In a population in Hardy-Weinberg equilibrium, one in a hundred people have the inherited ability to curl their tongue downwards in a particular fashion. If this trait is the recessive phenotype, what percentage of the population in the next generation is likely to be carriers of this allele, although unable to curl their tongues downwards?
A) 2 per cent
B) 10 per cent
C) 18 per cent
D) 90 per cent
E) 98 per cent

Question

Quizplus · Accepted Answer

Given that the ability to curl the tongue downwards is a recessive trait, let's denote the recessive allele as "a" and the dominant allele as "A". If 1 in 100 people can curl their tongues downwards, this means that the frequency of the homozygous recessive genotype (aa) is 0.01 (or 1%). According to the Hardy-Weinberg principle, the frequency of the recessive allele "a" can be found by taking the square root of the frequency of the homozygous recessive genotype, which is √0.01 = 0.1 or 10%. The frequency of the dominant allele "A" would then be 1 - 0.1 = 0.9 or 90%. The frequency of carriers (heterozygous genotype Aa) can be calculated as 2pq, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. Substituting the values, we get 2 * 0.9 * 0.1 = 0.18 or 18%. Therefore, 18% of the population is likely to be carriers of the allele but unable to curl their tongues downwards.

In a Population in Hardy-Weinberg Equilibrium, One in a Hundred