Jake L asked in Science & MathematicsBiology · 8 years ago

The frequency of individuals who cannot taste phenylthiocarbamide (PTC) is approximately 0.3. The inability to taste this bitter substance is due to a recessive allele. Assume that there are only two alleles in the population (namely, tasters, T, and nontasters, t) and that the population is in Hardy-Weinberg equilibrium. What is the frequency of the recessive allele in the population?

A. 0.45

B. 0.30

C. 0.15

D. 0.55

Relevance
• 8 years ago

This is the third time this question has been posted. Here is the answer:

D.

It's my best guess. We know that if frequency is 50% (everybody is a carrier), then 25% of their posterity will express the recessive trait. You are looking for something that would produce 30% (slightly greater than 25%), so D, or 55% makes sense.

Also, I want to say that the odds work something like .55 * .55 = 0.3025. That means that there is a 30% chance that two carriers will get together and both pass on the recessive trait to their offspring.

Source(s): Pharmacy student
• 8 years ago

f= (aa)=q^2

0.3 =q^2

square root of .3 rounds out to be 0.55

Source(s): http://en.wikipedia.org/wiki/Hardy–Weinberg_princi... look under derivation there is a list of three handy equations there you may want to memorize for the test. Basically you are solving the equation for one allele of the recessive trait.