Question

“Slowypoxyius” is a rare recessive disorder (its not, I made it up) that effects 4% of the population. If we have a population of 50,000 individuals, how many will have slowpoxyius? From the above population, how many heterozygotes will there be in the population?

Answer 1

Let population is in Hardy-Weinberg equilibrium. Then the frequency of individuals with the “Slowypoxyius” genotype is the square of the frequency of the “Slowypoxyius” allele.

Now, 4% of the population affected by “Slowypoxyius”, which is a recessive trait.

So, number of affected individuals by “Slowypoxyius” will be = 4% of 50000 = 2000

Now, frequency of “Slowypoxyius” (q²) = 0.04 (As affected individuals are 4%)

Therefore, frequency of “Slowypoxyius” allele (q) = = 0.2

Therefore, frequency of dominant allele (p) = 1 - 0.2 = 0.8 (As the population is in Hardy-Weinberg equilibrium, so p+q = 1)

Therefore, frequency of heterozygotes (2pq) = 2 x 0.8 x 0.2 = 0.32

So, number of heterozygotes will be = 50000 x 0.32 = 16000