DNA Microarrays

1
DNA Microarrays

• Patrick Schmid
• CSE 497
• Spring 2004

2
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

3
What is a DNA Microarray? (cont.)
Cheung et al. 1999
4
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

5
How do we manufacture a microarray?
Cheung et al. 1999
6
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)

7
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

8
Overview of Example
Brown Botstein, 1999
9
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

10
Reading an array (cont.)
Block Column Row Gene Name Red Green RedGreen 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 0.59
Campbell Heyer, 2003
11
Real DNA Microarray
Campbell Heyer, 2003
12
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

13
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 (11 ratio)
• Or
• Green control condition (e.g. aerobic)
• Red experimental condition (e.g. anaerobic)
• We only use ratio

Campbell Heyer, 2003
14
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

15
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?

16
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 0.125
What is the pattern?
Campbell Heyer, 2003
17
Example data log2 transformation
Name 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 -3
Campbell 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
19
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)(1.5
  8)(3)(1) 28.5564
• ?X 33.5843.583 17.16
• ?X2 323.582423.58232 59.6328
• ?Y 1.58221.581 8.16
• ?Y2 1.58222221.58212 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
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
Campbell Heyer, 2003
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Clustering of example
Campbell Heyer, 2003
23
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.)
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
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
C
D
E

• Average observations
• Gene D (0.940.84)/2 0.89
• Gene F (-0.40(-0.57))/2 -0.485
• Gene G (0.950.89)/2 0.92

F
1
G
C
E
27
Hierarchical Clustering (cont.)
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
D
F
1
2
G
C
E
G
D
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Hierarchical Clustering (cont.)
1 2 Gene F
1 0.905 -0.485
2 -0.225
Gene F
3
F
1
2
C
E
G
D
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Hierarchical Clustering (cont.)
3 Gene F
3 -0.355
Gene F
4
3
F
1
2
F
C
E
G
D
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Hierarchical Clustering (cont.)
Did this algorithm not look familiar?
4
Remember Neighbor-Joining?
3
1
2
F
C
E
G
D
31
Hierarchical Clustering (cont.)
Eisen et al., 1998
32
Hierarchical Clustering (cont.)

• We differentiate hierarchical clustering
  algorithms by how they agglomerate distances

• 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 ?

http//www.resample.com/xlminer/help/HClst/HClst_i
ntro.htm
33
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 xlt0.5, in
  another class for x0.5.

36
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

37
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 gt 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 gt 0
• Hyperplane

38
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, n3)
• x w1a13 w2a12a2 w3a1a22 w4a23
• 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.

39
Support Vector Machines (cont.)

• 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

support vectors
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 ? ai?i (a(i) a)
• i is support vector
• ?i is the class value of a(i)
• b and ai are numeric values to be determined
• Vector a represents a test instance
• a(i) are the support vectors
• Determining b and ai 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

41
Support Vector Machines (cont.)

• Did I not introduce Support Vector Machines by
  talking about non-linear class boundaries?
• x b ? ai?i (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 n1
  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

42
Demo Shneiderman
43
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