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Question

# If the gene frequency for PTC nontasters is 0.4, then what will be the number of heterozygotes tasters in a population of 4000 ?

A
1920
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B
1080
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C
2520
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D
1440
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Solution

## The correct option is A 1920Since there are only two alleles, we will label them ‘A’ (PTC taster) and ‘a’ (non-PTC taster). The frequency of the dominant allele (A) is designated by p. The frequency of the recessive allele (a) is designated by q. In the Hardy-Weinberg equation (p2+2pq+q2=1),q2 is the frequency of the ‘aa’ genotype (non-PTC taster homozygous, in this case). Once we know q, we know p, because p + q = 1 Therefore, p = 1 – q, or p = 1 – 0.4 = 0.6 Now we know p and q (p = 0.6 and q = 0.4) The Hardy-Weinberg equation is: p2+2pq+q2=1 Therefore, frequency of heterozygotes (Aa) in the population is 2pq Put the values of p and q into this and you get: Frequency of heterozygotes =2∗p∗q=2∗0.6∗0.4=0.48 To be complete, we predict all the genotype frequencies to be: Frequency of AA=p2=(0.6)2=0.36 Frequency of Aa=2pq=2∗0.6∗0.4=0.48 Frequency of aa=q2=(0.4)2=0.16 In a population of 4000, multiplying every term by 4000 p2+2pq+q2=4000 p2=1440 q2=640 2pq=1920 Hence, the frequency of heterozygote tasters in a population of 4000 is 1920. The correct answer is option B.

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