Application of Class Discovery and Class Prediction Methods to Microarray Data

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Application of Class Discovery and Class
Prediction Methods to Microarray Data

• Kellie J. Archer, Ph.D.
• Assistant Professor
• Department of Biostatistics
• kjarcher_at_vcu.edu

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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.

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Diffuse Large B-Cell Lymphoma
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Clinically Distinct DLBCL Subgroups
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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)

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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/datase
  ts.cgi (near bottom of page)

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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.

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Distinguishing AML vs. ALL

• Class comparison
• Neighborhood analysis
• Class prediction
• Weighted voting

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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)

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Class Discovery Distinguishing AML vs. ALL

• For each gene, compute the log of the expression
  values. For a given gene g,

For ALL
Let
represent the mean log expression value
represent the stdev log expression value.
Let
For AML
represent the mean log expression value
Let
represent the stdev log expression value.
Let
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Class Discovery Distinguishing AML vs.
ALLIllustration usingALL AML example.xls
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Class Discovery Distinguishing AML vs. ALL

• 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) gt r or
• P(g,c) lt -r. r was chosen to be 0.3

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Aside

• This differs from Pearsons correlation and is
  therefore not confined to -1,1 interval

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Aside Illustration usingCorrelation.xls
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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.

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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

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Permutation based methods
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Permutation based methods

• Permutation algorithm for the bth permutation,
  b1,,B
• 1) Permute the n labels of the data matrix X
• 2) Compute relative class separation P(g1,c)b,,
  P(gp,c)b for each gene gi.
• The permutation distribution of the relative
  class separation P(g,c) for gene gi, i1,,p is
  given by the empirical distribution of
  P(g,c)j,1,, P(g,c)j,B.

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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.

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Class Prediction Distinguishing AML vs. ALL

• For set of informative genes, each expression
  value xi votes for either ALL or AML, depending
  on whether its expression value is closer to µALL
  or µAML
• Let µALL represent the mean expression value for
  ALL
• Let µAML 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

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Class Prediction Distinguishing AML vs. ALL

• wi is a weighting factor that reflects how well
  the gene is correlated with class distinction
  wivi is the weighted vote
• For each sample, the weighted votes for each
  class are summed to get VALL and VAML
• The sample is assigned to the class with the
  higher total, provided the Prediction Strength
  (PS) gt 0.3 where
• PS (Vwin Vlose)/ (Vwin Vlose)

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Class Prediction Distinguishing AML vs. ALL
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Class Prediction Distinguishing AML vs. ALL

• Checking model adequacy
• Cross-validation of training dataset
• Applied model to an independent dataset of 34
  samples

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Class Discovery

• Determine whether the samples can be divided
  based only on gene expression without regard to
  the class labels
• Self-organizing maps

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Hypothesis Testing

• The hypothesis that two means ?1 and ?2 are equal
  is called a null hypothesis, commonly abbreviated
  H0.
• This is typically written as H0 ?1 ?2
• Its antithesis is the alternative hypothesis, HA
  ?1 ? ?2

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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 H0 if any
  discrepancy from H0 could readily be due to
  chance (i.e., sampling error).
• Data judged to be incompatible with H0 are taken
  as evidence in favor of HA.

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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 H0 is true
  if the chance difference in the sample means
  should not exceed a couple of standard errors.

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T-test

• In testing H0 ?1 ?2 against HA ?1 ? ?2 note
  that we could have restated the null hypothesis
  as
  
• H0 ?1 - ?2 0 and HA ?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)
  

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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)

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T-test Probe set 208680_at
Sample number ALL AML
1 2013.7 1974.6
2 2141.9 2027.6
3 2040.2 1914.8
4 1973.3 1955.8
5 2162.2 1963.0
6 1994.8 2025.5
7 1913.3 1865.1
8 2068.7 1922.4
2038.5 1956.1
s2 7051.284 3062.991
n 8 8
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T-test Probe set 208680_at
P0.039