intro to bioinfo statastics

Applied Statistics for Bioinformatics using R Wim P. Krijnen November 10, 2009ii Preface The purpose of this book is to give an introduction into statistics in order to solve some problems of bioinformatics. Statistics provides procedures to explore and visualize data as well as to test biological hypotheses. The book intends to be introductory in explaining and programming elementary statis- tical concepts, thereby bridging the gap between high school levels and the specialized statistical literature. After studying this book readers have a suf- ¯cient background for Bioconductor Case Studies (Hahne et al., 2008) and Bioinformatics and Computational Biology Solutions Using R and Biocon- ductor (Genteman et al., 2005). The theory is kept minimal and is always illustrated by several examples with data from research in bioinformatics. Prerequisites to follow the stream of reasoning is limited to basic high-school knowledge about functions. It may, however, help to have some knowledge of gene expressions values (Pevsner, 2003) or statistics (Bain & Engelhardt, 1992; Ewens & Grant, 2005; Rosner, 2000; Samuels & Witmer, 2003), and elementary programming. To support self-study a su±cient amount of chal- lenging exercises are given together with an appendix with answers. The programming language R is becoming increasingly important because it is not only very °exible in reading, manipulating, and writing data, but all its outcomes are directly available as objects for further programming. R is a rapidly growing language making basic as well as advanced statisti- cal programming easy. From an educational point of view, R provides the possibility to combine the learning of statistical concepts by mathematics, programming, and visualization. The plots and tables produced by R can readily be used in typewriting systems such as Emacs, LATEX, or Word. Chapter 1 gives a brief introduction into basic functionalities of R. Chap- ter 2 starts with univariate data visualization and the most important de- scriptive statistics. Chapter 3 gives commonly used discrete and continuous distributions to model events and the probability by which these occur. These distributions are applied in Chapter 4 to statistically test hypotheses from bioinformatics. For each test the statistics involved are brie°y explained and its application is illustrated by examples. In Chapter 5 linear models are ex- plained and applied to testing for di®erences between groups. It gives a basic approach. In Chapter 6 the three phases of analysis of microarray data (pre- processing, analysis, post processing) are brie°y introduced and illustrated by many examples bringing ideas together with R scrips and interpretation of results. Chapter 7 starts with an intuitive approach into Euclidian distanceiii and explains how it can be used in two well-known types of cluster analysis to ¯nd groups of genes. It also explains how principal components analysis can be used to explore a large data matrix for the direction of largest variation. Chapter 8 shows how gene expressions can be used to predict the diagnosis of patients. Three such prediction methods are illustrated and compared. Chapter 9 introduces a query language to download sequences e±ciently and gives various examples of computing important quantities such as alignment scores. Chapter 10 introduces the concept of a probability transition matrix which is applied to the estimation of phylogenetic trees and (Hidden) Markov Models. R commands come after its prompt >, except when commands are part of the ongoing text. Input and output of R will be given in verbatim typewriting style. To save space sometimes not all of the original output from R is printed. The end of an example is indicated by the box . In its Portable Document Format (PDF)1 there are many links to the Index, Table of Contents, Equations, Tables, and Figures. Readers are encouraged to copy and paste scripts from the PDF into the R system in order to study its outcome. Apart from using the book to study application of statistics in bioinformatics, it can also be useful for statistical programming. I would like to thank my colleges Joop Bouman, Sven Warris and Jan Peter Nap for their useful remarks on parts of an earlier draft. Many thanks also go to my students for asking questions that gave hints to improve clarity. Remarks to further improve the text are appreciated. Wim P. Krijnen Hanze University Institute for Life Science and Technology Zernikeplein 11 9747 AS Groningen The Netherlands w.p.krijnen@pl.hanze.nl Groningen October 2009 1°c This document falls under the GNU Free Document Licence and may be used freely for educational purposes.ivContents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii 1 Brief Introduction into Using R 1 1.1 Getting R Started on your PC . . . . . . . . . . . . . . . . . . 1 1.2 Getting help . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Calculating with R . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4 Generating a sequence and a factor . . . . . . . . . . . . . . . 4 1.5 Computing on a data vector . . . . . . . . . . . . . . . . . . . 5 1.6 Constructing a data matrix . . . . . . . . . . . . . . . . . . . 6 1.7 Computing on a data matrix . . . . . . . . . . . . . . . . . . . 8 1.8 Application to the Golub (1999) data . . . . . . . . . . . . . . 10 1.9 Running scripts . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.10 Overview and concluding remarks . . . . . . . . . . . . . . . . 13 1.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2 Data Display and Descriptive Statistics 17 2.1 Univariate data display . . . . . . . . . . . . . . . . . . . . . . 17 2.1.1 Pie and Frequency table . . . . . . . . . . . . . . . . . 17 2.1.2 Plotting data . . . . . . . . . . . . . . . . . . . . . . . 18 2.1.3 Histogram . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.1.4 Boxplot . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.1.5 Quantile-Quantile plot . . . . . . . . . . . . . . . . . . 22 2.2 Descriptive statistics . . . . . . . . . . . . . . . . . . . . . . . 24 2.2.1 Measures of central tendency . . . . . . . . . . . . . . 24 2.2.2 Measures of spread . . . . . . . . . . . . . . . . . . . . 25 2.3 Overview and concluding remarks . . . . . . . . . . . . . . . . 26 2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 vvi CONTENTS 3 Important Distributions 31 3.1 Discrete distributions . . . . . . . . . . . . . . . . . . . . . . . 31 3.1.1 Binomial distribution . . . . . . . . . . . . . . . . . . . 31 3.2 Continuous distributions . . . . . . . . . . . . . . . . . . . . . 34 3.2.1 Normal distribution . . . . . . . . . . . . . . . . . . . . 35 3.2.2 Chi-squared distribution . . . . . . . . . . . . . . . . . 37 3.2.3 T-Distribution . . . . . . . . . . . . . . . . . . . . . . . 39 3.2.4 F-Distribution . . . . . . . . . . . . . . . . . . . . . . . 40 3.2.5 Plotting a density function . . . . . . . . . . . . . . . . 41 3.3 Overview and concluding remarks . . . . . . . . . . . . . . . . 42 3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4 Estimation and Inference 47 4.1 Statistical hypothesis testing . . . . . . . . . . . . . . . . . . . 47 4.1.1 The Z-test . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.1.2 One Sample t-Test . . . . . . . . . . . . . . . . . . . . 51 4.1.3 Two-sample t-test with unequal variances . . . . . . . 54 4.1.4 Two sample t-test with equal variances . . . . . . . . . 56 4.1.5 F-test on equal variances . . . . . . . . . . . . . . . . . 57 4.1.6 Binomial test . . . . . . . . . . . . . . . . . . . . . . . 58 4.1.7 Chi-squared test . . . . . . . . . . . . . . . . . . . . . 59 4.1.8 Normality tests . . . . . . . . . . . . . . . . . . . . . . 63 4.1.9 Outliers test . . . . . . . . . . . . . . . . . . . . . . . . 64 4.1.10 Wilcoxon rank test . . . . . . . . . . . . . . . . . . . . 65 4.2 Application of tests to a whole set gene expression data . . . . 66 4.3 Overview and concluding remarks . . . . . . . . . . . . . . . . 68 4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 5 Linear Models 73 5.1 De¯nition of linear models . . . . . . . . . . . . . . . . . . . . 74 5.2 One-way analysis of variance . . . . . . . . . . . . . . . . . . . 77 5.3 Two-way analysis of variance . . . . . . . . . . . . . . . . . . 83 5.4 Checking assumptions . . . . . . . . . . . . . . . . . . . . . . 85 5.5 Robust tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5.6 Overview and concluding remarks . . . . . . . . . . . . . . . . 88 5.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88CONTENTS vii 6 Micro Array Analysis 91 6.1 Probe data . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 6.2 Preprocessing methods . . . . . . . . . . . . . . . . . . . . . . 94 6.3 Gene ¯ltering . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 6.4 Applications of linear models . . . . . . . . . . . . . . . . . . 100 6.5 Searching an annotation package . . . . . . . . . . . . . . . . 104 6.6 Using annotation to search literature . . . . . . . . . . . . . . 106 6.7 Searching GO numbers and evidence . . . . . . . . . . . . . . 107 6.8 GO parents and children . . . . . . . . . . . . . . . . . . . . . 108 6.9 Gene ¯ltering by a biological term . . . . . . . . . . . . . . . . 109 6.10 Signi¯cance per chromosome . . . . . . . . . . . . . . . . . . . 110 6.11 Overview and concluding remarks . . . . . . . . . . . . . . . . 112 6.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 7 Cluster Analysis and Trees 117 7.1 Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 7.2 Two types of Cluster Analysis . . . . . . . . . . . . . . . . . . 121 7.2.1 Single Linkage . . . . . . . . . . . . . . . . . . . . . . . 121 7.2.2 k-means . . . . . . . . . . . . . . . . . . . . . . . . . . 125 7.3 The correlation coe±cient . . . . . . . . . . . . . . . . . . . . 130 7.4 Principal Components Analysis . . . . . . . . . . . . . . . . . 133 7.5 Overview and concluding remarks . . . . . . . . . . . . . . . . 141 7.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 8 Classi¯cation Methods 145 8.1 Classi¯cation of microRNA . . . . . . . . . . . . . . . . . . . . 146 8.2 ROC types of curves . . . . . . . . . . . . . . . . . . . . . . . 147 8.3 Classi¯cation trees . . . . . . . . . . . . . . . . . . . . . . . . 150 8.4 Support Vector Machine . . . . . . . . . . . . . . . . . . . . . 160 8.5 Neural Networks . . . . . . . . . . . . . . . . . . . . . . . . . 162 8.6 Generalized Linear Models . . . . . . . . . . . . . . . . . . . . 164 8.7 Overview and concluding remarks . . . . . . . . . . . . . . . . 167 8.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 9 Analyzing Sequences 173 9.1 Using a query language . . . . . . . . . . . . . . . . . . . . . . 173 9.2 Getting information on downloaded sequences . . . . . . . . . 174 9.3 Computations on sequences . . . . . . . . . . . . . . . . . . . 176viii CONTENTS 9.4 Matching patterns . . . . . . . . . . . . . . . . . . . . . . . . 181 9.5 Pairwise alignments . . . . . . . . . . . . . . . . . . . . . . . . 182 9.6 Overview and concluding remarks . . . . . . . . . . . . . . . . 189 9.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 10 Markov Models 193 10.1 Random sampling . . . . . . . . . . . . . . . . . . . . . . . . . 193 10.2 Probability transition matrix . . . . . . . . . . . . . . . . . . . 194 10.3 Properties of the transition matrix . . . . . . . . . . . . . . . 199 10.4 Stationary distribution . . . . . . . . . . . . . . . . . . . . . . 201 10.5 Phylogenetic distance . . . . . . . . . . . . . . . . . . . . . . . 203 10.6 Hidden Markov Models . . . . . . . . . . . . . . . . . . . . . . 209 10.7 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 10.8 Overview and concluding remarks . . . . . . . . . . . . . . . . 214 10.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 A Answers to exercises 219 B References 257List of Figures 2.1 Plot of gene expression values of CCND3 Cyclin D3. . . . . . . 20 2.2 Stripchart of gene expression values of CCND3 Cyclin D3 for ALL and AML patients. . . . . . . . . . . . . . . . . . . . . . 20 2.3 Histogram of ALL expression values of gene CCND3 Cyclin D3. 21 2.4 Boxplot of ALL and AML expression values of gene CCND3 Cyclin D3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.5 Q-Q plot of ALL gene expression values of CCND3 Cyclin D3. 23 2.6 Boxplot with arrows and explaining text. . . . . . . . . . . . 29 3.1 Binomial probabilities with n = 22 and p = 0:7 . . . . . . . . 34 3.2 Binomial cumulative probabilities with n = 22 and p = 0:7. . . 34 3.3 Graph of normal density with mean 1.9 and standard deviation 0.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3.4 Graph of normal distribution with mean 1.9 and standard de- viation 0.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3.5 Â25-density. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.6 Â25distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.7 Density of T10 distribution. . . . . . . . . . . . . . . . . . . . . 39 3.8 Distribution function of T10. . . . . . . . . . . . . . . . . . . . 39 3.9 Density of F26;10. . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.10 Distribution of F26;10. . . . . . . . . . . . . . . . . . . . . . . . 41 4.1 Acceptance and rejection regions of the Z-test. . . . . . . . . . 50 4.2 Acceptance and rejection regions of the T5-test. . . . . . . . . 52 4.3 Rejection region of Â23-test. . . . . . . . . . . . . . . . . . . . . 59 5.1 Plot of SKI-like oncogene expressions for three patient groups. 81 5.2 Plot of Ets2 expression values for three patient groups. . . . . 81 ixx LIST OF FIGURES 6.1 Mat plot of intensity values for a probe of MLL.B. . . . . . . . 93 6.2 Density of MLL.B data. . . . . . . . . . . . . . . . . . . . . . . 93 6.3 Boxplot of the ALL1/AF4 patients. . . . . . . . . . . . . . . . 97 6.4 Boxplot of the ALL1/AF4 patients after median subtraction and MAD division. . . . . . . . . . . . . . . . . . . . . . . . . 97 6.5 Venn diagram of seleced ALL genes. . . . . . . . . . . . . . . . 100 6.6 Boxplot of the ALL1/AF4 patients after median subtraction and MAD division. . . . . . . . . . . . . . . . . . . . . . . . . 100 7.1 Plot of ¯ve points to be clustered. . . . . . . . . . . . . . . . . 122 7.2 Tree of single linkage cluster analysis. . . . . . . . . . . . . . . 122 7.3 Example of three without clusters. . . . . . . . . . . . . . . . 123 7.4 Three clusters with di®erent standard deviations. . . . . . . . 123 7.5 Plot of gene "CCND3 Cyclin D3" and "Zyxin" expressions for ALL and AML patients. . . . . . . . . . . . . . . . . . . . . . 124 7.6 Single linkage cluster diagram from gene "CCND3 Cyclin D3" and "Zyxin" expressions values. . . . . . . . . . . . . . . . . 124 7.7 K-means cluster analysis. . . . . . . . . . . . . . . . . . . . . . 126 7.8 Tree of single linkage cluster analysis. . . . . . . . . . . . . . . 126 7.9 Plot of kmeans (stars) cluster analysis on CCND3 Cyclin D3 and Zyxin discriminating between ALL (red) and AML (black) patients. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 7.10 Vectors of linear combinations. . . . . . . . . . . . . . . . . . . 135 7.11 First principal component with projections of data. . . . . . . 135 7.12 Scatter plot of selected genes with row labels on the ¯rst two principal components. . . . . . . . . . . . . . . . . . . . . . . 138 7.13 Single linkage cluster diagram of selected gene expression values.138 7.14 Biplot of selected genes from the golub data. . . . . . . . . . . 144 8.1 ROC plot for expression values of CCND3 Cyclin D3. . . . . . 149 8.2 ROC plot for expression values of gene Gdf5. . . . . . . . . . 149 8.3 Boxplot of expression values of gene a for each leukemia class. 151 8.4 Classi¯cation tree for gene for three classes of leukemia. . . . . 151 8.5 Boxplot of expression values of gene a for each leukemia class. 154 8.6 Classi¯cation tree of expression values from gene A, B, and C for the classi¯cation of ALL1, ALL2, and AML patients. . . . 154 8.7 Boxplot of expression values from gene CCND3 Cyclin D3 for ALL and AML patients . . . . . . . . . . . . . . . . . . . . . 156LIST OF FIGURES xi 8.8 Classi¯cation tree of expression values from gene CCND3 Cy- clin D3 for classi¯cation of ALL and AML patients. . . . . . 156 8.9 rpart on ALL B-cel 123 data. . . . . . . . . . . . . . . . . . . 159 8.10 Variable importance plot on ALL B-cell 123 data. . . . . . . 159 8.11 Logit ¯t to the CCND3 Cyclin D3 expression values. . . . . . 171 9.1 G + C fraction of sequence "AF517525.CCND3" along a win- dow of length 50 nt. . . . . . . . . . . . . . . . . . . . . . . . 178 9.2 Frequency plot of amino acids from accession number AF517525.CCND3.179 9.3 Frequency plot of amino acids from accession number AL160163.CCND3.179 10.1 Graph of probability transition matrix . . . . . . . . . . . . . 196 10.2 Evaluation of models by AIC . . . . . . . . . . . . . . . . . . . 216 10.3 Tree according to GTR model. . . . . . . . . . . . . . . . . . . 217xii LIST OF FIGURESList of Tables 2.1 A frequency table and its pie of Zyxin gene. . . . . . . . . . . 18 3.1 Discrete density and distribution function values of S3, with p = 0:6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.2 Built-in-functions for random variables used in this chapter. . 42 3.3 Density, mean, and variance of distributions used in this chapter. 43 7.1 Data set for principal components analysis. . . . . . . . . . . 134 8.1 Frequencies empirical p-values lower than or equal to 0.01. . . 146 8.2 Ordered expression values of gene CCND3 Cyclin D3, index 2 indicates ALL, 1 indicates AML, cuto® points, number of false positives, false positive rate, number of true positives, true positive rate. . . . . . . . . . . . . . . . . . . . . . . . . 170 9.1 BLOSUM50 matrix. . . . . . . . . . . . . . . . . . . . . . . . 186 xiiixiv LIST OF TABLESChapter 1 Brief Introduction into Using R To get started a gentle introduction to the statistical programming language R will be given (R Development Core Team, 2009), speci¯c for our purposes. This will solve the practical issues to follow the stream of reasoning. In particular, it is brie°y explained how to install R and Bioconductor, how to obtain help, and how to perform simple calculations. Since many computations are essentially performed on data vectors, sev- eral basic illustrations of this are given. With respect to gene expressions the data vectors are placed one beneath the other to form a data matrix with the genes as rows and the patients as columns. The idea of a data matrix is extensively explained and illustrated by several examples. A larger example consists of the classical Golub et al. (1999) data, which will be analyzed frequently to illustrate statistical procedures. 1.1 Getting R Started on your PC You can downloaded R freely from http://cran.r-project.org. Click on your favorite operating system (Windows, Linux or MacOS) and simply follow the instructions. After a little patience you should be able to start R (Ihaka & Gentleman, 1996) after which a screen is opened with the prompt >. The input and output of R will be displayed in verbatim typewriting style. All useful functions of R are contained in libraries which are called "pack- ages". The standard installation of R makes basic packages available such as base and stats. From the button Packages at cran.r-project.org it can be seen that R has a huge number of packages available for a wide scale 12 CHAPTER 1. BRIEF INTRODUCTION INTO USING R of statistical procedures. To download a speci¯c package you can use the following. > install.packages(c("TeachingDemos"),repo="http://cran.r-project.org", + dep=TRUE) This installs the package TeachingDemos developed by Greg Snow from the repository http://cran.r-project.org. By setting the option dep to TRUE the packages on which the TeachingDemos depend are also installed. This is strongly recommended! Alternatively, in the Windows application of R you can simply click on the Packages button at the top of your screen and follow the instructions. After installing you have to load the package in order to use its functions. For instance, to produce a nice plot of the outcome of throwing twelve times with a die, you can use the following. > library(TeachingDemos) > plot(dice(12,1)) In the sequel we shall often use packages from Bioconductor, a very useful open source software project for the analysis and comprehension of genomic data. To follow the book it is essential to install Bioconductor on your PC or network. Bioconductor is primarily based on R and can be installed, as follows. > source("http://www.bioconductor.org/biocLite.R") > biocLite() Then to download the ALL package from a repository to your system, to load it, and to make the ALL data (Chiaretti, et. al, 2004) available for usage, you can use the following. > biocLite("ALL") > library(ALL) > data(ALL) These data will be analyzed extensively later-on in Chapter 5 and 6. General help on loaded Bioconductor packages becomes available by openVignette(). For further information the reader is referred to www.bioconductor.org or to several other URL's1 . 1 http://mccammon.ucsd.edu/~bgrant/bio3d/user_guide/user_guide.html http://rafalab.jhsph.edu/software.html http://dir.gmane.org/gmane.science.biology.informatics.conductor1.2. GETTING HELP 3 In this and the following chapters we will illustrate many statistical ideas by the Golub et al. (1999) data, see also Section 1.8. The golub data become available by the following.2 > library(multtest) > data(golub) R is object-oriented in the sense that everything consists of objects belonging to certain classes. Type class(golub) to obtain the class of the object golub and str(golub) to obtain its structure or content. Type objects() or ls() to view the currently loaded objects, a list probably growing soon to be large. To prevent con°icting de¯nitions, it is wise to remove them all at the end of a session by rm(list=ls()). To quit a session, type q(), or simply click on the cross in the upper right corner of your screen. 1.2 Getting help All functionalities of R are well-organized in so-called packages. Use the func- tion library() to see which packages are currently installed on your oper- ating system. The packages stats and base are automatically installed, be- cause these contain many basic functionalities. To obtain an overview of the content of a package use ls(package:stats) or library(help="stats"). Help on the purpose of speci¯c functions can be obtained from the (package) manual by typing a question mark in front of a function. For instance, ?sum gives details on summation. In case you are seeking help on a function which uses if, simply type apropos("if"). When you are starting with a new con- cept such as "boxplot", it is convenient to have an example showing output (a plot) and programming code. Such is given by example(boxplot). The function history can be useful for collecting previously given commands. Type help.start() to launch an HTML page linking to several well- written R manuals such as: "An Introduction to R", "The R Language De¯- nition", "R Installation and Administration", and "R Data Import/Export". Further help can be obtained from http://cran.r-project.org. Its "con- tributed" page contains well-written freely available on-line books3 and use- ful reference charts4. At http://www.r-project.org you can use R site 2 Functions to read data into R are read.table or read.csv, see also the "The R Data Import/Export manual". 3"R for Beginners" by Emmanuel Paradis or the "The R Guide" by Jason Owen 4"R reference card" by Tom Short or by Jonathan Baron4 CHAPTER 1. BRIEF INTRODUCTION INTO USING R search, Rseek, or other useful search engines. There are a number of useful URL's with information on R.5 1.3 Calculating with R R can be used as a simple calculator. For instance, to add 2 and 3 we simply insert the following. > 2+3 [1] 5 In many calculations the natural base e = 2:718282 of exponential functions is used. Such type of functions can be called as follows. > exp(1) [1] 2.718282 To compute e2 = e ¢ e we use exp(2).6 So, indeed, we have ex =exp(x), for any value of x. The sum 1 + 2 + 3 + 4 + 5 can be computed by > sum(1:5) [1] 15 and the product 5! = 5 ¢ 4 ¢ 3 ¢ 2 ¢ 1 by > prod(1:5) [1] 120 1.4 Generating a sequence and a factor In order to compute so-called quantiles of distributions (see e.g. Section 2.1.4) or plots of functions, we need to generate sequences of numbers. The easiest way to construct a sequence of numbers is by > 1:5 [1] 1 2 3 4 5 5We mention in particular: http://faculty.ucr.edu/~tgirke/Documents/R_BioCond/R_BioCondManual.html 6The argument of functions is always placed between parenthesis ().1.5. COMPUTING ON A DATA VECTOR 5 This sequence can also be produced by the function seq, which allows for various sizes of steps to be chosen. For instance, in order to compute per- centiles of a distribution we may want to generate numbers between zero and one with step size equal to 0.1. > seq(0,1,0.1) [1] 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 For plotting and testing of hypotheses we need to generate yet another type of sequence, called a \factor". It is designed to indicate an experimen- tal condition of a measurement or the group to which a patient belongs.7 When, for instance, for each of three experimental conditions there are mea- surements from ¯ve patients, the corresponding factor can be generated as follows. > factor <- gl(3,5) > factor [1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 Levels: 1 2 3 The three conditions are often called \levels" of a factor. Each of these levels has ¯ve repeats corresponding to the number of observations (patients) within each level (type of disease). We shall further illustrate the idea of a factor soon because it is very useful for purposes of visualization. 1.5 Computing on a data vector A data vector is simply a collection of numbers obtained as outcomes from measurements. This can be illustrated by a simple example on expression values of a gene. Suppose that gene expression values 1; 1:5; and 1:25 from the persons "Eric", "Peter", and "Anna" are available. To store these in a vector we use the concatenate command c(), as follows. > gene1 <- c(1.00,1.50,1.25) > gene1 [1] 1.00 1.50 1.25 7 See e.g. Samuals & Witmer (2003, Chap. 8) for a full explanation of experiments and statistical principles of design.6 CHAPTER 1. BRIEF INTRODUCTION INTO USING R Now we have created the object gene1 containing three gene expression val- ues. To compute the sum, mean, and standard deviation of the gene expres- sion values we use the corresponding built-in-functions. > sum(gene1) [1] 3.75 > mean(gene1) [1] 1.25 > sum(gene1)/3 [1] 1.25 > sd(gene1) [1] 0.25 > sqrt(sum((gene1-mean(gene1))^2)/2) [1] 0.25 By de¯ning x1 = 1:00, x2 = 1:50, and x3 = 1:25, the sum of the weights can be expressed as Pni=1 xi = 3:75. The mathematical summation symbol P is in R language simply sum. The mean is denoted by x = P3i=1 xi=3 = 1:25 and the sample standard deviation as s =vuutX3 i=1 (xi ¡ x)2=(3 ¡ 1) = 0:25: 1.6 Constructing a data matrix In various types of spreadsheets it is custom to store data values in the form of a matrix consisting of rows and columns. In bioinformatics gene expression values (from several groups of patients) are stored as rows such that each row contains the expressions values of the patients corresponding to a particular gene and each column contains all gene expression values for a particular person. To illustrate this by a small example suppose that we have the following expression values on three genes from Eric, Peter, and Anna.8 > gene2 <- c(1.35,1.55,1.00) > gene3 <- c(-1.10,-1.50,-1.25) > gene4 <- c(-1.20,-1.30,-1.00) 8By the function data.entry you can open and edit a screen with the values of a matrix.1.6. CONSTRUCTING A DATA MATRIX 7 Before constructing the matrix it is convenient to add the names of the rows and the columns. To do so we construct the following list. > rowcolnames <- list(c("gene1","gene2","gene3","gene4"), + c("Eric","Peter","Anna")) After the last comma in the ¯rst line we give a carriage return for R to come up with a new line starting with + in order to complete a command. Now we can construct a matrix containing the expression values from our four genes, as follows. > gendat <- matrix(c(gene1,gene2,gene3,gene4), nrow=4, ncol=3, + byrow=TRUE, dimnames = rowcolnames) Here, nrow indicates the number of rows and ncol the number of columns. The gene vectors are placed in the matrix as rows. The names of the rows and columns are attached by the dimnames parameter. To see the content of the just created object gendat, we print it to the screen. > gendatEric Peter Anna gene1 1.00 1.50 1.25 gene2 1.35 1.55 1.30 gene3 -1.10 -1.50 -1.25 gene4 -1.20 -1.30 -1.00 A matrix such as gendat has two indices [i,j], the ¯rst of which refers to rows and the second to columns9. Thus, if you want to print the second element of the ¯rst row to the screen, then type gendat[1,2]. If you want to print the ¯rst row, then use gendat[1,]. For the second column, use gendat[,2]. It may be desirable to write the data to a ¯le for using these in a later stage or to send these to a college of yours. Consider the following script. > write.table(gendat,file="D:/data/gendat.Rdata") > gendatread <- read.table("D:/data/gendat.Rdata") > gendatread Eric Peter Anna gene1 1.00 1.50 1.25 9Indices referring to rows, columns, or elements are always between square brackets [].8 CHAPTER 1. BRIEF INTRODUCTION INTO USING R gene2 1.35 1.55 1.30 gene3 -1.10 -1.50 -1.25 gene4 -1.20 -1.30 -1.00 An alternative is to use write.csv.10 1.7 Computing on a data matrix Means or standard deviations of rows or columns are often important for drawing biologically relevant conclusions. Such type of computations on a data matrix can be accomplished by \for loops". However, it is much more convenient to use the apply functionality on a matrix. To do so we specify the name of the matrix, indicate rows or columns (1 for rows and 2 for columns), and the name of the function. To illustrate this we compute the mean of each person (column). > apply(gendat,2,mean) Eric Peter Anna 0.0125 0.0625 0.0750 Similarly, the mean of each gene (row) can be computed. > apply(gendat,1,mean) gene1 gene2 gene3 gene4 1.250000 1.400000 -1.283333 -1.166667 It frequently happens that we want to re-order the rows of a matrix according to a certain criterion, or, more speci¯cally, the values in a certain column vector. For instance, to re-order the matrix gendat according to the row means, it is convenient to store these in a vector and to use the function order. > meanexprsval <- apply(gendat,1,mean) > o <- order(meanexprsval,decreasing=TRUE) > o [1] 2 1 4 3 10For more see the "R Data import/Export" manual, Chapter 3 of the book "R for Beginners", or search the internet by the key "r wiki matrix".1.7. COMPUTING ON A DATA MATRIX 9 Thus gene2 appears ¯rst because it has the largest mean 1.4, then gene1 with 1.25, followed by gene4 with -1.16 and, ¯nally, gene3 with -1.28. Now that we have collected the order numbers in the vector o, we can re-order the whole matrix by specifying o as the row index.11 > gendat[o,] Eric Peter Anna gene2 1.35 1.55 1.30 gene1 1.00 1.50 1.25 gene4 -1.20 -1.30 -1.00 gene3 -1.10 -1.50 -1.25 Another frequently occurring problem is that of selecting genes with a certain property. We illustrate this by several methods to select genes with positive mean expression values. A ¯rst method starts with the observation that the ¯rst two rows have positive means and to use c(1,2) as a row index. > gendat[c(1,2),] Eric Peter Anna gene1 1.00 1.50 1.25 gene2 1.35 1.55 1.30 A second way is to use the row names as an index. > gendat[c("gene1","gene2"),] Eric Peter Anna gene1 1.00 1.50 1.25 gene2 1.35 1.55 1.30 A third and more advanced way is to use an evaluation in terms of TRUE or FALSE of logical elements of a vector. For instance, we may evaluate whether the row mean is positive. > meanexprsval > 0 gene1 gene2 gene3 gene4 TRUE TRUE FALSE FALSE Now we can use the evaluation of meanexprsval > 0 in terms of the values TRUE or FALSE as a row index. 11You can also use functions like sort or rank.10 CHAPTER 1. BRIEF INTRODUCTION INTO USING R > gendat[meanexprsval > 0,] Eric Peter Anna gene1 1.00 1.50 1.25 gene2 1.35 1.55 1.30 Observe that this selects genes for which the evaluation equals TRUE. This illustrates that genes can be selected by their row index, row name or value on a logical variable. 1.8 Application to the Golub (1999) data The gene expression data collected by Golub et al. (1999) are among the clas- sical in bioinformatics. A selection of the set is called golub and is contained in the multtest package, which is part of Bioconductor. The data consist of gene expression values of 3051 genes (rows) from 38 leukemia patients12. Twenty seven patients are diagnosed as acute lymphoblastic leukemia (ALL) and eleven as acute myeloid leukemia (AML). The tumor class is given by the numeric vector golub.cl, where ALL is indicated by 0 and AML by 1. The gene names are collected in the matrix golub.gnames of which the columns correspond to the gene index, ID, and Name, respectively. We shall ¯rst concentrate on expression values of a gene with manufacturer name "M92287_at", which is known in biology as "CCND3 Cyclin D3". The ex- pression values of this gene are collected in row 1042 of golub. To load the data and to obtain relevant information from row 1042 of golub.gnames, use the following. > library(multtest); data(golub) > golub.gnames[1042,] [1] "2354" "CCND3 Cyclin D3" "M92287_at" The data are stored in a matrix called golub. The number of rows and columns can be obtained by the functions nrow and ncol, respectively. > nrow(golub) [1] 3051 > ncol(golub) [1] 38 12The data are pre-processed by procedures described in Dudoit et al. (2002).1.8. APPLICATION TO THE GOLUB (1999) DATA 11 So the matrix has 3051 rows and 38 columns, see also dim(golub). Each data element has a row and a column index. Recall that the ¯rst index refers to rows and the second to columns. Hence, the second value from row 1042 can be printed to the screen as follows. > golub[1042,2] [1] 1.52405 So 1.52405 is the expression value of gene CCND3 Cyclin D3 from patient number 2. The values of the ¯rst column can be printed to the screen by the following. > golub[,1] To save space the output is not shown. We may now print the expression values of gene CCND3 Cyclin D3 (row 1042) to the screen. > golub[1042,] [1] 2.10892 1.52405 1.96403 2.33597 1.85111 1.99391 2.06597 1.81649 [9] 2.17622 1.80861 2.44562 1.90496 2.76610 1.32551 2.59385 1.92776 [17] 1.10546 1.27645 1.83051 1.78352 0.45827 2.18119 2.31428 1.99927 [25] 1.36844 2.37351 1.83485 0.88941 1.45014 0.42904 0.82667 0.63637 [33] 1.02250 0.12758 -0.74333 0.73784 0.49470 1.12058 To print the expression values of gene CCND3 Cyclin D3 to the screen only for the ALL patients, we have to refer to the ¯rst twenty seven elements of row 1042. A possibility to do so is by the following. > golub[1042,1:27] However, for the work ahead it is much more convenient to construct a factor indicating the tumor class of the patients. This will turn out useful e.g. for separating the tumor groups in various visualization procedures. The factor will be called gol.fac and is constructed from the vector golub.cl, as follows. > gol.fac <- factor(golub.cl, levels=0:1, labels = c("ALL","AML")) In the sequel this factor will be used frequently. Obviously, the labels corre- spond to the two tumor classes. The evaluation of gol.fac=="ALL" returns TRUE for the ¯rst twenty seven values and FALSE for the remaining eleven. This is useful as a column index for selecting the expression values of the ALL patients. The expression values of gene CCND3 Cyclin D3 from the ALL patients can now be printed to the screen, as follows.12 CHAPTER 1. BRIEF INTRODUCTION INTO USING R > golub[1042,gol.fac=="ALL"] For many types of computations it is very useful to combine a factor with the apply functionality. For instance, to compute the mean gene expression over the ALL patients for each of the genes, we may use the following. > meanALL <- apply(golub[,gol.fac=="ALL"], 1, mean) The speci¯cation golub[,gol.fac=="ALL"] selects the matrix with gene ex- pressions corresponding to the ALL patients. The 3051 means are assigned to the vector meanALL. After reading the classical article by Golub et al. (1999), which is strongly recommended, one becomes easily interested in the properties of certain genes. For instance, gene CD33 plays an important role in distinguishing lymphoid from myeloid lineage cells. To perform computations on the ex- pressions of this gene we need to know its row index. This can obtained by the grep function.13 > grep("CD33",golub.gnames[,2]) [1] 808 Hence, the expression values of antigen CD33 are available at golub[808,] and further information on it by golub.gnames[808,]. 1.9 Running scripts It is very convenient to use a plain text writer like Notepad, Kate, Emacs, or WinEdt for the formulation of several consecutive R commands as separated lines (scripts). Such command lines can be executed by simply using copy and paste into the command line editor of R. Another possibility is to execute a script from a ¯le. To illustrate the latter consider the following. > library(multtest); data(golub) > gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) > mall <- apply(golub[,gol.fac=="ALL"], 1, mean) > maml <- apply(golub[,gol.fac=="AML"], 1, mean) > o <- order(abs(mall-maml), decreasing=TRUE) > print(golub.gnames[o[1:5],2]) 13Indeed, several functions of R are inspired by the Linux operating system.1.10. OVERVIEW AND CONCLUDING REMARKS 13 [1] "CST3 Cystatin C (amyloid angiopathy and cerebral hemorrhage)" [2] "INTERLEUKIN-8 PRECURSOR" [3] "Interleukin 8 (IL8) gene" [4] "DF D component of complement (adipsin)" [5] "MPO Myeloperoxidase" The row means of the expression values per patient group are computed and stored in the object mall and maml, respectively. The absolute values of the di®erences in means are computed and their order numbers (from large to small) are stored in the vector o. Next, the names of the ¯ve genes with the largest di®erences in mean are printed to the screen. After saving the script under e.g. the name meandif.R in the directory D:\\Rscripts\\meandif.R, it can be executed by using source("D:\\Rscripts\\meandif.R"). Once the script is available for a typewriter it is easy to adapt it and to re-run it. Readers are strongly recommended to trial-and-error with respect to writ- ing programming scripts. To run these it is very convenient to have your favorite word processor available and to use, for instance, the copy-and-paste functionality. 1.10 Overview and concluding remarks It is easy to install R and Bioconductor. R has many convenient built-in- functions for statistical programming. Help and illustrations on many topics are available from various sources. With the reference charts, R manuals, (on-line) books and R Wiki at hand you have various sources of information to help you along with practical issues. Although there recently became several GUI's available, we shall concentrate on the command line editor because its range of possibilities is much larger. The above introduction is of course very brief. A more extensive in- troduction into R, assuming some background on biomedical statistics, is given by Dalgaard (2002). There are book length treatments combining R with statistics (Venables, & Ripley, 2002; Everitt & Hothorn, 2006). Other treatments go much deeper into programming aspects (Becker, Chambers, & Wilks, 1988; Venables & Ripley, 2000; Gentleman, 2008). For the sake of illustration we shall work frequently with the data kindly provided by Golub et al. (1999) and Chiaretti et al. (2004). The corre-14 CHAPTER 1. BRIEF INTRODUCTION INTO USING R sponding scienti¯c articles are freely available from the web. Having these available may further motivate readers for the computations ahead. 1.11 Exercises 1. Some questions to orientate yourself. (a) Use the function class to ¯nd the class to which the follow- ing objects belong: golub, golub[1,1],golub.cl, golub.gnames, apply, exp, gol.fac, plot, ALL. (b) What is the meaning of the following abbreviations: rm, sum, prod, seq, sd, nrow. (c) For what purpose are the following functions useful: grep, apply, gl, library, source, setwd, history, str. 2. gendat Consider the data in the matrix gendat, constructed in Sec- tion 1.6. Its small size has the advantage that you can check your computations even by a pocket calculator. 14 (a) Use apply to compute the standard deviation of the persons. (b) Use apply to compute the standard deviation of the genes. (c) Order the matrix according to the gene standard deviations. (d) Which gene has the largest standard deviation? 3. Computations on gene means of the Golub data. (a) Use apply to compute the mean gene expression value. (b) Order the data matrix according to the gene means. (c) Give the names of the three genes with the largest mean expression value. (d) Give the biological names of these genes. 4. Computations on gene standard deviations of the Golub data. (a) Use apply to compute the standard deviation per gene. 14Obtaining some routine with the apply functionality is quite helpful for what follows.1.11. EXERCISES 15 (b) Select the expression values of the genes with standard deviation larger than two. (c) How many genes have this property? 5. Oncogenes in Golub data. (a) How many oncogenes are there in the dataset? Hint: Use grep. (b) Find the biological names of the three oncogenes with the largest mean expression value for the ALL patients. (c) Do the same for the AML patients. (d) Write the gene probe ID and the gene names of the ten genes with largest mean gene expression value to a csv ¯le. 6. Constructing a factor. Construct factors that correspond to the follow- ing setting. (a) An experiment with two conditions each with four measurements. (b) Five conditions each with three measurements. (c) Three conditions each with ¯ve measurements. 7. Gene means for B1 patients. Load the ALL data from the ALL library and use str and openVignette() for a further orientation. (a) Use exprs(ALL[,ALL$BT=="B1"] to extract the gene expressions from the patients in disease stage B1. Compute the mean gene expressions over these patients. (b) Give the gene identi¯ers of the three genes with the largest mean.16 CHAPTER 1. BRIEF INTRODUCTION INTO USING RChapter 2 Data Display and Descriptive Statistics A few essential methods are given to display and visualize data. It quickly answers questions like: How are my data distributed? How can the frequen- cies of nucleotides from a gene be visualized? Are there outliers in my data? Does the distribution of my data resemble that of a bell-shaped curve? Are there di®erences between gene expression values taken from two groups of patients? The most important central tendencies (mean, median) are de¯ned and illustrated together with the most important measures of spread (standard deviation, variance, inter quartile range, and median absolute deviation). 2.1 Univariate data display To observe the distribution of data various visualization methods are made available. These are frequently used by practitioners as well as by experts. 2.1.1 Frequency table Discrete data occur when the values naturally fall into categories. A fre- quency table simply gives the number of occurrences within a category. Example 1. A gene consists of a sequence of nucleotides fA;C; G; Tg. The number of each nucleotide can be displayed in a frequency table. This 1718 CHAPTER 2. DATA DISPLAY AND DESCRIPTIVE STATISTICS will be illustrated by the Zyxin gene which plays an important role in cell adhesion (Golub et al., 1999). The accession number (X94991.1) of one of its variants can be found in a data base like NCBI (UniGene). The code below illustrates how to read the sequence "X94991.1" of the species homo sapiens from GenBank, , to construct a pie from a frequency table of the four nucleotides. install.packages(c("ape"),repo="http://cran.r-project.org",dep=TRUE) library(ape) table(read.GenBank(c("X94991.1"),as.character=TRUE)) pie(table(read.GenBank(c("X94991.1")))) From the resulting frequencies in Table 2.1 it seems that the nucleotides are not equally likely. A nice way to visualize a frequency table is by plotting a pie. Table 2.1: A frequency table and its pie of Zyxin gene. A C G T 410 789 573 394 a c g t 2.1.2 Plotting data An elementary method to visualize data is by using a so-called stripchart, by which the values of the data are represented as e.g. small boxes. Often,2.1. UNIVARIATE DATA DISPLAY 19 it is useful in combination with a factor that distinguishes members from di®erent experimental conditions or patients groups. Example 1. Many visualization methods will be illustrated by the Golub et al. (1999) data. We shall concentrate on the expression values of gene "CCND3 Cyclin D3", which are collected in row 1042 of the data matrix golub. To plot the data values one can simply use plot(golub[1042,]). In the resulting plot in Figure 2.1 the vertical axis gives the size of the expression values and the horizontal axis the index of the patients. It can be observed that the values for patient 28 to 38 are somewhat lower, but, indeed, the picture is not very clear because the groups are not plotted separately. To produce two adjacent stripcharts one for the ALL and one for the AML patients, we use the factor called gol.fac from the previous chapter. data(golub, package = "multtest") gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) stripchart(golub[1042,] ~ gol.fac, method="jitter") From the resulting Figure 2.2 it can be observed that the CCND3 Cyclin D3 expression values of the ALL patients tend to have larger expression values than those of the AML patients. 