bb
Lv 4
bb asked in Science & MathematicsBiology · 8 years ago

# PTC Genetics Question????????????????????

About 30% of people do not recognize the bitter taste of phenyl-thio-carbamate (PTC). Inability to taste it is due to a recessive allele denoted by allele a. In other words, AA and Aa are normal individuals who can recognize the bitter taste while an aa individual cannot. The tasting allele A is dominate over the non-tasting allele a. Let p be the frequency of allele A and q be the frequency of allele a. Given that the frequency of genotype aa is 0.30 and the population is in Hardy-Weinberg equilibrium,

(1) What is the frequency of the non-tasting allele, i.e., what is the q value?

(2) Among the people who can recognize the bitter taste, what is the probability of carriers for the non-tasting allele? Note that the probability of carrier is not the frequency of heterozygote.

For 1, i got .5477.

For 2, I'm not sure how to do it.

Relevance
• 8 years ago

Hardy-Weinberg equilibrium describes the allele frequencies in a non-evolving population. It uses this equation:

p^2 + 2pq + q^2 = 1

p^2 is the frequency of homozygous dominant individuals. (AA)

2pq is the frequency of heterozygous individuals. (Aa or aA)

q^2 is the frequency of homozygous recessive individuals. (aa)

You got the right answer for 1, since you know that q^2=0.30

For 2, you have to set up a ratio. The numerator is the frequency of heterozygotes, or 2pq. To find the denominator, you have to limit the number of individuals you're talking about to only those who can recognize the bitter taste, i.e., those who are either AA or Aa. This is equal to p^2+2pq.

Your answer, the probability of carriers among people who can recognize the bitter taste, will be:

(2pq)/(p^2+2pq)

That's equal to 0.495/0.70, or .7071.

I hope this helps!