1. Motif search Prof. William Stafford Noble Department of Genome Sciences Department of Computer Science and Engineering University of Washington thabangh@gmail.com

2. Outline One-minute response Revision Motifs – Definition and motivation – Representation as a PSSM – Scanning for motif occurrences Python

3. One-minute responses I would appreciate a little revision on the two types of tests. Class was not too fast today. The new method was fine. I understood about 50% of the Python part. The revision was very helpful. Need more explanation by chalk. Today was clear. I am OK with stats but struggling with the Python. We need to work more on Python. Can you explain how to control the FDR, the j* and that table of p- values? I would like to know more about Python execution.

4. Other comments We need extra Python tutorials with the tutors. Tutors should write an explanation of each program so that we understand what it does. Emile is complaining because our code looks like yours (no main function, not modular). What is the best way to code? – Emile’s comments will not affect your marks – they are stylistic suggestions only.

5. Revision You have searched a database of 1000 proteins with a single query sequence. What p-value threshold should you use if you want to apply Bonferroni correction and achieve 99% confidence? – 0.01 / 1000 = 0.00001 You have searched a database of 5000 proteins, and you observed a top-scoring p-value of 0.00002. What E-value does this correspond to? – 0.00002 * 5000 = 0.1 How do you decide whether to control the family-wise error rate or the false discovery rate? – If the conclusion or follow-up experiment involves a single result, then control family-wise error rate; otherwise, control false discovery rate.

6. Revision How many p-values in this list achieve an FDR of 10%? j = index α = 0.1 m = 1000 1 0.0000152 2 0.0001525 3 0.0057250 4 0.0017798 5 0.0025807 6 0.0057372 7 0.0080466 8 0.0104888 9 0.0111517 10 0.0113193 11 0.0121449 12 0.012389 13 0.0136114 14 0.0156425 15 0.0168226 16 0.0172086 17 0.0183523... 1000 0.999999 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007...

7. Motif Set of similar substrings, within a family of diverged sequences. Motif long DNA or protein sequence

8. Protein motifs Protein binding site Phosphorylation site Structural motif HAHU V.LSPADKTN..VKAAWGKVG.AHAGE..........YGAEAL.ERMFLSF..PTTKTYFPH.FDLS.HGSA HAOR M.LTDAEKKE..VTALWGKAA.GHGEE..........YGAEAL.ERLFQAF..PTTKTYFSH.FDLS.HGSA HADK V.LSAADKTN..VKGVFSKIG.GHAEE..........YGAETL.ERMFIAY..PQTKTYFPH.FDLS.HGSA HBHU VHLTPEEKSA..VTALWGKVN.VDEVG...........G.EAL.GRLLVVY..PWTQRFFES.FGDL.STPD HBOR VHLSGGEKSA..VTNLWGKVN.INELG...........G.EAL.GRLLVVY..PWTQRFFEA.FGDL.SSAG HBDK VHWTAEEKQL..ITGLWGKVNvAD.CG...........A.EAL.ARLLIVY..PWTQRFFAS.FGNL.SSPT MYHU G.LSDGEWQL..VLNVWGKVE.ADIPG..........HGQEVL.IRLFKGH..PETLEKFDK.FKHL.KSED MYOR G.LSDGEWQL..VLKVWGKVE.GDLPG..........HGQEVL.IRLFKTH..PETLEKFDK.FKGL.KTED IGLOB M.KFFAVLALCiVGAIASPLT.ADEASlvqsswkavsHNEVEIlAAVFAAY..PDIQNKFSQaFKDLASIKD GPUGNI A.LTEKQEAL..LKQSWEVLK.QNIPA..........HS.LRL.FALIIEA.APESKYVFSF.LKDSNEIPE GPYL GVLTDVQVAL..VKSSFEEFN.ANIPK...........N.THR.FFTLVLEiAPGAKDLFSF.LKGSSEVPQ GGZLB M.L.DQQTIN..IIKATVPVLkEHGVT...........ITTTF.YKNLFAK.HPEVRPLFDM.GRQ..ESLE

9. Transcription from DNA to RNA

10. Transcription factor binding A transcription factor is a protein that affects transcription by binding to DNA. Transcription factor binding sites

11. Why identify motifs? In proteins – Identify functionally important regions of a protein family – Find similarities to known proteins In DNA – Discover how genes are regulated

