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1、动态规划法双序列比对第1页，此课件共55页哦2/55习题4，求两条序列的最长共同子序列。【作业】v=TACGGGTATw=GGACGTACG第2页，此课件共55页哦3/550123456789000000000001020304050607080905GGACGTACGTACGGGTAT第3页，此课件共55页哦4Sequence AlignmentSequence Alignment第4页，此课件共55页哦5/55OutlineGlobal Alignment Scoring MatricesLocal AlignmentAlignment with Affine Gap Penalties第
2、5页，此课件共55页哦6/55From LCS to Alignment:Change up the ScoringThe Longest Common Subsequence(LCS)problemthe simplest form of sequence alignment allows only insertions and deletions(no mismatches).In the LCS Problem,we scored 1 for matches and 0 for indelsConsider penalizing indels and mismatches with ne
3、gative scoresSimplest scoring schema:+1:match premium -:mismatch penalty -:indel penalty-T G C A T -A -CA T -C -T G A T C第6页，此课件共55页哦7/55Simple ScoringWhen mismatches are penalized by,indels are penalized by,and matches are rewarded with+1,the resulting score is:#matches (#mismatches)(#indels)第7页，此课
4、件共55页哦8/55The Global Alignment ProblemFind the best alignment between two strings under a given scoring schemaInput:Strings v and w and a scoring schemaOutput:Alignment of maximum scorem:mismatch penalty:indel penalty第8页，此课件共55页哦9/55Scoring MatricesTo generalize scoring,consider a(4+1)x(4+1)scoring
5、matrix.In the case of an amino acid sequence alignment,the scoring matrix would be a(20+1)x(20+1)size.The addition of 1 is to include the score for comparison of a gap character“-”.This will simplify the algorithm as follows:第9页，此课件共55页哦10/55The Blosum62 Scoring Matrix第10页，此课件共55页哦11/55Measuring Sim
6、ilarityMeasuring the extent of similarity between two sequencesBased on percent sequence identityBased on conservation第11页，此课件共55页哦12/55Percent Sequence IdentityThe extent to which two nucleotide or amino acid sequences are invariantA C C T G A G A G A C C T G A G A G A C G T G G C A GA C G T G G C
7、A G70%identicalmismatchindel第12页，此课件共55页哦13/55Making a Scoring MatrixScoring matrices are created based on biological evidence.Alignments can be thought of as two sequences that differ due to mutations.Some of these mutations have little effect on the proteins function,therefore some penalties,(vi,w
8、j),will be less harsh than others.第13页，此课件共55页哦14/55Scoring Matrix:ExampleARNKA5-2-1-1R-7-13N-70K-6 Notice that although R and K are different amino acids,they have a positive score.Why?They are both positively charged amino acids will not greatly change function of protein.第14页，此课件共55页哦15/55Conserv
9、ationAmino acid changes that tend to preserve the physico-chemical properties of the original residuePolar to polaraspartate glutamateNonpolar to nonpolaralanine valineSimilarly behaving residuesleucine to isoleucine第15页，此课件共55页哦16/55Scoring matricesAmino acid substitution matricesPAMBLOSUMDNA subst
10、itution matricesDNA is less conserved than protein sequencesLess effective to compare coding regions at nucleotide level第16页，此课件共55页哦17/55PAMPoint Accepted Mutation(Dayhoff et al.)1 PAM=PAM1=1%average change of all amino acid positionsAfter 100 PAMs of evolution,not every residue will have changedso
11、me residues may have mutated several timessome residues may have returned to their original statesome residues may not changed at all第17页，此课件共55页哦18/55PAMXPAMx=PAM1xPAM250=PAM1250PAM250 is a widely used scoring matrix:Ala Arg Asn Asp Cys Gln Glu Gly His Ile Leu Lys.A R N D C Q E G H I L K .Ala A 13
12、6 9 9 5 8 9 12 6 8 6 7 .Arg R 3 17 4 3 2 5 3 2 6 3 2 9Asn N 4 4 6 7 2 5 6 4 6 3 2 5Asp D 5 4 8 11 1 7 10 5 6 3 2 5Cys C 2 1 1 1 52 1 1 2 2 2 1 1Gln Q 3 5 5 6 1 10 7 3 7 2 3 5.Trp W 0 2 0 0 0 0 0 0 1 0 1 0Tyr Y 1 1 2 1 3 1 1 1 3 2 2 1Val V 7 4 4 4 4 4 4 4 5 4 15 10第18页，此课件共55页哦19/55BLOSUMBlocks Subst
