Testing for Selective Neutrality Using the difference in estimates of polymorphism to detect deviation from neutrality. Tajima’ s Test (1989): P- K / a D = V(P- K/a) Normalizing factor Rationale: Pand K are differentially influenced by the frequency of alleles.

P K/a Few alleles at intermediate frequency > < Many low frequency, variable alleles D = 0 neutral prediction D > 0 balancing selection D < 0 directional selection

Gene genealogies under no selection, positive selection, balancing selection, and background selection. No Selection : 7 neutral mutations accumulate since the time of the last common ancestor. D = 0

Consider the Effects of Selection on Neutral Sites Linked to a Selected Site Positive Selection : neutral variation at linked sites will be eliminated (swept away) as the advantageous allele quickly is fixed in the population. This process is also called hitch-hiking. Time D < 0

Consider the Effects of Selection on Neutral Sites Linked to a Selected Site Balancing Selection : neutral variation at linked sites accumulates during the long period of time that both allele lineages are maintained. Time D > 0

Consider the Effects of Selection on Neutral Sites Linked to a Selected Site Background Selection : gene lineages become extinct not only by chance, but because of deleterious mutations to which they are linked, which eliminates some gene copies. Time D < 0

Problem: Background selection and hitchhiking are contrasting processes that lead to the same pattern. How to differentiate? Dramatic examples of reduced polymorphism=hitchhiking. Less dramatic examples=background selection.

Genetic Linkage

Two pops may have the same allele frequencies but different chromosome frequencies.

Conditions for Linkage Equilibrium (Two locus case) Frequency of B is the same on chromosome A and a Frequency of haplotypes AB, ab, Ab, aB can be calculated from allele frequencies A=p, a=q, B=t, and b=s. f(AB) = pt f(ab) = qs f(Ab) = ps f(aB) = pt 3) Coefficient of linkage disequilibrium (D) = 0 f(AB) x f(ab) - f(Ab) x f(aB) = D ps x qt - pt x qs = 0 (ranges from -0.25 to +0.25)

Hardy Weinberg Principle for Two Loci Chromosome frequencies remain unchanged across generations if loci are in linkage equilibria. If loci are in linkage disequilibria, the chromosome frequencies will move closer to linkage equilibrium each generation. What Creates Linkage Disequilibrium in a Population? selection, drift, and population admixture

Effects of Selection on Chromosome Frequencies ab/ab = body size of value 10 each additional A or B adds 1 value predators eat all individuals of size 12 or less 65.28% survive

Locus A and B are in disequilibrium among the survivors Freq a = (0.1536 + 0.1536) / .6528)/2 = 0.24 Freq b = (0.0576 + 0.0576) / .6528)/2 = 0.09 Freq ab = .24 x .09 = .02 however! 65.28% survive Locus A and B are in disequilibrium among the survivors