1. Application of Class Discovery and Class Prediction Methods to Microarray Data Kellie J. Archer, Ph.D. Assistant Professor Department of Biostatistics kjarcher@vcu.edu

2. Basis of Cancer Diagnosis Pathologist makes an interpretation based upon a compendium of knowledge which may include –Morphological appearance of the tumor –Histochemistry –Immunophenotyping –Cytogenetic analysis –etc.

4. Clinically Distinct DLBCL Subgroups

5. Improved Cancer Diagnosis: Identify sub-classes Divide morphologically similar tumors into different groups based on response. Application of microarrays: Characterize molecular variations among tumors by monitoring gene expression Goal: microarrays will lead to more reliable tumor classification and sub-classification (therefore, more appropriate treatments will be administered resulting in improved outcomes)

6. Distinguishing two types of acute leukemia (AML vs. ALL) Golub, T.R. et al 1999. Molecular classification of cancer: class discovery and class prediction by gene expression monitoring. Science 286: 531-537. http://www-genome.wi.mit.edu/cgi-bin/cancer/datasets.cgi (near bottom of page)http://www-genome.wi.mit.edu/cgi-bin/cancer/datasets.cgi

7. Distinguishing AML vs. ALL 38 BM samples (27 childhood ALL, 11 adult AML) were hybridized to Affymetrix GeneChips –GeneChip included 6,817 human genes. –Affymetrix MAS 4.0 software was used to perform image analysis. –MAS 4.0 Average Difference expression summary method was applied to the probe level data to obtain probe set expression summaries. –Scaling factor was used to normalize the GeneChips. –Samples were required to meet quality control criteria.

8. Distinguishing AML vs. ALL Class comparison –Neighborhood analysis Class prediction –Weighted voting

9. Class Discovery: Distinguishing AML vs. ALL The mean of a random variable X is a measure of central location of the density of X. The variance of a random variable is a measure of spread or dispersion of the density of X. Var(X)=E[(X-  ) 2 ] =∑(X -  ) 2 /(n-1) Standard deviation = =  (X)

10. Class Discovery: Distinguishing AML vs. ALL For each gene, compute the log of the expression values. For a given gene g, Let represent the mean log expression value; represent the stdev log expression value. represent the mean log expression value; represent the stdev log expression value. For AML Let For ALL Let

11. Class Discovery: Distinguishing AML vs. ALL Illustration using ALL AML example.xls

12. For each gene, compute a relative class separation (quasi-correlation measure) as follows Define neighborhoods of radius r about classes 1 and 2 such that P(g,c) > r or P(g,c) < -r. r was chosen to be 0.3 Class Discovery: Distinguishing AML vs. ALL

13. Aside This differs from Pearson’s correlation and is therefore not confined to [-1,1] interval

14. Aside Illustration using Correlation.xls

15. Class Discovery: Distinguishing AML vs. ALL A permutation test was used to calculate whether the observed number of genes in a neighborhood was significantly higher than expected.

16. Permutation based methods Permutation based adjusted p-values –Under the complete null, the joint distribution of the test statistics can be estimated by permuting the columns of the gene expression matrix –Permuting entire columns creates a situation in which membership to the Class 1 and Class 2 groups is independent of gene expression but preserves the dependence structure between genes

17. Permutation based methods

18. Permutation algorithm for the b th permutation, b=1,…,B –1) Permute the n labels of the data matrix X –2) Compute relative class separation P(g 1,c) b,…, P(g p,c) b for each gene g i. The permutation distribution of the relative class separation P(g,c) for gene g i, i=1,…,p is given by the empirical distribution of P(g,c) j,1,…, P(g,c) j,B.

19. Distinguishing AML vs. ALL Class comparisons using neighborhood analysis revealed approximately 1,100 genes were correlated with class (AML or ALL) than would be expected by chance.

20. Class Prediction: Distinguishing AML vs. ALL For set of informative genes, each expression value x i votes for either ALL or AML, depending on whether its expression value is closer to μ ALL or μ AML –Let μ ALL represent the mean expression value for ALL –Let μ AML represent the mean expression value for AML Informative genes were the n/2 genes with the largest P(g,c) and the n/2 genes with the smallest P(g,c) Golub et al choose n = 50

21. Class Prediction: Distinguishing AML vs. ALL w i is a weighting factor that reflects how well the gene is correlated with class distinction; w i v i is the weighted vote For each sample, the weighted votes for each class are summed to get V ALL and V AML The sample is assigned to the class with the higher total, provided the Prediction Strength (PS) > 0.3 where PS = (V win – V lose )/ (V win + V lose )

22. Class Prediction: Distinguishing AML vs. ALL

23. Checking model adequacy –Cross-validation of training dataset –Applied model to an independent dataset of 34 samples

24. Class Discovery Determine whether the samples can be divided based only on gene expression without regard to the class labels –Self-organizing maps

25. Hypothesis Testing The hypothesis that two means  1 and  2 are equal is called a null hypothesis, commonly abbreviated H 0. This is typically written as H 0 :  1 =  2 Its antithesis is the alternative hypothesis, H A :  1   2

26. Hypothesis Testing A statistical test of hypothesis is a procedure for assessing the compatibility of the data with the null hypothesis. –The data are considered compatible with H 0 if any discrepancy from H 0 could readily be due to chance (i.e., sampling error). –Data judged to be incompatible with H 0 are taken as evidence in favor of H A.

27. Hypothesis Testing If the sample means calculated are identical, we would suspect the null hypothesis is true. Even if the null hypothesis is true, we do not really expect the sample means to be identically equal because of sampling variability. We would feel comfortable concluding H 0 is true if the chance difference in the sample means should not exceed a couple of standard errors.

28. T-test In testing H 0 :  1 =  2 against H A :  1   2 note that we could have restated the null hypothesis as H 0 :  1 -  2 = 0 and H A :  1 -  2  0 To carry out the t-test, the first step is to compute the test statistic and then compare the result to a t- distribution with the appropriate degrees of freedom (df)

29. T-test Data must be independent random samples from their respective populations Sample size should either be large or, in the case of small sample sizes, the population distributions must be approximately normally distributed. When assumptions are not met, non- parametric alternatives are available (Wilcoxon Rank Sum/Mann-Whitney Test)

30. T-test: Probe set 208680_at Sample numberALLAML 12013.71974.6 22141.92027.6 32040.21914.8 41973.31955.8 52162.21963.0 61994.82025.5 71913.31865.1 82068.71922.4 2038.51956.1 s2s2 7051.2843062.991 n88

31. T-test: Probe set 208680_at P=0.039