13、itution Matrix Scores derived from observations of the frequencies of substitutions in blocks of local alignments in related proteinsMatrix name indicates evolutionary distanceBLOSUM62 was created using sequences sharing no more than 62%identity第19页，此课件共55页哦20/55The Blosum62 Scoring MatrixBLOSUM 62第
14、20页，此课件共55页哦21/55BLOSUM90PAM30低趋异度小鼠和大鼠RBPBLOSUM45PAM240高趋异度小鼠和细菌的lipocalinBLOSUM80PAM120BLOSUM62PAM180相似度越低的序列，在比对的时候，采用PAM矩阵时，后面的数字越大，采用BLOSUM矩阵时，后面的数字越小。第21页，此课件共55页哦22/55Local vs.Global AlignmentThe Global Alignment Problem tries to find the longest path between vertices(0,0)and(n,m)in the edit
15、graph.The Local Alignment Problem tries to find the longest path among paths between arbitrary vertices(i,j)and(i,j)in the edit graph.第22页，此课件共55页哦23/55Local vs.Global AlignmentThe Global Alignment Problem tries to find the longest path between vertices(0,0)and(n,m)in the edit graph.The Local Alignm
16、ent Problem tries to find the longest path among paths between arbitrary vertices(i,j)and(i,j)in the edit graph.In the edit graph with negatively-scored edges,Local Alignmet may score higher than Global Alignment第23页，此课件共55页哦24/55Local vs.Global Alignment(contd)Global AlignmentLocal Alignmentbetter
17、alignment to find conserved segment -T-CC-C-AGT-TATGT-CAGGGGACACGA-GCATGCAGA-GAC|AATTGCCGCC-GTCGT-T-TTCAG-CA-GTTATGT-CAGAT-C tccCAGTTATGTCAGgggacacgagcatgcagagac|aattgccgccgtcgttttcagCAGTTATGTCAGatc第24页，此课件共55页哦25/55Local Alignment:ExampleGlobal alignmentLocal alignmentCompute a“mini”Global Alignmen
18、t to get Local第25页，此课件共55页哦26/55Local Alignments:Why?Two genes in different species may be similar over short conserved regions and dissimilar over remaining regions.Example:Homeobox genes have a short region called the homeodomain that is highly conserved between species.A global alignment would no
19、t find the homeodomain because it would try to align the ENTIRE sequence第26页，此课件共55页哦27/55The Local Alignment ProblemGoal:Find the best local alignment between two stringsInput:Strings v,w and scoring matrix Output:Alignment of substrings of v and w whose alignment score is maximum among all possibl
20、e alignment of all possible substrings第27页，此课件共55页哦28/55The Problem with this ProblemLong run time O(n4):-In the grid of size n x n there are n2 vertices(i,j)that may serve as a source.-For each such vertex computing alignments from(i,j)to(i,j)takes O(n2)time.This can be remedied by giving free ride
21、s第28页，此课件共55页哦29/55Local Alignment:ExampleGlobal alignmentLocal alignmentCompute a“mini”Global Alignment to get Local第29页，此课件共55页哦30/55Local Alignment:Example第30页，此课件共55页哦31/55Local Alignment:Example第31页，此课件共55页哦32/55Local Alignment:Example第32页，此课件共55页哦33/55Local Alignment:Example第33页，此课件共55页哦34/55L
22、ocal Alignment:Example第34页，此课件共55页哦35/55Local Alignment:Running TimeLong run time O(n4):-In the grid of size n x n there are n2 vertices(i,j)that may serve as a source.-For each such vertex computing alignments from(i,j)to(i,j)takes O(n2)time.This can be remedied by giving free rides第35页，此课件共55页哦36/
23、55Local Alignment:Free RidesVertex(0,0)The dashed edges represent the free rides from(0,0)to every other node.Yeah,a free ride!第36页，此课件共55页哦37/55The Local Alignment RecurrenceThe largest value of si,j over the whole edit graph is the score of the best local alignment.The recurrence:0 si,j =max si-1,
24、j-1+(vi,wj)s i-1,j +(vi,-)s i,j-1+(-,wj)Notice there is only this change from the original recurrence of a Global Alignment第37页，此课件共55页哦38/55The Local Alignment RecurrenceThe largest value of si,j over the whole edit graph is the score of the best local alignment.The recurrence:0 si,j =max si-1,j-1+
