This free hardy weinberg worksheet builds confidence with the two equations of population genetics. Students identify which equilibrium condition is broken in four real-world scenarios, then complete a frequency table for four populations with plenty of space to show their work.
They then solve word problems about cystic fibrosis carriers, a mouse population and counting alleles directly, decide whether a lizard population is evolving, and explain what each part of the equations means.
- Grade 11 Biology
- 22 marks
- 40–50 minutes
- Worked answer key
What's in This Hardy Weinberg Worksheet
| Part | Skill practised | Marks |
|---|
| A | Which equilibrium condition is broken? | 2 |
| B | Frequency table for four populations | 8 |
| C | Three word problems | 6 |
| D | Is the population evolving? | 2 |
| E | Understanding the equations | 2 |
| F | Spot the mistake | 2 |
It supports the Ontario Grade 11 biology curriculum expectations in the Evolution unit and works as review for Grade 12 biology.
The Hardy-Weinberg Equations
For a gene with two alleles, p is the frequency of the dominant allele and q is the frequency of the recessive allele:
p + q = 1 p² + 2pq + q² = 1
The first equation describes alleles; the second describes genotypes: p² is the frequency of homozygous dominant individuals, 2pq of heterozygotes, and q² of homozygous recessive individuals. Every Hardy Weinberg worksheet comes back to these two lines. For the history and proof, see the Hardy-Weinberg principle on Wikipedia.
How to Solve a Problem
Example: 16% of a population shows the recessive phenotype.
1. Recessive phenotype = aa, so q² = 0.16
2. q = √0.16 = 0.4
3. p = 1 − 0.4 = 0.6
4. p² = 0.36 (AA) · 2pq = 2(0.6)(0.4) = 0.48 (Aa)
5. Check: 0.36 + 0.48 + 0.16 = 1 ✓
Always start with the recessive phenotype, because dominant individuals could be AA or Aa and you can't tell which by looking. Part B of the Hardy Weinberg worksheet repeats this for four populations:
| q² (aa) | q | p | p² (AA) | 2pq (Aa) |
|---|
| Population 1 | 0.16 | 0.4 | 0.6 | 0.36 | 0.48 |
| Population 2 | 0.09 | 0.3 | 0.7 | 0.49 | 0.42 |
| Population 3 | 0.25 | 0.5 | 0.5 | 0.25 | 0.5 |
| Population 4 | 0.04 | 0.2 | 0.8 | 0.64 | 0.32 |
The word problems on the Hardy Weinberg worksheet apply the same steps. For cystic fibrosis, q² = 1/2500, so q = 0.02 and about 3.9% of people (roughly 1 in 26) are carriers, even though only 0.04% are affected. That is why recessive disorders persist: most copies of the allele hide in healthy carriers. To review where these genotypes come from, see the Punnett square worksheet.
The Five Equilibrium Conditions
A population stays in equilibrium only if there is no mutation, mating is random, there is no natural selection, the population is very large, and there is no gene flow. Part A of the Hardy Weinberg worksheet gives a scenario for four of them:
| Scenario | Condition broken |
|---|
| Birds from a neighbouring valley fly in and breed with the population. | No gene flow (no migration) |
| Peahens choose to mate with the males that have the brightest tails. | Random mating |
| A storm kills all but 40 beetles, and by chance most survivors carry the same allele. | Very large population (genetic drift) |
| Moths whose colour matches the tree bark are eaten less by birds. | No natural selection |
In real populations these conditions are almost never all met, which is the point: Hardy-Weinberg gives a baseline, and any change in allele frequency over time is evidence of evolution. Natural selection, the most important of these forces, is the focus of the natural selection worksheet, and the evidence for evolution worksheet covers the bigger picture.
Hardy Weinberg Worksheet Practice Questions
These questions match the Hardy Weinberg worksheet PDF, which has the frequency table and boxes for showing your work.
Part A – Which Condition Is Broken?
- Birds from a neighbouring valley fly in and breed with the population.
- Peahens choose to mate with the males that have the brightest tails.
- A storm kills all but 40 beetles, and by chance most survivors carry the same allele.
- Moths whose colour matches the tree bark are eaten less by birds.
Part B – Calculate the Frequencies
- q² = 0.16: find the other values
- q² = 0.09: find the other values
- q² = 0.25: find the other values
- p = 0.8: find the other values
Part C – Word Problems
- Cystic fibrosis is a recessive disorder that affects about 1 in 2500 newborns in some populations. Find the frequency of carriers (heterozygotes). About 1 in how many people is a carrier? (2 marks)
- A population of 1000 mice has 840 brown mice (dominant, B) and 160 white mice (bb). Assuming Hardy–Weinberg equilibrium, how many mice are heterozygous, and how many are homozygous dominant? (2 marks)
- A population of 200 plants has 90 AA, 80 Aa and 30 aa. Calculate p and q by counting alleles. Then find how many aa plants you would expect if the population were in equilibrium. (2 marks)
Part D – Is the Population Evolving?
