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Reconstructing Ancestral Recombination Graphs - or Phylogenetic Networks with Recombination Dan Gusfield UC Davis Different parts of this work are joint with Satish Eddhu, Charles Langley, Dean Hickerson, Yun Song, Yufeng Wu.

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Reconstructing the Evolution of Binary Bio-Sequences Perfect Phylogeny (tree) model Phylogenetic Networks (DAG) with recombination Phylogenetic Networks with disjoint cycles: Galled-Trees Phylogenetic Networks with unconstrained cycles: Blobbed-Trees Combinatorial Structure and Efficient Algorithms Efficiently Computed Lower and Upper bounds on the number of recombinations needed

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Geneological or Phylogenetic Networks The major biological motivation comes from genetics and attempts to reconstruct the history of recombination in populations. Also relates to phylogenetic-based haplotyping. Some of the algorithmic and mathematical results have phylogenetic applications, for example in hybrid speciation, lateral gene transfer.

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Why binary? Single nucleotide polymorphisms are the key data that we use. SNPs imply that the sequences are binary, and also that the order of the sites is fixed (on a chromosome). This is in contrast to a set of taxonomic characters, which may be binary, but where the given order is arbitrary.

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The Perfect Phylogeny Model for binary sequences 00000 1 2 4 3 5 10100 10000 01011 00010 01010 12345 sites Ancestral sequence Extant sequences at the leaves Site mutations on edges The tree derives the set M: 10100 10000 01011 01010 00010

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The converse problem Given a set of sequences M we want to find, if possible, a perfect phylogeny that derives M. Remember that each site can change state from 0 to 1 only once. n will denote the number of sequences in M, and m will denote the length of each sequence in M. n m M 01101001 11100101 10101011

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Classic NASC: Arrange the sequences in a matrix. Then (with no duplicate columns), the sequences can be generated on a unique perfect phylogeny if and only if no two columns (sites) contain all four pairs: 0,0 and 0,1 and 1,0 and 1,1 This is the 4-Gamete Test When can a set of sequences be derived on a perfect phylogeny?

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A richer model 00000 1 2 4 3 5 10100 10000 01011 00010 01010 12345 10100 10000 01011 01010 00010 10101 added pair 4, 5 fails the three gamete-test. The sites 4, 5 ``conflict”. Real sequence histories often involve recombination.

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10100 01011 5 10101 The first 4 sites come from P (Prefix) and the sites from 5 onward come from S (Suffix). P S Sequence Recombination A recombination of P and S at recombination point 5. Single crossover recombination

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Network with Recombination 00000 1 2 4 3 5 10100 10000 01011 00010 01010 12345 10100 10000 01011 01010 00010 10101 new 10101 The previous tree with one recombination event now derives all the sequences. 5 P S

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Multiple Crossover Recombination 4-crossovers 2-crossovers = ``gene conversion”

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Elements of a Phylogenetic Network (single crossover recombination) Directed acyclic graph. Integers from 1 to m written on the edges. Each integer written only once. These represent mutations. A choice of ancestral sequence at the root. Every non-root node is labeled by a sequence obtained from its parent(s) and any edge label on the edge into it. A node with two edges into it is a ``recombination node”, with a recombination point r. One parent is P and one is S. The network derives the sequences that label the leaves.

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A Phylogenetic Network 00000 5 2 3 3 4S p P S 1 4 a:00010 b:10010 c:00100 10010 01100 d:10100 e:01100 00101 01101 f:01101 g:00101 00100 00010

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Which Phylogenetic Networks are meaningful? Given M we want a phylogenetic network that derives M, but which one? A: A perfect phylogeny (tree) if possible. As little deviation from a tree, if a tree is not possible. Use as little recombination or gene-conversion as possible.

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Minimizing recombinations Any set M of sequences can be generated by a phylogenetic network with enough recombinations, and one mutation per site. This is not interesting or useful. However, the number of (observable) recombinations is small in realistic sets of sequences. ``Observable” depends on n and m relative to the number of recombinations. Two algorithmic problems: given a set of sequences M, find a phylogenetic network generating M, minimizing the number of recombinations (Hein’s problem). Find a network generating M that has some biologically- motivated structural properties.

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Minimization is NP-hard The problem of finding a phylogenetic network that creates a given set of sequences M, and minimizes the number of recombinations, is NP- hard. (Wang et al 2000) (Semple 2004) Wang et al. explored the problem of finding a phylogenetic network where the recombination cycles are required to be node disjoint, if possible. They gave a sufficient but not a necessary condition to recognize cases when this is possible. O(nm + n^4) time.

