Question

Approximately one in 12,500 individuals in a given population have porphyria, an autosomal dominant condition. Approximately how many individuals in this population will be in the heterozygous state?

Answer 1

The people suffering from porphyria are 1 in 12,500. Also, porphyria is an autosomal dominant condition, which means that the homozygous dominants as well as heterozygotes will suffer from the disease. If we assume the population to be in Hardy-Weinberg equilibrium, then the combined frequency of homozygous dominants and heterozygotes will be 1 in 12,500, i.e., 1/12500 = 0.00008,

Hence, p² + 2pq = 0.00008

According to Hardy-Weinberg equilibrium, p² + q² + 2pq = 1, therefore, q² = 0.99992, hence q = 0.999959

Also, p + q = 1, therefore, p = 1- 0.999959 = 0.00004

Therefore, the frequency of heterozygotes will be 2pq = 2 x 0.00004 x 0.999959 = 0.0000799, hence the number of individuals in heterozygote stage in this population will be 0.0000799 x 12500 = 0.999959 = approximately 1.