Question

Phenylthiocarbamide (PTC), also known as phenylthiourea (PTU), has the unusual property that it either tastes very bitter or is virtually tasteless, depending on the genetic makeup of the taster. There is a single gene which codes for a protein found in our tongues. PTC will bind with the protein if it present and a person will taste it. If the protein is not present, PTC will not bind and a person cannot taste it. The ability to taste PTC is often treated as a dominant genetic trait. The dominant allele, T, causes a person to be able to taste PTC, and the recessive allele is represented as t. Assume that a population is in Hardy-Weinberg equilibrium and 16 individuals out of 100 are unable to taste PTC.

What are the allele frequencies of T and t in this population?

A. The allele frequency of T is 0.80, and the allele frequency of t is 0.20
B. The allele frequency of T is 0.84, and the allele frequency of tis 0.16.
C. The allele frequency of T is 0.00, and the allele frequency of t is 0.40
D. The allele frequency of T is 0.50, and the allele frequency of t is 0.50.

Answer 1

B. The allele frequency of T is 0.84, and the allele frequency of t is 0.16.

Step-by-step explanation:

The Hardy-Weinberg equation is:

p2 + 2pq + q2 = 1

Where:

p = the frequency of the dominant allele

q = the frequency of the recessive allele

1 = the total number of alleles

Given that 16 individuals out of 100 are unable to taste PTC, we can calculate that the frequency of the recessive allele, q, is 0.16. We can then use the Hardy-Weinberg equation to solve for the frequency of the dominant allele, p:

p2 + 2pq + q2 = 1

p2 + 2p(0.16) + (0.16)2 = 1

p2 + 0.32p + 0.0256 = 1

p2 + 0.32p - 0.9144 = 0

(p + 0.32)(p - 2.84) = 0

p = -0.32 or p = 2.84

Since the allele frequencies must add up to 1, we know that p cannot equal -0.32. This leaves us with p = 2.84. Therefore, the allele frequency of the dominant allele is 2.84, and the allele frequency of the recessive allele is 0.16.