DNA Microarrays

Patrick Schmid

CSE 497

Spring 2004

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What is a DNA Microarray?

Also known as DNA Chip Allows simultaneous measurement of the

level of transcription for every gene in a genome (gene expression)

Transcription? Process of copying of DNA into messenger RNA

(mRNA) Environment dependant!

Microarray detects mRNA, or rather the more stable cDNA

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What is a DNA Microarray? (cont.)

Cheung et al. 1999

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How do we manufacture a microarray? Start with individual genes, e.g. the ~6,200

genes of the yeast genome Amplify all of them using polymerase chain

reaction (PCR) “Spot” them on a medium, e.g. an ordinary

glass microscope slide Each spot is about 100 µm in diameter Spotting is done by a robot Complex and potentially expensive task

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How do we manufacture a microarray?

Cheung et al. 1999

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Example

Remember the flash animation? Yeast Grow in aerobic and anaerobic environment Different genes will be activated in order to

adapt to each environment Extract mRNA Convert mRNA into colored cDNA

(fluorescently labeled)

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Example (cont.)

Mix cDNA together Hybridize cDNA with array Each cDNA sequence hybridizes specifically

with the corresponding gene sequence in the array

Wash unhybridized cDNA off Read array with laser Analyze images

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Overview of Example

Brown & Botstein, 1999

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Reading an array

Laser scans array and produces images One laser for each color, e.g. one for green, one for

red Image analysis, main tasks:

Noise suppression Spot localization and detection, including the extraction of

the background intensity, the spot position, and the spot boundary and size

Data quantification and quality assessment Image Analysis is a book on its own:

Kamberova, G. & Shah, S. “DNA Array Image Analysis Nuts & Bolts“. DNA Press LLC, 2002

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Reading an array (cont.)Block Column Row Gene Name Red Green Red:Green

Ratio

1 1 1 tub1 2,345 2,467 0.95

1 1 2 tub2 3,589 2,158 1.66

1 1 3 sec1 4,109 1,469 2.80

1 1 4 sec2 1,500 3,589 0.42

1 1 5 sec3 1,246 1,258 0.99

1 1 6 act1 1,937 2,104 0.92

1 1 7 act2 2,561 1,562 1.64

1 1 8 fus1 2,962 3,012 0.98

1 1 9 idp2 3,585 1,209 2.97

1 1 10 idp1 2,796 1,005 2.78

1 1 11 idh1 2,170 4,245 0.51

1 1 12 idh2 1,896 2,996 0.63

1 1 13 erd1 1,023 3,354 0.31

1 1 14 erd2 1,698 2,896 0.59Campbell & Heyer, 2003

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Real DNA Microarray

Campbell & Heyer, 2003

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Y-fold

Biologists rather deal with folds than with ratios A fold is nothing else than saying “times” We express it either as a Y-fold repression, or a

Y-fold induction It is calculated by taking the inverse of the ratio

Ratio of 0.33 = 3-fold repression Ratio of 10 = 10-fold induction

Fractional ratios can cause problems with techniques of analyzing and comparing gene expression patterns

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Color Coding

Tables are difficult to read Data is presented with a color scale Coding scheme:

Green = repressed (less mRNA) gene in experiment Red = induced (more mRNA) gene in experiment Black = no change (1:1 ratio)

Or Green = control condition (e.g. aerobic) Red = experimental condition (e.g. anaerobic)

We only use ratio

Campbell & Heyer, 2003

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Logarithmic transformation

log2 is commonly used Sometimes log10 is used Example:

log2(0.0625) = log2(1/16) = log2(1) – log2(16) = -log2(16) = -4

log2 transformations ease identification of doublings or halvings in ratios

log10 transformations ease identification of order of magnitude changes

Key attribute: equally sized induction and repression receive equal treatment visually and mathematically

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Complication: Time Series

Biologists care more about the process of adaptation than about the end result

For example, measure every 2 hours for 10 hours (depletion of oxygen)

31,000 gene expression ratios Or 6,200 different graphs with five data points each Question: Are there any genes that responded in

similar ways to the depletion of oxygen?

