If we have the recessive genotypic frequency, we can calculate the phenotypic frequencies in a population in Hardy-Weinberg equilibrium. The percentage is 82%.
The Hardy-Weinberg equilibrium theory states that allelic and genotypic frequencies remain the same through generations in a population that is in equilibrium.
The allelic frequencies in a locus are represented as p and q. Assuming a diallelic gene,
Allelic frequencies
Genotypic frequencies after one generation are
• p² ⇒ H0m0zyg0us dominant genotypic frequency,
• 2pq ⇒ Heter0zyg0us genotypic frequency,
• q² ⇒ H0m0zyg0us recessive genotypic frequency.
The addition of the allelic frequencies equals 1
p + q = 1.
The sum of genotypic frequencies equals 1
p² + 2pq + q² = 1
Available data:
18% of individuals express the recessive trait of Syndactyly
Let us say that the dominant allele is A and the recessive allele is a.
Assuming this gene expresses complete dominance,
Their frequency in the population is p² + 2pq.
Their frequency in the population is q².
Individuals that express Syndactyly have webbed finger and their genotype is h0m0zyg0us recessive, aa. Their frequency in the population is q².
q² = 18% = 0.18
We can get the frequency of individuals with normal fingers by clearing the following equation,
p² + 2pq + q² = 1
p² + 2pq + 0.18 = 1
p² + 2pq = 1 - 0.18
p² + 2pq = 0.82
The frequency of individuals with normal fingers is p² + 2pq = 0.82 = 82%.
The percentage of the population are not Webbed is 82%.
You can learn more about the hardy-weinberg equilibrium at
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