# 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)?

Sep 29, 2016

The percent of heterozygous individuals in African American population, for sickle cell allele, is roughly 8.54%.

#### Explanation:

The Hardy-Weinberg equation is:

${p}^{2} + 2 p q + {q}^{2} = 1 \text{ }$ and also $\text{ } p + q = 1$

• 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

$p + q = 1$

$p = 1 - q$

$p = 0.9553$

${p}^{2} = 0.9126$

$2 p q = 2 \left(0.9553\right) \left(0.0447\right) = 0.0854$

Thus % of heterozygous individuals in the population is = 8.54%

Double Check:

$0.9126 + 0.0854 + 0.002 = 1$