12. Representing motifs as PSSMs AAGTGT TAATGT AATTGT AATTGA ATCTGT AATTGT TGTTGT AAATGA TTTTGT A 6 6 2 0 0 2 C 0 0 1 0 0 0 G 0 1 1 0 9 0 T 3 2 5 9 0 7 Convert these 9 6-letter sequences into a PSSM. Use uniform background probabilities (A=0.25, C=0.25, G=0.25, T=0.25) and a pseudocount weight of 1. A 6.25 6.25 2.25 0.25 0.25 2.25 C 0.25 0.25 1.25 0.25 0.25 0.25 G 0.25 1.25 1.25 0.25 9.25 0.25 T 3.25 2.25 5.25 9.25 0.25 7.25 A 0.625 0.625 0.225 0.025 0.025 0.225 C 0.025 0.025 0.125 0.025 0.025 0.025 G 0.025 0.125 0.125 0.025 0.925 0.025 T 0.325 0.225 0.525 0.925 0.025 0.725 A 2.5 2.5 0.9 0.1 0.1 1 C 0.1 0.1 0.5 0.1 0.1 0 G 0.1 0.5 0.5 0.1 3.7 0 T 1.3 0.9 2.1 3.7 0.1 3 A 1.32 1.32 -0.15 -3.32 -3.32 -0.15 C -3.32 -3.32 -1.00 -3.32 -3.32 -3.32 G -3.32 -1.00 -1.00 -3.32 1.89 -3.32 T 0.38 -0.15 1.07 1.89 -3.32 1.54

13. Scanning for motif occurrences Given: – a long DNA sequence, and TAATGTTTGTGCTGGTTTTTGTGGCATCGGGCGAGAATAGCGCGTGGTGTGAAAG – a DNA motif represented as a PSSM Find: – occurrences of the motif in the sequence A 1.32 1.32 -0.15 -3.32 -3.32 -0.15 C -3.32 -3.32 -1.00 -3.32 -3.32 -3.32 G -3.32 -1.00 -1.00 -3.32 1.89 -3.32 T 0.38 -0.15 1.07 1.89 -3.32 1.54

14. Scanning for motif occurrences A 1.32 1.32 -0.15 -3.32 -3.32 -0.15 C -3.32 -3.32 -1.00 -3.32 -3.32 -3.32 G -3.32 -1.00 -1.00 -3.32 1.89 -3.32 T 0.38 -0.15 1.07 1.89 -3.32 1.54 TAATGTTTGTGCTGGTTTTTGTGGCATCGGGCGAGAATAGCGCGTGGTGTGAAAG 0.38 + 1.32 – 0.15 + 1.89 + 1.89 + 1.54 = 6.87

15. Scanning for motif occurrences A 1.32 1.32 -0.15 -3.32 -3.32 -0.15 C -3.32 -3.32 -1.00 -3.32 -3.32 -3.32 G -3.32 -1.00 -1.00 -3.32 1.89 -3.32 T 0.38 -0.15 1.07 1.89 -3.32 1.54 TAATGTTTGTGCTGGTTTTTGTGGCATCGGGCGAGAATAGCGCGTGGTGTGAAAG 1.32 + 1.32 + 1.07 – 3.32 – 3.32 + 1.54 = -1.39

16. CTCF One of the most important transcription factors in human cells. Responsible both for turning genes on and for maintaining 3D structure of the DNA.

17. Searching human chromosome 21 with the CTCF motif

18. Significance of scores Motif scanning algorithm TTGACCAGCAGGGGGCGCCG 6.30 Low score = not a motif occurrence High score = motif occurrence How high is high enough? A 1.32 1.32 -0.15 -3.32 -3.32 -0.15 C -3.32 -3.32 -1.00 -3.32 -3.32 -3.32 G -3.32 -1.00 -1.00 -3.32 1.89 -3.32 T 0.38 -0.15 1.07 1.89 -3.32 1.54

19. Two way to assess significance 1.Empirical – Randomly generate data according to the null hypothesis. – Use the resulting score distribution to estimate p- values. 2.Exact – Mathematically calculate all possible scores – Use the resulting score distribution to estimate p- values.

