1. MLL translocations specify a distinct gene expression profile that distinguishes a unique leukemia Armstrong et al, Nature Genetics 30, 41-47 (2002)

2. E, C&N_log2E expression matrix E log2 E, center, normalize E. Canaani: A.L.L. LEUKEMIA 27 PATIENTS, 14 with MLL translocation, (“MLL”) 13 without (ALL) 3250 genes passed filter

3. E. Canaani: A.L.L. LEUKEMIA 27 PATIENTS, 14 with MLL translocation, (“MLL”) 13 without (ALL) MLLALL

4. AIM: FIND THE GENES THAT ARE EITHER ACTIVATED OR SUPPRESSED BY THE t(4,11) CHIMERIC MLL PROTEIN THESE GENES MAY BE THE CAUSE OF CANCER AND TARGETS FOR THERAPY Rozovskaia, Ravid, Getz,....., Canaani: PNAS (2003)

5. SUPERVISED ANALYSIS: HYPOTHESIS TESTING USING CLINICAL INFORMATION (MLL/ALL=NO TRANS.) IDENTIFY DIFFERENTIATING GENES Basic methodologies1 HYPOTHESIS: THE EXPRESSION LEVELS OF GENE g IN SAMPLES WITH THE MLL TRANSLOCATION AND WITHOUT – ARE DRAWN FROM THE SAME DISTRIBUTION t(4:11) MLL ALL without t(4:11) USE STANDARD STATISTICAL TESTS, FOR ONE GENE AT A TIME TO CALCULATE P g = probability that the 27 expression levels of gene g (14 from MLL, 13 ALL) are taken from the same distribution

6. gene10 -1.0522 -0.5557 -1.0522 -0.6297 -1.0522 -0.7643 -0.4200 -0.7931 -1.0522 -0.7548 -1.0522 -0.5799 0.0225 -0.8326 2.1513 0.5143 0.0709 0.0421 0.9885 1.1993 2.0765 -0.0126 1.2574 0.2553 0.5241 -0.2726 1.7740 Cluster Incl. U70321:Human herpes virus entry mediator mRNA MLL: ALL: mean = -0.76 std = 0.3 mean = 0.81 std = 0.82

7. histograms mean = -0.76 std = 0.3mean = 0.81 std = 0.82

8. NORMALIZED (FREQUENCIES) mean = -0.76 std = 0.3 mean = 0.81 std = 0.82

10. t-test T = -6.6 P = 6e-7

11. SUPERVISED ANALYSIS; t-TEST n SAMPLES; n A KNOWN TO BELONG TO GROUP A, n B TO B (27) (14) (MLL) (13) (ALL) CONSIDER ONE PARTICULAR GENE, g; WE HAVE IT’S n A EXPRESSION LEVELS E gs FOR SAMPLES OF GROUP A n B EXPRESSION LEVELS E gs FOR SAMPLES OF GROUP B T-TEST NULL HYPOTHESIS: ALL n NUMBERS WERE DRAWN FROM THE SAME DISTRIBUTION (GENE g HAS SIMILAR EXPRESSION IN MLL AND ALL) - MEANS, AND - STANDARD DEVIATIONS OF THE TWO GROUPS OF n A, n B NUMBERS. TABLES P-VALUE

12. SUPERVISED ANALYSIS; t-TEST ASSUMING THAT THE VARIABLES A B ARE NORMAL DISTRIBUTED, THE DISTRIBUTION OF CAN BE READ OFF STANDARD TABLES. IF WE FOUND THE VALUE t g FOR GENE g, WE ASK WHAT IS THE PROBABILITY P TO OBTAIN UNDER THE NULL HYPOTHESIS. e.g. IF P=0.01, THIS MEANS THAT SUCH A VALUE OF t COULD HAVE BEEN OBTAINED BY CHANCE WITH PROBABILITY 0.01. WHAT IF WE TESTED 2000 GENES AND FOUND 20 ? THE PROBLEM OF MULTIPLE COMPARISONS T-TEST2 RANDOM DATA t 0 Pr(t) p

13. SUPERVISED ANALYSIS; t-TEST ASSUMING THAT THE VARIABLES A B ARE NORMAL DISTRIBUTED, THE DISTRIBUTION OF CAN BE READ OFF STANDARD TABLES. IF WE FOUND THE VALUE t g FOR GENE g, WE ASK WHAT IS THE PROBABILITY P TO OBTAIN UNDER THE NULL HYPOTHESIS. e.g. IF P=0.01, THIS MEANS THAT SUCH A VALUE OF t COULD HAVE BEEN OBTAINED BY CHANCE WITH PROBABILITY 0.01. T-TEST2 t 0 Pr(t) p

