Midterm Test 2015

Mean after bonus = 18.1.

For each question, a brief explanation is provided, so that you can evaluate your own answer.  The number of students selecting various choices is given for answers chosen by 10 or more students (N = 79). 

1	Lamarck came up with evolution but attributed it to the effect of use and disuse.
2	Weismann showed that somatic cells have no influence on germ cells.
3	No evidence has ever been given for a vital force (22), and so it is not part of the current model.  The merging of Darwinian evolution with Mendelian genetics per se do not disprove a vital force (57).
4	While it is true that Darwin formulated pangenesis (14) and common descent (35), the reason he is recognized in the contet of the Modern Synthesis is his idea of natural selection, i.e., differential reproductive success (30).
5	The mean length of each pure line, not the whole population, is genetically based (18).  The population variance combined genetic and environmental variance (58).  This is the only question is the entire test that did not have a positive "point biserial correlation", a measure of whether a question predicts success on the test as a whole.
6	A self-cross of heterozygotes for two loci with additive alleles can produce five colour classes in their offspring, with a 1:4:6:4:1 ratio (64).  A cross between genotypes AABbCc and AaBBCc (14). would generate a similar distribution, but all genotypes would generate darker colours and there would be no whites.
7	Balance selectionists hold that variation is actively maintained (61).  Darwinians emphasize small, gradual changes (10).
8	We need a sample of at least 100 to conclude anything about P.99 (38) and this sample indicates that the population is monomorphic at the 0.95 level: both A and B are wrong (41).
9	The 0:1:49 ratio fits HW quite well (46) and so does not show a heterozygote deficit (27).
10	Heterozygosity here is 1/50 = 0.02 (51).  Given that ps is 1/100 = 0.01, the predicted number of slow/slow homozygotes  is 0.01^2 = 0.001.  In a sample of 100, that would be 0.1 individuals, not very likely (19).
11	χ^2 is not a probability (21).  It can be associated with the probability of a type I error (44).  Cochrane's rules apply to rejecting, but not accepting H0 (14).
12	The maximum heterozygosity with 2 alleles is 0.5 (27).  For 5 alleles, 3 of which have a frequency of 0.3, predicted homozygosity for these 3 alleles is 3 x 0.3^2 = 0.27 and for the remaining 2, it could approach 0.1^2 = 0.01.  The remainder is the predicted heterozygosity, 1 - 0.28 = 0.72 (37).  If one allele has p = 0.9, there should be at least 0.81 homozygotes, and no more than 0.19 heterozygotes (15).
13	1:8:16 is a perfect HW ratio.
14	This is a clear heterozygote deficit and could be due to assortative mating.
15	p = (nAA + nAa/2) / 200 = 2/200 = 0.01 (63), not 0.02 (14).
16	For perfect assortative mate choice to occur, there must be a direct correspondence between phenotype and genotype.  This can only occur when dominance is incomplete or when there is codominance (58).  With complete domiance (19), individuals with genotypes AA and Aa have the same phenotype.
17	The whole point of our lab on assortative mating is that it causes a loss of heterozygosity (53) without any change in allele frequencies (17).
18	The Wahlund effect is due to the fact that the heterozygosity predicted within demes and averaged out is always less than the heterozygosity predicted for the whole population (61).  Increased mating among related individuals could occur, but only if the demes are very small (13).
19	Equal p and q contribute to genotypic variance (15) and differences in fitness contribute to fitness variance (15).  A Hardy-Weinberg frequency of heterozygotes is irrelevant to variance in fitness (49).
20	The binomial equation ALWAYS adds up to 1.0 (14).  For selection to occur, mean population fitness must be less than 1.0 (57).
21	My 3466b students understand that fitness is not a tautology.  Too bad creationists don't seem to.
22	Codominance means that each genotype has a distinct phenotype (58).  Band may vary in intensity for various reasons, but that would not be relevant to codominance (20).
23	It is true that the relative fitnesses are 0.14, 0.14, and 1.0 (12).  Because the fitness of the heterozygote is the same as that of the homozygote, dominance is complete (27).  And so both A and B are correct: you were to select the single BEST choice (40).
24	The mutant allele is deleterious and recessive.
25	Selection due to overdominance is most rapid when the genetic variance is high but p is different from the equilibrium value (62).  When p is at equilibrium, the rate of selection is zero (10).
Bonus 26	Two thirds got it right.