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Course: CS 273, Fall 2009
School: Stanford
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Algorithms CS273: for Structure and Motion in Biology Stanford University Lecture #4: Topics: Scribe: 8 April 2004 Sequence Similarity Sonil Mukherjee Handout # 4 Thursday, 8 April 2004 1 Introduction When aligning multiple sequences, the general idea is that if the evolutionary tree is known, we just do pairwise alignment on the closest two sequences, or alignments of sequences, in the order specified by the...

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Algorithms CS273: for Structure and Motion in Biology Stanford University Lecture #4: Topics: Scribe: 8 April 2004 Sequence Similarity Sonil Mukherjee Handout # 4 Thursday, 8 April 2004 1 Introduction When aligning multiple sequences, the general idea is that if the evolutionary tree is known, we just do pairwise alignment on the closest two sequences, or alignments of sequences, in the order specified by the tree. We will discuss three systems used for aligning protein sequences: ClustalW, T-COFFEE, and ProbCons. 2 ClustalW ClustalW starts by finding the score of the pairwise alignment between each pair of sequences, using a scoring function that is appropriate for proteins. For example, since the core of a protein has less insertions and deletions, and hydrophobic regions are more likely than hydrophillic regions to be in the core, the scoring function has a lower gap penalty in hydorphillic regions that in hydrophobic regions. Now that it has all the scores, ClustalW can construct a tree by merging pairs of sequences with a higher alignment score before merging pairs with a lower score. However, it is not always correct to do this, as the two nodes with the highest alignment score, and thus the smallest distance in the correct tree, may still be merged with other nodes in the correct tree before they are merged with each other. 2.1 Neighbor joining Definition 1. An additive distance measure is a distance measure in which the distance between any pair of leaves is the sum of the lengths of the edges connecting them. If the distances are additive, neighbor joining is guaranteed to find the correct tree. It works as follows: Define Di j = di j - (ri + rj ) Where ri = (1/(|Leaf s| - 2)) k di k Theorem 2. The above calculations ensure that Di j is minimal iff i and j are neighbors. The proof of this theorem is omitted, but it shows that ClustalW can construct the tree correctly. 2 CS273: Handout # 4 2.2 Weighted progressive alignment In weighted progressive alignment, each leaf x has weight wx proportional to the tree length attributable to x. That is, if there is an edge of length l, and x is one of n nodes that can be reached by following l, the weight attributable to x from this edge is n/l. 2.3 Limitations Consider the following four sequences, shown as they would be optimally aligned: Seq Seq Seq Seq A GARFIELD THE LAST FA-T CAT B GARFIELD THE ---- FAST CAT C GARFIELD THE VERY FAST CAT D ----------- THE ---- FA-T CAT Since proteins often have different lengths, terminal gaps are preferred to internal gaps. Therefore, ClustalW will prefer to align sequences A and B like this: Seq A GARFIELD THE LAST FAT CAT Seq B GARFIELD THE FAST CAT ---- instead of like this: Seq A GARFIELD THE LAST FA-T CAT Seq B GARFIELD THE ---- FAST CAT Now the alignment of A to B is fixed and cannot be improved later. Therefore, it is impossible for ClustalW to find the optimal alignment of the four sequences. There are two methods that attempt to overcome this limitation: iterative refinement and consistency. 2.3.1 Iterative refinement The main idea is, given a multiple alignment, to search the space around it for a better alignment. Algorithm(Barton-Stenberg): 1. Align most similar x(i), x(j) 2. Align x(k) most similar to (x(i)x(j)) 3. Repeat 2 until (x(1) ... x(n)) are aligned 4. For j=1 to N Remove x(j) and realign it to x(1) ... x(j - 1)x(j + 1) ... x(n) CS273: Handout # 4 5. Repeat 4 until convergence 3 Each time step 4 is run, the alignment either gets better or stays the same, so the algorithm is guaranteed to converge. For example, say we align (x,y), (z,w), and (xy,zw) to get the following multiple alignment: x: GAAGTTA y: GACTTA z: GAACTGA w: GTACTGA When we remove sequence y and realign it to the rest, we will see that we gain 3 matches with this alignment: x: GAAGTTA y: GACTTA z: GAACTGA w: GTACTGA However, iterative refinement does not always work this well. Consider the following multiple alignment: x: GAAGTTA y: GACTTA v: GACTTA u: GACTTA z: GAACTGA w: GTACTGA We can see that we will get more matches if we move the gap from the fourth space to the second space for y, v, and u, but since we only realign one sequence at a time, the algorithm will never fix this. When we remove y, for example and realign it, we will keep the gap in the fourth place because u and v also have their gaps in the fourth place. 