2.1.3 Histogram Another method to visualize data is by dividing the range of data values into a number of intervals and to plot the frequency per interval as a bar. Such a plot is called a histogram. Example 1. A histogram of the expression values of gene "CCND3 Cyclin D3" of the acute lymphoblastic leukemia patients can be produced as follows. > hist(golub[1042, gol.fac=="ALL"]) The function hist divides the data into 5 intervals having width equal to 0.5, see Figure 2.3. Observe from the latter that one value is small and the other are more or less symmetrically distributed around the mean.20 CHAPTER 2. DATA DISPLAY AND DESCRIPTIVE STATISTICS 0 10 20 30 −0.5 0.0 0.5 1.0 1.5 2.0 2.5 Index golub[1042, ] Figure 2.1: Plot of gene ex- pression values of CCND3 Cyclin D3. ALL AML −0.5 0.0 0.5 1.0 1.5 2.0 2.5 Figure 2.2: Stripchart of gene expression values of CCND3 Cyclin D3 for ALL and AML patients. 2.1.4 Boxplot It is always possible to sort n data values to have increasing order x1 · x2 · ¢ ¢ ¢ · xn, where x1 is the smallest, x2 is the ¯rst-to-the smallest, etc. Let x0:25 be a number for which it holds that 25% of the data values x1; ¢ ¢ ¢ ; xn is smaller. That is, 25% of the data values lay on the left side of the number x0:25, reason for which it is called the ¯rst quartile or the 25th percentile. The second quartile is the value x0:50 such that 50% of the data values are smaller. Similarly, the third quartile or 75th percentile is the value x0:75 such that 75% of the data is smaller. A popular method to display data is by drawing a box around the ¯rst and the third quartile (a bold line segment for the median), and the smaller line segments (whiskers) for the smallest and the largest data values. Such a data display is known as a box-and-whisker plot. Example 1. A vector with gene expression values can be put into in- creasing order by the function sort. We shall illustrate this by the ALL2.1. UNIVARIATE DATA DISPLAY 21 expression values of gene "CCND3 Cyclin D3" in row 1042 of golub. > x <- sort(golub[1042, gol.fac=="ALL"], decreasing = FALSE) > x[1:5] [1] 0.458 1.105 1.276 1.326 1.368 The second command prints the ¯rst ¯ve values of the sorted data values to the screen, so that we have x1 = 0:458, x2 = 1:105, etc. Note that the mathematical notation xi corresponds exactly to the R notation x[i] Histogram of golub[1042, gol.fac == "ALL"] golub[1042, gol.fac == "ALL"] Frequency 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0 2 4 6 8 10 12 Figure 2.3: Histogram of ALL ex- pression values of gene CCND3 Cyclin D3. ALL AML −0.5 0.0 0.5 1.0 1.5 2.0 2.5 Figure 2.4: Boxplot of ALL and AML expression values of gene CCND3 Cyclin D3. Example 2. A view on the distribution of the expression values of the ALL and the AML patients on gene CCND3 Cyclin D3 can be obtained by constructing two separate boxplots adjacent to one another. To produce such a plot the factor gol.fac is again very useful. > boxplot(golub[1042,] ~ gol.fac) From the position of the boxes in Figure 2.4 it can be observed that the gene expression values for ALL are larger than those for AML. Furthermore, since the two sub-boxes around the median are more or less equally wide, the data are quite symmetrically distributed around the median.22 CHAPTER 2. DATA DISPLAY AND DESCRIPTIVE STATISTICS To compute exact values for the quartiles we need a sequence running from 0.00 to 1.00 with steps equal to 0.25. To construct such a sequence the function seq is useful. > pvec <- seq(0,1,0.25) > quantile(golub[1042, gol.fac=="ALL"],pvec) 0% 25% 50% 75% 100% 0.458 1.796 1.928 2.179 2.766 The ¯rst quartile x0:25 = 1:796, the second x0:50 = 1:928, and the third x0:75 = 2:179. The smallest observed expression value equals x0:00 = 0:458 and the largest x1:00 = 2:77. The latter can also be obtained by the function min(golub[1042, gol.fac=="ALL"]) and max(golub[1042, gol.fac=="ALL"]), or more brie°y by range(golub[1042, gol.fac=="ALL"]). Outliers are data values laying far apart from the pattern set by the majority of the data values. The implementation in R of the (modi¯ed) boxplot draws such outlier points separately as small circles. A data point x is de¯ned as an outlier point if x < x0:25 ¡ 1:5 ¢ (x0:75 ¡ x0:25) or x > x0:75 + 1:5 ¢ (x0:75 ¡ x0:25): From Figure 2.4 it can be observed that there are outliers among the gene expression values of ALL patients. These are the smaller values 0.45827 and 1.10546, and the largest value 2.76610. The AML expression values have one outlier with value -0.74333. To de¯ne extreme outliers, the factor 1.5 is raised to 3.0. Note that this is a descriptive way of de¯ning outliers instead of statistically testing for the existence of an outlier. 2.1.5 Quantile-Quantile plot A method to visualize the distribution of gene expression values is by the so-called quantile-quantile (Q-Q) plot. In such a plot the quantiles of the gene expression values are displayed against the corresponding quantiles of the normal (bell-shaped). A straight line is added representing points which correspond exactly to the quantiles of the normal distribution. By observing the extent in which the points appear on the line, it can be evaluated to what degree the data are normally distributed. That is, the closer the gene2.1. UNIVARIATE DATA DISPLAY 23 expression values appear to the line, the more likely it is that the data are normally distributed. −2 −1 0 1 2 0.5 1.0 1.5 2.0 2.5 Normal Q−Q Plot Theoretical Quantiles Sample Quantiles Figure 2.5: Q-Q plot of ALL gene expression values of CCND3 Cyclin D3. Example 1. To produce a Q-Q plot of the ALL gene expression values of CCND3 Cyclin D3 one may use the following. qqnorm(golub[1042, gol.fac=="ALL"]) qqline(golub[1042, gol.fac=="ALL"]) From the resulting Figure 2.5 it can be observed that most of the data points are on or near the straight line, while a few others are further away. The above example illustrates a case where the degree of non-normality is moderate so that a clear conclusion cannot be drawn. By making the24 CHAPTER 2. DATA DISPLAY AND DESCRIPTIVE STATISTICS exercises below, the reader will gather more experience with the degree in which gene expression values are normally distributed. 2.2 Descriptive statistics There exist various ways to describe the central tendency as well as the spread of data. In particular, the central tendency can be described by the mean or the median, and the spread by the variance, standard deviation, interquartile range, or median absolute deviation. These will be de¯ned and illustrated. 2.2.1 Measures of central tendency The most important descriptive statistics for central tendency are the mean and the median. The sample mean of the data values x1; ¢ ¢ ¢ ; xn is de¯ned as x = 1nXn i=1 xi = 1n(x1 + ¢ ¢ ¢ + xn): Thus the sample mean is simply the average of the n data values. Since it is the sum of all data values divided by the sample size, a few extreme data values may largely in°uence its size. In other words, the mean is not robust against outliers. The median is de¯ned as the second quartile or the 50th percentile, and is denoted by x0:50. When the data are symmetrically distributed around the mean, then the mean and the median are equal. Since extreme data values do not in°uence the size of the median, it is very robust against outliers. Robustness is important in bioinformatics because data are frequently con- taminated by extreme or otherwise in°uential data values. Example 1. To compute the mean and median of the ALL expression values of gene CCND3 Cyclin D3 consider the following. > mean(golub[1042, gol.fac=="ALL"]) [1] 1.89 > median(golub[1042, gol.fac=="ALL"]) [1] 1.932.2. DESCRIPTIVE STATISTICS 25 Note that the mean and the median do not di®er much so that the distribu- tion seems quite symmetric. 2.2.2 Measures of spread The most important measures of spread are the standard deviation, the in- terquartile range, and the median absolute deviation. The standard deviation is the square root of the sample variance, which is de¯ned as s2 = 1 n ¡ 1Xn i=1 (xi ¡ x)2 = 1 n ¡ 1 ¡(x1 ¡ x)2 + ¢ ¢ ¢ + (xn ¡ x)2¢ : Hence, it is the average of the squared di®erences between the data values and the sample mean. The sample standard deviation s is the square root of the sample variance and may be interpreted as the distance of the data values to the mean. The variance and the standard deviation are not robust against outliers. The interquartile range is de¯ned as the di®erence between the third and the ¯rst quartile, that is x0:75 ¡ x0:25. It can be computed by the function IQR(x). More speci¯cally, the value IQR(x)/1.349 is a robust estimator of the standard deviation. The median absolute deviation (MAD) is de¯ned as a constant times the median of the absolute deviations of the data from the median (e.g. Jure·ckov¶a & Picek, 2006, p. 63). In R it is computed by the function mad de¯ned as the median of the sequence jx1¡x0:50j; ¢ ¢ ¢ ; jxn¡x0:50j multiplied by the constant 1.4826. It equals the standard deviation in case the data come from a bell-shaped (normal) distribution (see Section 3.2.1). Because the interquartile range and the median absolute deviation are based on quantiles, these are robust against outliers. Example 1. These measures of spread for the ALL expression values of gene CCND3 Cyclin D3 can be computed as follows. > sd(golub[1042, gol.fac=="ALL"]) [1] 0.491 > IQR(golub[1042, gol.fac=="ALL"]) / 1.349 [1] 0.284 > mad(golub[1042, gol.fac=="ALL"]) [1] 0.36826 CHAPTER 2. DATA DISPLAY AND DESCRIPTIVE STATISTICS Due to the three outliers (cf. Figure 2.4) the standard deviation is larger than the interquartile range and the mean absolute deviation. That is, the absolute di®erences with respect to the median are somewhat smaller, than the root of the squared di®erences. 2.3 Overview and concluding remarks Data can be stored as a vector or a data matrix on which various useful functions are de¯ned. In particular, it is easy to produce a pie, histogram, boxplot, or Q-Q plot of a vector of data. These plots give a useful ¯rst impression of the degree of (non)normality of gene expression values. To construct the histogram used the default method to compute the num- ber of bars or breaks. If the data are distributed according to a bell-shaped curve, then this is often a good strategy. The number of bars can be chosen by the breaks option of the function hist. Optimal choices for this are dis- cussed by e.g. Venables and Ripley (2002). 2.4 Exercises Since the majority of the exercises are based on the Golub et al. (1999) data, it is essential to make these available and to learn to work with it. To stimulate self-study the answers are given at the end of the book. 1. Illustration of mean and standard deviation. (a) Compute the mean and the standard deviation for 1; 1:5; 2; 2:5; 3. (b) Compute the mean and the standard deviation for 1; 1:5; 2; 2:5; 30. (c) Comment on the di®erences. 2. Comparing normality for two genes. Consider the gene expression val- ues in row 790 and 66 of the Golub et al. (1999) data. (a) Produce a boxplot for the expression values of the ALL patients and comment on the di®erences. Are there outliers?2.4. EXERCISES 27 (b) Produce a QQ-plot and formulate a hypothesis about the normal- ity of the genes. (c) Compute the mean and the median for the expression values of the ALL patients and compare these. Do this for both genes. 3. E®ect size. An important statistic to measure the e®ect size which is de¯ned for a sample as x=s. It measures the mean relative to the standard deviation, so that is value is large when the mean is large and the standard deviation small. (a) Determine the ¯ve genes with the largest e®ect size of the ALL patients from the Golub et al. (1999) data. Comment on their size. (b) Invent a robust variant of the e®ect size and use it to answer the previous question. 4. Plotting gene expressions "CCND3 Cyclin D3". Use the gene expres- sions from "CCND3 Cyclin D3" of Golub et al. (1999) collected in row 1042 of the object golub from the multtest library. After using the function plot you produce an object on which you can program. (a) Produce a so-called stripchart for the gene expressions separately for the ALL as well as for the AML patients. Hint: Use a factor for appropriate separation. (b) Rotate the plot to a vertical position and keep it that way for the questions to come. (c) Color the ALL expressions red and AML blue. Hint: Use the col parameter. (d) Add a title to the plot. Hint: Use title. (e) Change the boxes into stars. Hint: Use the pch parameter. Hint: Store the ¯nal script you like the most in your typewriter in order to be able to use it e±ciently later on. 5. Box-and-Whiskers plot of "CCND3 Cyclin D3". Use the gene expres- sions "CCND3 Cyclin D3" of Golub et al. (1999) from row 1042 of the object golub of the multtest library. (a) Construct the boxplot in Figure 2.6.28 CHAPTER 2. DATA DISPLAY AND DESCRIPTIVE STATISTICS (b) Add text to the plot to explain the meaning of the upper and lower part of the box. (c) Do the same for the wiskers. (d) Export your plot to eps format. Hint 1: Use locator() to ¯nd coordinates of the position of the plot. Hint 2: Use xlim to make the plot somewhat wider. Hint 3: Use arrows to add an arrow. Hint 4: Use text to add information at a certain position. 6. Box-and-wiskers plot of persons of Golub et al. (1999) data. (a) Use boxplot(data.frame(golub)) to produce a box-and-wiskers plot for each column (person). Make a screen shot to save it in a word processor. Describe what you see. Are the medians of similar size? Is the inter quartile range more or less equal. Are there outliers? (b) Compute the mean and medians of the persons. What do you observe? (c) Compute the range (minimal and maximum value) of the standard deviations, the IQR and MAD of the persons. Comment of what you observe. 7. Oncogenes of Golub et al. (1999) data. (a) Select the oncogens by the grep facility and produce a box-and- wiskers plot of the gene expressions of the ALL patients. (b) Do the same for the AML patients and use par(mfrow=c(2,1)) to combine the two plots such that the second is beneath the ¯rst. Are there genes with clear di®erences between the groups? 8. Descriptive statistics for the ALL gene expression values of the Golub et al. (1999) data. (a) Compute the mean and median for gene expression values of the ALL patients, report their range and comment on it. (b) Compute the SD, IQR, and MAD for gene expression values of the ALL patients, report their range and comment on it.2.4. EXERCISES 29 0.5 1.0 1.5 2.0 2.5 Median Outlier Figure 2.6: Boxplot with arrows and explaining text.30 CHAPTER 2. DATA DISPLAY AND DESCRIPTIVE STATISTICSChapter 3 Important Distributions Questions that concern us in this chapter are: What is the probability to ¯nd fourteen purines in a microRNA of length twenty two? If expressions from ALL patients of gene CCND3 Cyclin D3 are normally distributed with mean 1.90 and standard deviation 0.5, what is the probability to observe expression values larger than 2.4? To answer such type of questions we need to know more about statis- tical distributions (e.g. Samuels & Witmer, 2003). In this chapter several important distributions will be de¯ned, explained, and illustrated. In par- ticular, the discrete distribution binomial and the continuous distributions normal, T, F, and chi-squared will be elaborated. These distributions have a wealth of applications to statistically testing biological hypotheses. Only when deemed relevant, the density function, the distribution function, the mean ¹ (mu), and the standard deviation ¾ (sigma), are explicitly de¯ned. 3.1 Discrete distributions The binomial distribution is fundamental and has many applications in medicine and bioinformatics. 3.1.1 Binomial distribution The binomial distribution ¯ts to repeated trials each with a dichotomous out- come such as succes-failure, healthy-disease, heads-tails, purine-pyrimidine, etc. When there are n trials, then the number of ways to obtain k successes 3132 CHAPTER 3. IMPORTANT DISTRIBUTIONS out of n is given by the binomial coe±cient n! k!(n ¡ k)! ; where n! = n ¢ (n ¡ 1) ¢ ¢ ¢ 1 and 0! = 1 (Samuels & Witmer, 2003). The binomial probability of k successes out of n consists of the product of this coe±cient with the probability of k successes and the probability of n ¡ k failures. Let p be the probability of succes in a single trial and X the (random) variable denoting the number of successes. Then the probability P of the event (X = k) that k successes occur out of n trails can be expressed as P(X = k) = n! k!(n ¡ k)!pk(1 ¡ p)n¡k; for k = 0; ¢ ¢ ¢ ; n: (3.1) The collection of these probabilities is called the probability density function. 1Example 1. To visualize the Binomial distribution, load the TeachingDemos package and use the command vis.binom(). Click on "Show Normal Ap- proximation" and observe that the approximation improves as n increases, taking p for instance near 0:5. Example 2. If two carriers of the gen for albinism marry, then each of the children has probability of 1/4 of being albino. What is the probability for one child out of three to be albino? To answer this question we take n = 3, k = 1, and p = 0:25 into Equation (3.1) and obtain P(X = 1) = 3! 1!(3 ¡ 1)!0:2510:752 = 3 ¢ 0:140625 = 0:421875: An elementary manner to compute this in R is by > choose(3,1)* 0.25^1* 0.75^2 where choose(3,1) computes the binomial coe±cient. It is more e±cient to compute this by the built-in-density-function dbinom(k,n,p), for instance to print the values of the probabilities. 1p For a binomially distributed variable np is the mean, np(1 ¡ p) the variance, and np(1 ¡ p) the standard deviation.3.1. DISCRETE DISTRIBUTIONS 33 > for (k in 0:3) print(dbinom(k,3,0.25)) Changing d into p yields the so-called distribution function with the cumula- tive probabilities. That is, the probability that the number of Heads is lower than or equal to two P(X · 2) is computed by pbinom(2,3,0.25). The values of the density and distribution function are summarized in Table 3.1. From the table we read that the probability of no albino child is 0.4218 and the probability that all three children are albino equals 0.0156. Table 3.1: Discrete density and distribution function values of S3, with p = 0:6. number of Heads k = 0 k = 1 k = 2 k = 3 density P(X = k) 0.4218 0.4218 0.1406 0.0156 distribution P(X · k) 0.4218 0.843 0.9843 1 Example 3. RNA consists of a sequence of nucleotides A, G, U, and C, where the ¯rst two are purines and the last two are pyrimidines. Suppose, for the purpose of illustration, that the length of a certain micro RNA is 22, that the probability of a purine equals 0.7, and that the process of placing purines and pyrimidines is binomially distributed. The event that our microRNA contains 14 purines can be represented by X = 14. The probability of this event can be computed by P(X = 14) = 22! 14!(22 ¡ 14)!0:7140:38 = dbinom(14; 22; 0:7) = 0:1423: This is the value of the density function at 14. The probability of the event of less than or equal to 13 purines equals the value of the distribution function at value 13, that is P(X · 13) = pbinom(13; 22; 0:7) = 0:1865: The probability of strictly more than 10 purines is P(X ¸ 11) = X22 k=11 P(S22 = k) = sum(dbinom(11 : 22; 22; 0:7)) = 0:9860: The binomial density function can be plotted by:34 CHAPTER 3. IMPORTANT DISTRIBUTIONS 0 5 10 15 20 0.00 0.05 0.10 0.15 x f(x) Figure 3.1: Binomial probabilities with n = 22 and p = 0:7 0 5 10 15 20 0.0 0.2 0.4 0.6 0.8 1.0 x F(x) Figure 3.2: Binomial cumulative probabilities with n = 22 and p = 0:7. > x <- 0:22 > plot(x,dbinom(x,size=22,prob=.7),type="h") By the ¯rst line the sequence of integers f1; 2; ¢ ¢ ¢ ; 22g is constructed and by the second the density function is plotted, where the argument h speci¯es pins. From Figure 3.1 it can be observed that the largest probabilities oc- cur near the expectation 15:4. The graph in Figure 3.2 illustrates that the distribution is an increasing step function, with x on the horizontal axis and P(X · x) on the vertical. A random sample of size 1000 from the binomial distribution with n = 22 and p = 0:7 can be drawn by the command rbinom(1000,22,0.7). This simulates the number of purines in 1000 microRNA's each with purine prob- ability equal to 0.7 and length 22. 3.2 Continuous distributions The continuous distributions normal, T, F, and chi-squared will be de¯ned, explained and illustrated.3.2. CONTINUOUS DISTRIBUTIONS 35 3.2.1 Normal distribution The normal distribution is of key importance because it is assumed for many (preprocessed) gene expression values. That is, the data values x1; ¢ ¢ ¢ ; xn are seen as realizations of a random variable X having a normal distribution. Equivalently one says that the data values are members of a normally dis- tributed population with mean ¹ (mu) and variance ¾2 (sigma squared). It is good custom to use Greek letters for population properties and N(¹; ¾2) for the normal distribution. The value of the distribution function is given by P(X · x), the probability of the population to have values smaller than or equal to x. Various properties of the normal distribution are illustrated by the examples below. Example 1. To view members of the normal distribution load the TeachingDemos package and give the command vis.normal() to launch an interactive display of bell-shaped curves. These bell-shaped curves are also called normal densities. The curves are symmetric around ¹ and attain a unique maximum at x = ¹. If x moves further away from the mean ¹, then the curves moves to zero so that extreme values occur with small probability. Move the Mean and the Standard Deviation from the left to the right to explore their e®ect on the shape of the normal distribution. In particular, when the mean ¹ increases, then the distribution moves to the right. If ¾ is small/large, then the distribution is steep/°at. Example 2. Suppose that the expression values of gene CCND3 Cyclin D3 can be represented by X which is distributed as N(1:90; 0:52). From the graph of its density function in Figure 3.3, it can be observed that it is symmetric and bell-shaped around ¹ = 1:90. A density function may very well be seen as a histogram with arbitrarily small bars (intervals). The probability that the expression values are less then 1.4 is P(X < 1:4) = pnorm(1:4; 1:9; 0:5) = 0:1586: Figure 3.4 illustrates the value 0:16 of the distribution function at x = 1:4. It corresponds to the area of the blue colored surface below the graph of the density function in Figure 3.3. The probability that the expression values are larger than 2.4 is P(X ¸ 2:4) = 1 ¡ pnorm(2:4; 1:9; 0:5) = 0:1586:36 CHAPTER 3. IMPORTANT DISTRIBUTIONS 0 1 2 3 4 0.0 0.2 0.4 0.6 0.8 x density f(x) 1.4 P(X<=1.4) = 0.16 Figure 3.3: Graph of normal den- sity with mean 1.9 and standard deviation 0.5. 0 1 2 3 4 0.0 0.2 0.4 0.6 0.8 1.0 x normal distribution F(x) 1.4 0.16 Figure 3.4: Graph of normal dis- tribution with mean 1.9 and stan- dard deviation 0.5. The probability that X is between 1.4 and 2.4 equals P(1:4 · X · 2:4) = pnorm(2:4; 1:9; 0:5) ¡ pnorm(1:4; 1:9; 0:5) = 0:9545: The graph of the distribution function in Figure 3.4 illustrates that it is strictly increasing. The exact value for the quantile x0:025 can be computed by > qnorm(0.025,1.9,0.5) [1] 0.920018 That is, the quantile x0:025 = 0:920018. Hence, it holds that the probability of values smaller than 0.920018 equals 0.025, that is P(X · 0:920018) = 0:025, as can be veri¯ed by pnorm(0:920018; 1:9; 0:5). When X is distributed as N(1:90; 0:52), then the population mean is 1:9 and the population standard deviation 0:5. To verify this we draw a random sample of size 1000 from this population by > x <- rnorm(1000,1.9,0.5) The estimate mean(x)=1.8862 and sd(x)=0.5071 are close to their popula- tion values ¹ = 1:9 and ¾ = 0:5. 2 2Use the function round to print the mean in a desired number a decimal places.3.2. CONTINUOUS DISTRIBUTIONS 37 For X distributed as N(¹; ¾2), it holds that (X¡¹)=¾ = Z is distributed as N(0; 1). Thus by subtracting ¹ and dividing the result with ¾ any normally distributed variable can be standardized into a standard normally distributed Z having mean zero and standard deviation one. 3.2.2 Chi-squared distribution The chi-squared distribution plays an important role in testing hypotheses about frequencies, see Chapter 4. To de¯ne it, let fZ1; ¢ ¢ ¢ ;Zmg be indepen- dent and standard normally distributed random variables. Then the sum of squares Â2m = Z21 + ¢ ¢ ¢ + Z2m=Xm i=1 Z2i ; is the so-called chi-squared distributed (random) variable with m degrees of freedom. Example 1. To view various members of the Â2 distribution load the TeachingDemos package. Use the command vis.gamma() to open an inter- active display of various distributions. Click on "Visualizing the gamma", "Visualizing the Chi-squared", and adapt "Xmax". Move the "Shape" but- ton to the right to increase the degrees of freedom. Observe that the graphs of chi-squared densities change from heavily skew to the right into more bell- shaped normal as the degrees of freedom increases. Example 2. Let's consider the chi-squared variable with 5 degrees of freedom; Â25= Z21 + ¢ ¢ ¢ + Z25 . To compute the probability of values smaller than eight we use the function pchisq, as follows. P ¡Â25· 8¢ = pchisq(8; 5) = 0:8437644: This yields the value of the distribution function at x = 8 (see Figure 3.6). This value corresponds to the area of the blue colored surface below the graph of the density function in Figure 3.5. Often we are interested in the value for the quantile x0:025, where P(Â25· x0:025) = 0:025. 3 Such can be computed by3If the distribution function is strictly increasing, then there exists an exact and unique solution for the quantiles.38 CHAPTER 3. IMPORTANT DISTRIBUTIONS 0 5 10 15 20 25 0.00 0.05 0.10 0.15 Chi−Squared Density f(x) 8 area=0.84 Figure 3.5: Â25-density. 0 5 10 15 20 0.0 0.2 0.4 0.6 0.8 1.0 Chi−Squared Distribution F(x) 8 0.84 x Figure 3.6: Â25distribution. > qchisq(0.025, 5, lower.tail=TRUE) [1] 0.8312 Example 3. The chi-squared distribution is frequently used as a so-called goodness of ¯t measure. With respect to the Golub et. al. (1999) data we may hypothesize that the expression values of gene CCND3 Cyclin D3 for the ALL patients are distributed as N(1:90; 0:502). If this indeed holds, then the sum of squared standardized values equals their number and the probability of larger values is about 1/2. In particular, let x1; ¢ ¢ ¢ ; x27 be the gene expression values. Then the standardized values are zi = (xi¡1:90)=0:50 and their sum of squares P27 1 z2i = 25:03312. The probability of larger values is P (Â227 ¸ 25:03312) = 0:5726, which indicates that this normal distribution ¯ts the data well. Hence, it is likely that the speci¯ed normal distribution is indeed correct. Using R the computations are as follows. library(multtest); data(golub) gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) x <- golub[1042,gol.fac=="ALL"] z <- (x-1.90)/0.50 sum(z^2) pchisq(sum(z^2),27, lower.tail=FALSE)3.2. CONTINUOUS DISTRIBUTIONS 39 3.2.3 T-Distribution The T-distribution has many useful applications for testing hypotheses about means of gene expression values, in particular when the sample size is lower than thirty. If the data are normally distributed, then the values of pn(x ¡ ¹)=s follow a T-distribution with n¡1 degrees of freedom. The T-distribution is approximately equal to the normal distribution when the sample size is thirty. −4 −2 0 2 4 0.0 0.1 0.2 0.3 0.4 x−axis Density f(x) Figure 3.7: Density of T10 distri- bution. −4 −2 0 2 4 0.0 0.2 0.4 0.6 0.8 1.0 x−axis Distribution F(x) Figure 3.8: Distribution function of T10. Example 1. Load the TeachingDemos and give vis.t() to explore a vi- sualization of the T-distribution. Click on "Show Normal Distribution" and increase the number of degrees of freedom to verify that df equal to thirty is su±cient for the normal approximation to be quite precise. Example 2. A quick NCBI scan makes it reasonable to assume that the gene Gdf5 has no direct relation with leukemia. For this reason we take ¹ = 0. The expression values of this gene are collected in row 2058 of the golub data. To compute the sample t-value pn(x ¡ ¹)=s use n <- 11 x <- golub[2058, gol.fac=="AML"]40 CHAPTER 3. IMPORTANT DISTRIBUTIONS t.value <- sqrt(n)*(mean(x)-0)/sd(x) t.value [1] 1.236324 From the above we know that this has a T10 distribution. The probability that T10 is greater than 1.236324 can be computed, as follows. P(T10 ¸ 1:236324) = 1 ¡ P(T10 · 1:236324) = 1 ¡ pt(1:236324; 10) = 0:12: This probability corresponds to the area of the blue colored surface below of the graph of the density function in Figure 3.7. The T distribution function with ten degrees of freedom is illustrated in Figure 3.8. The probability that the random variable T10 is between -2 and 2 equals P(¡2 · T11 · 2) = pt(2; 10) ¡ pt(¡2; 10) = 0:926612: The 2.5% quantile can be computed by qt(0.025,n-1)=-2.228139. 3.2.4 F-Distribution The F-distribution is important for testing the equality of two variances. It can be shown that the ratio of variances from two independent sets of nor- mally distributed random variables follows an F-distribution. More speci¯- cally, if the two population variances are equal (¾21 = ¾22), then s21=s22follows an F-distribution with n1 ¡ 1; n2 ¡ 1 degrees of freedom, where s21is the variance of the ¯rst set, s22that of the second, and n1 is the number of ob- servations in the ¯rst and n2 in the second.4 Example 1. For equal population variances the probability is large that that the ratio of sample variances is near one. With respect to the Golub et. al. (1999) data it is easy to compute the ratio of the variances of the expression values of gene CCND3 Cyclin D3 for the ALL patients and the AML patients. > var(golub[1042,gol.fac=="ALL"])/var(golub[1042,gol.fac=="AML"]) [1] 0.7116441 4It is more correct to de¯ne S21=S22 for certain random variables S21 and S22 , we shall , however, not border.3.2. CONTINUOUS DISTRIBUTIONS 41 0 2 4 6 8 10 0.0 0.2 0.4 0.6 0.8 F density f(x) 0.71 x 0.23 Figure 3.9: Density of F26;10. 0 2 4 6 8 10 0.0 0.2 0.4 0.6 0.8 1.0 F Distribution F(x) 0.23 0.71 Figure 3.10: Distribution of F26;10. Since n1 = 27 and n2 = 11 this ratio is a realization of the F26;10 distribution. Then, the probability that the ratio attains values smaller than 0.7116441 is P(X · 0:7116441) = pf(0:7116441; 26; 10) = 0:2326147: Figure 3.9 illustrates that this value corresponds to the area of the blue col- ored surface below the graph of the density function. Figure 3.10 gives the distribution function. To ¯nd the quantile x0:025 use qf(.025,26,10)=0.3861673. This subject is taken further in Section 4.1.5. 3.2.5 Plotting a density function 5 A convenient manner to plot a density function in by using the correspond- ing built-in-function. For instance to plot the bell-shaped density from the normally distributed variable use the function dnorm, as follows. > f<-function(x){dnorm(x,1.9,0.5)} > plot(f,0,4,xlab="x-axis",ylab="density f(x)") 5This subsection is solemly on plotting and can be skipped without loss of continuity.42 CHAPTER 3. IMPORTANT DISTRIBUTIONS This produces the graph of the density function in Figure 3.3. The speci¯ca- tion 0,4 de¯nes the interval on the horizontal axis over which f is plotted. The vertical axis is adapted automatically. We can give the surface under f running x from 0 to 1.4 a nice blue color by using the following. plot(f,0,4,xlab="x-axis",ylab="density f(x)") x<-seq(0,1.4,0.01) polygon(c(0,x,1.4), c(0,f(x),0), col="lightblue") The basic idea of plotting is to start with a plot and next to add colors, text, arrows, etc. In particular, the command polygon is used to give the surface below the graph the color "lightblue". The polygon (surface enclosed by many angles) is de¯ned by the sequence of points de¯ned as x and f(x). 3.3 Overview and concluding remarks For practical computations R has built-in-functions for the binomial, normal, t, F, Â2-distributions, where d stands for density, p for (cumulative) prob- ability distribution, q for quantiles, and r for drawing random samples, see Table 3.2. The density, expectation, and variance of most the distributions in this chapter are summarized in Table 3.3. Table 3.2: Built-in-functions for random variables used in this chapter. para- random Distribution meters density distribution quantiles sampling Bin n; p dbinom(x; n; p) pbinom(x; n; p) qbinom(®; n; p) rbinom(10; n; p) Normal ¹; ¾ dnorm(x; ¹; ¾) pnorm(x; ¹; ¾) qnorm (®; ¹; ¾) rnorm(10; ¹; ¾) Chi-squared m dchisq(x;m) pchisq(x;m) qchisq(®;m) rchisq(10;m) T m dt(x;m) pt(x;m) qt(®;m) rt(10;m) F m,n df(x; m; n) pf(x; m; n) qf(®; m; n) rf(10; m; n) Although for a ¯rst introduction the above distributions are without doubt among the most important, there are several additional distributions available such as the Poisson, Gamma, beta, or Dirichlet. Obviously, these can also be programmed by yourself. The freeware encyclopedia wikipedia of- ten gives a useful ¯rst, though technical, orientation. Note that a distribution acts as a population from which a sample can be drawn. Hence, distributions3.4. EXERCISES 43 can be seen as models of data generating procedures. For a more thorough treatment of distribution we refer the reader to Bain & Engelhardt (1992), Johnson et al. (1992), and Miller & Miller (1999). Table 3.3: Density, mean, and variance of distributions used in this chapter. Distribution parameters density expectation variance Binomial n; p n! k!(n¡k)!pk(1 ¡ p)n¡k np np(1 ¡ p) Normal ¹; ¾ 1 ¾p2¼ exp(¡12 ( x¡¹ ¾ )2) ¹ ¾2 Chi-squared df=m m 2m 3.4 Exercises It is importance to obtain some routine with the computation of probabilities and quantiles. 1. Binomial Let X be binomially distributed with n = 60 and p = 0:4. Compute the following. (a) P(X = 24), P(X · 24), and P(X ¸ 30). (b) P(20 · X · 30), P(20 · X). (c) P(20 · X or X ¸ 40), and P(20 · X and X ¸ 10). (d) Compute the mean and standard deviation of X. (e) The quantiles x0:025, x0:5, and x0:975. 2. Standard Normal. Compute the following probabilities and quantiles. (a) P(1:6 < Z < 2:3). (b) P(Z < 1:64). (c) P(¡1:64 < Z < ¡1:02). (d) P(0 < Z < 1:96). (e) P(¡1:96 < Z < 1:96). (f) The quantiles z0:025, z0:05, z0:5, z0:95, and z0:975. 3. Normal. Compute for X distributed as N(10; 2) the following proba- bilities and quantiles.44 CHAPTER 3. IMPORTANT DISTRIBUTIONS (a) P(X < 12). (b) P(X > 8). (c) P(9 < X < 10; 5). (d) The quantiles x0:025, x0:5, and x0:975. 4. T-distribution. Verify the following computations for the T6 distribu- tion. (a) P(T6 < 1). (b) P(T6 > 2). (c) P(¡1 < T6 < 1). (d) P(¡2 < T6 < ¡2). (e) The quantiles t0:025, t0:5, and t0:975. 5. F distribution. Compute the following probabilities and quantiles for the F8;5 distribution. (a) P(F8;5 < 3). (b) P(F8;5 > 4). (c) P(1 < F8;5 < 6). (d) The quantiles f0:025, f0:5, and f0:975. 6. Chi-squared distribution. Compute the following for the chi-squared distribution with 10 degrees of freedom. (a) P(Â210 < 3). (b) P(Â210 > 4). (c) P(1 < Â210 < 6). (d) The quantiles g0:025, g0:5, and g0:975. 7. MicroRNA. Suppose that for certain microRNA of size 20 the proba- bility of a purine is binomially distributed with probability 0.7. (a) What is the probability of 14 purines? (b) What is the probability of less than or equal to 14 purines?3.4. EXERCISES 45 (c) What is the probability of strictly more than 10 purines? (d) By what probability is of the number of purines between 10 and 15? (e) How many purines do you expect? In other words: What is the mean of the distribution? (f) What is the standard deviation of the distribution? 8. Zyxin. The distribution of the expression values of the ALL patients on the Zyxin gene are distributed according to N(1:6; 0:42). (a) Compute the probability that the expression values are smaller than 1.2? (b) What is the probability that the expression values are between 1.2 and 2.0? (c) What is the probability that the expression values are between 0.8 and 2.4? (d) Compute the exact values for the quantiles x0:025 and x0:975. (e) Use rnorm to draw a sample of size 1000 from the population and compare the sample mean and standard deviation with that of the population. 9. Some computations on Golub et al. (1999) data. (a) Take ¹ = 0 and compute the t-values for the ALL gene expression values. Find the three genes with largest absolute t-values. (b) Compute per gene the ratio of the variances for the ALL and the AML patients. How many are between 0.5 and 1.5? 10. Extreme value investigation. This (di±cult!) question aims to teach the essence of an extreme value distribution! An interesting extreme value distribution is given by Pevsner (2003, p.103). Take the maximum of a sample (with size 1000) from the standard normal distribution and repeat this 1000 times. So that you sampled 1000 maxima. Next, subtract from these maxima an and divide by bn, where an <- sqrt(2*log(n)) - 0.5*(log(log(n))+log(4*pi))*(2*log(n))^(-1/2) bn <- (2*log(n))^(-1/2)46 CHAPTER 3. IMPORTANT DISTRIBUTIONS Now plot the density from the normalized maxima and add the extreme value function f(x) from Pevsner his book, and add the density (dnorm) from the normal distribution. What do you observe?Chapter 4 Estimation and Inference Questions that we deal with in this chapter are related to statistically testing biological hypothesis. Does the mean gene expression over ALL patients di®er from that over AML patients? That is, does the mean gene expression level di®er between experimental conditions? Is the mean gene expression di®erent from zero? To what extent are gene expression values normally distributed? Are there outliers among a sample of gene expression values? How can an experimental e®ect be de¯ned? How can genes be selected with respect to an experimental e®ect? Other important questions are: How can it be tested whether the frequencies of nucleotide sequences of two genes are di®erent? How can it be tested whether outliers are present in the data? What is the probability of a certain micro RNA to have more than a certain number of purines? In the foregoing chapters many population parameters were used to de¯ne families of theoretical distributions. In any research (empirical) setting the speci¯c values of such parameters are unknown so that these must be esti- mated. Once estimates are available it becomes possible to statistically test biologically important hypotheses. The current chapter gives several basic examples of statistical testing and some of its background. Robust type of testing is brie°y introduced as well as an outlier test. 4.1 Statistical hypothesis testing Let ¹0 be a number representing the hypothesized population mean by a researcher on the basis of experience and knowledge from the ¯eld. With 4748 CHAPTER 4. ESTIMATION AND INFERENCE respect to the population mean the null hypothesis can be formulated as H0 : ¹ = ¹0 and the alternative hypothesis as H1 : ¹ 6= ¹0. These are two statements of which the latter is the opposite of the ¯rst: Either H0 or H1 is true. The alternative hypothesis is true if H1 : ¹ < ¹0 or H1 : ¹ > ¹0 holds true. This type of alternative hypothesis is called \two-sided". In case H1 : ¹ > ¹0, it is called \one-sided". Such a null hypothesis will be statistically tested against the alternative using a suitable distribution of a statistic (e.g. standardized mean). After conducting the experiment, the value of the statistic can be computed from the data. By comparing the value of the statistic with its distribution, the researcher draws a conclusion with respect to the null hypothesis: H0 is rejected or it is not. The probability to reject H0, given the truth of H0, is called the signi¯cance level which is generally denoted by ®. We shall follow the habit in statistics to use ® = 0:05, but it will be completely clear how to adapt the procedure in case other signi¯cance levels are desired. 4.1.1 The Z-test The Z-test applies to the situation where we want to test H0 : ¹ = ¹0 against H1 : ¹ 6= ¹0 and the standard deviation ¾ is known. Assuming that the gene expression values (x1; ¢ ¢ ¢ ; xn) are from a normal distribution we compute the standardized value z = pn(x ¡ ¹0)=¾. Next we de¯ne the so-called p- value as the standard normal probability of Z attaining values being more extreme than jzj, that is occurring to the left of ¡jzj or to the right of jzj.1 Accordingly, the p-value equals P(Z · ¡jzj) + P(Z ¸ jzj) = 2 ¢ P(Z · ¡jzj): The conclusion from the test is now as follows: If the p-value is larger than the signi¯cance level ®, then H0 is not rejected and if it is smaller than the signi¯cance level, then H0 is rejected. Example 1. To illustrate the Z-test we shall concentrate on the Gdf5 gene from the Golub et al. (1999) data2. The corresponding expression values are contained in row 2058. A quick search through the NCBI site 1Recall from a calculus course that j ¡ 2j = 2 and j2j = 2. 2We will work with golub throughout this chapter, so it is essential to load these data and to de¯ne the factor gol.fac.4.1. STATISTICAL HYPOTHESIS TESTING 49 makes it likely that this gene is not directly related to leukemia. Hence, we may hypothesize that the population mean of the ALL expression values equals zero. Accordingly, we test H0 : ¹ = 0 against H1 : ¹ 6= 0. For the sake of illustration we shall pretend that the standard deviation ¾ is known to be equal to 0.25. The z-value (=0.001116211) can be computed as follows. data(golub, package = "multtest") gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) sigma <- 0.25; n <- 27; mu0 <- 0 x <- golub[2058,gol.fac=="ALL"] z.value <- sqrt(n)*(mean(x) - mu0)/sigma The p-value can now be computed as follows. > 2*pnorm(-abs(z.value),0,1) [1] 0.9991094 Since it is clearly larger than 0:05, we conclude that the null hypothesis of mean equal to zero is not rejected (accepted). Note that the above procedure implies rejection of the null hypothesis when z is highly negative or highly positive. More precisely, if z falls in the region (¡1; z0:025] or [z0:975;1), then H0 is rejected. For this reason these intervals are called \rejection regions". If z falls in the interval (z0:025; z0:975), then H0 is not rejected and consequently this region is called "acceptance region". The situation is illustrated in Figure 4.1. The interval (z0:025; z0:975) is often named \con¯dence interval", because if the null hypothesis is true, then we are 95% con¯dent that the observed z-value falls in it. It is custom to rework the con¯dence interval into an interval with respect to ¹ (Samuels & Witmer, 2003, p. 186). In particular, the 95% con¯dence interval for the population mean ¹ is µx + z0:025 ¾ pn; x + z0:975 ¾ pn¶: (4.1) That is, we are 95% certain3 that the true mean falls in the con¯dence inter- val. Such an interval is standard output of statistical software. 3If we would repeat the procedure su±ciently often50 CHAPTER 4. ESTIMATION AND INFERENCE −4 −2 0 2 4 0.0 0.1 0.2 0.3 0.4 normal density z rejection area a 2 rejection area a 2 acceptance area 1 -a Figure 4.1: Acceptance and rejection regions of the Z-test. Example 2. Using the data from Example 1, the 95% con¯dence interval given by Equation 4.1 can be computed as follows.4 > mean(x)+qnorm(c(0.025),0,1)*sigma/sqrt(n) [1] -0.0942451 > mean(x)+qnorm(c(0.975),0,1)*sigma/sqrt(n) [1] 0.09435251 Hence, the rounded estimated 95% con¯dence interval is (¡0:094; 0:094). Since ¹0 = 0 falls within this interval, H0 is not rejected. It is instructive and convenient to run the Z-test from the TeachingDemos package, as follows. 