20. CTCF empirical null distribution

21. Poor precision in the tail

22. Converting scores to p-values Linearly rescale the matrix values to the range [0,100] and integerize. A -2.3 1.7 1.1 0.1 C 1.2 -0.3 0.4 -1.0 G -3.0 2.0 0.5 0.8 T 4.0 0.0 -2.1 1.5 A 10 67 59 44 C 60 39 49 29 G 0 71 50 54 T 100 43 13 64

23. Converting scores to p-values Find the smallest value. Subtract that value from every entry in the matrix. All entries are now non-negative. A -2.3 1.7 1.1 0.1 C 1.2 -0.3 0.4 -1.0 G -3.0 2.0 0.5 0.8 T 4.0 0.0 -2.1 1.5 A 0.7 4.7 4.1 3.1 C 4.2 2.7 3.4 2.0 G 0.0 5.0 3.5 3.8 T 7.0 3.0 0.9 4.5

24. Converting scores to p-values Find the largest value. Divide 100 by that value. Multiply through by the result. All entries are now between 0 and 100. A 0.7 4.7 4.1 3.1 C 4.2 2.7 3.4 2.0 G 0.0 5.0 3.5 3.8 T 7.0 3.0 0.9 4.5 100 / 7 = 14.2857 A 10.00 67.14 58.57 44.29 C 60.00 38.57 48.57 28.57 G 0.00 71.43 50.00 54.29 T 100.00 42.86 12.85 64.29

25. Converting scores to p-values Round to the nearest integer. A 10 67 59 44 C 60 39 49 29 G 0 71 50 54 T 100 43 13 64 A 10.00 67.14 58.57 44.29 C 60.00 38.57 48.57 28.57 G 0.00 71.43 50.00 54.29 T 100.00 42.86 12.85 64.29

26. Converting scores to p-values Say that your motif has N rows. Create a matrix that has N rows and 100N columns. The entry in row i, column j is the number of different sequences of length i that can have a score of j. A 10 67 59 44 C 60 39 49 29 G 0 71 50 54 T 100 43 13 64 0 1 2 3 4 … 400

27. Converting scores to p-values For each value in the first column of your motif, put a 1 in the corresponding entry in the first row of the matrix. There are only 4 possible sequences of length 1. A 10 67 59 44 C 60 39 49 29 G 0 71 50 54 T 100 43 13 64 0 1 2 3 4 … 10 60 100 400 1111

28. Converting scores to p-values For each value x in the second column of your motif, consider each value y in the zth column of the first row of the matrix. Add y to the x+zth column of the matrix. A 10 67 59 44 C 60 39 49 29 G 0 71 50 54 T 100 43 13 64 1 1 0 1 2 3 4 … 10 60 77 100 400 111

29. Converting scores to p-values For each value x in the second column of your motif, consider each value y in the zth column of the first row of the matrix. Add y to the x+zth column of the matrix. What values will go in row 2? – 10+67, 10+39, 10+71, 10+43, 60+67, …, 100+43 These 16 values correspond to all 16 strings of length 2. A 10 67 59 44 C 60 39 49 29 G 0 71 50 54 T 100 43 13 64 1 1 0 1 2 3 4 … 10 60 77 100 400 111

30. Converting scores to p-values In the end, the bottom row contains the scores for all possible sequences of length N. Use these scores to compute a p-value. A 10 67 59 44 C 60 39 49 29 G 0 71 50 54 T 100 43 13 64 1 1 0 1 2 3 4 … 10 60 77 100 400 111

31. Computing a p-value The probability of observing a score >4 is the area under the curve to the right of 4. This probability is called a p-value. p-value = Pr(data|null)

32. Sample problem #1 Given: – a file containing a length-n DNA sequence, and – a file containing a DNA motif represented as a PSSM of length n. Return: – the score of the motif versus the sequence A 1.32 1.32 -0.15 -3.32 -3.32 -0.15 C -3.32 -3.32 -1.00 -3.32 -3.32 -3.32 G -3.32 -1.00 -1.00 -3.32 1.89 -3.32 T 0.38 -0.15 1.07 1.89 -3.32 1.54

33. Sample problem #2 Given: – a file containing a DNA sequence, and – a file containing a DNA motif represented as a PSSM. Return: – For each position that scores greater than 0, print the position, the score and the matching sequence A 1.32 1.32 -0.15 -3.32 -3.32 -0.15 C -3.32 -3.32 -1.00 -3.32 -3.32 -3.32 G -3.32 -1.00 -1.00 -3.32 1.89 -3.32 T 0.38 -0.15 1.07 1.89 -3.32 1.54

34. Sample problem #3 Modify the previous program to print the same results, but in sorted order, with the greatest score first.