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Stubble bristles (Sb) is dominant to wildtype bristles (Sb+) in Drosophila melanogaster. In a population where there are 137 flies with wildtype bristles and 836 flies with stubble bristles, what are the allele frequencies for the Sb and Sb+ alleles (assuming the population is in Hardy-Weinberg equilibrium)? Please calculated your answers to 3 decimal places.
f(Sb)=
f(Sb+)=
If the population is in Hardy-Weinberg equilibrium, what proportion of the population would you expect to be heterozygous?

Respuesta :

marianaegarciaperedo

Populations in H-W equilibrium have the same allelic and genotypic frequencies generation after generation. f(Sb) = p = 0.626. f(Sb+) = q = 0.374. F(SbSb+) = 2pq = 0.468

What is Hardy-Weinberg Equilibrium?

The Hardy-Weinberg equilibrium is a theory that states that, assuming a diallelic gene,

⇒ The allelic frequencies in a locus are represented as p and q.

• 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 (Homozygous dominant genotypic frequency),

• 2pq (Heterozygous genotypic frequency),

• q² (Homozygous recessive genotypic frequency).

Allelic and genotypic frequencies remain the same through generations in a population that is in H-W equilibrium.

The addition of the allelic frequencies equals 1

p + q = 1.

The sum of genotypic frequencies equals 1

p² + 2pq + q² = 1

In the exposed example,

  • Sb is dominant
  • Sb+ is recessive ⇒ wild type

  • Population size ⇒ 137 + 836 = 973
  • Individuals with wildtype bristles ⇒ 137  ⇒ Sb+Sb+
  • Individuals with stubble bristles ⇒ 836 ⇒ SbSb and SbSb+

First, we will get the frequencies.

To do it, we will use the number of individuals. These values will tell us about phenotypic frequencies,

Phenotypic frequency

  • F(wildtype) = q² = 137/973 = 0.14
  • F(stubble) = p² + 2pq = 836/973 = 0.86

Now, we will use the frequency of the recessive phenotype to get its genotypic and allelic frequencies.

F(Sb+Sb+) = q² = 137/973 = 0.14 ⇒ Recessive Genotypic frequency

f(Sb+) = q = √0.14 = 0.374 ⇒ Recessive Allelic frequency

Now, Since this population is in H-W equilibrium, we will use q (recessive allelic frequency) to get the dominant allelic frequency and the dominant genotypic frequency. To do it, we will clear the following equation,

p + q = 1

p + 0.374 = 1

p = 1 - 0.374

p = 0.626 ⇒ Dominant Allelic frequency

F( SbSb) = p² = 0.626² = 0.392 ⇒ Dominant Genotypic frequency

Finally, we will get the heterozygous frequency. There are two options,

  • To susbtract the dominant genotypic frequency to the dominant phenotpic frequency,

F(SbSb + SbSb+) - F(SbSb) = 0.86 - 0.392 = 0.468

  • To clear the following equation,

p² + 2pq + q² = 1

0.392 + 2pq + 0.14 = 1

2pq = 1 - 0.392 - 0.14

2pq = 0.468

Answers:

- f(Sb) = p = 0.626 ⇒ Dominant Allelic frequency

- f(Sb+) = q = 0.374 ⇒ Recessive Allelic frequency

- F(SbSb+) = 2pq = 0.468 ⇒ Heterozygous frequency

You can learn more about Hardy Weinberg equilibrium at

https://brainly.com/question/16823644

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