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PROLOGUE: Pitfalls of standard alignments. Scoring a pairwise alignment. A: ALA E VLIRLIT K LYP B: ASA K HLNRLIT E LYP. Blosum62. Alignment of a family (globins). …………………………………………. Different positions are not equivalent. Sequence logos. http://weblogo.berkeley.edu/cache/file5h2DWc.png.

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## PROLOGUE: Pitfalls of standard alignments

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**PROLOGUE:**Pitfalls of standard alignments**Scoring a pairwise alignment**A: ALAEVLIRLITKLYP B: ASAKHLNRLITELYP Blosum62**Alignment of a family (globins)**…………………………………………. Different positions are not equivalent**Sequence logos**http://weblogo.berkeley.edu/cache/file5h2DWc.png The substitution score IN A FAMILY should depend on the position (the same for gaps) For modelling families we need more flexible tools**Probabilistic Models for Biological Sequences**• What are they?**Probabilistic models for sequences**• Generative definition: • Objects producing different outcomes (sequences) with different probabilities • The probability distribution over the sequences space determines the model specificity Probability Sequence space M Generates si with probability P(si | M) e.g.: M is the representation of the family of globins**Probabilistic models for sequences**• We don’t need a generator of new biological sequences • the generative definition is useful as operative definition • Associative definition: • Objects that, given an outcome (sequence), compute a probability value Probability M Sequence space Associates probability P(si | M) to si e.g.: M is the representation of the family of globins**Probabilistic models for sequences**Most useful probabilistic models are Trainable systems The probability density function over the sequence space is estimated from known examples by means of a learning algorithm Known examples Pdf estimate (generalization) Probability Probability Sequence space Sequence space e.g.: Writing a generic representation of the sequences of globins starting from a set of known globins**Probabilistic Models for Biological Sequences**• What are they? • Why to use them?**Seq1**Seq2 Seq3 Seq4 Seq5 Seq6 0.98 0.21 0.12 0.89 0.47 0.78 Modelling a protein family Probabilistic model Given a protein class (e.g. Globins), a probabilistic model trained on this family can compute a probability value for each new sequence This value measures the similarity between the new sequence and the family described by the model**Probabilistic Models for Biological Sequences**• What are they? • Why to use them? • Which probabilities do they compute?**P( s | M ) or P( M | s ) ?**A model M associates to a sequence s the probability P( s | M ) This probability answers the question: Which is the probability for a model describing the Globins to generate the sequence s ? The question we want to answer is: Given a sequence s, is it a Globin? We need to compute P( M | s ) !!**BayesTheorem**• P(X,Y) = P(X | Y) P(Y) = P(Y | X) P(X) Joint probability • So: P(X | Y) P(Y) P(Y | X) = P(X) P(s | M) P(M) A priori probabilities P(M | s) = P(s) P(s | M) P(M | s) Evidence M Conclusion s Conclusion M Evidence s**The A priori probabilities**P(s | M) P(M) A priori probabilities P(M | s) = P(s) P(M) is the probability of the model (i.e. of the class described by the model) BEFORE we know the sequence: can be estimated as the abundance of the class P(s) is the probability of the sequence in the sequence space. Cannot be reliably estimated!!