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Question

Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions.

11% of an African population is born with a severe form of sickle-cell anemia (ss), 

What is the frequency of the recessive allele?

Answer 1

This question is testing your understanding of Hardy-Weinberg Equilibrium. Hardy-Weinberg equilibrium includes two equations:

p+q=1, where p= frequency of the dominant allele (in this case = S), q= frequency of the recessive allele (in this case, s)

and

p²+2pq+q²=1 (can also be written as pp+2pq+qq=1), where:

p²= frequency of the homozygous dominant genotype (SS)

q²=frequency of the recessive genotype (ss)

2pq= frequency of the heterozygous genotype (Ss)

For this scenario, 11% of the African population expresses the homozygous recessive genotype, ss. Therefore, 11%( or 0.11) can be plugged into the second equation for q². To obtain the value of q, or the frequency of the recessive allele, all you need to do is find the √0.11 because the square root of q² is q. The square root of 0.11 is ~0.33 which represents the frequency of the recessive allele in the African population.

Note: A population is said to be in Hardy-Weinberg equilibrium when 5 conditions are met:

1. No mutations
2. No gene flow (no immigration/emigration)
3. Large population size(no genetic drift)
4. No selective forces
5. No non-random mating

From the information in the given scenario, there is no information that would suggest that these conditions were not met and so we can use these hardy-weinberg equations.