Assuming that the conditions of the Hardy-Weinberg equilibrium are met, determine the genotype and phenotype frequencies in the second generation of a population containing 64% tasters and 36% nontasters
q2 = _________ Tasters ____________
q = ¬¬¬¬__________ nontasters __________
p = __________ Homozygous dominant __________
p2 = _________ Heterozygous __________
2pq = ¬¬¬¬_______¬_ Homozygous recessive __________
Need answers
To determine the genotype and phenotype frequencies in the second generation of a population, we can use the Hardy-Weinberg equations:
1. Calculate allele frequencies:
- Let p represent the frequency of the dominant allele (Tasters).
- Let q represent the frequency of the recessive allele (Nontasters).
- Since the population contains 64% tasters and 36% nontasters, we can calculate the allele frequencies as follows:
p = sqrt(0.64) ≈ 0.8 (Tasters)
q = 1 - p ≈ 0.2 (Nontasters)
2. Determine genotype frequencies:
- In the Hardy-Weinberg equilibrium, the genotype frequencies are given by the following equations:
p^2 = frequency of homozygous dominant (Tasters)
2pq = frequency of heterozygous individuals (Carriers)
q^2 = frequency of homozygous recessive (Nontasters)
- Plugging in the values we calculated earlier, we get:
p^2 = (0.8)^2 = 0.64 (Homozygous dominant - Tasters)
2pq = 2(0.8)(0.2) = 0.32 (Heterozygous - Carriers)
q^2 = (0.2)^2 = 0.04 (Homozygous recessive - Nontasters)
3. Determine phenotype frequencies:
- The phenotype frequencies can be determined based on the genotype frequencies:
Tasters = p^2 + 0.5(2pq) = 0.64 + 0.5(0.32) = 0.8 (or 80%)
Nontasters = q^2 + 0.5(2pq) = 0.04 + 0.5(0.32) = 0.2 (or 20%)
Therefore, in the second generation of the population, we have the following frequencies:
- q2 = 0.04
- Tasters = 80%
- q = 0.2
- Homozygous dominant = 64%
- p2 = 0.64
- Heterozygous = 32%
- 2pq = 0.32
- Homozygous recessive = 4%