1. The second-simplest cDNA microarray data analysis problem Terry Speed, UC Berkeley Bioinformatic Strategies For Application of Genomic Tools to Environmental Health Research, March 5, 2001 NIEHS National Center for Toxicogenomics NCSU Bioinformatics Research Center

2. Biological question Differentially expressed genes Sample class prediction etc. Testing Biological verification and interpretation Microarray experiment Estimation Experimental design Image analysis Normalization Clustering Discrimination R, G 16-bit TIFF files (Rfg, Rbg), (Gfg, Gbg)

3. Some motherhood statements Important aspects of a statistical analysis include: Tentatively separating systematic from random sources of variation Removing the former and quantifying the latter, when the system is in control Identifying and dealing with the most relevant source of variation in subsequent analyses Only if this is done can we hope to make more or less valid probability statements

4. The simplest cDNA microarray data analysis problem is identifying differentially expressed genes using one slide This is a common enough hope Efforts are frequently successful It is not hard to do by eye The problem is probably beyond formal statistical inference (valid p-values, etc) for the foreseeable future, and here’s why

5. An M vs. A plot M = log 2 (R / G) A = log 2 (R*G) / 2

6. Background matters From Spot From GenePix

7. From the NCI60 data set (Stanford web site) No background correction With background correction

8. An experiment having within-slide replicates

9. Background makes a difference Background methodSegmentation methodExp1 Exp2 S.nbg66 Gp.nbg76 SA.nbg66 No backgroundQA.fix.nbg76 QA.hist.nbg76 QA.adp.nbg1414 S.valley1721 GP1111 Local surroundingSA1214 QA.fix1823 QA.hist98 QA.adp2726 OthersS.morph99 S.const1414 Medians of the SD of log 2 (R/G) for 8 replicated spots multiplied by 100 and rounded to the nearest integer.

10. Normalisation - lowess Global lowess (Matt Callow’s data, LNBL) Assumption: changes roughly symmetric at all intensities.

11. From the NCI60 data set (Stanford web site)

12. Ngai lab, UCB

13. Tiago’s data from the Goodman lab, UCB

14. From the Ernest Gallo Clinic & Research Center

15. From Peter McCallum Cancer Research Institute, Australia

16. Normalisation - print tip Assumption: For every print group, changes roughly symmetric at all intensities.

17. M vs A after print-tip normalisation

18. Normalization (ctd) Another data set After within slide global lowess normalization. Likely to be a spatial effect. Print-tip groups Log-ratios

19. Assumption: All print-tip-groups have the same spread in M True log ratio is  ij where i represents different print-tip-groups and j represents different spots. Observed is M ij, where M ij = a i  ij Robust estimate of a i is MAD i = median j { |y ij - median(y ij ) | } Taking scale into account

20. Normalization (ctd) That same data set Normalization (ctd) That same data set After print-tip location and scale normalization. Incorporate quality measures. Log-ratios Print-tip groups

21. Matt Callow’s Srb1 dataset (#5). Newton’s and Chen’s single slide method

22. Matt Callow’s Srb1 dataset (#8). Newton’s, Sapir & Churchill’s and Chen’s single slide method

23. Genomic DNA vs. Genomic DNA The approach of Roberts et al (Rosetta) Data from Bing Ren

24. The second simplest cDNA microarray data analysis problem is identifying differentially expressed genes using replicated slides There are a number of different aspects: First, between-slide normalization; then What should we look at: averages, SDs t- statistics, other summaries? How should we look at them? Can we make valid probability statements? A report on work in progress

25. Normalization (ctd) Yet another data set Between slides this time (10 here) Only small differences in spread apparent We often see much greater differences Slides Log-ratios

26. The “NCI 60” experiments (no bg)

27. Assumption: All slides have the same spread in M True log ratio is  ij where i represents different slides and j represents different spots. Observed is M ij, where M ij = a i  ij Robust estimate of a i is MAD i = median j { |y ij - median(y ij ) | } Taking scale into account

28. Which genes are (relatively) up/down regulated? Two samples. e.g. KO vs. WT or mutant vs. WT TC  n n For each gene form the t statistic: average of n trt Ms sqrt(1/n (SD of n trt Ms) 2 )  n n

29. Which genes are (relatively) up/down regulated? Two samples with a reference (e.g. pooled control) TC*  n n For each gene form the t statistic: average of n trt Ms - average of n ctl Ms sqrt(1/n (SD of n trt Ms) 2 + (SD of n ctl Ms) 2 ) C C*  n n

30. One factor: more than 2 samples Samples: Liver tissue from mice treated by cholesterol modifying drugs. Question 1: Find genes that respond differently between the treatment and the control. Question 2: Find genes that respond similarly across two or more treatments relative to control. T1 C T2T3T4 x 2

