Introduction to the design of cDNA Introduction to the design of cDNA microarray experimentsmicroarray experiments

Statistics 246, Spring 2002

Week 9, Lecture 1

Yee Hwa Yang

Some aspects of designSome aspects of design

Layout of the array– Which cDNA sequence to print?

• Library • Controls

– Spatial position

Allocation of samples to the slides – Different design layout

• A vs B : Treatment vs control• Multiple treatments• Time series• Factorial

– Replication• number of hybridizations• use of dye swap in replication• Different types replicates (e.g pooled vs unpooled material (samples))

– Other considerations• Physical limitations: the number of slides and the amount of material• Extensibility - linking

Issues that affect design of array Issues that affect design of array experimentsexperiments

Scientific• Aim of the experiment

Specific questions and priorities between them.How will the experiments answer the questions posed?

Practical (Logistic)• Types of mRNA samples:

reference, control, treatment 1, etc.• Amount of material.

Count the amount of mRNA involved in one channel of a hybridization as one unit.

• Number of slides available for experiment.Other Information• The experimental process prior to hybridization:

sample isolation, mRNA extraction, amplification , labelling.• Controls planned:

positive, negative, ratio, etc.• Verification method:

Northern, RT-PCR, in situ hybridization, etc.

Graphical representationGraphical representation

Natural design choiceNatural design choice

Case 1: Meaningful biological control (C)

Samples: Liver tissue from four mice treated by cholesterol modifying drugs.

Question 1: Genes that respond differently between the T and the C.

Question 2: Genes that responded similarly across two or more treatments relative to control.

Case 2: Use of universal reference.

Samples: Different tumor samples.

Question: To discover tumor subtypes.

C

T1 T2 T3 T4 T1

Ref

T2 Tn-1 Tn

Direct vs IndirectDirect vs Indirect

Two samples

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

T CT

CRef

Direct Indirect

2 /2 22

average (log (T/C)) log (T / Ref) – log (C / Ref )

I) Common Reference

II) Common reference

III ) Direct comparison

Number of Slides N = 3 N=6 N=3

Ave. variance 2 0.67

Units of material A = B = C = 1 A = B = C = 2 A = B = C = 2

Ave. variance 1 0.67

One-way layout: one factor, k levelsOne-way layout: one factor, k levels

C B

A

O

CBA

O

CBA

All pair-wise comparisons are of equal importance

Dye-swapDye-swap

C B

A

Design B1

C B

A

Design B2

- Design B1 and B2 have the same average variance- The direction of arrows potentially affects the bias of the estimate but not the variance-For k = 3, efficiency ratio (Design A1 / Design B) = 3-In general, efficiency ratio = (2k) / (k-1)

Design: how we sliced up the bulbDesign: how we sliced up the bulb

A

P D

V

M

L

Multiple direct comparisons between different samples (no common reference)

Different ways of estimating the same contrast:

e.g. A compared to P

Direct = log(A/P)

Indirect = log(A/M) + log((M/P) or

log(A/D) + log(D/P) or

log(A/L) – log((P/L)

How do we combine these?

LL

PPVV

DD

MM

AA

Linear model analysis

Define a matrix X so that E(Y)=Xba = log(A), p=log(P), d=log(D), v=log(V), m=log(M), l=log(L)

lm

lv

ld

lp

la

y

y

y

E

n 00011

00001

11000

2

1

MXXX ˆ 1

Pooled reference

T2 T4 T5 T6 T7T3T1

Ref

Compare to T1

t vs t+3t vs t+2t vs t+1

Time SeriesTime Series

Possible designs:1) All sample vs common pooled reference2) All sample vs time 0 3) Direct hybridization between times.

Design choices in time series t vs t+1 t vs t+2

T1T2 T2T3 T3T4 T1T3 T2T4 T1T4 Ave

N=3 A) T1 as common reference 1 2 2 1 2 1 1.5

B) Direct Hybridization 1 1 1 2 2 3 1.67

N=4 C) Common reference 2 2 2 2 2 2 2

D) T1 as common ref + more .67 .67 1.67 .67 1.67 1 1.06

E) Direct hybridization choice 1 .75 .75 .75 1 1 .75 .83

F) Direct Hybridization choice 2 1 .75 1 .75 .75 .75 .83

T2 T3 T4T1

T2 T3 T4T1

Ref

T2 T3 T4T1

T2 T3 T4T1

T2 T3 T4T1

T2 T3 T4T1

2 by 2 factorial – 2 by 2 factorial – two factors, each with two levelstwo factors, each with two levels

Example 1: Suppose we wish to study the joint effect of two drugs, A and B.

