1. March 2006Vineet Bafna CSE280b: Population Genetics Vineet Bafna/Pavel Pevzner www.cse.ucsd.edu/classes/sp06/cse280b

2. March 2006Vineet Bafna Population Sub-structure

3. March 2006Vineet Bafna Population sub-structure can increase LD Consider two populations that were isolated and evolving independently. They might have different allele frequencies in some regions. Pick two regions that are far apart (LD is very low, close to 0) 0.. 1 0.. 0 1.. 1 0.. 1 1.. 0 0.. 0 1.. 1 1.. 0 Pop. A Pop. B p 1 =0.1 q 1 =0.9 P 11 =0.1 D=0.01 p 1 =0.9 q 1 =0.1 P 11 =0.1 D=0.01

4. March 2006Vineet Bafna Recent ad-mixing of population If the populations came together recently (Ex: African and European population), artificial LD might be created. D = 0.15 (instead of 0.01), increases 10-fold This spurious LD might lead to false associations Other genetic events can cause LD to arise, and one needs to be careful 0.. 1 0.. 0 1.. 1 0.. 1 1.. 0 0.. 0 1.. 1 1.. 0 Pop. A+B p 1 =0.5 q 1 =0.5 P 11 =0.1 D=0.1-0.25=0.15

5. March 2006Vineet Bafna Determining population sub-structure Given a mix of people, can you sub-divide them into ethnic populations. Turn the ‘problem’ of spurious LD into a clue. – Find markers that are too far apart to show LD – If they do show LD (correlation), that shows the existence of multiple populations. – Sub-divide them into populations so that LD disappears.

6. March 2006Vineet Bafna Determining Population sub-structure Same example as before: The two markers are too similar to show any LD, yet they do show LD. However, if you split them so that all 0..1 are in one population and all 1..0 are in another, LD disappears 0.. 1 0.. 0 1.. 1 0.. 1 1.. 0 0.. 0 1.. 1 1.. 0

7. March 2006Vineet Bafna Iterative algorithm for population sub- structure Define N = number of individuals (each has a single chromosome) k = number of sub-populations. Z  {1..k} N is a vector giving the sub-population. – Z i =k’ => individual i is assigned to population k’ X i,j = allelic value for individual i in position j P k,j,l = frequency of allele l at position j in population k

8. March 2006Vineet Bafna Example Ex: consider the following assignment P 1,1,0 = 0.9 P 2,1,0 = 0.1 0.. 1 0.. 0 1.. 1 0.. 1 1.. 0 0.. 0 1.. 1 1.. 0 1111111111222222222211111111112222222222

9. March 2006Vineet Bafna Goal X is known. P, Z are unknown. The goal is to estimate Pr(P,Z|X) Various learning techniques can be employed. – max P,Z Pr(X|P,Z) (Max likelihood estimate) – max P,Z Pr(X|P,Z) Pr(P,Z) (MAP) – Sample P,Z from Pr(P,Z|X) Here a Bayesian (MCMC) scheme is employed to sample from Pr(P,Z|X). We will only consider a simplified version

10. March 2006Vineet Bafna Algorithm:Structure Iteratively estimate – (Z (0),P (0) ), (Z (1),P (1) ),.., (Z (m),P (m) ) After ‘convergence’, Z (m) is the answer. Iteration – Guess Z (0) – For m = 1,2,.. Sample P (m) from Pr(P | X, Z (m-1) ) Sample Z (m) from Pr(Z | X, P (m) ) How is this sampling done?

11. March 2006Vineet Bafna Example Choose Z at random, so each individual is assigned to be in one of 2 populations. See example. Now, we need to sample P (1) from Pr(P | X, Z (0) ) Simply count N k,j,l = number of people in pouplation k which have allele l in position j p k,j,l = N k,j,l / N 0.. 1 0.. 0 1.. 1 0.. 1 1.. 0 0.. 0 1.. 1 1.. 0 1221121212122112122112211212121221121221

12. March 2006Vineet Bafna Example N k,j,l = number of people in population k which have allele l in position j p k,j,l = N k,j,l / N k,j,* N 1,1,0 = 4 N 1,1,1 = 6 p 1,1,0 = 4/10 p 1,2,0 = 4/10 Thus, we can sample P (m) 0.. 1 0.. 0 1.. 1 0.. 1 1.. 0 0.. 0 1.. 1 1.. 0 1221121212122112122112211212121221121221

