1. Chap. 7. Building Trees

2. Variation Most organisms can increase exponentially
If all organisms survived and multiplied at the same rate, there will be no change in frequency of the variants, and thus no evolution
Limited by food, space, predators, etc.
When population size is limited, not all variants survive
A possibility of natural selection
Also, chance effects exist
Equal-sized populations with two variants will not stay the same even with the same degree of fitness
Called random drift, the chance effect will take over the whole population
This implies that evolution can occur even without natural selection, referred to as neutral evolution

3. Mutation Any change in a gene sequence that is passed on to offspring
Caused by
A damage to DNA moledule (from radiation, etc.)
Errors in replication
Point mutation – simplest form of mutation and occurs all over DNA sequences
Transition – mutation within purine (A,G) or pyrimidine (C,T/U)
Transversion – mutation between nt groups
Effects depend on where mutations occur
Non-coding region – no effect on proteins, and neutral
But may have significant effects if occurring in control region
Coding region
Synonymous substitution when a mutation does not change AA
Non-synonymous
AA is replaced by another
stop codon is introduced

4. Mutation Indel mutation Gene inversion
Small indels of a single base of a few bases are frequent
Caused by slippage during DNA replication
Particularly frequent with repeated sequences
GCGC…: insertion of extra GC or deletion cause slight slippage
CAG repeated region in huntingtin protein can expand, causing Huntington’s disease
Indels can cause frame shift, if indels are not multiples of three
Gene inversion
Whole genes are copied to offspring in reverse direction
Translocation
Whole genes can be deleted from one genome and inserted into another

5. Historical background: insulin
Mature insulin consists of an A chain and B chain heterodimer connected by disulphide bridges
The signal peptide and C peptide are cleaved, and their sequences display fewer functional constraints.

7. Note the sequence divergence in the
disulfide loop region of the A chain

8. 0.1 x 10-9
1 x 10-9
0.1 x 10-9
Number of nucleotide substitutions/site/year

9. Historical background: insulin
By the 1950s, it became clear that amino acid substitutions occur nonrandomly.
For example, Sanger and colleagues noted that most amino acid changes in the insulin A chain are restricted to a disulfide loop region.
Such differences are called “neutral” changes (Kimura, 1968; Jukes and Cantor, 1969)
Subsequent studies at the DNA level showed that rate of nucleotide (and of amino acid) substitution is about six-to ten-fold higher in the C peptide, relative to the A and B chains.

10. Historical background: insulin
Surprisingly, insulin from the guinea pig (and from the
related coypu) evolve seven times faster than insulin
from other species. Why?
The answer is that guinea pig and coypu insulin
do not bind two zinc ions, while insulin molecules from
most other species do. There was a relaxation on the
structural constraints of these molecules, and so
the genes diverged rapidly.

11. Guinea pig and coypu insulin have undergone an
extremely rapid rate of evolutionary change
Arrows indicate positions at which guinea pig
insulin (A chain and B chain) differs
from both human and mouse

12. Molecular clock hypothesis
In the 1960s, sequence data were accumulated for
small, abundant proteins such as globins,
cytochromes c, and fibrinopeptides. Some proteins
appeared to evolve slowly, while others evolved rapidly.
Linus Pauling, Emanuel Margoliash and others
proposed the hypothesis of a molecular clock:
For every given protein, the rate of molecular
evolution is approximately constant in all
evolutionary lineages

13. corrected amino acid changes
Dickerson
(1971)
corrected amino acid changes
per 100 residues (m)
Millions of years since divergence

14. Molecular clock hypothesis: implications
If protein sequences evolve at constant rates, they can be used to estimate the times that sequences diverged.
This is analogous to dating geological specimens by radioactive decay.
e.g. from fossil evidence, human and gorilla diverged 11 million years ago (MYA). Human globins have twice as many substitutions than gorilla globins.

15. K: number of substitutions per site
r: Rate of substitution per year
r = K/2T

16. Molecular phylogeny uses trees to depict evolutionary
relationships among organisms. These trees are based
upon DNA and protein sequence data. A B C D E F G H I
time 6 2 1 A 2 1 1 B 2 C 2 2 1 D 6
one unit E

17. Globin phylogeny by Dayhoff (1972)

18. Globin phylogeny by Dayhoff in evolutionary time (1972)

20. Tree nomenclature taxon taxon A B C D E F G H I A B C D E time 6 2 1 2
one unit E

21. Tree nomenclature: clades
Clade ABF (monophyletic group) 2 A F 1 1 G B 2 I H 2 C 1 D 6 E
time

22. Examples of clades
Lindblad-Toh et al., Nature 438: 803 (2005), fig. 10

23. Phylogenetic Methods
Family of related sequences evolved from a common ancestor is studied with phylogenetic trees showing the order of evolution
Want to have a tree representation showing
Divergence among species
Evolutionary distance
Usually unrooted B C A D B D C B D A C A