4 possible treatment combinations:– C: No treatment– A: drug A only.– B: drug B only.– A.B: both drug A and B.

Example 2: Our interest in comparing two strain of mice (mutant and wild-type) at two different times, postnatal and adult.

4 possible samples:– C: WT at postnatal– A: WT at adult (effect of time only)– B: MT at postnatal (effect of the mutation only)– A.B : MT at adult (effect of both time and the mutation).

Different ways of estimating parameters.

e.g. B effect.

1 = ( + b) - ()

= b

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

= (a) - (a + b)

= b

Factorial designFactorial design

a

b a+b+ab

AC

B AB

1

2

3

4

5

6

Factorial designFactorial design

4

1

2

1

2

1

4

1

2

1

2

1

4

1

2

1

2

1

a

b a+b+ab

AC

B AB

1

2

3

4

5

6

4

1

2

1

2

1

ab

abb

aba

2

12

1

Indirect A balance of direct and indirect

I) II) III) IV)

# Slides N = 6

Main effect A

0.5 0.67 0.5 NA

Main effect B

0.5 0.43 0.5 0.3

Interaction A.B

1.5 0.67 1 0.67

2 x 2 factorial2 x 2 factorial

C

A.BBA

B

C

A.B

A

B

C

A.B

A

B

C

A.B

A

Table entry: variance

Linear model analysis

Define a matrix X so that E(Y)=Xb

Use least squares estimate for a, b, ab

ab

b

a

y

y

y

y

y

y

E

101

011

111

110

001

010

6

5

4

3

2

1

MXXX ˆ 1

Common reference approach

Estimate (ab) with y3 - y2 - y1

y1 = log (A / C) = a

y2 = log (B / C) = b

y3 = log (AB / C) = a + b + ab C

A.BBA

y1 y2 y3

Indirect A balance of direct and indirect

I) II) III) IV)

# Slides N = 6

Main effect A

0.5 0.67 0.5 NA

Main effect B

0.5 0.43 0.5 0.3

Interaction A.B

1.5 0.67 1 0.67

2 x 2 factorial2 x 2 factorial

C

A.BBA

B

C

A.B

A

B

C

A.B

A

B

C

A.B

A

Table entry: variance

More general n by m factorial experimentMore general n by m factorial experiment

2 factors, one with n levels and the other with m levels

OE experiment (2 by 2):

interested in difference between zones, age and also zone.age interaction.

Further experiment (2 by 3):

only interested in genes where difference between treatment and controls changes with time.

0 12 24 0 12 24

treatmentcontrol controltreatment

WT.P11 + a1

MT.P21 + (a1 + a2) + b + (a1 + a2)b

MT.P11 +a1+b+a1.b

WT.P21 + a1 + a2

WT P1

MT.P1 + b

1

2

3

4

5

6

7

ba

ba

b

a

a

m

m

m

m

m

m

m

2

1

2

1

7

6

5

4

3

2

1

11100

10010

00010

01100

01001

00001

00100

Replication Replication

—Why replicate slides:– Provides a better estimate of the log-ratios– Essential to estimate the variance of log-ratios

—Different types of replicates:– Technical replicates

• Within slide vs between slides

– Biological replicates

Sample sizeSample size

Apo A1 Data Set

Technical replication - labellingTechnical replication - labelling

• 3 sets of self – self hybridization: (cerebellum vs cerebellum)

• Data 1 and Data 2 were labeled together and hybridized on two slides separately.

• Data 3 were labeled separately.

Data 1 Data 1

Dat

a 2

Dat

a 3

Technical replication - amplificationTechnical replication - amplificationOlfactory bulb experiment:• 3 sets of Anterior vs Dorsal performed on different days• #10 and #12 were from the same RNA isolation and

amplification• #12 and #18 were from different dissections and amplifications• All 3 data sets were labeled separately before hybridization

amplification

amplification

amplification

amplification

T1 T2

T1

T2

Original samples

Amplified samples

1

2

3

4

Replicate Design 2

Replicate Design 1

1

2

3

4

M6 = Lc.MT.P21 + (1 + 2) + + (1 + 2)*

Common reference approach

Estimate (1.) with M5 – M4 - M2 + M1

Estimate (1 + 2). with M6 – M4 – M3 + M1

M3 = Lc.WT.P21 + (1 + 2)

M2 = Lc.WT.P11 + 1

M4 = Lc.MT.P1 +

M5 = Lc.MT.P11 + 1 + + 1 *

M1 = Lc.MT.P1