2.3.2 Consistency The idea here is to prevent errors when performing the intial alignments instead of fixing them later. This is done by, when aligning two sequences, consulting a third sequence to resolve any conflicts. In the following example of aligning x and y, suppose a portion of x, xi aligns well with with two different protions of y, yj and yk . 4 x: -------------xi ---------- y: ------yj -------yk ---- CS273: Handout # 4 It is unclear which part to align xi to. But if we know that xi aligns to zk in a third sequence z, and zl aligns to yj , then we can choose to align xi to yj , knowing that this will be best when we align xy to z. 3 T-COFFEE This picture summarizes T-COFFEE: T-COFFEE starts with a users library, which contains the set of n sequences to be aligned. Its first step is to create the primary library. 3.1 Primary library The primary library is built by first builidng a ClustalW library and a Lalign(local alignment version of ClustalW). T-COFFEE calls ClustalW to construct all pairwise alignments the among n sequences, and then calls Lalign to construct a library of local alignments. Sequence identity is then used to assign a primary weight to each alignment in either library. For example, for the alignment: Seq A GARFIELD THE LAST FAT CAT Seq B GARFIELD THE FAST CAT ---- there are 18 letters in the shorter sequence, and only 2 of them do not match, so the sequence identity is 100*(16/18) = 88, so the primary weight for this alignment is 88. If an alignment is duplicated across two libraries, then it is merged into one entry with the new weight = sum of two previous weights. For example, if the alignment above is in both the ClustalW and Lalign libraries with a weight of 88 in each, then it becomes one entry with weight 88+88 = 176. This way, the ClustalW and Lalign libraries are megred into one primary library. 3.2 Library extension Lets say A and C are aligned with a score of 77, and C and B are aligned with a score of 100, as shown below: Seq A GARFIELD THE LAST FAT CAT Seq C GARFIELD THE VERY FAST CAT Seq B GARFIELD THE ---- FAST CAT CS273: Handout # 4 5 Now we have an alignment of A and B through C. The weight of this alignment W(A,B) = min(W(A,B), W(B,C)) = min(77, 100) = 77. To get the final W(A,B), we add this to W(A,B) from the primary library. So for this example, the final W(A,B) = 77 + 88 = 165. This value is put in the extended library. This approach requires examining all triplets of sequences, but not all will provide information. Consider the sequence D, whose alignments to A and B provide no information: Seq A GARFIELD THE LAST FA-T CAT Seq D ----------- THE ---- FA-T CAT Seq B GARFIELD THE ---- FAST CAT 3.3 Running time Letting L be the average sequence length and N be the number of sequences, the running time can be separated into four parts: 1. O(N 2 L2 ) time for pairwise library computation 2. O(N 3 L) time for library extension 3. O(N 3 ) time for computation of the neighbot joining tree 4. O(N L2 ) time for progressive alignment computation Therefore, the total running time is O(N 2 L2 ) + O(N 3 L) 4 ProbCons Unlike the ClustalW and T-COFFEE, ProbCons is not discrete. It is similar to TCOFFEE, but it is probabilistic. It uses a probabilistic model for alignment, probabilistic consistency, and it optimizes expected accuracy rather than the most likely alignment. 4.1 Hidden Markov Models Hidden Markov Models (HMMs) can be used to model alignments by having three states: 1. M corresponds to match, so when the HMM is in state M, a letter is emitted from both sequences. 2. I corresponds to a gap in the second sequence, so when the HMM is in state I, a letter is emitted from the first sequence only. 3. J corresponds to a gap in the first sequence, so when the HMM is in state I, a letter is emitted from the second sequence only. 6 CS273: Handout # 4 Therefore, an alignment of two sequences corresponds can only have one sequence of states in the HMM. An example is shown below: --AGGC TAG--C JJMIIM Probabilites are assigned to every transition arrow to determine which state to go to next, and to every emission to determine which letter(s) to emit in each state. When doing calculations, the logs of these probabilities will be used to prevent underflow. On...

