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A modeling Example

A+. C+. G+. T+. A-. C-. G-. T-. A modeling Example. CpG islands in DNA sequences. Methylation & Silencing. One way cells differentiate is methylation Addition of CH 3 in C-nucleotides Silences genes in region CG (denoted CpG) often mutates to TG, when methylated

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A modeling Example

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  1. A+ C+ G+ T+ A- C- G- T- A modeling Example CpG islands in DNA sequences

  2. Methylation & Silencing • One way cells differentiate is methylation • Addition of CH3 in C-nucleotides • Silences genes in region • CG (denoted CpG) often mutates to TG, when methylated • In each cell, one copy of X is silenced, methylation plays role • Methylation is inherited during cell division

  3. Example: CpG Islands CpG nucleotides in the genome are frequently methylated (Write CpG not to confuse with CG base pair) C  methyl-C  T Methylation often suppressed around genes, promoters  CpG islands

  4. Example: CpG Islands • In CpG islands, • CG is more frequent • Other pairs (AA, AG, AT…) have different frequencies Question: Detect CpG islands computationally

  5. A model of CpG Islands – (1) Architecture A+ C+ G+ T+ CpG Island A- C- G- T- Not CpG Island

  6. A model of CpG Islands – (2) Transitions How do we estimate parameters of the model? Emission probabilities: 1/0 • Transition probabilities within CpG islands Established from known CpG islands (Training Set) • Transition probabilities within other regions Established from known non-CpG islands (Training Set) Note: these transitions out of each state add up to one—no room for transitions between (+) and (-) states = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1

  7. Log Likehoods—Telling “CpG Island” from “Non-CpG Island” Another way to see effects of transitions: Log likelihoods L(u, v) = log[ P(uv | + ) / P(uv | -) ] Given a region x = x1…xN A quick-&-dirty way to decide whether entire x is CpG P(x is CpG) > P(x is not CpG)  i L(xi, xi+1) > 0

  8. A model of CpG Islands – (2) Transitions • What about transitions between (+) and (-) states? • They affect • Avg. length of CpG island • Avg. separation between two CpG islands 1-p Length distribution of region X: P[lX = 1] = 1-p P[lX = 2] = p(1-p) … P[lX= k] = pk-1(1-p) E[lX] = 1/(1-p) Geometric distribution, with mean 1/(1-p) X Y p q 1-q

  9. What if a new genome comes? • We just sequenced the porcupine genome • We know CpG islands play the same role in this genome • However, we have no known CpG islands for porcupines • We suspect the frequency and characteristics of CpG islands are quite different in porcupines How do we adjust the parameters in our model? LEARNING

  10. Learning Re-estimate the parameters of the model based on training data

  11. Two learning scenarios • Estimation when the “right answer” is known Examples: GIVEN: a genomic region x = x1…x1,000,000 where we have good (experimental) annotations of the CpG islands GIVEN: the casino player allows us to observe him one evening, as he changes dice and produces 10,000 rolls • Estimation when the “right answer” is unknown Examples: GIVEN: the porcupine genome; we don’t know how frequent are the CpG islands there, neither do we know their composition GIVEN: 10,000 rolls of the casino player, but we don’t see when he changes dice QUESTION: Update the parameters  of the model to maximize P(x|)

  12. 1. When the right answer is known Given x = x1…xN for which the true  = 1…N is known, Define: Akl = # times kl transition occurs in  Ek(b) = # times state k in  emits b in x We can show that the maximum likelihood parameters  (maximize P(x|)) are: Akl Ek(b) akl = ––––– ek(b) = ––––––– i AkicEk(c)

  13. 1. When the right answer is known Intuition: When we know the underlying states, Best estimate is the normalized frequency of transitions & emissions that occur in the training data Drawback: Given little data, there may be overfitting: P(x|) is maximized, but  is unreasonable 0 probabilities – BAD Example: Given 10 casino rolls, we observe x = 2, 1, 5, 6, 1, 2, 3, 6, 2, 3  = F, F, F, F, F, F, F, F, F, F Then: aFF = 1; aFL = 0 eF(1) = eF(3) = .2; eF(2) = .3; eF(4) = 0; eF(5) = eF(6) = .1

