Introduction to the design of cDNA microarray experiments Statistics 246, Spring 2002 Week 9, Lecture 1 Yee Hwa Yang

Some 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 experiments 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 representation

Natural 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 T2T3T4 T1T1 Ref T2T2 T n-1 TnTn

Direct vs Indirect Two samples e.g. KO vs. WT or mutant vs. WT TC T C Ref Direct Indirect  2 /22222 average (log (T/C))log (T / Ref) – log (C / Ref )

I) Common Reference II) Common reference III ) Direct comparison Number of SlidesN = 3N=6N=3 Ave. variance20.67 Units of materialA = B = C = 1A = B = C = 2 Ave. variance10.67 One-way layout: one factor, k levels CB A O CBA O CBA All pair-wise comparisons are of equal importance

Dye-swap CB A Design B1 CB 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 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? L P V D M A

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

Pooled reference T2T4T5T6T7T3T1 Ref Compare to T1 t vs t+3 t vs t+2 t vs t+1 Time Series Possible designs: 1)All sample vs common pooled reference 2)All sample vs time 0 3)Direct hybridization between times.

Design choices in time seriest vs t+1t vs t+2 T1T2T2T3T3T4T1T3T2T4T1T4Ave N=3A) T1 as common reference B) Direct Hybridization N=4C) Common reference D) T1 as common ref + more E) Direct hybridization choice F) Direct Hybridization choice T2 T3 T4 T1 T2 T3 T4 T1 Ref T2 T3 T4 T1 T2T3T4T1 T2T3T4T1 T2 T3 T4 T1

2 by 2 factorial – two 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 = ((  + a) - (  )) -((  + a)-(  + b)) = (a) - (a + b) = b Factorial design   a  b  a+b+ab AC BAB

Factorial design  a  b  a+b+ab AC B AB 

IndirectA balance of direct and indirect I)II)III)IV) # Slides N = 6 Main effect A NA Main effect B Interacti on A.B x 2 factorial C A.BBA B C A B C A B C A Table entry: variance

Linear model analysis Define a matrix X so that E(Y)=Xb Use least squares estimate for a, b, ab

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

IndirectA balance of direct and indirect I)II)III)IV) # Slides N = 6 Main effect A NA Main effect B Interacti on A.B x 2 factorial C A.BBA B C A B C A B C A Table entry: variance

More 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 treatment control treatment

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

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 size Apo A1 Data Set

Technical 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 2 Data 3

Technical replication - amplification Olfactory 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 T1 T2 T1 T2 Original samples Amplified samples Replicate Design 2 Replicate Design

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 