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:
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.
followed by unequal crossing over may result in which of the following?
A)nondisjunction and aneuploidy
E)none of the above
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).