1. 1 Introduction to molecular evolution Lecture 13, Statistics 246 March 4, 2004

2. 2 Evolution using molecules: implicit assumptions Our DNA is inherited from our parents more or less unchanged. Molecular evolution is dominated by mutations that are neutral from the standpoint of natural selection. Mutations accumulate at fairly steady rates in surviving lineages. We can study the evolution of (macro) molecules and reconstruct the evolutionary history of organisms using their molecules.

3. 3 Some important dates in history (billions of years ago) Origin of the universe15  4 Formation of the solar system 4.6 First self-replicating system 3.5  0.5 Prokaryotic-eukaryotic divergence 1.8  0.3 Plant-animal divergence 1.0 Invertebrate-vertebrate divergence 0.5 Mammalian radiation beginning 0.1 86 CSH Doolittle et al.

4. 4 The three kingdoms

5. 5 Two important early observations Different proteins evolve at different rates, and this seems more or less independent of the host organism, including its generation time. It is necessary to adjust the observed percent difference between two homologous proteins to get a distance more or less linearly related to the time since their common ancestor. ( Later we offer a rational basis for doing this.) An striking early version of these observations is next.

6. 6 Rates of macromolecular evolution

7. 7 Protein a PAMs/100 residues/10 8 years Theoretical lookback time b Pseudogenes 400 45 c Fibrinopeptides 90 200 c Lactalbumins 27 670 c Lysozymes 24 750 c Ribonucleases 21 850 c Hemoglobins 12 1.5 d Acid proteases 8 2.3 d Triosephosphate isomerase 3 6 d Phosphoglyceraldehyde dehydrogenase 2 9 d Glutamate dehydrogenase 1 18 d ______________________________________________________________________________________ a PAMs, Accepted point mutations (explained shortly). b Useful lookback time = 360 PAMs. c Million years. d Billion years. From Doolittle 1986 Different rates of change for different proteins

8. 8 Rates of change in protein families ProteinRate a ProteinRate Fibrinopeptides90Thyrotropin beta chain 7.4 Growth hormone37 Parathyrin 7.3 Ig kappa chain C region 37 Parvalbumin 7.0 Kappa casein33BPTI Protease inhibitors 6.2 Ig gamma chain C region 31Trypsin 5.9 Lutropin beta chain 30Melanotropin beta5.6 Ig lambda chain C region 27Alpha crystallin A chain 5.0 Complement C3a 27Endorphin 4.8 Lactalbumin 27Cytochrome b 5 4.5 Epidermal growth factor 26Insulin4.4 Somatotropin25Calcitonin 4.3 Pancreatic ribonuclease 21Neurophysin 23.6 Lipotropin beta21Plastocyanin 3.5 Haptoglobin alpha chain 20Lactate dehydrogenase 3.4 Serum albumin19Adenylate cyclase3.2 Phospholipase A 2 19Triosephosphate isomerase2.8 Protease inhibitor PST1 type 18Vasoactive intestinal peptide 2.6 Prolactin 17Corticotropin2.5 Pancreatic hormone 17Glyceraldehyde 3-P DH 2.2 Carbonic anydrase C 16Cytochrome C2.2 Lutropin alpha chain 16Plant ferredoxin1.9 Hemoglobin alpha chain 12Collagen1.7 Hemoglobin beta chain 12Troponin C, skeletal muscle 1.5 Lipid-binding protein A-II 10Alpha crystallin B-chain 1.5 Gastrin9.8Glucagon1.2 Animal lysozyme9.8Glutamate DH0.9 Myoglobin 8.9Histone H2B0.9 Amyloid A 8.7Histone H2A0.5 Nerve growth factor 8.5Histone H3 0.14 Acid proteases8.4Ubiquitin 0.1 Myelin basic protein 7.4Histone H4 0.1 a percent/100MyFrom (Nei, 1987; Dayhoff et al., 1978)

9. 9 Some terminology In evolution, homology (here of proteins), means similarity due to common ancestry. A common mode of protein evolution is by duplication. Depending on the relations between duplication and speciation dates, we have two different types of homologous proteins. Loosely, Orthologues: the “same” gene in different organisms;common ancestry goes back to a speciation event. Paralogues: different genes in the same organism; common ancestry goes back to a gene duplication. Lateral gene transfer gives another form of homology.

