1. Monen sekvenssin linjaus
Proteiinianalyysi 3
Monen sekvenssin linjaus

2. Multiple sequence alignment (msa)
A) define characters for phylogenetic analysis
B) search for additional family members
SeqA N ∙ F L S
SeqB N ∙ F – S
SeqC N K Y L S
SeqD N ∙ Y L S
NYLS NKYLS NFS NFLS
+K L
YF

3. Complexity of performing msa
Sum of pairs (SP) score
Extension of sequence pair alignment by dynamic programming
O(LN), where L is sequence length and N is number of sequences
MSA algorithm searches subspace bounded by
Heuristic multiple alignment (suboptimal)
Pairwise optimal alignments

4. Pairwise projection Pairwise optimal alignment
Greedy multiple alignment
Global msa optimum (sequences A,B,C,…) is probably found
in the area in-between
Sequence A
Sequence B

5. Dynamic programming 3 4 1 2 Maximal path sum BEGIN  END ?2 4
BEGIN
END D
Maximal path sum BEGIN  END ?
Enumerate every path brute force
Use induction: only one optimal path up to any node in graph.

6. Example: all paths leading to BC 3 3 3 8 3 4 1
BEGIN 7
END 1 2 1 D

7. Heuristic alignment algorithms
Motivation: existing alignment methods
Heuristic alignment algorithms

8. Progressive alignment
1. derive a guide tree from pairwise sequence distances
2. align the closest (groups of) sequences
Pluses: works with many sequences
Minuses: errors made early on are frozen in the final alignment
Programs: Clustal, Pileup, T-Coffee

9. Progressive pairwise alignment
Motivation: existing alignment methods
Progressive pairwise alignment
Evolution
Alignment
TACAGTA
CACCTA
TATA
TACGTC
TACAGTA
TACGTA
TACGTA
TACGTC
TACCTA
TATA
CACCTA
time
progress

10. Progressive pairwise alignment
Motivation: existing alignment methods
Progressive pairwise alignment 1.
TACAGTA
T A C A G T A
match
vertical gap
horizontal gap 1. T A C G
T A C A G T A
T A C G T C 2.
TACGTC 3.
TATA
CACCTA T A 2.
T A T A
progress C A T 3.
C A C G T A

11. Progressive pairwise alignment
Motivation: existing alignment methods
Progressive pairwise alignment 1.
T A C A G T A
T A C G T C 2.
T A T A 3.
C A C G T A
T A C A G T A
T A C A G T A
T A C G T C
T A T A
C A C G T A
T A C G T C
T A T A
C A C C T A

12. Profile HMMs Motivation: existing alignment methods
CAPNTAGNHPCNGFGTCVPGHGDGSAANNVF

13. Iterative methods Gotoh DIALIGN inner loop uses progressive alignment
outer loop recomputes distances and guide tree based on the msa obtained
DIALIGN
Identify high-scoring ungapped segment pairs
Build up msa using consistent segment pairs
Greedy algorithm DIALIGN2; exact DIALIGN3

14. Motifs
- several proteins are grouped together by similarity searches - they share a conserved motif - motif is stringent enough to retrieve the family members from the complete protein database  

15. Local msa
Profile analysis a portion of the alignment that is highly conserved and produces a type of scoring matrix called a profile. New sequences can be aligned to the profile using dynamic programming.
Block analysis scans a global msa for ungapped regions, called blocks, and these blocks are then used in sequence alignments.
Pattern-searching or statistical methods find recurrent regions of sequence similarity in a set of initially unaligned sequences.

16. TEIRESIAS
Finds all patterns with minimum support in input set of unaligned sequences
Maximum spacer length
For example L…G…………..A….L…L
Enumerative algorithm
Build up longer patterns by merging overlapping short patterns
For example A….L + L…L  A….L…L
Fewer instances by chance when more positions are specified
Pluses: exact
Minuses: biological significance of the patterns?

17. Expectation maximization (EM)
Initial guess of motif location in each sequence
E-step: Estimate frequencies of amino acids in each motif column. Columns not in motif provide background frequencies. For each possible site location, calculate the probability that the site starts there.
M-step: use the site probabilities as weights to provide a new table of expected values for base counts for each of the site positions.
Repeat E-step and M-step until convergence.

