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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Shweta Kumari posted an Question
February 25, 2023 • 15:04 pm
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30 points
30 points
- IIT JAM
- Biotechnology (BT)
Q.21 the inability in humans to taste capsaicin resides in a single gene difference between two alleles p and p. the allele p for tasting is dominant over the n
Q.21 The inability in humans to taste capsaicin resides in a single gene difference between two alleles P and p. The allele P for tasting is dominant over the nontasting allele. In a population of 400 individuals in Hardy-Weinberg equilibrium. 64 are nontasters. How many individuals are heterozygous for the gene? ptions 1 64 192 3. 128 4 144
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Nilanjana Sarkar![best-answer]()
Capsaicin taste gene possesses two alleles. 'P' is the dominant. Individuals possessing atleast one 'P' allele, i.e., the dominant allele shows the taster individuals. 'p' is the recessive allele. Homozygous individuals with 'p' allele are the nontasters for the capsaicin. It means 'pp' genotypes possessing individuals are the nontasters. In 400 individuals, 64 are the nontasters. so, pp= 64/400 so, p= 8/20 so, p=0.4 So, P=(1-0.4) So, P=0.6 So, the heterozygous individuals of this gene is = (2Pp)×400 ={(2×0.6×0.4)×400} =192. So, option 2 will be right answer. I'm attaching the supporting calculation here. If you still face problem feel free to ask.