1. Lecture 6: Sequence Alignment Statistics
CS 5263 Bioinformatics
Lecture 6: Sequence Alignment Statistics

2. Review of last lecture How to map gaps more accurately? GACGCCGAACG
||||| |||
GACGC---ACG
GACGCCGAACG
|||| | | ||
GACG-C-A-CG
Score = 8 x m – 3 x d
Score = 8 x m – 3 x d
Gaps usually occur in bunches
During evolution, chunks of DNA may be lost or inserted entirely
Aligning genomic sequences vs. cDNAs: cDNAs are spliced versions of the genomic seqs

3. Model gaps more accurately
Previous model:
Gap of length n incurs penalty nd
General:
Convex function
E.g. (n) = c * sqrt (n)
F(i-1, j-1) + s(xi, yj)
F(i, j) = max maxk=0…i-1F(k,j) – (i-k)
maxk=0…j-1F(i,k) – (j-k)
Running Time: O((M+N)MN) (cubic)
Space: O(NM)  n  n

4. Compromise: affine gaps
(n) = d + (n – 1)e
| |
gap gap
open extension e d
Match: 2
Gap open: -5
Gap extension: -1
GACGCCGAACG
||||| |||
GACGC---ACG
GACGCCGAACG
|||| | | ||
GACG-C-A-CG
8x2-5-2 = 9
8x2-3x5 = 1
We want to find the optimal alignment with affine gap penalty in
O(MN) time
O(MN) or better O(M+N) memory

5. Dynamic programming
Consider three sub-problems when aligning x1..xi and y1..yj
F(i,j): best alignment (score) of x1..xi & y1..yj if xi aligns to yj
Ix(i,j): best alignment of x1..xi & y1..yj if yj aligns to gap
Iy(i,j): best alignment of x1..xi & y1..yj if xi aligns to gap xi xi xi yj yj yj
F(i, j)
Ix(i, j)
Iy(i, j)

6. (-, yj) / e (xi,yj) /  (xi,yj) /  (-, yj) / d (xi,-) / d (xi,-) / e
Input
Output
(-, yj) / e
(xi,yj) /  Ix
(xi,yj) / 
(-, yj) / d F
(xi,-) / d Iy
(xi,-) / e
Start state
(xi,yj) / 
Current state
Input
Output
Next state F
(xi,yj) 
(-,yj) d Ix
(xi,-) Iy e …

7. (-, yj) / e (xi,yj) /  (-, yj) / d (xi,-) / d (xi,-) / eF Ix Iy
(xi,yj) / 
(xi,-) / d
(xi,-) / e
(-, yj) / d
(-, yj) / e
start
state
F-F-F-F
F-Iy-F-F-Ix
F-F-Iy-F-Ix
AAC
ACT
AAC
|||
ACT
AAC- ||
-ACT
AAC-
| |
A-CT
Given a pair of sequences, an alignment (not necessarily optimal) corresponds to a state path in the FSM.
Optimal alignment: a state path to read the two sequences such that the total output score is the highest

8. (-, yj)/e (xi,yj) / (xi,yj) / (-, yj) /d (xi,-) /d (xi,-)/eIx
(xi,yj) /
(-, yj) /d F
(xi,-) /d Iy
(xi,-)/e
(xi,yj) /
F(i-1, j-1) + (xi, yj)
F(i, j) = max Ix(i-1, j-1) + (xi, yj)
Iy(i-1, j-1) + (xi, yj) xi yj

9. F(i, j-1) + d Ix(i, j) = max Ix(i, j-1) + e (-, yj)/e (xi,yj) /
(-, yj) /d F
(xi,-) /d Iy
(xi,-)/e
(xi,yj) /
F(i, j-1) + d
Ix(i, j) = max
Ix(i, j-1) + e xi yj
Ix(i, j)

10. F(i-1, j) + d Iy(i, j) = max Iy(i-1, j) + e (-, yj)/e (xi,yj) /Ix
(xi,yj) /
(-, yj) /d F
(xi,-) /d Iy
(xi,-)/e
(xi,yj) /
F(i-1, j) + d
Iy(i, j) = max
Iy(i-1, j) + e xi yj
Iy(i, j)