14. 0.3449 1.1137 0.2628 -0.5576 0.3116 1.4970 0.0126 -1.2401 -1.2401 -1.2401 -0.6049 -1.2401 0.0520 -1.2401 1.5279 2.0361 -0.2097 0.4142 -1.2401 0.5464 0.8569 -0.9042 0.8856 0.5590 -0.4396 -1.2401 0.9760 Cluster Incl. AB005298:Homo sapiens BAI 2 mRNA mean = -0.27 std = 0.92 mean = 0.29 std = 1.00 MLL: ALL:

15. histograms

16. NORMALIZED (FREQUENCIES) mean = -0.27 std = 0.92mean = 0.29 std = 1.00

18. t-test T = -1.48 P = 0.15 85%

19. gene2000 Homo sapiens Human mRNA for beta-actin -0.9115 1.3851 -1.1715 0.3638 0.1726 -0.0786 -1.1979 -0.2578 1.3337 -0.8245 -1.2254 1.2949 1.9490 2.1157 0.3747 0.3895 0.0569 0.1268 0.1796 -0.1114 -0.0109 -0.0943 -0.1451 -0.1103 -0.0829 -0.1061 -0.1567 MLL: ALL: mean = 0.21 std = 1.2 mean = 0.18 std = 1.46

20. histograms mean = 0.21 std = 1.2mean = 0.18 std = 1.46

21. NORMALIZED (FREQUENCIES) mean = 0.21 std = 1.2mean = 0.18 std = 1.46

23. t-test T = 0.0535 P = 0.96

24. genes ordered by p-value 929 genes with p < 0.05 ordered by difference of means (ALL –MLL)

25. 929 genes with p < 0.05 MLL ALL (OUT OF 3250 GENES TESTED) 143 genes with p < 0.05 (OUT OF 3250 GENES TESTED) RANDOM DATA

26. (OUT OF 3250 GENES TESTED) 143 genes with p < 0.05 RANDOM DATA

27. 143 WITH P-value<0.05

28. SUPERVISED ANALYSIS: HYPOTHESIS TESTING USING CLINICAL INFORMATION (MLL/NO TRANS.) IDENTIFY DIFFERENTIATING GENES Basic methodologies1 HYPOTHESIS: THE EXPRESSION LEVELS OF GENE g IN SAMPLES WITH THE MLL TRANSLOCATION AND WITHOUT – ARE DRAWN FROM THE SAME DISTRIBUTION t(4:11) MLL ALL without t(4:11) USE STANDARD STATISTICAL TESTS, FOR ONE GENE AT A TIME TO CALCULATE P g = probability that the 27 expression levels of gene g (14 from MLL, 13 ALL) are taken from the same distribution BEWARE OF MULTIPLE COMPARISONS!!!

29. after ttest 0.05 order by diffmeans 929 genes with p < 0.05 Bonferroni – 30 pass at 0.05 This list of 30 genes is “error free” with prob. 0.95 BEWARE OF MULTIPLE COMPARISONS!!! Bonferroni: reject H 0 only for genes with p < 0.05 / N p < 0.05/3250 = 1.5 10 -5

30. genes ordered by p-value 929 genes with p < 0.05 ordered by difference of means (ALL –MLL)

31. 143 WITH P-value<0.05 929 genes with p < 0.05 RANDOM DATA

32. sorted p I=820 Q=0.15

33. how many out of 929 are false? FDR: 929*0.17=158 false separating genes

34. how many genes at FDR=0.05? 297*0.05=15 false separating genes

35. 15 out of 297 - false 15 - false

36. random data

37. 100separating (p<0.001), 1900 random

38. MLL translocations specify a distinct gene expression profile that distinguishes a unique leukemia Armstrong et al, Nature Genetics 30, 41-47 (2002) PROBLEM: HIDDEN, CONFOUNDING VARIABLES( (FACTORS)

39. Hematopoiesis: HSCs can be categorized into long-term self-renewing HSCs, short-term self-renewing HSCs and multipotent progenitors (red arrows indicate self renewal). HSC give rise to common lymphoid progenitors (CLPs) and to common myeloid progenitors (CMPs). CMPs mature into red blood cells, megakaryocyte (cells producing platelets), granulocytes, dendritic cells, and macrophages. The CLP differentiate into B and T cell lymphocytes, natural killer cells and dendritic cells. (adapted from Reya et al., 2001) HEMATOPOIESIS: Differentiation from STEM CELLS to mature cells LEUKEMIA Mature cells: finite lifetime, finite number of divisions fixed class (B stays B etc) STEM CELLS: immortal, (unlimited number of divisions) multipotent (differentiation into many targets) B T NK Red differentiation LEUKEMIA : MALIGNANCY, INDUCED BY MUTATIONS, TRANSLOCATION,.. ACUTE MYELOID LEUKEMIA (AML) ACUTE LYMPHOID LEUKEMIA (ALL)

40. LEUKEMIA : normal differentiation MALIGNANCY INDUCED BY MUTATION OR TRANSLOCATION UNCONTROLED PROLIFERATION OVERCROWDING & DEATH OF NORMAL CELLS PRO B-cell ALL DIFFERENTIATION ARREST OF DIFFERENTIATION PRE B-cell ALL PRE T-cell ALL EARLY LATE

