recessive phenotype (q²) at 0.16.
- When the principle applies: The principle describes a population that is not evolving.
- q2 = frequency of the recessive phenotype.
- q = √q2, then p = 1 − q.
- Carriers = 2pq; homozygous dominant = p2.
- Valid only if large, random mating, no selection, mutation or migration.
One picture every idea returns to
For a gene with a dominant allele A (frequency p) and a recessive allele a (frequency q): p + q = 1 (allele frequencies) and p2 + 2pq + q2 = 1 (genotype frequencies), where p2 = AA, 2pq = Aa and q2 = aa. Only the recessive phenotype reveals its genotype, so the usual method is: q2 = frequency of the recessive phenotype, q = √q2, p = 1 − q, then calculate 2pq (carriers) and p2. The principle predicts that allele frequencies stay constant from generation to generation only if the population is large, mating is random, there is no selection, no mutation and no migration (gene flow) in or out, and the organism is diploid and reproduces sexually. If observed frequencies differ from the prediction, one of these conditions is not being met.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Counting alleles in a population.. You cannot see which people carry a recessive allele, but if you know how many show the recessive phenotype you can calculate it. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Two equations, one method | Start with Recessive phenotype (q²) at 0.16. Project tab 5 of the lab (Selection and evolution lab, the Hardy-Weinberg principle) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: When the principle applies | The principle describes a population that is not evolving. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 19 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in When the principle applies, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Do not take the square root of the dominant phenotype frequency to find p. | People with the dominant phenotype are AA and Aa, so their frequency is p 2 + 2pq, not p 2. Always start from the recessive phenotype. |
| The Hardy-Weinberg principle does not say that the dominant allele becomes more common. | Without selection, a dominant allele stays at the same frequency, however rare it is. |
The Hardy-Weinberg principle: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
Cystic fibrosis is caused by a recessive allele. In one population, 1 in 2500 babies is born with cystic fibrosis. Use the Hardy-Weinberg equations to calculate the frequency of heterozygous carriers. Show your working.
In a population of 1500 moths, 540 have the recessive light wing colour. Calculate the number of moths expected to be homozygous dominant.
State four conditions that must be met for the Hardy-Weinberg principle to apply to a population.
A sample of 500 plants contains 245 with genotype RR, 210 with Rr and 45 with rr. Calculate the frequency of allele r and state whether the genotype frequencies fit the Hardy-Weinberg prediction.
Explain why, when a recessive allele is rare, most copies of that allele are found in heterozygous individuals.
Over 20 generations, the frequency of a recessive allele in a population of mice fell from 0.30 to 0.12. Suggest three reasons why the Hardy-Weinberg principle does not apply to this population.
Answer key and marking notes
- q2 = 1 ÷ 2500 = 0.0004, so q = √0.0004 = 0.02.
- p = 1 − 0.02 = 0.98.
- 2pq = 2 × 0.98 × 0.02 = 0.0392 (about 1 in 25 people).
- q2 = 540 ÷ 1500 = 0.36, so q = 0.6.
- p = 1 − 0.6 = 0.4, so p2 = 0.16.
- Number homozygous dominant = 0.16 × 1500 = 240.
- The population is very large (so there is no genetic drift).
- Mating is random.
- There is no selection (all genotypes are equally likely to survive and reproduce).
- There is no mutation and no migration into or out of the population (also: diploid organisms reproducing sexually).
- Number of r alleles = 210 + (2 × 45) = 300 out of 1000, so q = 0.3 (and p = 0.7).
- Expected: RR = 0.49 × 500 = 245; Rr = 0.42 × 500 = 210; rr = 0.09 × 500 = 45.
- Observed numbers match the expected numbers exactly, so the population fits the Hardy-Weinberg prediction.
- If q is small, q2 is very much smaller than q (for example q = 0.01 gives q2 = 0.0001).
- 2pq is approximately 2q when p is close to 1, so heterozygotes (0.0198) far outnumber homozygous recessives (0.0001).
- So the recessive allele is mostly hidden in carriers, where it is not exposed to selection.
- Selection against the recessive phenotype, for example aa mice are more visible to predators.
- Genetic drift, if the population is small.
- Migration of mice with a higher frequency of the dominant allele into the population, or non-random mating / mutation.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
In the Hardy-Weinberg equation p2 + 2pq + q2 = 1, what does p2 represent?
- Athe frequency of the homozygous dominant genotype
- Bthe frequency of the dominant allele
- Cthe frequency of the dominant phenotype
- Dthe frequency of heterozygous individuals
In a population, 4% of individuals show a recessive phenotype. What is the frequency of heterozygous carriers?