31. One factor: more than 2 samples Samples: tissues from different regions of the mouse olfactory bulb. Question 1: differences between different regions. Question 2: identify genes with a pre-specified patterns across regions. T3 T4 T2 T6 T1 T5

32. Two or more factors 6 different experiments at each time point. Dyeswaps. 4 time points (30 minutes, 1 hour, 4 hours, 24 hours) 2 x 2 x 4 factorial experiment. ctlOSM EGF OSM & EGF  4 times

33. Which genes have changed? When permutation testing possible 1. For each gene and each hybridisation (8 ko + 8 ctl), use M=log 2 (R/G). 2. For each gene form the t statistic: average of 8 ko Ms - average of 8 ctl Ms sqrt(1/8 (SD of 8 ko Ms) 2 + (SD of 8 ctl Ms) 2 ) 3. Form a histogram of 6,000 t values. 4. Do a normal Q-Q plot; look for values “off the line”. 5. Permutation testing. 6. Adjust for multiple testing.

34. Histogram & qq plot ApoA1

35. Apo A1: Adjusted and Unadjusted p-values for the 50 genes with the largest absolute t-statistics.

36. Which genes have changed? Permutation testing not possible Our current approach is to use averages, SDs, t-statistics and a new statistic we call B, inspired by empirical Bayes. We hope in due course to calibrate B and use that as our main tool. We begin with the motivation, using data from a study in which each slide was replicated four times.

37. Results from 4 replicates

38. B=LOR compared

39. M t t  M Results from the Apo AI ko experiment

40. M t t  M Results from the Apo AI ko experiment

41. Empirical Bayes log posterior odds ratio

42. M B t M  B t  B t  M  B Results from SR-BI transgenic experiment

43. M B t M  B t  B t  M  B Results from SR-BI transgenic experiment

44. Extensions include dealing with Replicates within and between slides Several effects: use a linear model ANOVA: are the effects equal? Time series: selecting genes for trends

45. Un-enriched DNA (Cy3) antibody-enriched DNA (Cy5) Rosetta once more: In vivo Binding Sites of Gal4p in Galactose P <0.001

46. Summary (for the second simplest problem) Microarray experiments typically have thousands of genes, but only few (1-10) replicates for each gene. Averages can be driven by outliers. Ts can be driven by tiny variances. B = LOR will, we hope –use information from all the genes –combine the best of M. and T –avoid the problems of M. and T

47. Acknowledgments UCB/WEHI Yee Hwa Yang Sandrine Dudoit Ingrid Lönnstedt Natalie Thorne David Freedman CSIRO Image Analysis Group Michael Buckley Ryan Lagerstorm Ngai lab, UCB Goodman lab, UCB Peter Mac CI, Melb. Ernest Gallo CRC Brown-Botstein lab Matt Callow (LBNL) Bing Ren (WI)

48. Some web sites: Technical reports, talks, software etc. http://www.stat.berkeley.edu/users/terry/zarray/Html/ Statistical software R “GNU’s S” http://lib.stat.cmu.edu/R/CRAN/ Packages within R environment: -- Spot http://www.cmis.csiro.au/iap/spot.htm -- SMA (statistics for microarray analysis) http://www.stat.berkeley.edu/users/terry/zarray/Software /smacode.html

49. Factorial Design Zone Effect A1P01 P04 A 4 1 2 3 4 5 Age Effect

50. Different ways of estimating parameters. e.g. Z effect. 1 = (  + z) - (  ) = z 2 - 5 = ((  + a) - (  )) -((  + a)-(  + z)) = (a) - (a + z) = z 4 + 3 - 5 =…= z Factorial design   a  z  z+a+za A1P01 P04A 4 1 2 3 4 5 How do we combine the information?

51. Regression analysis Define a matrix X so that E(M)=X  Use least squares estimate for z, a, za

52. Looking at effect of Z: log(zone 4 / zone1) gene A gene B

53. Estimate Log 2 (SE) Z effect  t =  / SE   t

54. Zone Age Zone  Age

55. Age 48 2 29 Zone. Age interaction Zone 19 Top 50 genes from each effect 0 0 19

56. T B t  M  B t  B

58. M t t  M

59. M B t M  B t  B t  M  B

60. Some statistical questions Image analysis: addressing, segmenting, quantifying Normalisation: within and between slides Quality: of images, of spots, of (log) ratios Which genes are (relatively) up/down regulated? Assigning p-values to tests/confidence to results.

61. Some statistical questions, ctd Planning of experiments: design, sample size Discrimination and allocation of samples Clustering, classification: of samples, of genes Selection of genes relevant to any given analysis Analysis of time course, factorial and other special experiments…………………………...& much more

62. The “NCI 60” experiments (bg)