41. TRANSLOCATION TRANSLOCATION: DURING DNA REPLICATION TWO STRANDS, FROM TWO DIFFERENT CHROMOSOMES, CROSS MLL – GENE ON BAND 23 OF CHROMOSOME 11 t(4,11)MLL fusion protein AF4 partner gene AF4 – GENE ON BAND 21 OF CHROMOSOME 4 MLL TRANSLOCATIONS ARE IMPLICATED IN 10% OF ALL AND IN UP TO 80% OF INFANT LEUKEMIA

43. AIM: FIND THE GENES THAT ARE EITHER ACTIVATED OR SUPPRESSED BY THE t(4,11) CHIMERIC MLL PROTEIN THESE GENES MAY BE THE CAUSE OF CANCER AND TARGETS FOR THERAPY Rozovskaia, Ravid, Getz,....., Canaani: PNAS (2003)

44. EXPRESSION DATA: 27 ALL samples, 14 with MLL TRANS. t(4,11) 13 without TRANS. 3060 GENES PASSED FILTER CANAANI USE STANDARD STATISTICAL TEST TO LOOK FOR GENES THAT SEPARATE t(4,11) MLL FROM ALL: PROBLEM OF MULTIPLE COMPARISONS SOLVED BY CONTROLLING THE FALSE DISCOVERY RATE (FDR) t(4:11) MLL ALL without t(4:11) There is another factor (differentiation) that separates these two groups of samples! Which of the 230 genes responds to the MLL translocation? 230 genes differentially expressed between ALL with t(4:11) and ALL without, at FDR = 0.05. t(4:11) MLL ALL without t(4:11) ??

45. EXPRESSION DATA: 27 ALL samples, 14 with MLL TRANS. t(4,11) 13 without TRANS. 3060 GENES PASSED FILTER CANAANI USE STANDARD STATISTICAL TEST TO LOOK FOR GENES THAT SEPARATE t(4,11) MLL FROM ALL: PROBLEM OF MULTIPLE COMPARISONS SOLVED BY CONTROLLING THE FALSE DISCOVERY RATE (FDR) t(4:11) MLL ALL without t(4:11) 3000 2000 1000 230 genes differentially expressed between ALL with t(4:11) and ALL without, at FDR = 0.05. t(4:11) MLL ALL without t(4:11) 100 200

46. two t(4,11) samples seem different – closer to ALL without translocations Mistaken diagnosis? NO!! the t(4,11) samples are known to be EARLY DIFFERENTIATING. Perhaps some of the “separating genes” are not directly sensitive to the presence of the MLL abnormality, but to other characteristics of these cells (such as EARLY DIFFERENTIATION)? Identify the MLL sensitive genes - HOW? USE the CD10- : early differentiating ALL without translocations ALL without t(4:11) 100 200 ? ALL t(4:11) MLL cd10-

47. MLL CD10- T, Pre-B ALL Late Differentiation CD10-Translocation Attribute Group No Yes MLL No Yes NoCD10- Yes No ALL Sensitive to Differentiation and/or CD10- Sensitive to Trasloc. and/or Differentiation MLL vs ALL FDR = 5% 448 genes MLL vs CD10- FDR = 12% 144 genes ALL vs CD10- FDR = 12% 167 genes Translocation sensitive 46 80 Differentiation sensitive Sensitive to Translocation and/or CD10- WHERE ARE THE MLLs THAT LOOK LIKE ALL?

48. AIM: FIND THE GENES THAT ARE EITHER ACTIVATED OR SUPPRESSED BY THE t(4,11) CHIMERIC MLL PROTEIN FINDING: WE IDENTIFIED 46 GENES THAT ARE ACTIVATED OR SUPPRESSED BY THE MLL ONCOGENE. TARGETS OF NEXT STAGE EXPERIMENTS ON MICE SPINOFF: SOME MLLS ARE LATE DIFFERENTIATORS Rozovskaia et al. PNAS 2003

49. separation E1E1 E2E2 ALL MLL E 1 -2E 2 = 0 = E 1 - 2E 2 < 0 = E 1 - 2E 2 > 0

50. projection 1 E1E1 E2E2 ALL MLL w +/- PROJECTIONS ON w – DO SEPARATE ALL FROM MLL

51. projection 2 E1E1 E2E2 ALL MLL +/- PROJECTIONS ON w’ – DO NOT SEPARATE ALL FROM MLL w’

52. projection 3 E1E1 E2E2 WELL SEPARATED CENTERS OF MASS - NO SEPARATION OF THE TWO CLOUDS

53. projection 4 E1E1 E2E2 WEAK SEPARATION OF CENTERS OF MASS – GOOD SEPARATION OF THE TWO CLOUDS

54. Fisher to perceptron E1E1 E2E2 ALL MLL OPTIMAL LINE TO PROJECT ON FISHER PERCEPTRON