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Allele T, the ability to taste for a particular chemical, is dominant over allele t, which makes an individual unable to taste that chemical. A group of researchers surveyed 500 students and found 80 were non-tasters. Assuming this population is in Hardy-Weinberg equilibrium, what is the expected frequency of heterozygous individuals in this population of university students (use up to two decimal places, e.g. 0.10)

Sagot :

marianaegarciaperedo

Assuming complete dominance, and according to the H-W equilibrium, the expected frequency of the heter0zyg0us individuals is F(Tt) = 2pq = 0.48.

What is the Hardy-Weinberg equilibrium?

The Hardy-Weinberg equilibrium theory states that allelic and genotypic frequencies in a population in equilibrium remain the same through generations.

The allelic frequencies in a locus are represented as p and q. Assuming a diallelic gene,

  • The frequency of the dominant allele p(X) is p
  • The frequency of the recessive allele p(x) is q

The 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:

  • T ⇒ Dominant allele ⇒ the ability to taste
  • t ⇒ Recessive allele ⇒ the unability to taste
  • From N =  500 students ⇒ 80 non-tasters
  • population is in Hardy-Weinberg equilibrium

We will assume that this diallelic gene espresses complete dominance, and the T allele hides the expression of t is heter0zyg0us state.

Let us first take the phenotypic frequencies, F(tasters) and F (non-tasters)

  • N = 500 individuals
  • 80 non-tasters
  • 420 tasters ⇒ 500 - 80

The frequency if individuals that are non-tasters is

F (non-tasters) = 80/500 = 0.16

The frequency if individuals that are tasters is

F (tasters) = 420/500 = 0.84  ⇒  These individuals include h0m0zyg0us  

                                                       dominant -TT- and heter0zyg0us -Tt-

                                                       individuals.

The frequency of the non-taster individuals is equal to the recessive genotypic frequency F(tt) = q ².

F(non-taster) = F(tt) = q ² = 0.16

From this value, we can take the frequency of the recessive allele, f(t) = q.

If F(tt) =  q ² = 0.16

f(t) = q = √0.16 = 0.4

q = 0.4

Now that we have the q value, by clearing the following equation, we can get the f(T) = p value.

p + q = 1

p + 0.4 = 1

p = 1 - 0.4

p = 0.6

If p = 0.6, then the frequency of the dominant alelle f(T) = p² is

p = 0.6

p² = 0.36

Finally, having both allelic frequencies, we can get the frequency of the heter0zyg0us genotype, F(Tt) = 2pq

F(Tt) = 2pq = 2 x 0.6 x 0.4 = 0.48

2pq = 0.48

If our calculations are right, the addition of all genotypic frequencies should be 1

p² + 2pq + q² = 1

0.36 + 0.48 + 0.16 = 1

You can learn more about the H-W equilibrium at

https://brainly.com/question/16823644

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