Name AP Biology Haury 4. widow\'s peak is dominant to lacking widow\'s peak. In
ID: 186378 • Letter: N
Question
Name AP Biology Haury 4. widow's peak is dominant to lacking widow's peak. In a population of 1000 people, 510 show the dominant phenotype. How many individuals would you expect of each of the three possible genotypes for this trait? 5. In the s a chemist in the Dupont Lab found that some individuals had the ability to taste a chemical called phenylthiocarbamide (PTC), while others did not. Later it was found that the ability to taste this chemical is determined by a dominant allele. In study of 1600 parents in Ohio it was found that 461 were non-tasters. What would you expect were the percentages of the three possible genotypes in this case?Explanation / Answer
Answer:
4).
Dominant phenotype = 510
Recessive phenotype = 1000-510 = 490
The frequency of recessive genotype (qq or q^2)= 490/1000 = 0.49
The frquency of recessive allele (q) = Sqrt of q^2 = 0.7
The frquency of dominant allele (p) = 1-q = 1-0.7 = 0.3
The frquency of homozygous dominant genotype (p^2) = 0.3 * 0.3 = 0.09
The frequency of heterozygoud dominant genotype (2pq) = 2 * 0.3 * 0.7 = 0.42
Homozygous dominant individuals = 0.09 * 1000 = 90
Heterozygous dominant individuals = 0.42 * 1000 = 420
Recessive individuals = 0.49 * 1000 = 490
4).
The frequency of non-tasters(recessive) = q^2 = 461/1600 = 0.29
The frequency of recessive allele (q) = Sqrt of 0.29 = 0.54
The frequency of dominant allele (p) = 1-0.54= 0.46
The frequency of homozygous (taster) dominant genotype (pp) = 0.21
The frequency of heterozygous (taster) dominant genotype (2pq) = 2 * 0.46 * 0.54 = 0.50
qq= 0.29 = 29%
pp = 0.21 = 21%
pq = 0.50 = 50%
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