**P(s | M1) P(M1)**P(M1 | s) P(s) P(M2 | s) P(s) P(s | M2) P(M2) P(s | M1) P(M1) P(s | M2) P(M2) Comparison between models We can overcome the problem comparing the probability of generating s from different models = = = Ratio between the abundance of the classes**Null model**Otherwise we can score a sequence for a model M comparing it to a Null Model: a model that generates ALL the possible sequences with probabilities depending ONLY on the statistical amino acid abundance P(s | M) S(M, s) = log P(s | N) Sequences NOT belonging to model M Sequences belonging to model M S(M, s) In this case we need a threshold and a statistic for evaluating the significance (E-value, P-value)**The simplest probabilistic models:**Markov Models • Definition**Markov Models Example: Weather**Register the weather conditions day by day: as a first hypothesis the weather condition in a day depends ONLY on the weather conditions in the day before. Define the conditional probabilities P(C|C), P(C|R),…. P(R|C)….. R C F S C: Clouds R: Rain F: Fog S: Sun The probability for the 5-days registration CRRCS P(CRRCS) = P(C)·P(R|C) ·P(R|R) ·P(C|R) ·P(S|C)**Markov Model**Stochastic generator of sequences in which the probability of state in position i depends ONLY on the state in position i-1 Given an alphabet C = {c1; c2; c3; ………cN } a Markov model is described with N×(N+2) parameters {art, aBEGIN t , ar END; r, tC} arq = P( s i= q| s i-1= r ) aBEGIN q = P( s 1= q ) ar END= P( s T= END | s T-1= r ) c2 c1 c3 END BEGIN t art + ar END = 1 r t aBEGIN t = 1 cN c4**T**aBEGIN s i=2as s as END = i T 1 i-1 Markov Models Given the sequence: s = s1s2s3s4s6 ……………sT with siC = {c1; c2; c3; ………cN } P( s | M ) = P( s1) i=2 P( s i| s i-1 ) = P(“ALKALI”)= aBEGIN A aA L aL K aK A aA L aL I aI END**??**0.3 0.2 0.4 0.3 0.3 C C R R 0.2 0.5 0.3 0.1 0.2 0.0 0.2 ?? 0.1 0.2 ?? 0.1 ?? 0.7 0.0 0.2 0.1 0.4 S S S S F F 0.2 1.0 0.3 0.0 0.8 0.2 0.4 0.0 Markov Models: Exercise 1) Fill the non defined values for the transition probabilities**0.1**0.3 0.2 0.4 0.3 0.3 C C R R 0.2 0.5 0.3 0.1 0.2 0.0 0.2 0.0 0.1 0.2 0.2 0.1 0.0 0.7 0.0 0.2 0.1 0.4 S S S S F F 0.2 1.0 0.3 0.0 0.8 0.2 0.4 0.0 Markov Models: Exercise 2) Which model better describes the weather in summer? Which one describes the weather in winter?**0.3**0.1 0.2 0.4 0.3 0.3 C C R R 0.2 0.5 0.1 0.3 0.2 0.0 0.2 0.0 0.1 0.2 0.1 0.2 0.7 0.0 0.2 0.0 0.1 0.4 S S S S F F 1.0 0.2 0.3 0.0 0.8 0.2 0.4 0.0 Markov Models: Exercise Winter 3) Given the sequence CSSSCFS which model gives the higher probability? [Consider the starting probabilities: P(X|BEGIN)=0.25] Summer**0.3**0.4 0.3 C R 0.5 0.3 0.2 0.2 0.2 0.2 0.0 0.2 0.1 S S F 0.2 0.3 0.2 0.4 0.1 0.2 0.3 C R 0.2 0.1 0.0 0.0 0.1 0.1 0.7 0.0 0.4 S S F 1.0 0.0 0.8 0.0 Markov Models: Exercise Winter P (CSSSCFS | Winter) = =0.25x0.1x0.2x0.2x0.3x0.2x0.2= =1.2 x 10-5 P (CSSSCFS | Summer) = =0.25x0.4x0.8x0.8x0.1x0.1x1.0= =6.4 x 10-4 4) Can we conclude that the observation sequence refers to a summer week? Summer**0.3**0.4 0.3 C R 0.5 0.3 0.2 0.2 0.2 0.2 0.0 P(Summer | Seq) 0.2 0.1 S S P(Winter | Seq) F 0.2 0.3 0.2 0.4 0.1 0.2 0.3 C R 0.2 0.1 0.0 0.0 0.1 0.1 0.7 0.0 0.4 S S F 1.0 0.0 0.8 0.0 Markov Models: Exercise Winter P (Seq | Winter) =1.2 x 10-5 P (Seq | Summer) =6.4 x 10-4 = Summer P(Seq |Summer) P(Summer) = P(Seq| Winter) P(Winter)**C**G A T Simple Markov Model for DNA sequences DNA: C = {Adenine, Citosine, Guanine, Timine } 16 transition probabilities (12 of which independent) + 4 Begin probabilities + 4 End probabilities. The parameters of the model are different in different zones of DNA They describe the overall composition and the couple recurrences**C**G A T P ( s | GpC ) ·P(GpC) P (GpC | s) = P (s | GpC) ·P(GpC) + P (s | nonGpC) ·P(nonGpC) C G A T Example of Markov Models: GpC Island GATGCGTCGC CTACGCAGCG GpC Islands Non-GpC Islands In the Markov Model of GpC Islands aGC is higher than in Markov Model Non-GpC Islands Given a sequence s we can evaluate**The simplest probabilistic models:**Markov Models • Definition • Training**Probabilistic training of a parametric method**Generally speaking, a parametric model M aims to reproduce a set of known data Model M Parameters T Modelled data Real data (D) How to compare them?