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mishaaus mishaaus
wrote...
Posts: 39
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12 years ago
A dominate allele, T, codes for the ability to taste the compound phenylthiocarbamide (PTC ). People who are homozygous for the recessive allele,t, are un able to taste PTC. In a Genetics class of 125 students, 88 students can taste PTC and 37 cannot.

A) calculate the expected frequencies for the T and t alleles in the student population.
B) How many students would you expect to be heterozygous for the tasting gene?
C) How many students would you expect to be homozygous donminant for the tasting gene?
D) How could you check your answers for parts (b) and (c)
 
I think you have to be way to smart to answer this one lol
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wrote...
Staff Member
12 years ago
It's not that hard, you have to use the Hardy-Weinberg equation and p + q = 1.

Let me know what you did. Thinking Face
- Master of Science in Biology
- Bachelor of Science
mishaaus Author
wrote...
12 years ago
a. T=.7o
t=.30


b. 53 students

c. 61 students
d 11 students b and c and subtract 125 should get you 11
wrote...
11 years ago
Hardy-Weinberg equilibrium equation:

p² + 2pq + q² = 1 (or TT + Tt + tt = 1)
p + q = 1 (T + t = 1)

In this problem: q² = tt =37/125 = 0.296

A) expected frequency for the t allele = q = √q² = √0.296 = 0.544
expected frequency for the T allele = p = 1 - q = 1 - 0.544 = 0.456

B) expected frequency for heterozygous genotype (Tt) = 2pq = 2*0.456*0.544=0.496
expected number of heterozygous students = 0.496 * 125 = 62 students

C) expected frequency for homozygous dominant genotype (TT) = p² = .456² = 0.208
expected number of homozygous dominant students = 0.208*125 = 26 students

D) You could check your answer for part b and c by mating the students who can taste PTC with the students who cannot to observe the phenotypes of the offspring. However, I personally think this is impossible.
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