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Recombination Cycles In a Phylogenetic Network, with a recombination node x, if we trace two paths backwards from x, then the paths will eventually meet. The cycle specified by those two paths is called a ``recombination cycle”.

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Galled-Trees A recombination cycle in a phylogenetic network is called a “gall” if it shares no node with any other recombination cycle. A phylogenetic network is called a “galled- tree” if every recombination cycle is a gall.

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4 1 3 25 a: 00010 b: 10010 d: 10100 c: 00100 e: 01100 f: 01101 g: 00101 A galled-tree generating the sequences generated by the prior network. 3 4 p s p s

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Sales pitch for Galled-Trees Galled-trees represent a small deviation from true trees. There are sufficient applications where it is plausible that a galled tree exists that generates the sequences. Observable recombinations tend to be recent; block structure of human DNA; recombination is sparse, so the true history of observable recombinations may be a galled-tree. The number of recombinations is never more than m/2. Moreover, when M can be derived on a galled-tree, the number of recombinations used is the minimum number over any phylogenetic network, even if multiple cross-overs at a recombination event are counted as a single recombination. A galled-tree for M is ``almost unique” - implications for reconstructing the correct history.

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Old (Aug. 2003) Results O(nm + n^3)-time algorithm to determine whether or not M can be derived on a galled-tree with all-0 ancestral sequence. Proof that the galled-tree produced by the algorithm is a “nearly-unique” solution. Proof that the galled-tree (if one exists) produced by the algorithm minimizes the number of recombinations used, over all phylogenetic- networks with all-0 ancestral sequence.

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New work We derive the galled-tree results in a more general setting that addresses unconstrained recombination cycles and multiple crossover recombination. This also solves the problem of finding the ``most tree-like” network when a perfect phylogeny is not possible. In this algorithm, no ancestral sequence is known in advance.

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Blobbed-trees: generalizing galled-trees In a phylogenetic network a maximal set of intersecting cycles is called a blob. Contracting each blob results in a directed, rooted tree, otherwise one of the “blobs” was not maximal. So every phylogenetic network can be viewed as a directed tree of blobs - a blobbed-tree. The blobs are the non-tree-like parts of the network.

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Ugly tangled network inside the blob. Every network is a tree of blobs. How do the tree parts and the blobs relate? How can we exploit this relationship?

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Incompatible Sites A pair of sites (columns) of M that fail the 4-gametes test are said to be incompatible. A site that is not in such a pair is compatible.

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0 0 0 1 0 1 0 0 1 0 0 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 2 3 4 5 abcdefgabcdefg 13 4 25 Two nodes are connected iff the pair of sites are incompatible, i.e, fail the 4-gamete test. Incompatibility Graph M THE MAIN TOOL: We represent the pairwise incompatibilities in a incompatibility graph.

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The connected components of G(M) are very informative The number of non-trivial connected components is a lower-bound on the number of recombinations needed in any network (Bafna, Bansal; Gusfield, Hickerson). When each blob is a single-cycle (galled-tree case) all the incompatible sites in a blob must come from a single connected component C, and that blob must contain all the sites from C. Compatible sites need not be inside any blob. (Gusfield et al 2003-5)

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Simple Fact If sites two sites i and j are incompatible, then the sites must be together on some recombination cycle whose recombination point is between the two sites i and j. (This is a general fact for all phylogenetic networks.) Ex: In the prior example, sites 1, 3 are incompatible, as are 1, 4; as are 2, 5.

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A Phylogenetic Network 00000 5 2 3 3 4S p P S 1 4 a:00010 b:10010 c:00100 10010 01100 d:10100 e:01100 00101 01101 f:01101 g:00101 00100 00010

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Simple Consequence of the simple fact All sites on the same (non-trivial) connected component of the incompatibility graph must be on the same blob in any blobbed-tree. Follows by transitivity. So we can’t subdivide a blob into a tree-like structure if it only contains sites from a single connected component of the incompatibility graph.

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Key Result about Galls: For galls, the converse of the simple consequence is also true. Two sites that are in different (non-trivial) connected components cannot be placed on the same gall in any phylogenetic network for M. Hence, in a galled-tree T for M each gall contains all and only the sites of one (non-trivial) connected component of the incompatibility graph. All compatible sites can be put on edges outside of the galls. This is the key to the galled-tree solution.

