Download presentation

Presentation is loading. Please wait.

1
Detecting Differentially Expressed Genes Pengyu Hong 09/13/2005

2
Background (Microarray) Cells Extract RNA

3
Background Cells Extract RNA

4
Background Cells Extract RNA

5
Background Cells Extract RNA

6
Background Cells Extract RNA 10 4 + genes

7
Background Cells Extract RNA 10 4 + genes

8
Background Cells Extract RNA 10 4 + genes

9
Background Biological sample RNA extraction (total RNA or mRNA) Amplification (in vitro transcription) Label samples Hybridization Washing and staining Scanning Microarrays are highly noisy Use replicated experiments to make inferences about differential expression for the population from which the biological samples originate biological variability technical variability

10
Background Normalization Calculate Gene Expression Index

11
An Example 5 normal sample and 9 myeloma (MM) samples 12558 genes (rows)

12
Genes of Interest Statistical significance: that the observed differential expression is unlikely to be due to chance. Scientific significance: that the observed level of differential expression is of sufficient magnitude to be of biological relevance.

13
Group 1 (N samples): X 1, X 2, … X N Group 2 (M samples): Y 1, Y 2, … Y M Statistical significance in the two group problem Assume Y j ~ Normal (μ 2, σ 2 ) X i ~ Normal (μ 1, σ 2 ) Null hypothesis: Group 1 is the “same” to Group 2 (i.e., μ 1 = μ 2 ) Parametric Test: t-test

14
Statistical significance in the two group problem Y j ~ Normal (μ 2, σ 2 )X i ~ Normal (μ 1, σ 2 ) Null hypothesis: μ 1 = μ 2 Test null hypothesis with test statistics: Parametric Test: t-test

15
If variances are unequal (1) When N+M > 30, this is approximately normal (2) When 1 >> 2, this is approximately t(df = N–1) (3) In general, Welch approximation: t’ ~ t(df’), where Y j ~ Normal (μ 2, σ 2 2 ) X i ~ Normal (μ 1, σ 1 2 ) σ1 σ2σ1 σ2

16
Wilcoxon rank sum test Consider row 7 of MM study 16 253 633 1008 708 36 72 28 14 33 19 49 58 23 13 4 3 1 2 8 5 10 14 9 12 7 6 11 --------------------------- rank sum = 23 This test is more appropriate than the t-tests when the underlying distribution is far from normal. (But it requires large group sizes)

17
P-value p-value = P(|T|>|t|) is calculated based on the distribution of T under the null hypothesis. p-value is a function of the test statistics and can be viewed as a random variable. –e.g. p-value = 2(1 - F(|t * |), F = cdf of t(N+M – 2). A small p-value represents evidence against the null hypothesis differentially expressed in our case.

18
Permutation test A non-parametric way of computation p-value for any test statistics. –In the MM-study, each gene has (14 choose 5) = 2002 different test values obtainable from permuting the group labels. Under the null hypothesis that the distribution for the two groups are identical, all these test values are equally probable. What is the probability of getting a test value at least as extreme as the observed one? This is the permutation p-value.

19
Permutation technique Condition 0Condition 1 Patient 4Patient 2Patient 3Patient 1Patient 5Patient 6 Condition 0Condition 1 Patient 1Patient 2Patient 5Patient 4Patient 3Patient 6 Condition 0Condition 1 Patient 1Patient 6Patient 3Patient 4Patient 5Patient 2 Condition 0Condition 1 Patient 1Patient 2Patient 3Patient 4Patient 5Patient 6 Compute TS 0 Compute TS 1 Compute TS 2 Compute TS 3 The set of TS i form the empirical distribution of the test statistic TS

20
Scientific Significance Fold change FC = May not be high when statistical significance is high. Not an appropriate measure if the dispersion is not taken into consideration.

21
Conservative fold change Conservative fold change (CFC) = Max (25 th percentile of sample 1 / 75 th percentile of sample 2, 25 th percentile of sample 2 / 75 th percentile of sample 1)

22
Sample 1: Normal (100, 1) Sample 2: Normal (103, 1) CFC = 1.0164

23
CFC=3.53 CFC=1.07 CFC=2.89 CFC=1.45

24
P-values and FC contains different information

25
Gene Selection and Ranking A high threshold of statistical significance Select genes with p-values smaller than a threshold The selected genes are ordered according to their scientific significance (i.e. ranked by fold-changes)

26
The False Positive Rate (FPR) If we select genes with p-value < 0.01, then the probability of making a positive call when the gene is in fact not differential is less than 0.01. Thus selection by p-value controls the FPR. However, if we have 12,000 genes in a microarray, then a FPR = 0.01 still allows up to 120 false positives. To make sensible decision, we must take multiple comparisons into consideration.

27
Dealing with Multiple Comparison Bonferroni inequality: To control the family-wise error rate for testing m hypotheses at level α, we need to control the FPR for each individual test at α/m Then P(false rejection at least one hypothesis) < α or P(no false rejection) > 1- α This is appropriate for some applications (e.g. testing a new drug versus several existing ones), but is too conservative for our task of gene selection.

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google