140 likes | 159 Views
Explore efficient methods for determining the minimum mosaic of recombinants in genetic sequences, offering practical solutions and insights, with a focus on founder mapping and breakpoint identification.
E N D
Improved Algorithms for Inferring the Minimum Mosaic of a Set of Recombinants Yufeng Wu and Dan Gusfield UC Davis CPM 2007
Suffix Prefix 11000 0000001111 Breakpoint Recombination • Recombination: one of the principle genetic forces shaping sequence variations within species. • Two equal length sequences generate a new equal length sequence. 110001111111001 000110000001111
Sampled sequences in current population 000000 111111 001111 111100 000000 001111 111100 011100 000000 111111 001111 Breakpoint Mosaic Founders and Mosaic • Current sequences are descendents of a small number of founders. • A current sequence is composed of blocks from the founders, due to recombination. • No mutations since formation of founders. 000000 001111 111100 011100 000000 111111 Founders
Four breakpoints: minimum for all possible three founders 1101101 1010001 0111111 0110100 1100011 1101111 1010001 0110100 Three Founders The Minimum Mosaic Problem • Given a set of aligned binary sequences in the current population and assume the number of founders is known to be Kf, find a set of founders and the mosaic with the minimum number of breakpoints. Assume Kf =3 1101101 1010001 0111111 0110100 1100011
Status of the Minimum Mosaic Problem • First studied by E. Ukkonen (WABI 2002). • Dynamic programming method. Not practical when the number of rows is more than 20 and Kf >2. • No polynomial-time algorithm was known even when Kf is small. No NP-completeness result is known. • Our results: • A simple polynomial-time algorithm for Kf = 2 case. • Exact and practical method for data of medium range for Kf 3 .
1111101 1010001 0111111 0110100 1100011 00 11 0? 1? 01 10 0? 1? 2 breakpoints between c1 and c2 2 breakpoints between c2 and c3 The Two-Founder Case 1111101 1010001 0111111 0110100 1100011 110111101 100100101 010111111 010101100 110000111 Remove uniform columns Study pairs of neighboring columns Founders Key: at columns 1 and 2, the founders are either or . There are two rows with 00/11, and three rows with 01/10. So, at least two breakpoints between columns 1 and 2 with founders as .
1 0 1 0 0 1 Founders The Two-Founder Case (Cont.) # breakpoints between two columns 2 2 1 2 2 2 Local founders c1 c2 c5 c7 c3 c4 c6 1 0 1 0 At least 2 + 2 + 2 +1 +2 +2 = 11 breakpoints are needed. On the other hand, we can construct two founders that use the same local optimal founders, and thus 11 breakpoints is global optimum. No matter which founder states are chosen for previous column, we can always choose the needed founders for current column.
1101101 Founder Mapping 1101101 Breakpoint! Founder 1 Founder 2 Three or More Founders: Assuming Known Founders Input Sequences Three Founders With known founders, can minimize breakpoints for each sequence, and thus also minimize the total number of breakpoints. For each input sequence, starting from the left, insert a breakpoint at the end of longest segments matching one founder. Founder mapping: at each position c in any input sequence s, which founder s[c] takes its value from. 1101101 1010001 0111111 0110100 1100011 1101111 1010001 0110100
Enumerating Founders for Founder-Unknown Case In reality, founders are not known. A straightforward way is to simply enumerate all possible sets of founders, and then run the previous method to find the minimum mosaic. 1 0 0 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 At each column, there are 2kf–2 founder settings. Let m be the number of columns, fully enumerate all possible sets of founders takes (2m*kf) time. Infeasible when m or Kf is large. Need more ideas to develop a practical method. First, we do the enumeration in the form of search paths in a search tree.
c2 c3 Founder settings up to column 3 On-line computation: Compute partial solution up to the current column for speedup. Num of tot. breakpoints up to current column 0 0 1 0 0 1 2 0 0 1 0 1 0 0 1 0 5 0 0 1 0 1 0 1 The founder-known method can be run with partially-known founders! 0 0 1 1 0 0 0 0 0 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 1 1 0 2 Search Paths and Search Tree Assume three founders Founder setting at column one 0 0 1 0 It works but there is exponential blowup of the search paths! Obvious idea to reduce search space: branch and bound (compute a lower bound and …). But we found a different idea is more useful. 0 1 1 0 c1
<= 5 bkpts 40 >= 0 bkpts An optimal search path following P2 Dropping Search Paths that are Beaten by Another Search Path Founder Config. P1 P1 and P2 are two search paths up to column 2. Can we say P1 is better than P2? Not really, because maybe P2 can lead to fewer breakpoints later on. But, suppose the number of input sequences is 5. We can then say P1 beats P2 (and so drop P2). Why? <=39 0 1 1 0 0 1 0 P2 1 0 1 0 1 1 6 Assume three founders
These two rows have the same founder mappings. Rows Match 1 2 If no bkpt at P2, no bkpt at p1 too. 3 P1 Row2 4 5 P2 Row2 Rows Match P1: No extra breakpoints at rows 2, 4 1 2 3 4 5 A More Powerful Beaten Rule Still five input rows. Now can not say P1 beats P2. But remember we have founder matching… P1 0 1 1 0 0 1 0 P2 0 0 1 0 1 1 4 So P1 beats P2 since at most 3 rows need extra breakpoints to get onto a path from P2, and P2 uses 4 more breakpoints than P1.
How Practical Is Our Method? Source of data and image: UNC Chapel Hill Five founders 20 rows, 36 columns UNC’s heuristic solution: 54 breakpoints Enumerating 2180 founder states is impossible! Our method takes 5 minutes to find the optimal solutions: 53 breakpoints. It is also practical for 50x50 matrix with four founders.
Open Problems and Software • Is the minimum mosaic problem NP-complete? • Is there a polynomial-time algorithm for the minimum mosaic problem for small (say three to ten) number of founders? • Software available at: http://wwwcsif.cs.ucdavis.edu/~wuyu • Thank you.