11. F(i, j) = (xi, yj) + max Ix(i – 1, j – 1) Iy(i – 1, j – 1)
F(i, j – 1) + d
Ix(i, j) = max
Ix(i, j – 1) + e
F(i – 1, j) + d
Iy(i, j) = max
Iy(i – 1, j) + e
Continuing alignment
Closing gaps in x
Closing gaps in y
Opening a gap in x
Gap extension in x
Opening a gap in y
Gap extension in y

12. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A-
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A
F: aligned on both
Iy: Insertion on y
y =
G C C
F(i-1, j-1)
Iy(i-1, j-1)
x = - -5 -6 -7
Iy(i-1,j)
(xi, yj)
F(i-1,j) G C A e
Ix(i-1, j-1) d
F(i, j)
F(i,j-1)
Iy(i,j) d
Ix(i,j)
Ix(i,j-1) e
Ix: Insertion on x

13. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = - -5 -6 -7
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = 2
Ix(i-1, j-1)
F(i, j) Ix

14. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = - -5 -6 -7
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = -2
Ix(i-1, j-1)
F(i, j) Ix

15. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4 G C A
F(i,j-1)
d = -5
Ix(i,j)
Ix(i,j-1)
e = -1 Ix

16. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = -2
Ix(i-1, j-1)
F(i, j) Ix

17. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = 2
Ix(i-1, j-1)
F(i, j) Ix

18. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
-12 -1 G C A
F(i,j-1)
d = -5
Ix(i,j)
Ix(i,j-1)
e = -1 Ix

19. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1
x = - -5 -6 -3 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
-12 -1
Iy(i-1,j) G C A
F(i-1,j)
e=-1
d=-5
Iy(i,j) Ix

20. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1 -5 -9 -6 1
x = - -5 -6 -3
-12
-13 -7 -8 -1 -2
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
F(i-1, j-1)
Iy(i-1, j-1)
x =
Iy(i-1,j) -5 -6 -7 - -3 -4
-12 -1
-13
-10
-14
-11
(xi, yj)
F(i-1,j) G C A e
Ix(i-1, j-1) d
F(i, j)
F(i,j-1)
Iy(i,j) d
Ix(i,j)
Ix(i,j-1) e Ix

21. GCAC || | GC-C y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1- 2 -7 -8 4 -1 -5 -9 -6 1
x = - -5 -6 -3
-12
-13 -7 -8 -1 -2
m = 2
s = -2
d = -5
e = -1 G C A G C A x
GCAC
|| |
GC-C y F Iy
y =
G C C
y =
G C C
x = -5 -6 -7 - -3 -4
-12 -1
-13
-10
-14
-11
x = G C A G C A Ix

22. Today: statistics of alignment
Where does (xi, yj) come from?
Are two aligned sequences actually related?

23. Probabilistic model of alignments
We’ll first focus on protein alignments without gaps
Given an alignment, we can consider two possible models
R: the sequences are related by evolution
U: the sequences are unrelated
How can we distinguish these two models?
How is this view related to amino-acid substitution matrix?

24. Model for unrelated sequences
Assume each position of the alignment is independently sampled from some distribution of amino acids
ps: probability of amino acid s in the sequences
Probability of seeing an amino acid s aligned to an amino acid t by chance is
Pr(s, t | U) = ps * pt
Probability of seeing an ungapped alignment between x = x1…xn and y = y1…yn randomly is i

25. Model for related sequences
Assume each pair of aligned amino acids evolved from a common ancestor
Let qst be the probability that amino acid s in one sequence is related to t in another sequence
The probability of an alignment of x and y is give by

26. Probabilistic model of Alignments
How can we decide which model (U or R) is more likely?
One principled way is to consider the relative likelihood of the two models (the odd ratios)
A higher ratio means that R is more likely than U

27. Log odds ratio Taking logarithm, we get
Recall that the score of an alignment is given by

28. Therefore, if we define
We are actually defining the alignment score as the log odds ratio between the two models R and U

29. How to get the probabilities?
ps can be counted from the available protein sequences
But how do we get qst? (the probability that s and t have a common ancestor)
Counted from trusted alignments of related sequences