4These computations only work together with those of Example 1, especially the de¯- nition of x.4.1. STATISTICAL HYPOTHESIS TESTING 51 > library(TeachingDemos) > z.test(x,mu=0,sd=0.25) One Sample z-test data: x z = 0.0011, n = 27.000, Std. Dev. = 0.250, Std. Dev. of the sample mean = 0.048, p-value = 0.9991 alternative hypothesis: true mean is not equal to 0 95 percent confidence interval: -0.09424511 0.09435251 sample estimates: mean of x 5.37037e-05 From the z-value, the p-value, and the con¯dence interval, the conclusion is not to reject the null-hypothesis of mean equal to zero. This illustrates that testing by either of these procedures yields equivalent conclusions. Example 3. To develop intuition with respect to con¯dence intervals load the package TeachingDemos and give the following command. > ci.examp(mean.sim =0, sd = 1, n = 25, reps = 100, + method = "z", lower.conf=0.025, upper.conf=0.975) Then 100 samples of size 25 from the N(0; 1) distribution are drawn and for each of these the con¯dence interval for the population mean is computed and represented as a line segment. Apart from sampling °uctuations, the con¯dence level corresponds to the percentage of intervals containing the true mean (colored in black) and that the signi¯cance level corresponds to intervals not containing it (colored in red or blue). 4.1.2 One Sample t-Test Indeed, in almost all research situations with respect to gene expression val- ues, the population standard deviation ¾ is unknown so that the above test is not applicable. In such cases t-tests are very useful for testing H0 : ¹ = ¹0 against H1 : ¹ 6= ¹0. The test is based on the t-value de¯ned by t =52 CHAPTER 4. ESTIMATION AND INFERENCE pn(x ¡ ¹0)=s. The corresponding p-value is de¯ned by 2 ¢ P(Tn¡1 · ¡jtj). Similar to the above, H0 is not rejected if the p-value is larger than the signif- icance level and H0 is rejected if the p-value is smaller than the signi¯cance level. Equivalently, if t falls in the acceptance region (t0:025;n¡1; t0:975;n¡1), then H0 is not rejected and otherwise it is. For n = 6 the acceptance and rejection regions are illustrated in Figure 4.2. The 95% con¯dence interval for the population mean is given by (x+t0:025 ¢ s=pn; x+t0:975 ¢ s=pn), where the expression s=pn gives the so-called \standard error of the mean". −4 −2 0 2 4 0.0 0.1 0.2 0.3 x−axis T densityrejection region a 2 rejection region a 2 acceptance region t0.025 t0.975 Figure 4.2: Acceptance and rejection regions of the T5-test. Example 1. Let's test H0 : ¹ = 0 against H1 : ¹ 6= 0 for the ALL population mean of the Gdf5 gene expressions. The latter are collected in row 2058 of the golub data. The t-value is computed as follows.4.1. STATISTICAL HYPOTHESIS TESTING 53 > x <- golub[2058,gol.fac=="ALL"]; mu0 <- 0; n <- 27 > t.value<-sqrt(n)*(mean(x) - mu0)/sd(x) > t.value [1] 0.001076867 The corresponding p-value can be computed by 2 ¢ P(T26 · ¡0:0010) = 2 ¤ pt(¡0:0010; 26) = 0:9991 > ®; so that the conclusion is not to reject the null hypothesis of mean equal to zero. To see whether the observed t-value belongs to the 95% con¯dence inter- val, we compute (t0:025;26; t0:975;26) = (qt(0:025; n ¡ 1); qt(0:975; n ¡ 1)) = (¡2:055; 2:055): Since this interval does contain the t-value, we do not reject the hypothesis that ¹ equals zero. The left boundary of the 95% con¯dence interval for the population mean can be computed, as follows. > mean(x)+qt(0.025,26)*sd(x)/sqrt(n) [1] -0.1024562 The 95% con¯dence interval equals (¡0:1025; 0:1025). Since it contains zero, we do not reject the null-hypothesis. In daily practice it is much more convenient to use the built-in-function t.test. We illustrate it with the current testing problem. > t.test(x,mu=0) One Sample t-test data: x t = 0.0011, df = 26, p-value = 0.9991 alternative hypothesis: true mean is not equal to 0 95 percent confidence interval: -0.1024562 0.1025636 sample estimates: mean of x 5.37037e-0554 CHAPTER 4. ESTIMATION AND INFERENCE This yields by one command line the observed t-value, the p-value, and the 95% con¯dence interval for ¹0. In the previous example the test is two-sided because H1 holds true if ¹ < ¹0 or ¹ > ¹0. If, however, the researcher desires to test H0 : ¹ = ¹0 against H1 : ¹ > ¹0, then the alternative hypothesis is one-sided and this makes the procedure slightly di®erent: H0 is accepted if P(Tn ¸ t) > ® and it is rejected if P(Tn ¸ t) < ®. We shall illustrate this by a variant of the previous example. Example 2. In Chapter 2 a box-and-whiskers plot revealed that the ALL gene expression values of CCND3 Cyclin D3 are positive. Hence, we test H0 : ¹ = 0 against H1 : ¹ > 0 by the built-in-function t-test. Recall that the corresponding gene expression values are collected in row 1042 of the golub data matrix (load it if necessary). > t.test(golub[1042,gol.fac=="ALL"],mu=0, alternative = c("greater")) One Sample t-test data: golub[1042, gol.fac == "ALL"] t = 20.0599, df = 26, p-value < 2.2e-16 alternative hypothesis: true mean is greater than 0 95 percent confidence interval: 1.732853 Inf sample estimates: mean of x 1.893883 The large t-value indicates that, relative to its standard error, the mean dif- fers largely from zero. Accordingly, the p-value is very close to zero, so that the conclusion is to reject H0. 4.1.3 Two-sample t-test with unequal variances Suppose that gene expression data from two groups of patients (experimen- tal conditions) are available and that the hypothesis is about the di®erence between the population means ¹1 and ¹2. In particular, H0 : ¹1 = ¹2 is to4.1. STATISTICAL HYPOTHESIS TESTING 55 be tested against H1 : ¹1 6= ¹2. These hypotheses can also be formulated as H0 : ¹1 ¡ ¹2 = 0 and H1 : ¹1 ¡ ¹2 6= 0. Suppose that gene expression data from the ¯rst group are given by fx1; ¢ ¢ ¢ xng and that of the second by fy1; ¢ ¢ ¢ ; ymg. Let x be the mean of the ¯rst and y that of the second, and s21 the variance of the ¯rst and s22that of the second. Then the t-statistic can be formulated as t = (x ¡ y) ¡ (¹1 ¡ ¹2) ps21=n + s22=m : (4.2) The decision procedure with respect to the null-hypothesis is completely sim- ilar to the above tests. Note that the t-value is large if the di®erence between x and y is large5, the standard deviations s1 and s2 are small, and the sample sizes are large. This test is known as the Welch two-sample t-test (Lehmann, 1999). Example 1. Golub et al. (1999) argue that gene CCND3 Cyclin D3 plays an important role with respect to discriminating ALL from AML patients. The boxplot in Figure 2.4 suggests that the ALL population mean di®ers from that of AML. The null hypothesis of equal means can be tested by the func- tion t.test and the appropriate factor and speci¯cation var.equal=FALSE. > t.test(golub[1042,] ~ gol.fac, var.equal=FALSE) Welch Two Sample t-test data: golub[1042, ] by gol.fac t = 6.3186, df = 16.118, p-value = 9.87e-06 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.8363826 1.6802008 sample estimates: mean in group ALL mean in group AML 1.8938826 0.6355909 The t-value is quite large, indicating that the two means x and y di®er largely from zero relative to the corresponding standard error (denominator in Equa- tion 4.2). Since the p-value is extremely small, the conclusion is to reject the null-hypothesis of equal means. The data provide strong evidence that the 5 Assuming ¹1 ¡ ¹2 = 0.56 CHAPTER 4. ESTIMATION AND INFERENCE population means do di®er. When the ¯rst group is an experimental group and the second a control group, then ¹1¡¹2 is the experimental e®ect in the population and x¡y that in the sample. The t-value is the experimental e®ect in the sample relative to the standard error. The size of the e®ect is measured by the p-value in the sense that it is smaller for larger e®ects. If the two population variances are equal, then the testing procedure simpli¯es considerably. This is the subject of the next paragraph. 4.1.4 Two sample t-test with equal variances Suppose exactly the same setting as in the previous paragraph, but now the variances ¾21 and ¾22 for the two groups are known to be equal. To test H0 : ¹1 = ¹2 against H1 : ¹1 6= ¹2, there is a t-test which is based on the so-called pooled sample variance s2p. The latter is de¯ned by the following weighted sum of the sample variances s21and s22, namely s2p = (n ¡ 1)s21+ (m ¡ 1)s22 n + m ¡ 2 : Then the t-value can be formulated as t = x ¡ y ¡ (¹1 ¡ ¹2) spq1n + 1m : Example 1. The null hypothesis for gene CCND3 Cyclin D3 that the mean of the ALL di®ers from that of AML patients can be tested by the two-sample t-test using the speci¯cation var.equal=TRUE. > t.test(golub[1042,] ~ gol.fac, var.equal = TRUE) Two Sample t-test data: golub[1042, ] by gol.fac t = 6.7983, df = 36, p-value = 6.046e-08 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.8829143 1.63366904.1. STATISTICAL HYPOTHESIS TESTING 57 sample estimates: mean in group ALL mean in group AML 1.8938826 0.6355909 From the p-value 6:046 ¢ 10¡8, the conclusion is to reject the null hypothesis of equal population means. Note that the p-value is slightly smaller than that of the previous test. In case of any uncertainty about the validity of the assumption of equal population variances, one may want to test this. 4.1.5 F-test on equal variances The assumption of the above t-test it that the two population variances are equal. Such an assumption can serve as a null hypothesis. That is, we desire to test H0 : ¾21 = ¾22 against H0 : ¾21 6= ¾22. This can be accomplished by the so-called F-test, as follows. From the sample variances s21and s22, the f-value f = s21=s22can be computed, which is Fn1¡1;n2¡1 distributed with n1 ¡ 1 and n2 ¡ 1 degrees of freedom. If P(Fn1¡1;n2¡1 < f) ¸ ®=2 for f < 1 or P(Fn1¡1;n2¡1 > f) ¸ ®=2 for f > 1, then H0 is not rejected and otherwise it is rejected. Example 1. The null hypothesis for gene CCND3 Cyclin D3 that the variance of the ALL patients equals that of the AML patients can be tested by the built-in-function var.test, as follows. > var.test(golub[1042,] ~ gol.fac) F test to compare two variances data: golub[1042, ] by gol.fac F = 0.7116, num df = 26, denom df = 10, p-value = 0.4652 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0.2127735 1.8428387 sample estimates: ratio of variances 0.711644158 CHAPTER 4. ESTIMATION AND INFERENCE From the p-value 0.4652, the null-hypothesis of equal variances is not re- jected. 4.1.6 Binomial test Suppose that for a certain micro RNA a researcher wants to test the hy- pothesis that the probability of a purine equals a certain value p0. However, another researcher has reason to believe that this probability is larger. In such a setting we want to test the null-hypothesis H0 : p = p0 against the one-sided alternative hypothesis H1 : p > p0. Suppose that sequencing re- veals that the micro RNA has k purines out of a total n. Assuming that the binomial distribution holds, the null-hypothesis can be tested by computing the p-value P(X ¸ k). If it is larger than the signi¯cance level ® = 0:05, then H0 is not rejected and otherwise it is. Example 1. A micro RNA of length 22 contains 18 purines. The null hypothesis H0 : p = 0:7 is to be tested against the one-sided H1 : p > 0:7. From P(X ¸ 18) = 1 ¡ pbinom(17; 22; 0:7) = 0:1645 ¸ 0:05 = ®; the conclusion follows not to reject the null-hypothesis. This test can also be conducted by the function binom.test as follows. > binom.test(18, 22, p = 0.7, alternative = c("greater"), + conf.level = 0.95) Exact binomial test data: 18 and 22 number of successes = 18, number of trials = 22, p-value = 0.1645 alternative hypothesis: true probability of success is greater than 0.7 95 percent confidence interval: 0.6309089 1.0000000 sample estimates: probability of success 0.8181818 The p-value 0.1645 is larger than the signi¯cance level 0.05, so that the null hypothesis is not rejected.4.1. STATISTICAL HYPOTHESIS TESTING 59 4.1.7 Chi-squared test It often happens that we want to test a hypothesis with respect to more than one probability. That is, the H0 : (¼1; ¢ ¢ ¢ ; ¼m) = (p1; ¢ ¢ ¢ ; pm) against H1 : (¼1; ¢ ¢ ¢ ; ¼m) 6= (p1; ¢ ¢ ¢ ; pm), where p1 to pm are given numbers corresponding to the hypothesis of a researcher. By multiplying the probabilities with the total number of observations we obtain the expected number of observations (ei = n ¢ pi). Now we can compute the statistic q = Pmi=1(oi ¡ ei)2=ei, where oi is the i-th observed and ei the i-th expected frequency. This statistic is chi-squared (Â2m¡1) distributed with m ¡ 1 degrees of freedom. The p-value of the chi-squared test is de¯ned as P(Â2m¡1 ¸ q). If it is larger than the signi¯cance level, then the null hypothesis is not rejected, and otherwise it is. 0 5 10 15 20 25 0.00 0.05 0.10 0.15 0.20 0.25 Chi−Squared Density f(x) rejection region acceptance region 7.8 q Figure 4.3: Rejection region of Â23-test.60 CHAPTER 4. ESTIMATION AND INFERENCE Example 1. Suppose we want to test the hypothesis that the nucleotides of Zyxin have equal probability. Let the probability of fA;C; G; Tg to occur in the sequence be given by (¼1; ¼2; ¼3; ¼4). Then the null hypothesis to be tested is (¼1; ¼2; ¼3; ¼4) = (1=4; 1=4; 1=4; 1=4). In particular, for the sequence "X94991.1" from Table 1.1 the total number of nucleotides is n = 2166, so that the expected frequencies ei are equal to 2166=4 = 541:5. Then, the q-value equals P4i=1(oi ¡ ei)2=ei = (410 ¡ 541:5)2 541:5 +(789 ¡ 541:5)2 541:5 +(573 ¡ 541:5)2 541:5 +(394 ¡ 541:5)2 541:5 = 187:0674 Since, P(Â2[3] ¸ 187:0674) is close to zero, the null hypothesis is clearly rejected. The nucleotides of Zyxin do not occur with equal probability. A more direct manner to perform the test is by using the built-in-function chisq.test, as follows. > library(ape) > zyxinfreq <- table(read.GenBank(c("X94991.1"),as.character=TRUE)) > chisq.test(zyxinfreq) Chi-squared test for given probabilities data: zyxinfreq X-squared = 187.0674, df = 3, p-value < 2.2e-16 The package ape is loaded, the Zyxin sequence "X94991.1" is downloaded, and the frequency table is constructed. The observed frequencies are given as input to chisq.test which has equal probabilities as the default option. The q-value equals X-squared and the degrees of freedom df = 3. From the corresponding p-value, the conclusion is to reject the null hypothesis of equal probabilities. The testing situation is illustrated in Figure 4.3, where the red colored surface corresponds to the rejection region (7:81;1). Remember from the previous chapter that the left bound of this rejection interval can by found by qchisq(0.95, 3). The observed q = 187:0674 obviously falls far into the right hand side of the rejection region, so that the corresponding p-value is very close to zero. Example 2. In the year 1866 Mendel observed in large number of exper- iments frequencies of characteristics of di®erent kinds of seed and their o®- spring. In particular, this yielded the frequencies 5474, 1850 the seed shape4.1. STATISTICAL HYPOTHESIS TESTING 61 of ornamental sweet peas. A crossing of B and b yields o® spring BB, Bb and bb with probability 0.25, 0.50, 0.25. Since Mendel could not distinguish Bb from BB, his observations theoretically occur with probability 0.75 (BB and Bb) and 0.25 (bb). To test the null hypothesis H0 : (¼1; ¼2) = (0:75; 0:25) against H1 : (¼1; ¼2) 6= (0:75; 0:25), we use the chi-squared test6, as follows. > pi <- c(0.75,0.25) > x <-c(5474, 1850) > chisq.test(x, p=pi) Chi-squared test for given probabilities data: x X-squared = 0.2629, df = 1, p-value = 0.6081 From the p-value 0.6081, we do not reject the null hypothesis. To further illustrate the great °exibility of the chi-squared test another example is given. Example 3. Given certain expression values for a healthy control group and an experimental group with a disease, we may de¯ne a certain cut o® value and classify e.g. smaller values to be healthy and larger ones to be infected. In such a manner cut-o® values can serve as a diagnostic instru- ment. The classi¯cation yields true positives (correctly predicted disease), false positives (incorrectly predicted disease), true negatives (correctly pre- dicted healthy) and false negatives (incorrectly predicted healty). For the sake of illustration suppose that among twenty patients there are 5 true pos- itives (tp), 5 false positives (fp), 5 true negatives (tn), and 5 false negatives (fn). These frequencies can be put is a two-by-two table giving the frequen- cies on two random variables: the true state of the persons and the predicted state of the persons (by the cut o® value). In the worst case the prediction by the cut-o® value is independent of the disease state of the patient. The null hypothesis of independence, can be tested by a chi-square test, as follows. > dat <- matrix(c(5,5,5,5),2,byrow=TRUE) > chisq.test(dat) 6For the sake of clarity the code is somewhat unelegant in using the symbol pi, the constant representing the ratio of a circle's circumference to its diameter.62 CHAPTER 4. ESTIMATION AND INFERENCE Pearson's Chi-squared test with Yates' continuity correction data: dat X-squared = 0.2, df = 1, p-value = 0.6547 Since the p-value is larger than the signi¯cance level, the null hypothesis of independence is not rejected. Suppose that for another cuto® value we obtain 8 true positives (tp), 2 false positives (fp), 8 true negatives (tn), and 2 false negatives (fn). Then testing independence yields the following. > dat <- matrix(c(8,2,2,8),2,byrow=TRUE) > chisq.test(dat) Pearson's Chi-squared test with Yates' continuity correction data: dat X-squared = 5, df = 1, p-value = 0.02535 Since the p-value is smaller than the signi¯cance level, the null hypothesis of independence is rejected. signi¯cant non-signi¯cant genes genes Chromosome 1 100 2000 genome 300 6000 Example 4. A related and frequently applied test in Bioinformatics is the Fisher exact test. In a two by two table with frequencies f11, f22, (f12, and f21), this test is based on the so-called odds ratio f11f22=(f12f21). Suppose that the number of signi¯cant onco type of genes in Chromosome 1 is f11 = 100 out of a total of f12 = 2000 and the number of signi¯cant genes in the whole genome is f21 = 300 out of a total of f22 = 6000. Then the odds ratio equals 100 ¢ 6000=(2000 ¢ 300) = 1 and the number of signi¯cant oncogenes in Chromosome 1 is exactly proportional to that in the genome. The null-hypothesis of the Fisher test is that the odds ratio equals 1 and the alternative hypothesis that it di®ers from 1. Suppose that the frequencies4.1. STATISTICAL HYPOTHESIS TESTING 63 of signi¯cant oncogenes for Chromosome 1 equals f11 = 300 out of a total of f12 = 500 and for the genome f21 = 3000 out of f22 = 6000. The hypothesis that the odd ratio equals one can now be tested as follows. > dat <- matrix(c(300,500,3000,6000),2,byrow=TRUE) > fisher.test(dat) Fisher's Exact Test for Count Data data: dat p-value = 0.01912 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 1.029519 1.396922 sample estimates: odds ratio 1.199960 Since the p-value is smaller than the signi¯cance level, the null hypothesis of odds ratio equal to one is rejected. There are more signi¯cant oncogenes in Chromosome 1 compared to that in the genome. Other examples of the Fisher test will be given in Chapter 6. 4.1.8 Normality tests Various procedures are available to test the hypothesis that a data set is normally distributed. The Shapiro-Wilk test is based on the degree of lin- earity in a Q-Q plot (Lehmann, 1999, p.347) and the Anderson-Darling test is based on the distribution of the data (Stephens, 1986, p.372). Example 1. To test the hypothesis that the ALL gene expression values of CCND3 Cyclin D3 from Golub et al. (1999) are normally distributed, the Shapiro-Wilk test can be used as follows. > shapiro.test(golub[1042, gol.fac=="ALL"]) Shapiro-Wilk normality test64 CHAPTER 4. ESTIMATION AND INFERENCE data: golub[1042, gol.fac == "ALL"] W = 0.947, p-value = 0.1774 Since the p-value is greater than 0.05, the conclusion is not to reject the null hypothesis that CCND3 Cyclin D3 expression values follow from a normal distribution. The Anderson-Darling test is part of the nortest package which probably needs to be installed and loaded ¯rst. Running the test on our CCND3 Cyclin D3 gene expression values comes down to the following. > library(nortest) > ad.test(golub[1042,gol.fac=="ALL"]) Anderson-Darling normality test data: scale(golub[1042, gol.fac == "ALL"]) A = 0.5215, p-value = 0.1683 Hence, the same conclusion is drawn as from the Shapiro-Wilk test. Note that the p-values from both tests are somewhat low. This con¯rms our obser- vation in Section 2.1.5 based on the Q-Q plot that the distribution resembles the normal. From the normality tests the conclusion is that the di®erences in the left tail are not large enough to reject the null-hypothesis that the CCND3 Cyclin D3 expression values are normally distributed. 4.1.9 Outliers test When gene expression values are not normally distributed, then outliers may appear with large probability. The appearance of outliers in gene expression data may in°uence the value of a (non-robust) statistic to a large extent. For this reason it is useful to be able to test whether a certain set of gene expression values is contaminated by an outlier or not. Accordingly, the null-hypothesis to be tested is that a set of gene expression values does not contain an outlier and the alternative is that it is contaminated with at least one outlier. Under the assumption that the data are realizations of one and the same distribution, such a hypothesis can be tested by the Grubbs (1950) test. This test is based on the statistic g = jsuspect value ¡ xj=s, where the suspect value is included for the computation of the mean x and the standard4.1. STATISTICAL HYPOTHESIS TESTING 65 deviation s. Example 1. From Figure 2.4 we have observed that expression values of gene CCND3 Cyclin D3 may contain outliers with respect to the left tail. This can actually be tested by the function grubbs.test of the outliers package, as follows. > library(outliers) > grubbs.test(golub[1042, gol.fac=="ALL"]) Grubbs test for one outlier data: golub[1042, gol.fac == "ALL"] G = 2.9264, U = 0.6580, p-value = 0.0183 alternative hypothesis: lowest value 0.45827 is an outlier Since the p-value is smaller than 0.05, the conclusion is to reject the null- hypothesis of no outliers. In case the data are normally distributed, the probability of outliers is small. Hence, extreme outliers indicate that the data are non-normally dis- tributed with large probability. Outliers may lead to such an increase of the standard error that a true experimental e®ect remains uncovered (false negatives). In such cases a robust test based on ranks may be preferred as a useful alternative. 4.1.10 Wilcoxon rank test In case the data are normally distributed with equal variance, the t-test is an optimal test for testing H0 : ¹1 = ¹2 against H1 : ¹1 6= ¹2 (Lehmann, 1999). If, however, the data are not normally distributed due to skewness or otherwise heavy tails, then this optimality does not hold anymore and there is no guarantee that the signi¯cance level of the test equals the intended level ® (Lehmann, 1999). For this reason rank type of tests are developed for which on beforehand no speci¯c distributional assumptions need to be made. In the below we shall concentrate on the two-sample Wilcoxon test because of its relevance to bioinformatics. We sustain with a brief description of the basic idea and refer the interested reader to the literature on non-parametric66 CHAPTER 4. ESTIMATION AND INFERENCE testing (e.g. Lehmann, 2006). To broaden our view we switch from hypothe- ses about means to those about distributions. An alternative hypothesis may then be formulated as that the distribution of a ¯rst group lays to the left of a second. To set the scene let the gene expression values of the ¯rst group (x1 to xm) have distribution F and those of the second group (y1 to yn) distribution G. The null hypothesis is that both distributions are equal (H0 : F = G) and the alternative that these are not. For example that the x's are smaller (or larger) than the y's. By the two-sample Wilcoxon test the data x1; ¢ ¢ ¢ ; xm; y1; ¢ ¢ ¢ ; yn are ranked and the rank numbers of the x's are summed to form the statistic W after a certain correction (Lehmann, 2006). The idea is that if the ranks of x's are smaller than those of the y's, then the sum is small. The distribution of the sum of ranks is known so that a p-value can be computed on the basis of which the null hypothesis is rejected if it is smaller than the signi¯cance level ®. Example 1. The null hypothesis that the expression values for gene CCND3 Cyclin D3 are equally distributed for the ALL patients and the AML patients can be tested by the built-in-function wilcox.test, as follows. > wilcox.test(golub[1042,] ~ gol.fac) Wilcoxon rank sum test data: golub[1042, ] by gol.fac W = 284, p-value = 6.15e-07 alternative hypothesis: true location shift is not equal to 0 Since the p-value is much smaller than ® = 0:05, the conclusion is to reject the null-hypothesis of equal distributions. 4.2 Application of tests to a whole set gene expression data Various tests are applied in the above to a single vector of gene expressions. In daily practice, however, we want to analyze a set of thousands of (row) vectors with gene expression values which are collected in a matrix. Such4.2. APPLICATION OF TESTS TO A WHOLE SET GENE EXPRESSION DATA67 can conveniently be accomplished by taking advantage of the fact that R stores the output of a test as an object in such a manner that we can extract information such as p-values. Recall that the smaller the p-value the larger the experimental e®ect. Hence, by collecting p-values in a vector we can select genes with large di®erences between patient groups. This and testing for normality will be illustrated by two examples. Example 1. Having a data matrix with gene expression values, a ques- tion one might ask is: What is the percentage of genes that passes a normality test? Such can be computed as follows. > data(golub,package="multtest") > gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) > sh <- apply(golub[,gol.fac=="ALL"], 1, function(x) shapiro.test(x)$p.value) > sum(sh > 0.05)/nrow(golub) * 100 [1] 58.27598 Hence, according to the Shapiro-Wilk test, 58.27% of the ALL gene ex- pression values is normally distributed (in the sense of non-rejection). For the AML expression values this is 60.73%. It can be concluded that about forty percent of the genes do not pass the normality test. Example 2. In case the gene expression data are non-normally dis- tributed the t-test may indicate conclusions di®erent from those of the Wilcoxon test. Di®erences between these can be investigated by collecting the p-values from both tests and seeking for the largest di®erences. > data(golub, package = "multtest"); > gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) > pt <- apply(golub, 1, function(x) t.test(x ~ gol.fac)$p.value) > pw <- apply(golub, 1, function(x) wilcox.test(x ~ gol.fac)$p.value) > resul <- data.frame(cbind(pw,pt)) > resul[pw<0.05 & abs(pt-pw)>0.2,] pw pt 456 0.04480288 0.2636088 1509 0.03215830 0.4427477 The p-value is extracted from the output of the t.test function and stored in the vector pt. The logical operator & is used to select genes for which the68 CHAPTER 4. ESTIMATION AND INFERENCE Wilcoxon p-value is smaller than 0.05 and the absolute di®erence with the p-value from the t-test is larger than 0.2. Since there are only two such genes we can draw the reassuring conclusion that the tests give similar results. 4.3 Overview and concluding remarks Statistical hypothesis testing consists of hypotheses, distributional assump- tions, and decisions (conclusions). The hypotheses pertain to the outcome of a biological experiment and are always formulated in terms of population values of parameters. Statistically, the outcomes of experiments are seen as realizations of random variables. The latter are assumed to have a certain suitable distribution which is seen as a statistical model for outcomes of an experiment. Then a statistic is formulated (e.g. a t-value) which is treated both as a function of the random variables and as a function of the data values. By comparing the distribution of the statistic with the value of the statistic, the p-value is computed and compared to the level of signi¯cance. A large p-value indicates that the model ¯ts the data well and that the as- sumptions as well as the null-hypothesis are correct with large probability. However, a low p-value indicates, under the validity of the distributional as- sumptions, that the outcome of the experiment is so unlikely that this causes a su±cient amount of doubt to the researcher to reject the null hypothesis. The quality of a test is often expressed in terms of e±ciency, which is usually directly related to the (asymptotic) variance of an estimator. The relative e±ciency is the ratio of the asymptotic variances. For Wilcoxon's test versus the t-test this equals .955, which means that in the optimal situation where the (gene expression) data are normally distributed, Wilcoxon's test is only a little worse than the t-test. In case, however, of a few outliers or a slightly heavier tail, the Wicoxon test can be far more e±cient than the t-test (Lehmann, 1999, p.176). E±ciency is directly related to power; the probability to reject a false hypothesis. The probability of drawing correct conclusions can always be improved by increasing the sample size. These considerations set the scene for making some recommendations, which obviously should not be followed blindly. If gene expression data pass a normality test, then the Welch type of t-test provides a general test with good power properties (Ramsey, 1980; Wang, 1971). In case normality does not hold and the sample size per group is at least least four, the Wilcoxon4.4. EXERCISES 69 test is recommended. Because the Wilcoxon p-values are based on ranks many of these are equal for di®erent genes, so that it is less suitable for ordering in case of small sample size. On the other hand, it is obviously questionable whether extremely small di®erences in p-values produced by the t-test contribute to biologically relevant gene discrimination. That is, extremely small di®erences should not be over-interpreted. 4.4 Exercises 1. Gene CD33. Use grep to ¯nd the index of the important gene CD33 among the list of characters golub.gnames. For each test below for- mulate the null hypothesis, the p-value and your conclusion. (a) Test the normality of the ALL and AML expression values. (b) Test for the equality of variances. (c) Test for the equality of the means by an appropriate t-test. (d) Is the experimental e®ect strong? 2. Gene "MYBL2 V-myb avian myeloblastosis viral oncogene homolog- like 2" has its expression values in row 1788. (a) Use a boxplot to construct a hypothesis about the experimental e®ect. (b) Test for the equality of means by an appropriate t-test. 3. HOXA9. Gene "HOXA9 Homeo box A9" with expression values in row 1391, can cause leukemia (Golub et al., 1999). (a) Test the normality of the expression values of the ALL patients. (b) Test for the equality of means by an appropriate t-test. 4. Zyxin. On NCBI there are various cDNA clones of zyxin. (a) Find the accession number of cDNA clone with IMAGE:3504464. (b) Test whether the frequencies of the nucleotides are equal for each nucleic acid.70 CHAPTER 4. ESTIMATION AND INFERENCE (c) Test whether the frequencies of "X94991.1" can be predicted by the probabilities of the cDNA sequence "BC002323.2". 5. Gene selection. Select the genes from the golub data with smallest two-sample t-test values for which the ALL mean is greater than the AML mean. Report the names of the best ten. Scan the Golub (1999) article for genes among the ten you found and discuss their biological function brie°y. 6. Antigenes. Antigenes play an important role in the development of cancer. Order the antigenes according to their p-values from the Welch two-sample t-test with respect to gene expression values from the ALL and AML patients of the Golub et al. (1999) data. 7. Genetic Model. A certain genetic model predicts that four phenotypes occur in ration 9:3:3:1. In a certain experiment the o®spring is observed with frequencies 930, 330, 290, 90. Do the data con¯rm the model? 8. Comparing two genes. Consider the gene expression values in row 790 and 66 of the Golub et al. (1999) data. (a) Produce a boxplot for the ALL expression values and comment on the di®erences. Are there outliers? (b) Compute the mean and the median for the ALL gene expression values for both genes. Do you observed di®erence between genes? (c) Compute three measures of spread for the ALL expression values for both genes. Do you observe di®erence between genes? (d) Test by Shapiro-Wilk and Anderson-Darling the normality for the ALL gene expression values for both genes. 9. Normality tests for gene expression values of the Golub et al. (1999) data. Perform the Shapiro-Wilk normality test separately for the ALL and AML gene expression values. What percentage passed the normal- ity test separately for the ALL and the AML gene expression values? What percentage passes both testes? 10. Two-sample tests on gene expression values of the Golub et al. (1999) data.4.4. EXERCISES 71 (a) Perform the two-sample Welch t-test and report the names of the ten genes with the smallest p-values. (b) Perform the Wilcoxon rank test and report the names of the ten genes with the smallest p-values. 11. Biological hypotheses. Suppose that the probability to reject a biolog- ical hypothesis by the results of a certain experiment is 0:05. Suppose that the experiment is repeated 1000 times. (a) How many rejections do you expect. (b) What is the probability of less than 10 rejections? (c) What is the probability of more than 5 rejections? (d) What is the probability that the number of rejections is between two and eight? 12. Programming some tests. (a) Program the two-sample t-test with equal variances and illustrate it with the expression values of row 1024 the of Golub et al. (1999) data. (b) The value of W in the two-sample Wilxoxon test equals the sum of the ranks of Group 1 minus n(n+ 1)=2, where n is the number of gene expression values in Group 1. Program this and illustrate it with the expression values of row 1024 of Golub et al. (1999) data. (c) The value of W in the two-sample Wilxoxon test equals the num- ber of values xi > yj , where xi; yj are values from Group 1 and 2, respectively. Program this and illustrate it with the expression values of row 1024 of Golub et al. (1999) data.72 CHAPTER 4. ESTIMATION AND INFERENCEChapter 5 Linear Models We have seen that the t-test can be used to discover genes with di®erent means in the population with respect to two groups of patients. In case, however, there are three groups of patients the question arises how genes can be selected having the largest di®erential expressions between group means (experimental e®ect)? A technique making this possible is an application of the linear model and is called analysis of variance. It is frequently applied bioinformatics. The validity of the technique is based on the assumption that the gene expression values are normally distributed and have equal variance across groups of patients. It is of importance to investigate these assumptions be- cause it either reassures our con¯dence in the conclusions or it indicates that alternative tests should be used. In this chapter the linear model will brie°y be explained. The main focus, however, is on application of the linear model for testing the hypothesis that three or more group means are equal. Several illustrations of analyzing gene expression data will be given. It will be explained how the assumptions about normality and equal variances (homogeneity) can be investigated and what alternatives can be used in case either of these does not hold. The somewhat technical concepts of \model matrix" and \contrast matrix" are explained because these are useful for several applications in the next chapter. 7374 CHAPTER 5. LINEAR MODELS 5.1 De¯nition of linear models Given a gene expression Yi, a basic form of the linear model is Yi = xi¯ + "i; for i = 1; ¢ ¢ ¢ ; n; where Yi is an observable variable, xi a ¯xed number, ¯ an unknown weight, "i a unobservable error variable. The ¯xed number xi follows from a sta- tistical \design", as we shall see. The xi value is part of the predictor, Yi the criterion, and "i the error of the model. The systematical part of the model xi¯ equals the mean of the gene expression Yi. The model is called "linear" because the degree of the coe±cient ¯ is one. For a linear model to be a statistical model there must be some assumption with respect to the distribution of the error variables. Frequently, it is assumed that the er- ror variables "1; ¢ ¢ ¢ ; "n are independent and normally distributed with zero mean, that is, according to N(0; ¾2). Then the mean of Yi equals xi¯ and its variance ¾2. Example 1. A common manner to introduce the linear model is by writing Yi = ¯1 + xi¯2 + "i; for i = 1; ¢ ¢ ¢ ; n; so that the model part represents a straight line with intercept ¯1 and slope ¯2. Given data points y1; ¢ ¢ ¢ ; yn and x1; ¢ ¢ ¢ ; xn, a best ¯tting line through the data can easily be computed by least squares estimation of the intercept and slope. A nice application to explore this is by the function put.points.demo() from the TeachingDemos package. It allows points to be added and deleted to a plot which interactively computes estimates for the slope and the intercept given the data. By choosing the points more or less on a horizontal line, the slope will be near zero. By choosing the points nearly vertical, the slope will be large. By choosing a few gross errors in the data it can be observed that the estimates are not robust against outliers. In order to handle gene expression data for three or more groups of pa- tients we need to extend the model. The idea simply is to increase the number of weights to the number of groups k, so that, we obtain the weights ¯1; ¢ ¢ ¢ ; ¯k and the corresponding design values xi1; ¢ ¢ ¢ ; xik. The system- atic part of the model consists of a weighted sum of these design values:5.1. DEFINITION OF LINEAR MODELS 75 xi1¯1 +¢ ¢ ¢+xik¯k. By adding measurement error to this systematic part we obtain the linear model Yi =Xk j=1 xij¯j + "i: The design values xij for Patient i in Group j are collected in the so-called "design" matrix denoted by X. In particular, the design value xij is chosen to be equal to 1 if Patient i belongs to Group j and zero if (s)he does not. By this choice it becomes possible to use linear model estimation for testing hypotheses about group means. This will be illustrated by an example. Example 2. Suppose we have the following arti¯cial gene expressing values 2,3,1,2, of Group 1, 8,7,9,8 of Group 2, and 11,12,13,12 of Group 3. We may assign these to a vector y, as follows. > y <- c(2,3,1,2, 8,7,9,8, 11,12,13,12) Next, we construct a factor indicating to which group each expression value belongs. In particular, the ¯rst four belong to Group 1, the second four to Group 2, and the third four to Group 3. We conveniently use the function gl to de¯ne the corresponding factor. > a <- gl(3,4) > a [1] 1 1 1 1 2 2 2 2 3 3 3 3 Levels: 1 2 3 The design matrix X is also called \model matrix". It is illuminating to print it to the screen. > model.matrix(y ~ a - 1) a1 a2 a3 1 1 0 0 2 1 0 0 3 1 0 0 4 1 0 0 5 0 1 0 6 0 1 0 7 0 1 0 8 0 1 076 CHAPTER 5. LINEAR MODELS 9 0 0 1 10 0 0 1 11 0 0 1 12 0 0 1 The notation y~a-1 represents a model equation, where -1 means to skip the intercept or general constant.1 In this situation, the weights (¯1; ¯2; ¯3) of the model specialize to the population means (¹1; ¹2; ¹3). The model for the ¯rst gene expression value of Group 1 is Y1 = ¹1 + "1, for the second expression value of Group 1 it is Y2 = ¹1 +"2, for the ¯rst member of Group 2 it is Y5 = ¹2 + "5, and for the ¯rst member of Group 3 it is Y9 = ¹3 + "9. Recall that population means are generally estimated by sample means. Similarly, in the current setting, estimation of the linear model comes down to estimation of group means for which there are one-sample t-type of tests available (see e.g. Rao & Toutenburg, 1995; Samuels & Witmer, 2003). To illustrate this we employ the estimation function lm and ask for a summary. > summary(lm(y ~ a - 1)) Coefficients: Estimate Std. Error t value Pr(>|t|) a1 2.0000 0.4082 4.899 0.000849 *** a2 8.0000 0.4082 19.596 1.09e-08 *** a3 12.0000 0.4082 29.394 2.98e-10 *** The output in the ¯rst column gives the estimated mean per group. The sec- ond gives the standard error of each mean, the third the t-value (the estimate divided by the standard error), and the last the corresponding p-values. From the p-values the conclusion follows to reject the null hypotheses H0 : ¹j = 0 for Group index j running from 1 to 3. Using the above design matrix, the model for the gene expression values from di®erent groups can be written as Yij = ¹j + "ij ; where "ij is distributed as N(0; ¾2); and Yij is the expression of Person i in Group j, ¹j the mean of Group j, and the "ij the error of Person i in Group j. The error is assumed to be normally distributed with zero mean and variance equal for di®erent persons. 1See also Chapter 11 of the manual "An Introduction to R".5.2. ONE-WAY ANALYSIS OF VARIANCE 77 The above illustrates that the linear model is useful for testing hypotheses about group means. In bioinformatics the linear model is applied to many sets of gene expressions, so that it is of great importance to have an overall test for the equality of means. 