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Example data: fold change (ratios)Name 0 hours 2 hours 4 hours 6 hours 8 hours 10 hours

Gene C 1 8 12 16 12 8

Gene D 1 3 4 4 3 2

Gene E 1 4 8 8 8 8

Gene F 1 1 1 0.25 0.25 0.1

Gene G 1 2 3 4 3 2

Gene H 1 0.5 0.33 0.25 0.33 0.5

Gene I 1 4 8 4 1 0.5

Gene J 1 2 1 2 1 2

Gene K 1 1 1 1 3 3

Gene L 1 2 3 4 3 2

Gene M 1 0.33 0.25 0.25 0.33 0.5

Gene N 1 0.125 0.0833 0.0625 0.0833 0.125Campbell & Heyer, 2003

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Example data: log2 transformationName 0 hours 2 hours 4 hours 6 hours 8 hours 10 hours

Gene C 0 3 3.58 4 3.58 3

Gene D 0 1.58 2 2 1.58 1

Gene E 0 2 3 3 3 3

Gene F 0 0 0 -2 -2 -3.32

Gene G 0 1 1.58 2 1.58 1

Gene H 0 -1 -1.60 -2 -1.60 -1

Gene I 0 2 3 2 0 -1

Gene J 0 1 0 1 0 1

Gene K 0 0 0 0 1.58 1.58

Gene L 0 1 1.58 2 1.58 1

Gene M 0 -1.60 -2 -2 -1.60 -1

Gene N 0 -3 -3.59 -4 -3.59 -3Campbell & Heyer, 2003

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Pearson Correlation Coefficient r Gene expression over time is a vector, e.g.

for gene C: (0, 3, 3.58, 4, 3.58, 3) Given two vectors X and Y that contain N

elements, we calculate r as follows:

Cho & Won, 2003

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Pearson Correlation Coefficient r (cont.) X = Gene C = (0, 3.00, 3.58, 4, 3.58, 3)

Y = Gene D = (0, 1.58, 2.00, 2, 1.58, 1) ∑XY = (0)(0)+(3)(1.58)+(3.58)(2)+(4)(2)+(3.58)(1.58)+(3)(1) = 28.5564 ∑X = 3+3.58+4+3.58+3 = 17.16 ∑X2 = 32+3.582+42+3.582+32 = 59.6328 ∑Y = 1.58+2+2+1.58+1 = 8.16 ∑Y2 = 1.582+22+22+1.582+12 = 13.9928 N = 6 ∑XY – ∑X∑Y/N = 28.5564 – (17.16)(8.16)/6 = 5.2188 ∑X2 – (∑X)2/N = 59.6328 – (17.16)2/6 = 10.5552 ∑Y2 – (∑Y)2/N = 13.9928 – (8.16)2/6 = 2.8952 r = 5.2188 / sqrt((10.5552)(2.8952)) = 0.944

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Example data: Pearson correlation coefficient

Gene C Gene D Gene E Gene F Gene G Gene H Gene I Gene J Gene K Gene L Gene M Gene N

Gene C 1 0.94 0.96 -0.40 0.95 -0.95 0.41 0.36 0.23 0.95 -0.94 -1

Gene D 0.94 1 0.84 -0.10 0.94 -0.94 0.68 0.24 -0.07 0.94 -1 -0.94

Gene E 0.96 0.84 1 -0.57 0.89 -0.89 0.21 0.30 0.43 0.89 -0.84 -0.96

Gene F -0.40 -0.10 -0.57 1 -0.35 0.35 0.60 -0.43 -0.79 -0.35 0.10 0.40

Gene G 0.95 0.94 0.89 -0.35 1 -1 0.48 0.22 0.11 1 -0.94 -0.95

Gene H -0.95 -0.94 -0.89 0.35 -1 1 -0.48 -0.21 -0.11 -1 0.94 0.95

Gene I 0.41 0.68 0.21 0.60 0.48 -0.48 1 0 -0.75 0.48 -0.68 -0.41

Gene J 0.36 0.24 0.30 -0.43 0.22 -0.21 0 1 0 0.22 -0.24 -0.36

Gene K 0.23 -0.07 0.43 -0.79 0.11 -0.11 -0.75 0 1 0.11 0.07 -0.23

Gene L 0.95 0.94 0.89 -0.35 1 -1 0.48 0.22 0.11 1 -0.94 -0.95

Gene M -0.94 -1 -0.84 0.10 -0.94 0.94 -0.68 -0.24 0.07 -0.94 1 0.94

Gene N -1 -0.94 -0.96 0.40 -0.95 0.95 -0.41 -0.36 -0.23 -0.95 0.94 1

Campbell & Heyer, 2003

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Example: Reorganization of data