25、(vi,wj)s i-1,j +(vi,-)s i,j-1+(-,wj)Power of ZERO:there is only this change from the original recurrence of a Global Alignment-since there is only one“free ride”edge entering into every vertex第38页，此课件共55页哦39/55http:/blast.ncbi.nlm.nih.gov/Blast.cgi NP_006735NP_000945第39页，此课件共55页哦40/55第40页，此课件共55页哦41
26、/55习题考虑序列v=TACGGGTAT和w=GGACGTACG。假设匹配奖励+1，错配和插缺罚分均为-1.【作业】填写序列v和w之间的全局联配的动态规划表（编辑图或相似度矩阵）。在各单元画出箭头以存储返回信息。全局最优联配的得分是多少？这个得分对应的联配又是什么？填写序列v和w之间的局部联配的动态规划表。在各单元画出箭头以存储返回信息。在这种情形下，局部最优联配的得分是多少？这个得分对应的联配又是什么？第41页，此课件共55页哦42/55-012345678900-1-2-3-4-5-6-7-8-91-12-23-34-45-56-67-78-899GGACGTACGTACGGGTAT全局比
27、对第42页，此课件共55页哦43/55局部比对-012345678900000000000102030405060708090GGACGTACGTACGGGTAT第43页，此课件共55页哦44/55Scoring Indels:Naive ApproachA fixed penalty is given to every indel:-for 1 indel,-2 for 2 consecutive indels-3 for 3 consecutive indels,etc.Can be too severe penalty for a series of 100 consecutive in
28、dels第44页，此课件共55页哦45/55Affine Gap PenaltiesIn nature,a series of k indels often come as a single event rather than a series of k single nucleotide events:Normal scoring would give the same score for both alignmentsThis is more likely.This is less likely.第45页，此课件共55页哦46/55Accounting for GapsGaps-conti
29、guous sequence of spaces in one of the rowsScore for a gap of length x is:-(+x)where 0 is the penalty for introducing a gap:gap opening penalty will be large relative to:gap extension penalty because you do not want to add too much of a penalty for extending the gap.第46页，此课件共55页哦47/55Affine Gap Pena
30、ltiesGap penalties:-when there is 1 indel-2 when there are 2 indels-3 when there are 3 indels,etc.-x(-gap opening-x gap extensions)Somehow reduced penalties(as compared to nave scoring)are given to runs of horizontal and vertical edges第47页，此课件共55页哦48/55Affine Gap Penalties and Edit GraphTo reflect a
31、ffine gap penalties we have to add“long”horizontal and vertical edges to the edit graph.Each such edge of length x should have weight -x*第48页，此课件共55页哦49/55Adding“Affine Penalty”Edges to the Edit GraphThere are many such edges!Adding them to the graph increases the running time of the alignment algor
32、ithm by a factor of n(where n is the number of vertices)So the complexity increases from O(n2)to O(n3)第49页，此课件共55页哦50/55Manhattan in 3 Layers第50页，此课件共55页哦51/55Affine Gap Penalties and 3 Layer Manhattan GridThe three recurrences for the scoring algorithm creates a 3-layered graph.The top level create
33、s/extends gaps in the sequence w.The bottom level creates/extends gaps in sequence v.The middle level extends matches and mismatches.第51页，此课件共55页哦52/55Switching between 3 LayersLevels:The main level is for diagonal edges The lower level is for horizontal edges The upper level is for vertical edgesA
34、jumping penalty is assigned to moving from the main level to either the upper level or the lower level(-)There is a gap extension penalty for each continuation on a level other than the main level(-)第52页，此课件共55页哦53/55The 3-leveled Manhattan Grid第53页，此课件共55页哦54/55Affine Gap Penalty RecurrencesContinu
35、e Gap in w(deletion)Start Gap in w(deletion):from middleContinue Gap in v(insertion)Start Gap in v(insertion):from middleMatch or MismatchEnd deletion:from topEnd insertion:from bottom第54页，此课件共55页哦55/55习题考虑序列v=TACGGGTAT和w=GGACGTACG。假设匹配奖励+1，错配和插缺罚分均为-1.填写序列v和w之间的全局联配的动态规划表（编辑图或相似度矩阵）。在各单元画出箭头以存储返回信息。全局最优联配的得分是多少？这个得分对应的联配又是什么？填写序列v和w之间的局部联配的动态规划表。在各单元画出箭头以存储返回信息。在这种情形下，局部最优联配的得分是多少？这个得分对应的联配又是什么？假设使用仿射缺口惩罚，其中引入缺口开放的罚分为-10，缺口扩展的罚分为-1。匹配和错配的得分不改变。在这种情形下，全局最优联配是什么？他的得分是多少？第55页，此课件共55页哦