30. Protein Substitution Matrices
Two popular sets of matrices for protein sequences
PAM matrices [Dayhoff et al, 1978]
Better for aligning closely related sequences
BLOSUM matrices [Henikoff & Henikoff, 1992]
For both closely or remotely related sequences

31. BLOSUM-N matrices Constructed from a database called BLOCKS
Contain many closely related sequences
Conserved amino acids may be over-counted
N = 62: the probabilities qst were computed using trusted alignments with no more than 62% identity
identity: % of matched columns
Using this matrix, the Smith-Waterman algorithm is most effective in detecting real alignments with a similar identity level (i.e. ~62%)

32. : Scaling factor to convert score to integer.
Important: when you are told that a
scoring matrix is in half-bits =>  = ½ ln2
Positive for chemically similar substitution
Common amino acids get low weights
Rare amino acids get high weights

33. BLOSUM-N matrices
If you want to detect homologous genes with high identity, you may want a BLOSUM matrix with higher N. say BLOSUM75
On the other hand, if you want to detect remote homology, you may want to use lower N, say BLOSUM50
BLOSUM-62: good for most purposes
Weak homology
Strong homology

34. For DNAs No database of trusted alignments to start with
Specify the percentage identity you would like to detect
You can then get the substitution matrix by some calculation

35. For example Suppose pA = pC = pT = pG = 0.25 We want 88% identity
qAA = qCC = qTT = qGG = 0.22, the rest = 0.12/12 = 0.01
(A, A) = (C, C) = (G, G) = (T, T)
= log (0.22 / (0.25*0.25)) = 1.26
(s, t) = log (0.01 / (0.25*0.25)) = for s ≠ t.

36. Substitution matrix A C G T
1.26
-1.83

37. A C G T 5 -7
Scale won’t change the alignment
Multiply by 4 and then round off to get integers

38. Arbitrary substitution matrix
Say you have a substitution matrix provided by someone
It’s important to know what you are actually looking for when you use the matrix

39. Which one should I use for my sequences?
NCBI-BLAST
WU-BLAST A C G T 1 -2 A C G T 5 -4
What’s the difference?
Which one should I use for my sequences?

40. We had
Scale it, so that
Reorganize:

41. Since all probabilities must sum to 1,
We have
Suppose again ps = 0.25 for any s
We know (s, t) from the substitution matrix
We can solve the equation for λ
Plug λ into to get qst

42. A C G T 1 -2 A C G T 5 -4 Translate: 95% identity
NCBI-BLAST
WU-BLAST A C G T 1 -2 A C G T 5 -4
 = 1.33
qst = 0.24 for s = t, and for s ≠ t
Translate: 95% identity
 = 0.19
qst = 0.16 for s = t, and 0.03 for s ≠ t
Translate: 65% identity

43. Details for solving  A C G T 1 -2
Known: (s,t) = 1 for s=t, and (s,t) = -2 for s t.
Since
and s,t qst = 1, we have
12 * ¼ * ¼ * e-2 + 4 * ¼ * ¼ * e = 1
Let e = x, we have
¾ x-2 + ¼ x = 1. Hence,
x3 – 4x2 + 3 = 0;
X has three solutions: 3.8, 1, -0.8
Only the first leads to a positive 
 = ln (3.8) = 1.33 A C G T 1 -2

44. Today: statistics of alignment
Where does (xi, yj) come from?
Are two aligned sequences actually related?

45. Statistics of Alignment Scores
Q: How do we assess whether an alignment provides good evidence for homology (i.e., the two sequences are evolutionarily related)?
Is a score 82 good? What about 180?
A: determine how likely it is that such an alignment score would result from chance

46. P-value of alignment p-value
The probability that the alignment score can be obtained from aligning random sequences
Small p-value means the score is unlikely to happen by chance
The most common thresholds are 0.01 and 0.05
Also depend on purpose of comparison and cost of misclaim

47. Statistics of global seq alignment
Theory only applies to local alignment
For global alignment, your best bet is to do Monte-Carlo simulation
What’s the chance you can get a score as high as the real alignment by aligning two random sequences?
Procedure
Given sequence X, Y
Compute a global alignment (score = S)
Randomly shuffle sequence X (or Y) N times, obtain X1, X2, …, XN
Align each Xi with Y, (score = Ri)
P-value: the fraction of Ri >= S

