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# Smart people please help with multiple choice best answer 10 points?

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

Please provide an explanation to your answer

### 4 Answers

- 8 years agoFavorite Answer
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- Login to reply the answers

- 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.- Login to reply the answers

- Anonymous8 years ago
I'm not smart so i can't answer your question.

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