Question
Is the hardy Weinberg equilibrium always equal to 1? True or false
Answers
GPT-4o mini
False. The Hardy-Weinberg equilibrium refers to a principle in population genetics that describes the expected frequencies of alleles and genotypes in a non-evolving population. The equations used in this principle are:
- \( p + q = 1 \) (where \( p \) is the frequency of one allele and \( q \) is the frequency of the other allele)
- \( p^2 + 2pq + q^2 = 1 \) (where \( p^2 \) is the frequency of the homozygous dominant genotype, \( 2pq \) is the frequency of the heterozygous genotype, and \( q^2 \) is the frequency of the homozygous recessive genotype)
The values \( p \) and \( q \) themselves are not always equal to 1; instead, they are the proportions of the alleles in the population, and they must sum to 1. The expected genotype frequencies calculated using these proportions also sum to 1, but that does not mean the frequencies themselves are always equal to 1.
- \( p + q = 1 \) (where \( p \) is the frequency of one allele and \( q \) is the frequency of the other allele)
- \( p^2 + 2pq + q^2 = 1 \) (where \( p^2 \) is the frequency of the homozygous dominant genotype, \( 2pq \) is the frequency of the heterozygous genotype, and \( q^2 \) is the frequency of the homozygous recessive genotype)
The values \( p \) and \( q \) themselves are not always equal to 1; instead, they are the proportions of the alleles in the population, and they must sum to 1. The expected genotype frequencies calculated using these proportions also sum to 1, but that does not mean the frequencies themselves are always equal to 1.