24. Phylogenetic Trees
Rooted tree provide direction of evolution and its distance
Unrooted tree is less informative
Finding a root
Use known species relationship
If not known, use mid-point method: finding a point on the tree with the mean distance among the tree is identical in either side – assumes the same evolution rate

25. Phylogenetic Trees
Rooted tree provide direction of evolution and its distance
Unrooted tree is less informative
Finding a root
Use known species relationship
If not known, use mid-point method: finding a point on the tree with the mean distance among the tree is identical in either side – assumes the same evolution rate

26. Tree Construction Multiple sequences are aligned
Use JC or other models to compute pair-wise evolutionary distances
From distance matrix, use a clustering method
Join the closest two clusters to form a larger one
Recompute distances between all clusters
Repeat two steps above until all species are connected

27. Tree-building methods
distance-based and character-based:
Distance-based methods involve a distance metric,
such as the number of amino acid changes between
the sequences, or a distance score. Examples of
distance-based algorithms are UPGMA and
neighbor-joining.
Character-based methods include maximum parsimony
and maximum likelihood. Parsimony analysis involves
the search for the tree with the fewest amino acid
(or nucleotide) changes that account for the observed
differences between taxa.

28. Example with Globin

29. Distance-based tree
Calculate the pairwise alignments;
if two sequences are related,
put them next to each other on the tree

30. Character-based tree: identify positions that best describe how characters (amino acids) are derived from common ancestors

31. Tree from Distance Matrix
Given a weighted tree, with weights on edges representing evolutionary distances
Additive distances
di,c + dc,j = Di,j
Find the nearest leaves – combine to the same parent
Not easy to find neighboring leaves

32. Reconstructing tree Shorten all hanging edges of a tree
Reduce length of every hanging edge by the same small amount δ, then distance matrix is reduced by 2δ
Find the leaf with 0 weight and remove the leaf

33. Additive matrix

34. Distance-based methods: UPGMA trees
UPGMA is a simple approach for making trees.
An UPGMA tree is always rooted.
An assumption of the algorithm is that the molecular
clock is constant for sequences in the tree. If there
are unequal substitution rates, the tree may be wrong.
While UPGMA is simple, it is less accurate than the
neighbor-joining approach (described next).
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35. UMPGMA Method
Distance between two clusters is defined as the mean of the distances between species in the two clusters
Human cluster vs. chimpanzee/pygmy cluster
Mean of human-chimpanzee and human-pygmy distances
Produces a rooted tree
Tree distance between chimpanzee and pygmy is /2
All species end at the right aligned (because the same molecular evolution is assumed in every species) – not used
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37. UMPGMA Example Pick the smallest distance Recompute distances
C to AB: (60+50)/2 = 55
D to AB: (100+90)/2 = 95
E to AB: (90+80)/2 = 85

38. UMPGMA Example Pick the smallest distance Recompute distancesAB C D E 55 95 40 85 50 30
Pick the smallest distance
Recompute distances
AB to DE: (95+85)/2 = 90 AB C DE 55 90 45

39. Neighbor-Joining (NJ) Method
Additive distance in an unrooted tree
Distance between two species is the sum of branch lengths connecting them
NJ Method
Construct an unrooted tree whose branch lengths are as close to the distance matrix among species
Algorithm
Join two neighbors, and replace them by a new internal node
Keep repeating the step until all species are covered

40. Making trees using neighbor-joining
The neighbor-joining
method of Saitou and Nei
(1987) Is especially useful
for making a tree having a
large number of taxa.
Begin by placing all the taxa in a star-like clusters.structure.
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41. Tree-building methods: Neighbor joining
Next, identify neighbors (e.g. 1 and 2) that are most closely related. Connect these neighbors to other OTUs (operational taxonomic units) via an internal branch, XY. At each successive stage, minimize the sum of the branch lengths.