10. 10 Beta-globins (orthologues) 10203040 M V H L T P E E K S A V T A L W G K V N V D E V G G E A L G R L L V V Y P W T QBG-human -........ N... T..........................BG-macaque - - M.. A... A.... F.... K....................BG-bovine -... S G G...... N...... I N. L................BG-platypus... W. A... Q L I. G....... A. C. A... A... I......BG-chicken -.. W S E V. L H E I. T T. K S I D K H S L. A K.. A. M F I..... TBG-shark 50607080 R F F E S F G D L S T P D A V M G N P K V K A H G K K V L G A F S D G L A H L DBG-human.......... S......................... N...BG-macaque........... A.... N............ D S.. N. M K...BG-bovine.... A..... S A G............ A... T S. G. A. K N..BG-platypus... A... N.. S. T. I L... M. R....... T S. G. A V K N..BG-chicken. Y. G N L K E F T A C S Y G - - - - -.. E. A... T.. L G V A V T.. GBG-shark 90100110120 N L K G T F A T L S E L H C D K L H V D P E N F R L L G N V L V C V L A H H F GBG-human....... Q................ K...............BG-macaque D...... A................ K....... V... R N..BG-bovine D...... K................ N R..... I V... R.. SBG-platypus. I. N.. S Q.................... D I. I I... A.. SBG-chicken D V. S Q. T D.. K K. A E E.... V. S. K.. A K C F. V E. G I L L KBG-shark 130140 K E F T P P V Q A A Y Q K V V A G V A N A L A H K Y HBG-human..... Q.....................BG-macaque..... V L.. D F............. R..BG-bovine. D. S. E.... W.. L. S... H.. G....BG-platypus. D... E C... W.. L. R V.. H... R...BG-chicken D K. A. Q T.. I W E. Y F G V. V D. I S K E..BG-shark. means same as reference sequence - means deletion

11. 11 Beta-globins: uncorrected pairwise distances DISTANCES between protein sequences, calculated over: 1 to 147 Below diagonal: observed number of differences Above diagonal: number of differences per 100 amino acids hum mac bov pla chi sha hum ---- 5 16 23 31 65 mac 7 ---- 17 23 30 62 bov 23 24 ---- 27 37 65 pla 34 34 39 ---- 29 64 chi 45 44 52 42 ---- 61 sha 91 88 91 90 87 ----

12. 12 Beta-globins: corrected pairwise distances DISTANCES between protein sequences, calculated over 1 to 147. Below diagonal: observed number of differences Above diagonal: estimated number of substitutions per 100 amino acids Correction method: Jukes-Cantor hum mac bov pla chi sha hum ---- 5 17 27 37 108 mac 7 ---- 18 27 36 102 bov 23 24 ---- 32 46 110 pla 34 34 39 ---- 34 106 chi 45 44 52 42 ---- 98 sha 91 88 91 90 87 ----