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4 1 3 25 a: 00010 b: 10010 d: 10100 c: 00100 e: 01100 f: 01101 g: 00101 A galled-tree generating the sequences generated by the prior network. 2 4 p s p s 13 4 25 Incompatibility Graph

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Motivated by the one-one correspondence between galls and non-trivial connected components, we ask: To what extent does this one-one correspondence hold in general blobbed-trees, i.e. with no constraints on how recombination cycles interweave?

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The Decomposition Theorem (Recomb 2005) For any set of sequences M, there is a blobbed-tree T(M) that derives M, where each blob contains all and only the sites in one non-trivial connected component of G(M). The compatible sites can always be put on edges outside of any blob. A blobbed-tree with this structure is called fully-decomposed.

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General Structure So, for any set of sequences M, there is a phylogenetic network T(M) that is fully decomposed. Moreover, the tree part of T(M) is unique. And it is easy to find the tree part.

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4 1 3 25 a: 00010 b: 10010 d: 10100 c: 00100 e: 01100 f: 01101 g: 00101 A fully-decomposed network for the sequences generated by the prior network. 2 4 p s p s 13 4 25 Incompatibility Graph

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Moreover Since all sites from a single connected component must be together on some blob in any phylogenetic network, no network is more decomposed than the fully decomposed network.

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Proof Ideas Let C be a connected component of G(M). Define M[C] as the sequences in M restricted to the sites in C.

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0 0 0 1 0 1 0 0 1 0 0 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 2 3 4 5 abcdefgabcdefg 13 4 M a 0 0 1 b 1 0 1 c 0 1 0 d 1 1 0 e 0 1 0 f 0 1 0 g 0 1 0 1 3 4 C1 25 C2 abcdefgabcdefg 0 1 0 1 0 1 2 5 B1 B2 M[C1]M[C2]

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Faux Proof Pick one site from each connected component C in G(M) to ``represent” C. No pair of those sites are incompatible, so by the NASC for a perfect phylogeny, there will be a perfect phylogeny T for the sites. Expand each node to a network generating the sequences in M[C]. Incorrect, because the structure of T can be wrong. We need to use information about all the sites in each C.

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a 0 0 1 b 1 0 1 c 0 1 0 d 1 1 0 e 0 1 0 f 0 1 0 g 01 0 1 3 4 abcdefgabcdefg 0 1 0 1 0 1 2 5 M[C1]M[C2] abcdefgabcdefg 1 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 W 1 2 3 4 5 6 7 8 12343331234333 55556785555678 Now for each connected component C in G(M), call each distinct sequence in M[C] a supercharacter, and let W be the indicator matrix for the supercharacters. So W indicates which rows of M contain which particular supercharacters.

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Proof Ideas Lemma: No pair of supercharacters are incompatible. So by the NASC for a Perfect Phylogeny, there is a unique perfect phylogeny T for W.

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Proof Ideas For each connected component C of G(M), all supercharacters that originate from C label edges in T that are incident with one single node v[C] in T. So, if we expand each node v[C] to be a network that generates the supercharacters from C (the sequences in M[C]), and connect each network correctly to the edges in T, the resulting network is a fully-decomposed blobbed-tree that generates M.

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Algorithmically, T is easy to find and is the tree resulting from contracting each blob in the fully-decomposed blobbed-tree T(M) for M. T can be constructed from M in O(nm^2) time.

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Broader Biological Applications Our major interest is in recombination, but the proof of the decomposition theorem does not explicitly use recombination. So it holds for whatever biological phenomena caused the incompatibility of sites. For example, back or recurrent mutation, gene- conversion, lateral gene transfer etc.

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The main open question The Decomposition Theorem says there is always a fully-decomposed blobbed-tree for any M, but Is there always a fully-decomposed blobbed-tree that minimizes the number of recombinations over all possible phylogenetic networks for M?

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We conjecture the answer is yes. If true, then we can decompose the problem of minimizing the total number of recombinations into separate problems on each connected component, and also find lower bounds on the needed number of recombinations, in each component separately, adding those bounds to get a valid overall lower bound for M. This computation of lower bounds is known to be correct for certain lower bounds (Bafna, Bansal 2004).

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Progress on Proving the Conjecture Definition: If N is a phylogenetic network for M, and a node v in N is labeled with a sequence in M, then v is said to be visible in N. Theorem: If every node in N is visible, then there is a fully-decomposed network for M where the number of recombinations is at most the number in N. Corollary: The conjecture is true for any M where the Haplotype or History lower bounds (S. Myers) on the number of recombinations needed to generate M, is tight.

