4.10.5Additional lessons
Allele Frequencies: Skills — IB Biology HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 88 of 101
Part 1 of 5Counting Alleles
Allele frequency is found by counting alleles, not individuals.
Diploid counting
- Each individual carries two alleles of a gene, so a population of N individuals has 2N alleles in total.
Homozygotes
- An AA individual contributes two A alleles, and an aa individual contributes two a alleles.
Heterozygotes
- An Aa individual contributes one A and one a allele to the count.
Frequency
- Divide the number of copies of an allele by 2N; the frequencies of all alleles of the gene add up to 1.
| Genotype | Individuals | A alleles | a alleles |
|---|---|---|---|
| AA | 98 | 196 | 0 |
| Aa | 84 | 84 | 84 |
| aa | 18 | 0 | 36 |
| Total | 200 | 280 | 120 |
| Total allele count check | 280 + 120 = 400 = 2N |
Part 2 of 5Calculation Method
1
Count the total number of individuals, N, and double it to find the total number of alleles, 2N.
2
Count copies of the allele of interest from each genotype class and add them.
3
Divide by 2N to find the frequency; here f(A) = 280 ÷ 400 = 0.70.
4
Find the other allele by subtraction, f(a) = 1 − 0.70 = 0.30, then check by direct count: 120 ÷ 400 = 0.30.
Part 3 of 5Interpreting Changes
Interpreting Changes
Selection
- A steady rise in one allele across generations suggests the allele increases fitness.
Drift
- Erratic changes that differ between small populations suggest random sampling effects.
Migration
- A shift towards the frequency in a neighbouring population points to gene flow.
Graphs
- Plot frequency against generation and describe the direction, the rate and any plateau.
Part 4 of 5Worked Example
Question
- A population of 200 has 98 AA, 84 Aa and 18 aa individuals.
- (a) Calculate f(A) and f(a).
- (b) Several generations later f(A) is 0.82;
- calculate the mean rise per generation over 4 generations.
- Total alleles = 2 × 200 = 400.
- Copies of A = (2 × 98) + 84 = 280, so f(A) = 280 ÷ 400 = 0.70.
- Copies of a = 84 + (2 × 18) = 120, so f(a) = 120 ÷ 400 = 0.30.
- (b) Change = 0.82 − 0.70 = 0.12, and 0.12 ÷ 4 = 0.03 per generation.
Answer
- (a) f(A) = 0.70 and f(a) = 0.30;
- (b) a rise of 0.03 per generation.
Part 5 of 5Exam Checklist & Tips
4.10.5 – Key Points for the Exam
- Allele frequency Number of copies of an allele divided by total number of alleles.
- Genotype frequency Proportion of individuals with a given genotype.
- Homozygous Having two identical alleles of a gene.
- Heterozygous Having two different alleles of a gene.
- Diploid Having two sets of chromosomes.
Common Mistakes
- Dividing by N rather than 2N when calculating allele frequency.
- Forgetting that heterozygotes contribute one copy of each allele.
- Giving frequencies as percentages without stating it when the question wants decimals.
- Failing to check that frequencies of all alleles sum to 1.
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