Campbell & Heyer, 2003

Name 0 hours 2 hours 4 hours 6 hours 8 hours 10 hours

Gene M 1 0.33 0.25 0.25 0.33 0.5

Gene N 1 0.125 0.0833 0.0625 0.0833 0.125

Gene H 1 0.5 0.33 0.25 0.33 0.5

Gene K 1 1 1 1 3 3

Gene J 1 2 1 2 1 2

Gene E 1 4 8 8 8 8

Gene C 1 8 12 16 12 8

Gene L 1 2 3 4 3 2

Gene G 1 2 3 4 3 2

Gene D 1 3 4 4 3 2

Gene I 1 4 8 4 1 0.5

Gene F 1 1 1 0.25 0.25 0.1

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Clustering of example

Campbell & Heyer, 2003

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Clustering of entire yeast genome

Campbell & Heyer, 2003

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Hierarchical Clustering

Algorithm: First, find the two most similar genes in the entire

set of genes. Join these together into a cluster. Now join the next two most similar objects (an object can be a gene or a cluster), forming a new cluster. Add the new cluster to the list of available objects, and remove the two objects used to form the new cluster. Continue this process, joining objects in the order of their similarity to one another, until there is only one object on the list – a single cluster containing all genes.

(Campbell & Heyer, 2003)

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Hierarchical Clustering (cont.)Gene C Gene D Gene E Gene F Gene G Gene H Gene I Gene J Gene K Gene L Gene M Gene N

Gene C 0.94 0.96 -0.40 0.95 -0.95 0.41 0.36 0.23 0.95 -0.94 -1

Gene D 0.84 -0.10 0.94 -0.94 0.68 0.24 -0.07 0.94 -1 -0.94

Gene E -0.57 0.89 -0.89 0.21 0.30 0.43 0.89 -0.84 -0.96

Gene F -0.35 0.35 0.60 -0.43 -0.79 -0.35 0.10 0.40

Gene G -1 0.48 0.22 0.11 1 -0.94 -0.95

Gene H -0.48 -0.21 -0.11 -1 0.94 0.95

Gene I 0 -0.75 0.48 -0.68 -0.41

Gene J 0 0.22 -0.24 -0.36

Gene K 0.11 0.07 -0.23

Gene L -0.94 -0.95

Gene M 0.94

Gene N

Campbell & Heyer, 2003

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Hierarchical Clustering (cont.)

F

C

G

D

E

Gene C Gene D Gene E Gene F Gene G

Gene C 0.94 0.96 -0.40 0.95

Gene D 0.84 -0.10 0.94

Gene E -0.57 0.89

Gene F -0.35

Gene G

C E

1

1 Gene D Gene F Gene G

1 0.89 -0.485 0.92

Gene D -0.10 0.94

Gene F -0.35

Gene G

Average observations

•Gene D: (0.94+0.84)/2 = 0.89

•Gene F: (-0.40+(-0.57))/2 = -0.485

•Gene G: (0.95+0.89)/2 = 0.92

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Hierarchical Clustering (cont.)

F

G

D

C E

1

1 Gene D Gene F Gene G

1 0.89 -0.485 0.92

Gene D -0.10 0.94

Gene F -0.35

Gene G

G D

2

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Hierarchical Clustering (cont.)

F

C E

1

G D

2

1 2 Gene F

1 0.905 -0.485

2 -0.225

Gene F

3

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Hierarchical Clustering (cont.)

F C E

1

G D

2

3

3 Gene F

3 -0.355

Gene F

4

F

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Hierarchical Clustering (cont.)

F C E

1

G D

2

3

4

Did this algorithm not look familiar?

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Hierarchical Clustering (cont.)

Eisen et al., 1998

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Hierarchical Clustering (cont.)

Single Linkage Shortest link between two clusters

Complete Linkage Longest link between two clusters

Average Linkage Average of distances between all

pairs of objects Average Group Linkage

Groups once formed are represented by their mean values, and then those are averaged

Which one did we use in the previous example ?

We differentiate hierarchical clustering algorithms by how they agglomerate distances:

http://www.resample.com/xlminer/help/HClst/HClst_intro.htm

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Clustering Overview

Different similarity measures Pearson Correlation Coefficient Cosine Coefficient Euclidean Distance Information Gain Mutual Information Signal to noise ratio Simple Matching for Nominals

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Clustering Overview (cont.)