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Optimality of Galled-Trees Theorem: (G,H,B,B) The minimum number of recombination nodes in any phylogenetic network for M is at least the number of non-trivial connected components of the incompatibility graph. Hence, when the sequences on each blob on T(M) can be generated with a single recombination node, the blobbed-tree minimizes the number of recombination nodes over all phylogenetic networks and all choices of ancestral sequence. This solves the root-unknown galled-tree problem in polynomial time. Code is on the web.

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The number of arrangements on a gall (all-0 ancestral sequence) By analysing the algorithm to layout the sites on a gall (not discussed here), one can prove that the number of arrangements of any gall is at most three, and this happens only if the gall has two sites. If the gall has more than two sites, then the number of arrangements is at most two. If the gall has four or more sites, with at least two sites on each side of the recombination point (not the side of the gall) then the arrangement is forced and unique.

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Computing close bounds on the minimum number of recombinations Dan Gusfield UCD Y. Song, Y. F. Wu, D. Gusfield (ISMB2005) D. Gusfield, D. Hickerson (Dis. Appl. Math 2005) D. Gusfield, V. Bansal (Recomb 2005)

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The grandfather of all lower bounds - HK 1985 Arrange the nodes of the incompatibility graph on the line in order that the sites appear in the sequence. This bound requires a linear order. The HK bound is the minimum number of vertical lines needed to cut every edge in the incompatibility graph. Weak bound, but widely used - not only to bound the number of recombinations, but also to suggest their locations.

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Justification for HK If two sites are incompatible, there must have been some recombination where the crossover point is between the two sites.

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1 2 3 4 5 HK Lower Bound

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1 2 3 4 5 HK Lower Bound = 1

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More general view of HK Given a set of intervals on the line, and for each interval I, a number N(I), define the composite problem: Find the minimum number of vertical lines so that every interval I intersects at least N(I) of the vertical lines. In HK, each incompatibility defines an interval I where N(I) = 1. The composite problem is easy to solve by a left-to-right myopic placement of vertical lines: Sort the intervals by right end-point; Process the intervals left to right in that order; when the right endpoint of an interval I is reached, place there (if needed) additional vertical so that N(I) lines intersect I.

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This general approach is called the Composite Method (Simon Myers 2002). If each N(I) is a ``local” lower bound on the number of recombinations needed in interval I, then the solution to the composite problem is a valid lower bound for the full sequences. The resulting bound is called the composite bound given the local bounds.

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Haplotype Bound (Simon Myers) Rh = Number of distinct sequences (rows) - Number of distinct sites (columns) -1 (folklore) Before computing Rh, remove any site that is compatible with all other sites. A valid lower bound results - generally increases the bound. Generally really bad bound, often negative, when used on large intervals, but Very Good when used as local bounds in the Composite Interval Method, and other methods.

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Composite Interval Method using RH bounds Compute Rh separately for each of the C(m,2) intervals of the m sites; let N(I) = Rh(I) be the local lower bound for interval I. Then compute the composite bound using these local bounds. Polynomial time and gives bounds that often double HK in our simulations.

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Composite Subset Method (Myers) Let S be subset of sites, and Rh(S) be the haplotype bound for subset S. If the leftmost site in S is L and the rightmost site in S is R, then use Rh(S) as a local bound N(I) for interval I = [S,L]. Compute Rh(S) on many subsets, and then solve the composite problem to find a composite bound.

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RecMin (Myers) World Champion Lower Bound Program (until now). Often RecMin gives a bound three times as large as HK. Computes Rh on subsets of sites, but limits the size and the span of the subsets. Default parameters are s = 6, w = 15 (s = size, w = span). Generally, impractical to set s = w = m, so generally one doesn’t know if increasing the parameters would increase the bound. (example: Myers bound of 70 on the LPL data).

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Optimal RecMin Bound (ORB) The Optimal RecMin Bound is the lower bound that RecMin would produce if both parameters were set to their maximum values (s = w= m). In general, RecMin cannot compute (in practical time) the ORB. Practical computation of the ORB is our first contribution.

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Computing the ORB Gross Idea: For each interval I, use ILP to find a subset of sites S that maximizes Rh(S) over all subsets in interval I. Call the result Opt(I). Set N(I) = Opt(I), and compute the composite bound using those local bounds. The composite bound is the ORB. -- the result one would get by using all 2^m subsets in RecMin, with s = w = m.