13. March 2006Vineet Bafna Sampling Z Pr[Z 1 = 1] = Pr[”01” belongs to population 1]? We know that each position should be in linkage equilibrium and independent. Pr[”01” |Population 1] = p 1,1,0 * p 1,2,1 =(4/10)*(6/10)=(0.24) Pr[”01” |Population 2] = p 2,1,0 * p 2,2,1 = (6/10)*(4/10)=0.24 Pr [Z 1 = 1] = 0.24/(0.24+0.24) = 0.5 Assuming, HWE, and LE

14. March 2006Vineet Bafna Sampling Suppose, during the iteration, there is a bias. Then, in the next step of sampling Z, we will do the right thing Pr[“01”| pop. 1] = p 1,1,0 * p 1,2,1 = 0.7*0.7 = 0.49 Pr[“01”| pop. 2] = p 2,1,0 * p 2,2,1 =0.3*0.3 = 0.09 Pr[Z 1 = 1] = 0.49/(0.49+0.09) = 0.85 Pr[Z 6 = 1] = 0.49/(0.49+0.09) = 0.85 Eventually all “01” will become 1 population, and all “10” will become a second population 0.. 1 0.. 0 1.. 1 0.. 1 1.. 0 0.. 0 1.. 1 1.. 0 1112121211222122122111121212112221221221

15. March 2006Vineet Bafna Allowing for admixture Define q i,k as the fraction of individual i that originated from population k. Iteration – Guess Z (0) – For m = 1,2,.. Sample P (m),Q (m) from Pr(P,Q | X, Z (m-1) ) Sample Z (m) from Pr(Z | X, P (m),Q (m) )

16. March 2006Vineet Bafna Estimating Z (admixture case) Instead of estimating Pr(Z(i)=k|X,P,Q), (origin of individual i is k), we estimate Pr(Z(i,j,l)=k|X,P,Q) i,1 i,2 j

17. March 2006Vineet Bafna Results on admixture prediction

18. March 2006Vineet Bafna Results: Thrush data For each individual, q(i) is plotted as the distance to the opposite side of the triangle. The assignment is reliable, and there is evidence of admixture.

19. March 2006Vineet Bafna Population Structure 377 locations (loci) were sampled in 1000 people from 52 populations. 6 genetic clusters were obtained, which corresponded to 5 geographic regions (Rosenberg et al. Science 2003) Africa EurasiaEast Asia America Oceania

20. March 2006Vineet Bafna Population sub-structure:research problem Systematically explore the effect of admixture. Can admixture be predicted for a locus, or for an individual The sampling approach may or may not be appropriate. Formulate as an optimization/learning problem: – (w/out admixture). Assign individuals to sub-populations so as to maximize linkage equilibrium, and hardy weinberg equilibrium in each of the sub-populations – (w/ admixture) Assign (individuals, loci) to sub-populations

21. March 2006Vineet Bafna Allowing for admixture Define q i,k as the fraction of individual i that originated from population k. Iteration – Guess Z (0) – For m = 1,2,.. Sample P (m),Q (m) from Pr(P,Q | X, Z (m-1) ) Sample Z (m) from Pr(Z | X, P (m),Q (m) )

22. March 2006Vineet Bafna Estimating Z (admixture case) Instead of estimating Pr(Z(i)=k|X,P,Q), (origin of individual i is k), we estimate Pr(Z(i,j,l)=k|X,P,Q) i,1 i,2 j

23. March 2006Vineet Bafna Results on admixture prediction: simulated data

24. March 2006Vineet Bafna Results: Thrush data For each individual, q(i) is plotted as the distance to the opposite side of the triangle. The assignment is reliable, and there is evidence of admixture.

25. March 2006Vineet Bafna Population Structure 377 locations (loci) were sampled in 1000 people from 52 populations. 6 genetic clusters were obtained, which corresponded to 5 geographic regions (Rosenberg et al. Science 2003) Africa EurasiaEast Asia America Oceania

26. March 2006Vineet Bafna NJ versus Structure:thrush data Objective function is different in standard clustering algorithms!

27. March 2006Vineet Bafna Population sub-structure:research problem Systematically explore the effect of admixture. Can admixture be predicted for a locus, or for an individual The sampling approach may or may not be appropriate. Formulate as an optimization/learning problem: – (w/out admixture). Assign individuals to sub-populations so as to maximize linkage equilibrium, and hardy weinberg equilibrium in each of the sub-populations – (w/ admixture) Assign (individuals, loci) to sub-populations

28. March 2006Vineet Bafna Admixture mapping

29. March 2006Vineet Bafna Estimating Recombination Rates

30. March 2006Vineet Bafna Recombination in human chromosome 22 (Mb scale) Q: Can we give a direct count of the number of the recombination events? Dawson et al. Nature 2002

31. March 2006Vineet Bafna Recombination hot-spots (fine scale)

32. March 2006Vineet Bafna Recombination rates (chimp/human) Fine scale recombination rates differ between chimp and human The six hot-spots seen in human are not seen in chimp

33. March 2006Vineet Bafna Combinatorial Bounds for estimating recombination rate Recall that expected #recombinations =  log n Procedure Generate N random ARGs that results in the given sample Compute mean of the number of recombinations Alternatively, generate a summary statistic s from the population. For each , generate many populations, and compute the mean and variance of s (This only needs to be done once). Use this to select the most likely  What is the correct summary statistic? Today, we talk about the min. number of recombination events as a possible summary statistic. It is not the most natural, but it is the most interesting computationally.

