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kirstens kirstens
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Posts: 7
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9 years ago
A dominant allele, T, codes for the ability to taste the compound phenylthiocarbamide (PTC). People who are homozygous for the recessive allele, t, are unable to taste PTC. In a genetics class of 125 students, 28 cannot taste the PTC. How many students would you expect to be heterozygous for the tasting gene?

I answered : 0.83

A dominant allele, T, codes for the ability to taste the compound phenylthiocarbamide (PTC). People who are homozygous for the recessive allele, t, are unable to taste PTC. In a genetics class of 105 students, and 9 cannot taste the PTC. Calculate the expected frequencies of the t allele in the student population. 1 mark
Your Answer:
0.09

A dominant allele, T, codes for the ability to taste the compound phenylthiocarbamide (PTC). People who are homozygous for the recessive allele, t, are unable to taste PTC. In a genetics class of 165 students, and 19 cannot taste the PTC. Calculate the expected frequencies of the T allele in the student population. 1 mark
Your Answer: 0.88

In a randomly mating population of mice, 2 out of every  100 mice born have white fur, a recessive trait. Calculate the heterozygous frequency for the population.  1 mark
Your Answer:
0.24



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wrote...
Educator
9 years ago
This topic should help: [Solved] A dominant allele, T, codes for the ability to taste the compound PTC...
wrote...
9 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 dominant for the tasting gene?
D) How could you check your answers for parts (b) and (c)

Answer:

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.
wrote...
8 years ago
Is the fourth question correct? The one with the mice?
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