42. Tree-building methods: Neighbor joining
Define the distance from X to Y by
dXY = 1/2(d1Y + d2Y – d12)

44. NJ Example A B C D E 5 4 7 6 10 9 F 8 11 C(A) = (5+4+7+6+8)/4 = 7.55 4 7 6 10 9 F 8 11
C(A) = ( )/4 = 7.5
C(B) = ( )/4 = 10.5
C(C) = ( )/4 = 8
C(D) = ( )/4 = 9.5
C(E) = ( )/4 = 8.5
C(F) = ( )/4 = 11
D(A,B) – C(A) – C(B) = -13
D(D,E) – C(D) – C(E) = -13 …
Pick the smallest (eg. A,B)

45. NJ Example A B C D E 5 4 7 10 6 9 F 8 11 Join A and B to U1
C(AU1) = D(A,B)/2 + (C(A)-C(B))/2 = 1
C(BU1) = D(A,B)/2 + (C(B)-C(A))/2 = 4

46. NJ Example U1 C D E 3 6 7 5 F 8 9

47. Making trees using character-based methods
The main idea of character-based methods is to find
the tree with the shortest branch lengths possible.
Thus we seek the most parsimonious (“simple”) tree.
Identify informative sites. For example, constant
characters are not parsimony-informative.
Construct trees, counting the number of changes
required to create each tree. For about 12 taxa or
fewer, evaluate all possible trees exhaustively;
for >12 taxa perform a heuristic search.
Select the shortest tree (or trees).
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48. As an example of tree-building using maximum
parsimony, consider these four taxa:
AAG
AAA
GGA
AGA
How might they have evolved from a
common ancestor such as AAA?

49. Tree-building methods: Maximum parsimony1
AAA
AAA
AAA
AAA
AGA
AAA
AAA
AAA
AAA 1 1 1 1 2 1 2 1
AAG
AAA
GGA
AGA
AAG
AGA
AAA
GGA
AAG
GGA
AAA
AGA
Cost = 3
Cost = 4
Cost = 4
In maximum parsimony, choose the tree(s) with the
lowest cost (shortest branch lengths).

50. Parsimony
Use simplest possible explanation of the data, or with fewest assumptions
Binary states: 0 for ancestral, 1 derived character
0 may be ancestral tetrpod forelimb bone structure
1 may be the bone structure in the bird wing
C,D posses a derived character, not possessed by A, B
Tree A: the character must have evolved on the + branch
Tree b: evolved once (+), and lost (*)
Tree c: evolved independently (+) on two branches
Parsimony criterion – tree a is the simplest (single state change)

52. Small parsimony Problem
Find the most parsimonious labeling of the internal vertices
input: Tree T with each leaf labeled by an m-character string
output: Labeling of internal vertices of T minimizing the parsimony score
Characters in the string are independent, and the problem can be solved independently for each character
Assume that each leaf is labeled by a single character
Solve for a more general problem
Length of an edge is defined as the Hamming distance
For k-letter alphabet,
dH(v,w) = 0 if v=w; =1, otherwise

53. Weighted Small Parsimony Problem
Find the minimal weighted parsimony score labeling of internal vertices
input: Tree T with each leaf labeled by a k-letter alphabet,
and kxk scoring matrix
output: Labeling of internal vertices of T minimizing the weighted
parsimony score
David Sankoff Dynamic Programming, 1975
Internal vertex v with offsprings u and w
si(v) = min{si(u)+δi,t} + min{si(w)+ δj,t}

54. Parsimony Example
Five species, scored for 6 characters with 0 or 1 state each
Calculate how many changes of state are needed in a tree, for example
Species 1 2 3 4 5 6
Alpha
Beta
Gamma
Delta
Epsilon
alpha
delta
gamma
beta
epsilon

55. Reconstruction of Character 1
alpha
delta
gamma
beta
epsilon
Species 1
Alpha
Beta
Gamma
Delta
Epsilon
Red: state of 1
Regular: state of 0
alpha
delta
gamma
beta
epsilon

56. Reconstruction of Character 2
alpha
delta
gamma
beta
epsilon
Species 2
Alpha
Beta
Gamma 1
Delta
Epsilon
alpha
delta
gamma
beta
epsilon
alpha
delta
gamma
beta
epsilon

57. Reconstruction of Character 3
alpha
delta
gamma
beta
epsilon
Species 3
Alpha
Beta 1
Gamma
Delta
Epsilon
alpha
delta
gamma
beta
epsilon

58. Reconstruction of character 4, 5
alpha
delta
gamma
beta
epsilon
Species 4 5
Alpha 1
Beta
Gamma
Delta
Epsilon
alpha
delta
gamma
beta
epsilon

59. Reconstruction of character 6
Species 6
Alpha
Beta
Gamma
Delta 1
Epsilon
alpha
delta
gamma
beta
epsilon

60. Reconstruction with All Changes
Total # of changes, taking random choice for more than one trees
= = 9
alpha
delta
gamma
beta
epsilon
2,6 5 4
2,5 4
1,3

