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### Brief introduction to the evolution of proteins

Higher-order HMMs

- How do we model “memory” larger than one time point?
- P(i+1 = l | i = k) akl
- P(i+1 = l | i = k, i -1 = j) ajkl
- …
- A second order HMM with K states is equivalent to a first order HMM with K2 states

aHHT

state HH

state HT

aHT(prev = H)

aHT(prev = T)

aHTH

state H

state T

aHTT

aTHH

aTHT

state TH

state TT

aTH(prev = H)

aTH(prev = T)

aTTH

Similar Algorithms to 1st Order

- P(i+1 = l | i = k, i -1 = j)
- Vlk(i) = maxj{ Vkj(i – 1) + … }
- Time? Space?

Modeling the Duration of States

1-p

Length distribution of region X:

E[lX] = 1/(1-p)

- Geometric distribution, with mean 1/(1-p)
This is a significant disadvantage of HMMs

Several solutions exist for modeling different length distributions

X

Y

p

q

1-q

Solution 1: Chain several states

p

1-p

X

Y

X

X

q

1-q

Disadvantage: Still very inflexible

lX = C + geometric with mean 1/(1-p)

Solution 2: Negative binomial distribution

Duration in X: m turns, where

- During first m – 1 turns, exactly n – 1 arrows to next state are followed
- During mth turn, an arrow to next state is followed
m – 1 m – 1

P(lX = m) = n – 1 (1 – p)n-1+1p(m-1)-(n-1) = n – 1 (1 – p)npm-n

p

p

p

1 – p

1 – p

1 – p

Y

X(n)

X(1)

X(2)

……

Example: genes in prokaryotes

- EasyGene:
Prokaryotic

gene-finder

Larsen TS, Krogh A

- Negative binomial with n = 3

Solution 3: Duration modeling

Upon entering a state:

- Choose duration d, according to probability distribution
- Generate d letters according to emission probs
- Take a transition to next state according to transition probs
Disadvantage: Increase in complexity of Viterbi:

Time: O(D)

Space: O(1)

where D = maximum duration of state

F

d<Df

xi…xi+d-1

Pf

Warning, Rabiner’s tutorial claims O(D2) & O(D) increases

Viterbi with duration modeling

emissions

emissions

Recall original iteration:

Vl(i) = maxk Vk(i – 1) akl el(xi)

New iteration:

Vl(i) = maxk maxd=1…DlVk(i – d) Pl(d) akl j=i-d+1…iel(xj)

F

L

d<Df

d<Dl

Pl

Pf

transitions

xi…xi + d – 1

xj…xj + d – 1

Precompute cumulative values

A state model for alignment

M

(+1,+1)

Alignments correspond 1-to-1 with sequences of states M, I, J

I

(+1, 0)

J

(0, +1)

-AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---

TAG-CTATCAC--GACCGC-GGTCGATTTGCCCGACC

IMMJMMMMMMMJJMMMMMMJMMMMMMMIIMMMMMIII

Let’s score the transitions

s(xi, yj)

M

(+1,+1)

Alignments correspond 1-to-1 with sequences of states M, I, J

s(xi, yj)

s(xi, yj)

-d

-d

I

(+1, 0)

J

(0, +1)

-e

-e

-AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---

TAG-CTATCAC--GACCGC-GGTCGATTTGCCCGACC

IMMJMMMMMMMJJMMMMMMJMMMMMMMIIMMMMMIII

Alignment with affine gaps – state version

Dynamic Programming:

M(i, j): Optimal alignment of x1…xi to y1…yjending in M

I(i, j): Optimal alignment of x1…xi to y1…yj ending in I

J(i, j): Optimal alignment of x1…xi to y1…yjending in J

The score is additive, therefore we can apply DP recurrence formulas

Alignment with affine gaps – state version

Initialization:

M(0,0) = 0;

M(i, 0) = M(0, j) = -, for i, j > 0

I(i,0) = d + ie; J(0, j) = d + je

Iteration:

M(i – 1, j – 1)

M(i, j) = s(xi, yj) + max I(i – 1, j – 1)

J(i – 1, j – 1)

e + I(i – 1, j)

I(i, j) = max

d + M(i – 1, j)

e + J(i, j – 1)

J(i, j) = max

d + M(i, j – 1)

Termination:

Optimal alignment given by max { M(m, n), I(m, n), J(m, n) }

Protein sequence and structure

Protein classification

Phylogeny trees

Substitution matrices

Actin sequence

- Actin is ancient and abundant
- Most abundant protein in cells
- 1-2 actin genes in bacteria, yeasts, amoebas
- Humans: 6 actin genes
- -actin in muscles; -actin, -actin in non-muscle cells
- ~4 amino acids different between each version
MUSCLE ACTIN Amino Acid Sequence

