The second-simplest cDNA The second-simplest cDNA microarray data analysis problemmicroarray data analysis problem

Terry Speed, UC Berkeley

Bioinformatic Strategies For Application of Genomic Tools to Environmental Health

Research, March 5, 2001

NIEHS National Center for Toxicogenomics NCSU Bioinformatics Research Center

Biological questionDifferentially expressed genesSample class prediction etc.

Testing

Biological verification and interpretation

Microarray experiment

Estimation

Experimental design

Image analysis

Normalization

Clustering Discrimination

R, G

16-bit TIFF files

(Rfg, Rbg), (Gfg, Gbg)

Some motherhood statementsSome motherhood statementsImportant 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

The simplest cDNA microarray The simplest cDNA microarray data analysis problem is data analysis problem is identifying differentially identifying differentially

expressed genes using one slideexpressed 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

An M vs. A plotAn M vs. A plotM = log2(R / G)A = log2(R*G) / 2

Background mattersBackground matters

From Spot From GenePix

From the NCI60 data set (Stanford web site)

No background correction With background correction

An experiment having within-slide replicates

Background makes a differenceBackground makes a difference

Background method Segmentation method Exp1 Exp2S.nbg 6 6Gp.nbg 7 6SA.nbg 6 6

No background QA.fix.nbg 7 6QA.hist.nbg 7 6QA.adp.nbg 14 14S.valley 17 21GP 11 11

Local surrounding SA 12 14QA.fix 18 23QA.hist 9 8QA.adp 27 26

Others S.morph 9 9S.const 14 14

Medians of the SD of log2(R/G) for 8 replicated spots multiplied by 100and rounded to the nearest integer.

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

From the NCI60 data set (Stanford web site)

Ngai lab, UCB

Tiago’s data from the Goodman lab, UCB

From the Ernest Gallo Clinic & Research Center

From Peter McCallum Cancer Research Institute, Australia

Normalisation - print tipNormalisation - print tipAssumption: For every print group, changes roughly symmetric at all intensities.

M vs A after print-tip normalisationM vs A after print-tip normalisation

Normalization (ctd) Another data setNormalization (ctd) Another data set

• After within slide global lowess normalization.• Likely to be a spatial effect.

Print-tip groups

Lo

g-r

ati o

s

Assumption:

All print-tip-groups have the same spread in M

True log ratio is ij where i represents different print-tip-groups and j represents different spots.

Observed is Mij, where

Mij = ai ij

Robust estimate of ai is

MADi = medianj { |yij - median(yij) | }

Taking scale into accountTaking scale into account

MADi

MADii=1

I∏I

Normalization (ctd) That same data setNormalization (ctd) That same data set

• After print-tip location and scale normalization.• Incorporate quality measures.

Lo

g-r

ati o

s

Print-tip groups

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

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

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10000

100000

10 100 1000 10000 100000

Genomic DNA vs. Genomic DNA

The approach of Roberts et al (Rosetta)

X =a1 −a2

(σ12 +σ2

2 ) + f2 (a12 +a2

2 )

P=2(1−Erf(|X |))

Data from Bing Ren

The second simplest cDNA microarray The second simplest cDNA microarray data analysis problem is identifying data analysis problem is identifying differentially expressed genes using differentially expressed genes using

replicated slidesreplicated 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

Normalization (ctd) Yet another data set

• Between slides this time (10 here)

• Only small differences in spread apparent

• We often see much greater differences

Slides

Lo

g-r

ati o

s

The “NCI 60” experiments (no bg)

Assumption: All slides have the same spread in M

True log ratio is ij where i represents different slides and j represents different spots.

Observed is Mij, where

Mij = ai ij

Robust estimate of ai is

MADi = medianj { |yij - median(yij) | }

Taking scale into accountTaking scale into account

MADi

MADii=1

I∏I

Which genes are (relatively) up/down Which genes are (relatively) up/down regulated?regulated?

Two samples.

e.g. KO vs. WT or mutant vs. WT

T C n

For each gene form the t statistic: average of n trt Ms

sqrt(1/n (SD of n trt Ms)2)

n

Which genes are (relatively) up/down Which genes are (relatively) up/down regulated?regulated?

Two samples with a reference (e.g. pooled control)

T C* n

• 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)

C C* n

One factor: more than 2 samplesOne factor: more than 2 samples

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.