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We have moved from doing an exponential number of simple computations (computing Rh for each subset), to solving a quadratic number of (possibly expensive) ILPs. Is this a good trade-off in practice? Our experience - very much so!

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How to compute Opt(I) by ILP Create one variable Xi for each row i; one variable Yj for each column j in interval I. All variables are 0/1 variables. Define S(i,i’) as the set of columns where rows i and i’ have different values. Each column in S(i,i’) is a witness that rows i and i’ differ. For each pair of rows (i,i’), create the constraint: Xi + Xi’ <= 1 + ∑ [Yj: j in S(i,i’)] Objective Function: Maximize ∑ Xi - ∑ Yj -1

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Alternate way to compute Opt(I) by ILP Create one variable Yj for each column j in interval I. All variables are 0/1 variables. S(i,i’) as before. For each pair of rows (i,i’) create the constraint: 1 <= ∑ [Yj: j in S(i,i’)] Objective Function: Maximize N - ∑(Yj) -1 Finds the smallest number of columns to distinguish the rows. First remove any duplicate rows. Let N be the number of rows remaining.

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Two critical tricks 1)Use the second ILP formulation to compute Opt(I). It solves much faster than the first (why?) 2) Reduce the number of intervals where Opt(I) is computed: If the solution to Opt(I) uses columns that only span interval L, then there is no need to find Opt(I’) in any interval I’ containing L. Each ILP computation directly spawns at most 4 other ILPs. Apply this idea recursively. I L

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With the second trick we need to find Opt(I) for only 0.5% - 35% of all the C(m,2) intervals (empirical result). Surprisingly fast in practice (with either the GNU solver or CPLEX).

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Bounds Higher Than the Optimal RecMin Bound Often the ORB underestimates the true minimum, e.g. Kreitman’s ADH data: 6 v. 7 How to derive sharper bound? Idea: In the composite method, check if each local bound N(I) = Opt(I) is tight, and if not, increase N(I) by one. Small increases in local bounds can increase the composite bound.

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Bounds Sharper Than Optimal RecMin Bound A set of sequences M is called self-derivable if there is a network generating M where the sequence at every node (leaf and intermediate) is in M. Observation: The haplotype bound for a set of sequences is tight if and only if the sequences are self-derivable. So for each interval I where Opt(I) is computed, we check self-derivability of the sequences induced by columns S*, where S* are the columns specified by Opt(I). Increase N(I) by 1 if the sequences are not self-derivable.

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Algorithm for Self-Derivability Solution is easy when there are no mutations --only recombinations are allowed and one initial pair of sequences is chosen as ``reached”. Two reached sequences S1 and S2 can reach a third sequence S3, if S3 can be created by recombining S1 and S2. Do BFS to see if all sequences can be reached by successive application of the ``reach operation”. Clearly polynomial time and can be optimized with suffix- trees etc. (Kececiouglu, Gusfield)

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Self-derivability Test with mutations allowed For each site i, construct set MUT(i) of sequence pairs (S1, S2) in M where S1 and S2 differ at only site. Try each sequence in M as root (which is the only reached sequence initially). For each root, enumerate all ways of choosing exactly one ordered pair of sequences (Sp, Sq) from each MUT(i). Sp is allowed to ``reach” Sq. Run the prior self-derivability algorithm with these new permitted reach relations, to test if all sequences in M can be reached. If so, then M is self-derivable, otherwise it is not.

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Checking if N(I) should be increased by two If the set of sequences is not self-derivable, we can test if adding one new sequence makes it self-derivable. the number of candidate sequences is polynomial and for each one added to M we check self-derivability. N(I) should be increased by two if none of these sets of sequences is self-derivable.

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Self-derivability Test Algorithm Number of choices = n x Π site i |MUT(i)| = O(n m+1 ), but usually much less To get even better bounds, allow, say 1, new sequence to be added into M, and test self-derivability. Can increase bound by 2. This new sequence must be either recombinant of two existing sequences, or a mutant of one existing sequence. Thus their number is polynomial size.

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Program HapBound HapBound –S. Checks each Opt(I) subset for self- derivability. Increase N(I) by 1 or 2 if possible. This often beats ORB and is still practical for most datasets. HapBound –M. Explicitly examine each interval directly for self-derivability. Increase local bound if possible. Derives lower bound of 7 for Kreitman’s ADH data in this mode.