5.2 One-way analysis of variance A frequent problem is that of testing the null hypothesis that three or more population means are equal. By comparing two types of variances, this is made possible by a technique called analysis of variance (ANOVA). To set the scene, let three groups of patients be available with measurements in the form of gene expression values. The null-hypothesis to be tested is H0 : ¹1 = ¹2 = ¹3. In statistical language such groups are called levels of a factor. Let the data for Group 1 be represented by y11; y21; ¢ ¢ ¢ ; yn1 those of Group 2 by y12; y22; ¢ ¢ ¢ ; yn2 and those of Group 3 by y13; y23; ¢ ¢ ¢ ; yn3, where n is the number of expression values in each group. The three sample means per patient group can be expressed by y1 = 1nXn i=1 yi1; y2 = 1n Xn i=1 yi2; and y3 = 1n Xn i=1 yi3: The total number of measurements N = 3n, so that the overall mean y is equal to y = 1 N à Xn i=1 yi1 + Xn i=1 yi2 + Xn i=1 yi3!: For the de¯nition of the overall test on the equality of means there are two sums of squares of importance. The sum of squares within (SSW) is the sum of the squared deviation of the measurements to their group mean, that is SSW = Xg j=1 Xn i=1 (yij ¡ yj)2; where g is the number of groups. The sum of squares between (SSB) is the sum of squares of the deviances of the group mean with respect to the total mean, that is SSB = Xg j=1 Xn i=1 (yj ¡ y)2 = n Xg j=1(yj ¡ y)2:78 CHAPTER 5. LINEAR MODELS Now the f-value is de¯ned by f = SSB=(g ¡ 1) SSW=(N ¡ g) : If the data are normally distributed, then this f-value follows the Fg¡1;N¡g distribution, where g ¡ 1 and N ¡ g are the degrees of freedom (Rao, 1973, p.245). If P(Fg¡1;N¡g > f) ¸ ®, then H0 : ¹1 = ¹2 = ¹3 is not rejected, and, otherwise it is. The idea behind the test is that, under the null-hypothesis of equal group means, the value for SSB will tend to be small, so that the observed f-value will be small and H0 is accepted. Example 1. Let's continue with the data from the previous example. Recall that the data of Group 1 are 2, 3, 1, 2, those of Group 2 are 8, 7, 9, 8, and of Group 3 are 11, 12, 13, 12. The number of expression values per group n = 4, the total number of data values N = 12, and the number of groups g = 3. To load the data, to construct the corresponding factor, and to compute the group means one may use the following. > y <- c(2,3,1,2, 8,7,9,8, 11,12,13,12) > a <- gl(3,4) > gm <- as.numeric(tapply(y, a, mean)) > gm [1] 2 8 12 Thus we ¯nd that y1 = 2, y2 = 8, and y3 = 12. These group means are now collected in the vector gm. The grand mean y can be computed by mean(y)=7.333333. An elementary manner to compute the sums of squares between SSB is by gm <- as.numeric(tapply(y, a, mean)) g <- 3; n <- 4; N <-12; ssb <- 0 for (j in 1:g) {ssb <- ssb + (gm[j]- mean(y))^2} SSB <- n*ssb This results in SSB = 202:6667. In a similar manner the sums of squares within (SSW) and the f-value can be computed, as follows. > SSW <- 0 > for (j in 1:g) {SSW <- SSW + sum((y[a==j]-gm[j])^2)} > f <- (SSB/(g-1))/(SSW/(N-g))5.2. ONE-WAY ANALYSIS OF VARIANCE 79 This results in SSW = 6 and an observed f-value equal to 152. Hence, the overall p-value is P(F2;9 > 152) = 1 ¡ P(F2;9 < 152) = 1 ¡ pf(152; 2; 9) = 1:159156 ¢ 10¡7: Since this is smaller than the signi¯cance level 0.05, the conclusion is to reject the null hypothesis of equal means. The built-in-function anova can be used to extract the so-called analysis of variance table from an lm object. > anova(lm(y ~ a)) Analysis of Variance Table Response: xDf Sum Sq Mean Sq F value Pr(>F) fact 2 202.667 101.333 152 1.159e-07 *** Residuals 9 6.000 0.667 This gives the degrees of freedom g ¡ 1 = 2 and N ¡ g = 9, the sums of squares between (202.667), the sums of squares within (6.0), the f-value 152 and the overall p-value 1:159 ¢ 10¡7. Example 2. By the previous analysis of variance it is concluded that there are di®erences in population means. It is, however, not clear which of the means di®er. A way to clarify this is by estimating the mean of Group 1 (Level 1) and then computing the di®erence between Group 2 and Group 1, and the di®erence between Group 3 and Group 1. Such corresponds to the following contrast matrix C = 24 1 1 1 0 ¡1 0 0 0 ¡1 35 : This contrast matrix is by default implemented by the model speci¯cation y~a, as follows. > summary(lm(y ~ a)) Coefficients:Estimate Std. Error t value Pr(>|t|) (Intercept) 2.0000 0.4082 4.899 0.000849 ***80 CHAPTER 5. LINEAR MODELS factLevel 2 6.0000 0.5774 10.392 2.60e-06 *** factLevel 3 10.0000 0.5774 17.321 3.22e-08 *** Residual standard error: 0.8165 on 9 degrees of freedom Multiple R-Squared: 0.9712, Adjusted R-squared: 0.9649 F-statistic: 152 on 2 and 9 DF, p-value: 1.159e-07 The estimated intercept is the mean of Group 1 (Level 1). The factLevel 2 is the di®erence in means between Group 2 (Level 2) and Group 1 and factLevel 3 is the di®erence in means between Group 3 and Group 1. By t-tests the null-hypothesis is tested that the mean of Group 1 is zero, the di®erence in means between Group 2 and Group 1 is zero and the di®erence in means between Group 3 and Group 1 is zero. That is, the null-hypotheses are H0 : ¹1 = 0, H0 : ¹2 ¡ ¹1 = 0, and H0 : ¹3 ¡ ¹2 = 0. Since the p-values that correspond to the t-values are smaller than the signi¯cance level 0:05, all null-hypotheses are rejected. The last line of the output gives the f-value, the degrees of freedom, and the corresponding overall p-value. The latter equals that of ANOVA. Before we analyze real gene expression data it seems well to give an ex- ample where the means do not di®er. Example 3. Let's sample data from the normal distribution with mean 1.9 and standard deviation 0:5 corresponding to three groups of patients that do not possess any type of di®erences between groups. > y <- rnorm(12,1.9,0.5) > round(x,2) [1] 1.75 1.82 1.35 1.61 2.08 1.27 2.50 2.40 2.13 0.71 2.80 2.00 > a <- gl(3,4) > anova(lm(y ~ a))$Pr[1] [1] 0.6154917 Note that by the $Pr[1] operator extracts the p-value from the list generated by the anova function. The p-value implies the conclusion not to reject the null-hypotheseis of equal means, which is consistent with the data generation process.5.2. ONE-WAY ANALYSIS OF VARIANCE 81 B1 B2 B3 3.5 4.0 4.5 5.0 5.5 Figure 5.1: Plot of SKI-like onco- gene expressions for three patient groups. B1 B2 B3 6.0 6.2 6.4 6.6 6.8 7.0 7.2 Figure 5.2: Plot of Ets2 expression values for three patient groups. Example 4. B-cell ALL: 1866_g_at. To illustrate analysis of variance by real data we shall use the ALL data from the ALL package, see Section 1.1. Speci¯cally, expression levels from B-cell ALL patients in stage B1, B2, and B3 are selected with row name 1866_g_at, which refers to an SKI-like oncogene related to oncoproteins. From the plot of the data in Figure 5.1 it can be observed that the expression levels di®er between the disease stages. The null hypothesis is tested that the expression means in each stage are equal or in other words that there are no experimental e®ects. It is brie°y indicated how the data are loaded. > library(ALL);data(ALL) > ALLB123 <- ALL[,ALL$BT %in% c("B1","B2","B3")] > y <- exprs(ALLB123)["1866_g_at",] > summary(lm(y ~ ALLB123$BT)) Estimate Std. Error t value Pr(>|t|) (Intercept) 4.58222 0.08506 53.873 < 2e-16 ***82 CHAPTER 5. LINEAR MODELS ALLB123$BTB2 -0.43689 0.10513 -4.156 8.52e-05 *** ALLB123$BTB3 -0.72193 0.11494 -6.281 2.00e-08 *** Residual standard error: 0.3707 on 75 degrees of freedom Multiple R-squared: 0.3461, Adjusted R-squared: 0.3287 F-statistic: 19.85 on 2 and 75 DF, p-value: 1.207e-07 From the overall p-value 1:207 ¢ 10¡7 of the f-test the conclusion follows to reject the hypothesis of equal means. From the t-tests we conclude that the mean of B1 di®ers from zero and the di®erences between B2 and B1 as well as between B3 and B2 are unequal to zero. That is, the population means of Group B1, B2, and B3 do di®er. Example 5. B-cell ALL: 1242_at. To illustrate a case where the means do not di®er we selected the expression values for probe 1242_at of the B- cell ALL patients in stage B1, B2, and B3 from the ALL data. This probe corresponds to the Ets2 repressor factor which plays a role in telomerase regulation in human cancer cells. From the plot of the data in Figure 5.2, however, it can be observed that the expression values hardly di®er between disease stages. The data are extracted from the ALL object and collected in the vector y. The corresponding factor is given by ALLB123$BT. > library(ALL); data(ALL) > ALLB123 <- ALL[,ALL$BT %in% c("B1","B2","B3")] > y <- exprs(ALLB123)["1242_at",] > summary(lm(y ~ ALLB123$BT)) Estimate Std. Error t value Pr(>|t|) (Intercept) 6.55083 0.05673 115.483 <2e-16 *** ALLB123$BTB2 0.03331 0.07011 0.475 0.636 ALLB123$BTB3 -0.04675 0.07665 -0.610 0.544 Residual standard error: 0.2473 on 75 degrees of freedom Multiple R-squared: 0.01925, Adjusted R-squared: -0.006898 F-statistic: 0.7362 on 2 and 75 DF, p-value: 0.4823 From the overall p-value 0.4823, the conclusion is not to reject the null hy- pothesis of equal means. More speci¯cally, the null-hypotheses H0 : ¹1 = 0 is rejected, but from the p-value 0.636 the H0 : ¹2 ¡ ¹1 = 0 is not rejected,5.3. TWO-WAY ANALYSIS OF VARIANCE 83 and from p-value 0.544 the H0 : ¹3 ¡ ¹2 = 0 is not rejected either. Example 6. An interesting question is of course for how many genes of the ALL data the hypothesis of equal means is rejected by the overall ANOVA p-value? Such can be answered by collecting the p-values in a vector. > pano <- apply(exprs(ALLB123),1,function(x) anova(lm(x~ALLB123$BT))$Pr[1]) > sum(pano<0.05) [1] 2526 Thus the hypothesis of equal means is rejected for 2526 out of a total of 12625 genes (probes). 5.3 Two-way analysis of variance Having some experience with one way analysis of variance, the question may arise whether the model for means of groups can be extended from one factor to more factors. This is indeed possible. The model would then be equal to Yijk = ®i + ¯j + (®¯)ij + "ijk; where ®i is the mean of Group i indicated by the ¯rst factor, ¯j the mean of Group j indicated by the second factor, (®¯)ij the interaction e®ect and "ijk the error which is distributed according to N(0; ¾2). If the means of the i groups di®er, then there is a main e®ect of the ¯rst factor which is expressed in a p-value smaller than 0.05. Similarly, in case the means of the j groups di®er, there is a main e®ect of the second factor, expressed in a p-value smaller than 0.05. Two-way analysis of variance will brie°y be illus- trated. Example 5. A two-way approach. It case of the ALL data from Chiaretty et al. (2004) we may aggregate the B cell patients into two groups: B, B1 and B2 in the ¯rst and B3 and B4 in the second. For the second group we select from the molecular biology of the patients assigned to BCR/ABL and NEG. We shall perform the analysis on the expression values of NEDD4 binding protein 1 with probe id 32069_at. This can be computed as follows. library("ALL"); data(ALL) ALLBm <- ALL[,which(ALL$BT %in% c("B","B1","B2","B3","B4") & ALL$mol.biol %in% c("BCR/84 CHAPTER 5. LINEAR MODELS facmolb <- factor(ALLBm$mol.biol) facB <- factor(ceiling(as.integer(ALLBm$BT)/3),levels=1:2,labels=c("B012","B34> anova(lm(exprs(ALLBm)["32069_at",] ~ facB * facmolb)) Analysis of Variance Table Response: exprs(ALLBm)["32069_at", ] Df Sum Sq Mean Sq F value Pr(>F) facB 1 1.1659 1.1659 4.5999 0.0352127 * facmolb 1 3.2162 3.2162 12.6891 0.0006433 *** facB:facmolb 1 1.1809 1.1809 4.6592 0.0340869 * Residuals 75 19.0094 0.2535 First the patients are selected with B-cell ALL and assigned molecular biology of the cancer to BCR/ABL or NEG. Next the factors are constructed to group the patients. From the p-values in the analysis of variance table it can be concluded that there two main e®ects as well as an interaction e®ect. One may also ask for a summary of the individual e®ects. > summary(lm(exprs(ALLBm)["32069_at",] ~ facB * facmolb)) Call: lm(formula = exprs(ALLBm)["32069_at", ] ~ facB * facmolb) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 6.7649 0.1073 63.026 < 2e-16 *** facBB34 -0.5231 0.1686 -3.103 0.0027 ** facmolbNEG -0.6020 0.1458 -4.128 9.4e-05 *** facBB34:facmolbNEG 0.5016 0.2324 2.159 0.0341 * Residual standard error: 0.5034 on 75 degrees of freedom Multiple R-squared: 0.2264, Adjusted R-squared: 0.1954 F-statistic: 7.316 on 3 and 75 DF, p-value: 0.0002285 In bioinformatics the question often arises how many probes there are with have signi¯cant e®ects. In this case we may compute the number of probes with signi¯cant main as well as interaction e®ects. > pval <- apply(exprs(ALLBm), 1, function(x) anova(lm(x ~ facB * facmolb))$Pr[5.4. CHECKING ASSUMPTIONS 85 > pvalt <- data.frame(t(pval)) > colnames(pvalt) <- c("maineffectB","maineffectmolbiol","interaction") > sum(pvalt$maineffectB < 0.05 & pvalt$maineffectmolbiol < 0.05 & pvalt$interaction < [1] 47 The three p-values per probe are collected in a matrix. This matrix is trans- posed so that the columns correspond to the p-values and the rows to the probes. Using the logical AND (&) operator and summing the TRUE values yield 47 probes with signi¯cant main and interaction e®ects. 5.4 Checking assumptions When the linear model is applied for analysis of variance there are in fact two assumptions made. First, the errors are assumed to be independent and normally distributed, and, second, the error variances are assumed to be equal for each level (patient group). The latter is generally known as the homoscedasticity assumption. The normality assumption can be tested as a null hypothesis by applying the Shapiro-Wilk test on the residuals. The homoscedasticity assumption can be tested as a hypothesis by the Breusch and Pagan (1979) test on the residuals. This latter test may very well be seen as a generalization of the F-test for equal variances. Example 1. Testing normality of the residuals. From Figure 5.1 it can be observed that there are outliers being far apart from the bulk of the other expression values. This raises the question whether the normality assumption holds. The normality of the residuals from the estimated linear model on the B-cell ALL data from 1866_g_at, can be tested as follows. > data(ALL,package="ALL");library(ALL) > ALLB123 <- ALL[,ALL$BT %in% c("B1","B2","B3")] > y <- exprs(ALLB123)["1866_g_at",] > shapiro.test(residuals(lm(y ~ ALLB123$BT))) Shapiro-Wilk normality test data: residuals(lm(y ~ ALLB123$BT)) W = 0.9346, p-value = 0.000598986 CHAPTER 5. LINEAR MODELS From the p-value 0.0005989, the conclusion is to reject the null-hypothesis of normally distributed residuals. Example 2. Testing homoscedasticity of the residuals. From Figure 5.1 it can be observed that the spread of the expression values around their mean di®ers between groups of patients. In order to test the homoscedasticity assumption we use the function bptest from the lmtest package. > library(ALL); data(ALL); library(lmtest) > ALLB123 <- ALL[,ALL$BT %in% c("B1","B2","B3")] > y <- exprs(ALLB123)["1866_g_at",] > bptest(lm(y ~ ALLB123$BT),studentize = FALSE) Breusch-Pagan test data: lm(y ~ ALLB123$BT) BP = 8.7311, df = 2, p-value = 0.01271 From the p-value 0.01271, the conclusion follows to reject the null hypothesis of equal variances (homoscedasticity). 5.5 Robust tests In case departures from normality or homoscedasticity are large enough to cause concern with respect to the actual signi¯cance level or to the power of the test, an alternative testing procedure is called for. In case only ho- moscedasticity is violated, we are in a situation quite similar to that of t- testing with unequal variances. That is, the null hypothesis H0 : ¹1 = ¹2 = ¹3 of equal means can be tested without assuming equal variances by a test proposed by Welch (1951). Example 1. In Example 2 of the previous section the hypothesis of equal variances was rejected. To apply analysis of variance without assuming equal variances (homoscedasticity) one may use the function oneway.test, as follows. > data(ALL,package="ALL");library(ALL)5.5. ROBUST TESTS 87 > ALLB123 <- ALL[,ALL$BT %in% c("B1","B2","B3")] > y <- exprs(ALLB123)["1866_g_at",] > oneway.test(y ~ ALLB123$BT) One-way analysis of means (not assuming equal variances) data: y and ALLB123$BT F = 14.1573, num df = 2.000, denom df = 36.998, p-value = 2.717e-05 From the p-value 2:717 ¢ 10¡5, the conclusion follows to reject the hypothesis of equal means. In case normality is violated a rank type of test is more appropriate. In particular, to test the null-hypothesis of equal distributions of groups of gene expression values, the Kruskal-Wallis rank sum test is recommended. This test can very well be seen as a generalization of the Wilcoxon test for testing the equality of two distributions. Because it is based on ranking the data, it is highly robust against non-normality, it, however, does not estimate the size of experimental e®ects. Example 2. In Example 1 of the previous section we rejected the hypoth- esis of normally distributed residuals. We use the function kruskal.test to perform a non-parametric test. > data(ALL,package="ALL");library(ALL) > ALLB123 <- ALL[,ALL$BT %in% c("B1","B2","B3")] > y <- exprs(ALLB123)["1866_g_at",] > kruskal.test(y ~ ALLB123$BT) Kruskal-Wallis rank sum test data: y by ALLB123$BT Kruskal-Wallis chi-squared = 30.6666, df = 2, p-value = 2.192e-07 From the p-value 2:192 ¢ 10¡7, the null-hypothesis of equal distributions of expression values between patient groups is rejected. By the apply functionality the p-values can easily be computed for all 12625 gene expression values of the ALL data.88 CHAPTER 5. LINEAR MODELS 5.6 Overview and concluding remarks By applying the above normality and homogeneity tests to complete sets of gene expression values it can quickly be seen to what extent the assumptions for the classical analysis of variance test are violated. Based on these it can be decided to add rank type of testing in order to reduce the amount of false positives and false negatives. Here, false positives are signi¯cant p-values for equal populations means and false negatives are non-signi¯cant p-values for unequal populations means. In the next chapter it will brie°y be indicated how to combine two factors into a single analysis of variance. For instance, one may want to combine B-cell stage with age groups of persons. The interested reader is referred to Faraway (2004) and Venables & Ripley (2002) for more information on using linear models in R and for a general treatment of linear models to Rao & Toutenburg (1995). The p-values from overall tests of equality of means or distributions are important tools to order genes according to their experimental e®ect with respect to di®erent patient groups. More examples are given in the next chapter where several functionalities of Bioconductor will be used for the analysis of microarray data. 5.7 Exercises 1. Analysis of gene expressions of B-cell ALL patients. (a) Construct a data frame containing the expression values for the B-cell ALL patients in stage B, B1, B2, B3, B4 from the ALL data. (b) How many patients are in each group. (c) Test the normality of the residuals from the linear model used for analysis of variance for all gene expression values. Collect the p-values in a vector. (d) Do the same for the homoscedasticity assumption. (e) How many gene expressions are normally distributed and how many homoscedastic? For how many do both hold? 2. Further analysis of gene expressions of B-cell ALL patients. Continue with the previous data frame containing the expression values for the5.7. EXERCISES 89 B-cell ALL patients in stage B, B1, B2, B3, B4 from the ALL data. (a) Collect the overall p-values from ANOVA in a vector. (b) Use featureNames() to report the a®ymetrix id's of the genes with smaller p-values than 0.000001. (c) Collect the overall p-values from the Kruskal-Walles test in a vec- tor. (d) Use featureNames() to report the a®ymetrix id's of the genes with smaller p-values than 0.000001. (e) Brie°y comment on the di®erences you observe. That is, how many genes have p-values smaller than 0.001 from both ANOVA and Krusal-Wallis? How many only from one type of test? Hint: Collect TRUE/FALSES in logical vectors and use table. 3. Finding the ten best best genes among gene expressions of B-cell ALL patients. Continue with the previous data frame containing the expres- sion values for the B-cell ALL patients in stage B, B1, B2, B3, B4 from the ALL data. (a) Print the p-values and the corresponding (a±metrix) gene identi- ¯ers of the ten best from ANOVA. (b) Do the same for the p-values from the Kruskal-Wallis test. (c) Use the function intersect to ¯nd identi¯ers in both sets. 4. A simulation study on gene expression values. (a) Construct a data matrix with 10000 rows and 9 columns with data from the normal distribution with mean zero and variance equal to one. Such a matrix simulates gene expressions without di®erences between groups (sometimes called negatives). (b) Construct a factor for three groups each with three values. (c) How many p-values are smaller than the signi¯cance level ® = 0:05? (d) If the p-value is smaller than the signi¯cance level, then the con- clusion is that there an experimental e®ect (a positive). How many false positives do you expect and how many did you observe?90 CHAPTER 5. LINEAR MODELS (e) Construct a matrix with 10000 rows and 9 columns with normally distributed data with mean zero, one and two and variance equal to one. Assume again that there three groups each with three data values. This data matrix simulates gene expressions with di®er- ences between groups (sometimes called positives). Use ANOVA and kruskal-Wallis to ¯nd the number of signi¯cant genes (true positives). report the number of true positives and false nega- tives.Chapter 6 Micro Array Analysis The analysis of gene expression values is of key importance in bioinformatics. The technique makes it possible to give an initial answer to many important genetic type of questions. In this chapter you learn how to preprocess probe data, ¯lter genes, to program various visualizations, to use gene ontology identi¯ers, to load public available gene expression data, as well as how to summarize results in html output. 1 6.1 Probe data The microarray technique takes advantage of hybridization properties of nu- cleic acids. That is, to give a rough idea, complementary molecules are attached and labeled on a solid surface in order for a specialized scanner measure the intensity of target molecules. Per gene there are about twenty such measures obtained for each probe (gene). Per probe these measures come in pairs. The intensity of the perfect match (PM) intends to measure the amount of transcripts from the gene. The intensity of the mismatch (MM) is related to non-speci¯c binding and is often seen as a background type of noise. The raw data from the A®ymetrix scanner is stored in so-called DAT ¯les, which are processed to so-called CEL ¯les, where we will work with. The package affy has facilities to read data from a vector specifying several CEL ¯les produced by the A®ymetrix scanner. 1It may be convenient to explore the possibilities of the limmaGUI. Our approach, however, will be to concentrate on the programming aspects using the commandline. 9192 CHAPTER 6. MICRO ARRAY ANALYSIS Example 1. We will start with a built-in data set called MLL.B from the ALLMLL package. To load it and to retrieve basic information use > library(affy) > data(MLL.B, package = "ALLMLL") > MLL.B It is very useful to print the structure of the object str(MLL.B) and its slot names. > slotNames(MLL.B) [1] "cdfName" "nrow" "ncol" [4] "assayData" "phenoData" "featureData" [7] "experimentData" "annotation" ".__classVersion__" Additional information become available from str(MLL.B). The raw probe intensities are available from exprs(MLL.B), which extracts the probe in- tensities from the MLL.B object. The number of rows and columns of the expression values of MLL.B can be obtained by the dim function. > dim(exprs(MLL.B)) [1] 506944 20 The annotation can be extracted as follows. > annotation(MLL.B) [1] "hgu133b" To print the ¯rst 10 names of the probes use > probeNames(MLL.B)[1:10] [1] "200000_s_at" "200000_s_at" "200000_s_at" "200000_s_at" "200000_s_at" [6] "200000_s_at" "200000_s_at" "200000_s_at" "200000_s_at" "200000_s_at" Note that the probe names are the same as those obtained by geneNames. The PM and MM values are collected by the functions pm and mm. To print the PM values of the ¯rst four out of the sixteen rows of the probe with identi¯er 200000_s_at we may use the following.6.1. PROBE DATA 93 > pm(MLL.B,"200000_s_at")[1:4,1:3] JD-ALD009-v5-U133B.CEL JD-ALD051-v5-U133B.CEL JD-ALD052-v5-U133B.CEL 200000_s_at1 661.5 321.5 312.5 200000_s_at2 838.8 409.3 395.3 200000_s_at3 865.3 275.5 341.3 200000_s_at4 425.8 253.5 196.8 By function matplot a quick view on the variability of the data within and between probes can be obtained. > matplot(pm(MLL.B,"200000_s_at"),type="l", xlab="Probe No.", + ylab="PM Probe intensity") From the resulting plot in Figure 6.1 it can be observed that the variability is substantial. Density plots of the log of the probe values can be obtained by hist(MLL.B). From the density plot of the log of the intensity data in Figure 6.2 it can be seen that these are quite skew to the right. The script to program such plots 2 4 6 8 10 500 1000 1500 2000 Probe No. PM Probe intensity Figure 6.1: Mat plot of intensity values for a probe of MLL.B. 6 8 10 12 14 0.0 0.2 0.4 0.6 0.8 1.0 1.2 log intensity density Figure 6.2: Density of MLL.B data. is quite brief. > MAplot(MLL.B,pairs=TRUE, plot.method= "smoothScatter") > image(MLL.B)94 CHAPTER 6. MICRO ARRAY ANALYSIS 6.2 Preprocessing methods From various visualization methods it is clear that preprocessing of probe intensities is necessary for making biologically relevant conclusions. Biocon- ductor gives facilities for various preprocessing methods. Here we will only sketch what the main methods are and how these can be implemented. It should be noted that the topic of optimal preprocessing currently is a ¯eld of intense research (probably for the coming years), so that de¯nitive recom- mendations are not mandatory. Preprocessing consists of three major steps: Background correction, normalization, and summarization. To obtain the available background and pm correction methods use the following. > bgcorrect.methods [1] "mas" "none" "rma" "rma2" The mas background is part of the MAS A®ymetrix software and is based on the 2% lowest probe values. RMA uses only the PM values, neglects the MM values totally, and is based on conditional expectation and the normality assumption of probes values. There are also a number of correction methods available for the PM values: > pmcorrect.methods [1] "mas" "pmonly" "subtractmm" The following normalization methods are available: > normalize.methods(MLL.B) [1] "constant" "contrasts" "invariantset" "loess" [5] "qspline" "quantiles" "quantiles.robust" Constant is a scaling method equivalent to linear regression on a reference array although without intercept term. More general are the non-linear nor- malization methods such as loess, qspline, quantiles, and robust quantiles. Loess is a nonlinear method based on local regression of MA plots. The methods of contrasts is based on loess regression. Quantile normalization is an inverse transformation of the empirical distribution with respect to an averaged sample quantile in order to impose one and the same distribution to each array. The method qspline uses quantiles from each array and a target array to ¯t a system of cubic splines. The target should be the mean (geo- metric) or median of each probe, but could also be the name of a particular group.6.2. PREPROCESSING METHODS 95 The ¯nal step of preprocessing is to aggregate multiple probe intensities into a gene expression value. The available methods are: > express.summary.stat.methods [1] "avgdiff" "liwong" "mas" "medianpolish" "playerout" The ¯rst is the simplest as it is based on averaging. There is no single best method for all preprocessing problems. It seems, however, wise to use methods robust against outliers together with non-linear normalization methods. Example 1. The three pre-processing steps can be employed one after the other by the function expresso. To combine the background correction RMA with constant normalization and to use average di®erences for the computation of gene expression values, we may use the following. eset <- expresso(MLL.B,bgcorrect.method="rma", normalize.method="constant",pmcorrect.method="pmonly", summary.method="avgdiff") Example 2. Another frequently applied preprocessing method is RMA. It combines convolution background correction, quantile normalization, and summarization based on multi-array model ¯t in a robust manner by a so- called median polish algorithm. > library(affy) > data(MLL.B, package = "ALLMLL") > eset3 <- rma(MLL.B) Background correcting Normalizing Calculating Expression > boxplot(data.frame(exprs(eset3))) The three stages of preprocessing by rma are part of the output. Before a box-and-whiskers plot can be constructed the expression values need to be extracted from the object eset3. After the foregoing it is often desirable to further preprocess the data in order to remove patient speci¯c means or medians. When the patient me- dian is zero, for instance, testing for a gene to have mean expression value96 CHAPTER 6. MICRO ARRAY ANALYSIS di®erent from zero becomes meaningful. Example 3. In the sequel we shall frequently work with the ALL data from the ALL package of Bioconductor. Here the data set is brie°y introduced (see also Section 1.1) and further processing steps are illustrated. The raw data have been jointly normalized by RMA and are available in the form of an exprSet object. 12625 gene expression values are available from microarrays of 128 di®erent persons su®ering from acute lymphoblastic leukemia (ALL). A number of interesting phenotypical co-variates are available. For instance, the ALL$mol variable has TRUE/FALSE values for each of the 128 patients depending on whether a reciprocal translocation occurred between the long arms of Chromosome 9 and 22. This is casually related to chronic and acute leukemia. One can also ask for table(ALL$BT) to obtain an overview of the numbers of patients which are in certain phases of a disease. See also the general help ?ALL for further information on the data or the article by Chiaretti et al. (2004). > data(ALL, package = "ALL") > slotNames(ALL) [1] "assayData" "phenoData" "featureData" [4] "experimentData" "annotation" ".__classVersion__" > row.names(exprs(ALL))[1:10] [1] "1000_at" "1001_at" "1002_f_at" "1003_s_at" "1004_at" "1005_at" [7] "1006_at" "1007_s_at" "1008_f_at" "1009_at" By feno <- pData(ALL) phenotypical information from the patients is stored in a data frame, which is useful for further analysis. In case the gene expres- sion values over the patients are non-normally distributed one may want to subtract the median and divide by the MAD. An e±cient manner to do so is to use an apply function to compute the column mad and median, and sweep to subtract the median from each column entry and, next, to divide each column entry by the MAD. ALL1pp <- ALL1 <- ALL[,ALL$mol == "ALL1/AF4"] mads <- apply(exprs(ALL1), 2, mad) meds <- apply(exprs(ALL1), 2, median) dat <- sweep(exprs(ALL1), 2, meds) exprs(ALL1pp) <- sweep(dat, 2, mads, FUN="/")6.3. GENE FILTERING 97 By this script the patients are selected with assigned molecular biology equal to ALL1/AF4. Then ALL1 is copied in order to overwrite the expression values in a later stage. The median and the MAD are computed per column by the speci¯cation 2 (column index) in the apply function. Then the ¯rst sweep function subtracts the medians from the expression values and second divides these by the corresponding MAD. By comparing the box plots in Figure 6.3 and 6.4 the e®ect of preprocessing can be observed. The medians of the preprocessed data are equal to zero and the variation is smaller due to the division by their MAD. Note that by box plotting a data frame a fast overview of the distributions of columns in a data frame is obtained. X04006 X16004 X24005 X28028 X31007 2 4 6 8 10 12 14 Figure 6.3: Boxplot of the ALL1/AF4 patients. X04006 X16004 X24005 X28028 X31007 −1 0 1 2 3 4 Figure 6.4: Boxplot of the ALL1/AF4 patients after median subtraction and MAD division. 6.3 Gene ¯ltering A few important methods to ¯lter genes are illustrated here. It is wise to keep in mind that there are statistical as well as and biological criteria for ¯ltering genes and that a combination of these often gives the most satisfac- tory results. The examples stress the importance of careful thinking.98 CHAPTER 6. MICRO ARRAY ANALYSIS Example 1. Filtering by the coe±cient of variation. A manner to ¯lter genes is by the coe±cient of variation, which is de¯ned as the standard deviation divided by the absolute value of the mean: cv = ¾=j¹j. If cv = 1, then the standard deviation equals the mean, so that the experimental e®ect is small relative to the precision of measurement. If, however, cv < 0:2, then the mean is ¯ve times larger than the standard deviation, so that both the experimental e®ect and the measurement precision are large. Let's compute the coe±cient of variation per gene for the ALL1pp data of the previous section. > cvval <- apply(exprs(ALL1pp),1,function(x){sd(x)/abs(mean(x))}) Now using sum(cvval<0.2) yields 4751 genes with a coe±cient of variation smaller than 0.2. These genes can be selected by ALL1pp[cvval<0.2,]. Example 2. Combining several ¯lters. It is often desired to combine several ¯lters. Of course it is possible to program ¯lters completely on your own, however, we may conveniently use the function filterfun to combine several ¯lters. The script in this example is useful when several functions are to be applied to a single data set. library("genefilter") f1 <- function(x)(IQR(x)>0.5) f2 <- pOverA(.25, log2(100)) f3 <- function(x) (median(2^x) > 300) f4 <- function(x) (shapiro.test(x)$p.value > 0.05) f5 <- function(x) (sd(x)/abs(mean(x))<0.1) f6 <- function(x) (sqrt(10)* abs(mean(x))/sd(x) > qt(0.975,9)) ff <- filterfun(f1,f2,f3,f4,f5,f6) library("ALL"); data(ALL) selected <- genefilter(exprs(All[,ALL$BT=="B"]), ff) After running this script and using sum(selected) one obtains 317 genes that pass the combined ¯lter. The ¯rst function returns TRUE if the in- terquartile range is larger than 0.5, the second if 25% of the gene expression values is larger than 6.643856, the third if the median of the expression values taken as powers to the base two is larger than 300, the fourth if it passes the Shapiro-Wilk normality test, the ¯fth if the coe±cient of variation is smaller than 0.1, and the sixth if the one-sample t-value is signi¯cant. The ¯lter6.3. GENE FILTERING 99 functions are combined by filterfun and the function genefilter returns a logical vector indicating whether the gene passed all the ¯lters or failed at least one of them. In order to use these ¯lter steps properly it is well to think them through because several ¯lters focus on similar properties. In particular, since the IQR divided by 1.349 is a robust estimator of the stan- dard deviation, the ¯rst ¯lter selects genes with a certain minimal standard deviation. With respect to the third ¯lter note that 2x > 300 is equivalent to x > 2 log(300) ¼ 8:228819, which is highly similar to the second ¯lter. Furthermore, s=jxj < 0:1 is equivalent to p10jxj=s > 1=p10, so that the last two ¯lters are highly similar. Example 3. Filtering by t-test and normality. One may also want to select genes with respect to p-values of a two-sample t-test over B-cell ALL versus T-cell ALL. This can be combined with a normality test in the sense that only those genes are ¯ltered which pass the Shapiro-Wilk normality test. The latter will be applied separately for the B-cell ALL patients and for the T-cell ALL patients. For this we write a function that will be used twice. First, however, we create a logical factor patientB indicating patients with B-cell ALL (TRUE) and with T-cell ALL (FALSE). The ¯lter de¯ned selects genes that have their p-value from the Welch two-sample t-test smaller than the signi¯cance level 0.05. A logical variable named selected is de¯ned which attains TRUE only if sel1, sel2, as well as sel3 have the value TRUE. library("genefilter");library("ALL"); data(ALL) patientB <- factor(ALL$BT %in% c("B","B1","B2","B3","B4")) f1 <- function(x) (shapiro.test(x)$p.value > 0.05) f2 <- function(x) (t.test(x ~ patientB)$p.value < 0.05) sel1 <- genefilter(exprs(ALL[,patientB==TRUE]), filterfun(f1)) sel2 <- genefilter(exprs(ALL[,patientB==FALSE]), filterfun(f1)) sel3 <- genefilter(exprs(ALL), filterfun(f2)) selected <- sel1 & sel2 & sel3 ALLs <- ALL[selected,] This gives 1817 genes which pass the three ¯lters. For these genes it holds that the expression values for B-cell ALL patients as well as for T-cell ALL patients are normally distributed (in the sense of non-rejection). A fundamental manner to visualize how the genes are divided among ¯lters is100 CHAPTER 6. MICRO ARRAY ANALYSIS by construction of a Venn diagram. This can conveniently be done by using functions from the limma package (Smyth, 2005). library(limma) x <- matrix(as.integer(c(sel1,sel2,sel3)),ncol = 3,byrow=FALSE) colnames(x) <- c("sel1","sel2","sel3") vc <- vennCounts(x, include="both") vennDiagram(vc) From the resulting Venn diagram in Figure 6.5 it can be seen that 1817 genes pass all three ¯lters, 1780 genes pass none, 3406 genes pass the normality tests but not the t-test ¯lter, etc. sel1 sel2 sel3 1780 920 2151 1366 826 3593406 1817 Figure 6.5: Venn diagram of se- leced ALL genes. X04006 X16004 X24005 X28028 X31007 −1 0 1 2 3 4 Figure 6.6: Boxplot of the ALL1/AF4 patients after median subtraction and MAD division. 6.4 Applications of linear models The limma package is frequently used for analyzing microarray data by linear models, such as ANOVA.6.4. APPLICATIONS OF LINEAR MODELS 101 Example 1. Analysis of variance. We select patients with B-cell leukemia in a beginning stage B and in more progressive stages B1 and B2. The type of analysis is speci¯ed by using a factor that de¯nes the model (design) matrix. Then the linear model is ¯tted to the data and an empirical Bayes procedure is used to adapt the gene speci¯c variances with a global variance estimator (Smyth, 2004)2. library("ALL"); library("limma"); data(ALL, package = "ALL") allB <- ALL[,which(ALL$BT %in% c("B","B1","B2"))] design.ma <- model.matrix(~ 0 + factor(allB$BT)) colnames(design.ma) <- c("B","B1","B2") fit <- lmFit(allB, design.ma) fit <- eBayes(fit) > toptab <- topTable(fit, coef=2,5,adjust.method="fdr") > print(toptab[,1:5],digits=4) ID logFC AveExpr t P.Value 12586 AFFX-hum_alu_at 13.42 13.50 326.0 3.165e-99 2488 32466_at 12.68 12.70 306.3 1.333e-97 2773 32748_at 12.08 12.11 296.3 9.771e-97 5328 35278_at 12.44 12.45 295.5 1.146e-96 4636 34593_g_at 12.64 12.58 278.0 4.431e-95 By topTable the ¯ve genes are selected with the smallest p-values adjusted for the false discovery rate. Let's call the mean of the B patients ¹, that of B1 ¹1, and that of B2 ¹2. In the current case we are not so much interested in the hypothesis H0 : ¹ ¡ ¹2, because this is the di®erence between Stage 0 and Stage 3. Rather, we are interested in the hypothesis H0 : ¹ ¡ ¹1 and H0 : ¹1 ¡ ¹2. Such a speci¯c hypothesis can be tested by using a contrast matrix, which can be speci¯ed as follows. > cont.ma <- makeContrasts(B-B1,B1-B2, levels=factor(allB$BT)) > cont.ma Contrasts Levels B - B1 B1 - B2 B 1 0 B1 -1 1 B2 0 -1 2To obtain the appropriate number of levels we make a factor of ALLB$BT.102 CHAPTER 6. MICRO ARRAY ANALYSIS Observe that the contrast matrix speci¯es the di®erence between the levels B and B1 as well as between B1 and B2. It can be implemented as follows. fit1 <- contrasts.fit(fit, cont.ma) fit1 <- eBayes(fit1) toptabcon <- topTable(fit, coef=2,5,adjust.method="fdr") print(toptabcon[,1:5],digits=4) > toptabcon <- topTable(fit1, coef=2,5,adjust.method="fdr") > print(toptabcon[,1:5],digits=4) ID logFC AveExpr t P.Value 3389 33358_at 1.4890 5.260 7.374 5.737e-10 419 1389_at -1.7852 9.262 -7.081 1.816e-09 1016 1914_at 2.0976 4.939 7.019 2.315e-09 6939 36873_at 1.8646 4.303 6.426 2.361e-08 7542 37471_at 0.8701 6.551 6.106 8.161e-08 Here, we have applied a method called \false discovery rate" (fdr) which in- creases the p-values somewhat in order to reduce the number false positives. The number of genes requested equals 5. A very convenient manner to summarize, collect, and communicate vari- ous types of results is in the form of an HTML ¯le. Example 2. Summarizing output in HTML format. It is often desired to combine the typical output from a function like topTable with that of an HTML output page containing various types of information. To illustrate this we proceed with the object toptabcon of the previous example. library("annaffy");library("hgu95av2.db") anntable <- aafTableAnn(as.character(toptabcon$ID), "hgu95av2.db", aaf.handlesaveHTML(anntable, "ALLB123.html", title = "B-cell 012 ALL") By the function aafTableAnn various types of information are gathered from the output topTable of the estimated linear model, the annotation package, and the aaf.handler functionality. The information collected contains the following: Probe, Symbol, Description, Function, Chromosome, Chromo- some Location, GenBank, LocusLink, Cytoband, UniGene, PubMed, Gene Ontology, and Pathway. The resulting anntable is saved in HTML format in the working directory or the Desktop. It contains a wealth of information6.4. APPLICATIONS OF LINEAR MODELS 103 on e.g. Chromosome location, KEGG mappings, summaries from Pubmed articles, etc. Example 3. Using basic R functions. It is also possible to summarize results in an HTML table on the basis of p-values from e.g. analysis of variance (ANOVA). That is, the selected genes can directly be used as input for aafTableAnn. library("multtest"); library("annaffy"); library("hgu95av2.db") library("ALL"); data(ALL, package = "ALL") ALLB <- ALL[,which(ALL$BT %in% c("B","B1","B2"))] panova <- apply(exprs(ALLB), 1, function(x) anova(lm(x ~ ALLB$BT))$Pr[1]) genenames <- featureNames(ALLB)[panova<0.000001] atab <- aafTableAnn(genenames, "hgu95av2.db", aaf.handler()[c(1:3,8:9,11:13)]) saveHTML(atab, file="ANOVAonB-cellGroups.html") hgu95av2.db is a meta data annotation package connecting the requested information by the call to aaf.handler. The meaning of the columns can be obtained from the help page of the function. The resulting table is saved as an HTML ¯le in the working directory (getwd()) or desktop. In a similar manner the p-values from the Kruskal-Wallis test can be used to select genes. Bioconductor has a useful facility to download publicly available microar- ray data sets from NCBI. Example 4. Analyzing public available data. The GDS1365 data con- tain primed macrophage response to IFN-gamma restimulation after di®erent time periods. The purpose is to gain insight into the in°uence of IFN-gamma priming on IFN-gamma induced transcriptional responses. Among the phe- notypical covariates of the data there is a factor time with levels 0, 3 and 24 hours and a factor protocol with the levels "IFN-gamma primed" and "un- primed", which can be extracted by the function pData. Since researchers are often interested in the interaction between factors, we shall select genes with a signi¯cant interaction e®ect. library(GEOquery); library(limma); library(hgu95av2.db);library(annaffy) gds <- getGEO("GDS1365") eset <- GDS2eSet(gds,do.log2=T)104 CHAPTER 6. MICRO ARRAY ANALYSIS prot <- pData(eset)$protocol time <- pData(eset)$time pval <- apply(exprs(eset)[1:12625,], 1, function(x) anova(lm(x ~ prot * time))$Pr[1:3]) pvalt <- data.frame(t(pval)) colnames(pvalt) <- c("meffprot","mefftime","interaction") genenames <- featureNames(eset)[pvalt$meffprot< 0.01 & pvalt$mefftime < 0.01 & pvalt$interaction < 0.01] atab <- aafTableAnn(genenames,"hgu95av2.db",aaf.handler()[c(1:3,8:9,11:13)]) saveHTML(atab, file="Two-way ANOVA protocol by time.html") By getGEO the data are downloaded to the disk and next these can be loaded into the R system. By GDS2eSet these are transformed to an expression set so that these can be analyzed statistically. The function pData extracts the factors from the expression set eset. The function anova extracts the p- value of the interaction e®ect from the estimated linear model. We restrict the analysis to the ¯rst 12625 rows because the additional ones contain not available values. The resulting html ¯le seems to contain may interesting genes. 