13. 13 UPGMA tree 1 BG-bovine BG-humanBG-macaque BG-platypus BG-chicken BG-shark

14. 14 Human globins (paralogues) 102030 - V L S P A D K T N V K A A W G K V G A H A G E Y G A E A L E R M F L S F P T Talpha-human V H. T. E E. S A. T. L.... - - N V D. V. G... G. L L V V Y. W.beta-human V H. T. E E.. A. N. L.... - - N V D A V. G... G. L L V V Y. W.delta-human V H F T A E E. A A. T S L. S. M - - N V E. A. G... G. L L V V Y. W.epsilon-human G H F T E E.. A T I T S L.... - - N V E D A. G. T. G. L L V V Y. W.gamma-human - G.. D G E W Q L. L N V.... E. D I P G H. Q. V. I. L. K G H. E.myo-human 40506070 K T Y F P H F - D L S H G S A - - - - - Q V K G H G K K V A D A L T N A V A H Valpha-human Q R F. E S. G... T P D. V M G N P K.. A..... L G. F S D G L.. Lbeta-human Q R F. E S. G... S P D. V M G N P K.. A..... L G. F S D G L.. Ldelta-human Q R F. D S. G N.. S P.. I L G N P K.. A..... L T S F G D. I K N Mepsilon-human Q R F. D S. G N.. S A.. I M G N P K.. A..... L T S. G D. I K. Lgamma-human L E K. D K. K H. K S E D E M K A S E D L. K.. A T. L T.. G G I L K K Kmyo-human 8090100110 D D M P N A L S A L S D L H A H K L R V D P V N F K L L S H C L L V T L A A H Lalpha-human. N L K G T F A T.. E.. C D.. H... E.. R.. G N V. V C V.. H. Fbeta-human. N L K G T F. Q.. E.. C D.. H... E.. R.. G N V. V C V.. R N Fdelta-human. N L K P. F A K.. E.. C D.. H... E..... G N V M V I I.. T. Fepsilon-human.. L K G T F A Q.. E.. C D.. H... E..... G N V. V T V.. I. Fgamma-human G H H E A E I K P. A Q S.. T. H K I P V K Y L E F I. E. I I Q V. Q S K Hmyo-human 120130140 P A E F T P A V H A S L D K F L A S V S T V L T S K Y R - - - - - -alpha-human G K.... P. Q. A Y Q. V V. G. A N A. A H.. H......beta-human G K.... Q M Q. A Y Q. V V. G. A N A. A H.. H......delta-human G K.... E. Q. A W Q. L V S A. A I A. A H.. H......epsilon-human G K.... E. Q.. W Q. M V T A. A S A. S. R. H......gamma-human. G D. G A D A Q G A M N. A. E L F R K D M A. N. K E L G F Q Gmyo-human

15. 15 Human globins: uncorrected pairwise distances DISTANCES between protein sequence, calculated over 1 to 154. Below diagonal: observed number of differences Above diagonal: number of differences per 100 amino acids alpha beta delta epsil gamma myo alpha ---- 55 55 60 57 74 beta 82 ---- 7 25 27 75 delta 82 10 ---- 27 29 74 epsil 89 35 39 ---- 20 77 gamma 85 39 42 29 ---- 76 myo 116 117 116 119 118 ----

16. 16 Human globins: corrected pairwise distances DISTANCES between protein sequences,calculated over 1-141 Below diagonal: observed number of differences Above diagonal: estimated number of substitutions per 100 amino acids Correction method: Jukes-Cantor alpha beta delta epsil gamma myo alpha ---- 281 281 281 313 208 beta 82 ---- 7 30 31 1000 delta 82 10 ---- 34 33 470 epsil 89 35 39 ---- 21 402 gamma 85 39 42 29 ---- 470 myo 116 117 116 119 118 ----

17. 17 102030405060 C C G A C A G G C A C G G T G G C T C A C A C C T G T A A T C C C A G T A C T T T G G G A G G C T G A G G C G A G A G Ghum-1.............................. hum-2.............................. hum-3....... A................... C....... C................... G....chimp....... A................... C....... C................... G....bonob........... A....................... C.......... C........ G....goril. G..... A............. G............. C............ C.... T. G. C..orang 708090100110120 A T C A C C T G A G G T C G G G A G T T T G A G A C C A G C C T G A C C A A T A T G G A G A A A C C C C A G T T A T A Chum-1......................... T..................................hum-2...................................................... C.....hum-3.................... C.......................................chimp.................... C.......................................bonob.................... C........................... T...........goril............ T....... C.. A................................ C...orang 130140150160170180 T A A A A A T A C A A A A T T A G C T G G G T G T G G T G G C G C A T G C C T G T A A T C C T A G C T A C T A G G A A Ghum-1.............................. hum-2.............................. hum-3.......................... C................... C.......... G..chimp.............................................. C.......... G..bonob........................ G..................... C.......... G..goril...................... C..................... T. C.......... G..orang 190200210220230240 G C T G A G G C A G G A G A A T C G C T T G A A C C C G G G A G G T G G A G G T T G A G G T G A G C T G A G A T C A C Ghum-1.............................. hum-2.............................. hum-3............ A...............................................chimp............ A...............................................bonob................. A............. A......... T................. Agoril.......................................... T.............. G..orang 250260270280290300 C C A T T G C A C T C C A G C C T G G G C A A C A A G A G C A A A A C T C C G T C T C A A A A A A T A A A T A A A T A Ahum-1.............................. hum-2.............................. hum-3... C.................................. A................ C....chimp... C.................................. A.....................bonob...................................... A.....................goril...................................... A... T.................orang Alu sequences (a-globin2 Alu 1, Knight et al., 1996)

