Presentation is loading. Please wait.

Presentation is loading. Please wait.

Evolutionary Models CS 498 SS Saurabh Sinha. Models of nucleotide substitution The DNA that we study in bioinformatics is the end(??)-product of evolution.

Similar presentations


Presentation on theme: "Evolutionary Models CS 498 SS Saurabh Sinha. Models of nucleotide substitution The DNA that we study in bioinformatics is the end(??)-product of evolution."— Presentation transcript:

1 Evolutionary Models CS 498 SS Saurabh Sinha

2 Models of nucleotide substitution The DNA that we study in bioinformatics is the end(??)-product of evolution Evolution is a very complicated process Very simplified models of this process can be studied within a probabilistic framework Allows testing of various hypotheses about the evolutionary process, from multi-species data Source: Ewens and Grant, Chapter 14.

3 Diversity in a population There IS genetic variation between individuals in a population But relatively little variation at nucl. level E.g., two humans differ at the nucl. level at one in 500 to 1000 nucls. Roughly speaking, a single nucleotide dominates the population at a particular position in the genome

4 Substitution Over long time periods, the nucleotide at a given position remains the same But periodically, this nucleotide changes (over the entire population) This is called “substitution”, i.e., replacement of the predominant nucl. for that position with another predominant nucl.

5 Markov Chain to model substitution Markov chain to describe the substitution process at a position States are “a”, “c”, “g”, “t” The chain “runs” in certain units of time, i.e., the state may change from one time point to the next time point The unit of time (difference between successive time points) may be arbitrary, e.g., 20000 generations.

6 Markov Chain to model substitution A symbol such as “p ag ” is the probability of a change from “a” to “g” in one unit of time When studying two extant species, the evolutionary model has to provide the joint probability of the two species’ data Sometimes, this is done by computing probability of the ancestor, starting from one extant species, and then the probability of the other extant species, starting from the ancestor If we want to do this, the evolutionary process (model) must be “time reversible”: P(x)P(x->y) = P(y)P(y->x)

7 Jukes Cantor Model Markov chain with four states: a,c,g,t Transition matrix P given by: agct a 1-3  g   c   t 

8 Jukes Cantor Model  is a parameter depending on what a “time unit” means. If time unit represents more #generations,  will be larger  must be less than 1/3 though

9 Jukes Cantor Model Whatever the current nucl is, each of the other three nucls are equally likely to substitute for it

10 Understanding the J-C Model Consider a transition matrix P, and a probability vector v (a row vector) What does w = v P represent ? If v is the probability distribution of the 4 nucls (at a position) now, w is the prob. distr. at the next time step.

11 Understanding the J-C model Suppose we can find a vector  such that  P =  If the probability distribution is , it will continue to remain  at future times This is called the stationary distribution of the Markov Chain

12 Understanding the J-C model Check that  = (0.25, 0.25, 0.25, 0.25) satisfies  P =  Therefore, if a position evolves as per this model, for long enough, it will be equally likely to have any of the 4 nucls! This is the very long term prediction, but can we write down what the position will be as a function of time (steps) ?

13 Spectral Decomposition Recall that we found a  such that  P =  Such a vector is called an “eigenvector” of P, and the corresponding “eigenvalue” is 1. In general, if v P = v (for scalar ), is called an eigenvalue, and v is a left eigenvector of P

14 Spectral decomposition Similarly, if P u T = u T, then u is called a right eigenvector In general, there may be multiple eigenvalues j and their corresponding left and right eigenvectors v j and u j We can write P as

15 Spectral decomposition Then, for any positive integer, it is true that Why is P n interesting to us ? Because it tells us what the probability distribution will be after n time steps If we started with v, then P n v will be the prob. distr. after n steps

16 Back to the J-C model We reasoned that  = (.25,.25,.25,.25) is a left eigenvector for the eigenvalue 1. Actually, the J-C transition matrix has this eigenvalue and the eigenvalue (1-4  ), and if we do the math we get the spectral decomposition of P as:

17 Back to the J-C model So, if we started with (1,0,0,0), i.e., an “a”, the probability that we’ll see an “a” at that position after n time steps is: 0.25+0.75(1-4  ) n And the probability that the “a” would have mutated to say “c” is: 0.25 - 0.25(1-4  ) n

18 Substitution probability As a function of time n, we therefore get Pr(x -> y) = 0.25 + 0.75 (1-4  ) n if x = y and = 0.25 - 0.25 (1-4  ) n otherwise If n -> , we get back our (0.25, 0.25, 0.25, 0.25) calculation

19 More advanced models The J-C model made highly “symmetric” assumptions, in its formulation of the transition matrix P In reality, for example, “transitions” are more common than “transversions” –What are these? Purine = A or G. Pyrimidine = C or T. Transition is substitution in the same category; transversion is substitution across categories –Purines are similarly sized, and pyrimidines are similarly sized. More likely to be replaced by similar sized nucl. The “Kimura” model captures this transition/transversion bias

20 Kimura model agct a 1-  -2  g   c   t  This of course is the transition probability matrix P of the Markov chain Two parameters now, instead of one.

21 Kimura model Again, one of the eigenvalues is 1, and the left eigenvector corresponding to it is  = (.25,.25,.25,.25) So again, the stationary distribution is uniform P(x -> x) =.25+.25(1-4  ) n +.5(1-2(  +  )) n P(x -> y) =.25+.25(1-4  ) n +.5(1-2(  +  )) n if x is a purine and y is the other purine

22 Even more advanced models Get to greater levels of realism Kimura model still has a uniform stationary distribution, which is not true of real data One extension: purine to pyrimidine subst. prob. is different from pyrimidine to purine subst. prob. –This leads to a non-uniform stationary probability

23 Felsenstein models agct a 1-u+u  a ugug ucuc utut g uaua 1-u+u  g ucuc utut c uaua ugug 1-u+u  c utut t uaua ugug ucuc 1-u+u  t Transition probability proportional to the stationary probability of the target nucleotide. Stationary distribution is (  a,  g,  c,  t )

24 Reversible models Many inference procedures require that the evolutionary model be time reversible What does this mean?

25 Reversible Markov Chain Source: Wikipedia Looks like time has been reversed. That is, if we can find a  such that The models we have seen today all have this property.


Download ppt "Evolutionary Models CS 498 SS Saurabh Sinha. Models of nucleotide substitution The DNA that we study in bioinformatics is the end(??)-product of evolution."

Similar presentations


Ads by Google