1. Testing the Neutral Mutation Hypothesis The neutral theory predicts that polymorphism within species is correlated positively with fixed differences between species Genes that exhibit many interspecific differences will also have high levels of intraspecific polymorphism. i.e.

2. Nonsynonymous Synonymous Fixed Differences Polymorphisms 21 26 45% 2 36 5.3% nonsynonymous fixed synonymous fixed nonsynonymous polymorphism synonymous polymorphism = McDonald-Krietman Test G6PDH from D. melanogaster and D. simulans. Eanes et al. 1993 Neutral Prediction: % nonsynonymous

3. Frequency 1.0 0 If most nonsynonymous substitutions are adaptive, then they will increase in frequency and be fixed more rapidly than neutral alleles. Time As a result, they spend less time in a polymorphic state, therefore contribute less to within species polymorphism. neutral allele advantageous allele

4. Nonsynonymous Synonymous Fixed Differences Polymorphisms 7 17 29% 2 42 4.5% Adh from D. melanogaster, D. simulans, and D. Yakuba MacDonald and Kreitman 1991 % nonsynonymous Another example (N = 6-12 alleles per species for the coding region).

5. t2t2 t3t3 t4t4 t5t5 Coalescent Process t m is time for coalescence from m to m-1 sequences Gene Tree

6. ab cd e fgh Coalescent Process The geneology of n sequences has 2(n-1) branches. Gene Tree n = number of external branches. n-2 are internal

7. How long will the coalescence process take? Simplest case: If pick two random gene copies, probability that the second is the same as the first is 1 / (2N). This is the probability that two alleles coalesce in previous generation. It follows that 1 - 1 / (2N) is the probability that two sequences were derived from different sequences in the preceding generation. Therefore, the probability that 2 sequences derived from the same ancestor 2 generations ago (grandparent) is 1 - 1 / (2N) x 1 / (2N). It can be shown that the probability that two sequences were derived from the same ancestor t generations ago is: [1 - 1 / (2N) t x (1 / (2N)] ~ (1 / (2N e (-t/(2N)

8. [1 - 1 / (2N) g-1 x (1 / (2N)] Because N is in denominator, the probability will depend on sample size Consider probability of common ancestry for: Generations agoProb(N=5) Prob(N=10) 1 0.400 0.200 2 0.320 0.182 3 0.256 0.162 It can be shown that the average time back to common ancestry of a pair of genes in a diploid population is 2Ne, and the average time back to common ancestry of all gene copies is 4Ne generations.

9. Small pop Large pop Time back to common ancestor

10. The average degree of relationship increases with time. All of the gene copies in a population can be traced back to a single ancestral gene. A population will eventually become monomorphic for one allele or another, with this probability determined by initial allele frequencies. Coalescence with no mutation

11. Coalescence with mutation If each lineage experiences  mutations per generation, then the number of base pair differences between them will be #dif = 2  t ca. If the average time to coalescence is 2N for two randomly chosen gene copies, then #dif = 2  (2N). Therefore, expect the average number of base pair differences between gene copies to be greater in a larger population.

12. Total length of branches of gene tree I + L = J Internal branches External branches + = Total time length Now consider mutation among branches during the coalescent process.  i ) +  e ) =  Mutations internal branches + = Total number of mutations in gene tree Mutations external branches In theory: total number of mutations  equals the number of segregating sites (K)

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

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

15. Fu and Li’s Test (1993):  i -  e / (a - 1) V[  i -  e / (a - 1) D = Using the difference in # mutations in gene tree to detect deviation from neutrality. Rationale: An equivalent number of mutations is expected between interior verses exterior branches of a neutral gene tree.

16. D = 0 neutral prediction D > 0balancing selection D < 0directional selection Few alleles at intermediate frequency Many low frequency, variable alleles  i  e ><><

17. 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

18. 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

19. 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

20. 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

21. 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.