If we have the recessive genotypic frequency, we can calculate the phenotypic frequencies in a population in Hardy-Weinberg equilibrium. The percentage is <u>82%</u>.
<h3>What si the Hardy-Weinberg equilibrium theory?</h3>
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
- The frequency of the dominant allele f(X) is p
- The frequency of the recessive allele f(x) is q
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
<u>Available data</u>:
18% of individuals express the recessive trait of Syndactyly
Let us say that the dominant allele is A and the recessive allele is a.
- The dominant allele A codes for normal finger.
- The recessive allele a codes for webbed finger.
Assuming this gene expresses complete dominance,
- H0m0zyg0us dominant and heter0zyg0us -AA and Aa- individuals have normal finger.
Their frequency in the population is p² + 2pq.
- h0m0zyg0us recessive -aa- individuals have webbed finger.
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
<u>p² + 2pq = 0.82</u>
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The frequency of individuals with normal fingers is <u>p² + 2pq = 0.82 = 82%.</u>
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The percentage of the population are not Webbed is <u>82%</u>.
You can learn more about the hardy-weinberg equilibrium at
brainly.com/question/16823644
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