6.5 Searching an annotation package Detailed information on microarray experiments is stored in an annotation package. > library("ALL"); data(ALL) > annotation(ALL) [1] "hgu95av2" Hence, the annotation package we need is hgu95av2.db. Let's load it and obtain an overview of its functionality. > library(hgu95av2.db) > ls("package:hgu95av2.db") [1] "hgu95av2" "hgu95av2_dbconn" "hgu95av2_dbfile" [4] "hgu95av2_dbInfo" "hgu95av2_dbschema" "hgu95av2ACCNUM" [7] "hgu95av2ALIAS2PROBE" "hgu95av2CHR" "hgu95av2CHRLENGTHS" [10] "hgu95av2CHRLOC" "hgu95av2CHRLOCEND" "hgu95av2ENSEMBL"6.5. SEARCHING AN ANNOTATION PACKAGE 105 [13] "hgu95av2ENSEMBL2PROBE" "hgu95av2ENTREZID" "hgu95av2ENZYME" [16] "hgu95av2ENZYME2PROBE" "hgu95av2GENENAME" "hgu95av2GO" [19] "hgu95av2GO2ALLPROBES" "hgu95av2GO2PROBE" "hgu95av2MAP" [22] "hgu95av2MAPCOUNTS" "hgu95av2OMIM" "hgu95av2ORGANISM" [25] "hgu95av2PATH" "hgu95av2PATH2PROBE" "hgu95av2PFAM" [28] "hgu95av2PMID" "hgu95av2PMID2PROBE" "hgu95av2PROSITE" [31] "hgu95av2REFSEQ" "hgu95av2SYMBOL" "hgu95av2UNIGENE" [34] "hgu95av2UNIPROT" The annotation package contains environments with di®erent types of infor- mation. An easy manner to make the content of an environment available is by converting it into a list and to print part of it to the screen. > ChrNrOfProbe <- as.list(hgu95av2CHR) > ChrNrOfProbe[1] $`1000_at` [1] "16" We recognize the manufacturers identi¯er of genes and the corresponding chromosome. Asking information by ?hgu95av2CHR reveals that it is an environment (hash table) which provides mappings between identi¯ers and chromosomes. From these we obtain various types of information on the basis of the manufacturer's identi¯er such as "1389_at". Below we obtain, respectively, the GenBank accession number, the Entrez Gene identi¯er, the gene abbreviation, gene name, brief summaries of functions of the gene prod- ucts, and the UniGene identi¯er. For this we use the get function in order to search an environment for a name. > get("1389_at", env = hgu95av2ACCNUM) [1] "J03779" > get("1389_at", env = hgu95av2ENTREZID) [1] 4311 > get("1389_at", env = hgu95av2SYMBOL) [1] "MME" > get("1389_at", env = hgu95av2GENENAME) [1] "membrane metallo-endopeptidase (neutral endopeptidase, enkephalinase, CALLA, CD10)" > get("1389_at", env = hgu95av2SUMFUNC) [1] NA106 CHAPTER 6. MICRO ARRAY ANALYSIS > get("1389_at", env = hgu95av2UNIGENE) [1] "Hs.307734" Let's use the GenBank accession number to search its nucleotide data base. > library(annotate) > genbank("J03779",disp="browser") From this we obtain the corresponding GI:179833 number, which can be used to obtain a complete XML document. > genbank(1430782,disp="data",type="uid") Obviously, probes correspond to genes and frequently we are interested in their chromosome location, and, speci¯cally, in starting position(s). > get("1389_at", env = hgu95av2CHRLOC) 3 3 3 156280152 156280327 156280748 Its cytoband location can also be obtained. > get("1389_at", env = hgu95av2MAP) [1] "3q25.1-q25.2" Hence, we see that the gene is on Chromosome 3 at q arm band 25 sub- band 1 and 2. In case we have a LocusLink ID, e.g. 4121, available the corresponding GO terms can be obtained and stored in a list. ll1<-GOENTREZID2GO[["4121"]] 6.6 Using annotation to search literature Given the manufactures probe identi¯er it is possible to search literature by collecting Pubmed ID's and to use these to collect relevant articles. > library(hgu95av2.db);library(annotate); library(ALL); data(ALL) > pmid <- get("1389_at",env=hgu95av2PMID) > pubmed(pmid,disp="browser") Another possibility is to collect a list containing PubMed ID, authors, ab- stract text, title, journal, and publication date.6.7. SEARCHING GO NUMBERS AND EVIDENCE 107 > absts <- pm.getabst("1389_at", "hgu95av2") > pm.titles(absts) The list can obviously be searched for regular expressions. ne <- pm.abstGrep("neutral endopeptidase",absts[[1]]) Another possibility is to construct an HTML table with the titles. > pmAbst2HTML(absts[[1]],filename="pmon1389_at.html") 6.7 Searching GO numbers and evidence By the phrase \ontology" we mean a structured language about some con- ceptual domain. The gene ontology consortium de¯nes three ontologies: A Molecular Function (MF) describes a phenomenon at the biochemical level such as \enzyme", \transporter", or \ligand". A Biological Process (BP) may coordinate various related molecular functions such as \DNA replica- tion" or \signal transduction". A Cellular Component (CC) is a unit within a part of the cell such as \chromosome", \nucleus", or \ribosome". Each term is identi¯ed by a unique GO number. To ¯nd GO numbers and their dependencies we use get to extract a list from the annotation ¯les hgu95av2GO for example. From the latter we extract a list and use an apply type of function to extract another list containing GO identi¯cation numbers. > go1389 <- get("1389_at", env = hgu95av2GO) > idl <- lapply(go1389,function(x) x$GOID) > idl[[1]] [1] "GO:0006508" The list idl contains 8 members of which only the ¯rst is printed to the screen. By changing GOID into Ontology more speci¯c information pertaining to ontology is extracted. From the annotate package we may now select the GO numbers which are related to a biological process. > library(annotate) > getOntology(go1389,"BP") [1] "GO:0006508" "GO:0007267"108 CHAPTER 6. MICRO ARRAY ANALYSIS There are various types of evidence such as: inferred from genetic interaction (IGI), inferred from electronic annotation (IEA), traceable author statement (TAS), etc. Per GO identi¯er the type of evidence can be obtained. > getEvidence(go1389) GO:0004245 GO:0005886 GO:0005887 GO:0006508 GO:0007267 GO:0008237 GO:0008270 "IEA" "TAS" "TAS" "TAS" "TAS" "TAS" "IEA" GO:0046872 "IEA" When we now want to select the GO numbers with evidence of a traceable author statement we can use the subset function to create a list. go1389TAS <- subset(go1389,getEvidence(go1389)=="TAS") A manner to extract information from this list is by using an apply type of function. > sapply(go1389TAS,function(x) x$GOID) > sapply(go1389TAS,function(x) x$Evidence) > sapply(go1389TAS,function(x) x$Ontology) We shall use this list in the below. 6.8 GO parents and children The term \transmembrane receptor protein-tyrosine kinase" is more speci¯c and therefore a 'child' of the more general term parent term \transmembrane receptor" (Gentleman, et. al, 2005). Example 1. Collecting GO information. There are functions to obtain parents and children from a GO identi¯er. > GOMFPARENTS$"GO:0003700" isa isa "GO:0003677" "GO:0030528" > GOMFCHILDREN$"GO:0003700" isa "GO:0003705"6.9. GENE FILTERING BY A BIOLOGICAL TERM 109 In case of a list of GO identi¯ers you may want to collect the ontology, parents, and children identi¯ers in a vector. go1389 <- get("1389_at", env = hgu95av2GO) gonr <- getOntology(go1389, "BP") gP <- getGOParents(gonr) gC <- getGOChildren(gonr) gPC <- c(gonr,gP,gC) pa <- sapply(gP,function(x) x$Parents) ch <- sapply(gC,function(x) x$Children) gonrc <- c(gonr,unlist(pa),unlist(ch)) Example 2. Probe selection by GO. A research strategy may be to start with a probe number, to ¯nd the GO identi¯ers of the biological process, to obtains its parents, and next to transform these to probes. library(GO); library(annotate); library("ALL"); data(ALL) go1389 <- get("1389_at", env = hgu95av2GO) gonr <- getOntology(go1389, "BP") gP <- getGOParents(gonr) pa <- sapply(gP,function(x) x$Parents) probes <- mget(pa,hgu95av2GO2ALLPROBES) probeNames <- unlist(probes) ALLpr <- ALL[probeNames,] > dim(exprs(ALLpr)) [1] 7745 128 Indeed, you may end up with many genes, useful for further analysis. 6.9 Gene ¯ltering by a biological term An application of working with GO numbers is to ¯lter for genes which are related to a biological term. Example 1. Filter gene by a term. From a biological point of view it is most interesting to select genes which are related to a certain biolog- ical process to be speci¯ed by a term such as "transcriptional repression".110 CHAPTER 6. MICRO ARRAY ANALYSIS We combine this with the previous ¯lter. For this we need the annota- tion package used in the stage of data collection. This can be obtained by annotation(ALL). First we de¯ne a function (Gentleman, et al., 2005, p. 123) to collect appropriate GO numbers from the environment GOTERM. library("GO"); library("annotate"); library("hgu95av2.db") GOTerm2Tag <- function(term) { GTL <- eapply(GOTERM, function(x) {grep(term, x@Term, value=TRUE)}) Gl <- sapply(GTL, length) names(GTL[Gl>0]) }> GOTerm2Tag("transcriptional repressor") [1] "GO:0016564" "GO:0016565" "GO:0016566" "GO:0017053" The functions eapply and sapply search an environment like GOTERM by grep for matches of a speci¯ed term. A precaution is taken to select only those names which are not empty. This gives the GO terms which can now be translated to probe of the ALLs data. tran1 <- hgu95av2GO2ALLPROBES$"GO:0016564" tran2 <- hgu95av2GO2ALLPROBES$"GO:0016566" tran3 <- hgu95av2GO2ALLPROBES$"GO:0017053" tran <- c(tran1,tran2,tran3) inboth <- tran %in% row.names(exprs(ALLs)) ALLtran <- ALLs[tran[inboth],] The GO translated probe names are intersected with the row names of the data giving the logical variable inboth. The variable tran[inboth] gives the ids by which genes can be selected. Next, gene ids for which inboth equals TRUE are selected and the corresponding data are collected in the data frame ALLtran. More information can be obtained by GOTERM$"GO:0016564. By dim(exprs(ALLtran)) it can be observed that 26 genes which passed the normality ¯lter are related to "transcriptional repression". 6.10 Signi¯cance per chromosome After a statistical analysis to ¯lter and order genes it is often quite useful to do post analysis on the results. In particular, after collecting p-values from6.10. SIGNIFICANCE PER CHROMOSOME 111 a t-test one may wonder whether genes with signi¯cant p-values occur more often within a certain chromosome. To test for such over or under represen- tation the Fisher test is very useful (see Section 4.1.7). Example 1. On the expression values of the ALL data we perform a two sample t-test using the patient group for which remission was achieved and for which it was not achieved. Per chromosome it can be tested whether the odds ratio di®ers from 1 or, equivalently, whether there is independence. The data for the test consist of the number of signi¯cant probes on Chromosome 19, the number of non-signi¯cant probes on Chromosome 19, the number of remaining signi¯cant probes, and the number of remaining non-signi¯cant probes. > library("ALL"); data(ALL); library("hgu95av2.db") > rawp <- apply(exprs(ALL), 1, function(x) t.test(x ~ ALL$remission)$p.value) > xx <- as.list(hgu95av2CHR) > AffimIDChr <- names(xx[xx=="19"]) > names(rawp) <- featureNames(ALL) > f <- matrix(NA,2,2) > f[1,1] <- sum(rawp[AffimIDChr]<0.05); f[1,2] <- sum(rawp[AffimIDChr]>0.05) > f[2,1] <- sum(rawp<0.05) - f[1,1] ; f[2,2] <- sum(rawp>0.05) - f[1,2] > print(f) [,1] [,2] [1,] 106 638 [2,] 832 11049 > fisher.test(f) Fisher's Exact Test for Count Data data: f p-value = 4.332e-11 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 1.757949 2.748559 sample estimates: odds ratio 2.206211112 CHAPTER 6. MICRO ARRAY ANALYSIS > chisq.test(f) Pearson's Chi-squared test with Yates' continuity correction data: f X-squared = 52.3803, df = 1, p-value = 4.573e-13 The number of signi¯cant probes is larger for Chromosome 19 resulting in an odds ratio of 2.2. The hypothesis of independence is rejected by both tests. 6.11 Overview and concluding remarks Many examples are given on using analysis of variance or T-tests for select- ing genes with large experimental e®ects on di®erent patient groups. The above statistical methods seem to cover the majority of problems occurring in practice. 6.12 Exercises 1. Gene ¯ltering on normality per group of B-cell ALL patients. (a) Use genefilter to program the Shapiro normality test separately for each gene of the groups "B1","B2","B3","B4". (b) How many pass the ¯lter? (c) Compute a Venn diagram for group "B2", "B3", and "B4", plot it, and give a correct interpretation for each number. 2. Analysis of gene expressions of B-cell ALL patients using Limma. (a) Construct a data frame containing the expression values for the B-cell ALL patients in stage B, B1, B2, B3, B4 from the ALL data. (b) Construct the design matrix and an appropriate contrast matrix. (c) Compute the twenty best genes by topTable. (d) Collect information on the twenty best gene s in an HTML page.6.12. EXERCISES 113 3. Finding a row number. Use grep to ¯nd the row number of gene 1389_at. Hint: Use row.names or featureNames. 4. Remission (genezing) from acute lymphocytic leukemia (ALL). With respect to the ALL data from the ALL library there is a phenotypical vari- able called remission indicating complete remission CR or refractory REF meaning improvement from disease and less or no improvement, respectively. (a) How many persons are classi¯ed as CR and REF, respectively? Hint: Use pData to extract a data frame with phenotypical data. (b) Program the two-sample t-test not assuming equal variances to select genes with p-values smaller than 0.001. Hint: You may have to select the persons with values on remission, excluding not available data. (c) Collect and give the manufactures probe names of the genes with p-values smaller than 0.001. (d) Use the probe names to ¯nd the corresponding gene names. Give the code. (e) Is the famous protein p53 is among them? (f) How many unique gene names are there? 5. Remission achieved. For the ALL data from its ALL library the patients are checked for achieving remission. The variable ALL$CR has values CR (became healthy) and REF (did not respond to therapy; remain ill). (a) Construct a separate data frame consisting of only those gene expression values from patients that have values CR or REF. (b) How many genes have a p-value smaller than 0.0001 from the two- sample T-test not assuming equal variances? Hint: Use the apply functionality to program the test. (c) Give the a®ymetrix names (symbols) of the genes the pass the selection criterion of p-value smaller than 0.0001. (d) Use the latter to ¯nd the biological names. (e) How many oncogenes are there is total?114 CHAPTER 6. MICRO ARRAY ANALYSIS (f) Do the Fisher test on the number of oncogenes out of the total versus the number of signi¯cant oncogenes out of the selected. 6. Gene ¯ltering of ALL data. The data are in the library called "ALL". The persons with T-cell leukemia which are in stage T2 and T3 can be selected by the variable ALL$BT. You may use the function "table" to ¯nd the frequencies of the patient types and leukemia stages. To answer the questions below functions from the library "gene¯lter" are helpful. (a) Program a gene ¯lter step separately for T2 and T3 patients such that only those genes pass which are normally distributed. (b) Program a second ¯lter step which passes only those genes with a signi¯cant p-value from the two sample T-test. (c) How many genes pass all ¯lter steps? (d) How many genes pass normality? 7. Stages of B-cell ALL in the ALL data. Use the limma package to answer the questions below. (a) Select the persons with T-cell leukemia which are in stage B1, B2, B3, and B4. (b) What type of contrast matrix would you like to suggest in this situation? Give its code. (c) Perform analysis of variance to test the hypothesis of equal pop- ulation means. Use the Benjamini & Hochberg (1995) ("BH") adjustment method for the false discovery rate and topTable to report the ¯ve best genes. (d) For how many genes is the null-hypothesis to be rejected? 8. Analysis of public micro array data on rheumatoid arthritis. (a) Download GDS486 and transform it into eset form. Here we meet a missing data problem. A manner to solve it is as follows. Use the function function(x) sum(is.na(x)) in apply on the rows to count the number of missing values per row. Select the rows without missing value to perform a two-sample t-test with the6.12. EXERCISES 115 groups in cell.line. Overwrite the vector with the number of missing values with the p-values in a suitable manner. (b) Download GDS711 and repeat the above using ANOVA p-values with the covariate disease.state to indicate the groups. (c) Download GDS2126 and repeat the above using ANOVA p-values with the covariate disease.state to indicate the groups. (d) Compute the symbols of the twenty best genes in the sense of having smallest summed p-values. (e) Summarize information of the twenty best genes in an HTML table. Does p53 play a role in the path way of the best gene? 9. Analysis of genes from a GO search. (a) Select the patients on the covariate mol.biol with values ALL1/AF4, BCR/ABL, and NEG. (b) Collect the ANOVA p-values with contrast between NEG and ALL1/AF4, and between NEG and BCR/ABL. Report the number of signi¯cant a®y ID's and the total. Hint: Re-order the columns into "NEG", "ALL1/AF4", and "BCR/ABL". (c) Find the GO ID's refereing to the term "protein-tyrosine kinase" since it mediates many steps due to BCR/ABL translocation. (d) Select the a®y ID's corresponding to the GO ID's and report its number and the number of signi¯cant genes. (e) Perform Fisher exact to test the odds ratio equal to one hypoth- esis.116 CHAPTER 6. MICRO ARRAY ANALYSISChapter 7 Cluster Analysis and Trees Given the expression values of several genes, a problem which often arises is to ¯nd genes which are similar or close. Genes with expressions in small distance may have similar functions and may be potentially interesting for further research. In order to discover genes which form a group there are sev- eral methods developed called cluster analysis. These methods are based on a distance function and an algorithm to join data points to clusters. The so- called single linkage cluster analysis is intuitively appealing and often applied in bioinformatics. By this method several clusters of genes can be discov- ered without specifying the number of clusters on beforehand. The latter is necessary for another method called k-means cluster analysis. Each analysis produces a tree which represents similar genes as close leaves and dissimilar ones on di®erent edges. An other measure to investigate similarity or dependency of pairs of gene expressions is the correlation coe±cient. Various examples of applications will be given. It prepares the way to searching a data set for directions of large variance. That is, since gene expression data sets tend to be large, it is of importance to have a method available which discovers important \directions" in the data. A frequently used method to ¯nd such directions is that by principal components analysis. Its basic properties will be explained as well as how it can be applied in combination with cluster analysis. In applications where it is di±cult to formulate distributional assumptions of the statistic it may still be of importance to construct a con¯dence interval. It will be illustrated by several examples how the bootstrap can be applied to construct 95% con¯dence intervals. Many examples are given to clarify the application of cluster analysis and principal components analysis. 117118 CHAPTER 7. CLUSTER ANALYSIS AND TREES 7.1 Distance The concept of distance plays a crucial role in all types of cluster analysis. For real numbers a and b a distance function d is de¯ned as the absolute value of their di®erenced(a; b) = ja ¡ bj = p(a ¡ b)2: The properties of a distance function should be in line with our intuition. That is, if a = b, then d(a; a) = 0 and if a 6= b, then d(a; b) > 0. Hence, the distance measure should be de¯nitive in the sense that d(a; b) = 0 if and only if a = b. Since the square is symmetric, it follows that d(a; b) = ja ¡ bj = p(a ¡ b)2 = p(b ¡ a)2 = jb ¡ aj = d(b; a): In other words, d(a; b) = d(b; a), the distance between a and b equals that between b and a. Furthermore, it holds for all points c between a and b that d(a; b) = d(a; c)+d(c; b). For all points c not between a and b, it follows that d(a; b) < d(a; c) + d(c; b). The latter two notions can be summarized by the so-called triangle inequality. That is, for all real c it holds that d(a; b) · d(a; c) + d(c; b): Directly going from a to b is shorter than via c. Finally, the distance between two points a and b should increase as these move further apart. Example 1. Let a = 1 and b = 3. Then, obviously, the distance d(1; 3) = 2. The number c = 2 is between a and b, so that d(1; 3) = 2 = 1 + 1 = d(1; 2) + d(2; 3) and the triangle inequality becomes an equality. For the situation where gene expression values for several patients are available, it is of importance to de¯ne a distance for vectors of gene expres- sions such as a = (a1; ¢ ¢ ¢ ; an) and b = (b1; ¢ ¢ ¢ ; bn). We shall concentrate mainly on the Euclidian distance, which is de¯ned as the root of the sum of the squared di®erences d(a; b) =vuutXn i=1 (ai ¡ bi)2:7.1. DISTANCE 119 The distance measure satis¯es the above properties of de¯niteness, symme- try, and triangle inequality. Although many other, but often highly similar, distance functions are available we shall mainly concentrate on Euclidian dis- tance because it is applied most frequently in bioinformatics. Example 2. Suppose that a = (a1; a2) = (1; 1) and b = (b1; b2) = (4; 5). Then d(a; b) = p(a1 ¡ b1)2 + (a2 ¡ b2)2 = p(1 ¡ 4)2 + (1 ¡ 5)2 = p9 + 16 = 5: Since the di®erences are squared it is immediate that d(a; b) = d(b; a), the distance from a to b equals that from b to a. For c = (c1; c2) = (2; 2) we have that d(a; c) = p2, d(b; c) = p22 + 32 = p13. Hence, d(a; b) = 5 < p2 + p13 = d(a; c) + d(b; c); so that the triangle inequality is strict. This is in line with our intuitive idea that the road directly from a to b is shorter than from a to b via c. Example 3. To compute the Euclidian distance between two vectors one may use the following. > a <- c(1,1); b <- c(4,5) > sqrt(sum((a-b)^2)) [1] 5 Example 4. Distances between Cyclin gene expressions. By the build-in- function dist the Euclidian distance between two vectors of gene expression values can be computed. To select genes related to the biological term "Cy- clin" and to compute the Euclidian distance between the gene expression values of the Golub et al. (1999) data, we may use the following. > library(multtest); data(golub) > index <- grep("Cyclin",golub.gnames[,2]) > golub.gnames[index,2] [1] "CCND2 Cyclin D2" [2] "CDK2 Cyclin-dependent kinase 2" [3] "CCND3 Cyclin D3" [4] "CDKN1A Cyclin-dependent kinase inhibitor 1A (p21, Cip1)" [5] "CCNH Cyclin H"120 CHAPTER 7. CLUSTER ANALYSIS AND TREES [6] "Cyclin-dependent kinase 4 (CDK4) gene" [7] "Cyclin G2 mRNA" [8] "Cyclin A1 mRNA" [9] "Cyclin-selective ubiquitin carrier protein mRNA" [10] "CDK6 Cyclin-dependent kinase 6" [11] "Cyclin G1 mRNA" [12] "CCNF Cyclin F" > dist.cyclin <- dist(golub[index,],method="euclidian") > diam <- as.matrix(dist.cyclin) > rownames(diam) <- colnames(diam) <- golub.gnames[index,3] > diam[1:5,1:5] D13639_at M68520_at M92287_at U09579_at U11791_at D13639_at 0.000000 8.821806 11.55349 10.056814 8.669112 M68520_at 8.821806 0.000000 11.70156 5.931260 2.934802 M92287_at 11.553494 11.701562 0.00000 11.991333 11.900558 U09579_at 10.056814 5.931260 11.99133 0.000000 5.698232 U11791_at 8.669112 2.934802 11.90056 5.698232 0.000000 By the grep function the order numbers of the genes with the phrase "Cy- clin" in their names are assigned to the vector called index. The euclidian distances are assigned to the matrix called diam. Its diagonal has distances between identical genes which are, of course, zero. The distance between the ¯rst (CCND2 Cyclin D2) and the third (CCND3 Cyclin D3) is relatively small, which is in line with the fact the these genes have related functions. Note, however, that there are genes with in smaller distance. Example 5. Finding the ten closest genes to a given one. After selecting certain genes it often happens that one wants to ¯nd genes which are close to the selected ones. This can be done with the genefinder functionality by specifying either an index or a name (consistent with the geneNames of the exprSet). To ¯nd genes from the ALL data (Chiaretti et al., 2004) close to the MME expression values of the probe with identi¯er 1389_at, we may use the following. library("genefilter"); library("ALL"); data(ALL) closeto1389_at <- genefinder(ALL, "1389_at", 10, method = "euc") closeto1389_at[[1]]$indices round(closeto1389_at[[1]]$dists,1) featureNames(ALL)[closeto1389_at[[1]]$indices]7.2. TWO TYPES OF CLUSTER ANALYSIS 121 The function genefilter produces a list from which the selected row num- bers can be extracted as well as the probe names can be found.1 If desired, these can be used for further analysis. From the output it can be observed that the gene expressions of row 2653 with probe identi¯er 32629_f_at has the smallest distance (12.6) to those of 1389_at. 7.2 Two types of Cluster Analysis Some important types of cluster analysis are de¯ned and illustrated here. 7.2.1 Single Linkage A cluster I is simply a set of data points I = fxig, where xi is the i-th vector with gene expressions. In single linkage cluster analysis the distance between clusters I and J is de¯ned as the smallest distance over all pairs of points of the two clusters: d(I; J) = min i;j fd(xi; xj ; ) : xi in I and xj in Jg : Hence, the distance between the two clusters is the same as that of the nearest neighbors. The algorithm of single linkage cluster analysis starts with creat- ing as many clusters as data points. Next, the nearest two are determined and these are merged into one cluster. Then the next two nearest clusters are determined and merged into one cluster. This process continuous until all points belong to one cluster. Example 1. An explanatory example. To illustrate single linkage cluster analysis suppose we the following ¯ve gene expressions g1 = (1; 1), g2 = (1; 1:2), g3 = (3; 2), g4 = (3; 2:2), and g5 = (5; 5), from two persons. The expressions for each gene can be seen as coordinates on two perpendicular axis p1 and p2. The script below produces Figure 7.1 which illustrates the idea. It computes the the distances between the genes and performs a single linkage cluster analysis. names <- list(c("g1","g2","g3","g4","g5"),c("p1","p2")) 1For information on lists, see Chapter 6 of the manual "An Introduction to R".122 CHAPTER 7. CLUSTER ANALYSIS AND TREES 1 2 3 4 5 1 2 3 4 5 a1 a2 xx12 x3 x4 x5 Figure 7.1: Plot of ¯ve points to be clustered. x5 x1 x2x3 x4 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Cluster Dendrogram hclust (*, "single") dist(sl.clus.dat, method = "euclidian") Height Figure 7.2: Tree of single linkage cluster analysis. sl.clus.dat <- matrix(c(1,1,1,1.1,3,2,3,2.3,5,5),ncol = 2, byrow = TRUE,dimnames = names) plot(sl.clus.dat,type="n", xlim=c(0,6), ylim=c(0,6)) text(sl.clus.dat,labels=row.names(sl.clus.dat)) > print(dist(sl.clus.dat,method="euclidian"),digits=3) x1 x2 x3 x4 x2 0.10 x3 2.24 2.19 x4 2.39 2.33 0.30 x5 5.66 5.59 3.61 3.36 > sl.out<-hclust(dist(sl.clus.dat,method="euclidian"),method="single") > plot(sl.out) At the start each data point is seen as a separate cluster. Then the nearest two points (genes) from the Euclidian distance matrix are g1 and g2, having d(g1; g2) = 0:10. These two data points are merged into one cluster, say I = fg1; g2g. In Figure 7.2 this is illustrated by the horizontal line at height 0.10 in the tree. The other three data points g3; g4; g5 are seen as three di®erent clusters. Next, the minimal distance between clusters can be read from the Euclidian distance matrix. Since the smallest is d(g3; g4) = 0:30, the new cluster J = fg3; g4g, corresponding to the horizontal line at height7.2. TWO TYPES OF CLUSTER ANALYSIS 123 0.30. Now there are three clusters, I, J, and K = fx5g. From the Euclidian distance matrix, it can be observed that the distance between cluster I and J is 2.19, see the corresponding horizontal line at this height. Hence, the cluster I and J are merged into one. Finally, the distance between cluster fg1; g2; g3; g4g, and the data point g5 equals d(g4; g5) = 3:36, see the corre- sponding horizontal line at this height. 9 2 1 15 354 18 8 14 20 10 17 11 16 19 7 13 6 12 0.0 0.2 0.4 0.6 0.8 1.0 Cluster Dendrogram hclust (*, "single") dist(rnorm(20, 0, 1), method = "euclidian") Height Figure 7.3: Example of three with- out clusters. 22 25 28 30 24 21 29 23 26 27 219 7 1056 438 16 19 2017 15 13 1812 11 14 0 1 2 3 4 5 Cluster Dendrogram hclust (*, "single") dist(x, method = "euclidian") Height Figure 7.4: Three clusters with dif- ferent standard deviations. Example 2. Relating data generation processes to cluster trees. It is of importance to have some experience with data that does and does not con- tain clusters. To illustrate this we perform single linkage cluster analysis on twenty data points from the standard normal population. sl.out<-hclust(dist(rnorm(20,0,1),method="euclidian"),method="single") plot(sl.out) From the resulting tree in Figure 7.3 one might get the impression that there are ¯ve separate clusters in the data. Note, however, that there is no underlying data generation process which produces separate clusters from di®erent populations. If, however, the data are generated by di®erent normal distributions, then there are di®erent processes producing separate clusters. To illustrate124 CHAPTER 7. CLUSTER ANALYSIS AND TREES this, ten data points were sampled from the N(0; 0:1) population, ten from N(3; 0:5), and ten from N(10; 1). x <- c(rnorm(10,0,0.1),rnorm(10,3,0.5),rnorm(10,10,1.0)) plot(hclust(dist(x,method="euclidian"),method="single")) plot(sl.out) From the tree in Figure 7.4, it can be observed that there clearly exist three clusters. These examples illustrate that results from cluster analysis may very well reveal population properties, but that some caution is indeed in order. −0.5 0.0 0.5 1.0 1.5 2.0 2.5 −1 0 1 2 CCND3 Cyclin D3 Zyxin ALL AML Figure 7.5: Plot of gene "CCND3 Cyclin D3" and "Zyxin" expres- sions for ALL and AML patients. 21 35 29 16 20 10 19 5 12 26 4 13 15 22 24 1 36 11 23 8 2779 25 17 182 14 34 30 37 32 36 38 33 28 31 0.0 0.2 0.4 0.6 0.8 1.0 1.2 Cluster Dendrogram hclust (*, "single") dist(clusdata, method = "euclidian") Height Figure 7.6: Single linkage cluster diagram from gene "CCND3 Cy- clin D3" and "Zyxin" expressions values. Example 3. Application to the Golub (1999) data. Recall that the ¯rst twenty seven patients belong to ALL and the remaining eleven to AML and that we found earlier that the expression values of the genes "CCND3 Cyclin D3" and "Zyxin" di®er between the patient groups ALL and AML. Figure 7.5 illustrates that the patient groups di®er with respect to gene expression values. How to produce this plot and a single linkage cluster analysis is shown by the script below.7.2. TWO TYPES OF CLUSTER ANALYSIS 125 data(golub, package="multtest") clusdata <- data.frame(golub[1042,],golub[2124,]) colnames(clusdata)<-c("CCND3 Cyclin D3","Zyxin") gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) plot(clusdata, pch=as.numeric(gol.fac)) legend("topright",legend=c("ALL","AML"),pch=1:2) plot(hclust(dist(clusdata,method="euclidian"),method="single")) Figure 7.6 gives the tree from single linkage cluster analysis. Apart from three expressions the tree shows two clusters corresponding to the two pa- tient groups. 7.2.2 k-means K-means cluster analysis is a popular method in bioinfomatics. It is de¯ned by minimizing the within cluster sum of squares over K clusters. That is, given the data points x1; ¢ ¢ ¢ ; xn the method seeks to minimize the function XK k=1 Xnk i2Ik d2(xi; ak) over all possible points a1; ¢ ¢ ¢ aK. This is accomplished by an algorithm (Hartigan & Wong, 1979) which starts by partitioning the data points into K initial clusters, either at random or using some heuristic device. It then computes the cluster means (step 1) and constructs a new partition by asso- ciating each point with the closest cluster mean (step 2). The latter yields new clusters of which the means are calculated (step 1). Then it constructs a new partition by associating each point with the closest cluster mean (step 2). These two steps are repeated until convergence. The latter occurs when the data points no longer change clusters. The iterative algorithm is fast in the sense that it often converges in less iterations than the number of points n, but it need not to attain the global minimum. For the optimal points a1; ¢ ¢ ¢ aK, it holds that these are equal to the mean per cluster, that is ak = xk for each cluster k. When the data points are independent and identically distributed, then the cluster means converge in probability to the corresponding population means (Pollard, 1981).126 CHAPTER 7. CLUSTER ANALYSIS AND TREES −1 0 1 2 3 −1 0 1 2 3 data[,1] data[,2] Figure 7.7: K-means cluster anal- ysis. x5 x1 x2x3 x4 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Cluster Dendrogram hclust (*, "single") dist(sl.clus.dat, method = "euclidian") Height Figure 7.8: Tree of single linkage cluster analysis. Example 1. Relating a data generation process to k-means cluster analysis. To illustrate k-means cluster analysis we shall simulate gene expressions from two di®erent normal populations. That is, we randomly take ¯fty gene ex- pressions for two persons from the N(0; 0:5) population and ¯fty expressions for two persons from the N(2; 0:5) population. The data points are collected in two matrices of order ¯fty by two which are placed one above the other. On the total of one hundred data points a (k =)2-means cluster analysis is performed. > data <- rbind(matrix(rnorm(100,0,0.5), ncol = 2), + matrix(rnorm(100,2,0.5), ncol = 2)) > cl <- kmeans(data, 2) K-means clustering with 2 clusters of sizes 50, 50 Cluster means: [,1] [,2] 1 1.87304978 2.01940342 2 0.01720177 0.07320413 Clustering vector: [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 7.2. TWO TYPES OF CLUSTER ANALYSIS 127 [38] 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 [75] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Within cluster sum of squares by cluster: [1] 22.60733 20.54411 Available components: [1] "cluster" "centers" "withinss" "size" The output of k-means cluster analysis is assigned to a list called cl. Observe that the cluster means are fairly close to the population means (0; 0) and (2; 2). The Clustering vector indicates to which cluster each data point (gene) belongs and that these correspond exactly to the two populations from which the data are sampled. The variable cl$cluster contains cluster membership and can be used to specify the color of each data point a plot, as follows. > plot(data, col = cl$cluster) > points(cl$centers, col = 1:2, pch = 8, cex=2) The data points are plotted in red and black circles and the cluster means by a star, see Figure 7.7. The sum of the within cluster sum of squares equals the minimal function value obtained by the algorithm. Before performing a k-means cluster analysis a plot from a single linkage cluster analysis may reveal the number of clusters. If the number of clusters is not at all clear, then it becomes questionable whether k-means is appropriate. For cases where the number of clusters is only moderately clear, the algorithm is more sensible to get stuck into a solution which is only locally optimal. Such solutions are of limited scienti¯c value. To cope with the danger of suboptimal solutions one may simply run the algorithm repeatedly by using the nstart option. Another possibility is to use rational initial starting values for the cluster means. In particular, the sample means of potential clusters or the hypothesized population means can be used. > initial <- matrix(c(0,0,2,2), nrow = 2, ncol=2, byrow=TRUE) > cl<- kmeans(data, initial, nstart = 10) The so-called bootstrap (Efron, 1979) can be used to estimate 95% con¯dence intervals around cluster means. The idea is to re-sample with replacement from the given sample one thousand times with replacement and to compute quantiles for the corresponding con¯dence intervals.128 CHAPTER 7. CLUSTER ANALYSIS AND TREES n <- 100; nboot<-1000 boot.cl <- matrix(0,nrow=nboot,ncol = 4) for (i in 1:nboot){ dat.star <- data[sample(1:n,replace=TRUE),] cl <- kmeans(dat.star, initial, nstart = 10) boot.cl[i,] <- c(cl$centers[1,],cl$centers[2,]) }> quantile(boot.cl[,1],c(0.025,0.975)) 2.5% 97.5% -0.1098886 0.1627979 > quantile(boot.cl[,2],c(0.025,0.975)) 2.5% 97.5% -0.04830563 0.19721732 > quantile(boot.cl[,3],c(0.025,0.975)) 2.5% 97.5% 1.730495 2.009014 > quantile(boot.cl[,4],c(0.025,0.975)) 2.5% 97.5% 1.898407 2.162019 From the bootstrap con¯dence intervals the null hypothesis that the cluster population means are equal to (0; 0) and (2; 2) are accepted. Example 2. Application to the Golub (1999) data. In the above we found that the expression values of the genes "CCND3 Cyclin D3" and "Zyxin" are closely related to the distinction between ALL and AML. Hence, a 2-means cluster analysis of these gene expression values is appropriate here. > data <- data.frame(golub[1042,],golub[2124,]) > colnames(data)<-c("CCND3 Cyclin D3","Zyxin") > cl <- kmeans(data, 2,nstart = 10) > cl K-means clustering with 2 clusters of sizes 11, 27 Cluster means: CCND3 Cyclin D3 Zyxin 1 0.6355909 1.5866682 2 1.8938826 -0.29479267.2. TWO TYPES OF CLUSTER ANALYSIS 129 Clustering vector: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 27 28 29 30 31 32 33 34 35 36 37 38 2 1 1 1 1 1 1 1 1 1 1 1 Within cluster sum of squares by cluster: [1] 4.733248 19.842225 The two clusters discriminate exactly the ALL patients from the AML pa- tients. This can also be seen from Figure 7.9, where expression values of CCND3 Cyclin D3 are depicted on the horizontal axis and those of Zyxin on the vertical, and the ALL patients are in red and the AML patients in black. By the bootstrap the cluster means and their con¯dence intervals can be estimated. > mean(data.frame(boot.cl)) X1 X2 X3 X4 0.6381860 1.5707477 1.8945878 -0.2989426 > quantile(boot.cl[,1],c(0.025,0.975)) 2.5% 97.5% 0.2548907 0.9835898 > quantile(boot.cl[,2],c(0.025,0.975)) 2.5% 97.5% 1.259608 1.800581 > quantile(boot.cl[,3],c(0.025,0.975)) 2.5% 97.5% 1.692813 2.092361 > quantile(boot.cl[,4],c(0.025,0.975)) 2.5% 97.5% -0.60802142 -0.02420802 The di®erence between the bootstrap means and the k-means from the orig- inal data gives an estimate of the estimation bias. It can be observed that the bias is small. The estimation is quite precise because the 95% bootstrap con¯dence intervals are fairly small.130 CHAPTER 7. CLUSTER ANALYSIS AND TREES −0.5 0.0 0.5 1.0 1.5 2.0 2.5 −1 0 1 2 CCND3 Cyclin D3 Zyxin Figure 7.9: Plot of kmeans (stars) cluster analysis on CCND3 Cyclin D3 and Zyxin discriminating between ALL (red) and AML (black) patients. 7.3 The correlation coe±cient A frequently used coe±cient to express the degree of linear relationship between two sets of gene expression values is the correlation coe±cient ½. For two sequences of gene expressions such as x = (x1; ¢ ¢ ¢ ; xn) and y = (y1; ¢ ¢ ¢ ; yn), the correlation coe±cient ½ is estimated by b½ = Pni=1(xi ¡ xi)(yj ¡ yj) qPni=1(xi ¡ xi)2Pni=1(yj ¡ yj)2 : The value of the correlation coe±cient is always between minus one and plus one. If the value is close to either of these values, then the variables are7.3. THE CORRELATION COEFFICIENT 131 linearly related in the sense that the ¯rst is a linear transformation of the second. That is, there are constants a and b such that axi + b = yi for all i. By the function cor.test, the null hypothesis H0 : ½ = 0 can be tested against the alternative H0 : ½ 6= 0. Example 1. Teaching demonstration. To develop intuition with respect to the correlation coe±cient the function run.cor.examp(1000) of the TeachingDemos package is quite useful. It launches an interactive plot with 1000 data points on two random variables X and Y . When the correlation is near zero, then the data points are distributed along contours of circles. By moving the slider slowly from the left to the right it can be observed that all points are approximately on a straight line. If the sign of the correlation coe±cient is positive, then small/large values of X tend to go together with small/large values of Y . Example 2. Another teaching demonstration. By the function put.points.demo() it is possible to add and delete points to a plot which interactively re- computes the value for the correlation coe±cient. By ¯rst creating a few points that lie together on a circle the corresponding correlation coe±cient will be near zero. By next adding one outlier, it can be observed that the correlation coe±cient changes to nearly §1. This illustrates that the corre- lation coe±cient is not robust against outliers. Example 3. Application to the Golub (1999) data. We shall illustrate the correlation coe±cient by two sets of expression values of the MCM3 gene of the Golub et al. (1999) data. This gene encodes for highly conserved mini-chromosome maintenance proteins (MCM) which are involved in the initiation of eukaryotic genome replication. Here, we ¯nd its row numbers, collect the gene expression value in vectors x and y, and compute the value of the correlation coe±cient by the function cor(x,y). > library(multtest); data(golub) > x <- golub[2289,]; y <- golub[2430,] > cor(x,y) [1] 0.6376217 The value is positive which means that larger values of x occur together with larger values of y and vice versa. This can also be observed by plot(x,y). By132 CHAPTER 7. CLUSTER ANALYSIS AND TREES the function cor.test, the null hypothesis H0 : ½ = 0 can be tested against the alternative H0 : ½ 6= 0. It also estimates a 95% con¯dence interval for ½. > cor.test(x,y) Pearson's product-moment correlation data: x and y t = 4.9662, df = 36, p-value = 1.666e-05 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.3993383 0.7952115 sample estimates: cor 0.6376217 The test is based on the normality assumption and prints therefore a t-value. Since the corresponding p-value is very small, we reject the null hypothesis of zero correlation. The left bound of the con¯dence interval falls far to the right hand side of zero. Example 4. Con¯dence interval by the bootstrap. Another method to con- struct a 95% con¯dence interval is by the bootstrap. The idea (Efron, 1979) is to obtain thousand samples from the original sample with replacement and to compute the correlation coe±cient for each of these. This yields thousand coe±cients from which the quantiles for the 95% con¯dence interval can be computed. > nboot <- 1000; boot.cor <- matrix(0,nrow=nboot,ncol = 1) > data <- matrix(c(x,y),ncol=2,byrow=FALSE) > for (i in 1:nboot){ + dat.star <- data[sample(1:nrow(data),replace=TRUE),] + boot.cor[i,] <- cor(dat.star)[2,1]} > mean(boot.cor) [1] 0.6534167 > quantile(boot.cor[,1],c(0.025,0.975)) 2.5% 97.5% 0.2207915 0.92048657.4. PRINCIPAL COMPONENTS ANALYSIS 133 Observe that the 95% con¯dence interval is larger than that found by cor.test. This indicates that the assumption of normality may not be completely valid here. Since the con¯dence interval does not contain zero, we reject the null- hypothesis of zero correlation. Example 5. Application to the Golub (1999) data. The ALL and AML patients of the Golub et al. (1999) data are indicated by zero and ones of the binary vector golub.cl. A manner to select genes it by the correlation of the expression values with this binary vector. Such can be computed by using the apply functionality. > library(multtest); data(golub) > corgol<- apply(golub, 1, function(x) cor(x,golub.cl)) > o <- order(corgol) By golub.gnames[o[3041:3051],2] it can be seen that various of these genes seem indeed to have important cell functions referred to by Golub et al. (1999). In particular, Interleukin 8 is recently related to in°ammatory cytokine production in myeloid cells (Tessarz et al., 2007). 7.4 Principal Components Analysis To make the basic ideas behind principal components analysis explicit, it is wise to start with a small arti¯cial example. Suppose that for six genes the standardized expression values on two patients (variables) became available as these are given in Table 7.1. The data are collected in a 6 by 2 data matrix Z, where e.g. element z21 is expression value -0.40 which belongs to the second gene of the ¯rst patient. The whole idea of principal components analysis is to ¯nd new directions in the data along which there is maximal variation. A direction is de¯ned as a linear combination Zk of the data Z by a vector k with weights. The i-th element of the linear combination is the weighted sum P2j=1 zijkj . The direc- tion of maximal variation is de¯ned as the linear combination with maximal variance. To ¯nd this direction the correlation matrix plays an important role. The latter contains the correlations between each pair of patients (vari- ables). In our case correlations between the columns (patients) in Table 7.1134 CHAPTER 7. CLUSTER ANALYSIS AND TREES Table 7.1: Data set for principal components analysis. Var 1 Var 2 gene 1 1.63 1.22 gene 2 ¡0.40 0.79 gene 3 0.93 0.97 gene 4 ¡1.38 ¡1.08 gene 5 ¡0.17 ¡0.96 gene 6 ¡0.61 ¡0.93 can be placed in a matrix R, which has ones on the diagonal and the value 0.8 elsewhere. To illustrate a direction let's try the linear combination k = (2; 1)2 of the sample correlation matrix R. This gives Rk = · 1 0:8 0:8 1 ¸ · 21 ¸ = · 2:8 2:6 ¸ : Both vectors k and Rk can be plotted in the xy-plane. The vector (2; 1) is plotted by drawing an arrow from (0,0) to the point with x = 2 and y = 1. This is done completely similar for (2:8; 2:6) in Figure 7.10. It can be observed that the two vectors (arrows) do not fall on the same line and therefore have di®erent directions. The crux of principal components analysis is that a linear combination with the same direction as the weights represent the direction of maximum variation. Such is the case if Rk di®ers from k only by a constant of multiplication, that is there exists a constant d such that Rk = dk. We shall determine such a constant by ¯nding the weights vector ¯rst. To do so observe from our correlations matrix that the sum of both rows equals 1.8. Taking k = (1; 1) yields Rk = · 1 0:8 0:8 1 ¸ · 11 ¸ = · 1:8 1:8 ¸ = 1:8 · 11 ¸ = 1:8k: So that we obtain d = 1:8. A similar result follows by observing that the di®erences per row are equal in absolute value. That is, taking k = (1;¡1) 2For the sake of simple notation we shall not use the transposition operator T to indicate rows.7.4. PRINCIPAL COMPONENTS ANALYSIS 135 yields Rk = · 1 0:8 0:8 1 ¸ · 1 ¡1 ¸ = · 0:2 ¡0:2 ¸ = 0:2 · 1 ¡1 ¸ = 0:2k: A vector k for which Rk = dk holds is called an eigenvector corresponding to the eigenvalue d. Eigenvectors are often re-scaled by dividing by their Euclid- ian length. Since the Euclidian length of (1; 1) is p12 + 12 = p2, we obtain the new eigenvector k1 = (1=p2; 1=p2) t (0:71; 0:71). Since the length of eigenvector (1;¡1) also equals p2 the re-scaled second eigenvector equals k2 = (1=p2;¡1=p2) t (0:71;¡0:71). Now the ¯rst principal component is de¯ned as Zk1 and the second as Zk2. In practical applications the actual computation of eigenvectors and eigenvalues is performed by well-designed numerical methods (Golub & Van Loan, 1983). 