Question

In a certain population, the allele causing sickle cell anemia has an allele frequency of 0.2. if the population is in genetic equilibrium for this allele, what fraction of the population would be carriers for the allele (heterozygous individuals)?

Answer 1

Aa=.32

Step-by-step explanation:

Allele freqs:

AA: .2

aa: 1- AA or 1-.2=.8

Genotype freqs for heterozygotes:

2pr or 2(.2)(.8)

Aa=.32

Answer 2

We solve the problem using the Hardy-Wineberg equation.

Let us start with the homozygous recessive percentage of sickle cell anaemia trait first,

∴ q² = 0.2

We have the frequency for sickle cell anaemia allele as q = √0.2 = 0.45

Solving for the frequency of normal allele that is not causing anaemia by using the fact that,

p + q = 1

∴ p = 1 - q = 1 - 0.45 = 0.55

Now we can calculate 2pq in p² + 2pq + q², which is the heterozygous alleles frequency.

∴ 2pq = 2 x 0.45 x 0.55 = 0.50

Hence 50% of the population is heterozygous.