The frequency of tasters and nontasters of PTC (see Problem 10) v...

Question

The frequency of tasters and nontasters of PTC (see Problem 10) varies among populations. In population A, 64% of people are tasters (an autosomal dominant trait) and 36% are nontasters. In population B, tasters are 75% and nontasters 25%. In population C, tasters are 91% and nontasters are 9%.

Assuming that Hardy–Weinberg conditions apply, determine the genotype frequencies in each population.

Accepted Answer

Hello, everyone and welcome to today's video. So suppose that in a population of mice, black color is dominant over brown color. If the percentage of Homo Sago's recessive genotype is 25% what is the recessive allele frequency? Assuming that the population is under Hardy Weinberg equilibrium? As answer choice A we have 75% as answer choice B we have 50% as answer choice C we have 25%. And as answer choice D we have 100%. Now, in order to solve this problem, we really need to go back and look at our Hardy Weinberg equilibrium equation which is going to be that P square plus two PQ plus Q square is going to be equal to one. No, we are told in the problem that the percentage or the frequency of the homo cycles recessive genotype is 25%. This is going to be represented in the Hardy Weber equilibrium equation by Q squared. So we know that Q square is 25% or 0.25. If we take the square root of both sides, we're going to see that Q is going to be equal to 0.50 or 50%. Now, in the problem, we're asked that is that what is the frequency of the recessive allele in the population? Well, this Q is going to be representing the frequency of the recessive allele in the population. So the value is going to be 50% in our answer choices. This is going to be b 50%. I really hope this helped and I hope to see you on the next video.

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