18. EM procedure Eqn 1 Eqn 2 Eqn 1 Eqn 2
Let the observed variables be known as y and the latent variables as z. Together, y and z form the complete data. Assume that p is a joint model of the complete data with parameters θ: p(y,z | θ). An EM algorithm will then iteratively improve an initial estimate θ0 and construct new estimates θ1 through θN. An individual re-estimation step that derives from takes the following form (shown for the discrete case; the continuous case is similar):
Eqn 1
In other words, θn + 1 is the value that maximizes (M) the expectation (E) of the complete data log-likelihood with respect to the conditional distribution of the latent data under the previous parameter value. This expectation is usually denoted as Q(θ):
Eqn 2
Speaking of an expectation (E) step is a bit of a misnomer. What is calculated in the first step are the fixed, data-dependent parameters of the function Q. Once the parameters of Q are known, it is fully determined and is maximized in the second (M) step of an EM algorithm.
It can be shown that an EM iteration does not decrease the observed data likelihood function, and that the only stationary points of the iteration are the stationary points of the observed data likelihood function. In practice, this means that an EM algorithm will converge to a local maximum of the observed data likelihood function.
EM is particularly useful when maximum likelihood estimation of a complete data model is easy. If closed-form estimators exist, the M step is often trivial. A classic example is maximum likelihood estimation of a finite mixture of Gaussians, where each component of the mixture can be estimated trivially if the mixing distribution is known.
"Expectation-maximization" is a description of a class of related algorithms, not a specific algorithm; EM is a recipe or meta-algorithm which is used to devise particular algorithms. The Baum-Welch algorithm is an example of an EM algorithm applied to hidden Markov models. Another example is the EM algorithm for fitting a mixture density model.
An EM algorithm can also find maximum a posteriori (MAP) estimates, by performing MAP estimation in the M step, rather than maximum likelihood.
There are other methods for finding maximum likelihood estimates, such as gradient descent, conjugate gradient or variations of the Gauss-Newton method. Unlike EM, such methods typically require the evaluation of first and/or second derivatives of the likelihood function.
Eqn 1
Eqn 2

19. Exercise Seq1 C CAG A Seq2 G TTA A Seq3 G TAC C Seq4 T TAT T
Seq5 C AGA T
Seq6 T TTT G
Seq7 A TAC T
Seq8 C TAT G
Seq9 A GCT C
Seq10 G TAG A
Analyze the following ten DNA sequences by the expectation maximization algorithm. Assume that the background base frequencies are each 0.25 and that the middle three positions are a motif. The size of the motif is a guess that is based on a molecular model. The alignment of the sequences is also a guess.

20. (a) Calculate the observed frequency of each base at each of the three middle positions
Seq1 C CAG A
Seq2 G TTA A
Seq3 G TAC C
Seq4 T TAT T
Seq5 C AGA T
Seq6 T TTT G
Seq7 A TAC T
Seq8 C TAT G
Seq9 A GCT C
Seq10 G TAG A
Base
1st
2nd
3rd A
0.1
0.6
0.2 C G T
0.7
0.4

21. (b) Calculate the odds likelihood of finding the motif at each of the possible locations in sequence 5
Seq5 CAGAT
CAGAT:
0.1*0.6*0.2/0.25/0.25/0.25=0.768
0.1*0.1*0.2/0.25/0.25/0.25=0.128
0.1*0.6*0.4/0.25/0.25/0.25=1.536
Non-motif sites: 0.25/0.25=1
Base
1st
2nd
3rd A
0.1
0.6
0.2 C G T
0.7
0.4

22. (c) Calculate the probability of finding the motif at each position of sequence 5
Seq5 CAGAT
CAGAT:
0.1*0.6*0.2*0.25*0.25=
0.25*0.1*0.1*0.2*0.25=
0.25*0.25*0.1*0.6*0.4=0.0015
Non-motif sites: p=0.25
Base
1st
2nd
3rd A
0.1
0.6
0.2 C G T
0.7
0.4

23. (d) Calculate what change will be made to the base count in each column of the motif table as a result of matching the motif to sequence 5.
CAGAT: 0.1*0.6*0.2*0.25*0.25= , rel. weight 0.32
CAGAT: 0.25*0.1*0.1*0.2*0.25= , rel. weight 0.05
CAGAT: 0.25*0.25*0.1*0.6*0.4=0.0015, rel. weight 0.63
Base
1st
2nd
3rd A
0.05
=0.95 C
0.32 G
0.63 T
Updated counts from sequence 5 (initial location guess shaded).

24. (e) What other steps are taken to update or maximize the table values?
The weighted sequence data from the remaining sequences are also added to the counts table
The base frequencies in the new table are used as an updated estimate of the site residue composition.
The expectation and maximization steps are repeated until the base frequencies do not change.

25. Gibbs sampler
Start from random msa; realign one sequence against profile derived from the other sequences; iterate
Finds the most probable pattern common to all of the sequences by sliding them back and forth until the ratio of the motif probability to the background probability is a maximum.