34. March 2006Vineet Bafna The Infinite Sites Assumption & the 4 gamete condition 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 Consider a history without recombination. No pair of sites shows all four gametes 00,01,10,11. A pair of sites with all 4 gametes implies a recombination event

35. March 2006Vineet Bafna Hudson & Kaplan Any pair of sites (i,j) containing 4 gametes must admit a recombination event. Disjoint (non-overlapping) sites must contain distinct recombination events, which can be summed! This gives a lower bound on the number of recombination events. Based on simulations, this bound is not tight.

36. March 2006Vineet Bafna Myers and Griffiths’03: Idea 1 Let B(i,j) be a lower bound on the number of recombinations between sites i and j. 1=i 1 i 2 i 3 i 4 i 5 i 6 i k =n Can we compute max P R(P) efficiently?

37. March 2006Vineet Bafna The R m bound

38. March 2006Vineet Bafna Improved lower bounds The R m bound also gives a general technique for combining local lower bounds into an overall lower bound. In the example, R m =2, but we cannot give any ARG with 2 recombination events. Can we improve upon Hudson and Kaplan to get better local lower bounds? 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

39. March 2006Vineet Bafna Myers & Griffiths: Idea 2 Consider the history of individuals. Let H t denote the number of distinct haplotypes at time t One of three things might happen at time t: – Mutation: H t increase by at most 1 – Recombination: H t increase by at most 1 – Coalescence: H t does not increase

40. March 2006Vineet Bafna The R H bound 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 Ex: R>= 8-3-1=4

41. March 2006Vineet Bafna R H bound In general, R H can be quite weak: – consider the case when S>H However, it can be improved – Partitioning idea: sum R H over disjoint intervals – Apply to any subset of columns. Ex: Apply R H to the yellow columns 000000000000000 000000000000001 000000010000000 000000010000001 100000000000000 100000000000001 100000010000000 111111111111111 (BB’05)

42. March 2006Vineet Bafna The R s bound Compute the minimum number of recombination events R in any ARG. Note that, we do not explicitly construct the ARG. Consider a matrix with M with H rows and S columns. – The rows correspond to haplotypes. – Columns correspond to sites.

43. March 2006Vineet Bafna R s bound: Observation I abc 000::1000::1 a s  Non-informative column: If a site contains at most one 1, or one 0, then in any history, it can be obtained by adding a mutation to a branch.  EX: if a is the haplotype containing a 1, It can simply be added to the branch without increasing number of recombination events  R(M) = R(M-{s})

44. March 2006Vineet Bafna R s bound: Observation 2 Redundant rows: If two rows h 1 and h 2 are identical, then – R(M) = R(M-{h 1 }) r1r1 r2r2 c

45. March 2006Vineet Bafna Rs bound: Observation 3 Suppose M has no non- informative columns, or redundant rows. – Then, at least one of the haplotypes is a recombinant. – There exists h s.t. R(M) = R(M-{h})+1 – Which h should you choose?

46. March 2006Vineet Bafna R s bound (Procedural) Procedure Compute_R s (M) If  non-informative column s return (Compute_R s (M-{s})) Else if  redundant row h return (Compute_R s (M-{h})) Else return (1 + min h (Compute_R s (M-{h}))

47. March 2006Vineet Bafna Results

48. March 2006Vineet Bafna Additional results/problems Using dynamic programming, R s can be computed in 2^n poly(mn) time. Also, Rs can be augmented to handle intermediates. Are there poly. time lower bounds? – The number of connected components in the conflict graph is a lower bound (BB’04). Fast algorithms for computing ARGs with minimum recombination. – Poly. Time to get ARG with 0 recombination – Poly. Time to get ARGs that are galled trees (Gusfield’03)

49. March 2006Vineet Bafna Underperforming lower bounds Sometimes, R s can be quite weak An R I lower bound that uses intermediates can help (BB’05)

50. March 2006Vineet Bafna LPL data set 71 individuals, 9.7Kbp genomic sequence – R m =22, R h =70

51. March 2006Vineet Bafna