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Chap. 7 Genome Rearrangements Introduction to Computational Molecular Biology Chap. 7.2.5 ~ 7.3.2
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Bad Components Hurdle do not separate any other two bad components Non-hurdle separate any pair of bad component Super hurdle if its removal would cause some non-hurdle to become hurdle Simple hurdles All other hurdles Bad Components NonhurdlesHurdles Simple HurdlesSuper Hurdles “Bad component” consists entirely of bad cycles Bad component hierarchy
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Bad Components Fortress A permutation which contains odd number of hurdles and all of them are super hurdles Lower Bound on the Reversal Distance Non-Hurdle
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Special kind of reversal Hurdle cutting Reversal on convergent edges Not change c( ) and decreases h( ) When h( ) is odd, the hurdle is simple Hurdle merging Reversal on edges of different cycles Decrease c( ) and decrease h( ) by two When h( ) is even, merging opposite hurdles Two hurdles become good components, as well as any non- hurdle that separates them
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Algorithm Input: Permutations and Output: sorting reversal for with target If there is a good component then Pick two divergent edges e, f. and Reverse if result creates not a bad component Return Else if the number of hurdles is even then Merge 2 opposite hurdles Return Else If there is a simple hurdle then cut it. Return Else merge two hurdles. Return
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Unoriented Blocks Unoriented permutation Mapping from {1,2,…,n} to a set L of n labels In this book, Identity permutation is target Reversal Same as oriented permutation but except flipping arrow
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Strips Strip A sequence of consecutive labels surrounded by breakpoints but with no internal breakpoints Increasing or Decreasing strip A single label is an increasing and decreasing strip ‘L’, ‘R’ are always part of a single increasing strip L 1 2 8 7 3 5 6 4 R Increasing Strip Decreasing Strip Increasing and Decreasing Strip
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Theorem 7.4 If Label k belongs to a decreasing strip k-1 belongs to an increasing strip then there is a reversal that removes at least one breakpoint Theorem 7.5 If Label k belongs to a decreasing strip K+1 belongs to an increasing strip then there is a reversal that removes at least one breakpoint … (k-1) … k … … k … (k-1) … … (k+1) … k … … k … (k+1) … Theorem
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Theorem 7.6 Let be a permutation with a decreasing strip If No breakpoint-removing reversal leaves a decreasing strip then there is a reversal that removes two breakpoint from Theorem 7.7 The number of iterations in algorithm Sorting Unoriented Permutation is less than or equal to the number of breakpoints in the initial permutation … (k-1) … k … … l … (l+1) … Theorem
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Algorithm Input: Permutations Output: sorting reversal for with target If there is a good component then Pick two divergent edges e, f. and Reverse if result creates not a bad component Return Else if the number of hurdles is even then Merge 2 opposite hurdles Return Else If there is a simple hurdle then cut it. Return Else merge two hurdles. Return
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