26. A. Estimate the amino acid frequencies in the motif columns of all but one sequence.
Also obtain background.
Random start Motif
positions chosen ↓
xxxMxxxxx xxxMxxxxx
xxxxxxMxx xxxxxxMxx
xxxxxMxxx xxxxxMxxx
xMxxxxxxx xMxxxxxxx
Xxxxxxxxx Xxxxxxxxx
Mxxxxxxxx Mxxxxxxxx
xxxxMxxxx xxxxMxxxx
xxxxxxxxM xxxxxxxxM
B. Use the estimates from A to calculate the ratio of probability of motif to background
score at each position in the left-out sequence. This ratio for each possible location
is the weight of the position.
xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx
M M M M

27. C. Choose a new location for the motif in the left-out sequence by a random selection
using the weights to bias the choice.
xxxxxxxMxx Estimated location of the motif in left-out sequence.
D. Repeat steps A to C many (>100) times.

28. Exercise: Gibbs sampler
Seq1 C CAG A
Seq2 G TTA A
Seq3 G TAC C
Seq4 T TAT T
Seq5 C AGA T
Seq6 T TTT G
Seq7 A TAC T
Seq8 C TAT G
Seq9 A GCT C
Seq10 G TAG A
Analyze the left-hand-side DNA sequences by the Gibbs sampling algorithm.

29. (a) Assuming that the background base frequencies are 0
(a) Assuming that the background base frequencies are 0.25, calculate a log odds matrix for the central three positions.
Base
1st
2nd
3rd A
0.1
0.6
0.2 C G T
0.7
0.4
Base
1st
2nd
3rd A
-1.32
1.26
-0.32 C G T
1.49
0.68

30. GTTTG: -1.32 + -0.32 + 0.68 = -1.00 GTTTG: 1.49 + -0.32 + 0.68 = 1.81
(b) Assuming that another sequence GTTTG is the left-out sequence, slide the log-odds matrix along the left-out sequence and find the log-odds score at each of three possible positions.
Base
1st
2nd
3rd A
-1.32
1.26
-0.32 C G T
1.49
0.68
GTTTG: = -1.00
GTTTG: = 1.81
GTTTG: = 0.85

31. log-odds=log2(p/bg), p=2s bg GTTTG: 2-1.00/64=0.078, 0.1*0.2*0.4=0.008
(c) Calculate the probability of a match at each position in the left-out sequence
Base
1st
2nd
3rd A
0.1
0.6
0.2 C G T
0.7
0.4
log-odds=log2(p/bg), p=2s bg
GTTTG: /64=0.078, 0.1*0.2*0.4=0.008
GTTTG: /64=0.055, 0.7*0.2*0.4=0.056
GTTTG: 20.85/64=0.028, 0.7*0.2*0.2=0.028

32. (d) How do we choose a possible location for the motif in the left-out sequence?
=0.092
Normalised weights:
GTTTG: /0.092=0.09
GTTTG: /0.092=0.61
GTTTG: 0.028/0.092=0.30

33. Modelling protein families

34. PSSM
The PSSM is constructed by a logarithmic transformation of a matrix giving the frequency of each amino acid in the motif.
If a good sampling of sequences is available, the number of sequences is sufficiently large, and the motif structure is not too complex, it should, in principle, be possible to produce a PSSM that is highly representative of the same motif in other sequences also.
Pseudocounts
Sequence logo, information content

35. Exercise: construct a PSSM
Base
Column 1 frequency
Column 2 frequency
Column 3 frequency
Column 4 frequency A
0.6
0.1
0.2 C
0.7 G T

36. Exercise cntd
(a) Assuming the background frequency is 0.25 for each base, calculate a log odds score for each table position, i.e. log to the base 2 of the ratio of each observed value to the expected frequency.

37. Observed/expected frequencies
Base
Column 1 frequency
Column 2 frequency
Column 3 frequency
Column 4 frequency A
0.6/0.25=2.4
0.1/0.25=0.4
0.2/0.25=0.8 C
0.7/0.25=2.8 G T

38. Log-odds scores A C G T Base Column 1 frequency Column 2 frequency

39. Exercise cntd
(b) Align the matrix with each position in the sequence TCACGTAA starting at position 1,2, etc., and calculate the log odds score for the matrix to that position.

40. A C G T TCACGTAA: -0.32 + 1.49 + -0.32 + -1.32 = -0.47
Base
Column 1 frequency
Column 2 frequency
Column 3 frequency
Column 4 frequency A
1.26
-1.32
-0.32 C
1.49 G T
TCACGTAA: = -0.47
TCACGTAA: = -5.28
TCACGTAA: = 4.5
TCACGTAA: = -2.81
TCACGTAA: = -4.28

41. (c) Calculate the probability of the best matching position.
p(TCACGTAA): 0.6*0.7*0.6*0.7=0.1764

42. Alignment of sequences with a structure: hidden Markov models
HMMs suit well on describing correlation among neighbouring sites
probabilistic framework allows for realistic modelling
well developed mathematical methods; provide
best solution (most probable path through the model)
confidence score (posterior probability of any single solution)
inference of structure (posterior probability of using model states)
we can align multiple sequence using complex models and simultaneously predict their internal structure