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The second-simplest cDNA microarray data analysis problem

The second-simplest cDNA microarray data analysis problem

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The second-simplest cDNA microarray data analysis problem

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  1. The second-simplest cDNA microarray data analysis problem Terry Speed, UC Berkeley Fred Hutchinson Cancer Research Center March 9, 2001

  2. excitation scanning cDNA clones (probes) laser 2 laser 1 emission PCR product amplification purification printing mRNA target) overlay images and normalise 0.1nl/spot Hybridise target to microarray microarray analysis

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

  4. 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

  5. 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

  6. An M vs. A plot M = log2(R / G) A = log2(R*G) / 2

  7. Background matters From Spot From GenePix

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

  9. An experiment having within-slide replicates

  10. Background makes a difference Background method Segmentation method Exp1 Exp2 S.nbg 6 6 Gp.nbg 7 6 SA.nbg 6 6 No background QA.fix.nbg 7 6 QA.hist.nbg 7 6 QA.adp.nbg 14 14 S.valley 17 21 GP 11 11 Local surrounding SA 12 14 QA.fix 18 23 QA.hist 9 8 QA.adp 27 26 Others S.morph 9 9 S.const 14 14 Medians of the SD of log2(R/G) for 8 replicated spots multiplied by 100 and rounded to the nearest integer.

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

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

  13. Ngai lab, UCB

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

  15. From the Ernest Gallo Clinic & Research Center

  16. From Peter McCallum Cancer Research Institute, Australia

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

  18. M vs A after print-tip normalisation

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

  20. Taking scale into account Assumption: All print-tip-groups have the same spread in M True log ratio is mij where i represents different print-tip-groups and j represents different spots. Observed is Mij, where Mij = aimij Robust estimate of ai is MADi = medianj { |yij - median(yij) | }

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

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

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

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

  25. 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

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

  27. Lowess Normalized M Apo A1 Experiments

  28. Lowess Normalized M Srb1 Experiments

  29. Tiago’s Experiments: mutant fly vs. WT

  30. The “NCI 60” experiments (no bg)

  31. Taking scale into account Assumption: All slides have the same spread in M True log ratio is mij where i represents different slides and j represents different spots. Observed is Mij, where Mij = aimij Robust estimate of ai is MADi = medianj { |yij - median(yij) | }

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

  33. Which genes are (relatively) up/down regulated? Two samples with a reference (e.g. pooled control) n T C* n C* C • 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)

  34. 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. One factor: more than 2 samples T2 T3 T4 T1 x 2 x 2 x 2 x 2 C

  35. 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. One factor: more than 2 samples T6 T1 T5 T2 T4 T3

  36. 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. ctl OSM  4 times OSM & EGF EGF

  37. Which genes have changed?When permutation testing possible 1. For each gene and each hybridisation (8 ko + 8 ctl), use M=log2(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.

  38. Histogram & qq plot ApoA1

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

  40. 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.

  41. Results from 4 replicates

  42. B=LOR compared

  43. M • t • t M Results from the Apo AI ko experiment

  44. M • t • t M Results from the Apo AI ko experiment

  45. EmpiricalBayes log posterior odds ratio

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

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

  48. 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

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

  50. 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