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?

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

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,

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

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,

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

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

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