1 / 32

L4: Counting Recombination events

This algorithm iteratively estimates the structure of populations by counting recombination events and allows for admixture mapping. It uses a sampling approach to estimate the admixture probability and assigns individuals to sub-populations based on linkage equilibrium and Hardy Weinberg equilibrium. The algorithm also provides estimates of recombination rates and uses lower bounds to compute the minimum number of recombination events in a population. The results on simulated and real data show the effectiveness of the algorithm in predicting population structure and admixture.

juanstewart
Download Presentation

L4: Counting Recombination events

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. L4: Counting Recombination events

  2. 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?

  3. Allowing for admixture • Define qi,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))

  4. 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

  5. Results on admixture prediction: simulated data

  6. 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.

  7. 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) Oceania Eurasia East Asia America Africa

  8. NJ versus Structure:thrush data • Objective function is different in standard clustering algorithms!

  9. 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

  10. Admixture mapping

  11. Estimating Recombination Rates

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

  13. Recombination hot-spots (fine scale)

  14. 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

  15. 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.

  16. The Infinite Sites Assumption & the 4 gamete condition 0 0 0 0 0 0 0 0 3 0 0 1 0 0 0 0 0 5 8 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 1 • 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

  17. 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.

  18. 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=i1 i2 i3 i4 i5 i6 ik=n • Can we compute maxP R(P) efficiently?

  19. The Rm bound

  20. Improved lower bounds • The Rm bound also gives a general technique for combining local lower bounds into an overall lower bound. • In the example, Rm=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

  21. Hudson and Kaplan: Idea 2 • Consider the history of individuals. Let Ht denote the number of distinct halotypes at time t • One of three things might happen at time t: • Mutation: Ht increase by at most 1 • Recombination: Ht increase by at most 1 • Coalescence: Ht does not increase

  22. The RH 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

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

  24. The Rs 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.

  25. a b c Rs bound: Observation I 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}) 0 0 0 : : 1 a

  26. Rs bound: Observation 2 • Redundant rows: If two rows h1 and h2 are identical, then • R(M) = R(M-{h1}) c r1 r2

  27. 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?

  28. Rs bound (Procedural) Procedure Compute_Rs(M) If  non-informative column s return (Compute_Rs(M-{s})) Else if  redundant row h return (Compute_Rs(M-{h})) Else return (1 + minh(Compute_Rs(M-{h}))

  29. Results

  30. Additional results/problems • Using dynamic programming, Rs 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)

  31. Underperforming lower bounds • Sometimes, Rs can be quite weak • An RI lower bound that uses intermediates can help (BB’05)

  32. LPL data set • 71 individuals, 9.7Kbp genomic sequence • Rm=22, Rh=70

More Related