- Before a new predator arrived, the frequency of a recessive allele in a lizard population was 0.30. Ten generations later it is 0.42. Is the population in Hardy–Weinberg equilibrium? Which condition is most likely being broken? (2 marks)
Part E – Understanding the Equations
- Why must p + q always equal 1?
- What does each term in p² + 2pq + q² = 1 represent?
Part F – Spot the Mistake
- A student finds q² = 0.36 and writes q = 0.36 ÷ 2 = 0.18. Explain the error and correct it.
- A student writes: “A rare recessive allele will soon disappear from a population.” Explain the error and correct it.
Hardy Weinberg Worksheet Answer Key
Open each part of the Hardy Weinberg worksheet answers to check your work. The printable key is at the end of the PDF.
Part A answers (Q1–4)
- No gene flow (no migration)
- Random mating
- Very large population (genetic drift)
- No natural selection
Part B answers (Q5–8)
- q² = 0.16, q = 0.4, p = 0.6, p² = 0.36, 2pq = 0.48
- q² = 0.09, q = 0.3, p = 0.7, p² = 0.49, 2pq = 0.42
- q² = 0.25, q = 0.5, p = 0.5, p² = 0.25, 2pq = 0.5
- q² = 0.04, q = 0.2, p = 0.8, p² = 0.64, 2pq = 0.32
Part C answers (Q9–11)
- q² = 1/2500 = 0.0004, so q = 0.02 and p = 0.98. 2pq = 2(0.98)(0.02) = 0.0392 ≈ 3.9%, about 1 in 26 people.
- q² = 160/1000 = 0.16 → q = 0.4, p = 0.6. Heterozygous: 2pq × 1000 = 480; homozygous dominant: p² × 1000 = 360 (check: 480 + 360 = 840 brown).
- p = (2×90 + 80) ÷ 400 = 0.65; q = 0.35. Expected aa = q² × 200 = 0.1225 × 200 = 24.5. There are more aa plants (30) than expected, so the population may not be in equilibrium.
Part D answer (Q12)
- No: the allele frequency changed, so the population is evolving. The predator suggests natural selection is favouring the recessive phenotype.
Part E answers (Q13–14)
- There are only two alleles for this gene, so their frequencies must add up to 100% of the gene pool.
- p² = frequency of homozygous dominant, 2pq = heterozygous, q² = homozygous recessive individuals.
Part F answers (Q15–16)
- To find q from q², take the square root: q = √0.36 = 0.6.
- Without evolutionary forces, allele frequencies stay the same. Rare recessive alleles hide in heterozygotes (2pq), where selection can't act on them.
5 Common Mistakes on a Hardy Weinberg Worksheet
- Halving instead of square rooting. q = √q², not q² ÷ 2.
- Starting from the dominant phenotype. Dominant individuals include both AA and Aa.
- Forgetting the 2 in 2pq. There are two ways to be heterozygous (Aa and aA).
- Mixing up allele and genotype frequencies. p and q are alleles; p², 2pq and q² are genotypes.
- Assuming equilibrium means no change ever. It is a model; real populations usually break at least one condition.
How to Use This Hardy Weinberg Worksheet in Class
- Lesson 1: Parts A and B of the Hardy Weinberg worksheet, after a bead or card simulation of a gene pool.
- Lesson 2: Parts C–F, then rerun the simulation with selection or a "bottleneck" to watch frequencies change.
Review with the Hardy-Weinberg equilibrium quiz.
Frequently Asked Questions
Is this hardy weinberg worksheet free?
Yes. The worksheet and its answer key are free to download and print as a PDF, with no account needed.
What do p and q stand for in the Hardy-Weinberg equation?
p is the frequency of the dominant allele and q is the frequency of the recessive allele in a population's gene pool. Because there are only two alleles, p + q = 1.
Why do you start with q² in Hardy-Weinberg problems?
Only homozygous recessive individuals show the recessive phenotype, so their frequency is the one you can observe directly. Take the square root of q² to get q, then p = 1 − q.
What are the five conditions for Hardy-Weinberg equilibrium?
No mutation, random mating, no natural selection, a very large population, and no gene flow (migration). If any condition is broken, allele frequencies can change and the population evolves.
How do you find the frequency of carriers?
Carriers are heterozygous, so their frequency is 2pq. For a disorder affecting 1 in 2500 people, q = 0.02, p = 0.98 and 2pq ≈ 0.039, or about 1 in 26 people.
More Free Grade 11 Biology Resources
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