Um, technically you need a population of infinite size of randomly breeding individuals for HE equilibrium, but we'll overlook that for the moment... (usually 10,000 is the considered minimum effective size for HE -- otherwise you start running into genetic drift issues.)
The frequency of phenotypes of a simple dominant and recessive gene in a population are p squared + 2 pq + q squared... you can see that ratio when you do a simple heterozygous cross with a punnet square;
| A | a
---------------
A | AA | Aa
---------------
a | Aa | aa
Note that the frequency of A and a in this population is 0.5 -- p=0.5 and q =0.5. If the frequency of A is 0.5, then the frequency of the AA genotype is 0.5 x 0.5 = 0.25 (or 1/4) (p squared) The frequency of the heterozygote is 2 x p x q = 2 x 0.5 x 0.5 = 0.5 (or 1/2). The frequency of aa is q squared, or 0.5 x 0.5 = 0.25
Your problem involves a population... 4000 individuals, of which you know 360 are homozygous for bent antennae -- that number is your q squared value.
360/4000 = .09 = frequency of q squared = frequency of homozygous recessive in the population.
To find the frequency of the allele, you need to take the square root of q squared, which in this case is [sqrt 0.09] = 0.3 -- so 30% of the alleles in this population are the recessive allele, a.
Now, how are you going to figure out the frequency of the allele A in the population? If 30% of the alleles are a, then 70% must be A, right? (1 - 0.3 = 0.7)
Last part of the question: when there is no selection in a population at Hardy Weinberg equilibrium, what happens to the allele frequency? Nothing! After 6 generations, q still equals 0.3
(Or at least it would if the population had a larger effective size...)
More here:
http://en.wikipedia.org/wiki/Hardy-weinberg_equilibrium and
http://www.uwyo.edu/dbmcd/molmark/lect07/lect7.html
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