AP Biology 20 Views 1 Answers
How do I calculate allele frequency?
How do I calculate allele frequency?
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) andq
as the frequency of the other allele (e.g., N), knowing thatp + q = 1
. - Step 2: Calculate
p
andq
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
Did this page help you?