Question

Sickle cell anemia is caused by a recessive allele. Roughly 1 out of every 500 african americans is afflicted with sickle cell anemia. How would you use the hardy weinberg equation to calculate the % of heterozygotes for sickle cell allele (.002=q^2)?

Answer 1

Did you mean: Answer: The percent of heterozygous individuals in African American population, for sickle cell allele, is roughly 8.54 % . Explanation: The Hardy-Weinberg equation is: p 2 + 2 p q + q 2 = 1 and also p + q = 1 where p 2 is the percentage of homozygous dominant phenotype where 2 p q is the percentage of heterozygous dominant phenotype where q 2 is the percentage of homozygous recessive phenotype One double recessive afflicted individual means there are two individuals among 1000 . Thus, q 2 = 0.002 q = 0.0447 Then p + q = 1 p = 1 − q = 0.9553 p 2 = 0.9126 2 p q = 2 ( 0.9553 ) ( 0.0447 ) = 0.0854 Thus % of heterozygous individuals in the population is = 8.54 % Double Check: 0.9126 + 0.0854 + 0.002 = 1