Different Clustering Methods Unsupervised

Hierarchical Clustering k-means Clustering (k nearest neighbors)

Thursday Self-organizing map

Thursday Supervised

Support vector machine Ensemble classifier

Data Mining

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Support Vector Machines

Linear regression: x = w0 + w1a1 + w2a2 + … + wkak

x is the class, ai are the attribute values and wj are the weights

Given a distance vector Y with distances ai in which class x does Y belong?

What do we mean by a class x? Primitive method: Y is in one class if x<0.5, in

another class for x≥0.5.

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Support Vector Machines (cont.) Multi-response linear regression:

Set output to 1 for training instances that belong to a class

Set output to 0 for training instances that do not belong to that class

Result is a linear expression for each class Classification of unknown example:

Compute all linear expressions Choose the one that gives the largest output value

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Support Vector Machines (cont.) This means… Two pairs of classes Weight vector for class 1:

w0(1) + w1

(1)a1 + w2(1)a2 + … + wk

(1)ak

Weight vector for class 2: w0

(2) + w1(2)a1 + w2

(2)a2 + … + wk(2)ak

An instance will be assigned to class 1 rather than class 2 if w0

(1) + w1(1)a1 + w2

(1)a2 + … + wk(1)ak > w0

(2) + w1(2)a1 + w2

(2)a2 + … + wk(2)ak

We can rewrite this as (w0

(1) - w0(2)) + (w1

(1) - w1(2)) a1 + … + (wk

(1) - wk(2)) ak > 0

Hyperplane

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Support Vector Machines (cont.) We can only represent linear boundaries between classes so far Trick: Transform the input using a nonlinear mapping, then construct

a linear model in the new space Example: Use all products of n factors (2 attributes, n=3):

x = w1a13 + w2a1

2a2 + w3a1a22 + w4a2

3

Then use multi-response linear regression However, for 10 attributes and including all products with 5 factors,

we would need to determine more than 2000 coefficients Linear regression is O(n3) in time Problem: Training is infeasible Another problem: Overfit. The resulting model will be “too

nonlinear”, because there are just too many parameters in the model.

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Support Vector Machines (cont.)

support vectors

Convex hull of points is the tightest enclosing polygon

Maximum margin hyperplane

Instances closest to hyperplane are called support vectors

Support vectors define maximum margin hyperplane uniquely

Witten & Frank, 2000

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Support Vector Machines (cont.) We only need set of support vectors, everything else is irrelevant A hyperplane separating two classes can then be written as

x = w0 + w1a1 + w2a2 Or

x = b + ∑ αiγi (a(i) ∙ a) i is support vector γi is the class value of a(i) b and αi are numeric values to be determined Vector a represents a test instance a(i) are the support vectors

Determining b and αi is a constrained quadratic optimization problem that can be solved with off-the-shelf software packages

Support Vector Machines do not overfit, because there are usually only a few support vectors

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Support Vector Machines (cont.) Did I not introduce Support Vector Machines by

talking about non-linear class boundaries? x = b + ∑ αiγi (a(i) ∙ a)n

n is the number of factors (x ∙ y)n is called a polynomial kernel A good way of choosing n is by starting with n=1

and incrementing it until estimated error ceases to improve

If you want to know more: SVMs in general: Witten & Frank, 2000 (lecture material

based on this) Application to cancer classification: Cho & Won, 2003

Demo – Shneiderman

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References

Brown, P., Botstein, D. “Exploring the new world of the genome with DNA microarrays” Nature genetics supplement, vol. 21, January 1999

Campbell A. & Heyer L. “discovering Genomics, Proteomics, & Bioninformatics” Benjamin Cummings, 2003.

Cheung, V., Morley, M., Aguilar, F., Massimi, A., Kucherlapati, R. & Childs, G. “Making and reading microarrays” Nature genetics supplement, vol. 21, January 1999

Cho, S. & Won, H. “Machine Learning in DNA Microarray Analysis for Cancer Classification” Proceedings of the First Asia-Pacific bioinformatics conference on Bioinformatics 2003 - Volume 19, Australian Computer Society Inc.

Eisen, M., Spellman, P., Brown, P. & Botstein, D. “Cluster analysis and display of genome-wide expression patterns” Proc. Natl. Acad. Sci. USA. Vol 95, pp. 14 863-14868, December 1998. Genetics

Seo, J. & Sheiderman, B. “Interactively Exploring Hierarchical Clustering Results” IEEE Computer, July 2002

Witten, I. & Frank, E. “Data Mining” Morgan Kaufmann Publishers, 2000