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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)?

Answer 1

Madeline Andrews

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

The formula for Hardy-Weinberg is:

#p^2 + 2pq + q^2 = 1" "# and also #" "p+q=1#

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

#q^2=0.002#

#q = 0.0447#

#p + q = 1#

#p = 1 - q#

#p = 0.9553#

#p^2 = 0.9126#

#2pq = 2(0.9553)(0.0447) = 0.0854#

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

Verify Again:

#0.9126 + 0.0854 + 0.002 = 1#

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Answer 2

Nicholas Ahn

To calculate the percentage of heterozygotes for the sickle cell allele using the Hardy-Weinberg equation, you can use the equation:

[2pq = 2 \times 0.02 \times 0.98 = 0.0392]

So, approximately 3.92% of African Americans would be heterozygous carriers of the sickle cell allele.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.