1. 1 Population Genetics Basics

2. 2 Terminology review Allele Locus Diploid SNP

3. 3 Single Nucleotide Polymorphisms 00000101011 10001101001 01000101010 01000000011 00011110000 00101100110 Infinite Sites Assumption: Each site mutates at most once

4. 4 What causes variation in a population? Mutations (may lead to SNPs) Recombinations Other genetic events (gene conversion) Structural Polymorphisms

5. 5 Recombination 00000000 11111111 00011111

6. 6 Gene Conversion Gene Conversion versus crossover – Hard to distinguish in a population

7. 7 Structural polymorphisms Large scale structural changes (deletions/insertions/inversions) may occur in a population.

8. 8 Topic 1: Basic Principles In a ‘stable’ population, the distribution of alleles obeys certain laws – Not really, and the deviations are interesting HW Equilibrium – (due to mixing in a population) Linkage (dis)-equilibrium – Due to recombination

9. 9 Hardy Weinberg equilibrium Consider a locus with 2 alleles, A, a p (respectively, q) is the frequency of A (resp. a) in the population 3 Genotypes: AA, Aa, aa Q: What is the frequency of each genotype If various assumptions are satisfied, (such as random mating, no natural selection), Then P AA =p 2 P Aa =2pq P aa =q 2

10. 10 Hardy Weinberg: why? Assumptions: – Diploid – Sexual reproduction – Random mating – Bi-allelic sites – Large population size, … Why? Each individual randomly picks his two chromosomes. Therefore, Prob. (Aa) = pq+qp = 2pq, and so on.

11. 11 Hardy Weinberg: Generalizations Multiple alleles with frequencies – By HW, Multiple loci?

12. 12 Hardy Weinberg: Implications The allele frequency does not change from generation to generation. Why? It is observed that 1 in 10,000 caucasians have the disease phenylketonuria. The disease mutation(s) are all recessive. What fraction of the population carries the mutation? Males are 100 times more likely to have the “red’ type of color blindness than females. Why? Conclusion: While the HW assumptions are rarely satisfied, the principle is still important as a baseline assumption, and significant deviations are interesting.

13. 13 Recombination 00000000 11111111 00011111

14. 14 What if there were no recombinations? Life would be simpler Each individual sequence would have a single parent (even for higher ploidy) The relationship is expressed as a tree.

15. 15 The Infinite Sites Assumption 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 1 0 0 0 3 8 5 The different sites are linked. A 1 in position 8 implies 0 in position 5, and vice versa. Some phenotypes could be linked to the polymorphisms Some of the linkage is “destroyed” by recombination

16. 16 Infinite sites assumption and Perfect Phylogeny Each site is mutated at most once in the history. All descendants must carry the mutated value, and all others must carry the ancestral value i 1 in position i 0 in position i

17. 17 Perfect Phylogeny Assume an evolutionary model in which no recombination takes place, only mutation. The evolutionary history is explained by a tree in which every mutation is on an edge of the tree. All the species in one sub-tree contain a 0, and all species in the other contain a 1. Such a tree is called a perfect phylogeny.

18. 18 The 4-gamete condition A column i partitions the set of species into two sets i 0, and i 1 A column is homogeneous w.r.t a set of species, if it has the same value for all species. Otherwise, it is heterogenous. EX: i is heterogenous w.r.t {A,D,E} i A 0 B 0 C 0 D 1 E 1 F 1 i0i0 i1i1

19. 19 4 Gamete Condition – There exists a perfect phylogeny if and only if for all pair of columns (i,j), either j is not heterogenous w.r.t i 0, or i 1. – Equivalent to – There exists a perfect phylogeny if and only if for all pairs of columns (i,j), the following 4 rows do not exist (0,0), (0,1), (1,0), (1,1)

20. 20 4-gamete condition: proof Depending on which edge the mutation j occurs, either i 0, or i 1 should be homogenous. (only if) Every perfect phylogeny satisfies the 4- gamete condition (if) If the 4-gamete condition is satisfied, does a prefect phylogeny exist? i0i0 i1i1 i

21. 21 An algorithm for constructing a perfect phylogeny We will consider the case where 0 is the ancestral state, and 1 is the mutated state. This will be fixed later. In any tree, each node (except the root) has a single parent. – It is sufficient to construct a parent for every node. In each step, we add a column and refine some of the nodes containing multiple children. Stop if all columns have been considered.

22. 22 Inclusion Property For any pair of columns i,j – i < j if and only if i 1  j 1 Note that if i<j then the edge containing i is an ancestor of the edge containing i i j

23. 23 Example 1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 1 0 D 0 0 1 0 1 E 1 0 0 0 0 r A BCDE Initially, there is a single clade r, and each node has r as its parent

24. 24 Sort columns Sort columns according to the inclusion property (note that the columns are already sorted here). This can be achieved by considering the columns as binary representations of numbers (most significant bit in row 1) and sorting in decreasing order 1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 1 0 D 0 0 1 0 1 E 1 0 0 0 0

25. 25 Add first column In adding column i – Check each edge and decide which side you belong. – Finally add a node if you can resolve a clade r A B C D E 1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 1 0 D 0 0 1 0 1 E 1 0 0 0 0 u

26. 26 Adding other columns Add other columns on edges using the ordering property r E B C D A 1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 1 0 D 0 0 1 0 1 E 1 0 0 0 0 1 2 4 3 5

27. 27 Unrooted case Switch the values in each column, so that 0 is the majority element. Apply the algorithm for the rooted case

28. 28 Handling recombination A tree is not sufficient as a sequence may have 2 parents Recombination leads to loss of correlation between columns

29. 29 Linkage (Dis)-equilibrium (LD) Consider sites A &B Case 1: No recombination – Pr[A,B=0,1] = 0.25 Linkage disequilibrium Case 2:Extensive recombination – Pr[A,B=(0,1)=0.125 Linkage equilibrium AB0101000010101010AB010100001010101000

30. 30 Handling recombination A tree is not sufficient as a sequence may have 2 parents Recombination leads to loss of correlation between columns

31. 31 Recombination, and populations Think of a population of N individual chromosomes. The population remains stable from generation to generation. Without recombination, each individual has exactly one parent chromosome from the previous generation. With recombinations, each individual is derived from one or two parents. We will formalize this notion later in the context of coalescent theory.