**nik**aik = Sjnij Training of Markov Models • Let M be the set of parameters of model M. • During the training phase, M parameters are estimated from the set of known data D • Maximum Likelihood Extimation (ML) • ML = argmaxP( D | M, ) • It can be proved that: Frequency of occurrence as counted in the data set D Maximum A Posteriori Extimation (MAP) MAP = argmax P( | M, D ) = argmax[P( D | M, ) P( ) ]**d P(D|M)**= 0 d p d P(D|M) = ph -1(1- p)t-1(h(1-p)-tp) = 0 d p Example (coin-tossing) Given N tossing of a coin (our data D), the outcomes are h heads and t tails (N=t+h) ASSUME the model P(D|M)= ph (1- p)t Computing the maximum likelihood ofP(D|M) We obtain that our estimate of p is p = h / (h+t) = h / N**Example (Error measure)**Suppose you think that your data are affected by a Gaussian error So that they are distributed according to F(xi)=A*exp-[(xi – m)2 /2s 2] With A=1/sqrt(2 s) If your measures are independent the data likelihood is P(Data| model) = PiF(xi) Find m and sthat maximize the P(Data| model)**Maximum Likelihood training: Proof**Given a sequence s contained in D: s = s1s2s3s4s6 ……………sT We can count the number of transitions between any to states j and k: njk Where states 0 and N+1 are BEGIN and END Normalisation contstraints are taken into account using the Lagrange multipliers lk**Hidden Markov Models**• Preliminary examples**Loaded dice**We have 99 regular dice (R) and 1 loaded die (L). P(1) P(2) P(3) P(4) P(5) P(6) R 1/6 1/6 1/6 1/6 1/6 1/6 L 1/10 1/10 1/10 1/10 1/10 1/2 Given a sequence: 4156266656321636543662152611536264162364261664616263 We don’t know the sequence of dice that generated it. RRRRRLRLRRRRRRRLRRRRRRRRRRRRLRLRRRRRRRRLRRRRLRRRRLRR**Loaded dice**Hypothesis: We chose a different die for each roll Two stochastic processes give origin to the sequence of observations. 1) Choosing the die ( R o L ). 2) Rolling the die The sequence of dice is hidden The first process is assumed to be Markovian (in this case a 0-order MM) The outcome of the second process depends only on the state reached in the first process (that is the chosen die)**Casinò**• Model • Each state(R and L) generates a character of the alphabet • C = {1, 2, 3, 4, 5, 6 } • The emission probabilities depend only on the state. • The transition probabilities describe a Markov model that generates a state path: the hidden sequence (p) • The observationssequence (s) is generated by two concomitant stochastic processes 0.01 R L 0.01 0.99 0.99**The observationssequence (s) is generated by two concomitant**stochastic processes 4156266656321636543662152611 RRRRRLRLRRRRRRRLRRRRRRRRRRRR 0.01 R L 0.01 0.99 0.99 Choose the State : R Probability= 0.99 Chose the Symbol: 1 Probability= 1/6 (given R) 4156266656321636543662152611 RRRRRLRLRRRRRRRLRRRRRRRRRRRR**The observationssequence (s) is generated by two concomitant**stochastic processes 415626665632163654366215261 RRRRRLRLRRRRRRRLRRRRRRRRRRR 0.01 R L 0.01 0.99 0.99 Choose the State : L Probability= 0.99 Chose the Symbol: 5 Probability= 1/10 (given L) 41562666563216365436621526115 RRRRRLRLRRRRRRRLRRRRRRRRRRRRL**0.01**R L 0.01 0.99 0.99 Loaded dice • Model • Each state(R and L) generates a character of the alphabet • C = {1, 2, 3, 4, 5, 6 } • The emission probabilities