Question

16% of a particular human population is unable to taste the bitterness compound, PTC, in Brussel sprouts. This “non-taster” phenotype is a recessive trait caused by the n allele.

What percentage of this population is heterozygous for the n allele? Assume Hardy-Weinberg equilibrium.

a) 24%

b) 36%

c) 48%

d) 62%

Answer 1

As per Hardy weinberg equillibrium, p² + 2pq + q² = 1 and p + q = 1

p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p² = percentage of homozygous dominant individuals
q² = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals

The frequency of the non taster "nn" genotype = 16% ( given )

The frequency of nn is 16%, it means q² =0.16

If q² = 0.16, then q = 0.4,

Since, p + q =1

p=1 - q = 1 - 0.4 = 0.6

As per formula, 2pq = percentage of heterozygous individuals for n allele

so, 2pq= 2 x 0.6 x 0.4 = 0.48

The percentage of population heterozygous for n allele is 48%.

The frequency of "NN" is equal to p², and the frequency of "nn" is equal to q². So, the frequency of "Nn" is equal to 2pq which is 48%. Therefore, C) 48% is the correct option.

Correct option: c) 48%