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I had asked the question day before in a population of 8000,80people can not recognize the taste of phenyl thio carb amide find out the number of homozygous tasters in a but the correct answer is 6480 how please explain me 

Asked by Sherbalatha2015 10th March 2016, 10:09 AM
Answered by Expert
Answer:

Hi,

Given is,

Population – 8000 and Non-tasters – 80

Allele for ability to taste PTC is dominant ‘T’.

Hence, tasters will be of two genotypes – Homozyogous dominant (TT) and Heterozygous dominant (Tt).

Non-tasters will have one genotype = Homozygous recessive (tt).

begin mathsize 9px style According space to space Hardy minus Weinberg space Equation comma space straight p squared space plus space 2 pq space plus space straight q squared space equals space 1 space space where comma space straight p space equals space frequency space of space homozygous space dominant space genotype space left parenthesis TT right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 2 pq space equals space frequency space of space heterozygous space dominant space genotype space left parenthesis Tt right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight q space equals space frequency space of space homozygous space recessive space genotype space left parenthesis tt right parenthesis space Non minus tasters space left parenthesis tt right parenthesis space equals space straight q 2 space equals space 80 over 8000 space equals space 0.01 space space space space space space space space space space space space space space space space space space space space space space space space space space space space square root of straight q space space equals space 0.1 because straight p space plus space straight q space equals space 1 straight p space equals space 1 space minus space straight q space equals space 1 space minus space 0.1 space equals space 0.9 straight p squared space equals space 0.81  2 pq space equals space 2 space cross times space 0.9 space cross times space 0.1 space equals space 0.18  therefore space straight p squared space plus space 2 pq space plus space straight q squared space equals space 1 space space space space 0.81 space plus space 0.18 space plus space 0.1 space equals space 1 space space space space 1 space equals space 1  end style

Out of 8000, frequency of non-tasters (tt) = 0.01

frequency of homozygous tasters (TT) = 0.81

frequency of heterozygous tasters (Tt) = 0.18

0.01 of 8000 are non-tasters = 8000

0.81 of 8000 are homozygous tasters = 8000 x 0.81 = 6480

0.18 of 8000 are heterozygous tasters = 8000 x 0.18 = 1440

Answered by Expert 10th March 2016, 6:08 PM
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