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CAP5510 – Bioinformatics Substitution Patterns Tamer Kahveci CISE Department University of Florida 1 Goals • Understand how mutations occur • Learn models for predicting the number of mutations • Understand why scoring matrices are used and how they are derived • Learn major scoring matrices 2 Why Substitute Patterns ? • Mutations happen because of mistakes in DNA replication and repair. • Our genetic code changes due to mutations – Insert, delete, replace • Three types of mutations – Advantageous – Disadvantageous – Neutral • We only observe substitutions that passed selection process 3 Mutation Rates Parent Organism T time Organism A R = K/(2T) Organism B K: number of substitutions 4 Functional Constraints • Functional sites are less likely to mutate – Noncoding = 3.33 (subs/109 yr) – Coding = 1.58 (subs/109 yr) • Indels about 10 times less likely than substitutions 5 Nucleotide Substitutions and Amino Acids • Synonymous substitutions do not change amino acids • Nonsynonymous do change • Degeneracy – Fourfold degenerate: gly = {GGG, GGA, GGU, GGC} – Twofold degenerate: asp = {GAU, GAC}, glu = {GAA, GAG} – Non-degenerate: phe = UUU, leu = CUU, ile = AUU, val = GUU • Example substitution rates in human and mouse – Fourfold degenerate: 2.35 – Twofold degenerate: 1.67 – Non-degenerate: 0.56 6 Predicting Substitutions How can we count the true number of substitutions ? 7 Jukes-Cantor Model • Each nucleotide can change into another one with the same probability P(A->A’, 1) = x, for each A’ P(A->A, 1) = 1 – 3x Compute P(A->A’, 2) & P(A->A, 2) x A x G C x T P(A->A, t+1) = 3 P(A->A’, t) P(A’->A, 1) + P(A->A, t) P(A->A, 1) P(A->A, t) ~ ¼ + (3/4)e-4ft K = num. subst. = -¾ ln(1 – f4/3), f = fraction of observed substitutions Oversimplification 8 Two Parameter Model • Transition: – purine->purine (A, G), pyrimidine->pyrimidine (C, T) Purine • Transversion: – purine <-> pyrimidine • Transitions are more likely than transversions. • Use different probabilities for transitions and transversions. Pyrimidine 9 Two Parameter Model •P(AA,1) = 1-x-2y •Compute P(AA,2) y A x G C y T P(AA,2) = (1-x-2y) P(AA,1) + x P(AG,1) + y P(AC,1) + y P(AT,1) P(AA,t) = ¼ + ¼ e-4yt + ½ e-2(x+y)t K = ½ ln(1/(1-2P-Q)) + ¼ ln(1/(1-2Q)) P,Q: fraction of transitions and transversions observed. 10 More Parameters ? • Assign a different probability for each pair of nucleotides • Not harder to compute than simpler models • Not necessarily better than simpler models 11 Amino Acid substitutions (1) • Harder to model than nucleotides – An amino acid can be substituted for another in more than one ways – The number of nucleotide substitutions needed to transform one amino acid to another may differ • Pro = CCC, leu = CUC, ile = AUC – The likelihood of nucleotide substitutions may differ • Asp = GAU, asn = AAU, his = CAU – Amino acid substitutions may have different effects on the protein function 12 Amino Acid substitutions (2) • Mutation rates may vary greatly among genes – Nonsynonymous substitution may affect functionality with smaller probability in some genes • Molecular clock (Zuckerlandl, Paulding) – Mutation rates may be different for different organisms, but it remains almost constant over the time. 13 Scoring Matrices 14 What is it & why ? • Let alphabet contain N letters – N = 4 and 20 for nucleotides and amino acids • N x N matrix • (i,j) shows the relationship between ith and jth letters. – Positive number if letter i is likely to mutate into letter j – Negative otherwise – Magnitude shows the degree of proximity • Symmetric 15 The BLOSUM45 Matrix A R N D C Q E G H I L K M F P S T W Y V A 5 -2 -1 -2 -1 -1 -1 0 -2 -1 -1 -1 -1 -2 -1 1 0 -2 -2 0 R -2 7 0 -1 -3 1 0 -2 0 -3 -2 3 -1 -2 -2 -1 -1 -2 -1 -2 N -1 0 6 2 -2 0 0 0 1 -2 -3 0 -2 -2 -2 1 0 -4 -2 -3 D -2 -1 2 7 -3 0 2 -1 0 -4 -3 0 -3 -4 -1 0 -1 -4 -2 -3 C -1 -3 -2 -3 12 -3 -3 -3 -3 -3 -2 -3 -2 -2 -4 -1 -1 -5 -3 -1 Q -1 1 0 0 -3 6 2 -2 1 -2 -2 1 0 -4 -1 0 -1 -2 -1 -3 E -1 0 0 2 -3 2 6 -2 0 -3 -2 1 -2 -3 0 0 -1 -3 -2 -3 G 0 -2 0 -1 -3 -2 -2 7 -2 -4 -3 -2 -2 -3 -2 0 -2 -2 -3 -3 H -2 0 1 0 -3 1 0 -2 10 -3 -2 -1 0 -2 -2 -1 -2 -3 2 -3 I -1 -3 -2 -4 -3 -2 -3 -4 -3 5 2 -3 2 0 -2 -2 -1 -2 0 3 L -1 -2 -3 -3 -2 -2 -2 -3 -2 2 5 -3 2 1 -3 -3 -1 -2 0 1 K -1 3 0 0 -3 1 1 -2 -1 -3 -3 5 -1 -3 -1 -1 -1 -2 -1 -2 M -1 -1 -2 -3 -2 0 -2 -2 0 2 2 -1 6 0 -2 -2 -1 -2 0 1 F -2 -2 -2 -4 -2 -4 -3 -3 -2 0 1 -3 0 8 -3 -2 -1 1 3 0 P -1 -2 -2 -1 -4 -1 0 -2 -2 -2 -3 -1 -2 -3 9 -1 -1 -3 -3 -3 S 1 -1 1 0 -1 0 0 0 -1 -2 -3 -1 -2 -2 -1 4 2 -4 -2 -1 T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -1 -1 2 5 -3 -1 0 W -2 -2 -4 -4 -5 -2 -3 -2 -3 -2 -2 -2 -2 1 -3 -4 -3 15 3 -3 Y -2 -1 -2 -2 -3 -1 -2 -3 2 0 0 -1 0 3 -3 -2 -1 3 8 -1 V 0 -2 -3 -3 -1 -3 -3 -3 -3 3 1 -2 1 0 -3 -1 0 -3 -1 5 16 Scoring Matrices for DNA A A A C G T 1 0 0 0 A C C 0 1 0 T 0 0 0 0 identity 1 0 -3 -3 1 G -3 -3 T -3 -3 1 1 -3 -3 -3 BLAST C G T A 1 -5 -1 -5 C -5 1 -5 -1 G -1 -5 1 -5 T -5 -1 -5 1 -3 -3 0 T A -3 0 G G 1 C 1 Transitions & transversions 17 Scoring Matrices for Amino Acids • Chemical similarities – – – – Non-polar, Hydrophobic (G, A, V, L, I, M, F, W, P) Polar, Hydrophilic (S, T, C, Y, N, Q) Electrically charged (D, E, K, R, H) Requires expert knowledge • Genetic code: Nucleotide substitutions – E: GAA, GAG – D: GAU, GAC – F: UUU, UUC • Actual substitutions – PAM – BLOSUM 18 Scoring Matrices: Actual Substitutions • • • • Manually align proteins Look for amino acid substitutions Entry ~ log(freq(observed)/freq(expected)) Log-odds matrices 19 PAM Matrices (Dayhoff 1972) 20 PAM • PAM = “Point Accepted Mutation” interested only in mutations that have been “accepted” by natural selection • An accepted mutation is a mutation that occurred and was positively selected by the environment; that is, it did not cause the demise of the particular organism where it occurred. 21 Interpretation of PAM matrices • PAM-1 : one substitution per 100 residues (a PAM unit of time) • “Suppose I start with a given polypeptide sequence M at time t, and observe the evolutionary changes in the sequence until 1% of all amino acid residues have undergone substitutions at time t+n. Let the new sequence at time t+n be called M’. What is the probability that a residue of type j in M will be replaced by i in M’?” • PAM-K : K PAM time units 22 PAM Matrices (1) • Starts with a multiple sequence alignment of very similar (>85% identity) proteins. Assumed to be homologous • Compute the relative mutability, mi, of each amino acid – e.g. mA = how many times was alanine substituted with anything else on the average? 23 Relative Mutability • ACGCTAFKI GCGCTAFKI ACGCTAFKL GCGCTGFKI GCGCTLFKI ASGCTAFKL ACACTAFKL • Across all pairs of sequences, there are 28 A X substitutions • There are 10 ALA residues, so mA = 2.8 24 Pam Matrices (2) • Construct a phylogenetic tree for the sequences in the alignment ACGCTAFKI AG GCGCTAFKI AG GCGCTGFKI FG,A = 3 IL ACGCTAFKL AL GCGCTLFKI CS ASGCTAFKL GA ACACTAFKL • Calculate substitution frequencies FX,X • Substitutions may have occurred either way, so A G also counts as G A. 25 Mutation Probabilities • Mi,j represents the probability of J I substitution. ACGCTAFKI AG IL GCGCTAFKI AG AL GCGCTGFKI M ij ACGCTAFKL GCGCTLFKI m j Fij F ij i CS ASGCTAFKL M G, A GA ACACTAFKL 2.8 ´ 3 = 2.1 4 26 The PAM Matrix • The entries of the scoring matrix are the Mi,j values divided by the frequency