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Exercise 1.88-X is length of human pregnancies (in days). X is assumed to follow N(266,16). Z follows N(0,1).(a) Standardization + table/computer lookup:Pr(X&lt;240) = Pr(X-266)/16 &lt; (240-266)/16) = Pr(Z&lt;-1.625) = 0.052 = 5.2%.(b) Same procedur
UPenn - VHM - 801
Exercise 1.124-Minitab commands and output:MTB &gt; # Opening worksheet from file: C:\.\DATASETS\APPENDIX\CHEESE.DATMTB &gt; # File was last modified on 6/3/98Current worksheet: CHEESE.DATMTB &gt; name c1='case' c2='taste' c3='acetic' c4='H2S' c5=
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Exercise 2.51-MTB &gt; # Opening worksheet from file: H:\VHM801\DATASETS\CHAP02\Ex02_051.datMTB &gt; # File was last modified on 6/3/98Current worksheet: Ex02_051.datMTB &gt; name c1='age'MTB &gt; name c2='weight'MTB &gt; %Fitline 'weight' 'age';SUBC&gt;
UPenn - VHM - 801
Exercise 2.94-Minitab commands and output (exercise can also be done by calculator):MTB &gt; # Opening worksheet from file: C:\DATA.AVC\teaching\vhm801\9P\02_094.datMTB &gt; # File was last modified on 10/29/102MTB &gt; name c1='Alaska'MTB &gt; name c2='A
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Exercise 4.10-Simulation of draws of opinions on husbands' participation in household work.Probability of women expressing 'yes' (husbands should do their share) is 0.73.(a)The following commands in Minitab will simulate 100 draws of 20 women:
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Exercise 4.12-Answers:(a) S = {germination, non-germination}(b) Measured in days after treatment, a reasonable samplingspace would be S = {0,1,2,.,M}, where M is a large number (e.g., 100*365)In practice one would usually use the unbounde
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Exercise 4.40-X = number of persons in an American household.Answers:(a) The probabilities sum to 1, and are all &gt;=0, so the probabilitydistribution is legitimate. For a plot, enter the inhabitants andprobabilities as columns c1 and c2 in Mi
UPenn - VHM - 801
Exercise 4.69-Minitab commands for computation of mean and standard deviation:MTB &gt; Set c1DATA&gt; 1( 540 545 550 555 560 )1DATA&gt; End.MTB &gt; name c1 'temp_C'MTB &gt; Set c2DATA&gt; 1( .1 .25 .3 .25 .1 )1DATA&gt; End.MTB &gt; name c2 'prob'MTB &gt;
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Exercise 4.70-Two scales for measuring weight. For an item with true weightof 2g, the scales give results: X with mean 2.000g and std.dev. 0.002g, Y with mean 2.001g and std.dev. 0.001g.Also, X and Y are independent.(a) Z = Y-X. EZ = E
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Exercise 6.48-Measurements of radon concentrations by a particular type of detectors.Let X_1,.,X_12 denote the 12 measurements.We assume that the observations are i.i.d. (independent and identicallydistributed) with mean mu and standard deviati
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Exercise 6.52-A farmer &quot;recognizes&quot; water in 4 trials out of 5.(a) The hypothesis of interest is H0: p=0.5 against the alternative Ha: p&gt;0.5. (It is difficult to imagine that p&lt;0.5, unless there is someerror in the experiment or the farmer mes