1 EEEQTALVCD NGSGLVKAGF AGDDAPRAVF PSIVRPRHQG VMVGMGQKDS YVGDEAQSKR

61 GILTLKYPIE HGIITNWDDM EKIWHHTFYN ELRVAPEEHP VLLTEAPLNP KANREKMTQI

121 MFETFNVPAM YVAIQAVLSL YASGRTTGIV LDSGDGVSHN VPIYEGYALP HAIMRLDLAG

181 RDLTDYLMKI LTERGYSFVT TAEREIVRDI KEKLCYVALD FEQEMATAAS SSSLEKSYEL

241 PDGQVITIGN ERFRGPETMF QPSFIGMESS GVHETTYNSI MKCDIDIRKD LYANNVLSGG

301 TTMYPGIADR MQKEITALAP STMKIKIIAP PERKYSVWIG GSILASLSTF QQMWITKQEY

361 DESGPSIVHR KCF

Protein Phylogenies

- Proteins evolve by both duplication and species divergence

Structure Determines Function

The Protein Folding Problem

- What determines structure?
- Energy
- Kinematics

- How can we determine structure?
- Experimental methods
- Computational predictions

Primary Structure: Sequence

- The primary structure of a protein is the amino acid sequence

Primary Structure: Sequence

- Twenty different amino acids have distinct shapes and properties

Primary Structure: Sequence

A useful mnemonic for the hydrophobic amino acids is "FAMILY VW"

Secondary Structure: , , & loops

- helices and sheets are stabilized by hydrogen bonds between backbone oxygen and hydrogen atoms

PDB Growth

New PDB structures

Protein classification

- Number of protein sequences grows exponentially
- Number of solved structures grows exponentially
- Number of new folds identified very small (and close to constant)
- Protein classification can
- Generate overview of structure types
- Detect similarities (evolutionary relationships) between protein sequences
- Help predict 3D structure of new protein sequences

Classification of 27,599 protein structures in PDB

Protein structure classification

Protein fold

Protein superfamily

Protein family

Morten Nielsen,CBS, BioCentrum, DTU

Structure Classification Databases

- SCOP
- Manual classification (A. Murzin)
- scop.berkeley.edu

- CATH
- Semi manual classification (C. Orengo)
- www.biochem.ucl.ac.uk/bsm/cath

- FSSP
- Automatic classification (L. Holm)
- www.ebi.ac.uk/dali/fssp/fssp.html

Morten Nielsen,CBS, BioCentrum, DTU

Major classes in SCOP

- Classes
- All a proteins
- All b proteins
- a and b proteins (a/b)
- a and b proteins (a+b)
- Multi-domain proteins
- Membrane and cell surface proteins
- Small proteins
- Coiled coil proteins

Morten Nielsen,CBS, BioCentrum, DTU

All a: Hemoglobin (1bab)

Morten Nielsen,CBS, BioCentrum, DTU

All b: Immunoglobulin (8fab)

Morten Nielsen,CBS, BioCentrum, DTU

a/b:Triosephosphate isomerase (1hti)

Morten Nielsen,CBS, BioCentrum, DTU

a+b: Lysozyme (1jsf)

Morten Nielsen,CBS, BioCentrum, DTU

Families

- Proteins whose evolutionarily relationship is readily recognizable from the sequence
(>~25% sequence identity)

- Families are further subdivided into Proteins
- Proteins are divided into Species
- The same protein may be found in several species

Fold

Superfamily

Family

Proteins

Morten Nielsen,CBS, BioCentrum, DTU

Superfamilies

- Proteins which are (remotely) evolutionarily related
- Sequence similarity low
- Share function
- Share special structural features

- Relationships between members of a superfamily may not be readily recognizable from the sequence alone

Fold

Superfamily

Family

Proteins

Morten Nielsen,CBS, BioCentrum, DTU

Folds

- >~50% secondary structure elements arranged in the same order in sequence and in 3D
- No evolutionary relation

Fold

Superfamily

Family

Proteins

Morten Nielsen,CBS, BioCentrum, DTU

Substitutions of Amino Acids

Mutation rates between amino acids have dramatic differences!

