With me so far?
So to get your A allele frequency, count up the number of A alleles in your population and divide by the total number of alleles.
P = (2 * 50 + 1 * 100) / 400 = 0.5
> Q= frequency of a
Do the same analysis we did for the A allele above.
You end up with
Q = (1 * 100 + 2 * 50) / 400 = 0.5
> What are frequencies of genotypes?
> What are frequencies of alleles?
Um, we already did those.
> Is this kennel right now in Hardy-Weinberg balance? why or why not?
Um, not a great question. To be in H-W equilibrium you need a huge population to prevent genetic drift from happening. If you have a huge population, you don't get random mating (you are more likely to marry someone from your home town than you are to marry someone from New Delhi). So no population is ever really in H-W equilibrium in real life.
But that's not what your teacher wants you to do.
Your teacher wants you to plug in the p and q you found above into the Hardy-Weinberg equations.
So let's do that.
p + q = 1
0.5 + 0.5 = 1, Check!
p^2 + 2pq + q^2 = 1
0.25 + 0.5 + 0.25 = 1 Check!
Using these equations, it does indeed appear that your population is in H-W equilibrium for the gene you're interested in.
> What will be the frequencies of genotypes in next generation under random breeding conditions and other Hardy-Weinberg assumptions?
I think you'll still have
25% AA, 50% Aa, and 25% aa
since you know that in H-W equilibrium, allele frequencies do not change in the population, and random mating should give you this. I think.