61. Most Parsimonious Tree
alpha
delta
gamma
beta
epsilon
alpha
delta
gamma
beta
epsilon
2,6 5 4
4,5
2,5 6
4,5 4 2
1,3
1,3
9 8

62. Most Parsimonious Trees
alpha
delta
gamma
beta
epsilon
4,5 6
4,5
Identical, if unrooted 2
alpha
1,3
gamma
beta
4,5
4,5
1,3 2 6
delta
gamma
delta
alpha
beta
epsilon
epsilon 6
4,5
4,5
1,3 2

63. How to determine Branch Lengths
Given an unrooted tree, want to determine the number of changes in each branch
Ambiguous as to where the changes are
=> use the average over all possible reconstructions of each character
alpha
delta
gamma
beta
epsilon
gamma
alpha
beta
2,6 5 4
1.5
0.5 1
2,5 4
2.5 1 1
1.5
1,3
epsilon
delta

64. Large Parsimony Problem
Find a tree with n leaves having the minimal parsimony score
input: An nxm matrix M describing n species, each represented by
m-character string
output: A tree T with n leaves labeled by n rows of matrix M,
and a labeling of the internal vertices with minimal
parsimony score over all possible trees and labelings
NP-complete
Greedy heuristics
Start with an arbitrary tree
Move from one tree to another if it lowers parsimony score by nearest neighbor interchange

65. Probabilistic Models Likelihood ratios
Example: predict helices and loops in a protein
Known info: helices have a high content of hydrophobic residues
ph and pl: frequencies of AA being in the helix or loop
Lh and Ll : likelihoods that a sequence of N AAs are in a helix or a loop
Lh = ∏N ph , Ll = ∏N pl
Rather than likelihoods, their ratios have more info
Lh/Ll : is sequence more or less likely to be a helical or loop region
S = ln(Lh/Ll) = ∑ N ln(ph/pl): positive for helical region
Partition a sequence into N-AA segments (N=300)

66. Prior and Posterior Probs.
Previous example has two hypotheses (Helix or Loop)
The sequence is described by models 0 and 1
Models 0 and 1 are defined by ph and pl
Generalize to k hypotheses: Mk models (k=0,1,2,…)
Given a test dataset D, what is the prob. that D is described by each of the models ?
Known info: prior probs., Pprior(Mk) for each model from other info sources
Compute likelihood of D according to each of the models: L(D|Mk)
Of interest is not the prob of D arising from Mk but the prob of D being described by Mk
Namely, Ppost(Mk| D) ∞ L(D|Mk) Pprior(Mk) : posterior prob.
Ppost(Mk| D) = L(D|Mk) Pprior(Mk)/∑iL(D|ii) Pprior(Mi)
=> Bayesian prob.

67. Bayesian Prob. Basic principles Special case: two models
We make inference using posterior probs.
If a posterior prob. of one model is higher, it can be the best model with confidence
Special case: two models
Two prior probs.: Pprior0 , Pprior1
Pposti = Li Ppriori/(L0 Pprior0 + L1 Pprior1)
Log-odd score:
S΄ = ln(L1Pprior1/L0Pprior0) = ln(L1/L0) + ln(Pprior1/Pprior0)
= S + ln(Pprior1/Pprior0)
Difference between S΄and S is simply the additive constant, and ranking will be identical whether we use S΄or S
Warning: if Pprior1 is small, S has to be high to make S΄positive
When Pprior0 = Pprior1, S΄= S
Ppost1 = 1/(1 + L0 Pprior0 /L1 Pprior1) = 1/(1 + exp(- S΄))
S΄=0 →Ppost1 =1/2; S΄is large and negative → Ppost1 ≈1

68. Maximum Likelihood (ML) Phylogeny
Given a model of sequence evolution and a proposed tree structure, compute the likelihood that the known sequences would have evolved on that tree
ML chooses the tree that maximizes this likelihood
Three parameters
Tree toplogy
Branch lengths
Values of the parameters in the rate matrix

69. What is Likelihood in ML Tree ?
Given a model of sequence evolution at a site
Likelihood of ancestor X: L(X) = PXA(t1) PXG(t2)
L(Y) = PYG(t4) ∑X L(X) PYX(t3)
L(W) = ∑y ∑Z L(Y)PWY(t5)L(Z)PWZ(t6)
Total likelihood for the site:
L = ∑W P W L(W)
P W: equilibrium prob.
Is equal to posterior prob. of different clades W t5 Y t6 t3 X Z t4 t1 t2 A G G T T

