Download Policy: Content on the Website is provided to you AS IS for your information and personal use only and may not be sold or licensed nor shared on other sites. SlideServe reserves the right to change this policy at anytime. While downloading, If for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
1. Alignment IIIPAM Matrices
2. 2 PAM250 scoring matrix
3. 3 Scoring Matrices S = [sij] gives score of aligning character i with character j for every pair i, j.
4. 4 Scoring with a matrix Optimum alignment (global, local, end-gap free, etc.) can be found using dynamic programming
No new ideas are needed
Scoring matrices can be used for any kind of sequence (DNA or amino acid)
5. 5 Types of matrices PAM
PAM, Gonnet, JTT, and DNA PAM matrices are based on an explicit evolutionary model; BLOSUM matrices are based on an implicit model
6. 6 PAM matrices are based on a simple evolutionary model
7. 7 Log-odds scoring What are the odds that this alignment is meaningful? X1X2X3?? Xn Y1Y2Y3?? Yn
Random model: Were observing a chance event. The probability is where pX is the frequency of X
Alternative: The two sequences derive from a common ancestor. The probability is where qXY is the joint probability that X and Y evolved from the same ancestor.
8. 8 Log-odds scoring Odds ratio:
Log-odds ratio (score):whereis the score for X, Y. The s(X,Y)s define a scoring matrix
9. 9 PAM matrices: Assumptions Only mutations are allowed
Sites evolve independently
Evolution at each site occurs according to a simple (first-order) Markov process
Next mutation depends only on current state and is independent of previous mutations
Mutation probabilities are given by a substitution matrix M = [mXY], where mxy = Prob(X?? Y mutation) = Prob(Y|X)
10. 10 PAM substitution matrices and PAM scoring matrices Recall that
Probability that X and Y are related by evolution: qXY = Prob(X)?? Prob(Y|X) = px ? mXY
11. 11 Mutation probabilities depend on evolutionary distance Suppose M corresponds to one unit of evolutionary time.
Let f be a frequency vector (fi = frequency of a.a. i in sequence). Then
M?f = frequency vector after one unit of evolution.
If we start with just amino acid i (a probability vector with a 1 in position i and 0s in all others) column i of M is the probability vector after one unit of evolution.
After k units of evolution, expected frequencies are given by Mk ? f.
12. 12 PAM matrices Percent Accepted Mutation: Unit of evolutionary change for protein sequences [Dayhoff78].
A PAM unit is the amount of evolution that will on average change 1% of the amino acids within a protein sequence.
13. 13 PAM matrices Let M be a PAM 1 matrix. Then,
Reason: Miis are the probabilities that a given amino acid does not change, so (1-Mii) is the probability of mutating away from i.
14. 14 The PAM Family
15. 15 Generating PAM matrices Idea: Find amino acids substitution statistics by comparing evolutionarily close sequences that are highly similar
Easier than for distant sequences, since only few insertions and deletions took place.
Computing PAM 1 (Dayhoffs approach):
Start with highly similar aligned sequences, with known evolutionary trees (71 trees total).
Collect substitution statistics (1572 exchanges total).
Let mij = observed frequency (= estimated probability) of amino acid Ai mutating into amino acid Aj during one PAM unit
Result: a 20 20 real matrix where columns add up to 1.
16. 16 Dayhoffs PAM matrix