Substitution Matrices

BLOSUM matrices:

- Start from BLOCKS database (curated, gap-free alignments)
- Cluster sequences according to > X% identity
- Calculate Aab: # of aligned a-b in distinct clusters, correcting by 1/mn, where m, n are the two cluster sizes
- Estimate
P(a) = (b Aab)/(c≤d Acd); P(a, b) = Aab/(c≤d Acd)

Probabilistic interpretation of an alignment

An alignment is a hypothesis that the two sequences are related by evolution

Goal:

Produce the most likely alignment

Assert the likelihood that the sequences are indeed related

A Pair HMM for alignments

Model M

1 – 2

This model generates two sequences simultaneously

Match/Mismatch state M:

P(x, y) reflects substitution frequencies between pairs of amino acids

Insertion states I, J:

P(x), P(y) reflect frequencies of each amino acid

: set so that 1/2 is avg. length before next gap

:set so that 1/(1 – ) is avg. length of a gap

M

P(xi, yj)

1 –

1 –

I

P(xi)

J

P(yj)

optional

A Pair HMM for unaligned sequences

Model R

Two sequences are independently generated from one another

P(x, y | R) = P(x1)…P(xm) P(y1)…P(yn) = i P(xi) j P(yj)

1

1

J

P(yj)

I

P(xi)

To compare ALIGNMENT vs. RANDOM hypothesis

1 – 2

Every pair of letters contributes:

M

- (1 – 2) P(xi, yj) when matched
- P(xi) P(yj) when gapped
R

- P(xi) P(yj) in random model
Focus on comparison of

P(xi, yj) vs. P(xi) P(yj)

M

P(xi, yj)

1 –

1 –

I

P(xi)

J

P(yj)

1

1

J

P(yj)

I

P(xi)

To compare ALIGNMENT vs. RANDOM hypothesis

1 – 2

Every pair of letters contributes:

M

- (1 – 2) P(xi, yj) when matched
- P(xi) P(yj) when gapped
R

- P(xi) P(yj) in random model
Focus on comparison of

P(xi, yj) vs. P(xi) P(yj)

M

P(xi, yj)

1 – 2

1 – 2

(1 – )

-----------

(1 – 2)

I

P(xi)

J

P(yj)

Equivalent!

1

1

J

P(yj)

I

P(xi)

To compare ALIGNMENT vs. RANDOM hypothesis

Idea:

We will divide alignment score by the random score, and take logarithms

Let

P(xi, yj)

s(xi, yj) = log ––––––––– + log (1 – 2)

P(xi) P(yj)

(1 – ) P(xi)

d = – log –––––––––––––

(1 – 2) P(xi)

P(xi)

e = – log ––––––

P(xi)

=Defn substitution score

=Defn gap initiation penalty

=Defn gap extension penalty

The meaning of alignment scores

- The Viterbi algorithm for Pair HMMs corresponds exactly to global alignment DP with affine gaps
VM(i, j) = max { VM(i – 1, j – 1), VI( i – 1, j – 1) – d, Vj( i – 1, j – 1) } + s(xi, yj)

VI(i, j) = max { VM(i – 1, j) – d, VI( i – 1, j) – e }

VJ(i, j) = max { VM(i – 1, j) – d, VI( i – 1, j) – e }

- s(.,.) (1 – 2) ~how often a pair of letters substitute one another
- 1/mean length of next gap
- (1 – ) / (1 – 2) 1/mean arrival time of next gap

The meaning of alignment scores

Match/mismatch scores:

P(xi, yj)

s(a, b) log –––––––––– (ignore log(1 – 2) for the moment)

P(xi) P(yj)

Example:

DNA regions between human and mouse genes have average conservation of 80%

- What is the substitution score for a match?
P(a, a) + P(c, c) + P(g, g) + P(t, t) = 0.8 P(x, x) = 0.2

P(a) = P(c) = P(g) = P(t) = 0.25

s(x, x) = log [ 0.2 / 0.252 ] = 1.163

- What is the substitution score for a mismatch?
P(a, c) +…+P(t, g) = 0.2 P(x, yx) = 0.2/12 = 0.0167

s(x, y x) = log[ 0.0167 / 0.252 ] = -1.322

- What ratio matches/(matches + mism.) gives score 0?
x(#match) – y(#mism) = 0

1.163 (#match) – 1.322 (#mism) = 0

#match = 1.137(#mism)

matches = 53.2%

Substitution Matrices

BLOSUM matrices:

- Start from BLOCKS database (curated, gap-free alignments)
- Cluster sequences according to > X% identity
- Calculate Aab: # of aligned a-b in distinct clusters, correcting by 1/mn, where m, n are the two cluster sizes
- Estimate
P(a) = (b Aab)/(c≤d Acd); P(a, b) = Aab/(c≤d Acd)

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