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HapBound vs. RecMin on LPL from Clark et al. ProgramLower BoundTime RecMin (default)593s RecMin –s 25 –w 25757944s RecMin –s 48 –w 48No result5 days HapBound7531s HapBound -S781643s 2 Ghz PC

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Example where RecMin has difficulity in Finding Optimal Bound on a 25 by 376 Data Matrix ProgramBoundTime RecMin default361s RecMin –s 30 –w 30423m 25s RecMin –s 35 –w 354324m 2s RecMin –s 40 –w 40432h 9m 4s RecMin –s 45 –w 454310h 20m 59s HapBound442m 59s HapBound -S4839m 30s

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Frequency of HapBound –S Bound Sharper Than Optimal RecMin Bound ms param.Rho=1Rho=5Rho=10Rho=20 Theta=10.0%0.4%0.5%1.5% Theta=50.7%4.0%10.4%27.0% Theta=101.4%9.2%17.8%40.4% Theta=201.4%10.5%27.8%45.4% For every ms parameters, 1000 data sets are used.

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Computing Upper Bounds The method is an adaptation of the ``history” lower bound (Myers). A non-informative column is one with fewer than two 0’s or fewer than two 1’s.

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Single History Computation 1)Set W = 0 2)Collapse identical rows together, and remove non- informative columns. Repeat until neither is possible. 3)Let A be the data at this point. If A is empty, stop, else remove some row r from A, and set W = W + 1. Go to step 2). Note that the choice of r is arbitrary in Step 3), so the resulting W can vary.

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History Lower Bound Theorem (Myers) Let W* be the minimum W obtained from all possible single history computations. Then W* is a valid lower bound on the number of recombinations needed. Naïve time: theta(n!) (RecMin), but can be reduced to theta(2^n) (Bafna, Bansal).

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Converting the History Lower Bound to an Upper Bound Given a set of rows A and a single row r, define w(r | A - r) as the minimum number of recombinations needed to create r from A-r (well defined in our application). w(r | A-r) can be computed in linear time by a greedy-type algorithm.

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Upper Bound Computation 1)Set W = 0 2)Collapse identical rows together, and remove non- informative columns. Repeat until neither is possible. 3)Let A be the data at this point. If A is empty, stop, else remove some row r from A, and set W = W + W(r | A-r). Go to step 2). Note that the choice of r is arbitrary in Step 3), so the resulting W can vary. This is the Single History Computation, with a change in step 3).

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Note, even a single execution of the upper bound computation gives a valid upper bound, and a way to construct a network. This is in contrast to the History Bound which requires finding the minimum W over all histories. However, we also would like to find the lowest upper bound possible with this approach. This can be done in O(2^n) time by DP. In practice, we can use branch and bound to find the lowest upper bound, but we have also found that branching on the best local choice, or randomizing gives good results.

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Branch and Bound (Branching) In Step 3) choose r to minimize w(r | A-r) + L(A-r), where L(A-r) is some fast lower bound on the number of recombinations needed for the set A-r. Even HK is good for this purpose. (Bounding) Let C be the min for an full solution found so far; If W + L(A) >= C, then backtrack.

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Kreitman’s 1983 ADH Data 11 sequences, 43 segregating sites Both HapBound and SHRUB took only a fraction of a second to analyze this data. Both produced 7 for the number of detected recombination events Therefore, independently of all other methods, our lower and upper bound methods together imply that 7 is the minimum number of recombination events.

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A minimal ARG for Kreitman’s data SHRUB produces code that can be input to an open source program to display the constructed ARG

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The Human LPL Data (Nickerson et al. 1998) Our new lower and upper bounds Optimal RecMin Bounds (We ignored insertion/deletion, unphased sites, and sites with missing data.) (88 Sequences, 88 sites)

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Study on simulated data: Exact- Match frequency for varying parameters = Scaled mutation rate = Scaled recombination rate n = Number of sequences n = 25 For < 5, our lower and upper bounds match more than 95% of the time. Used Hudson’s MS to generate1000 simulated datasets for each pair of and n = 15

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Exact-Match frequency for varying number of sequences Match frequency does not depend on n as much as it does on or

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A closer look at the deviation Average ratio of lower bound to upper bound when they do not match For n = 25: The numerical difference between lower and upper bounds grows as or increases, but their ratio is more stable.

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Papers and Software wwwcsif.cs.ucdavis.edu/~gusfield/

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