1. Distinguishing active from non active genes: Main principle: DNA hybridization -DNA hybridizes due to base pairing using H-bonds -A/T and C/G and A/U possible RNA: AUGCAUGCUGCUAGCUACGUAUGCAUGCUGCUAGCUACGU cDNA: TACGTACGACGATCGATGCATACGTACGACGATCGATGCA Probe: GCTACGTATGCAT Mix probe with cDNA: probe will find complementary DNA sequence and bind to it. TACGTACGACGATCGATGCATACGTACGACGATCGATGCA GCTACGTATGCAT

2. Expression microarray:

3. Statistical analysis of Microarrays: An Introduction Why do replication of arrays? controltreatment

4. Biological Replication

5. Technical Replication RNA mixed probe pool

6. Dye Swap Design What type of replication ?

7. Background subtraction

8. Transformation using logarithmic values Assume red and green signal are the same: log 2 (1/1) => 0 (by definition) Assume red signal is twice of green signal: log 2 (2/1) => log 2 (2) =1(b/c 2 1 =2) Assume red signal is half of green signal: log 2 (1/2) => log 2 (1) - log 2 (2) =-1(= 0-1 => -1)

9. Using logarithmic values 1 1 2 0.5 Normal scale Logarithmic scale log 2 (2) =1 log 2 (0.5) =-1 unequal arrow distances equal arrow distances, same absolute values for the same-fold up or down regulation

10. Graphing all array values: the MA plot M: the greater distance from 0= the greater the R/G ratio A: the greater the distance from 0 the darker the spot on the microarray (redder or greener).

11. Using logarithmic values Two values used in Microarray analysis: M= ratio of red value/green value A= overall spot intensity

12. The Dye-swap Why? To account for dye bias (Cy5, the red dye fluoresces brighter than Cy3, the green dye. This is unfortunate but impossible to change due to differences in chemical structures of the two dyes).

13. Normalizing Why? A mathematical way to account for the systematic error due to dye intensity differences. Example: Gene X is 2-fold up-regulated by drought stress R/G :2.0 for gene X (drought/normal) G/R :should be 2.0 as well after swapping the dyes and RNA samples, but let’s say it is 1.9 for gene X (drought/normal).

14. Normalizing, cont’d Remember : Bottom line: M i is the average of 2 dye-swap array slides for each spot

15. How do you analyze replicated results? Mean Median Stand Dev (average) (value in middle) (spread around average) X= each data point, x (bar) = average, I= # of data points

16. Is a gene differentially expressed? In other words: Is the R/G ratio = 0 or not? The test statistic _ x = average of n samples s = SD

17. Example: Six observations of the same gene: average = -1.15 SD= 1.28 N=6 Look up p-value for the calculated t-statistic. Here: 9.21% are in the red shaded area.  p= 0.09  Accept null hypothesis: Treatment and control are NOT different, M = 0 Null hypothesis: treatment and control show equal gene expression (M=0) (see next slide, too)

18. The null hypothesis

19. Bonferroni Correction Assume you do a stats test for more than one gene: Each time you accept  = 0.05 (5%) uncertainty. That means you accept false positives 5% of the time for each gene. If you accept the same error for two genes it is 1 - (1- 0.05) 2 = 0.1 (10% uncertainty). You accept that out of the 2 genes in 10% of cases one is a false positive.. For an array with n= 1000 genes, this means: 1 - (1- 0.05) 1000 = 0.999 This means in 99.99% you WILL make an error in at least one gene. Assume 1000 genes and desired Bonferroni correction of 10%: Use only those genes with a p value = 0.10/1000 = 0.0001

20. False Discovery Rate (FDR) Correction Here: 1. 0.05 * (1/6)=0.008 --> under 0.05? YES, significant 2. 0.05* (2/6)=0.016 --> under 0.05? NO, not significant Why use FDR? Can use instead of Bonferroni. How? Sort all p-values low to high. Decide on your desired FDR rate (e.g 5%) Rank the genes (here: 1-6) Calculate 0.05 * (i/N) i= rank (here 1-6) N= total number of genes (here 6) If the p-value is < than 0.05*(i/N) then it is a significant gene.