PTC tasting observations # Students Tasters 14 Non-tasters 7 Total 21 Fraction o
ID: 183694 • Letter: P
Question
PTC tasting observations
# Students
Tasters
14
Non-tasters
7
Total
21
Fraction of non-tasters in the class (decimal form)
number of non-tasters ÷ total number of students = ___________ = tt (homozygous recessive) = q2
t = square root of q2 = q = ___________ T (dominant) = 1 – q = p = _____________
The Hardy-Weinberg equation is: p2 + 2pq + q2 = 1
In this equation, p is the frequency of the T allele and q is the frequency of the t allele Which means that:
p2 is the frequency of the homozygous genotype TT 2pq is the frequency of the heterozygous genotype Tt q2 is the frequency of the homozygous genotype tt
Verify the Hardy-Weinberg equation for the PTC tasting results:
_______ + ________ + ________ = __________
(p2) (2pq) (q2) (1)
PTC tasting observations
# Students
Tasters
14
Non-tasters
7
Total
21
Explanation / Answer
In population,PTC tasting is dominant character while non tasters are recessive in population.
Fraction of non tasters=7÷21=0.33(tt)
q2=0.33
q=0.33=0.574(t)
p=1-0.574=0.426(T)
2pq=2*0.574*0.426=.489(Tt)
p2=0.4262=.181(TT)
According to hardy Weinberg principle
p2+2pq+q2=1
0.181+0.489+0.33=1
Hence hardy Weinberg principle is verified
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