- A0.32
- B0.16
- C0.64
- D0.04
A sample of 1000 snails contains 360 AA, 480 Aa and 160 aa. What is the frequency of allele a?
- A0.40
- B0.16
- C0.60
- D0.24
Which is a condition that must be met for the Hardy-Weinberg principle to apply?
- Athere is no migration of individuals into or out of the population
- Bthe population is small, so every individual can find a mate
- Cindividuals choose mates with a phenotype similar to their own
- Done allele gives a selective advantage over the other
In a population, 91% of individuals show the dominant phenotype. What percentage are homozygous dominant?
- A49%
- B95%
- C42%
- D70%
A recessive condition affects 1 in 400 people. How many heterozygous carriers would be expected in a town of 5000 people?
- A475
- B13
- C250
- D238
The table shows observed genotype numbers in a plant population. Compare them with the Hardy-Weinberg expected numbers. What do the data suggest?
- Athere are fewer heterozygotes than expected (400 v 500), suggesting non-random mating such as self-pollination
- Bthe numbers fit the prediction exactly, because p and q are both 0.5
- Cthere are more heterozygotes than expected, suggesting heterozygotes have an advantage
- Dthe prediction cannot be made, because heterozygotes cannot be identified
A recessive allele causes a disease that kills homozygous children before they reproduce, yet its frequency stays at about 0.02 for many generations. Which is the best explanation?
- Aalmost all copies of the allele are in heterozygous carriers, who are not selected against
- Bthe Hardy-Weinberg principle shows that allele frequencies can never change
- Crecessive alleles are favoured by natural selection because they are hidden
- Dthe dominant allele mutates into the recessive allele in most individuals each generation
Quiz key
p2 is the frequency of individuals with two dominant alleles (AA). The dominant allele frequency is p itself. The dominant phenotype includes AA and Aa, so its frequency is p2 + 2pq. Heterozygotes are 2pq.
q2 = 0.04, so q = 0.2 and p = 0.8. 2pq = 2 × 0.8 × 0.2 = 0.32. 0.16 is pq, forgetting the factor of 2. 0.64 is p2, the homozygous dominant frequency. 0.04 is q2, the affected individuals, not the carriers.
Number of a alleles = (2 × 160) + 480 = 800, out of 2 × 1000 = 2000 alleles, so q = 0.40. 0.16 is the frequency of aa individuals (q2), not of the allele. 0.60 is p, the frequency of A. 0.24 is 480 ÷ 2000, which counts only the a alleles carried by heterozygotes and leaves out the 320 in the aa snails.
Migration brings in or removes alleles (gene flow), so the principle requires none. The population must be large, not small, to avoid genetic drift. Choosing mates by phenotype is non-random (assortative) mating, which breaks the condition of random mating. A selective advantage would change allele frequencies, so there must be no selection.
Recessive phenotype = 9%, so q2 = 0.09, q = 0.3, p = 0.7 and p2 = 0.49, or 49%. 95% comes from √0.91, wrongly treating the dominant phenotype as p2. 42% is 2pq, the heterozygotes. 70% is p, an allele frequency, not a genotype frequency.
q2 = 0.0025, so q = 0.05 and p = 0.95. 2pq = 2 × 0.95 × 0.05 = 0.095; 0.095 × 5000 = 475. 13 (12.5) is q2 × 5000, the number of affected people. 250 is q × 5000, an allele frequency used as a genotype frequency. 238 (237.5) is pq × 5000, missing the factor of 2.
Number of A alleles = (2 × 300) + 400 = 1000 of 2000, so p = q = 0.5. Expected: AA 250, Aa 500, aa 250. Observed has 100 fewer heterozygotes and 50 more of each homozygote, the pattern produced by inbreeding such as self-pollination (non-random mating). p = q = 0.5 does not make the observed numbers fit; the expected values differ. There are fewer, not more, heterozygotes than expected. Genotypes are given in the table, so the comparison can be made.
With q = 0.02, q2 = 0.0004 but 2pq is about 0.039, so roughly 99% of copies of the allele are in healthy carriers, where selection cannot act on them; only the rare homozygotes are removed, so the frequency falls extremely slowly (and new mutations may replace some losses). The Hardy-Weinberg principle predicts constant frequencies only when there is no selection, which is not true here. Being hidden protects the allele from selection but does not favour it. Mutation rates are far too low for most individuals to gain a new mutation each generation.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.