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 V[,1] V[,2] Figure 7.10: Vectors of linear com- binations. −2 −1 0 1 2 −2 −1 0 1 2 Z[,1] Z[,2] Figure 7.11: First principal com- ponent with projections of data. Example 1. Using R on the above data. It is convenient to store the data of the ¯rst two columns of Table 7.1 as a matrix object called Z. The correlations matrix can be computed by the built-in-function cor and the eigenvectors and eigenvalues by the built-in-function eigen, as follows. Z <- matrix(c( 1.63, 1.22, -0.40, 0.79, 0.93, 0.97, -1.38,136 CHAPTER 7. CLUSTER ANALYSIS AND TREES -1.08, -0.17, -0.96, -0.61, -0.93), nrow=6, byrow=TRUE) K <- eigen(cor(Z)) The output is stored as an object called K which can be printed to the screen in two digits. > print(K,digits=2) $values [1] 1.8 0.2 $vectors [,1] [,2] [1,] 0.71 0.71 [2,] 0.71 -0.71 The eigenvalues are assigned to K$values and the eigenvectors are the columns of K$vectors. To compute the principal components we use the matrix mul- tiplication operator %*%. Then the ¯rst principal component is de¯ned as the linear combination of the data with the ¯rst eigenvector, Z%*%K$vec[,1]. To print the scores on the ¯rst and the second principal component one can use the following. > print(Z %*% K$vec, digits=2) [,1] [,2] [1,] 2.02 0.290 [2,] 0.28 -0.841 [3,] 1.34 -0.028 [4,] -1.74 -0.212 [5,] -0.80 0.559 [6,] -1.09 0.226 To illustrate the ¯rst principal component the six data points from the Z matrix are plotted as small circles in Figure 7.11. Gene 1, for instance, has x coordinate 1.63 and y coordinate 1.22 and appears therefore in the right upper corner. A convenient manner to perform principal components analysis is by using the built-in-function princomp, as follows. pca <- princomp(Z, center = TRUE, cor=TRUE, scores=TRUE) pca$scores7.4. PRINCIPAL COMPONENTS ANALYSIS 137 The scores are the component scores and the loadings from princomp are the eigenvectors. The eigenvalues represent an amount of variance related to the compo- nent. In the previous example the ¯rst component has variance 1.8 and the second 0.2, so that the ¯rst represents 1:8=2 = 0:9 or 90% of the variance. On the basis of the eigenvalues the number of interesting directions in the data can be evaluated by two rules of thumb. The ¯rst is that each eigenvalue should represent more variance than that of any of the observed variables. The second is the so-called elbow rule saying that when the ¯rst few eigen- values are large and the remaining considerably smaller, then the ¯rst few are the most interesting. Principal components analysis is a descriptive method to analyze depen- dencies (correlations) between variables. If there are a few large eigenvalues, then there are equally many directions in the data which summarize the most important variation among the gene expressions. Then it may be use- ful to explore simultaneously a two dimensional visualization of the genes and the patients. Furthermore, it can be rewarding to study the weights of the eigenvectors because these may reveal a structure in the data otherwise gone unnoticed. Finally, the principal components contain less (measure- ment) error than the individual variables. For this reason, cluster analysis on the values on the principal components may be useful. Example 2. Application to the Golub (1999) data. The ¯rst ¯ve eigenvalues from the correlation matrix of golub can be printed by the following. > eigen(cor(golub))$values[1:5] [1] 25.4382629 2.0757158 1.2484411 1.0713373 0.7365232 Because the eigenvalues are arranged in decreasing order the sixth to the 38th are smaller than one. Reason for which these will be neglected. The ¯rst eigenvalue is by far the largest, indicating that the persons are dependent to a large extent. Applying the previous bootstrap methods to estimate 95% con¯dence intervals for the eigenvalues we obtain the following intervals. data <- golub; p <- ncol(data); n <- nrow(data) ; nboot<-1000 eigenvalues <- array(dim=c(nboot,p)) for (i in 1:nboot){dat.star <- data[sample(1:n,replace=TRUE),] eigenvalues[i,] <- eigen(cor(dat.star))$values}138 CHAPTER 7. CLUSTER ANALYSIS AND TREES > for (j in 1:p) print(quantile(eigenvalues[,j],c(0.025,0.975))) 2.5% 97.5% for (j in 1:5) cat(j,as.numeric(quantile(eigenvalues[,j], + c(0.025,0.975))),"\n" ) 1 24.83581 26.00646 2 1.920871 2.258030 3 1.145990 1.386252 4 0.9917813 1.154291 5 0.6853702 0.7995948 The cat function allows for much control in printing. The null hypothesis of eigenvalue being equal to one is accepted for the fourth component and rejected for the ¯rst three and the ¯fth. Thus the fourth represents less variance than an individual variable, reason for which it is neglected. The percentages of variance explained by the ¯rst two components can be computed by sum(eigen(cor(golub))$values[1:2])/38*100, which yields the amount 72.4052%. Thus the ¯rst two components represent more than 72% of the variance in the data. Hence, the data allow for a reduction in dimensions from thirthy eight to two. −10 −5 0 5 10 15 −10 −5 0 5 10 X1 X2 68182 313 504 792808 885 893 892 1101 1616 1737 1754 1756 1798 18821910 1911 2233 2321 2350 2397 2459 2611 2653 2673 2761 2749 450 2874 849 316 2289 2430 Figure 7.12: Scatter plot of se- lected genes with row labels on the ¯rst two principal components. 1910 28742350 1737 1911 2749 2761 792 2321 68 808 2611 450 504 313 1756 893 182 2653 2289 2430 316 1616 1101 2673 2459 892 1882 885 849 2397 1754 1798 2233 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Cluster Dendrogram hclust (*, "single") dist(leu, method = "euclidian") Height Figure 7.13: Single linkage cluster diagram of selected gene expres- sion values. It can be checked that all correlations between the patients are positive.7.4. PRINCIPAL COMPONENTS ANALYSIS 139 This implies that large expression values on gene i co-vary positively with large deviations of gene j. The positivity of the correlations also implies that the weights of the ¯rst eigenvector have the same sign, so that these can be taken to be positive for all patients (Horn & Johnson, 1985). Unfortunately, this is not automatic in R so that caution is in order with respect to inter- pretation of the components. By using -eigen(cor(golub))$vec[,1:2] to print the weights to the screen it can be observed that those that correspond to the ¯rst component are positive. All weights of the ¯rst eigenvector are positive and have very similar size (all are between 0.13 and 0.17). Thus the ¯rst component is almost equal to the sum of the variables (the correlation equals 0.9999). The weights of the second component have a very interesting pattern. Namely, almost all of the ¯rst 27 weights are positive and the last 11 weights are negative. Thus the second component contrasts the ALL patients with the AML patients. By contrasting ALL patients with AML patients a second to the largest amount of variance is explained in the data. Hence, the AML-ALL distinction is discovered by the second component, which is in line with ¯ndings of Golub et al. (1999). Obviously the genes with the largest expression values from the ¯rst com- ponent can be printed. We shall, however, concentrate on the second compo- nent because it appears to be more directly related to the research intentions of Golub et. al. (1999). The ¯rst and the last ten gene names with respect to the values on the second component can be printed by the following. > pca <- princomp(golub, center = TRUE, cor=TRUE, scores=TRUE) > o <- order(pca$scores[,2]) > golub.gnames[o[1:10],2] > golub.gnames[o[3041:3051],2] Many of these genes are related to leukemia (Golub, et al., 1999). Example 3. Biplot. A useful manner to plot both genes (cases) and patients (variables) is the biplot, which is based on a two-dimensional approximation of the data very similar to principal components analysis. Here, we illustrate how it can be combined with principal components analysis. > biplot(princomp(data,cor=TRUE),pc.biplot=TRUE,cex=0.5,expand=0.8) The resulting plot is given by Figure 7.14. The left and bottom axis refer to the component scores and the top and right to the patient scores, which140 CHAPTER 7. CLUSTER ANALYSIS AND TREES are scaled to unit length by the speci¯cation cor. It can be seen that the patients are clearly divided in two groups corresponding to ALL and AML. Example 4. Critical for S-phase. Golub et al. (1999) mention that among genes which are useful for tumor class prediction there are genes that encode for proteins critical for S-phase cell cycle progression such as Cyclin D3, Op18, and MCM3. We select genes which carry "CD", "Op", or "MCM" in their names and collect the corresponding row numbers. data(golub, package = "multtest") factor <- factor(golub.cl) o1 <- grep("CD",golub.gnames[,2]) o2 <- grep("Op",golub.gnames[,2]) o3 <- grep("MCM",golub.gnames[,2]) o <- c(o1,o2,o3) This yields 110 genes. In order to select those that do have an experimental e®ect, we use a two-sample t-test. pt <- apply(golub, 1, function(x) t.test(x ~ gol.fac)$p.value) oo <- o[pt[o]<0.01] This yields 34 genes, of which the row numbers are selected in the vector oo. In order to identify genes in directions of large variation we use the scores on the ¯rst two principal components. Z <- as.matrix(scale(golub, center = TRUE, scale = TRUE)) K <- eigen(cor(Z)) P <- Z %*% -K$vec[,1:2] leu <- data.frame(P[oo,], row.names= oo) attach(leu) The scores on the ¯rst two principal components of the selected genes are stored in the data frame leu. From the plotted component scores in Figure 7.12, it seems that there are several sub-clusters of genes. The genes that belong to these clusters can be identi¯ed by hiearchical cluster analysis. cl <- hclust(dist(leu,method="euclidian"),method="single") plot(cl)7.5. OVERVIEW AND CONCLUDING REMARKS 141 From the tree (dendrogram) in Figure 7.13 various clusters of genes are apparent that also appear in Figure 7.12. 3 The ordered genes can be obtained from the object cl as follows. > a <- as.integer(rownames(leu)[cl$order]) > for (i in 1:length(a)) cat(a[i],golub.gnames[a[i],2],"\n") 1910 FCGR2B Fc fragment of IgG, low affinity IIb, receptor for (CD32) 2874 GB DEF = Fas (Apo-1, CD95) The cluster with rows 504, 313, 1756, and 893 consists of antigenes. The genes MCM3 Minichromosome maintenance de¯cient (S. cerevisiae) 3 with row numbers 2289 and 2430 appear adjacent to each other. This illustrates that genes with similar functions may indeed be close with respect to their gene expression values. 7.5 Overview and concluding remarks Single linkage cluster analysis can be applied to explore for groups in a set of gene expressions. When groups are present a k-means cluster analysis can be applied in combination with the bootstrap to estimate con¯dence intervals for the cluster means. The correlation coe±cient measures the degree of dependency between pairs of gene expression values. It can also be used to ¯nd gene expressions which are highly dependent with a phenotypical variable. It is reassuring to ¯nd in applications that the con¯dence interval for a correlation coe±cient is small. Principal components analysis is very useful for ¯nding directions in the data where the gene expression values vary maximally, see Jolli®e (2002) for a complete treatment of the principal component analysis. When these di- rections can be represented well by the ¯rst two components a biplot helps to simultaneously visualize genes and patients. Principal components analysis can be useful in identifying clusters of genes in a lower dimensional space. 3Unfortunately, some row numbers of genes are less readable because the points are very close.142 CHAPTER 7. CLUSTER ANALYSIS AND TREES 7.6 Exercises 1. Cluster analysis on the "Zyxin" expression values of the Golub et al. (1999) data. (a) Produce a chatter plot of the gene expression values using showing di®erent symbols for the two groups. (b) Use single linkage cluster analysis to see whether the three indi- cates two di®erent groups. (c) Use k-means cluster analysis. Are the two clusters according to the diagnosis of the patient groups? (d) Perform a bootstrap on the cluster means. You will have to modify the code here and there. Do the con¯dence intervals for the cluster means overlap? 2. Close to CCND3 Cyclin D3. Recall that we did various analysis on the expression data of the CCND3 Cyclin D3 gene of the Golub (1999) data. (a) Use genefilter to ¯nd the ten closed genes to the expression values of CCND3 Cyclin D3. Give their probe as well as their biological names. (b) Produce of combined boxplot separately for the ALL and the AML expression values. Compare it with that on the basis of CCND3 Cyclin D3 and comment of the similarities. (c) Compare the smallest distances with those among the Cyclin genes computed above. What is your conclusion? 3. MCM3. In the example on MCM3 a plot shows that there is an outlier. (a) Plot the data and invent a manner to ¯nd the row number of the outlier. (b) Remove the outlier, test the correlation coe±cient. Compare the results to those above. (c) Perform the bootstrap to construct a con¯dence interval. 4. Cluster analysis on part of Golub data.7.6. EXERCISES 143 (a) Select the oncogenes from the Golub data and plot the tree from a single linkage cluster analysis. (b) Do you observe meaningful clusters. (c) Select the antigenes and answer the same questions. (d) select the receptor genes and answer the same questions. 5. Principal Components Analysis on part of the ALL data. (a) Construct an expression set with the patients with B-cell in stage B1, B2, and B3. Compute the corresponding ANOVA p-values of all gene expressions. Construct the expression set with the p- values smaller than 0.001. Report the dimensionality of the data matrix with gene expressions. (b) Are the correlations between the patients positive? (c) Compute the eigenvalues of the correlation matrix. Report the largest ¯ve. Are the ¯rst three larger than one? (d) Program a bootstrap of the largest ¯ve eigenvalues. Report the bootstrap 95% con¯dence intervals and draw relevant conclusions. (e) Plot the genes in a plot of the ¯rst two principal components. 6. Some correlation matrices. · 1 ¡0:8 ¡0:8 1 ¸ ;24 1 0:8 0:8 0:8 1 0:8 0:8 0:8 1 35 ;24 1 ¡0:5 ¡0:5 ¡0:5 1 ¡0:5 ¡0:5 ¡0:5 1 35 ; (a) Verify that the eigenvalues of the matrices are 1.8, 0.2, 2.6, 0.2, 0.2, and 1.500000e+00, 1.500000e+00, -7.644529e-17. (b) How much variance represents the ¯rst component corresponding to the second matrix? (c) Verify that the ¯rst eigen vector of the second correlation matrix has identical signs.144 CHAPTER 7. CLUSTER ANALYSIS AND TREES −3 −2 −1 0 1 2 −3 −2 −1 0 1 2 Comp.1 Comp.2 23 68 96 108 182 126 202 244 246 259 313 323 329 345 357376 377 378 394 422 462 494 522 523 546 561 563 566 571 621 648 703 713 704 717 725 735 738 746 766 786 801 803 808 829 838 839 866 896 888 922 937 932 938 968 984 1006 1030 1037 1042 10451060 1066 1069 1081 1086 1109 1110 1145 1162 1206 1245 1271 1334 1348 1368 1396 1413 1445 1448 1455 11552442 1556 1585 1598 1638 1640 1642 1652 1653 1665 1732 1754 1774 1778 1817 1829 1834 1856 1869 1882 1887 1901 1909 1911 1920 1916 1939 1959 1978 1995 20022020 2065 2122 2124 2172 2179 2198 2265 2266 2307 2289 2343 2347 2356 2386 2418 2410 2459 2466 2489 2553 2589 2593 2600 2616 2627 2645 2656 2663 2664 2673 2702 2734 2749 2786 2801 2794 2813 2821 2829 2851 2889 222999222102 2937 2939 2950 2955 3046 −1.0 −0.5 0.0 0.5 −1.0 −0.5 0.0 0.5 1 2 3 45 678 9 10 11 12 13 14 116517 18 19 20 21 22 23 24 252627 28 29 30 31 32 33 34 35 3376 38 Figure 7.14: Biplot of selected genes from the golub data.Chapter 8 Classi¯cation Methods In medical settings groups of patients are often diagnosed into classes corre- sponding to types of diseases. In bioinformatics the question arises whether the diagnosis of a patient can be predicted by gene expression values? Re- lated is the question which genes play an important role in the prediction of class membership. A similar question is the prediction of micro RNA's from values of folding energy. More generally, for objects like proteins, mRNA's, or microRNA's it may be of importance to classify these on the basis of certain measurements. Many classi¯cation methods have been developed for various scienti¯c purposes. In bioinformatics, methods such as recursive partitioning, support vector machine and neural network are frequently applied to solve classi¯ca- tion problems. In this chapter you learn what recursive partitioning is and how to use it. To evaluate the quality of prediction the fundamental concepts of sensitivity and speci¯city are frequently used. The speci¯city can be summarized in a single number by the area under the curve of a receiver operator curve. This will be explained and illustrated. Two other methods to predict disease class from gene expression data are the support vector machine and the neural network. It will brie°y be explained what these methods are about and how these can be applied. A validation set will be used to evaluate the predictive accuracy. 145146 CHAPTER 8. CLASSIFICATION METHODS 8.1 Classi¯cation of microRNA The subject of making a correct medical diagnosis is highly similar to that of correctly classifying microRNA. Example 1. Classi¯cation of Micro RNA. MicroRNA are small RNA molecules with important functions in cell growth and disease development. In order to identify microRNA's from arbitrary sequences its characterizing proper- ties are used to distinguish non-microRNA from microRNA molecules. One of these properties is that microRNA's have the capacity to fold in a cer- tain hairpin type of structure. Such a structure typically exhibits a small minimum folding energy (Zuker, 2003; Zuker & Stiegler, 1981). This prop- erty can be used as a test to discriminate microRNA's from non-microRNA's (Bonnet, et al., 2004), as follows. Given a set of 3424 di®erent microRNA's the minimum folding energy was computed for each of these. Next, for each microRNA the order of the nucleotides was shu²ed with replacement 1000 times. This yielded per microRNA 1000 di®erently shu²ed sequences of nu- cleotides for which the minimum folding energy is computed.1 Per microRNA the 1001 energy values were arranged to have increasing order, similar as for empirical distributions in the previous chapter. Then the number of mini- mum folding energies below that of the original microRNA is counted and divided by 1001 as the p-value. If the minimum folding energie of the original microRNA is the smallest, then the empirical p-value is zero. This proce- dure yielded a total of 3424 p-values. The number of sequences with p-values below the threshold value 0.01 is given in Table 8.1. The same procedure is conducted for non-microRNA molecules which were taken as sequences with similar length and nucleotide percentages. Table 8.1: Frequencies empirical p-values lower than or equal to 0.01. test positive test negative total p · 0:01 p > 0:01 microRNA 2973 451 3424 non microRNA 33 3391 3424 total 3006 3842 6848 1I am obliged to Sven Warris for computing the minimum energy values.8.2. ROC TYPES OF CURVES 147 From the frequency Table 8.1, the sensitivity, the speci¯city, and the predictive power can be computed in order to evaluate the quality of the test. The sensitivity is the probability that the test is positive given that the sequence is a microRNA (true positive). Thus sensitivity = P(true positive) = P(test positivejmicroRNA) = 2973 3424 = 0:8682: The speci¯city is the probability that the test is negative given that the sequence is not a microRNA (true negative). Thus speci¯city = P(true negative) = P(test negativejno microRNA) = 3391 3424 = 0:9903: For practical applications of a test the predictive power is of crucial impor- tance. In particular, the predictive value positive is the probability that the sequence is a microRNA given that the test is positive. That is, Predictive value positive = PV + = P(microRNAjtest positive) = 2973 3006 = 0:9890 Thus when the test is positive we are 98.90% certain that the sequence is indeed a microRNA. The predictive value negative is the probability that the sequence is not a microRNA given that the test is negative. Predictive value negative = PV ¡ = P(no microRNAjtest negative) = 3391 3842 = 0:8826: Thus when the test is negative we are 88.26% certain that the sequence is not a microRNA. From the estimated conditional probabilities it can be concluded that the test performs quite well in discriminating between mi- croRNA's from non-microRNA's. 8.2 ROC types of curves In Chapter 2 we have observed with respect to the Golub et al. (1999) data that the expression values of gene CCND3 Cyclin D3 tend to be greater for ALL patients. We may therefore use these as a test for predicting ALL using a certain cuto® value. In particular, for gene expression values larger than a cuto® we declare the test \positive" in the sense of indicating ALL. By doing148 CHAPTER 8. CLASSIFICATION METHODS so the corresponding true and false positives can be computed for each cuto® value. To brie°y indicate the origin of the terminology, imagine that the test results are a characteristic received by an operator. The receiver operator characteristic (ROC) is a curve where the false positive rates are depicted horizontally and the true positive rates vertically. The larger the area under the ROC curve, the better the test is because then low false positive rates go together with large true positive rates.2 These ideas are illustrated by several examples. Example 1. For the sake of illustration we consider the prediction of ALL from the expression values for gene CCND3 Cyclin D3 from Golub et al. (1999) in row 1042 of the matrix golub. Now consider cuto® point 1.27. For such a cuto® point we can produce a table with TRUE/FALSE frequencies of predicting ALL/not ALL. > data(golub, package = "multtest") > gol.true <- factor(golub.cl,levels=0:1,labels= c(" ALL","not ALL")) > gol.pred <- factor(golub[1042,]>1.27,levels=c("TRUE","FALSE"), labels=c("ALL","notALL")) > table(gol.pred,gol.true) gol.true gol.pred ALL not ALL ALL 25 1 notALL 2 10 There are 25 ALL patients with expression values greater than or equal to 1.27, so that the true positive rate is 25/27=0.93. For this cuto® value there is one false positive because one patient without ALL has a score larger than 1.27. Hence, the false positive rate is 1=11 = 0:09. Example 2. The expression values for gene CCND3 Cyclin D3 from the Golub et al. (1999) data are sorted in decreasing order, see Table 8.2. The procedure to draw the ROC curve starts with cuto® point in¯nity. Obviously, there are no expression values equal to in¯nity, so there is no patient tested positive. Next, the cut o® point 2.77 is taken and values greater than or equal to 2.77 are tested as positive. This yields one true positive implying a 2More detailed information can be obtained from a wikipedia search using "ROC curve".8.2. ROC TYPES OF CURVES 149 true positive rate of 1/27, see second row of Table 8.2. For this cuto® value there are no negatives so that the false positive rate is zero. Now consider cuto® point 1.52. There are 22 ALL patients with expres- sion values greater than or equal to 1.52, so that the true positive rate is 22/27=0.81. For this cuto® value there are no false positives because all pa- tients without ALL have expression values lower than 1.51. Hence, the false positive rate is 0 and the true positive rate is 0.81. To indicate this there is a vertical line drawn in the ROC curve from point (0; 0) to point (0; 0:81) in Figure 8.1. Now consider the next cuto® point 1.45. There are 22 ALL patients with expression values greater than or equal to 1.45, so that the true positive rate is again 22/27=0.81. However, there is one patient with- out ALL having expression value 1.45, whom receives therefore a positive test. Hence, the number of false positives increases from zero to one, which implies a false positive rate of 1/11=0.09. In the ROC curve this is indicated by point (0:09; 0:81) and the horizontal line from (0; 0:81) to (0:09; 0:81), see Figure 8.1. This process goes on (see Table 8.2) until the smallest data point -0.74 is taken as cuto® point. For this point all patients are tested positive, so that the false positive rate is 11/11 and the true positive rate is 27/27. This is indicated by the end point (1; 1) in the plot at the top on the right hand side. False positive rate True positive rate 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Figure 8.1: ROC plot for expres- sion values of CCND3 Cyclin D3. False positive rate True positive rate 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Figure 8.2: ROC plot for expres- sion values of gene Gdf5.150 CHAPTER 8. CLASSIFICATION METHODS It is obviously helpful to use a computer for producing an ROC such as in Figure 8.1. To do so we construct an appropriate factor with the value TRUE for ALL and FALSE for not ALL gol.true and use functions from the ROCR package. library(ROCR) gol.true <- factor(golub.cl,levels=0:1,labels= c("TRUE","FALSE")) pred <- prediction(golub[1042,], gol.true) perf <- performance(pred, "tpr", "fpr" ) plot(perf) It seems clear that the expression values are better in testing for ALL when the curve is very steep in the beginning and attains its maximum value soon. In such a case the true positive rate is large for a small false positive rate. A manner to express the predictive accuracy of a test in a single number is by the area under the curve. Using the function performance(pred,"auc") we obtain that the area under the curve is 0.96, which is large. Hence, the ex- pression values of CCND3 Cyclin D3 are suitable for discrimination between ALL and not ALL (AML). The ROC curve for the expression values of gene Gdf5 is given by Figure 8.2. It can be observed that the true positive rate is much lower as one moves on the horizontal axis from the left to the right. This corresponds to the area under the curve of 0.35, which is small. This illustrates that genes may express large di®erences with respect to prediction of the disease status of patients. In practical applications one is often interested in a single optimal cut-o® value and in combining several predictors in a decision scheme. 8.3 Classi¯cation trees The purpose of classi¯cation is to allocate organisms into classes on the ba- sis of measurements on attributes. For instance, in case of the Golub et al. (1999) data the organisms are 38 patients which have measurements on 3051 genes. The classes consist of diagnosis of patients into the ALL class (27 patients) and the AML class (11 patients). A tree model resembles that of a linear model, where the criterion is the factor indicating class membership and the predictor variables are the gene expression values. In case of, for in- stance, the Golub et al. (1999) data the gene expression values fx1; ¢ ¢ ¢ ; x38g8.3. CLASSIFICATION TREES 151 can serve as predictors to form a decision tree. For instance, if xj < t, then patient j is AML, and otherwise if xj ¸ t, then patient j is ALL. Obvi- ously, the threshold value t on which the decision is based should be optimal given the predictor. Such can be estimated by a regression tree (Breiman et al., 1984; Chambers & Hastie, 1992; Venables, & Ripley, 2000), which is implemented in the rpart package (Therneau & Atkinson, 1997). A training set is used to estimate the threshold values that construct the tree. When many predictor variables are involved, 3051 for instance, then we have a tremendous gene (variable) selection problem. The rpart package au- tomatically selects genes which are important for classi¯cation and neglects others. A further problem is that of over¯tting where additional nodes of a tree are added to increase prediction accuracy. When such nodes are speci¯c for the training sample set, these can not be generalized to other samples so that these are of limited scienti¯c value. Prevention of such over¯tting is called pruning and is automatically done by the rpart function. Many basic ideas are illustrated by an elementary example. ALL1 ALL2 AML 0 1 2 3 4 Figure 8.3: Boxplot of expression values of gene a for each leukemia class. genea< 0.9371 genea< 3.025 ALL1 10/0/0 ALL2 0/10/0 AML 0/0/10 Figure 8.4: Classi¯cation tree for gene for three classes of leukemia. Example 1. Optimal gene expressions. Suppose microarray expres- sion data are available with respect to patients su®ering from three types of152 CHAPTER 8. CLASSIFICATION METHODS leukemia abbreviated as ALL1, ALL2, and AML. Gene A has expression val- ues from the populations (patient groups) N(0; 0:52) for ALL1, N(2; 0:52) for ALL2, and N(4; 0:52) for AML. The script below generates thirty expression values for gene A, the patients of the three disease classes, and the estimates of the classi¯cation tree. set.seed(123); n<-10 ; sigma <- 0.5 fac <- factor(c(rep(1,n),rep(2,n),rep(3,n))) levels(fac) <- c("ALL1","ALL2","AML") geneA <- c(rnorm(10,0,sigma),rnorm(10,2,sigma),rnorm(10,4,sigma)) dat <- data.frame(fac,geneA) library(rpart) rp <- rpart(fac ~ geneA, method="class",data=dat) plot(rp, branch=0,margin=0.1); text(rp, digits=3, use.n=TRUE) From the boxplot in Figure 8.3 it can be observed that there is no overlap of gene expressions between classes. This makes gene A an ideal predictor for separating patients into classes. By the construction of the gene expression values x1; ¢ ¢ ¢ ; x30 we expect the following partition. If xi < 1, then ALL1, if xi is in interval [1; 3], then ALL2, and if xi > 3, then AML. From the estimated tree in Figure 8.4 it can be observed that the estimated splits are close to our expectations: If xi < 0:971, then ALL1, if xi is in [0:9371; 3:025], then ALL2, and if xi > 3:025, then AML. The tree consists of three leaves (nodes) and two splits. The prediction of patients into the three classes per- fectly matches the true disease status. Obviously, such an ideal gene need not exist because the expression values overlap between the disease classes. In such a case more genes may be used to build the classi¯cation tree. Example 2. Gene selection. Another situation is where Gene A discrim- inates between ALL and AML and Gene B between ALL1 patients and ALL2 or AML patients and Gene C does not discriminate at all. To simulate this setting we generate expression values for Gene A from N(0; 0:52) for both ALL1 and ALL2, and from N(2; 0:52) for AML patients. Next, we generate expression values for Gene B from N(0; 0:52) for ALL1 and from N(2; 0:52) for ALL2 and AML. Finally, we generate for Gene C from N(1; 0:52) for ALL1, ALL2, and AML. For this and for estimating the tree, we use the following script.8.3. CLASSIFICATION TREES 153 set.seed(123) n<-10 ; sigma <- 0.5 fac <- factor(c(rep(1,n),rep(2,n),rep(3,n))) levels(fac) <- c("ALL1","ALL2","AML") geneA <- c(rnorm(20,0,sigma),rnorm(10,2,sigma)) geneB <- c(rnorm(10,0,sigma),rnorm(20,2,sigma)) geneC <- c(rnorm(30,1,sigma)) dat <- data.frame(fac,geneA,geneB,geneC) library(rpart) rp <- rpart(fac ~ geneA + geneB + geneC, method="class",data=dat) Note the addition in the model notation for the rpart function.3 It is con- venient to collect the data in the form of a data frame.4 From the boxplot in Figure 8.5 it can be seen that Gene A discriminates well between ALL and AML, but not between ALL1 and ALL2. The expres- sion values for Gene B discriminate well between ALL1 and ALL2, whereas those of Gene C do not discriminate at all. The latter can also be seen from the estimated tree in Figure 8.6, where Gene C plays no role at all. This il- lustrates that rpart automatically selects the genes (variables) which play a role in the classi¯cation tree. Expression values on Gene A larger than 1.025 are predicted as AML and smaller ones as ALL. Expression values on Gene B smaller than 0.9074 are predicted as ALL1 and larger as ALL2. Hence, Gene A separates well within the ALL class. Example 3. Prediction by CCND3 Cyclin D3 gene expression values. From various visualizations and statistical testing in the previous chapters, it can be conjectured that CCND3 Cyclin D3 gene expression values form a suitable predictor for discriminating between ALL and AML patients. Note, however, from Figures 2.2 and 8.7 that there is some overlap between the expression values from the ALL and the AML patients, so that a perfect classi¯cation is not possible. By the function rpart the regression partition- ing can be computed as follows. > library(rpart);data(golub); library(multtest) > gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) > gol.rp <- rpart(gol.fac ~ golub[1042,] , method="class") 3See Chapter 11 of the manual "An Introduction to R" for more on model notation. 4See Chapter 6 of the manual "An Introduction to R" for more on data frames.154 CHAPTER 8. CLASSIFICATION METHODS ALL1 ALL2 AML −1 0 1 2 Figure 8.5: Boxplot of expression values of gene a for each leukemia class. genea< 1.025 geneb< 0.9074 ALL1 10/0/0 ALL2 0/10/0 AML 0/0/10 Figure 8.6: Classi¯cation tree of expression values from gene A, B, and C for the classi¯cation of ALL1, ALL2, and AML patients. > predictedclass <- predict(gol.rp, type="class") > table(predictedclass, gol.fac) gol.fac predictedclass ALL AML ALL 25 1 AML 2 10 Note that (25 + 10)=38 ¢ 100% = 92:10% of the ALL/AML patients are cor- rectly classi¯ed by gene CCND3 Cyclin D3. By the function predict(gol.rp,type="class") the predictions from the regression tree of the patients in the two classes can be obtained. The factor gol.fac contains the levels ALL and AML corre- sponding to the diagnosis to be predicted. The predictor variable consists of the expression values of gene CCND3 Cyclin D3. The output of recursive partitioning is assigned to an object called gol.rp, a list from which fur- ther information can be extracted by suitable functions. A summary can be obtained as follows. > summary(gol.rp) Call:8.3. CLASSIFICATION TREES 155 rpart(formula = gol.fac ~ golub[1042, ], method = "class") n= 38 CP nsplit rel error xerror xstd 1 0.7272727 0 1.0000000 1.0000000 0.2541521 2 0.0100000 1 0.2727273 0.5454545 0.2043460 Node number 1: 38 observations, complexity param=0.7272727 predicted class=ALL expected loss=0.2894737 class counts: 27 11 probabilities: 0.711 0.289 left son=2 (26 obs) right son=3 (12 obs) Primary splits: golub[1042, ] < 1.198515 to the right, improve=10.37517, (0 missing) Node number 2: 26 observations predicted class=ALL expected loss=0.03846154 class counts: 25 1 probabilities: 0.962 0.038 Node number 3: 12 observations predicted class=AML expected loss=0.1666667 class counts: 2 10 probabilities: 0.167 0.833 26 [1] 0.03846154 The expected loss in prediction accuracy of Node number 2 is 1/26 and that of Node number 3 is 2/12. This equals the probabilities from the class counts. The primary splits gives the estimated threshold value. To predict the class of the individual patients one may use the function predict, as follows. > predict(gol.rp,type="class") 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ALL ALL ALL ALL ALL ALL ALL ALL ALL ALL ALL ALL ALL ALL ALL ALL AML ALL ALL ALL 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 AML ALL ALL ALL ALL ALL ALL AML ALL AML AML AML AML AML AML AML AML AML Levels: ALL AML156 CHAPTER 8. CLASSIFICATION METHODS Hence, Patient 17 and 21 are erroneously predicted as AML and Patient 29 is erroneously predicted in the ALL class. A more precise output is obtained by asking for the probability of class membership. > predict(gol.rp, type="prob") ALL AML 1 0.9615385 0.03846154 2 0.9615385 0.03846154 etc. Based on this the probability of patient 21 to have ALL is 0.16 and that to have AML is 0.83. ALL AML −0.5 0.0 0.5 1.0 1.5 2.0 2.5 Figure 8.7: Boxplot of expression values from gene CCND3 Cyclin D3 for ALL and AML patients golub[1042, ]>=1.199 ALL 25/1 AML 2/10 Figure 8.8: Classi¯cation tree of expression values from gene CCND3 Cyclin D3 for classi¯ca- tion of ALL and AML patients. Example 4. Gene selection of the Golub (1999) data. By recursive partitioning it is possible to select among the genes of Golub et al. (1999) those which give the best partitioning. For the latter to work we have to specify the gene expressions as the variables (columns). For this we use the transposition operator t. To facilitate reading the output we add gene 1 to gene 3051 as column names.8.3. CLASSIFICATION TREES 157 library(rpart);data(golub); library(multtest) row.names(golub)<- paste("gene", 1:3051, sep = "") goldata <- data.frame(t(golub[1:3051,])) gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) gol.rp <- rpart(gol.fac~., data=goldata, method="class", cp=0.001) plot(gol.rp, branch=0,margin=0.1); text(gol.rp, digits=3, use.n=TRUE) golub.gnames[896,] Inspection of the plot yields gene "FAH Fumarylacetoacetate" as the predic- tor by which the two classes of patients can be predicted perfectly. In order to further illustrate possibilities of classi¯cation methods we use the ALL data collected by Chiaretti, et al. (2004), see also Chapter 6. Example 5. Application to the Chiaretti (2004) data. With respect to the ALL data we want to predict from the gene expressions the diagnosis of B- cell State B1, B2, and B3. Since the complete set of 12625 gene expressions is too large, we select the genes with di®erent means over the patients groups. It is obvious that only these gene can contribute to the prediction of the disease states. In particular we select the gene with ANOVA p-value is smaller than 0.000001. library("hgu95av2.db");library(ALL);data(ALL) ALLB123 <- ALL[,ALL$BT %in% c("B1","B2","B3")] pano <- apply(exprs(ALLB123), 1, function(x) anova(lm(x ~ ALLB123$BT))$Pr[1]) names <- featureNames(ALL)[pano<0.000001] symb <- mget(names, env = hgu95av2SYMBOL) ALLBTnames <- ALLB123[names, ] probedat <- as.matrix(exprs(ALLBTnames)) row.names(probedat)<-unlist(symb) The probe symbols are extracted from the hgu95av2SYMBOL environment and used as row names to facilitate readability of the resulting tree. There are 78 patients selected and 29 probes. The recursive partitioning to ¯nd the tree can be performed by the following script. > diagnosed <- factor(ALLBTnames$BT) > tr <- rpart(factor(ALLBTnames$BT) ~ ., data = data.frame(t(probedat))) > plot(tr, branch=0,margin=0.1); text(tr, digits=3, use.n=TRUE)158 CHAPTER 8. CLASSIFICATION METHODS > rpartpred <- predict(tr, type="class") > table(rpartpred,diagnosed) diagnosed rpartpred B1 B2 B3 B1 17 2 0 B2 1 33 5 B3 1 1 18 The rows to the left of the table give the frequencies of the predicted B cell stages and the columns on top the diagnosed B cell stages from the factor. The matrix with frequencies of the predicted and true patient status is often called a \confusion table". The resulting tree in Figure 8.9 should be read as follows. If gene expression MME is strictly smaller than the cuto® value 8.395, then the patient is predicted to be in state (class) B1. If the expression of LSM6 smaller than 4.192, then the predicted state is B2, and if it is larger than the predicted state it is B3. The misclassi¯cation rate is 10/78=0.1282051, which is low, but not zero. It may happen that the probability of the predicted class is close to that of the diagnosed. An overview of the latter can be obtained as follows. predicted.class <- predict(tr, type="class") predicted.probabilities <- predict(tr, type="prob") out <- data.frame(predicted.probabilities,predicted.class, diagnosis=factor(ALLBTnames$BT)) > print(out,digits=2) B1 B2 B3 predicted.class diagnosis 01005 0.026 0.85 0.13 B2 B2 01010 0.026 0.85 0.13 B2 B2 04006 0.895 0.11 0.00 B1 B1 04007 0.026 0.85 0.13 B2 B2 04008 0.895 0.11 0.00 B1 B1 04010 0.050 0.05 0.90 B3 B1 04016 0.895 0.11 0.00 B1 B1 06002 0.026 0.85 0.13 B2 B2 08001 0.026 0.85 0.13 B2 B2 08011 0.026 0.85 0.13 B2 B3 08012 0.026 0.85 0.13 B2 B3 08018 0.050 0.05 0.90 B3 B3 08024 0.895 0.11 0.00 B1 B28.3. CLASSIFICATION TREES 159 09008 0.026 0.85 0.13 B2 B3 ... For instance, the sixth patient is with probability .90 in class B3 and with probability .05 in class B1, which is the diagnosed disease state. MME< 8.395 B1 LSM6< 4.192 17/2/0 B2 1/33/5 B3 1/1/18 Figure 8.9: rpart on ALL B-cel 123 data. 40480_s_at 1173_g_at 37320_at 40729_s_at 36829_at 32716_at 32116_at 32977_at 34378_at 36711_at 37544_at 307_at 40440_at 38032_at 1389_at 0.35 0.45 0.55 0.65 MeanDecreaseAccuracy 32977_at 32716_at 307_at 34333_at 34347_at 35769_at 37043_at 37544_at 34891_at 36829_at 40493_at 40440_at 38032_at 36711_at 1389_at 0.0 0.2 0.4 0.6 MeanDecreaseGini rf1 Figure 8.10: Variable importance plot on ALL B-cell 123 data. Note the reduction in variables from twenty nine to two in the actual construction of the tree. In a construction like this the gene expressions (variables) are linearly dependent in the sense that once the ¯rst gene is selected for the ¯rst split, then highly similar ones are not selected anymore. It can be instructive to leave out the variables selected from the data and to redo the analysis. A generally applied manner to evaluate an estimated model is by its pre- dictive accuracy with respect to a future data set. When such a future data set is not available, it is common practice to split the available data in two parts: A training set and a validation set. Then the model is estimated from the training set and this is used to predict the class of the patients in the validation set. Then a confusion matrix is constructed with the frequencies of true classes against predicted classes. Next, the misclassi¯cation rate can be computed to evaluate the predictive accuracy. This can very well be seen160 CHAPTER 8. CLASSIFICATION METHODS as a method to detect for over ¯tting where the model estimates are so data speci¯c that generalization to future data sets is in danger. Example 6. Training and validation. In the setting of B-cell ALL data with State 1, 2, and 3 the manner to split the data centers around randomly splitting the patients in two halves. The 78 patients in State 1, 2 or 3 can be split in two halves, as follows. i <- sample(1:78, 39, replace = FALSE) noti <- setdiff(1:78,i) df <- data.frame(Y = factor(ALLBTnames$BT), X =t(probedat)) rpart.est <- rpart(Y ~ ., data = df, subset=i) rpart.pred.t <- predict(rpart.est, df[i,], type="class") > table(rpart.pred.t,factor(ALLBTnames$BT[i])) rpart.pred.t B1 B2 B3 B1 11 1 0 B2 0 12 0 B3 0 1 14 > rpart.pred.v <- predict(rpart.est,df[noti,], type="class") > table(rpart.pred.v,factor(ALLBTnames$BT[noti])) rpart.pred.v B1 B2 B3 B1 6 1 0 B2 1 19 3 B3 1 2 6 The misclassi¯cation rate in the training set is 2=39 = 0:05 and in the val- idation set is 7=39 = 0:18. Note that the di®erences mainly occur between State 2 and 3. Generally the prediction of disease state from the training set is better because the model is estimated from these data. The same split of the data into training and validation set will be used for other methods as well. 8.4 Support Vector Machine A support vector machine ¯nds separating lines (hyper planes) between groups of points. This works like a classi¯cation problem where the classes8.4. SUPPORT VECTOR MACHINE 161 of patients are to be predicted from gene expression values. If such sepa- rating lines do exist in the data, then a linear support vector machine will ¯nd these. This is because the optimization method behind it is based on quadratic programming by iterative algorithms which ¯nd the globally opti- mal solution with certainty. Support vector machines do not automatically select variables and are designed for continuous predictor variables. Since the mathematical details are beyond the current scope, we shall con¯ne with illustrating applications to gene expression data. Example 1. Application to the Chiaretti (2004) data. The parameters for the support vector machine can be determined by the function svm from the e1071 package, as follows. library(e1071) df <- data.frame(Y = factor(ALLBTnames$BT), X =t(probedat)) Y <- factor(ALLBTnames$BT);X <- t(probedat) svmest <- svm(X, Y, data=df, type = "C-classification", kernel = "linear") svmpred <- predict(svmest, X, probability=TRUE) > table(svmpred, factor(ALLBTnames$BT)) svmpred B1 B2 B3 B1 19 0 0 B2 0 36 1 B3 0 0 22 The confusion matrix shows that the misclassi¯cation rate of the three classes of B-cell ALL is 1/78=0.0128 is very small, so that the prediction is almost perfect. Note, however, from summary(svmest) that the number of support vectors per class equals 20, 9, and 11, for class B1, B2, and B3, respectively. These have values for all input variables (genes) as can be obtained from dim(svmest$SV) and the coe±cient vectors dim(svmest$coefs). Hence, the excellent prediction properties are obtained by a very large number of estimated parameters. Example 2. Training and validation. A generally applied manner to evaluate the predictive quality of an estimated model is by splitting the data into a training and a validation set. The model is estimated by the training set and then the class of the patients in the validation set is predicted. We shall use the same split as in Example 6 of the previous section.162 CHAPTER 8. CLASSIFICATION METHODS > Yt <- factor(ALLBTnames$BT)[i]; Yv <- factor(ALLBTnames$BT)[noti] > X <- t(probedat); Xt <- X[i,]; Xv <- X[noti,] > svmest <- svm(Xt, Yt, type = "C-classification", kernel = "linear") > svmpredt <- predict(svmest, Xt, probability=TRUE) > table(svmpredt, Yt) Yt svmpredt B1 B2 B3 B1 11 0 0 B2 0 14 0 B3 0 0 14 > svmpredv <- predict(svmest, Xv, probability=TRUE) > table(svmpredv, Yv) Yv svmpredv B1 B2 B3 B1 5 0 0 B2 1 19 4 B3 2 3 5 The predictions of the disease states of the patients from the training set per- fectly match the diagnosed states. The predictions, however, of the classes of the patients from the validation set have misclassi¯cation rate 10/39=0.25 and are therefore less accurate. Hence, the parameter estimates from the training set are sample speci¯c and do not generalize with the same accuracy to the validation set. 8.5 Neural Networks Neural networks are nonlinear models consisting of nonlinear hyperplanes around classes of objects given a set of prediction variables (Ripley, 1996). We con¯ne with illustrating the method by two examples. Example 1. Application to the Chiaretti (2004) data. The models can be estimated by the function nnet from the package that goes under the same name. To avoid having to many variables we randomly select a subset of 20 genes. > Y <- factor(ALLBTnames$BT);X <- t(probedat)8.5. NEURAL NETWORKS 163 > library(nnet) > df <- data.frame(Y = Y, X = X[, sample(ncol(X), 20)]) > nnest <- nnet(Y ~ .,data = df, size = 5, maxit = 500, decay = 0.01, + MaxNWts = 5000) > pred <- predict(nnest, type = "class") > table(pred, Y) # prints confusion ma Y pred B1 B2 B3 B1 19 0 0 B2 0 36 0 B3 0 0 23 The confusion matrix shows that zero out of 78 patients are mis-classi¯ed. Example 2. Training and validation. The results from cross validation on the neural networks are as follows. > nnest.t <- nnet(Y ~ ., data = df,subset=i, size = 5,decay = 0.01, + maxit=500) > prednnt <- predict(nnest.t, df[i,], type = "class") > table(prednnt,Ytrain=Y[i]) Ytrain prednnt B1 B2 B3 B1 11 0 0 B2 0 14 0 B3 0 0 14 > prednnv <- predict(nnest.t, df[noti,], type = "class") > table(prednnv, Yval= Y[noti]) Yval prednnv B1 B2 B3 B1 4 1 0 B2 4 17 4 B3 0 4 5 The predictions on the training set have misclassi¯cation rate zero and that on the validation set 13/39=0.33.164 CHAPTER 8. CLASSIFICATION METHODS 8.6 Generalized Linear Models Within the framework of generalized linear models the diagnosis of a patient is seen as a response. In case the response has the values healthy or disease for which it may be assumed that the binomial distribution holds with a succes probability pi. Recall from Chapter 3 that for a binomially distributed variable with yi successes out of ni it holds that the probability of yi successes out of ni equals P(Yi = yi) = ni! yi!