Combinatorial Approaches to Haplotype Inference

1. Combinatorial Approaches to Haplotype Inference Dan Gusfield CS, UC Davis

2. Three topics • Super-charging Clark’s method of 1990 - Two ideas: Min haps selection + Consensus • Follow-on to Haplotyping By Perfect Phylogeny - a more biological view, and phase transitions, recombinations (?) • Computing True Parsimony Solutions, Clark/Parsimony hybrids - Integer Programming

3. Topic I: Supercharging Clark’s Method S. Orzack, D. Gusfield, V. Stanton Genetics, in revision

4. Generic Clark Method Basic Step: Given a known haplotype H (original homozygote or single-site heterozygote, or previously inferred), and an unresolved genotype G, if G can be explained by H and another vector H’, then call H’ a known haplotype, available for additional inferrals. example: H 0 1 0 0 1 G 2 1 0 2 2 G is “resolved” by H and H’ ------------------ H’ 1 1 0 1 0 In a single run, repeat the basic step until stuck - resolve as many genotypes as possible in the data. Clark (1990) Randomize choices, and do the computations many times to find an execution (run) that explains the most genotypes.

5. Many variations of Clark • Variations based on which parts are randomized. • We closely examine eight variations on a real data set. Variation 1 randomizes every decision - probably more than Clark originally intended. • Truth in advertising - we implemented our own Clark versions - did not actually use Clark’s software.

6. Super-charging Clark’s method Why? The reported accuracy is significantly less than that of PHASE and HAPLOTYPER. Answers: 1) Scale 2) Deeper insight

7. Supercharging Clark’s Method • How? Run many times (10,000) to get (generally) a huge range of solutions, often 10,000 different solutions. This is a good thing! • Select the (few - in the 10’s) runs that use the smallest number of distinct haplotypes over the 10,000 runs. Let S be that set of runs. • Vote - for each individual find the haplotype pair used most often in S. This gives the Consensus solution.

8. APOE locus 80 individuals, 9 polymorphic sites, 47 ambiguous after homozygotes and single-site heterozygotes identified - lab determined haplotypes Results on Lab Determined Data

9. Variation 1 Average correct over all 10,000 runs: 29.0 24 runs that use 20 haplotypes (smallest observed) or 21 distinct haplotypes Average correct over those 24 runs: 36.1 0.9987 percentile of all the 10,000 runs Correct in the Consensus solution from those 24 runs: 39 which is equal to the best of the 10,000 executions

10. Variation 4 Average over all 10,000 runs: 18.3 Consensus number of correct: 42

11. Consensus Thresholds The calls that are made with high frequency have high reliability. We observed few or no false calls among the haplotype pairs that were called with high frequency (85% up). This identifies a subset of calls that can be believed with high confidence, and a subset that has to be determined by other means. Bottom-up method to molecularly determine the haplotypes.

12. Comparison with Phase and Haplotyper Phase consistently gets 42 correct. 1 false call with confidence value > 0.9 Haplotyper: high variation.

13. distribution of high confidence wrong calls distribution of calls that don’t explain the genotype distribution of accuracy

14. Topic 2: Haplotyping as Perfect Phylogeny

15. The Perfect Phylogeny Model We assume that the evolution of extant haplotypes can be displayed on a rooted, directed tree, with the all-0 haplotype at the root, where each site changes from 0 to 1 on exactly one edge, and each extant haplotype is created by accumulating the changes on a path from the root to a leaf, where that haplotype is displayed. In other words, the extant haplotypes evolved along a perfect phylogeny with all-0 root.

16. The Perfect Phylogeny Model sites 12345 Ancestral haplotype 00000 1 4 Site mutations on edges 3 00010 2 10100 5 10000 01010 01011 Extant haplotypes at the leaves

17. Justification for Perfect Phylogeny Model • In the absence of recombination each haplotype of any individual has a single parent, so tracing back the history of the haplotypes in a population gives a tree. • Recent strong evidence for long regions of DNA with no recombination. Key to the NIH haplotype mapping project. (See NYT October 30, 2002) • Mutations are rare at selected sites, so are assumed non-recurrent. • Connection with coalescent models.

18. Solving the Haplotype Phylogeny Problem (PPH) in nearly linear time • Finds if there is a solution. • Counts the number of solutions. • Implicitly represents all the solutions Gusfield, RECOMB, April 2002

19. Program PPH • Program PPH solves the perfect phylogeny haplotyping problem using the graph realization approach. It solves problems with 50 sites and 100 individuals in about 1 second. • Program PPH can be obtained at www.cs.ucdavis.edu/~gusfield

20. An alternative solution Recently, we developed an algorithm that is not based on graph realization, and which is much easier to understand. However, it runs in O(nm^2) time. V. Bafna, D. Gusfield, G. Lancia, S. Yooseph See Gusfield’s website.

21. The implicit representation: A column partition The algorithm partitions the sites (columns) into classes, such that within any class, the phase relationship (in, or out of phase) of any two columns is INVARIANT over all perfect phylogenies for the data. Between any two classes, the phase relationship can be set ARBITRARILLY. All solutions can be genetated in this way. That is the representation of the set of all solutions. The number of solutions is always a power of two.

