Question

Sickle-cell disease is caused by a recessive allele. Roughly one out of every 500 African Americans (0.2%) is afflicted with sickle-cell disease. Use the Hardy-Weinberg equation to calculate the percentage of African Americans who are carriers of the sickle-cell allele. Hint: 0.002 = q2. Show your work.

Answer 1

Answer:

heterozygous (Aa) are 0.085 = or 8.54%

Step-by-step explanation:

The heterozygous individuals are carriers of the sickle cell trait. They have a genotype of Aa and are represented by the 2pq term

in the H-W equilibrium equations.

According to the question 0.2% of the population is affected with sickle cell anemia, thus q^2

= 0.2% = 0.002 in decimal. So, q =

sqr(q^2)

or sqr(0.002) =

0.04472

and p + q = 1, thus p = 1 – q = 1 – 0.04472 = 0.96

Thus, A allele has a frequency of 0.96 and the a allele has a frequency of 0.04472. Therefore, the

percentage of the population that is heterozygous (Aa) and are carriers is = 2pq = 2× 0.04472× 0.96 =0.085 = or 8.54%