of occurrence, fi, of residue i. • fG = 10 GLY / 63 residues = 0.1587 • RG,A = log(2.1/0.1587) = log(12.760) = 1.106 • Log-odds matrix • Diagonal entries are Mjj = 1– mj 27 Computation of PAM-K • Assume that changes at time T+1 are independent of the changes at time T. • Markov chain • P(A-->B) = X P(A->X) P(X->B) • PAM-K = (PAM-1)K • PAM-250 is most commonly used 28 PAM - Discussion • Smaller K, PAM-K is better for closely related sequences, large K is better for distantly related sequences • Biased towards closely related sequences since it starts from highly similar sequences (BLOSUM solves this) • If Mi,j is very small, we may not have a large enough sample to estimate the real probability. When we multiply the PAM matrices many times, the error is magnified. • Mutation rate may change from one gene to another 29 BLOSUM Matrices Henikoff & Henikoff 1992 30 BLOSUM Matrix • Begin with a set of protein sequences and obtain blocks. – ~2000 blocks from 500 families of related proteins – More data than PAM • A block is the ungapped alignment of a highly conserved region of a family of proteins. • MOTIF program is used to find blocks • Substitutions in these blocks are used to compute BLOSUM matrix block 1 block 2 WWYIR WFYVR WYYVR WYFIR CASILRKIYIYGPV CASILRHLYHRSPA AAAVARHIYLRKTV AASICRHLYIRSPA … block 3 … GVSRLRTAYGGRKNRG GVGSITKIYGGRKRNG GVGRLRKVHGSTKNRG GIGSFEKIYGGRRRRG 31 Constructing the Matrix • Count the frequency of occurrence of each amino acid. This gives the background distribution pa • Count the number of times amino acid a is aligned with amino acid b: fab – A block of width w and depth s contributes ws(s-1)/2 = np pairs • Compute the occurrence probability of each pair – qab = fab/ np • Compute the probability of occurrence of amino acid a i – pa = qaa + Σ qab /2 • Compute the expected probability of occurrence of each pair a≠b • – eab = 2papb, if a ≠ b papb otherwise Compute the log likelihood ratios, normalize, and round. – 2* log2 qab / eab 32 Constructing the Matrix: Example • fAA = 36, fAS = 9 • Observed frequencies of pairs A A A A S … A … A A A A – qAA = fAA/(fAA+fAS) = 36/45 = 0.8 – qAS = 9/45 = 0.2 • Expected frequencies of letters – pA = qAA + qAS/2 = 0.9 – pS = qAS/2 = 0.1 • Expected frequencies of pairs – eAA = pA x pA = 0.81 – eAS = 2 x pA x pS = 0.18 • Matrix entries – MAA = 2x log2(qAA/eAA) = -0.04 ~ 0 – MAS = 2 x log2(qAS/eAS) = 0.3 ~ 0 9A, 1S 33 Computation of BLOSUM-K • Different levels of the BLOSUM matrix can be created by differentially weighting the degree of similarity between sequences. For example, a BLOSUM62 matrix is calculated from protein blocks such that if two sequences are more than 62% identical, then the contribution of these sequences is weighted to sum to a b one. In this way the contributions of multiple entries of closely related sequences is reduced. • Larger numbers used to measure recent divergence, default is BLOSUM62 34 BLOSUM 62 Matrix Check scores for MILV -small hydrophobic NDEQ -acid, hydrophilic HRK -basic FYW -aromatic STPAG -small hydrophilic C -sulphydryl 35 PAM vs. BLOSUM Equivalent PAM and BLOSSUM matrices: PAM100 PAM120 PAM160 PAM200 PAM250 = = = = = Blosum90 Blosum80 Blosum60 Blosum52 Blosum45 BLOSUM62 is the default matrix to use. 36 PAM vs. BLOSUM PAM Built from global alignments Built from small amout of Data Counting is based on minimum replacement or maximum parsimony Perform better for finding global alignments and remote homologs Higher PAM series means more divergence BLOSUM Built from local alignments Built from vast amout of Data Counting based on groups of related sequences counted as one Better for finding local alignments Lower BLOSUM series means more divergence 37