IB Biology How to Solve: IB Biology – Hardy-Weinberg Equilibrium Calculations

How to Solve: IB Biology – Hardy-Weinberg Equilibrium Calculations

Complete Guide

Introduction

"Mastering Hardy-Weinberg calculations can earn you 4-6 marks in IB Biology Paper 2—enough to push you from a 6 to a 7. It’s also the key to predicting genetic disease frequencies in real-world populations, like sickle-cell anemia in malaria-prone regions."

WHAT YOU NEED TO KNOW FIRST

Before tackling Hardy-Weinberg, you must understand:
1. Alleles vs. genotypes – An allele is a version of a gene (e.g., A or a); a genotype is the pair (e.g., AA, Aa, aa).
2. Dominant vs. recessive traits – Dominant alleles mask recessive ones in heterozygous individuals.
3. Basic probability – The chance of two independent events both happening is their probabilities multiplied.

KEY TERMS & FORMULAS

Key Terms

• Gene pool: All alleles in a population.
• Allele frequency: Proportion of a specific allele in the gene pool (e.g., frequency of A = p).
• Genotype frequency: Proportion of a specific genotype in the population (e.g., frequency of AA = p²).
• Hardy-Weinberg equilibrium: A model where allele frequencies do not change between generations if:
• No mutations
• No migration
• Large population
• No natural selection
• Random mating

Formulas

1. Allele frequency equation p + q = 1
2. p = frequency of the dominant allele (A)
3. q = frequency of the recessive allele (a)

4. MEMORISE THIS

5. Genotype frequency equation p² + 2pq + q² = 1

6. p² = frequency of homozygous dominant (AA)
7. 2pq = frequency of heterozygous (Aa)
8. q² = frequency of homozygous recessive (aa)
9. MEMORISE THIS

STEP-BY-STEP METHOD

Follow these steps for every Hardy-Weinberg problem:

1. Identify the given information
2. Is it an allele frequency (p or q)?
3. Is it a genotype frequency (p², 2pq, or q²)?

4. Is it a phenotype (e.g., "25% of the population has the recessive trait")?

5. Write down the known value(s)

6. Example: If 9% of the population has the recessive phenotype, then q² = 0.09.

7. Use the correct formula to find the missing variable

8. If you have q², take the square root to find q.
9. If you have q, use p + q = 1 to find p.

10. If you need genotype frequencies, use p², 2pq, or q².

11. Check if the question asks for a percentage or decimal

12. Convert to percentage if needed (e.g., 0.09 → 9%).

13. Verify your answer makes sense

14. p + q must equal 1.
15. p² + 2pq + q² must equal 1.

Worked Example (Using the Steps)

Question: In a population, 16% of individuals have a recessive genetic disorder. What is the frequency of carriers (Aa) in the population?

Step 1: Identify the given information. - Recessive phenotype = aa = q² = 16% = 0.16

Step 2: Write down the known value. - q² = 0.16

Step 3: Find q (recessive allele frequency). - q = √0.16 = 0.4

Step 4: Find p (dominant allele frequency). - p + q = 1 - p = 1 – 0.4 = 0.6

Step 5: Find carrier frequency (2pq). - 2pq = 2 × 0.6 × 0.4 = 0.48

Step 6: Convert to percentage if needed. - 0.48 = 48%

Final Answer: The frequency of carriers is 48%.

What we did and why: We started with the recessive phenotype (q²), found q, then p, and finally calculated the heterozygous frequency (2pq) because carriers are Aa.

WORKED EXAMPLES

Example 1 – Basic

Question: In a population, the frequency of the dominant allele (A) is 0.7. What is the frequency of homozygous recessive individuals (aa)?

Solution:
1. Given: p = 0.7
2. Find q: p + q = 1 → q = 1 – 0.7 = 0.3
3. Find q²: q² = (0.3)² = 0.09
4. Convert to percentage: 0.09 = 9%

Answer: The frequency of aa individuals is 9%.

What we did and why: We used p to find q, then squared q to get the homozygous recessive frequency.

Example 2 – Medium

Question: In a population, 36% of individuals are homozygous recessive (aa). What percentage of the population is heterozygous (Aa)?

Solution:
1. Given: q² = 0.36
2. Find q: q = √0.36 = 0.6
3. Find p: p = 1 – 0.6 = 0.4
4. Find 2pq: 2 × 0.4 × 0.6 = 0.48
5. Convert to percentage: 0.48 = 48%

Answer: The frequency of heterozygous individuals is 48%.

What we did and why: We started with q², found q, then p, and calculated 2pq because heterozygotes are Aa.

Example 3 – Exam-Style

Question: A genetic disorder is caused by a recessive allele. In a population of 10,000, 400 individuals have the disorder. What is the expected number of carriers in the population?

Solution:
1. Find q²: 400/10,000 = 0.04
2. Find q: q = √0.04 = 0.2
3. Find p: p = 1 – 0.2 = 0.8
4. Find 2pq: 2 × 0.8 × 0.2 = 0.32
5. Find number of carriers: 0.32 × 10,000 = 3,200

Answer: The expected number of carriers is 3,200.

What we did and why: We converted the number of affected individuals to q², found q and p, then calculated 2pq to find carriers.

COMMON MISTAKES

1. Mistake: Confusing q and q².
2. Why it happens: Students forget that q is the allele frequency, while q² is the genotype frequency.

3. Correct approach: Always ask: "Is this an allele or a genotype?"

4. Mistake: Forgetting to take the square root when given q².

5. Why it happens: Students rush and skip steps.

6. Correct approach: If given q², always take the square root to find q.

7. Mistake: Using p² or 2pq when the question asks for q².

8. Why it happens: Misreading the question.

9. Correct approach: Circle what the question is asking for (AA, Aa, or aa).

10. Mistake: Not converting between decimals and percentages.

11. Why it happens: Forgetting that 0.09 = 9%.

12. Correct approach: Check if the answer should be a decimal or percentage.

13. Mistake: Assuming p is always the dominant allele.

14. Why it happens: Overgeneralizing.
15. Correct approach: p is just the first allele—it could be dominant or recessive.

EXAM TRAPS

1. Trap: Giving a phenotype frequency instead of a genotype frequency.
2. How to spot it: The question says "X% of the population shows the recessive trait."

3. How to avoid it: Remember that the recessive phenotype = q² (genotype aa).

4. Trap: Asking for the number of individuals, not the frequency.

5. How to spot it: The question gives a population size (e.g., "in a population of 500").

6. How to avoid it: Calculate the frequency first, then multiply by the population size.

7. Trap: Using the wrong formula for heterozygotes.

8. How to spot it: The question asks for "carriers" or "heterozygous individuals."
9. How to avoid it: Always use 2pq for heterozygotes.

1-MINUTE RECAP

"Hardy-Weinberg is all about two equations: p + q = 1 and p² + 2pq + q² = 1. If you’re given the recessive phenotype, that’s q². Take the square root to get q, then find p by subtracting from 1. If you need carriers, use 2pq. Always check if the answer should be a decimal or percentage. And remember—p and q are allele frequencies, while p², 2pq, and q² are genotype frequencies. Practice a few problems, and you’ll nail this on exam day!