22. 1 2 3 4 5 6 7 a An example. Each row starts duplicated for simplicity. a b b c c d d e e

23. 1 2 3 4 6 5 7 Starting from a PPH Solution, if all shaded cells in a class switch value, then the result is also a PPH solution, and any PPH solution can be obtained in this way, i.e. by choosing in each block whether to switch or not. a a b b c c d d e e

24. Secondary information and optimization • The partition shows explicitly what added information is useful and what is redundant. Information about the relationship of a pair of columns is redundant if and only if they are in the same class of the column partition. Apply this successively as additional information is obtained. • Problem: Minimize the number of haplotype pairs (individuals) that need be laboratory determined in order to find the correct tree. • Minimize the number of (individual, site1, site2) triples whose phase relationship needs to be determined, in order to find the correct tree.

25. The implicit representation of all solutions provides a framework for solving these secondary problems, as well as other problems involving the use of additional information, and specific tree-selection criteria.

26. A Phase-Transition Problem, as the ratio of sites to genotypes changes, how does the probability that the PPH solution is unique change? For greatest utility, we want genotype data where the PPH solution is unique. Intuitively, as the ratio of genotypes to sites increases, the probability of uniqueness increases.

27. Frequency of a unique solution with 50 and 100 sites, 5% rule and 2500 datasets per entry # geno. Frequency of unique solution

28. Data with recombination If the genotype data does not fit a perfect phylogeny, we decompose the sites (greedily left to right) into maximal intervals that do fit a unique perfect phylogeny. Experimental results: number of intervals reduces to about a tenth. The solutions are highly accurate in each interval. Then we have a phase problem again between the intervals. The final haplotype determination can be made with one site per interval, greatly reducing any lab effort.

29. For a set of genotypes, find a Smallest set H of haplotypes, such that each genotype can be explained by a pair of haplotypes in H. Topic 3: The True Parsimony Objective No Approximations - find the true smallest, and know for sure you have it !

30. Example of Parsimony 00100 01110 02120 3 distinct haplotypes set S has size 3 01110 10110 22110 00100 10110 20120

31. Pure Parsimony is NP-hard Earl Hubbel (Affymetrix) showed that Pure Parsimony is NP-hard. However, for a range of parameters of current interest (50 sites and 50 genotypes) a True Parsimony solution can be computed efficiently, using Integer Linear Programming, and one speed-up trick. For larger parameters (100 sites and 50 genotypes) A near-parsimony solution can be found efficiently.

32. How Fast? How Good? Depends on the level of recombination in the underlying data. True Parsimony can be computed in seconds to minutes for most cases with 50 genotypes and up to 60 sites, faster as the level of recombination increases. As the level of recombination increases, the accuracy of the True Parsimony Solution falls, but remains within 10% of the quality of PHASE (for comparison).

33. The APOE Data There are 17 distinct haplotypes in the real data. The IP finds a True Parsimony Solution with 15 distinct haplotypes. PHASE and HAPLOTYPER each use 15 haplotypes. Over the 10,000 executions, Clark variation 1 used a minimum of 20 haplotypes.

34. Clark/Parsimony Hybrid For low recombination, large (>60) sites Find an execution of Clark’s method that maximizes the number of genotypes resolved minimizes the number of distinct haplotypes used We can do this by mixing the Digraph View of Clark’s method (Gusfield 2001) with the parsimony criteria, and truly find an execution of Clark’s method that minimizes the number of distinct haplotypes used. On datasets where we can compute True Parsimony, this hybrid does only a bit worse than True Parsimony.

35. Other uses of IP On datasets where we know the solution, find the best that a Clark method can ever do. IP can find the best possible execution. On the APOE data, Clark’s method can get all get 47 correct! In fact in a huge number of ways. (But the best we found by actually running Clark’s method was 42 correct). This kind of test is not possible for Statistical methods.

36. The Conceptual Integer Programming Formulation For each genotype (individual) j, create one integer programming variable Yij for each pair of haplotypes whose merge creates genotype j. If j has k 2’s, then This creates 2^(k-1) Y variables. Create one integer programming variable Xq for Each distinct haplotype q that appears in one of the pairs for a Y variable.

37. Conceptual IP For each genotype, create an equality that says that exactly one of its Y variables must be set to 1. For each variable Yij, whose two haplotypes are given variables Xq and Xq’, include an inequality that says that if variable Yij is set to 1, then both variables Xq and Xq’ must be set to 1. Then the objective function is to Minimize the sum of the X variables.

38. Example 02120 Creates a Y variable Y1 for pair 00100 X1 01110 X2 and a Y variable Y2 for pair 01100 X3 00110 X4 Include the following (in)equalities into the IP The objective function will include the subexpression X1 + X2 + X3 + X4 But any X variable is included exactly once no matter how many Y variables it is associated with. Y1 + Y2 = 1 Y1 - X1 <= 0 Y1 - X2 <= 0 Y2 - X3 <= 0 Y2 - X4 <= 0

39. Efficiency Tricks Ignore any Y variable and its two X variables if those X variables are associated with no other Y variable. The Resulting IP is much smaller, and can be used to find the optimal to the conceptual IP. Also, we need not enumerate all X pairs for a given genotype, but can efficiently recognize the pairs we need.

40. Avoiding Enumeration of unneeded haplotypes For each pair of genotypes, G1, G2 it is easy to find all the haplotypes that appear in an explanation for G1 and in an explanation for G2. Example: 0 2 1 1 0 2 0 2 0 1 1 1 2 2 0 2 0 1 1 1 0 V 0 2 V and then generate all combinations of 0,1’s over the V sites. So the time is O(m x # haps in both explanation sets)

41. Recombination Helps! As the level of recombination increases, the number of haps in two explanation sets decreases, so the size of the IP falls.

42. Full paper on Parsimony See the technical report: Haplotyping by Pure Parsimony available at wwwcsif.cs.ucdavis.edu/~gusfield/

43. Thanks to Andy, Sorin and Mike.