T1

C

T2 T3 T4

x 2x 2x 2 x 2

One factor: more than 2 samplesOne factor: more than 2 samples

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.

T3 T4

T2

T6T1

T5

Two or more factorsTwo 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

EGF OSM & EGF

4 times

Which genes have changed?Which genes have changed?When permutation testing possibleWhen 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 Mssqrt(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.

Histogram & qq plotHistogram & qq plot

ApoA1

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

Which genes have changed?Which genes have changed?Permutation testing not possiblePermutation 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.

Results from 4 replicatesResults from 4 replicates

B=LOR comparedB=LOR compared

•M •t•t M

Results from the Apo AI ko experiment

•M •t•t M

Results from the Apo AI ko experiment

B=const+log

2an

+s2 +M•2

2an

+s2 +M•

2

1+nc

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

Empirical Bayes log posterior odds ratio

•M •B•t•M B•t B•t M B

Results from SR-BI transgenic experiment

•M •B•t•M B•t B•t M B

Results from SR-BI transgenic experiment

Extensions include dealing withExtensions 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

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10 100 1000 10000 100000 1000000

Galactose

PCL10GAL80

GAL1/10

GAL2

GAL3

GAL7

GCY1

MTH1

WCE-DNA (Cy3)

IP-DNA (Cy5)

Un

-en

rich

ed D

NA

(C

y3)

antibody-enriched DNA (Cy5)

Rosetta once more: In vivo Binding Sites of Gal4p in Galactose

P <0.001

Summary (for the second simplest problem)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

AcknowledgmentsAcknowledgments

UCB/WEHIUCB/WEHI

Yee Hwa YangYee Hwa Yang

Sandrine DudoitSandrine Dudoit

Ingrid Lönnstedt

Natalie Thorne Natalie Thorne

David FreedmanDavid Freedman

CSIRO Image Analysis Group

Michael BuckleyMichael Buckley

Ryan Lagerstorm

Ngai lab, UCB

Goodman lab, UCB

Peter Mac CI, Melb.

Ernest Gallo CRC

Brown-Botstein lab

Matt Callow (LBNL)

Bing Ren (WI)Bing Ren (WI)

Some web sites:

Technical reports, talks, software etc.

http://www.stat.berkeley.edu/users/terry/zarray/Html/

Statistical software R “GNU’s S” http://lib.stat.cmu.edu/R/CRAN/

Packages within R environment:

-- Spot http://www.cmis.csiro.au/iap/spot.htm

-- SMA (statistics for microarray analysis) http://www.stat.berkeley.edu/users/terry/zarray/Software /smacode.html

Factorial DesignFactorial Design

Zone Effect

A1P01

P04 A 4

1

2

3

4

5

Age Effect

Different ways of estimating parameters.

e.g. Z effect.

1 = ( + z) - ()

= z

2 - 5 = (( + a) - ()) -(( + a)-( + z))

= (a) - (a + z)

= z

4 + 3 - 5 =…= z

Factorial designFactorial design

a

z z+a+za

A1P01

P04 A 4

1

2

3

4

5

How do we combine the information?

Regression analysis

Define a matrix X so that E(M)=X

Use least squares estimate for z, a, za

E

m1

m2

m3

m4

m5

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

=

1 0 0

0 1 0

−1 0 −1

1 1 1

−1 1 0

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

•

z

a

za

⎛

⎝

⎜ ⎜

⎞

⎠ ⎟ ⎟

ˆ = X' X( )−1X'M

Looking at effect of Z: log(zone 4 / zone1)

gene A

gene B

EstimateEstimate

Log2(SE)

Z e

ffec

t

•t = / SE t

ZoneAgeZone Age

Age

48

229

Zone . Age interaction

Zone

19

Top 50 genesfrom each effect

0

0

19

•T •B•t M B• t B

•M •t•t M

•M •B•t•M B•t B•t MB

Some statistical questionsSome statistical questions

Image analysis: addressing, segmenting, quantifying

Normalisation: within and between slides

Quality: of images, of spots, of (log) ratios

Which genes are (relatively) up/down regulated?

Assigning p-values to tests/confidence to results.

Some statistical questions, ctdSome statistical questions, ctd

Planning of experiments: design, sample size

Discrimination and allocation of samples

Clustering, classification: of samples, of genes

Selection of genes relevant to any given analysis

Analysis of time course, factorial and other special experiments…………………………...& much more

The “NCI 60” experiments (bg)