About 70% of the human population can taste the bitter chemical phenylthiocarbamide (PTC), which is found in foods like broccoli. The allele, T, for tasting PTC is dominant over the allele, t, for not tasting it. Calculate the allele frequencies using the Hardy-Weinberg equation, where p represents the dominant allele frequency, and q represents the recessive allele frequency. Which of the following is closest to your results?
To calculate the allele frequencies using the Hardy-Weinberg equation, we need to know the proportion of individuals in the population who can taste PTC (70%). Let's represent the allele for tasting PTC as T and the allele for not tasting it as t.
In the Hardy-Weinberg equation, the frequencies of the dominant and recessive alleles add up to 1.
The equation is: p^2 + 2pq + q^2 = 1
Given that the proportion of individuals who can taste PTC is 70%, or 0.7, we can assume that this represents the frequency of the dominant trait (TT and Tt genotypes). Thus, p^2 + 2pq = 0.7.
To solve for p and q, we can use algebra.
Let's assume that p represents the frequency of the T allele (dominant) and q represents the frequency of the t allele (recessive).
Since p + q = 1, we can subtract p from both sides of the equation to obtain q.
q = 1 - p
Substituting q in the original equation, we have:
p^2 + 2p(1 - p) = 0.7
Expanding and simplifying:
p^2 + 2p - 2p^2 = 0.7
Rearranging:
-p^2 + 2p + 0.7 = 0
Multiplying the equation by -1:
p^2 - 2p - 0.7 = 0
This is a quadratic equation that can be solved using the quadratic formula. The solutions to this equation will give us the values of p.
p = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = -2, and c = -0.7.
Using the quadratic formula, we get two possible values for p: p ≈ 0.877 or p ≈ 1.12.
Since the allele frequency cannot exceed 1, p must be approximately 0.877.
Therefore, the allele frequency for the dominant allele (T) is closest to 0.877, and the recessive allele frequency (t) is closest to 0.123.