Answer
This population could be evolving as the described genotype distribution at this locus is not at Hardy-Weinberg equilibrium.
Work Step by Step
To determine this from the given data, we calculate whether the population is at Hardy-Weinberg equilibrium. Equilibrium would indicate that the population is not evolving.
120 organisms with two alleles each is a total of 240 alleles.
124 are V alleles.
116 are v alleles.
For V:
$p = \frac{124}{240}=0.52$
From v we can do two things: we can calculate the frequency as above, or we can use the above-calculated frequency of V. Since the frequencies of both V and v must add to one:
$1-0.52=0.48$
We can then use these allelic frequencies to determine if the population is at Hardy-Weinberg equilibrium. To do this, we use the above-calculated frequencies to determine the expected equilibrium genotype distribution.
For VV:
$p^2=0.52\times0.52=0.27$
$120\times0.27=32$ organisms of genotype VV
For Vv:
$2pq=0.52\times0.48=0.5$
$120\times0.5=60$ organisms of genotype Vv
For vv:
$q^2=0.48\times0.48=0.23$
$120\times0.23=28$ organisms of genotype vv
Since the actual observed genotypes in this population do not match the expected numbers at equilibrium, this population is not at Hardy-Weinberg equilibrium and should be evolving.