(ni ¡ yi)!pyi i (1 ¡ pi)ni¡yi ; for k = 0; ¢ ¢ ¢ ; ni: The value of pi is closely related to one or more predictor variables x1 and x2 via a linear combination. That is, the linear model holds such that ´i = ¯0+¯1xi1+¯2xi2. The predictors are linked to the succes probability via the so-called logit link pi = e´i e´i + 1 = exp(¯0 + ¯1xi1 + ¯2xi2) 1 + exp(¯0 + ¯1xi1 + ¯2xi2) : Rather than going deeper into the details, the usefulness of generalized linear models will be illustrated with two examples. Example 1. CCND3 Cyclin D3. In the Golub et al. (1999) data we may model Yi = 1 if the patient is diagnosed as ALL and Yi = 0 if (s)he is diagnosed as AML. We use the CCND3 Cyclin D3 gene expression values as predictor. To will be convenient to compute the response by -golub.cl + 1. This yield 1 for ALL and 0 for not ALL5. library(faraway) logitmod <- glm((-golub.cl + 1) ~ golub[1042,], family=binomial(link = "logit")) pchisq(deviance(logitmod),df.residual(logitmod),lower=FALSE) plot((-golub.cl + 1) ~ golub[1042,], xlim=c(-2,5), ylim = c(0,1), xlab="CCND3 expression values ", ylab="Probability of ALL") x <- seq(-2,5,.1) lines(x,ilogit(-4.844124 + 4.439953*x)) pchisq(deviance(logitmod),df.residual(logitmod),lower=FALSE) 5One may also conveniently use a factor as response variable8.6. GENERALIZED LINEAR MODELS 165 > summary(logitmod) Call: glm(formula = (-golub.cl + 1) ~ golub[1042, ], family = binomial(link = "logit")) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -4.844 1.849 -2.620 0.00880 ** golub[1042, ] 4.440 1.488 2.984 0.00284 ** Null deviance: 45.728 on 37 degrees of freedom Residual deviance: 18.270 on 36 degrees of freedom AIC: 22.270 From Figure 8.11 it can be seen that the logit curve ¯ts the data fairly well. From the summary of the output it can be seen that the estimated intercept is -4.844 and the estimated slope is 4.440. Both coe±cients are signi¯cantly di®erent from zero. The goodness-of-¯t value of the model is computed from the chi-square distribution and equals .99. The model ¯ts the data well. The predictive accuracy of the model may be obtained as follows. > pred <- predict(logitmod,type="response") > 0.5 > pred.fac <- factor(pred,levels=c(TRUE,FALSE),labels=c("ALL","not ALL")) > table(pred.fac,gol.fac) gol.fac pred.fac ALL AML ALL 26 2 not ALL 1 9 The diagnosis of the majority of patients is predicted correctly. Example 2. Application to the Chiaretti (2004) data. With respect to the ALL data we want model the diagnosis of B-cell State B1, B2, and B3 as a response. The factor representing these levels can be used as input for the response. Here we use the gene expressions with greatest importance from the classi¯cation tree in Section 8.3. We assign the biological name to the predictor variables and estimate the generalized linear model. library(nnet);library("hgu95av2.db");library(ALL);data(ALL)166 CHAPTER 8. CLASSIFICATION METHODS probe.names <- c("1389_at","35991_at","40440_at") ALLB123 <- ALL[,ALL$BT %in% c("B1","B2","B3")] probedat <- exprs(ALLB123)[probe.names,] row.names(probedat) <- unlist(mget(probe.names, env = hgu95av2SYMBOL)) fac <- factor(ALLB123$BT,levels=c("B1","B2","B3")) dat <- data.frame(fac,t(probedat)) mnmod <- multinom(fac ~ ., family=binomial(link = "logit"),data=dat) > summary(mnmod) Call: multinom(formula = fac ~ ., data = dat, family = binomial(link = "logit")) Coefficients: (Intercept) MME LSM6 SERBP1 B2 14.36158 4.14002 -0.8494635 -5.104337 B3 -12.90584 4.94908 4.7415802 -5.655420 Std. Errors: (Intercept) MME LSM6 SERBP1 B2 16.36959 1.367058 1.716513 2.222486 B3 17.97896 1.424526 1.744425 2.313700 Residual Deviance: 59.88298 AIC: 75.88298 Apart from the intercepts, the estimated coe±cients are signi¯cantly di®erent from zero. > predmn <- predict(mnmod,type="class") > table(predmn,fac) fac predmn B1 B2 B3 B1 17 2 1 B2 1 31 5 B3 1 3 17 The model predict the diagnosed classes quite well. Generalized linear models are statistical models which have to be esti- mated by an iterative process which may need some computation time. It8.7. OVERVIEW AND CONCLUDING REMARKS 167 has the advantage that con¯dence intervals or the signi¯cance of the param- eters can be estimated. On the other hand, using statistical model building procedures may not designed to handle large amounts of predictor variables. Hence, the researcher does need to have some idea on which gene expressions (s)he want to use as predictors. Indeed, better models than those estimated in the above example may certainly exist. 8.7 Overview and concluding remarks Central themes in prediction methods are the face validity (clarity) of the model, the size of the model, and predictive accuracy on a validation set. For many researchers it is of crucial importance to have a clear idea on what a method is essentially doing. Some models and their estimation procedures are mathematically intricate and seem to be recollected in the mind of many researchers as black boxes. Even from a more pragmatic point of view such need not be devastating if the predictive accuracy is excellent. However, support vector machines and neural networks typically use a large number of parameters to predict well on a test set, but less well on validation sets. It is, furthermore, questionable whether a zero misclassi¯cation rate is ra- tional since patients may be misclassi¯ed by the diagnosis or very close to transferring from one state to the other. Recursive partitioning to estimate a classi¯cation tree performs very well on variable selection and pruning in order to discover as few variables (gene expressions) as possible for maximum predictive accuracy. In addition, it seems obvious that classi¯cation trees have great clarity, see e.g. the CART package (Breiman et al., 1984) for further types of recursive trees. Note that several methods have di®erent misclassi¯cation rates with respect to the whole sample, but comparable rates on the validation sets. It should, however, be clear that when there are non-linear relationships between pre- dictor variables and classes, then nonlinear models should outperform linear ones6. 8.8 Exercises 1. Classi¯cation tree of Golub data. Use recursive partitioning in rpart 6Some people may want to use the ade4TkGUI()168 CHAPTER 8. CLASSIFICATION METHODS (a) Find a manner to identify an optimal gene with respect the Golub data to prediction of the ALL AML patients. (b) Explain what the code does. (c) Use rpart to construct the classi¯cation tree with the genes that you found. Does it have perfect predictions? (d) Find the row number of gene Gdf5, which is supposed not to have any relationship with leukemia. Estimate a classi¯cation tree and report the probability of misclassi¯cation. Give explanations of the results. 2. Sensitivity versus speci¯city. (a) Produce a sensitivity versus speci¯city plot for the gene expression values of CCND3 Cyclin D3. (b) In what sense does it resemble Figure 8.2. (c) Compute the area under the curve for sensitivity versus speci¯city curve. 3. Comparing Classi¯cation Methods. To obtain an idea on the misclas- si¯cation rate when there is no relation between the predictors and the factor indicating groups, we perform a small simulation study. (a) Construct a factor with 100 values one and two and a matrix with predictor variables of 500 by 4 with values from the normal distribution. Use the ¯rst four letters of the alphabet for the column names. (b) Use rpart to construct a recursive tree and report the misclassi- ¯cation rate. Comment on the results. (c) Do the same for support vector machines. (d) Do the same for neural networks. (e) Think through your results and comment on these. 4. Prediction of achieved remission. For the ALL data from its ALL library the patients are checked for achieving remission. The variable ALL$CR has values CR (became healthy) and REF (did not respond to therapy; remain ill).8.8. EXERCISES 169 (a) Construct an expression set containing the patients with values on the phenotypical variable remission and the gene expressions with a signi¯cant p-value on the t-test with the patient groups CR or REF. (b) Use recursive partitioning to predict the remission. Report the misclassi¯cation rate and the names of the genes that play a role in the tree. 5. Gene selection by area under the curve. A strategy of selecting genes is to compute the auc for each gene and to use the best 10 for further investigation. Compute the auc for each row with gene expressions of the Golub at al. (1999) data. Collect these in a vector and select the ten best. Is "CCND3 Cyclin D3" among these? 6. Classi¯cation Tree for Ecoli. The ecoli data can be download by the following: (Hint: Copy two separated lines into one before running it.) ecoli <- read.table( "http://www.grappa.univ-lille3.fr/~torre/Recherche/Datasets/ downloads/ecoli/ecoli.data",sep=",",header = TRUE) colnames(ecoli) <- c("SequenceName","mcg","gvh","lip","chg", "aac","alm1","alm2","ecclass") (a) Use ecclass to construct a factor containing the "cp","im",and "pp". (b) Construct a classi¯cation tree using the variables "mcg","gvh","lip","aac","alm1","alm2". Give the code. Hint: Use the addition notation. (c) Plot the tree and report the variables that play a role in the con- structed tree. (d) Predict the class by the tree. Report the code and the miss- classi¯cation rate. (e) Leaf out the upper variable in the classi¯cation tree and re-estimate the tree. Report the miss-classi¯cation rate. Is it much worse?170 CHAPTER 8. CLASSIFICATION METHODS Table 8.2: Ordered expression values of gene CCND3 Cyclin D3, index 2 indicates ALL, 1 indicates AML, cuto® points, number of false positives, false positive rate, number of true positives, true positive rate. data index cuto® fp fpr tp tpr 1 Inf 0 0.00 0 0.00 2 2.77 2 2.77 0 0.00 1 0.04 3 2.59 2 2.59 0 0.00 2 0.07 4 2.45 2 2.45 0 0.00 3 0.11 ... 22 1.78 2 1.78 0 0.00 21 0.78 23 1.52 2 1.52 0 0.00 22 0.81 24 1.37 2 1.45 1 0.09 22 0.81 25 1.33 2 1.37 1 0.09 23 0.85 26 1.28 2 1.33 1 0.09 24 0.89 27 1.11 2 1.28 1 0.09 25 0.93 28 0.46 2 1.12 2 0.18 25 0.93 29 1.45 1 1.11 2 0.18 26 0.96 30 1.12 1 1.02 3 0.27 26 0.96 31 1.02 1 0.89 4 0.36 26 0.96 32 0.89 1 0.83 5 0.45 26 0.96 33 0.83 1 0.74 6 0.55 26 0.96 34 0.74 1 0.64 7 0.64 26 0.96 35 0.64 1 0.49 8 0.73 26 0.96 36 0.49 1 0.46 8 0.73 27 1.00 37 0.43 1 0.43 9 0.82 27 1.00 38 0.13 1 0.13 10 0.91 27 1.00 39 ¡0.74 1 ¡0.74 11 1.00 27 1.008.8. EXERCISES 171 −2 −1 0 1 2 3 4 5 0.0 0.2 0.4 0.6 0.8 1.0 CCND3 expression values Probability of ALL Figure 8.11: Logit ¯t to the CCND3 Cyclin D3 expression values.172 CHAPTER 8. CLASSIFICATION METHODSChapter 9 Analyzing Sequences For many purposes in bioinformatics nucleotide or amino acid sequences are analyzed. The idea is that highly similar sequences may have identical bi- ological functions. For expressing the similarity of sequences it is necessary to compute ¯rst their optimal alignment. It will be explained and illustrated how optimal pairwise alignment can be obtained. Furthermore, it is of im- portance to compute quantities for DNA sequences such as the CG fraction, or, for amino acid sequences, the isoelectric point or the hydropathy score. It will be explained and illustrated how such quantities can be computed. In this chapter you learn how to query online data bases, to translate RNA into protein sequences, to match patterns, and to program pairwise align- ments. We will start, however, with a query language in order to download sequences. 9.1 Using a query language It will be illustrated how the query language from the seqinr package can be used for various types of searches. However, before we download anything, it is important to know which banks can be chosen. > library(seqinr) > choosebank() [1] "genbank" "embl" "emblwgs" "swissprot" "ensembl" [6] "refseq" "nrsub" "hobacnucl" "hobacprot" "hovergendna" [11] "hovergen" "hogenom" "hogenomdna" "hogennucl" "hogenprot" [16] "hoverclnu" "hoverclpr" "homolens" "homolensdna" "greview" 173174 CHAPTER 9. ANALYZING SEQUENCES [21] "polymorphix" "emglib" "HAMAPnucl" "HAMAPprot" "hoppsigen" [26] "nurebnucl" "nurebprot" "taxobacgen" There are many possibilities to use the query language e.g. for answering questions about sequences from online data bases (Gouy, et al. 1984). We give a few examples to illustrate some of its possibilities. For this we shall temporary use the option virtual=TRUE to save time by preventing actual downloading.1 We may ask: How many ccnd sequences has genbank? > choosebank("genbank") > query("ccnd","k=ccnd",virtual=TRUE)$nelem [1] 147 More speci¯c: How many sequences ccnd sequences has genbank for the species homo sapiens. > query("ccnd3hs","sp=homo sapiens AND k=ccnd3",virtual=TRUE)$nelem [1] 9 For many other combinations of search options we refer to the manual of the seqinr package and for a book length treatment with many examples to Charif et al. (2008). 9.2 Getting information on downloaded se- quences After sequences are downloaded in binary format it is essential to obtain information with respect to their accession number, length, actual elements, translation to amino acids, and annotation. How to do this will brie°y be illustrated by an example. Example 1. Let's download sequences related to the species homo sapi- ens and a gene name like "CCND3". > choosebank("genbank") > query("ccnd3hs","sp=homo sapiens AND k=ccnd3@") > ccnd3hs$nelem [1] 9 1The results below are obviously time dependent.9.2. GETTING INFORMATION ON DOWNLOADED SEQUENCES 175 The sequences are downloaded in binary format. The symbol @ acts as a wildcard for any zero or other characters. There are a number of useful functions available to obtain further information. Some of these are getName, getLength, getSequence, getTrans, and getAnnot. To use these on a list containing sets of sequences the functionality sapply is very convenient. This is illustrated by extracting the NCBI accession numbers. > sapply(ccnd3hs$req, getName) [1] "AF517525.CCND3" "AL160163.CCND3" "AL160163.PE5" "AL161651" [5] "BC011616.CCND3" "CR542246" "HUMCCND3A.CCND3" "HUMCCND3PS.PE1" [9] "HUMCCNDB04.CCND3" "HUMCYCD3A.CCND3" The length of the sequences can be obtained by the getLength function. > sapply(ccnd3hs$req, getLength) [1] "879" "879" "729" "211627" "879" "879" "879" "537" "559" "879" Let's obtain the ¯rst sequence and print its ¯rst ¯fteen nucleotides to the screen. 2 > getSequence(ccnd3hs$req[[1]])[1:15] [1] "a" "t" "g" "g" "a" "g" "c" "t" "g" "c" "t" "g" "t" "g" "t" Its translation into amino acids can be obtained > getTrans(ccnd3hs$req[[1]])[1:15] [1] "M" "E" "L" "L" "C" "C" "E" "G" "T" "R" "H" "A" "P" "R" "A" as well as its annotation from the corresponding web page: > getAnnot(ccnd3hs$req[[1]]) [1] " CDS join(1051..1248,2115..2330,5306..5465,6005..6141," [2] " 6593..6760)" [3] " /gene=\"CCND3\"" [4] " /codon_start=1" [5] " /product=\"cyclin D3\"" [6] " /protein_id=\"AAM51826.1\"" [7] " /db_xref=\"GI:21397158\"" [8] " /translation=\"MELLCCEGTRHAPRAGPDPRLLGDQRVLQSLLRLEERYVPRASY" 2Use double brackets to extract a sequence from a list.176 CHAPTER 9. ANALYZING SEQUENCES [9] " FQCVQREIKPHMRKMLAYWMLEVCEEQRCEEEVFPLAMNYLDRYLSCVPTRKAQLQLL" [10] " GAVCMLLASKLRETTPLTIEKLCIYTDHAVSPRQLRDWEVLVLGKLKWDLAAVIAHDF" [11] " LAFILHRLSLPRDRQALVKKHAQTFLALCATDYTFAMYPPSMIATGSIGAAVQGLGAC" [12] " SMSGDELTELLAGITGTEVDCLRACQEQIEAALRESLREAAQTSSSPAPKAPRGSSSQ" [13] " GPSQTSTPTDVTAIHL\"" 9.3 Computations on sequences A basic quantity to compute are the nucleotide and the dinucleotide frequen- cies.Example 1. Frequencies of (di)nucleotides. We shall continue with the ¯rst result from the CCND3 (Cyclin D3) search with accession num- ber "AF517525.CCND3". To compute the frequencies we may extract the sequence from a list in order to use the basic function table, as follows. > table(getSequence(ccnd3hs$req[[1]])) a c g t 162 288 267 162 This table can also be computed by the seqinr function count, which is more general in the sense that frequencies of dinucleotides can be computed. > count(getSequence(ccnd3hs$req[[1]]),2) aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt 25 44 64 29 68 97 45 78 52 104 76 34 16 43 82 21 This will be quite useful in the next chapter. Indeed, changing 2 into 3 makes it possible to count trinucleotides. Example 2. G + C percentage. We are often interested in the fraction G plus C in general (GC), or starting from the ¯rst position of the codon bases (GC1), the second (GC2), or third (GC3). > GC(getSequence(ccnd3hs$req[[1]])) [1] 0.63139939.3. COMPUTATIONS ON SEQUENCES 177 > GC1(getSequence(ccnd3hs$req[[1]])) [1] 0.6484642 > GC2(getSequence(ccnd3hs$req[[1]])) [1] 0.4641638 > GC3(getSequence(ccnd3hs$req[[1]])) [1] 0.78157 Hence, the G + C percentage is largest when started at position three. It is also possible to compute the G + C fraction in a window of length 50 nt, say, and to plot it along the sequence. GCperc <- double() n <- length(ccnd3[[1]]) for (i in 1:(n - 50)) GCperc[i] <- GC(ccnd3[[1]][i:(i+50)]) plot(GCperc,type="l") By double() we ¯rst create a vector. From Figure 9.1 it can be seen that the G + C fraction changes drastically along a window of 50 nucleotides. With respect to over or under representation of dinucleotides there is a func- tion ½ (rho) available, which is de¯ned as ½(xy) = fxy fx ¢ fy ; where fxy, fx, and fy are the frequencies of the (di)nucleotide xy, x, and y, respectively. The z-score is computed by subtracting the mean and dividing by the standard deviation (Palmeira, et al., 2006). The latter is somewhat more sensitive for over and under representation. Example 3. Rho and z-scores. The coe±cient rho and the corresponding z-scores will be computed from the sequence with NCBI accession number "AF517525.CCND3". > round(rho(getSequence(ccnd3hs$req[[1]])),2) aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt 0.84 0.83 1.30 0.97 1.28 1.03 0.51 1.47 1.06 1.19 0.94 0.69 0.54 0.81 1.67 0.70 > round(zscore(getSequence(ccnd3hs$req[[1]]),modele='base'),2)178 CHAPTER 9. ANALYZING SEQUENCES 0 200 400 600 800 0.5 0.6 0.7 0.8 0.9 Index GCperc Figure 9.1: G + C fraction of sequence "AF517525.CCND3" along a window of length 50 nt. aa ac ag at ca cc cg ct ga gc gg gt ta -1.08 -1.67 2.81 -0.18 2.78 0.42 -6.63 4.64 0.54 2.60 -0.80 -2.87 -3.10 tc tg tt -1.86 6.22 -1.98 The rho value for CG is not extreme, but its z-score certainly is. In case we have an amino acid sequence it may be useful to obtain a plot of the amino acid frequencies. When we have translated the nucleotide sequence into an amino acid sequence, it may be interesting to construct a9.3. COMPUTATIONS ON SEQUENCES 179 plot expressing their frequencies. Such can be useful for a ¯rst impression on sequence similarity. Example 4. Comparing Amino acid frequencies. We continue with the ¯rst result from the CCND3 (Cyclin D3) search, translate and order it, and, next, produce a dotchart with amino acid frequencies. tab <- table(getTrans(ccnd3hs$req[[1]])) taborder <- tab[order(tab)] names(taborder) <- aaa(names(taborder)) dotchart(taborder,pch=19,xlab="Stop and amino-acid-counts") abline(v=1,lty=2) The script was run on both sequences AF517525.CCND3 and AL160163.CCND3 resulting in Figure 9.2 and 9.3, respectively. The two sequences are highly similar with respect to amino acid frequencies. Stp Asn Trp Phe His Tyr Met Ile Lys Cys Asp Gly Val Gln Pro Thr Ser Glu Arg Ala Leu 0 10 20 30 40 Stop and amino−acid−counts Figure 9.2: Frequency plot of amino acids from accession num- ber AF517525.CCND3. Stp Asn Trp Phe His Tyr Met Ile Lys Cys Asp Gly Val Gln Pro Thr Ser Glu Arg Ala Leu 0 10 20 30 40 Stop and amino−acid−counts Figure 9.3: Frequency plot of amino acids from accession num- ber AL160163.CCND3. For amino acid sequences it may be of importance to compute the the- oretical isoelectric point or the molecular weight of the corresponding protein.180 CHAPTER 9. ANALYZING SEQUENCES Example 5. Isoelectric point. The function computePI computes the theoretical isoelectric point of a protein, which is the pH at which the protein has a neutral charge (Gasteiger, et al. 2005). > computePI(getTrans(ccnd3hs$req[[1]])) [1] 6.657579 The protein molecular weight can be computed as follows. > pmw(getTrans(getSequence(ccnd3hs$req[[1]]))) [1] 32503.38 Note that it is easy to compute these for all downloaded proteins and to compare these. Another important quantity is hydropathy score (Kyte & Doolittle, 1982) of proteins, which is de¯ned as a weighted sum P20 i=1 ®ifi of amino acid co- e±cients ®i and the relative frequencies fi. An example will illustrate how it can be computed. Example 6. Hydropathy score. The coe±cients ®1; ¢ ¢ ¢ ; ®20 are available as KD data from the EXP list of the seqinr package. The unique names are lexicographically ordered and stored in the object kdc. The scale is changed by the minus sign below so that hydrophilic proteins are positive, but smaller than one. A function is de¯ned to compute the hydropathy score for a set of amino acid sequences. ccnd3 <- sapply(ccnd3hs$req, getSequence) ccnd3transl <- sapply(ccnd3, getTrans) data(EXP) names(EXP$KD) <- sapply(words(), function(x) translate(s2c(x))) kdc <- EXP$KD[unique(names(EXP$KD))] kdc <- -kdc[order(names(kdc))] linform <- function(data, coef) { #data are sequences f <- function(x) { freq <- table(factor(x, levels = names(coef)))/length(x) return(coef %*% freq) } res <- sapply(data, f) names(res) <- NULL9.4. MATCHING PATTERNS 181 return(res) }kdath <- linform(ccnd3transl, kdc) > print(kdath,digits=3) [1] 0.0874 0.0962 0.0189 0.1496 0.0962 0.0874 0.0874 0.2659 0.2220 Indeed, the largest score is still much smaller than one, so the conclusion is that there are no hydrophilic proteins among our sequences. The data set aaindex of the seqinr library contains more than ¯ve hun- dred sets of coe±cients for computing speci¯c quantities with respect to proteins. 9.4 Matching patterns A manner to investigate a long sequence is to search for identical patterns, eventually allowing for a speci¯ed number of mismatches. There are many relevant examples such as seeking for one of the stop codons UAG, UGA UAA in RNA, or recognition sequences of enzymes (e.g. Roberts, et al., 2007). We sustain with a brief example. Example 1. Pattern match. In the sequence with NCBI accession number "AF517525.CCND3", we seek the pattern "cccggg" with zero mismatches as well as those with a single mismatch. By the function c2s a sequence of characters is converted into a single string. library(seqinr) choosebank("genbank") query("ccnd3hs","sp=homo sapiens AND k=ccnd3@") ccnd3 <- sapply(ccnd3hs$req, getSequence) ccnd3nr1 <- c2s(ccnd3[[1]]) > ccnd3nr1 [1] "atggagctgctgtgttgcgaaggcacccggcacgcgccccgggccgggccggacccgcgg"... > subseq <- "cccggg" > countPattern(subseq, ccnd3nr1, mismatch = 0) [1] 2 > matchPattern(subseq, ccnd3nr1, mismatch = 0) Views on a 879-letter BString subject182 CHAPTER 9. ANALYZING SEQUENCES Subject: atggagctgctgtgttgcgaaggcacccggcacg...actcctacagatgtcacag Views: start end width [1] 38 43 6 [cccggg] [2] 809 814 6 [cccggg] > matchPattern(subseq, ccnd3nr1, mismatch = 1) Views on a 879-letter BString subject Subject: atggagctgctgtgttgcgaaggcacccggcacg...actcctacagatgtcacag Views:start end width [1] 26 31 6 [cccggc] [2] 37 42 6 [ccccgg] [3] 38 43 6 [cccggg] [4] 43 48 6 [gccggg] [5] 54 59 6 [cccgcg] [6] 119 124 6 [cccgcg] [7] 236 241 6 [ccctgg] [8] 303 308 6 [cctggg] [9] 512 517 6 [cccgtg] [10] 612 617 6 [cacggg] [11] 642 647 6 [cctggg] [12] 661 666 6 [tccggg] [13] 662 667 6 [ccgggg] [14] 808 813 6 [ccccgg] [15] 809 814 6 [cccggg] [16] 810 815 6 [ccgggg] The number of counted patterns allowing two mismatches is much larger. 9.5 Pairwise alignments Among the basic questions about genes or proteins is to what extent a pair of sequences are similar. To ¯nd this out these are aligned in a certain man- ner after which a similarity score can be computed. In order to understand sequence alignment it is fundamental to have some idea about recursion.9.5. PAIRWISE ALIGNMENTS 183 Example 1. Basic recursion. The idea of recursion is to generate a sequence by de¯ning the current value as a function of the previous. Suppose that the ¯rst element is one, x1 = 1, and that the sequence is de¯ned by xi = xi¡1 + 1: Then we obtain x1 = 1, x2 = 2, x3 = 3, etc, so that the sequence becomes 1; 2; 3; ¢ ¢ ¢. Indeed, this is as fundamental as counting. Another manner to de¯ne a sequence is by multiplying the previous value by a constant. For example, let xi = 2xi¡1 with x1 = 1. Then the values of the sequence are x1 = 1, x2 = 2, x3 = 4, x3 = 8, etc. Also we see that in fact xn = 2n, so that a value of the sequence can be computed without actually computing all previous elements. Another example would be xi = 2xi¡1 ¡ 10, with x1 = 1. In order to compute the value x10 we may use R, as follows. > x<-double();x[1]<-1 > for (i in 2:10) {x[i]<- 2*x[i-1]-10} > x[10] [1] -4598 This illustrates basic ideas about recursively de¯ned sequences. Suppose we want to compute an alignment score for two small DNA sequences GAATTC and GATTA (Durbin et. al., 1998, p.18). We agree that a match between two letters should have the score +2 and a mismatch the score -1. A gap at a certain position of the sequences should be punished by subtracting a score by d = 2. A possible alignment is GG AA AT TT T¡ CA, where the minus sign indicates a gap. Then the alignment consists of a match, match, mismatch, match, gap, mismatch, respectively, so that the score is 2 + 2 ¡ 1 + 2 ¡ 2 ¡ 1 = 2. Now the question is whether this alignment is optimal in the sense that the score is maximal? The answer is: No! To see this, consider the alignment GG AA A¡ TT TT CA. Then we have a match, match, gap, match, match, mismatch, respectively, so that the score is 2+2¡2+2+2¡1 = 5. This is better, but still we do not know whether this alignment is optimal. In order to ascertain that the alignment is optimal we have to build an alignment score matrix F(i; j). To do so it is convenient to start with building the (mis)match score matrix s(i; j). Its (i; j)th element s(i; j) has the value 2 in case of a match and the value -1 is case of a mismatch. Note that for184 CHAPTER 9. ANALYZING SEQUENCES each step we can choose between a gap, a match, or a mismatch. Building up the matrix F(i; j) recursively, means that we de¯ne its elements on the basis of the values of its preceding elements. That is, given the values of the previous elements F(i ¡ 1; j ¡ 1), F(i ¡ 1; j), and F(i; j ¡ 1), we will be able to ¯nd the best consecutive value for F(i; j). In particular, in case of a match or a mismatch, we take F(i; j) = F(i ¡ 1; j ¡ 1) + s(xi; yj) and in case of a gap we take F(i; j) = F(i ¡ 1; j) ¡ d or F(i; j) = F(i; j ¡ 1) ¡ d. The famous Needleman-Wunsch alignment algorithm consists of taking the maximum out of these possibilities at each step (e.g, Durbin et. al., 1998, p.21). Their algorithm can be summarized, as follows. F(i; j) = max8< : F(i ¡ 1; j ¡ 1) + s(i; j) F(i ¡ 1; j) ¡ d F(i; j ¡ 1) ¡ d Note, however, that this will not yet work because we have not de¯ned any initial values. In fact we will agree to start with F(0; 0) = 0 and due to the gap penalties we take F(i; 0) = ¡id for the ¯rst column and F(0; j) = ¡jd for the ¯rst row. Then, the ¯nal score F(n;m) is the optimal score and the values of the matrix F(i; j) indicates the optimal path. By informaticians this recursive scheme is often called a \dynamic programming algorithm". Example 2. Dynamic programming of DNA sequences. Consider again the DNA sequences GAATTC, GATTA, the score +2 for a match, -1 for a mismatch, and the gap penalty d = 2. It is clarifying to ¯rst construct the score matrix s(i; j). For this we use the string-to-character function s2c, a for loop, and an if else statement. library(seqinr) x <- s2c("GAATTC"); y <- s2c("GATTA"); d <- 2 s <- matrix(data=NA,nrow=length(y),ncol=length(x)) for (i in 1:(nrow(s))) for (j in 1:(ncol(s))) {if (y[i]==x[j]) s[i,j]<- 2 else s[i,j]<- -1 } rownames(s) <- c(y); colnames(s) <- c(x) > sG A A T T C G 2 -1 -1 -1 -1 -1 A -1 2 2 -1 -1 -1 T -1 -1 -1 2 2 -19.5. PAIRWISE ALIGNMENTS 185 T -1 -1 -1 2 2 -1 A -1 2 2 -1 -1 -1 To initialize the ¯rst row and column of the matrix F(i; j), it is convenient to use the function seq. The purpose of the max function seems obvious. F <- matrix(data=NA,nrow=(length(y)+1),ncol=(length(x)+1)) rownames(F) <- c("",y); colnames(F) <- c("",x) F[,1] <- -seq(0,length(y)*d,d); F[1,] <- -seq(0,length(x)*d,d) for (i in 2:(nrow(F))) for (j in 2:(ncol(F))) {F[i,j] <- max(c(F[i-1,j-1]+s[i-1,j-1],F[i-1,j]-d,F[i,j-1]-d))} > F G A A T T C 0 -2 -4 -6 -8 -10 -12 G -2 2 0 -2 -4 -6 -8 A -4 0 4 2 0 -2 -4 T -6 -2 2 3 4 2 0 T -8 -4 0 1 5 6 4 A -10 -6 -2 2 3 4 5 From the lower corner to the right hand side we see that the optimal score is indeed 5. Optimal alignment for pairs of amino acid sequences are often considered to be more relevant because these are more closely related to biological func- tions. For this purpose we may modify the previous scheme by changing the gap penalty d and the (mis)match scores s(i; j). In particular, we shall use the gap penalty d = 8 and for the (mis)match the scores from the so-called BLOSUM50 matrix. Example 3. Programming Needleman-Wunsch. For the two sequences "PAWHEAE" and "HEAGAWGHEE" (see, Durbin et. al., 1998, p.21) we seek the Needleman-Wunsch optimal alignment score, using the BLOSUM50 (mis)match score matrix and gap penalty d = 8. You can either directly read a BLOSUM matrix from NCBI > file <- "ftp://ftp.ncbi.nih.gov/blast/matrices/BLOSUM50" > BLOSUM50 <- as.matrix(read.table(file, check.names=FALSE))186 CHAPTER 9. ANALYZING SEQUENCES Table 9.1: BLOSUM50 matrix. A R N D C Q E G H I L K M F P S T W Y V A 5 -2 -1 -2 -1 -1 -1 0 -2 -1 -2 -1 -1 -3 -1 1 0 -3 -2 0 R -2 7 -1 -2 -4 1 0 -3 0 -4 -3 3 -2 -3 -3 -1 -1 -3 -1 -3 N -1 -1 7 2 -2 0 0 0 1 -3 -4 0 -2 -4 -2 1 0 -4 -2 -3 D -2 -2 2 8 -4 0 2 -1 -1 -4 -4 -1 -4 -5 -1 0 -1 -5 -3 -4 C -1 -4 -2 -4 13 -3 -3 -3 -3 -2 -2 -3 -2 -2 -4 -1 -1 -5 -3 -1 Q -1 1 0 0 -3 7 2 -2 1 -3 -2 2 0 -4 -1 0 -1 -1 -1 -3 E -1 0 0 2 -3 2 6 -3 0 -4 -3 1 -2 -3 -1 -1 -1 -3 -2 -3 G 0 -3 0 -1 -3 -2 -3 8 -2 -4 -4 -2 -3 -4 -2 0 -2 -3 -3 -4 H -2 0 1 -1 -3 1 0 -2 10 -4 -3 0 -1 -1 -2 -1 -2 -3 2 -4 I -1 -4 -3 -4 -2 -3 -4 -4 -4 5 2 -3 2 0 -3 -3 -1 -3 -1 4 L -2 -3 -4 -4 -2 -2 -3 -4 -3 2 5 -3 3 1 -4 -3 -1 -2 -1 1 K -1 3 0 -1 -3 2 1 -2 0 -3 -3 6 -2 -4 -1 0 -1 -3 -2 -3 M -1 -2 -2 -4 -2 0 -2 -3 -1 2 3 -2 7 0 -3 -2 -1 -1 0 1 F -3 -3 -4 -5 -2 -4 -3 -4 -1 0 1 -4 0 8 -4 -3 -2 1 4 -1 P -1 -3 -2 -1 -4 -1 -1 -2 -2 -3 -4 -1 -3 -4 10 -1 -1 -4 -3 -3 S 1 -1 1 0 -1 0 -1 0 -1 -3 -3 0 -2 -3 -1 5 2 -4 -2 -2 T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 2 5 -3 -2 0 W -3 -3 -4 -5 -5 -1 -3 -3 -3 -3 -2 -3 -1 1 -4 -4 -3 15 2 -3 Y -2 -1 -2 -3 -3 -1 -2 -3 2 -1 -1 -2 0 4 -3 -2 -2 2 8 -1 V 0 -3 -3 -4 -1 -3 -3 -4 -4 4 1 -3 1 -1 -3 -2 0 -3 -1 5 or load a BLOSUM matrix from the Biostrings package. For the sake of clarity we shall conveniently construct the matrix s(i; j) without any concern about computer memory. library(seqinr);library(Biostrings);data(BLOSUM50) x <- s2c("HEAGAWGHEE"); y <- s2c("PAWHEAE"); s <- BLOSUM50[y,x]; d <- 8 F <- matrix(data=NA,nrow=(length(y)+1),ncol=(length(x)+1)) F[1,] <- -seq(0,80,8); F[,1] <- -seq(0,56,8) rownames(F) <- c("",y); colnames(F) <- c("",x) for (i in 2:(nrow(F))) for (j in 2:(ncol(F))) {F[i,j] <- max(c(F[i-1,j-1]+s[i-1,j-1],F[i-1,j]-d,F[i,j-1]-d))}9.5. PAIRWISE ALIGNMENTS 187 > F H E A G A W G H E E 0 -8 -16 -24 -32 -40 -48 -56 -64 -72 -80 P -8 -2 -9 -17 -25 -33 -41 -49 -57 -65 -73 A -16 -10 -3 -4 -12 -20 -28 -36 -44 -52 -60 W -24 -18 -11 -6 -7 -15 -5 -13 -21 -29 -37 H -32 -14 -18 -13 -8 -9 -13 -7 -3 -11 -19 E -40 -22 -8 -16 -16 -9 -12 -15 -7 3 -5 A -48 -30 -16 -3 -11 -11 -12 -12 -15 -5 2 E -56 -38 -24 -11 -6 -12 -14 -15 -12 -9 1 Hence, from the lower-right corner we observe that the optimal score equals one. Example 4. Needleman-Wunsch. We may also conveniently use the pairwiseAlignment function from the Biostrings package to ¯nd the optimal Needleman-Wunsch aligment score for the sequences PAWHEAE" and "HEAGAWGHEE" (see, Durbin et. al., 1998, p.21). library(Biostrings);data(BLOSUM50) > pairwiseAlignment(AAString("PAWHEAE"), AAString("HEAGAWGHEE"), + substitutionMatrix = "BLOSUM50",gapOpening = 0, gapExtension = -8, + scoreOnly = FALSE) Global Pairwise Alignment 1: --P-AW-HEAE 2: HEAGAWGHE-E Score: 1 Hence, we obtain the optimal score 1 as well as a representation of the opti- mal alignment. An obvious question is whether in the previous example the obtained score 1 is to be evaluated as being \large" or not. A manner to answer this question is by comparing it with the alignment score of random sequences. That is, we may compute the probability of alignment scores larger than 1. Example 5. Comparing with random sequences. To illustrate how the probability of alignment scores larger than 1 can be computed we sample randomly from the names of the amino acids, seven for y and 10 for x and188 CHAPTER 9. ANALYZING SEQUENCES compute the maximum alignment score. This is repeated 1000 times and the probability of optimal alignment scores greater than 1 is estimated by the corresponding proportion. library(seqinr);library(Biostrings);data(BLOSUM50) randallscore <- double() for (i in 1:1000) { x <- c2s(sample(rownames(BLOSUM50),7, replace=TRUE)) y <- c2s(sample(rownames(BLOSUM50),10, replace=TRUE)) randallscore[i] <- pairwiseAlignment(AAString(x), AAString(y), substitutionMatrix = "BLOSUM50",gapOpening = 0, gapExtension = -8, scoreOnly = TRUE) }> sum(randallscore>1)/1000 [1] 0.003 By the option scoreOnly = TRUE the optimal score is written to the vector randallscore. The probability of scores larger than 1 equals 0.003 and is therefore small and the alignment is stronger than expected from randomly constructed sequences. Example 6. Sliding window on Needleman-Wunsch scores. We may also program a sliding window such that for each the Needleman-Wunsch align- ment score is computed. Then the maximum can be found and localized. choosebank("genbank"); library(seqinr) query("ccnd3hs","sp=homo sapiens AND k=ccnd3@") ccnd3 <- sapply(ccnd3hs$req, getSequence) ccnd3transl <- sapply(ccnd3, getTrans) x <- c2s(ccnd3transl[[1]]) y <- c2s(ccnd3transl[[1]][50:70]) nwscore <- double() ; n <- length(ccnd3transl[[1]]) for (i in 1:(n-21)) nwscore[i] <- pairwiseAlignment(AAString(c2s(ccnd3transl[[1]][i:(i+20)])), AAString(y),substitutionMatrix = "BLOSUM50",gapOpening = 0, gapExtension = -8, scoreOnly = TRUE) > pairwiseAlignment(AAString(y), AAString(y), substitutionMatrix = "BLOSUM50", gapOpening = 0, gapExtension = -8,9.6. OVERVIEW AND CONCLUDING REMARKS 189 + scoreOnly = TRUE) [1] 152 > max(nwscore) [1] 152 > which.max(nwscore) [1] 50 Note that the maximum occurs when the subsequences are identical. The value of the maximum is 152 which occurs at position 50. 9.6 Overview and concluding remarks It was illustrated how the query language of the seqinr library can be used to download sequences, to translate these and to compute relevant quanti- ties such as the isoelectric point or the hydropathy score. Furthermore, it was illustated how patterns can be matched and how algorithms for optimal pairwise alignment can be programmed. Further applications are given by the exercises below. The package Biostrings contains the various PAM matrices for optimal alignment, as well as facilities to ¯nd palindromes, and to read and write data in FASTA format (readFASTA). 9.7 Exercises 1. Writing to a FASTA ¯le. Read, similar to the above, the ccnd3 se- quences using the query language and write the ¯rst sequence to a ¯les in FASTA format. Also try to write them all to FASTA format. 2. Dotplot of sequences. Use the function dotPlot of the seqinr package and par(mfrow=c(1,2)) to produce two adjacent plots. (a) Construct two random sequence of size 100 and plot the ¯rst against second and the ¯rst against the ¯rst. (b) Construct a plot of the ¯rst against the ¯rst and the ¯rst against the ¯rst in reverse order.190 CHAPTER 9. ANALYZING SEQUENCES (c) Download the sequences related to the species homo sapiens and a gene name like "CCND3 Cyclin D3". Construct a dotplot of the most similar and the least similar sequences. Report your observations. 3. Local alignment. The Smith-Waterman algorithm seeks maximum lo- cal alignment between subsequences of sequences. Their algorithm can be summarized (Durbin et al., 2005, p.22), as follows. F(i; j) = max8< : F(i ¡ 1; j ¡ 1) + s(i; j) F(i ¡ 1; j) ¡ d F(i; j ¡ 1) ¡ d The algorithm allows the score zero if the others have negative values. The idea is that the maximum alignment can occur anywhere in the matrix, optimal alignment is de¯nes as the maximum over the whole matrix. Program the Smith-Waterman algorithm and ¯nd the optimal local alignment of the sequences PAWHEAE" and "HEAGAWGHEE". 4. Probability of more extreme alignment score. Sample x and y randomly from the names of the amino acids, seven for y and 10 for x. repeat this 1000 times and compute the optimal alignment score and use it to evaluate the signi¯cance of the previously obtained score. 5. Prochlorococcus marinus. Each of three strains of P. marinus is ex- posed to di®erent intensities of UV radiation because these live in dif- ferent depths in water. The MIT 9313 strain lives at depth 135 m, SS120 at 120 m, and MED4 at 5 m. The latter strain is considered to be high-light-adapted. The residual intensities of 260-nm UVb ir- radiation corresponding to the given depths is 0.00007%, 0.0002% and 70%, respectively. It is hypothesized that the G + C content depends on the amount of radiation. The accession numbers of Gen bank are AE017126, BX548174, and BX548175, respectively. (a) Use the operator OR together with the accession numbers to download the sequences of the bacteria strains. (b) Compte the GC fraction of each of the sequences. (c) Is there a relation between UVb radiation and GC fraction? (d) Formulate a relevant hypothesis and test it.9.7. EXERCISES 191 6. Sequence equality. Download the sequences "AF517525.CCND3" and "AL160163.CCND3". Hint: These are the ¯rst two from the query "ccnd3" within homo sapiens. (a) Compute the length of the sequences. (b) Translate the sequences into amino acids and compare their fre- quencies. (c) Are they equal or, if not, in what position do they di®er? 7. Conserved region. At http://blocks.fhcrc.org there are blocks of highly conserved regions for proteins in PROSITE. Find PR00851A which contains blocks of protein related to a human gene responsible for DNA-repair defect xeroderma pigmentosum (sensitivity to ultravi- olet light) Perform a pairwise alignment with these subsequences and report the ones most and least similar. Use BLOSUM50. 8. Plot of CG proportion from Celegans. (a) Produce a plot of the CG proportion of the chromosome I of Cel- egans (Celegans.UCSC.ce2) along a window of 100 nucleotides. Take the ¯rst 10,000 nucleotides. (b) A binding sequence of the enzyme EcoRV is the subsequence GATATC. How many exact matches has Chromosome I of Cel- egans. How many do you expect by chance? 9. Plot of codon usage. Go to the seqinr help page on dotchart.uco. (a) Redo the example and brie°y describe its usage. (b) Use the query language to ¯nd192 CHAPTER 9. ANALYZING SEQUENCESChapter 10 Markov Models The idea of a Markov process forms the basis of many important models in bioinformatics such as (Hidden) Markov Models, models for sequence align- ment, and models for phylogenetic trees. By the latter it is possible to estimate distances between several sequences and to visualize these in a tree. Classical matrices for sequence alignment such as BLOSUM and PAM are constructed on the basis of a Markov process. By (Hidden) Markov Mod- els the speci¯c repetitive order of DNA sequences can be modeled so that predictions of families becomes possible. In this chapter you learn what a probability transition matrix is and which role it plays in a Markov process to construct speci¯c sequences. Various models for phylogenetic trees are explained in terms of the rate matrix as well as the probability transition matrix. The basic ideas of the Hidden Markov Model are brie°y explained and illustrated by an example1. 10.1 Random sampling Models to predict and classify DNA type of sequences make it possible to draw a sample from a population. The latter is the same as a distribution with certain properties. Recall from Chapter 3 that a discrete distribution is a set of values with certain probabilities that add up to one. Two basic examples illustrate this point. 1This chapter is somewhat more technical in its notation with respect to e.g. conditional probability. This is, however, inevitable for the understanding of Markov processes. 193194 CHAPTER 10. MARKOV MODELS Example 1. Throwing a coin. A fair coin X attains Head and Tail with probability 1/2. Thus we may write P(X = H) = 0:5 and P(X = T) = 0:5. With such a random variable there always correspond a population as well as a sampling scheme which can be simulated on a computer (e.g. Press, et al., 1992). > sample(c("H","T"),30,rep=TRUE,prob=c(0.5,0.5)) [1] "H" "H" "T" "T" "T" "H" "H" "T" "T" "H" "H" "H" "T" "T" "H" "T" [20] "H" "T" "T" "T" "H" "T" "H" "T" "T" "T" "T" Thus the sampled values Head and Tail correspond to the process of actu- ally throwing with a fair coin. The function sample randomly draws thirty times one of the values c("H","T") with replacement (rep=TRUE) and equal probabilities (prob=c(0.5,0.5)). Example 2. Generating a sequence of nucleotides. Another example is that of a random variable X which has the letters of the nucleotides as its values. So the events are X = A, X = C, X = G, and X = T. These events may occur in a certain DNA sequence with probabilities P(X = A) = 0:1, P(X = G) = 0:4, P(X = C) = 0:4, and P(X = T) = 0:1, respectively. Then the actual placement of the nucleotides along a sequence can be simulated. > sample(c("A","G","C","T"),30,rep=TRUE,prob=c(0.1,0.4,0.4,0.1)) [1] "G" "C" "T" "G" "C" "G" "G" "G" "T" "C" "T" "T" "C" "C" "C" [20] "G" "G" "C" "G" "G" "G" "C" "C" "C" "G" "C" Of course, if you do this again, then the resulting sequence will di®er due to the random nature of its generation. For these sampling schemes it holds that the events occur independently from the previous. 