48. Human HEXA
Fly HEXO1
Score = -74

49. -74
Distribution of the alignment scores between fly HEXO1 and 200 randomly shuffled human HEXA sequences
There are 88 random sequences with alignment score >= -74.
So: p-value = 88 / 200 = 0.44 => alignment is not significant

50. ……………………………………………………
Mouse HEXA
Human HEXA
Score = 732
……………………………………………………

51. No random sequences with alignment score >= 732
Distribution of the alignment scores between mouse HEXA and 200 randomly shuffled human HEXA sequences
732
No random sequences with alignment score >= 732
So: the P-value is less than 1 / 200 = 0.05
To get smaller p-value, have to align more random sequences
Very slow
Unless we can fit a distribution (e.g. normal distribution)
Such distribution may not be generalizable
No theory exists for global alignment score distribution

52. Statistics for local alignment
Elegant theory exists
Score for ungapped local alignment follows extreme value distribution (Gumbel distribution)
Normal distribution
Extreme value distribution
An example extreme value distribution:
Randomly sample 100 numbers from a normal distribution, and compute max
Repeat 100 times.
The max values will follow extreme value distribution

53. Statistics for local alignment
Given two unrelated sequences of lengths M, N
Expected number of ungapped local alignments with score at least S can be calculated by
E(S) = KMN exp[-S]
Known as E-value
: scaling factor as computed in last lecture
K: empirical parameter ~ 0.1
Depend on sequence composition and substitution matrix

54. P-value for local alignment score
P-value for a local alignment with score S
when P is small.

55. Example You are aligning two sequences, each has 1000 bases
m = 1, s = -1, d = -inf (ungapped alignment)
You obtain a score 20
Is this score significant?

56.  = ln3 = 1.1 (computed as discussed on slide #41)
E(S) = K MN exp{- S}
E(20) = 0.1 * 1000 * 1000 * 3-20 = 3 x 10-5
P-value = 3 x 10-5 << 0.05
The alignment is significant
Distribution of 1000 random sequence pairs 20

57. Multiple-testing problem
Searching a 1000-base sequence against a database of 106 sequences (each of length 1000)
How significant is a score 20 now?
You are essentially comparing 1000 bases with 1000x106 = 109 bases (ignore edge effect)
E(20) = 0.1 * 1000 * 109 * 3-20 = 30
By chance we would expect to see 30 matches
The P-value (probability of seeing at least one match with score >= 30) is 1 – e-30 =
The alignment is not significant
Caution: it does NOT mean that the two sequences are unrelated. Rather, it simply means that you have NO confidence to say whether the two sequences are related.

58. Score threshold to determine significance
You want a p-value that is very small (even after taking into consideration multiple-testing)
What S will guarantee you a significant p-value?
E(S)  P(S) << 1
=> KMN exp[-S] << 1
=> log(KMN) -S < 0
=> S > T + log(MN) /  (T = log(K) / , usually small)

59. Score threshold to determine significance
In the previous example
m = 1, s = -1, d = -inf =>  = 1.1
Aligning 1000bp vs 1000bp
S > log(106) / 1.1 = 13.
So 20 is significant.
Searching 1000bp against 106 x 1000bp
S > log(1012) / 1.1 = 25.
so 20 is not significant.

60. Statistics for gapped local alignment
Theory not well developed
Extreme value distribution works well empirically
Need to estimate K and  empirically
Given the database and substitution matrix, generate some random sequence pairs
Do local alignment
Fit an extreme value distribution to obtain K and 

61. In summary How to obtain a substitution matrix?
Obtain qst and ps from established alignments (for DNA: from your knowledge)
Computing score:
How to understand arbitrary substitution matrix?
Solve function to obtain  and target qst
Which tells you what percent identity you are expecting
How to understand alignment score?
probability that a score can be expected from chance.
Global alignment: Monte-Carlo simulation
Local alignment: Extreme Value Distribution
Estimate p-value from a score
Determine a score threshold without computing a p-value