AP Biology 20 Views 1 Answers
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Sourav PanOctober 25, 2024

How do I calculate allele frequency?

How do I calculate allele frequency?

Sourav Pan
Sourav PanOctober 25, 2024

Answered step-by-step

General Formula:
Allele Frequency = (Number of copies of the allele) / (Total number of copies of all alleles at that locus in the population)

Methods to Calculate Allele Frequency:

1. Direct Counting (Discrete Data)

  • Step 1: Count the number of individuals with each genotype.
  • Step 2: Determine the total number of alleles in the population (2 times the total number of individuals, since each individual has 2 alleles).
  • Step 3: For each allele, count the number of its copies:
    • For homozygous genotypes, count 2 copies per individual (e.g., MM contributes 2 M alleles).
    • For heterozygous genotypes, count 1 copy per individual (e.g., MN contributes 1 M allele and 1 N allele).
  • Step 4: Calculate the allele frequency for each allele using the general formula.

Example from the Text:

Genotype # of Individuals
MM 1787
MN 3039
NN 1303
Total 6129
  • Total Alleles: 12,258 (2 * 6129)
  • Allele M Frequency (p): f(M) = [(2*1787) + 3039] / 12,258 = 0.5395
  • Allele N Frequency (q): f(N) = [(2*1303) + 3039] / 12,258 = 0.4605

2. Using Genotypic Frequencies (when genotypes are known)

  • Given: Frequencies of genotypes (e.g., MM, MN, NN)
  • Step 1: Assign p as the frequency of one allele (e.g., M) and q as the frequency of the other allele (e.g., N), knowing that p + q = 1.
  • Step 2: Calculate p and q using the genotype frequencies:
    • p = f(MM) + ½ f(MN)
    • q = f(NN) + ½ f(MN)

Example from the Text (Greenland):

Genotype Frequency
MM 0.835
MN 0.156
NN 0.009
  • p (M allele): 0.835 + ½(0.156) = 0.913
  • q (N allele): 0.009 + ½(0.156) = 0.087

3. From Molecular Data (e.g., fragment sizes)

  • Step 1: Count the number of individuals with each fragment size.
  • Step 2: Assuming each individual has two alleles, calculate the total number of alleles.
  • Step 3: Calculate the frequency of each allele as in the direct counting method.

Example from the Text (Molecular Data):

Fragment Size # of Individuals
6.5 kb 30 (homozygous)
6.5 kb & 3.0 kb 50 (heterozygous)
3.0 kb 20 (homozygous)
Total Individuals 100
  • Total Alleles: 200 (2 * 100)
  • Allele 6.5 kb Frequency (p): [(2*30) + 50] / 200 = 0.55
  • Allele 3.0 kb Frequency (q): [(2*20) + 50] / 200 = 0.45

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