18. 18 Alu sequences:uncorrected pairwise distances DISTANCES between nucleic acid sequences calculated over: 1 to 300, considering all base positions Below diagonal: observed number of differences Above diagonal: number of differences per 100 bases hum-1 hum-2 hum-3 chimp bonob goril orang hum-1 ---- 0 0 4 3 5 7 hum-2 1 ---- 1 4 4 5 7 hum-3 1 2 ---- 4 4 5 7 chimp 12 13 13 ---- 1 5 6 bonob 10 11 11 2 ---- 4 5 goril 14 15 15 14 12 ---- 7 orang 20 21 21 18 16 22 ----

19. 19 Alu sequences: corrected pairwise distances DISTANCES between nucleic acid sequences, calculated over: 1 to 300, considering all base positions Below diagonal: observed number of differences Above diagonal: estimated number of substitutions per 100 bases Correction method: Jukes-Cantor hum-1 hum-2 hum-3 chimp bonob goril orang hum-1 ---- 0 0 4 3 5 7 hum-2 1 ---- 1 4 4 5 7 hum-3 1 2 ---- 4 4 5 7 chimp 12 13 13 ---- 1 5 6 bonob 10 11 11 2 ---- 4 6 goril 14 15 15 14 12 ---- 8 orang 20 21 21 18 16 22 ----

20. 20 Correcting distances between DNA and protein sequences We mentioned earlier that it is necessary to adjust observed percent differences to get a distance measure which scales linearly with time. This is because we can have multiple and back substitutions at a given position along a lineage. All of the correction methods (with names like Jukes-Cantor, 2-parameter Kimura, etc) are justified by simple probabilistic arguments involving Markov chains whose basis is worth mastering. The same molecular evolutionary models are used in scoring sequence alignments.

21. 21 Markov chain State space = {A,C,G,T}. p(i,j) = pr(next state S j | current state S i ) Markov assumption: p(i,j) = pr(next state S j | current state S i & any configuration of states before this) Only the present state, not previous states, affects the probs of moving to next states.

22. 22 A simple 4-state Markov chain: the Kimura 2-parameter model for nucleotide change A G T C Transition probabilities: Horizontal: a Diagonal & vertical: b Self: c =  a  2b c a b b a c b b b b c a b b a c A G C T AGCTAGCT

23. 23 The multiplication rule pr(state after next is S k | current state is S i ) = ∑ j pr(state after next is S k, next state is S j | current state is S i ) [addition rule] = ∑ j pr(next state is S j | current state is S i ) x pr(state after next is S k | current state is S i, next state is S j ) [multiplication rule] = ∑ j p(i,j) x p(j,k) [Markov assumption] = (i,k)-element of P 2, where P=(p(i,j)). More generally, pr(state t steps from now is S k | current state is S i ) = i,k element of P t

24. 24 Continuous-time version For any s,t write p ij (t) = pr(S j at time t+s | S i at time s) for the stationary (time-homogeneous) transition probabilities. Write P(t) = (p ij (t)) for the matrix of p ij (t)’s. Then for any t,u: P(t+u) = P(t) P(u). It follows that P(t) = exp(Qt), where Q = P’(0) is the derivative of P(t) at t = 0. Q is called the infinitesimal matrix of P(t), and satisfies P’(t) = QP(t) = P(t)Q.

