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)?

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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)?

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Answer 1 · 2016-09-29T06:04:23

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

Explanation:

The Hardy-Weinberg equation is:

$p^{2} + 2 p q + q^{2} = 1$ $p^2 + 2pq + q^2 = 1" "$ and also $p + q = 1$ $" "p+q=1$

where $p^{2}$ $p^2$ is the percentage of homozygous dominant phenotype
where $2 p q$ $2pq$ is the percentage of heterozygous dominant phenotype
where $q^{2}$ $q^2$ is the percentage of homozygous recessive phenotype

One double recessive afflicted individual means there are two individuals among $1000$ $1000$ . Thus,

$q^{2} = 0.002$ $q^2=0.002$

$q = 0.0447$ $q = 0.0447$

Then

$p + q = 1$ $p + q = 1$

$p = 1 - q$ $p = 1 - q$

$p = 0.9553$ $p = 0.9553$

$p^{2} = 0.9126$ $p^2 = 0.9126$

$2 p q = 2 (0.9553) (0.0447) = 0.0854$ $2pq = 2(0.9553)(0.0447) = 0.0854$

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

Double Check:

$0.9126 + 0.0854 + 0.002 = 1$ $0.9126 + 0.0854 + 0.002 = 1$

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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)?

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