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CSE182-L10

CSE182-L10. Gene Finding. HMM. T[j,k ]. Q is a set of states T is a matrix of transition probabilities T[j,k ]: probability of moving from state j to state k Σ is a set of symbols e j (S ) is the probability of emitting S while in state j. Automaton M=(Q,T, π,Σ,e )

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CSE182-L10

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  1. CSE182-L10 Gene Finding

  2. HMM T[j,k] • Q is a set of states • T is a matrix of transition probabilities • T[j,k]: probability of moving from state j to state k • Σ is a set of symbols • ej(S) is the probability of emitting S while in state j. • Automaton M=(Q,T, π,Σ,e) • At first, M goes to initial state j with probability πj • In state j, M emits a symbol from Σ according to ej, and moves to state k with probability T[j,k]. k j i

  3. Viterbi and sum algorithm for testing membership S1…Si-1 • Pmax(i,j|M) = max k Pmax(i-1,k) T[k,j] ej(Si) (Viterbi) • Psum(i,j|M) = ∑ k (Psum(i-1,k) T[k,j]) ej(Si) k Si j

  4. HMM ‘fair-coin’ example 0.6 0.6 1 0.4 2 3 1 0.4 EF(H)=0.5 EL(H)=0.1 M=(Q,T, π,Σ,e)

  5. 0.6 0.6 1 0.4 • H H T T T is the observed sequence 0.4 EF(H)=0.5 EL(H)=0.1 P_max(2,F) 0.6 1.5e-1 4.5e-2 1.3e-2 5.8e-3 0.5 1 0 0.4 4.86e-3 0 2e-2 5.4e-2 1.6e-2

  6. Summary • HMMs allow us to model position specific gap penalties, and allow for automated training to get a good alignment. • Patterns/Profiles/HMMs allow us to represent families and foucs on key residues • Each has its advantages and disadvantages, and needs special algorithms to query efficiently.

  7. Protein sequence analysis • Input: a protein sequence of unknown function. • To get function: • Compare against a database of protein sequences with known function. • Create a database of multiple alignment of diverged family members. • Search using patterns such as regular expressions • Search using profiles • Search using HMMs • In all cases, search domains, not the entire sequence

  8. Protein Domain databases • A number of databases capture proteins (domains) using various representations • Each domain is also associated with structure/function information, parsed from the literature. • Each database has specific query mechanisms that allow us to compare our sequences against them, and assign function HMM 3D

  9. Gene Finding What is a Gene? CSE 182

  10. Biology • In our discussion of BLAST, we alternated between looking at DNA, and protein sequences, treating them as strings. • DNA, RNA, and proteins are the 3 important molecules • What is the relation between the three?

  11. Gene • We define a gene as a location on the genome that codes for proteins. • The genic information is used to manufacture proteins through transcription, and translation. • There is a unique mapping from triplets to amino-acids CSE 182

  12. Transcription

  13. Translation

  14. Transcription start ATAGATGATGTACGATGAGAATGTGATTAATG Translation start Donor Acceptor

  15. Translation • The ribosomal machinery reads mRNA. • Each triplet is translated into a unique amino-acid until the STOP codon is encountered. • There is also a special signal where translation starts, usually at the ATG (M) codon. CSE 182

  16. Translation • The ribosomal machinery reads mRNA. • Each triplet is translated into a unique amino-acid until the STOP codon is encountered. • There is also a special signal where translation starts, usually at the ATG (M) codon. • Given a DNA sequence, how many ways can you translate it? CSE 182

  17. Gene Features • The gene can lie on any strand (relative to the reference genome) • The code can be in one of 3 frames. Frame 1 Frame 2 Frame 3 S R V * W R V Q Y S G * S I V D AGTAGAGTATAGTGGACG TCATCTCATATCACCTGC -ve strand

  18. Eukaryotic gene structure CSE 182

  19. ATG 5’ UTR 3’ UTR exon intron Translation start Acceptor Donor splice site Transcription start Gene Features CSE 182

  20. Gene identification • Eukaryotic gene definitions: • Location that codes for a protein • The transcript sequence(s) that encodes the protein • The protein sequence(s) • Suppose you want to know all of the genes in an organism. • This was a major problem in the 70s. PhDs, and careers were spent isolating a single gene sequence. • All of that changed with better reagents and the development of high throughput methods like EST sequencing CSE 182

  21. History: the drug and its target • Proteins are the molecular machinery of the cell. • Drugs “target” proteins, binding to activate/inhibit. http://www-medchem.ch.cam.ac.uk/2ifb.jpg http://www.uchsc.edu/pharmacology/img/resImg/cellBiol.jpg

  22. Rational drug design • Only a few (protein) targets were known in1970s • The focus was on designing drugs that interact with the target.

