Hardy-Weinberg Equilibrium - ScienceAid
a
q=0.04; p=0.16
b
q=0.8; p=0.2
c
q=0.2; p=0.8
d
q=0.1; p=1.0
The Hardy-Weinberg Equilibrium is a principle in population genetics that states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.
To determine if a population is in Hardy-Weinberg Equilibrium, we can calculate the expected genotype frequencies using the allele frequencies.
The equation to calculate genotype frequencies under Hardy-Weinberg Equilibrium is:
p^2 + 2pq + q^2 = 1
Where:
- p^2 represents the frequency of the homozygous dominant genotype (AA)
- 2pq represents the frequency of the heterozygous genotype (Aa)
- q^2 represents the frequency of the homozygous recessive genotype (aa)
- 1 represents the total frequency of all genotypes in the population
Let's calculate the expected genotype frequencies for each scenario:
a) q=0.04; p=0.16
- p^2 = 0.16 x 0.16 = 0.0256
- 2pq = 2 x 0.16 x 0.04 = 0.0128
- q^2 = 0.04 x 0.04 = 0.0016
b) q=0.8; p=0.2
- p^2 = 0.2 x 0.2 = 0.04
- 2pq = 2 x 0.2 x 0.8 = 0.32
- q^2 = 0.8 x 0.8 = 0.64
c) q=0.2; p=0.8
- p^2 = 0.8 x 0.8 = 0.64
- 2pq = 2 x 0.8 x 0.2 = 0.32
- q^2 = 0.2 x 0.2 = 0.04
d) q=0.1; p=1.0
- p^2 = 1.0 x 1.0 = 1.0
- 2pq = 2 x 1.0 x 0.1 = 0.2
- q^2 = 0.1 x 0.1 = 0.01
Based on these calculations, we can see that scenarios (a), (b), and (c) do not meet Hardy-Weinberg Equilibrium conditions because the genotype frequencies do not add up to 1. Scenario (d) does meet Hardy-Weinberg Equilibrium conditions because the genotype frequencies add up to 1.