Q.88 The population data present in an island is as follows
Genotype Number
𝐴𝐴 300
𝐴𝑎 500
𝑎𝑎 200
Total 1000
The allele frequency of 𝑨 (upto two decimals) will be ____
Allele frequency calculation determines how common a specific gene variant
is within a population. This is a core concept in population genetics and
frequently appears in exams such as NEET, CSIR-UGC, GATE, and other biology
entrance tests.
For the given island population data, the allele frequency of A is
0.55. Below is a complete step-by-step explanation.
Given Genotype Data
- AA = 300 individuals
- Aa = 500 individuals
- aa = 200 individuals
- Total population (N) = 1000 individuals
Step-by-Step Calculation
Allele Frequency Formula
The standard formula for calculating the frequency of allele A (p) is:
p = [(2 × AA) + Aa] / (2 × Total Individuals)
Step 1: Count Total A Alleles
- AA individuals contribute 2 A alleles each:
2 × 300 = 600 - Aa individuals contribute 1 A allele each:
1 × 500 = 500
Total A alleles = 600 + 500 = 1100
Step 2: Count Total Alleles in the Population
Since the population is diploid:
Total alleles = 2 × 1000 = 2000
Step 3: Calculate Allele Frequency of A
p = 1100 / 2000 = 0.55
Final Answer
The allele frequency of A in the population is:
p = 0.55
The allele frequency of a (q) can also be verified:
q = [(2 × 200) + 500] / 2000 = 900 / 2000 = 0.45
Since p + q = 0.55 + 0.45 = 1, the calculation is correct.
Why This Formula Works
Each individual carries two alleles. Homozygous dominant (AA) individuals
contribute two A alleles, heterozygotes (Aa) contribute one A allele, and homozygous recessive
(aa) individuals contribute none.
This approach aligns with Hardy–Weinberg principles, where:
- p² = frequency of AA
- 2pq = frequency of Aa
- q² = frequency of aa
Observed genotype frequencies here are:
AA = 0.30, Aa = 0.50, aa = 0.20.
Common Mistakes to Avoid
| Mistake | Wrong Calculation | Correct Approach |
|---|---|---|
| Ignoring heterozygotes | p = 300 / 1000 = 0.30 | Add Aa contribution: +500 A alleles |
| Dividing by individuals | p = 1100 / 1000 = 1.10 | Divide by total alleles (2000) |
| Using only homozygotes | p = 600 / 2000 = 0.30 | Include heterozygotes: 1100 / 2000 |
Exam Tips
- Always calculate allele counts first, not frequencies.
- Remember: p + q = 1 for autosomal loci.
- For allele a, use: q = [(2 × aa) + Aa] / (2N).
- Do not round until the final step.
This method is widely used in evolutionary biology, population studies,
and exam-based genetics problems.
