Now, let's say there are only two possibilities (brown or blue). We will call the frequency of the brown allele in the population p and the frequency of the blue allele in the population q. Obviously, because there are only two possibilities, p + q must equal one (or 100%). If p is .1, then q must be .9, and so forth.
However, you get two chromosomes, so you have two alleles. Because there are only 2 alleles, you could only be BB, Bb, or bb. To express this mathematically, (p + q) = 1, therefore (p + q)^2 =
1^2, which in expanded form is p^2 + 2 pq + q^2 = 1. p squared (p^2) is the probability that an individual will have two B alleles ( be BB), q squared (q^2) is the probability that an individual will have two b alleles (bb) and 2 pq is the probability that an individual will be a hybrid (Bb).
Let's look at an example. If the frequency of the brown eyed allele and the blue eyed allele are the same, they would be 0.5. (Because 0.5 + 0.5 = 1). Expanded, you would have a population of about 0.25 BB, 0.5 Bb and 0.25 bb.
If given population numbers for each genotype, if you have to figure out the frequency of either gene, you would take the square root of the population numbers of individuals with either of the homozygous traits.
Let's look at your teacher's example. Is there any combination of values for p and q in which you could get a value of 2pq that is less than p^2 and less than q^2? Remember that p + q must equal 1. If p is greater than q, then p^2 must be larger than 2 pq. If q is greater than p, then q^2 must also be larger than 2 pq. And (as in our example above), if p = q, then 2 pq MUST be larger than either homozygous state.
There's a good you tube video on the Hardy Weinberg equilibrium that explains all the assumptions necessary. Look at: http://www.youtube.com/watch?v=4Kbruik_L…