70. Computing Posterior Prob.
ML tree maximizes the total likelihood of the data given the tree, i.e., L(data|tree)
We want to compute posterior prob: P(tree|data)
From Bayes theorem,
Pt(tree|data)= L(data|tree)*Pr(tree)/ ∑L(data|tree)*Pr(tree) (summation is over all possible trees)
Namely, posterior prob. ∞ L(data|tree)*Pr(tree)
Problem is the summation over all possible trees
Moreover, what we really want is, given the data, the posterior prob. that a particular clade of interest is present
Ppost(clade|data)= ∑P(data|tree) for trees containing the clade
= ∑cladeL(data|tree)*Pprior(tree)/ ∑all treesL(data|tree)*Pprior(tree)
In practice,
Ppsot(clade|data) = # of trees containing clade/total # of trees in the sample

71. Making trees using maximum likelihood
Maximum likelihood is an alternative to maximum
parsimony. It is computationally intensive. A likelihood
is calculated for the probability of each residue in
an alignment, based upon some model of the
substitution process.
What are the tree topology and branch lengths that have the greatest likelihood of producing the observed data set?
ML is implemented in the TREE-PUZZLE program,
as well as PAUP and PHYLIP.
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72. Bayesian inference of phylogeny with MrBayes
Calculate:
Pr [ Tree | Data] =
Pr [ Data | Tree] x Pr [ Tree ]
Pr [ Data ]
Pr [ Tree | Data ] is the posterior probability distribution of trees. Ideally this involves a summation over all possible trees. In practice, Monte Carlo Markov Chains (MCMC) are run to estimate the posterior probability distribution.
Notably, Bayesian approaches require you to specify prior assumptions about the model of evolution.

73. Evaluating trees: bootstrapping
Bootstrapping is a commonly used approach to
measuring the robustness of a tree topology.
Given a branching order, how consistently does
an algorithm find that branching order in a
randomly permuted version of the original data set?
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74. Evaluating trees: bootstrapping
Bootstrapping is a commonly used approach to
measuring the robustness of a tree topology.
Given a branching order, how consistently does
an algorithm find that branching order in a
randomly permuted version of the original data set?
To bootstrap, make an artificial dataset obtained by
randomly sampling columns from your multiple
sequence alignment. Make the dataset the same size
as the original. Do 100 (to 1,000) bootstrap replicates.
Observe the percent of cases in which the assignment
of clades in the original tree is supported by the
bootstrap replicates. >70% is considered significant.
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75. In 61% of the bootstrap
resamplings, ssrbp and btrbp
(pig and cow RBP) formed a
distinct clade. In 39% of the
cases, another protein joined
the clade (e.g. ecrbp), or one
of these two sequences joined
another clade.

76. Bootstrapping Bootstrapping
A method of assessing the reliability of trees
Numbers in the rooted tree – called bootstrap percentages
Distances according to models are not realistic due to chance fluctuations
Boostrapping
addresses the question on if these fluctuations are influencing the tree configuration
Boostrapping deliberately construct sequence data sets that differe by some small random fluctuations from real sequences
And check if the same tree topology is obtained
Randomized sequences are constructed by sampling columns

77. Bootstrapping Generate 100 or 1,000 randomized sequences
And compute what percentage of randomized trees contain the same group
77% boostrap value is considered to be reliable
e.g.
24% -- doubtful if they form a clade
71% - human/chimpanzee/pygmy chimapnzee
Between two high figures
Chimpanzee/pygmy always form a clade
Gorilla/human/chimpanzee/pygmy always form a clade
(Gorilla.(human,chimpanzees)) appear more frequently than (human,(gorilla, chimpanzees)) or (chimpanzees, (gorilla,human))
Thus, can conclude (human, chimpanzees) is more reliable
Can construct a consensus tree
Frequency of each possible clade is determined
Construct a consensus tree by adding clades from more frequent clades

78. ML vs. Parsimony
Parsimony is fast – ML requires each tree topology to be optimized
ML is model-based
parsimony’s model is equal substitution
Parsimony can incorporate models, but not clear what the weights have to be
Parsimony tries to minimize the number of substitutions, irrespective of the branch lengths
ML allows for changes more likely to happen on longer branches
On a long branch, no reason to try to minimize the number of substitutions
Parsimony is strong for evaluating trees based on qualitative characters

79. Phylogeny programs Joseph Felsenstein
Phylogeny.fr