### Dummy

If I could remember the names of all these particles, I'd be a botanist... Attributed to Enrico Fermi.
I've mentioned that I'm taking a course in evolution. I've learned some things, including that I'm not cut out to be a Botanist. The prof favors a focus on minutiae on his exams, like the following:

Paracentric inversions
followed by unequal crossing over may result in which of the following?
A)nondisjunction and aneuploidy
B)replication slippage
C)homoploidy
D)reticulation
E)none of the above
I have included helpful links for those interested in parsing this. One minute per question is allowed, just in case it might take you a while to sort through the possibilities.

The correct answer, btw, is E)none of the above.

Yeah, I missed this one, among many others. My bad. See title above.

He also has some peculiar ideas about logic. Or maybe I mean that he is always right, even when he is wrong. Consider two alleles, or variant copies of the same gene, present with with frequencies p and q, respectively, with p + q = 1. Obviously, it follows that (p+q)^2 = p^2 + 2pq + q^2 = 1.

Here is another question:

Under what circumstances is a population in Hardy-Weinberg equilibrium?
A)when p^2+2pq+q^2 != 1
B)when p^2+2pq+q^2 = 1
C)when p^2+2pq+q^2 = observed allele frequencies
D)when p^2+2pq+q^2 = observed genotype frequencies
E)when p^2+2pq+q^2 != observed genotype frequencies.

Here I have used != to mean not equal since I'm not smart enough to figure out how to make ≠ sign. Oops! learned something

I say A) is trivially false, B)trivially true, and C), D), and E) are nonsensical.

What the Hardy-Weinberg equilibrium actually means for the frequencies of diploid genotypes with alleles A and a is that freq(AA) = p^2, freq(Aa) = 2 pq, and freq (aa) = q^2. This one I got "right", meaning that I correctly deduced prof's particular delusion D).