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Hardy–Weinberg allele frequencies: Class 12 practice

In a population in equilibrium, p + q = 1 and p² + 2pq + q² = 1, where p and q are allele frequencies and p², 2pq and q² are genotype frequencies. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Square roots

The idea behind hardy–weinberg allele frequencies

In a population in equilibrium, p + q = 1 and p² + 2pq + q² = 1, where p and q are allele frequencies and p², 2pq and q² are genotype frequencies. If the recessive phenotype frequency is known, q = √(q²).

Step-by-step method

  1. Find q² from the recessive phenotype.
  2. q = √q², p = 1 − q.
  3. Carriers = 2pq.

Worked example 1 (Basic)

In a population, q = 0.3. Find p and the frequency of heterozygotes.

  1. p = 1 − 0.3 = 0.7
  2. 2pq = 0.42
  3. Answer: p = 0.7, 2pq = 0.42

Worked example 2 (Standard)

4% of a population show a recessive trait. What percentage are carriers?

  1. q² = 0.04 → q = 0.2
  2. p = 0.8; 2pq = 0.32
  3. Answer: 32%

Worked example 3 (Challenge)

In 10000 people, 225 have a recessive condition. Estimate how many are homozygous dominant.

  1. q² = 0.022, q = 0.15, p = 0.85
  2. p²N = 7225
  3. Answer: 7225

Common mistakes and misconceptions

  • Using q² as q
  • Forgetting the 2 in 2pq

Practice questions

  1. (Basic) In a population, q = 0.2. Find p and the frequency of heterozygotes.
  2. (Basic) In a population, q = 0.4. Find p and the frequency of heterozygotes.
  3. (Standard) 6.25% of a population show a recessive trait. What percentage are carriers?
  4. (Standard) 1% of a population show a recessive trait. What percentage are carriers?
  5. (Challenge) In 2000 people, 5 have a recessive condition. Estimate how many are homozygous dominant.
  6. (Challenge) In 10000 people, 25 have a recessive condition. Estimate how many are homozygous dominant.

Answer key with full working

  • Q1: p = 1 − 0.2 = 0.8 → 2pq = 0.32 → Answer: p = 0.8, 2pq = 0.32
  • Q2: p = 1 − 0.4 = 0.6 → 2pq = 0.48 → Answer: p = 0.6, 2pq = 0.48
  • Q3: q² = 0.063 → q = 0.25 → p = 0.75; 2pq = 0.375 → Answer: 37.5%
  • Q4: q² = 0.01 → q = 0.1 → p = 0.9; 2pq = 0.18 → Answer: 18%
  • Q5: q² = 0.003, q = 0.05, p = 0.95 → p²N = 1805 → Answer: 1805
  • Q6: q² = 0.003, q = 0.05, p = 0.95 → p²N = 9025 → Answer: 9025

Syllabus note

This skill is listed under "Evolution" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is hardy–weinberg allele frequencies in?

"Evolution", Class 12 Biology (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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