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4.10.5Additional lessons

Allele Frequencies: Skills — IB Biology HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 88 of 101

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Part 1 of 5Counting Alleles

Allele frequency is found by counting alleles, not individuals.

Diploid counting
  • Diploid counting Each individual carries two alleles of a gene, so a population of N individuals has 2N alleles in total.
Homozygotes
  • Homozygotes An AA individual contributes two A alleles, and an aa individual contributes two a alleles.
Heterozygotes
  • Heterozygotes An Aa individual contributes one A and one a allele to the count.
Frequency
  • Frequency Divide the number of copies of an allele by 2N;
    • the frequencies of all alleles of the gene add up to 1.
GenotypeIndividualsA allelesa alleles
AA981960
Aa848484
aa18036
Total200280120
Total allele count check280 + 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
  • Selection A steady rise in one allele across generations suggests the allele increases fitness.
Drift
  • Drift Erratic changes that differ between small populations suggest random sampling effects.
Migration
  • Migration A shift towards the frequency in a neighbouring population points to gene flow.
Graphs
  • 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.

  1. Total alleles = 2 × 200 = 400.
  2. Copies of A = (2 × 98) + 84 = 280, so f(A) = 280 ÷ 400 = 0.70.
  3. Copies of a = 84 + (2 × 18) = 120, so f(a) = 120 ÷ 400 = 0.30.
  4. (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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