Introduction to the design of cDNA microarray experiments
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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
Introduction to the design of cDNA microarray experiments
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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.
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. T2 T3 T4 T1 T2 Tn Tn-1 T1 Ref Natural design choice C
Direct vs Indirect Two samples e.g. KO vs. WT or mutant vs. WT Indirect Direct T Ref T C C average (log (T/C)) log (T / Ref) – log (C / Ref ) 2 /2 22
A B C A A B C C B O O One-way layout: one factor, k levels All pair-wise comparisons are of equal importance
Dye-swap A A C B C B Design B2 Design B1 • - 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 D P L V M
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) D A M L P V How do we combine these?
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)
Time Series T1 T2 T3 T4 T5 T6 T7 Ref • Possible designs: • All sample vs common pooled reference • All sample vs time 0 • Direct hybridization between times. Pooled reference Compare to T1 t vs t+1 t vs t+2 t vs t+3
T2 T4 T1 T3 T2 T4 T1 T3 T2 T4 T1 T3 Ref T1 T2 T3 T4 T1 T2 T3 T4 T2 T4 T1 T3
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).
Factorial design m m+a Different ways of estimating parameters. e.g. B effect. 1 = (m + b) - (m) = b 2 - 5 = ((m + a) - (m)) -((m + a)-(m + b)) = (a) - (a + b) = b 2 C A 4 1 3 5 6 B AB m+b m+a+b+ab
C A 2 4 1 3 5 AB B 6 Factorial design m m+a m+a+b+ab m+b
C A C A C A A B A.B A.B A.B B B A.B B C 2 x 2 factorial Table entry: variance
Linear model analysis Define a matrix X so that E(Y)=Xb Use least squares estimate for a, b, ab
A B A.B y2 y3 y1 C y1 = log (A / C) = a y2 = log (B / C) = b y3 = log (AB / C) = a + b + ab Common reference approach Estimate (ab) with y3 - y2 - y1
C A C A C A A B A.B A.B A.B B B A.B B C 2 x 2 factorial 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 control treatment 0 12 24 0 12 24
WT.P21 + a1 + a2 WT P1 WT.P11 + a1 2 5 7 4 1 MT.P21 + (a1 + a2) + b + (a1 + a2)b MT.P1 + b MT.P11 +a1+b+a1.b 3 6
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 3 Data 2 Data 1 Data 1
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
T1 T2 Replicate Design 1 amplification 1 2 3 4 T1 amplification Replicate Design 2 amplification T2 1 2 3 4 amplification Amplified samples Original samples
M1 = Lc.MT.P1 M2 = Lc.WT.P11 + 1 M3 = Lc.WT.P21 + (1 + 2) M4 = Lc.MT.P1 + M5 = Lc.MT.P11 + 1 + + 1 * 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