depend only on the state. • The transition probabilities describe a Markov model that generates a state path: the hidden sequence (p) • The observationssequence (s) is generated by two concomitant stochastic processes**Some not so serious example**1) DEMOGRAPHY Observable: Number of births and deaths in a year in a village. Hidden variable: Economic conditions (as a first approximation we can consider the success in business as a random variable, and by consequence, the wealth as a Markov variable ---> can we deduce the economic conditions of a village during a century by means of the register of births and deaths? 2) THE METEREOPATHIC TEACHER Observable: Average of the marks that a metereopathic teacher gives to their students during a day. Hidden variable: Weather conditions ---> can we deduce the weather conditions during a years by means of the class register?**To be more serious**1) SECONDARY STRUCTURE Observable: protein sequence Hidden variable: secondary structure ---> can we deduce (predict) the secondary structure of a protein given its amino acid sequence? 2) ALIGNMENT Observable: protein sequence Hidden variable: position of each residue along the alignment of a protein family ---> can we align a protein to a family, starting from its amino acid sequence?**Hidden Markov Models**• Preliminary examples • Formal definition**Formal definition of Hidden Markov Models**• A HMM is a stochastic generator of sequences characterised by: • N states • A set of transition probabilities between two states {akj} • akj = P( (i) = j | (i-1) = k ) • A set of starting probabilities {a0k} • a0k = P( (1) = k ) • A set of ending probabilities {ak0} ak0 = P( p (i) = END | (i-1) = k ) • An alphabet C with M characters. • A set of emission probabilities for each state {ek (c)} • ek (c) = P( s i= c | (i) = k ) • Constraints: ka0 k = 1 ak0 + jak j = 1 k cCek (c) = 1 k s: sequence p: path through the states**Choose the initial state p (1) following the probabilities**a0k i = 1 Choose the character s i from the alphabet C following the probabilities ek(c) Choose the next state following the probabilities ak j and ak0 No Yes ii +1 End Is the END state choosed? Generating a sequence with a HMM**BEGIN**a0Y= 0.2 a0N = 0.8 Gpc Island Non- Gpc Island aYN = 0.2 aNN = 0.8 Y N aYY = 0.7 aNY = 0.1 eY (A) = 0.1 eY (G) = 0.4 eY (C) = 0.4 eY (T) = 0.1 eN (A) = 0.25 eN (G) = 0.25 eN (C) = 0.25 eN (T) = 0.25 aN0 = 0.1 aY0 = 0.1 END GpC Island, simple model • s :AGCGCGTAATCTG • p :YYYYYYYNNNNNN • P( s, p | M ) can be easily computed**BEGIN**a0Y= 0.2 a0N = 0.8 GpC Island Non- GpC Island aYN = 0.2 aNN = 0.8 Y N aYY = 0.7 aNY = 0.1 eY (A) = 0.1 eY (G) = 0.4 eY (C) = 0.4 eY (T) = 0.1 eN (A) = 0.25 eN (G) = 0.25 eN (C) = 0.25 eN (T) = 0.25 aN0 = 0.1 aY0 = 0.1 END P( s, p | M ) can be easily computed s : A G C G C G T A A T C T G p : Y Y Y Y Y Y Y N N N N N N Emission: 0.1 0.4 0.4 0.4 0.4 0.4 0.10.250.250.250.250.250.25 Transition: 0.2 0.7 0.7 0.7 0.7 0.7 0.7 0.20.8 0.8 0.80.8 0.8 0.1 Multiplying all the probabilities gives the probability of having the sequence AND the path through the states**Evaluation of the joint probability of the sequence ad the**path**Hidden Markov Models**• Preliminary examples • Formal definition • Three questions**BEGIN**a0Y= 0.2 a0N = 0.8 GpC Island Non- GpC Island aYN = 0.2 aNN = 0.8 Y N aYY = 0.7 aNY = 0.1 eY (A) = 0.1 eY (G) = 0.4 eY (C) = 0.4 eY (T) = 0.1 eN (A) = 0.25 eN (G) = 0.25 eN (C) = 0.25 eN (T) = 0.25 aN0 = 0.1 aY0 = 0.1 END GpC Island, simple model • s :AGCGCGTAATCTG • p:????????????? • P( s, p | M ) can be easily computed • How to evaluate P ( s | M )?

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