  23. Expressed Sequence Tags • It is possible to extract all of the mRNA from a cell. • However, mRNA is unstable • An enzyme called reverse transcriptase is used to make a DNA copy of the RNA. • Use DNA polymerase to get a complementary DNA strand. • Sequence the (stable) cDNA from both ends. • This leads to a collection of transcripts/expressed sequences (ESTs). • Many might be from the same gene AAAA TTTT AAAA TTTT CSE 182

  24. EST sequencing • The expressed transcript (mRNA) has a poly-A tail at the end, which can be used as a template for Reverse Transcriptase. • This collection of DNA has only the spliced message! • It is sampled at random and sequenced from one (3’/5’) or both ends. • Each message is sampled many times. • The resulting collection of sequences is called an EST database AAAA TTTT AAAA TTTT CSE 182

  25. EST Sequencing • Often, reverse transcriptase breaks off early. Why is this a good thing? • The 3’ end may not have a much coding sequence. • We can assemble the 5’ end to get more of the coding sequence CSE 182

  26. EST sequencing meets next gen sequencing • Newer methods like RNA-seq offer a more comprehensive sampling of the set of transcripts: • They can be used for gene finding, but,… • Differences in expression/abundance of transcripts • The gene sequence is in small pieces • The fragments must still be mapped back to the genome to get the coordinates. • There are other features of the gene that are not revealed by transcript sequencing

  27. ATG 5’ UTR 3’ UTR exon intron Translation start Acceptor Donor splice site Transcription start Computational Gene Finding • Given Genomic DNA, identify all the coordinates of the gene • TRIVIA QUIZ! What is the name of the FIRST gene finding program? (google testcode)

  28. Gene Finding: The 1st generation • Given genomic DNA, does it contain a gene (or not)? • Key idea: The distributions of nucleotides is different in coding (translated exons) and non-coding regions. • Therefore, a statistical test can be used to discriminate between coding and non-coding regions.

  29. Coding versus non-coding • You are given a collection of exons, and a collection of intergenic sequence. • Count the number of occurrences of ATGATG in Introns and Exons. • Suppose 1% of the hexamers in Exons are ATGATG • Only 0.01% of the hexamers in Intergenic are ATGATG • How can you use this idea to find genes?

  30. Generalizing Frequencies (X10-5) I E X AAAAAA AAAAAC AAAAAG AAAAAT 10 10 5 5 20 10 • Compute a frequency count for all hexamers. • Exons, Intergenic and the sequence X are all vectors in a multi-dimensional space • Use this to decide whether a sequence X is exonic/intergenic.

  31. A geometric approach (2 hexamers) • Plot the following vectors • E= [10, 20] • I= [10, 5] • V3 = [6, 10] • V4 = [9, 15] • Is V3 more like E or more like I? E 20 15 V3 10 I 5 5 10 15

  32. E V3 I Choosing between Intergenic and Exonic • Normalize V’ = V/||V|| • All vectors have the same length (lie on the unit circle) • Next, compute the angle to E, and I. • Choose the feature that is ‘closer’ (smaller angle.

  33. Coding versus non-coding signals • Fickett and Tung (1992) compared various measures • Measures that preserve the triplet frame are the most successful. • Genscan uses a 5th order Markov Model

  34. 5th order markov chain Exon Intron • PrEXON[AAAAAACGAGAC..] =T[AAAAA,A] T[AAAAA,C] T[AAAAC,G] T[AAACG,A]…… = (20/78) (50/78)………. A AAAAAA 20 1 AAAAAC 50 10 AAAAAG 5 30 AAAAAT 3 .. Tot C AAAAC AAAAA G AAAAG

  35. Scoring for coding regions • The coding differential can be computed as the log odds of the probability that a sequence is an exon vs. and intron. • In Genscan, separate transition matrices are trained for each frame, as different frames have different hexamer distributions

  36. Coding differential for 380 genes

  37. Coding region can be detected • Plot the coding score using a sliding window of fixed length. • The (large) exons will show up reliably. • Not enough to predict gene boundaries reliably Coding

  38. Other Signals • Signals at exon boundaries are precise but not specific. Coding signals are specific but not precise. • When combined they can be effective ATG AG GT Coding

  39. Combining Signals • We can compute the following: • E-score[i,j] • I-score[i,j] • D-score[i] • A-score[i] • Goal is to find coordinates that maximize the total score i j

  40. The second generation of Gene finding • Ex: Grail II. Used statistical techniques to combine various signals into a coherent gene structure. • It was not easy to train on many parameters. Guigo & Bursett test revealed that accuracy was still very low. • Problem with multiple genes in a genomic region

  41. Combining signals using D.P. • An HMM is the best way to model and optimize the combination of signals • Here, we will use a simpler approach which is essentially the same as the Viterbi algorithm for HMMs, but without the formalism.

  42. i1 i2 i3 i4 IIIIIEEEEEEIIIIIIEEEEEEIIIIEEEEEE IIIII Hidden states & gene structure • Identifying a gene is equivalent to labeling each nucleotide as E/I/intergenic etc. • These ‘labels’ are the hidden states • For simplicity, consider only two states E and I

  43. i1 i2 i3 i4 IIIIIEEEEEEIIIIIIEEEEEEIIIIEEEEEE IIIII Gene finding reformulated • Given a labeling L, we can score it as • I-score[0..i1-1] + E-score[i1..i2] + D-score[i2+1] + I-score[i2+1..i3-1] + A-score[i3-1] + E-score[i3..i4] + ……. • Goal is to compute a labeling with maximum score.

  44. Optimum labeling using D.P. (Viterbi) • Define VE(i) = Best score of a labeling of the prefix 1..i such that the i-th position is labeled E • Define VI(i) = Best score of a labeling of the prefix 1..i such that the i-th position is labeled I • Why is it enough to compute VE(i) & VI(i) ?

  45. Optimum parse of the gene j i j i

  46. Generalizing • Note that we deal with two states, and consider all paths that move between the two states. E I i

  47. Generalizing IG • We did not deal with the boundary cases in the recurrence. • Instead of labeling with two states, we can label with multiple states, • Einit, Efin, Emid, • I, IG (intergenic) I Efin Emid Note: all links are not shown here Einit

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