The given problem shows that the different genotypic frequencies for the trait: 0.0023 for H^{S}H^{S}, 0.7560 for H^{B}H^{B}, and 0.241 for H^{S}H^{B}. In a Hardy-Weinberg equilibrium, the frequency of the dominant allele can be represented by p, and the frequency of the recessive allele can be represented by q. For a trait that only has two alleles, p + q = 1. Furthermore, these values can be used to compute for the genotypic frequencies: p^{2} is for the homozygous dominant, 2pq is for the heterozygotes and q^{2} for the homozygous recessive.

Sickle cell anemia is caused by the deleterious H^{S} allele. The WT allele is the H^{B} allele. In a population of 200,000 individuals, the observed frequency of H^{S}H^{S} is 0.0023, H^{B}H^{B} is 0.7560, and H^{S}H^{B} is 0.241. How many are the heterozygous individuals?

A. 24,100

B. 8,290

C. 36,500

D. 48,200

E. 829,200

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