A population shows 81% have the dominant phenotype. If Hardy-Weinberg conditions are met, what will be the genotypic frequencies in the next generation?
______Homozygous dominant
_____ Heterozygous
______Homozygous recessive
A population shows 81% have the dominant phenotype. If Hardy-Weinberg conditions are met, what will be the genotypic frequencies in the next generation?
To determine the genotypic frequencies in the next generation, we need to use the Hardy-Weinberg equation. The Hardy-Weinberg equation states that in a population that meets certain conditions (such as no selection, migration, mutation, or genetic drift), the frequencies of alleles and genotypes remain constant from generation to generation.
The equation is given as follows:
p^2 + 2pq + q^2 = 1
Where:
- p^2 represents the frequency of homozygous dominant individuals,
- 2pq represents the frequency of heterozygous individuals, and
- q^2 represents the frequency of homozygous recessive individuals.
In this case, we are given that 81% of the population has the dominant phenotype. This means that the frequency of the dominant allele (p) is √(0.81) ≈ 0.9, and the frequency of the recessive allele (q) is 1 - p = 1 - 0.9 = 0.1.
Now, let's substitute these allele frequencies into the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
(0.9)^2 + 2(0.9)(0.1) + (0.1)^2 = 1
0.81 + 0.18 + 0.01 = 1
So, in the next generation, the genotypic frequencies will be approximately:
- Homozygous dominant: (0.9)^2 = 0.81 or 81%
- Heterozygous: 2(0.9)(0.1) = 0.18 or 18%
- Homozygous recessive: (0.1)^2 = 0.01 or 1%
Therefore, the genotypic frequencies in the next generation will be:
- Homozygous dominant: 81%
- Heterozygous: 18%
- Homozygous recessive: 1%