A population is in Hardy-Weinberg equilibrium if the genotype frequencies of the population are equal to the expected frequencies of the next generation, that is, if:
The frequencies of the population are:
P= f(SS)= 0.9025
H= f(Ss)= 0.095
Q= f(ss)= 0.0025
The population of frogs has 8000 individuals, 20 of which are homozygous recessive for the trait "being spotted"
Assuming Hardy-Weinberg equilibrium you can calculate the frequency of the recessive allele as follows:
The genotypic frequency of the homozygous recessive is
Apply the square root to determine the value of q:
The allelic frequencies are complementary, which means that they add up to 1, knowing the value of q you can calculate the frequency of the dominant allele p:
Predict the genotype frequencies of the offspring on equilibrium:
Homozygous dominant SS
Heterozygous Ss
Homozygous recessive:
Compare the results:
As you can see the genotype frequencies of the current population are equal to the genotype frequencies of the original population, you can say that the population is in Hardy-Weinberg equilibrium.
Once the population reaches equilibrium it does not change, i.e. the frequencies are the same from one generation to the other, unless some outer factors, like deforestation, create a drastic change in the genetic pool (for example, by eliminating/ adding individuals or isolating small portions of the population)
On the other hand, if the population is evolving, i.e. the genotypic frequencies are changing, and the observed genotype and allele frequencies will change from one generation to the next one.
If you concluded that the population is in equilibrium, then it is not possible for it to be evolving as the frequencies do not change for a population in equilibrium as they do when the population is evolving.
This means that the population is not evolving.