25. 25 Interpretation of Q Roughly, q(i,j) is the rate of transitions of i to j, while q(i,i) =   j q(i,j), so each row sum is 0. If under some initial conditions, we have a Markov chain evolving in continuous time with infinitesimal matrix Q, and p j (t) = pr(S j at time t), then p j (t+h) =  i pr(S i at t, S j at t+h) =  i pr(S i at t)pr(S j at t+h | S i at t) = p j (t)x(1+hq jj ) +  i j p i (t)x hq ij i.e., h -1 [p j (t+h) - p j (t)] = p j (t)q(j,j) +  i j p i (t)q(i,j) which becomes P’ = QP as h  0. Important approximation: when t is small, P(t)  I + Qt.

26. 26 Q = P(t) = The Jukes-Cantor model (1969) -3       rsss srss ssrs sssr r = (1+3e  4  t )/4, s = (1  e  4  t )/4. Common ancestor of human and orang. t time units human (now) Consider e.g. the 2nd position in a-globin2 Alu1.

27. 27 Let P(t) = exp(Qt). Then the A,G element of P(t) is pr(G now | A then) = (1  e  4  t )/4. Same for all pairs of different nucleotides. Overall rate of change k = 3  t. When k =.01, described as 1 PAM PAM = accepted point mutation Put t =.01/3  = 1/300 . Then the resulting P = P(1/300  ) is called the PAM(1) matrix. Why use PAMs? Definition of PAM

28. 28 Evolutionary time, PAM Since sequences evolve at different rates, it is convenient to rescale time so that 1 PAM of evolutionary time corresponds to 1% expected substitutions. For Jukes-Cantor, k = 3  t is the expected number of substitutions in [0,t], so is a distance. (Show this.) Set 3  t = 1/100, or t = 1/300 , so 1 PAM = 1/300  years.

29. 29 Distance adjustment For a pair of sequences, k = 3  t is the desired metric, but not observable. Instead, pr(different) is observed. So we use a model to convert pr(different) to k. This is completely analogous to the conversion of  = pr(recombination) to genetic (map) distance (= expected number of crossovers) using the Haldane map function  = 1/2  (1  e -2d ), assuming the no-interference (Poisson) model.

30. 30 common ancestor G orang C human still 2 nd position in a-globin Alu 1 Assume that the common ancestor has A, G, C or T with probability 1/4. Then the chance of the nt differing p ≠ = 3/4  (1  e  8  t ) = 3/4  (1  e  4k/3 ), since k =2  3  t t 3/4 Towards Jukes-Cantor adjustment

31. 31 If we suppose all nucleotide positions behave identically and independently, and n ≠ differ out of n, we can invert this, obtaining =  3/4  log(1  4/3 n ≠ /n). This is the corrected or adjusted fraction of differences (under this simple model).  100 to get PAMs The analogous simple model for amino acid sequences has =  19/20  log(1  20/19 n ≠ /n).  100 for PAM. Jukes-Cantor adjustment

32. 32 Illustration 1. Human and bovine beta-globins are aligned with no deletions at 145 out of 147 sites. They differ at 23 of these sites. Thus n ≠ /n = 23/145, and the corrected distance using the Jukes-Cantor formula is (natural logs)  19/20  log(1  20/19  23/145) = 17.3  10 -2. 2. The hum-1 and gorilla sequences are aligned without gaps across all 300 bp, and differ at 14 sites. Thus n ≠ /n = 14/300, and the corrected distance using the Jukes-Cantor formula is  3/4  log(1  4/3  14/300) = 4.8  10 -2.

33. 33 Correspondence between observed a.a. differences and the evolutionary distance ( Dayhoff et al., 1978 ) Observed Percent DifferenceEvolutionary Distance in PAMs 1 5 11 17 23 30 38 47 56 67 80 94 112 133 159 195 246 1 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85328