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Use the following information to calculate the correct frequencies: In a population of
Pennsylvanians, only 18% of individuals express the recessive trait of Syndactyly or webbed finger
(formulas: p+q=1 and p²+2pq+q2²=1)
a. What percentage of the population are not Webbed?

Sagot :

marianaegarciaperedo

If we have the recessive genotypic frequency, we can calculate the phenotypic frequencies in a population in Hardy-Weinberg equilibrium. The percentage is 82%.

What si the Hardy-Weinberg equilibrium theory?

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

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.

  • 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

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

https://brainly.com/question/16823644

#SPJ1  

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