1. Scoring Matrices June 19, 2008 Learning objectives- Understand how scoring matrices are constructed. Workshop-Use different BLOSUM matrices in the Dotter Program to determine their effects on output when comparing squid p53 and human p53. Create your own scoring matrix and use it to compare two protein sequences. Explain to the instructor the rationale behind your scoring matrix.

2. Scoring Matrices Scoring matrices appear in all analyses involving sequence comparisons. Scoring matrices implicitly represent a particular theory of relationships. Understanding theories underlying a given scoring matrix can aid in making proper choice of scoring matrix.

3. Scoring Matrices When we consider scoring matrices, we encounter the convention that matrices have numeric indices corresponding to the rows and columns of the matrix. For example, M 12 refers to the entry at the first row and the second column. In general, M ij refers to the entry at the ith row and the jth column.

4. Two major scoring matrices for amino acid sequence comparisons PAM-derived from sequences known to be closely related (Eg. Chimpanzee and human). Generally ranges from PAM 1 to PAM 500 BLOSUM-derived from sequences not closely related (Eg. E. coli and human). Ranges from BLOSUM 10-BLOSUM 100

5. The Point-Accepted-Mutation (PAM) model of evolution and the PAM scoring matrix Started by Margaret Dayhoff, 1978 A series of matrices describing the extent to which two amino acids have been interchanged in evolution PAM 1 scoring matrix was obtained by aligning very similar sequences. Other PAMs were obtained by mathematical extrapolation 1) neighbor independence; 2) positional independence; and 3) historical independence. Dayhoff, M. O., Atlas of Protein Sequence and Structure Natl. Biomed. Res. Found., Silver Spring MD, 1978.

6. Protein families used to construct Dayhoff’s scoring matrix ProteinPAMs per 100 mil yrs IgG kappa C region37 Kappa casein33 Serum Albumin26 Cytochrome C0.9 Histone H30.14 Histone H40.10

7. Calculation of relative mutability of amino acid Find frequency of amino acid change at a certain position in protein. Divide this “change frequency” by the frequency that the amino acid occurs in all proteins. This gives the mutability of the amino acid. Multiply the alanine mutability by a factor to get the value 100. Multiply the 19 other a.a. mutabilities by the same factor. Result: Relative Mutabilities

8. Relative mutabilities of amino acids Asn134 Ser120 Asp106 Glu102 Ala100 Thr 97 Ile 96 Met 94 Gln 93 Val 74 His66 Arg65 Lys56 Pro56 Gly49 Tyr41 Phe41 Leu40 Cys20 Trp18

9. Why are the mutabilities different? High mutabilities because a similar amino acid can replace it. (Asp for Glu) Conversely, the low mutabilities are unique, can’t be replaced.

10. Tally all pairwise replacements Next, tally a.a. replacements "accepted" by natural selection, in all pair-wise sequence comparisons.

11. A R N D C Q E G H I L K M F P S T W Y V A R 30 N 109 17 D 154 0 532 C 33 10 0 0 Q 93 120 50 76 0 E 266 0 94 831 0 422 G 579 10 156 162 10 30 112 H 21 103 226 43 10 243 23 10 I 66 30 36 13 17 8 35 0 3 L 95 17 37 0 0 75 15 17 40 253 K 57 477 322 85 0 147 104 60 23 43 39 M 29 17 0 0 0 20 7 7 0 57 207 90 F 20 7 7 0 0 0 0 17 20 90 167 0 17 P 345 67 27 10 10 93 40 49 50 7 43 43 4 7 S 772 137 432 98 117 47 86 450 26 20 32 168 20 40 269 T 590 20 169 57 10 37 31 50 14 129 52 200 28 10 73 696 W 0 27 3 0 0 0 0 0 3 0 13 0 0 10 0 17 0 Y 20 3 36 0 30 0 10 0 40 13 23 10 0 260 0 22 23 6 V 365 20 13 17 33 27 37 97 30 661 303 17 77 10 50 43 186 0 17 Numbers of accepted point mutations, multiplied by 10 Original amino acids Replacement amino acids

12. Creation of a mutation probability matrix Used accepted point mutation data from previous slide and the mutability of each amino acid to create a mutation probability matrix. M ij =(m j *A ij )/(sum_over_all_i A ij ) M ij shows the probability that an original amino acid j (in columns) will be replaced by amino acid i (in rows) over a defined evolutionary interval. For PAM 1, an average of 1% of aa’s were changed.

13. PAM1 mutational probability matrix Values of each column will sum to 10,000 Orig. aa Replacement aa

14. The Point-Accepted-Mutation (PAM) model of evolution and the PAM scoring matrix Observed % aa Difference Evolutionary Distance in PAMs 1 5 10 20 40 50 60 70 80 1 5 11 23 56 80 112 159 246

15. Final Scoring Matrix is the Log- Odds Score Matrix S (a,b) = 10 log 10 (M ab /P b ) Original amino acid Replacement amino acid Mutational probability matrix number (from PAM 250) Normalized frequency of amino acid b S(a,alanine) = 10 log(0.13/0.087)=1.7 (round to 2)

16. At this evolution- ary distance, there is a 13% chance that the second sequence will also have an alanine.

17. Summary of PAM Scoring Matrix PAM = a unit of evolution (1 PAM = average of 1 point mutation/100 amino acids) Accepted Mutation means fixed point mutation Comparison of 71 groups of closely related proteins yielding 1,572 changes. (>85% identity) Different PAM matrices are derived from the PAM 1 matrix by matrix multiplication. The matrices are converted to log odds matrices.

18. BLOSUM Matrix (BLOcks SUbstitution Matrices) Blocks Sum-created from BLOCKS database A series of matrices describing the extent to which two amino acids are interchangeable in conserved structures of proteins The number in the series represents the threshold percent similarity between sequences, for consideration for calculation (Eg. BLOSUM62 means 62% of the aa’s were similar)

19. BLOSUM BLOSUMs are built from distantly related sequences within conserved blocks of sequences BLOSUMs are built from the BLOCKS database (the BLOCKS database is a secondary database that derives information from the PROSITE Family database)

20. BLOSUM (cont.1) Version 8.0 of the Blocks Database consists of 2884 blocks based on 770 protein families documented in PROSITE. Hypothetical entry in red box in BLOCK record: AABCDA...BBCDA DABCDA.A.BBCBB BBBCDABA.BCCAA AAACDAC.DCBCDB CCBADAB.DBBDCC AAACAA...BBCCC

21. Building BLOSUM Matrices 1. To build the BLOSUM 62 matrix one must eliminate sequences that are identical in more than 62% of their amino acid sequences. This is done by either removing sequences from the BLOCK or by finding a cluster of similar sequences and replacing the cluster with a single representative sequence. 2. Next, the probability for a pair of amino acids to be placed in the same column is calculated. In the previous page this would be the probability of replacement of A with A, A with B, A with C, and B with C. This gives the value q ij 3. Next, one calculates the frequency that the replacement amino acid exists in nature, f i.

22. Building BLOSUM Matrices (cont.) 4. Finally, we calculate the log odds ratio s i,j = log 2 (q ij /f i ). This value is entered into the matrix. Which BLOSUM to use? BLOSUM Identity 80 80% 62 62% (usually default value) 35 35% If you are comparing sequences that are very similar, use BLOSUM 80.

23. Which Scoring Matrix to use? PAM-1 BLOSUM-100 Small evolutionary distance High identity within short sequences PAM-250 BLOSUM-20 Large evolutionary distance Low identity within long sequences

24. Workshop