  14. Pseudocounts Solution for small training sets: Add pseudocounts Akl = # times kl transition occurs in  + rkl Ek(b) = # times state k in  emits b in x + rk(b) rkl, rk(b) are pseudocounts representing our prior belief Larger pseudocounts  Strong priof belief Small pseudocounts ( < 1): just to avoid 0 probabilities

  15. Pseudocounts Example: dishonest casino We will observe player for one day, 600 rolls Reasonable pseudocounts: r0F = r0L = rF0 = rL0 = 1; rFL = rLF = rFF = rLL = 1; rF(1) = rF(2) = … = rF(6) = 20 (strong belief fair is fair) rL(1) = rL(2) = … = rL(6) = 5 (wait and see for loaded) Above #s are arbitrary – assigning priors is an art

  16. 2. When the right answer is unknown We don’t know the true Akl, Ek(b) Idea: • We estimate our “best guess” on what Akl, Ek(b) are • Or, we start with random / uniform values • We update the parameters of the model, based on our guess • We repeat

  17. 2. When the right answer is unknown Starting with our best guess of a model M, parameters : Given x = x1…xN for which the true  = 1…N is unknown, We can get to a provably more likely parameter set  i.e.,  that increases the probability P(x | ) Principle: EXPECTATION MAXIMIZATION • Estimate Akl, Ek(b) in the training data • Update  according to Akl, Ek(b) • Repeat 1 & 2, until convergence

  18. Estimating new parameters To estimate Akl: (assume “|CURRENT”, in all formulas below) At each position i of sequence x, find probability transition kl is used: P(i = k, i+1 = l | x) = [1/P(x)]  P(i = k, i+1 = l, x1…xN) = Q/P(x) where Q = P(x1…xi, i = k, i+1 = l, xi+1…xN) = = P(i+1 = l, xi+1…xN | i = k) P(x1…xi, i = k) = = P(i+1 = l, xi+1xi+2…xN | i = k) fk(i) = = P(xi+2…xN | i+1 = l) P(xi+1 | i+1 = l) P(i+1 = l | i = k) fk(i) = = bl(i+1) el(xi+1) akl fk(i) fk(i) akl el(xi+1) bl(i+1) So: P(i = k, i+1 = l | x, ) = –––––––––––––––––– P(x | CURRENT)

  19. Estimating new parameters • So, Akl is the E[# times transition kl, given current ] fk(i) akl el(xi+1) bl(i+1) Akl = i P(i = k, i+1 = l | x, ) = i ––––––––––––––––– P(x | ) • Similarly, Ek(b) = [1/P(x | )]{i | xi = b} fk(i) bk(i) fk(i) bl(i+1) akl k l xi+2………xN x1………xi-1 el(xi+1) xi xi+1

  20. The Baum-Welch Algorithm Initialization: Pick the best-guess for model parameters (or arbitrary) Iteration: • Forward • Backward • Calculate Akl, Ek(b), given CURRENT • Calculate new model parameters NEW : akl, ek(b) • Calculate new log-likelihood P(x | NEW) GUARANTEED TO BE HIGHER BY EXPECTATION-MAXIMIZATION Until P(x | ) does not change much

  21. The Baum-Welch Algorithm Time Complexity: # iterations  O(K2N) • Guaranteed to increase the log likelihood P(x | ) • Not guaranteed to find globally best parameters Converges to local optimum, depending on initial conditions • Too many parameters / too large model: Overtraining

  22. Alternative: Viterbi Training Initialization: Same Iteration: • Perform Viterbi, to find * • Calculate Akl, Ek(b) according to * + pseudocounts • Calculate the new parameters akl, ek(b) Until convergence Notes: • Not guaranteed to increase P(x | ) • Guaranteed to increase P(x | , *) • In general, worse performance than Baum-Welch

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