10.2 Probability transition matrix In order to build a model that produces speci¯c sequences we will consider a certain type of random variable. In particular, we will consider a sequence fX1;X2; ¢ ¢ ¢ g with values from a certain state space E. The latter is simply a set containing the possible values or states of a process. If, for instance, Xn = i, then the process is in state i at time n. Similarly, the expression10.2. PROBABILITY TRANSITION MATRIX 195 P(X1 = i) denotes the probability that the process is in state i at time point 1. The event that the process changes its state from i to j (transition) between time point one and two corresponds to the event (X2 = jjX1 = i), where the bar means "given that". The probability for this event to happen is denoted by P(X2 = jjX1 = i). In general, the probability of the transition from i to j between time point n and n+1 is given by P(Xn+1 = jjXn = i). These probabilities can be collected in a probability transition matrix P with elements pij = P(Xn+1 = jjXn = i): We will assume that the transition probabilities are the same for all time points so that there is no time index needed on the left hand side. Given that the process Xn is in a certain state, the corresponding row of the tran- sition matrix contains the distribution of Xn+1, implying that the sum of the probabilities over all possible states equals one. The probability transition matrix contains a (conditional) discrete probability distribution on each of its rows. For a Markov process it holds that the state at time point n+1 de- pends upon the state at time point n, but not on states at earlier time points. Example 1. Using the probability transition matrix to generate a Markov sequence. Suppose Xn has two states: 1 for a pyrimidine and 2 for a purine. A sequence can now be generated, as follows. If Xn = 1, then we throw with a fair die: If the outcome is lower than or equal to 5, then Xn+1 = 1 and, otherwise, (outcome equals 6) Xn+1 = 2. If Xn = 2, then we throw with a fair coin: If the outcome equals Tail, then Xn+1 = 1, and otherwise Xn+1 = 2. For this process the two by two probability transition matrix equals from to1 2 1 p11 p12 2 p21 p22 ; where p21 is the probability that the process changes from 2 to 1. This transition matrix can also be written as P = µ p11 p12 p21 p22 ¶ = µ P(X1 = 1jX0 = 1) P(X1 = 2jX0 = 1) P(X1 = 1jX0 = 2) P(X1 = 2jX0 = 2) ¶ = µ 56 16 12 12 ¶: Any matrix probability transition matrix P can be visualized by a transi- tion graph, where the transition probabilities are visualized by an arrow from196 CHAPTER 10. MARKOV MODELS state i to state j and the value of pij . For the current example the transition graph is given by Figure 10.12. The values 1 and 2 of the process are written within the circles and the transition probabilities are written near the arrows. To actually generate a sequences with values equal to 1 and 2 according the 5/6 1/6 1/2 1/2 0 1 Figure 10.1: Graph of probability transition matrix transition matrix we may use the following. markov1 <- function(x,P,n){ seq <- x for(k in 1:(n-1)){ seq[k+1] <- sample(x, 1, replace=TRUE, P[seq[k],])} return(seq) } P <- matrix(c(1/6,5/6,0.5,0.5), 2, 2, byrow=TRUE) rownames(P) <- colnames(P) <- StateSpace <- x <- c(1,2) > markov1(x,P,30) [1] 1 2 1 2 1 2 2 1 2 1 2 2 1 2 1 2 2 1 2 1 2 1 2 1 2 2 2 2 2 2 In the function markov1 the actual sampling is conducted by sample. We sample one element from the set containing 1 and 2 according to the prob- abilities in row seq[k] of the matrix P. This makes the probabilities of the states dependent on the corresponding row of the transition matrix. We con- veniently use the fact that R adds an element to the sequence; we do not have to declare its length on beforehand (although we could!). The sequence has a ¯xed start at State 1 and thereafter the ¯rst row in the probability transition matrix. Note that without the return command the function does 2The values 1 and 2 are erroneously depicted as 0 and 1, respectively10.2. PROBABILITY TRANSITION MATRIX 197 not give any output. Example 2. A sequence with a large frequency of C and G. To illustrate that certain probability transition matrices imply a large frequency of C and G residues, we use the following. markov2 <- function(StateSpace,P,pi0,n){ seq <- character(n) seq[1] <- sample(StateSpace, 1, replace=TRUE, pi0) for(k in 1:(n-1)){ seq[k+1] <- sample(StateSpace, 1, replace=TRUE, P[seq[k],])} return(seq) }P <- matrix(c( 1/6,5/6,0,0, 1/8,2/4,1/4,1/8, 0,2/6,3/6,1/6, 0,1/6,3/6,2/6),4,4,byrow=TRUE) rownames(P) <- colnames(P) <- StateSpace <- c("a","c","g","t") pi0 <- c(1/4,1/4,1/4,1/4) x <- markov2(StateSpace,P,pi0,1000) > table(x) x a c g t 72 409 378 141 The function starts with sampling just once from the distribution with equal probabilities pi0. It conveniently uses the the column and row names of the probability transition matrix for the sampling. The probabilities to go from "a" or "t" to "c" or "g" are large and as well as that to stay within "c" or "g". From the frequency table it can be observed that the majority of residues are "c" or "g". Example 3. A sequence with high phenylalanine frequency. Now it is possible to construct a sequence which produces the amino acid phenylalanine (F) with high probability. Recall that it is coded by the triple TTT or TTC. We use the function getTrans of the seqinr package to translate nucleotide triplets into amino acids.198 CHAPTER 10. MARKOV MODELS pi0 <- c(1/4,1/4,1/4,1/4) P <- matrix(c(.01,.01,.01,.97, .01,.01,.01,.97, .01,.01,.01,.97, .01,.28,.01,0.70),4,4,byrow=T) rownames(P) <- colnames(P) <- StateSpace <- c("a","c","g","t") x <- markov2(StateSpace,P,pi0,30000) > table(getTrans(x)) * A C D F H I L M N P R S T V 2 1 75 2 5205 24 76 2260 1 2 19 26 2154 1 91 Y 60 From the table it is clear that the F frequency is the largest among the gen- erated amino acids. Example 4. To illustrate estimation of the probability transition matrix we proceed with the sequence produced by the previous example. nr <- count(x,2) names(nr) A <- matrix(NA,4,4) A[1,1]<-nr["aa"]; A[1,2]<-nr["ag"]; A[1,3]<-nr["ac"]; A[1,4]<-nr["at"] A[2,1]<-nr["ga"]; A[2,2]<-nr["gg"]; A[2,3]<-nr["gc"]; A[2,4]<-nr["gt"] A[3,1]<-nr["ca"]; A[3,2]<-nr["cg"]; A[3,3]<-nr["cc"]; A[3,4]<-nr["ct"] A[4,1]<-nr["ta"]; A[4,2]<-nr["tg"]; A[4,3]<-nr["tc"]; A[4,4]<-nr["tt"] rowsumA <- apply(A, 1, sum) Phat <- sweep(A, 1, rowsumA, FUN="/") rownames(Phat) <- colnames(Phat) <- c("a","g","c","t") > round(Phat,3) a g c t a 0.011 0.000 0.007 0.982 g 0.017 0.003 0.010 0.969 c 0.010 0.011 0.012 0.967 t 0.009 0.009 0.279 0.703 The number of transitions are counted and divided by the row totals. The estimated transition probabilities are quite close to the true transition proba- bilities. The zero transition probabilities are exactly equal to the true because10.3. PROPERTIES OF THE TRANSITION MATRIX 199 these do not occur. This estimation procedure can easily be applied to DNA sequences. 10.3 Properties of the transition matrix In the above, the sequence was started at a certain state. Often, however, the probabilities of the initial states are available. That is, we have a vec- tor ¼0 with initial probabilities ¼10 = P(X0 = 1) and ¼20 = P(X0 = 2). Furthermore, if the transition matrix P = µ p11 p12 p21 p22 ¶ = µ P(X1 = 1jX0 = 1) P(X1 = 2jX0 = 1) P(X1 = 1jX0 = 2) P(X1 = 2jX0 = 2) ¶; then the probability that the process is in State 1 at time point 1 can be written as P(X1 = 1) = ¼10p11 + ¼20p21 = ¼T0p1; (10.1) where p1 is the ¯rst column of P, see Section 10.7. Note that the last equality holds by de¯nition of matrix multiplication. In a similar manner, it can be shown that P(X1 = 2) = ¼T0p2, where p2 is column 2 of the transition matrix P = (p1; p2). It can be concluded that ¼T0P = ¼T1, where ¼T1= (P(X1 = 1); P(X1 = 2)); the probability at time point 1 that the process is in State 1, State 2, respectively. This holds in general for all time points n, that is ¼TnP = ¼Tn+1: (10.2) Thus to obtain the probabilities of the states at time point n + 1, we can simply use matrix multiplication 3. Example 1. Matrix multiplication to compute probabilities. Suppose the following initial distribution and probability matrix ¼0 = ¡ 23 13 ¢ ;P = µ 56 16 12 12 ¶; for State 1 and 2, respectively. Then P(X1 = 1) and P(X1 = 2) collected in ¼T1= (P(X1 = 1); P(X1 = 2)) can be computed as follows. ¼T1= ¼T0P = ¡ 23 13 ¢µ 56 16 12 12 ¶ = ¡ 23 ¢ 56 + 13 ¢ 12 23 ¢ 16 + 13 ¢ 12 ¢ = ¡ 13 18 5 18 ¢ 3The transposition sign T simply transforms a column into a row.200 CHAPTER 10. MARKOV MODELS Using R its operator %*% for matrix multiplication, the product ¼T0P can be computed as follows. > P <- matrix(c(5/6,1/6,0.5,0.5),2,2,byrow=T) > pi0 <- c(2/3,1/3) > pi0 %*% P[,1] [,2] [1,] 0.7222222 0.2777778 Yet, another important property of the probability transition matrix deals with the probability of being in state 1 given that the process is in state 1 two time points before. In particular, it holds (see Section 10.7) that P(X2 = 1jX0 = 1) = p211; (10.3) where the latter is element (1; 1) of the matrix4 P2. In general, we have that P(Xn = jjX0 = i) = pnij ; which is element i; j of Pn. Example 3. Given the probability matrix of the previous example, the values P(X2 = jjX0 = i) for all of i; j can be computed by matrix multipli- cation. P2 = µ 56 16 12 12 ¶ ¢ µ 56 16 12 12 ¶ = µ ( 56 )2 + 16 12 56 16 + 16 12 12 56 + ( 12 )2 12 16 + ( 12 )2 ¶ = µ 28 36 8 36 24 36 12 36 ¶: Obviously, such matrix multiplications can be accomplished much more con- venient on a personal computer. > P %*% P [,1] [,2] [1,] 0.7777778 0.2222222 [2,] 0.6666667 0.3333333 Larger powers of P can be computed more e±ciently by methods given be- low. 4For a brief de¯nition of matrix multiplication, see Pevsner (2003, p.56) or wikipedia using the search string "wiki matrix multiplication".10.4. STATIONARY DISTRIBUTION 201 10.4 Stationary distribution A probability distribution ¼ satisfying ¼T = ¼TP is stationary because the transition matrix does not change the probabilities of the states of the process. Such a distribution usually exists, is unique, and plays an essential role in the long term behavior of the process. It sheds light on the question: What is the probability P(Xn = 1jX0 = 1) = pn11, as n increases without bound. That is: What is the probability that the process is in State 1, given that it started in State 1, as time increases without bound? To answer such a question we need large powers of the probability transition matrix. To compute these we need the eigen-decomposition of the probability transition matrix P = V ¤V ¡1; where V is the eigenvector matrix and ¤ the diagonal matrix with eigen- values. The latter are usually sorted in decreasing order so that the ¯rst (left upper) is the largest. Now the third power of the probability transition matrix can be computed, as follows P3 = V ¤V ¡1V ¤V ¡1V ¤V ¡1 = V ¤¤¤V ¡1 = V ¤3V ¡1: So that, indeed, in general Pn = V ¤nV ¡1: The latter is a computationally convenient expression because we only have to take the power of the eigenvalues in ¤ and to multiply by the left and right eigenvector matrices. This will be illustrated below. In the long term the Markov process tends to a certain value (Br¶emaud, 1999, p.197) because a probability transition matrix has a unique largest eigenvalue equal to 1 with corresponding eigenvectors 1 and ¼ (or rather normalized versions of these). It follows that, as n increases without bound, then Pn tends to 1¼T . In other words, P(Xn = jjX0 = i) = pnij tends to element (i; j) of 1¼T , which is equal to element j of ¼. For any initial dis- tribution ¼0, it follows that ¼00Pn tends to ¼T . Example 1. Stationary distribution. To compute the eigen-decomposition of the probability transition matrix P as well as powers of it, we may use the function eigen.202 CHAPTER 10. MARKOV MODELS > P <- matrix(c(1/6,5/6,0.5,0.5),2,2,byrow=T) > V <- eigen(P,symmetric = FALSE) > V$values [1] 1.0000000 -0.3333333 > V$vectors[,1] [,2] [1,] -0.7071068 -0.8574929 [2,] -0.7071068 0.5144958 The output of the function eigen is assigned to the list V from which the eigenvalues and eigenvectors can be extracted and printed to the screen. Now we can compute P16; the probability transition matrix raised to the power sixteen. > V$vec %*% diag(V$va)^(16) %*% solve(V$vec) [,1] [,2] [1,] 0.375 0.625 [2,] 0.375 0.625 So that the stationary distribution ¼T equals (0:375; 0:625). Example 2. Diploid. Suppose A is a dominant gene, a a recessive and that we start with a heterozygote aA. From the latter we obtain the initial state probability ¼T = (0; 1; 0) for the events (AA; aA; aa). When we consider pure self-fertilization, then the o®spring from AA is AA with probability (1; 0; 0), that of aa is aa with probability (0; 0; 1), and that of aA is (AA; aA; aa) with probability 1=4; 1=2; 1=4, respectively. Hence, the probability transition matrix becomes P = 0@ 1 0 0 1=4 1=2 1=4 0 0 1 1A We can now compute the transition probability matrix after ¯ve generations. P <- matrix(c(1,0,0, 1/4,1/2,1/4,0,0,1),3,3,byrow=T) V <- eigen(P,symmetric = FALSE) > V$vec %*% diag(V$va)^(5) %*% solve(V$vec) [,1] [,2] [,3] [1,] 1.000000 0.00000 0.00000010.5. PHYLOGENETIC DISTANCE 203 [2,] 0.484375 0.03125 0.484375 [3,] 0.000000 0.00000 1.000000 Hence, the distribution we obtain can be read from the second row which is highly homozygotic. A little more precise, using Equation 10.2, it can be shown that ¼Tn+1 = µ12 ¡ µ12¶n ;µ12¶n ; 12 ¡ µ12¶n¶; so that the distribution converges to (1=2; 0; 1=2). Note that this method of raising the transition probability matrix to a large power can easily be applied to determine the stationary distribution. The idea of taking a transition matrix to a certain power is also used to construct the PAM250 matrix given the PAM1 matrix (Pevsner, 2003, p.53) and for the construction of various BLOSUM matrices (Pevsner, 2003, p.50- 59; Deonier, et al. 2005, 187-190). 10.5 Phylogenetic distance Phylogenetic trees are constructed on the basis of distances between DNA sequences. These distances are computed from substitution models which are de¯ned by a matrix representing the rate of substitutions of one state to the other. The latter is usually expressed as a matrix Q. The rates of staying in a state are given by a negative number on the diagonal of the substitution matrix. The probability transition matrix P can be computed by matrix exponentiation P = exp(Q). How to do this in practice will be illustrated by an example. Example 1. From a rate matrix to a probability transition matrix. Suppose the rate matrix Q = AGCT 2 A G C T 664 ¡0:60 0:20 0:20 0:20 0:20 ¡0:60 0:20 0:20 0:20 0:20 ¡0:60 0:20 0:20 0:20 0:20 ¡0:60 3775: Thus within a certain time period a proportion of 0.20 A changes into G, 0.20 A into C, and 0.20 A into T. Consequently, a proportion of 0.60 of the204 CHAPTER 10. MARKOV MODELS residues goes back to A. Given this rate matrix, we can ¯nd the probabil- ity transition matrix P = exp(Q) by using the function expm(Q) from the package Matrix. library(Matrix) Q <- 0.2 * Matrix(c(-3,1,1,1,1,-3,1,1,1,1,-3,1,1,1,1,-3),4) rownames(Q) <- colnames(Q) <- c("A","G","C","T") P <- as.matrix(expm(Q)) > round(P,2) A G C T A 0.59 0.14 0.14 0.14 G 0.14 0.59 0.14 0.14 C 0.14 0.14 0.59 0.14 T 0.14 0.14 0.14 0.59 Thus the probability that the state changes from A to A is 0.59, from A to G is 0.14, etc. Because all phylogenetic models are de¯ned in terms of rate matrices, we shall concentrate on these. For instance, the rate matrix for the Jukes and Cantor (1969) (JC69) model can be written as QJC69 = AGCT 2A G C T 664 ¢ ® ® ® ® ¢ ® ® ® ® ¢ ® ® ® ® ¢ 3775: The sum of each row of a rate matrix equals zero, so that from this require- ment the diagonal elements of the JC69 model are equal to ¡3®. Further- more, the non-diagonal substitution rates of the JC69 model all have the same value ®. That is, the change from i to j equals that from j to i, so that the rate matrix is symmetric. Also the probability that the sequence equals one of the nucleotides is 1/4. This assumption, however, is unrealistic is many cases. Transitions are substitutions of nucleotides within types of nucleotides, thus purine to purine or pyrmidine to pyrmidine (A $ G or C $ T). Transversions are substitutions between nucleotide type (A $ T, G $ T,A $ C, and C $ G). In the JC69 model a transition is assumed to10.5. PHYLOGENETIC DISTANCE 205 happen with equal probability as a transversion. That is, it does not account for the fact that transitions are more common that transversions. To cover this for more general type of models are proposed by Kimura (1980, 1981), which are commonly abbreviated by K80 and K81. In terms of the rate matrix these models can we written as QK80 = 2664 ¢ ® ¯ ¯ ® ¢ ¯ ¯ ¯ ¯ ¢ ® ¯ ¯ ® ¢ 3775; QK81 = 2664 ¢ ® ¯ ° ® ¢ ° ¯ ¯ ° ¢ ® ° ¯ ® ¢ 3775: In the K80 model a change within type (transition) occurs at rate ® and between type (transversion) at rate ¯. In the K81 model all changes occur at a di®erent though symmetric rate; the rate of change A ! G is ® and equals that of A à G. If ® is large, then the amount of transitions is large; if both ¯ and ° are very small, then the number of transversions is small. A model is called \nested" if it is a special case of a more general model. For instance, the K80 model is nested in the K81 model because when we take ° = ¯, then we obtain the K80 model. Similarly, the JC69 model is nested in the K80 model because if we take ¯ = ®, then we obtain the JC69 model. Some examples of models with even more parameters are the Hasegawa, Kishino, and Yano (1985) (HKY85) model and the General Time-Reversable Model (GTR) model QHKY 85 = 2664 ¢ ®¼G ¯¼C ¯¼T ®¼A ¢ ¯¼C ¯¼T ¯¼A ¯¼G ¢ ®¼T ¯¼A ¯¼G ®¼C ¢ 3775; QGTR = 2664 ¢ ®¼G ¯¼C °¼T ®¼A ¢ ±¼C ²¼T ¯¼A ±¼G ¢ ³¼T °¼A ²¼G ³¼C ¢ 3775: The distance between DNA sequences is de¯ned on the basis of these models. From these distances the phylogenetic tree is computed by a neighbor-joining algorithm such that it has the smallest total branch length. Example 2. The K81 model. To compute the rate matrix of the K81 model with ® = 3=6, ¯ = 2=6, ° = 1=6 we may use the following. alpha <- 3/6; beta <- 2/6; gamma<- 1/6; Q <- matrix(data=NA,4,4) Q[1,2] <- Q[2,1] <- Q[3,4] <- Q[4,3] <- alpha206 CHAPTER 10. MARKOV MODELS Q[1,3] <- Q[3,1] <- Q[2,4] <- Q[4,2] <- beta Q[1,4] <- Q[4,1] <- Q[2,3] <- Q[3,2] <- gamma > diag(Q) <- -(alpha + beta + gamma) > Q [,1] [,2] [,3] [,4] [1,] -1.0000000 0.5000000 0.3333333 0.1666667 [2,] 0.5000000 -1.0000000 0.1666667 0.3333333 [3,] 0.3333333 0.1666667 -1.0000000 0.5000000 [4,] 0.1666667 0.3333333 0.5000000 -1.0000000 > Q <- Matrix(Q) > P <- as.matrix(expm(Q)) > P [,1] [,2] [,3] [,4] [1,] 0.4550880 0.2288517 0.1767105 0.1393498 [2,] 0.2288517 0.4550880 0.1393498 0.1767105 [3,] 0.1767105 0.1393498 0.4550880 0.2288517 [4,] 0.1393498 0.1767105 0.2288517 0.4550880 By raising the power of the probability transition matrix to a su±ciently large number, it can be observed that the stationary distribution ¼T = (0:25; 0:25; 0:25; 0:25). Example 3. Stationarity for the JC69 model. Let's take ® = 1=5 as in Example 1 and compute the rate matrix Q of the JC69 model, the corresponding probability transitionmatrix P, and raise it to the power 50. library(Matrix) alpha <- 1/5; Q <- matrix(rep(alpha,16),4,4) diag(Q) <- -3 * alpha Q <- Matrix(Q) P <- as.matrix(expm(Q)) V <- eigen(P,symmetric = FALSE) > V$vec %*% diag(V$va)^(50) %*% solve(V$vec) [,1] [,2] [,3] [,4] [1,] 0.25 0.25 0.25 0.25 [2,] 0.25 0.25 0.25 0.25 [3,] 0.25 0.25 0.25 0.25 [4,] 0.25 0.25 0.25 0.2510.5. PHYLOGENETIC DISTANCE 207 Hence, the stationary distribution is ¼T = (0:25; 0:25; 0:25; 0:25) (cf. Ewens & Grant, 2005, p. 477). Example 4. Distance between two sequences according to the JC69 model. In case of the JC69 model, the distance between sequences is a function of the proportion of di®erent nucleotides. Namely, d = ¡34 log(1 ¡ 4p=3); where p is the proportion of di®erent nucleotides of the two sequences. The pairwise distances between DNA sequences can be computed by the function dist.dna from the ape package. > library(ape);library(seqinr) > accnr <- paste("AJ5345",26:27,sep="") > seqbin <- read.GenBank(accnr, species.names = TRUE, as.character = FALSE) > dist.dna(seqbin, model = "JC69") AJ534526 AJ534527 0.1326839 Hence, the distance is 0.133. Over a total of 1143 nucleotides there are 139 di®erences, som that the proportion of di®erent nucleotides 139=1143 = p. Inserting this into the previous distance formula gives the distance. This can be veri¯ed as follows. > seq <- read.GenBank(accnr, species.names = TRUE, as.character = TRUE) > p <- sum(seq$AJ534526==seq$AJ534527)/1143 > d <- -log(1-4*p/3)*3/4 > d [1] 0.1326839 Example 5. Phylogenetic tree of a series of downloaded sequences. To further illustrate distances between DNA sequences we shall download the Chamaea fasciata mitochondrial cytb gene for cytochrome b for 10 species of warblers of the genus sylvia (Paradis, 2006). The function paste is used to quickly de¯ne the accession numbers and read.GenBank to actually down- load the sequences. The species names are extracted and attached to the sequences. We shall use the dist.dna function with the K80 model.208 CHAPTER 10. MARKOV MODELS library(ape);library(seqinr) accnr <- paste("AJ5345",26:35,sep="") seq <- read.GenBank(accnr) names(seq) <- attr(seq, "species") dist <- dist.dna(seq, model = "K80") plot(nj(dist)) Obviously, in this manner various trees can be computed and their plots compared. When various di®erent models are de¯ned the question becomes appar- ent which of these ¯ts best to the data relative to the number of parameters (symbols) of the model. When the models are estimated by maximum likeli- hood, then the Akaike information criterion (AIC = -2 ¢ loglik + 2 ¢ number of free parameters) can be used to select models. The best model is the one with the smallest AIC value. Example 6. A program called PHYML (Guindon & Gascuel, 2003) can be downloaded from http://atgc.lirmm.fr/phyml/ and run by the R function phymltest, if the executable is available at the same directory. We ¯rst write the sequences to the appropriate directory. The output from the program is written to the object called out for which the functions plot(out) and summary(out) can be used to extract more detailed information. > setwd("/share/home/wim/bin") > write.dna(seq,"seq.txt", format ="interleaved") > out <-phymltest("seq.txt",format = "interleaved", execname ="phyml_linux") > print(out) nb.free.para loglik AIC JC69 1 -4605.966 9213.931 JC69+I 2 -4425.602 8855.203 JC69+G 2 -4421.304 8846.608 JC69+I+G 3 -4421.000 8848.001 K80 2 -4423.727 8851.455 K80+I 3 -4230.539 8467.079 K80+G 3 -4224.457 8454.915 K80+I+G 4 -4223.136 8454.272 F81 4 -4514.331 9036.66210.6. HIDDEN MARKOV MODELS 209 F81+I 5 -4309.600 8629.199 F81+G 5 -4304.530 8619.060 F81+I+G 6 -4303.760 8619.519 F84 5 -4351.164 8712.328 F84+I 6 -4112.006 8236.012 F84+G 6 -4106.568 8225.135 F84+I+G 7 -4105.500 8225.001 HKY85 5 -4333.086 8676.171 HKY85+I 6 -4102.262 8216.524 HKY85+G 6 -4097.401 8206.802 HKY85+I+G 7 -4096.624 8207.248 TN93 6 -4323.291 8658.581 TN93+I 7 -4097.099 8208.198 TN93+G 7 -4091.461 8196.922 TN93+I+G 8 -4090.790 8197.580 GTR 9 -4293.398 8604.795 GTR+I 10 -4084.522 8189.043 GTR+G 10 -4079.010 8178.020 GTR+I+G 11 -4078.149 8178.299 The notation "+I" and "+G" indicates the presence of invariant sites and/or a gamma distribution of substitution rates. It can be seen that the smallest AIC corresponds to model 27 called GTR+G. To plot it, we have to read the trees, and, next, to extract the 27th, see Figure 10.3. tr <- read.tree("seq.txt_phyml_tree.txt") plot(tr[[27]]) add.scale.bar(length=0.01) In case similar sequences have slightly di®erent lengths, these have to be aligned by programs such as clustalx or clustalw before these can be used. 10.6 Hidden Markov Models In a Hidden Markov Model (HMM) there are two probability transition ma- trices. There is an emission matrix E and a transition matrix A. The210 CHAPTER 10. MARKOV MODELS generation of an observable sequence goes in two steps. First, there is a tran- sition from a Markov process of a hidden state and given this value there is an emission of an observable value. We shall illustrate this by the clas- sical example of the occasionally dishonest casino (Durbin et. al., 1998, p.18). Example 1. Occasionally dishonest casino. A casino uses a fair die most of the time, however, occasionally it switches to an unfair die. The state with respect to fairness is hidden for the observer. The observer can only observe the values of the die and not its hidden state with respect to its fairness. It is convenient to denote fair by 1 and unfair by 2. The transition probabilities of the hidden states are by the emission matrix E = · P(Di = 1jDi¡1 = 1) P(Di = 2jDi¡1 = 1) P(Di = 1jDi¡1 = 2) P(Di = 2jDi¡1 = 2) ¸ = · 0:95 0:05 0:10 0:90 ¸ : Thus the probability is 0.95 that the die is fair at time point i, given that it is fair at time point i ¡ 1. The probability that it will switch from fair to unfair is 0.05. The probability that it will switch from loaded to fair is 0.10 and that it stays loaded is 0.90. With this emission matrix we can generate a sequence of hidden states, where the values 1 and 2 indicate whether the die is fair (1) or loaded (2). Given the fairness of the die we de¯ne the probability transition matrix. A = · P(Oi = 1jDi = 1) P(Oi = 2jDi = 1) P(Oi = 3jDi = 1) ¢ ¢ ¢ P(Oi = 1jDi = 2) P(Oi = 2jDi = 2) P(Oi = 3jDi = 2) ¢ ¢ ¢ ¸ = · 1=6 1=6 1=6 1=6 1=6 1=6 1=10 1=10 1=10 1=10 1=10 1=2 ¸ : (10.4) Thus given that the die is fair, the probability of any outcome equals 1/6. However, given that the die is unfair (loaded), the probability of outcome 6 equals 1/2 and that of any other outcome equals 1/10. The HMM with this transition and emission matrix can be programmed. After sampling the hidden states from a Markov chain and the outcomes of the die are sampled according to the value of the hidden state (die type). hmmdat <- function(A,E,n){ observationset <- c(1:6) hiddenset <- c(1,2) x <- h <- matrix(NA,nr=n,nc=1)10.6. HIDDEN MARKOV MODELS 211 h[1]<-1; x[1]<-sample(observationset,1,replace=T,E[h[1],]) h <- markov(hiddenset,A,n) for(k in 1:(n-1)){x[k+1] <- sample(observationset,1,replace=T,E[h[k],])} out <- matrix(c(x,h),nrow=n,ncol=2,byrow=FALSE) return(out) } E <- matrix(c(rep(1/6,6),rep(1/10,5),1/2),2,6,byrow=T) #emission matrix A <- matrix(c(0.95,0.05,0.1,0.9),2,2,byrow=TRUE) #transition matrix dat <- hmmdat(A,E,100) colnames(dat) <- c("observation","hidden_state") rownames(dat) <- 1:100 > t(dat) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 observations 5 2 3 1 6 1 3 1 1 5 6 6 2 2 3 5 4 6 1 2 4 4 3 2 3 hidden_states 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 observations 4 3 2 4 1 6 6 6 6 6 5 5 3 6 1 6 5 2 4 1 4 2 hidden_states 1 1 1 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 observations 5 6 5 2 3 3 1 3 3 5 6 6 2 4 5 4 6 1 6 5 2 6 hidden_states 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 observations 1 1 4 4 1 5 6 4 3 5 4 2 6 1 3 6 5 2 2 6 6 1 hidden_states 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 1 1 92 93 94 95 96 97 98 99 100 observations 4 1 6 5 5 6 5 3 4 hidden_states 1 1 1 1 1 1 1 1 1 In certain applications to bioinformatics, it is of most importance to es- timate the value of the hidden state given the data. The Viterbi algorithm is developed to predict the hidden state given the data and the (estimated) transition and emission matrix. The algorithm builds up a matrix v(i; l), where i runs from one to the number of observations and l from one to the number of states. The initial values are v(1; 1) = 1, and v(1; l) = 0 for all l. Then the values for v(i; l) are recursively de¯ned by v(i; l) = e(l; x(i)) ¢ max k fv(i ¡ 1; k)a(k; l)g :212 CHAPTER 10. MARKOV MODELS For each row of the matrix the maximum is taken as the best predictor of the hidden state. Example 2. The viterbi algorithm can be programmed and applied to the hidden states of the data generated with respect to the occasionally dishonest casino. viterbi <- function(A,E,x) { v <- matrix(NA, nr=length(x), nc=dim(A)[1]) v[1,] <- 0; v[1,1] <- 1 for(i in 2:length(x)) { for (l in 1:dim(A)[1]) {v[i,l] <- E[l,x[i]] * max(v[(i-1),] * A[l,])} } return(v) }vit <- viterbi(A,E,dat[,1]) vitrowmax <- apply(vit, 1, function(x) which.max(x)) hiddenstate <- dat[,2] > table(hiddenstate, vitrowmax) vitrowmax hiddenstate 1 2 1 72 11 2 15 2 datt <- cbind(dat,vitrowmax) colnames(datt) <- c("observation","hidden_state","predicted state") > t(datt) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 observation 5 2 3 1 6 1 3 1 1 5 6 6 2 2 3 5 4 6 1 2 4 4 3 2 hidden_state 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 predicted state 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 observation 3 4 3 2 4 1 6 6 6 6 6 5 5 3 6 1 6 5 2 4 1 hidden_state 1 1 1 1 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 predicted state 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 observation 4 2 5 6 5 2 3 3 1 3 3 5 6 6 2 4 5 4 6 1 6 hidden_state 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 110.7. APPENDIX 213 predicted state 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 observation 5 2 6 1 1 4 4 1 5 6 4 3 5 4 2 6 1 3 6 5 2 hidden_state 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 predicted state 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 88 89 90 91 92 93 94 95 96 97 98 99 100 observation 2 6 6 1 4 1 6 5 5 6 5 3 4 hidden_state 2 2 1 1 1 1 1 1 1 1 1 1 1 predicted state 1 1 1 1 1 1 1 1 1 1 1 1 1 The misclassi¯cation rate is 0.27 which is quite large given the fact that we used the true transition and emission matrix. An important observation is that after a transition of a hidden state, it takes a few values for the predic- tion to change. This is caused by the recursive nature of the algorithm. 10.7 Appendix The probability that the process is in State 1 at time point 1 can be computed as follows. P(X1 = 1) = P(X1 = 1;X0 = 1) + P(X1 = 1;X0 = 2) = P(X1 = 1jX0 = 1) ¢ P(X0 = 1) + P(X1 = 1jX0 = 2) ¢ P(X0 = 2) = ¼10p11 + ¼20p21 = ¼T0p1; where p1 is the ¯rst column of P.214 CHAPTER 10. MARKOV MODELS In particular, it holds that P(X2 = 1jX0 = 1) = P(X2 = 1;X1 = 1jX0 = 1) + P(X2 = 1;X1 = 2jX0 = 1) = X2 k=1 P(X2 = 1;X1 = kjX0 = 1) = X2 k=1 P(X2 = 1jX1 = k;X0 = 1) ¢ P(X1 = kjX0 = 1) = X2 k=1 P(X2 = 1jX1 = k) ¢ P(X1 = kjX0 = 1) = p11p11 + p21p12 = row 1 of P times column 1 of P = P211; where the latter is element (1; 1) of the matrix P2 = P ¢ P. 10.8 Overview and concluding remarks The probability transition matrix is extensively explained and illustrated because it is a cornerstone to many ideas in bioinformatics. A thorough treatment of phylogenetics is given by Paradis (2006) and of Hidden Markov Models by Durbin et. al (2005). 10.9 Exercises 1. Visualize by a transition graph the following transition matrices. For the process with four states take the names of the nucleotides in the order A, G, T, and C. µ 13 23 34 14 ¶;µ 1 0 0 1 ¶;µ 0 1 1 0 ¶;0BB@ 14 24 0 14 16 26 26 16 0 27 57 0 18 18 28 48 1CCA;0BB@ 14 34 0 0 16 56 0 0 0 0 57 27 0 0 38 58 1CCA: 2. Computing probabilities. Given the states 0 and 1 and the following initial distribution and probability matrix ¼0 = ¡ 12 12 ¢ ;P = µ 34 14 12 12 ¶:10.9. EXERCISES 215 (a) Compute P(X1 = 0). (b) Compute P(X1 = 1). (c) Compute P(X2 = 0jX0 = 0). (d) Compute P(X2 = 1jX0 = 0). 3. Programming GTR. Use ¼A = 0:15, ¼G = 0:35, ¼C = 0:35, ¼T = 0:15, ® = 4, ¯ = 0:5, ° = 0:4, ± = 0:3, ² = 0:2, and ³ = 4. (a) Program the rate matrix in such a manner that it is simple to adapt for other values of the parameters. (b) Is the transversion rate larger or smaller then the transition rate? (c) Compute the corresponding the probability transition matrix. (d) Try to argue whether you expect a large frequency of transversions or translations. (e) Generate a sequence of 99 nucleotide residues according to the markov model. 4. Distance according to JC69. (a) Down load the sequences AJ534526 and AJ534527. Hint: Use as.character = TRUE in the read.GenBank function. (b) Compute the proportion of di®erent nucleotides. (c) Use this proportion to verify the distances between these sequences according to the JC69 model.216 CHAPTER 10. MARKOV MODELS Akaike information criterion for phymlout 8200 8400 8600 8800 9000 9200 GTR +G GTR + I +G GTR + I TN93 +G TN93 + I +G HKY85 +G HKY85 + I +G TN93 + I HKY85 + I F84 + I +G F84 +G F84 + I K80 + I +G K80 +G K80 + I GTR F81 +G F81 + I +G F81 + I TN93 HKY85 F84 JC69 +G JC69 + I +G K80 JC69 + I F81 JC69 Figure 10.2: Evaluation of models by AIC .10.9. EXERCISES 217 Chamaea fasciata Sylvia nisoria Sylvia layardi Sylvia subcaeruleum Sylvia boehmi Sylvia buryi Sylvia lugens Sylvia leucomelaena Sylvia hortensis Sylvia crassirostris 0.01 Figure 10.3: Tree according to GTR model.218 CHAPTER 10. MARKOV MODELSAppendix A Answers to exercises Answers to exercises of Chapter 1: Brief Introduction to R 1. Some questions to orientate yourself. (a) matrix, numeric, numeric, matrix, function, function, factor, stan- dardGeneric, ExpressionSet. (b) remove, summation, product, sequence, standard deviation, num- ber of rows, (c) Use R its help or use the internet search key "r wiki grep" to ¯nd the following answers: searching regular expressions, return a vector from a function on the rows or columns of a matrix, generate a factor by specifying the pattern of levels, load add- on packages, make R reading input from a ¯le or URL, set the working directory to a certain map, print the last ¢ commands given from the command line, give the structure of an object. 2. gendat (a) apply(gendat,2,sd). (b) apply(gendat,1,sd). (c) To order the data frame according to the gene standard deviations. sdexprsval <- apply(gendat,1,sd) o <- order(sdexprsval,decreasing=TRUE) gendat[o,] 219220 APPENDIX A. ANSWERS TO EXERCISES (d) gene1 3. Computations on gene means of the Golub data. (a) Computation of mean gene expression values. data(golub, package = "multtest") meangol <- apply(golub,1,mean) (b) To order the data frame use o <- order(meangol,decreasing=TRUE) and golub[o,] (c) Give the names of the three genes with the largest mean expression value. > golub.gnames[o[1:3],3] [1] "U43901_rna1_s_at" "M13934_cds2_at" "X01677_f_at" (d) Give their biological names. > golub.gnames[o[1:3],2] [1] "37 kD laminin receptor precursor/p40 ribosome associated protein [2] "RPS14 gene (ribosomal protein S14) extracted from Human ribosoma[3] "GAPD Glyceraldehyde-3-phosphate dehydrogenase" 4. Computations on gene standard deviations of the Golub data. (a) The standard deviation per gene can be computed by sdgol <- apply(golub,1,sd). (b) The gene with standard deviation larger than 0.5 can be selected by golubsd <- golub[sdgol>0.5,]. (c) sum(sdgol>0.5) gives that the number of genes having sd larger than 0.5 is 1498. 5. Oncogenes in Golub data. (a) length(agrep("^oncogene",golub.gnames[,2])) gives 42. (b) By the script below the "Cellular oncogene c-fos is found. data(golub, package="multtest") rowindex <- agrep("^oncogene",golub.gnames[,2]) oncogol <- golub[rowindex,]221 oncogolub.gnames <- golub.gnames[rowindex,] gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) meangol <- apply(oncogol[,gol.fac=="ALL"],1,mean) o <- order(meangol,decreasing=TRUE) > oncogolub.gnames[o[1:3],2] [1] "PIM1 Pim-1 oncogene" "JUNB Jun B proto-oncogene" [3] "Proto-oncogene BCL3 gene" (c) meangol <- apply(oncogol[,gol.fac=="AML"],1,mean) o <- order(meangol,decreasing=TRUE) > oncogolub.gnames[o[1:3],2] [1] "PIM1 Pim-1 oncogene" "JUNB Jun B proto-oncogene" [3] "Proto-oncogene BCL3 gene" (d) Writing results to a csv ¯le. Be aware of the correct column separation. x <- oncogolub.gnames[o[1:10],c(3,2)] colnames(x) <- c("probe ID","gene name") write.csv(x,file="goluboutcsv") write.table(x,file="goluboutnorowname",row.names=FALSE) 6. Constructing a factor. (a) gl(2,4). (b) gl(5,3). (c) gl(3,5). 7. Gene means for B1 patients. library(ALL); data(ALL) meanB1 <- apply(exprs(ALL[,ALL$BT=="B1"]),1, mean) o <- order(meanB1,decreasing=TRUE) > meanB1[o[1:3]] AFFX-hum_alu_at 31962_at 31957_r_at 13.41648 13.16671 13.15995 Answers to exercises of Chapter 2: Descriptive Statistics and Data Dis- play222 APPENDIX A. ANSWERS TO EXERCISES 1. Illustration of mean and standard deviation. (a) Use x<- c(1,1.5,2,2.5,3) and mean(x) and sd(x) to obtain the mean is 2 and the standard deviation is 0.7905694. (b) Now the mean is 7.4 and dramatically increased the standard de- viation 12.64615. (c) The outlier increased the mean as well as the standard deviation. 2. Comparing two genes. Take i <- 66 or i <- 790. (a) Use boxplot(golub[i,]~gol.fac) to observe that 790 has three outliers and 66 has no. (b) Use qqnorm(golub[i,gol.fac=="ALL") and qqline(golub[i,gol.fac=="ALL"]) to observe that nearly all values of 66 are on the line, where as for 790 the three outliers are way of the normality line. Hypothesis: The expression values of 66 are normally distributed, but those of row 790 are not. (c) Use mean(golub[i,gol.fac=="ALL"]) and median(golub[i,gol.fac=="ALL"]). The mean (-1.174024) is larger than the median (-1.28137) due to outliers on the right hand side. For the gen in row 66 the mean is 1.182503 and the median 1.23023. The di®erences are smaller. 3. E®ect size. (a) The size 11 is large, because the mean is eleven times larger than the standard deviation. data(golub, package="multtest") gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) efs <- apply(golub[,gol.fac=="ALL"],1,function(x) mean(x)/sd(x)) o <- order(efs,decreasing=TRUE) > efs[o[1:5]] [1] 11.138128 10.638308 9.155108 8.954115 8.695353 > golub.gnames[o[1:5],2] [1] "YWHAZ Tyrosine 3-monooxygenase/tryptophan 5-monooxygenase activa[2] "ZNF91 Zinc finger protein 91 (HPF7, HTF10)" [3] "HnRNP-E2 mRNA" [4] "54 kDa protein mRNA" [5] "Immunophilin homolog ARA9 mRNA"223 (b) The robust variant can be de¯ned by dividing the median by the MAD. An alternative would be to divide the median by the IQR. This gives other best genes indicating that the some genes may have outliers that in°uence the outcome. refs <- apply(golub[,gol.fac=="ALL"],1,function(x) median(x)/mad(x)) o <- order(refs,decreasing=TRUE) > refs[o[1:5]] [1] 14.51217 13.57425 13.27698 13.14419 12.91608 > golub.gnames[o[1:5],2] [1] "COX6B gene (COXG) extracted from Human DNA from overlapping chromosome 1F25451, and R31076 containing COX6B and UPKA, genomic sequence" [2] "AFFX-HSAC07/X00351_M_at (endogenous control)" [3] "ATP5A1 ATP synthase, H+ transporting, mitochondrial F1 complex, alpha suisoform 1, cardiac muscle" [4] "ATP SYNTHASE GAMMA CHAIN, MITOCHONDRIAL PRECURSOR" [5] "YWHAZ Tyrosine 3-monooxygenase/tryptophan 5-monooxygenase activation prozeta polypeptide" 4. Plotting gene expressions "CCND3 Cyclin D3". The answers in the script below. data(golub, package = "multtest") gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) stripchart(golub[1042,] ~ gol.fac,method="jitter") stripchart(golub[1042,] ~ gol.fac,method="jitter",vertical = TRUE) stripchart(golub[1042,] ~ gol.fac,method="jitter",col=c("red", "blue"), vertical = TRUE) stripchart(golub[1042,] ~ gol.fac,method="jitter",col=c("red", "blue"), pch="*",vertical = TRUE) title("CCND3 Cyclin D3 expression value for ALL and AMl patients") 5. Box-and-Whiskers plot of "CCND3 Cyclin D3".. locator() x11() x <- data(golub, package = "multtest") gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) boxplot(x,xlim=c(0,4))224 APPENDIX A. ANSWERS TO EXERCISES arrows(2.0,1.93,1.24,1.93);text(2.5,1.93,"Median") arrows(2.0,1.1,1.24,1.1) ;text(2.5,1.1,"Outlier") arrows(2.0,1.79,1.24,1.79);text(2.5,1.79,"first quartile") arrows(2.0,2.17,1.24,2.17);text(2.5,2.17,"third quartile") arrows(2.0,1.27,1.24,1.27);text(2.5,1.27,"lower wisker") arrows(2.0,2.59,1.24,2.59);text(2.5,2.59,"upper wisker") dev.copy2eps(device=x11,file="BoxplotWithExplanation.eps") boxplot.stats(x, coef = 1.5, do.conf = TRUE, do.out = TRUE) #finds value6. Box-and-wiskers plot of persons of Golub et al. (1999) data.. (a) The medians are all around zero, the inter quartile range di®er only slightly, the minimal values are all around minus 1.5. All persons have outliers near three. (b) The means are very close to zero. The medians are all between (¡0:15383; 0:06922), so these are also close to zero. personmean <- apply(golub,2,mean) personmedian <- apply(golub,2,median) (c) The data seem preprocessed to have standard deviation equal to one. The re-scaled IQR and MAD have slightly larger range. > range(apply(golub,2,sd)) [1] 0.9999988 1.0000011 > range(apply(golub,2,function(x) IQR(x)/1.349)) [1] 0.9599036 1.3361527 > range(apply(golub,2,mad)) [1] 0.9590346 1.2420185 7. Oncogenes of Golub et al. (1999) data. (a) Note that we need the transpose operator t to change rows into columns. The script below will do. data(golub, package="multtest") gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) rowindex <- agrep("^oncogene",golub.gnames[,2]) oncogol <- golub[rowindex,] oncogolub.gnames <- golub.gnames[rowindex,]225 row.names(oncogol) <- oncogolub.gnames[,3] boxplot(data.frame(t(oncogol[,gol.fac=="ALL"]))) (b) The plot gives a nice overview of the distributions of the gene expressions values of the onco gene separately for the ALL and the AML patients. Several genes behave similarly for ALL and AML. Some are clearly distributed around zero, but others not. Also, some have a small inter quartile ranges, while for others this is large. A similar statement holds for outliers, some do not have outliers, but others certainly have. Some gene show distinct distributions between patient groups. For instance, the sixth has ALL expressions around zero, but those for AML are larger than zero. par(mfrow=c(2,1)) boxplot(data.frame(t(oncogol[,gol.fac=="ALL"]))) title("Box and wiskers plot for oncogenes of ALL patients ") boxplot(data.frame(t(oncogol[,gol.fac=="AML"]))) title("Box and wiskers plot for oncogenes of AML patients ") par(mfrow=c(1,1)) 8. Descriptive statistics for the ALL gene expression values of the Golub et al. (1999) data. (a) The ranges indicate strong di®erence in means. The range of the mean and of the median are similar. The bulk of the data seems symmetric. > range(apply(golub[,gol.fac=="ALL"],1,mean)) [1] -1.330984 3.278551 > range(apply(golub[,gol.fac=="ALL"],1,median)) [1] -1.36832 3.35455 (b) The range of the standard deviation is somewhat smaller than of the re-scaled IQR and MAD. > range(apply(golub[,gol.fac=="ALL"],1,sd)) [1] 0.1336206 2.0381309 > range(apply(golub[,gol.fac=="ALL"],1,function(x) IQR(x)/1.349)) [1] 0.1153113 2.7769867 > range(apply(golub[,gol.fac=="ALL"],1,mad))226 APPENDIX A. ANSWERS TO EXERCISES [1] 0.1056649 2.9656744 Answers to exercises of Chapter 3: Important Distributions 1. Binomial (a) P(X = 24) = 0:1046692, P(X · 24) = 0:5557756, and P(X ¸ 30) = 0:0746237. (b) P(20 · X · 30) = 0:83856, P(20 · X) = 0:8830403. (c) P(20 · XorX ¸ 40) = 0:8830403, and P(20 · XandX ¸ 10) = 0:999975. (d) E(X) = 24, var(X) = 3:794733 Use: sqrt(60 * 0.4 *0.6) (e) x0:025 = 17, x0:5 = 24, and x0:975 = 32. 2. Standard Normal. (a) P(1:6 < Z < 2:3) = 0:04408. (b) P(Z < 1:64) = 0:9495. (c) P(¡1:64 < Z < ¡1:02) = 0:1034. (d) P(0 < Z < 1:96) = 0:4750. (e) P(¡1:96 < Z < 1:96) = 0:9500. (f) z0:025 = ¡1:959964, z0:05 = ¡1:644854, z0:5 = 0, z0:95 = 1:644854, and z0:975 = 1:959964. 3. Normal. (a) P(X < 12) = 0:8413. (b) P(X > 8) = 0:8413. (c) P(9 < X < 10; 5) = 0:2917. (d) The quantiles x0:025 = 6:080072, x0:5 = 10, and x0:975 = 13:91993. 4. T-distribution. (a) P(T6 < 1) = 0:8220412. (b) P(T6 > 2) = 0:04621316. (c) P(¡1 < T6 < 1) = 0:6440823.227 (d) P(¡2 < T6 < ¡2) = 0:9075737. (e) t0:025 = ¡2:446912, t0:5 = 0, and t0:975 = 2:446912. 5. F distribution. (a) P(F8;5 < 3) = 0:8792198. (b) P(F8;5 > 4) = 0:07169537. (c) P(1 < F8;5 < 6) = 0:4931282. (d) The quantiles f0:025 = 0:2075862, f0:5 = 1:054510, and f0:975 = 6:757172. 6. Chi-squared distribution. (a) P(Â210 < 3) = 0:01857594. (b) P(Â210 > 4) = 0:947347. (c) P(1 < Â210 < 6) = 0:1845646. (d) The quantiles g0:025 = 3:246973, g0:5 = 9:341818, and g0:975 = 20:48318. 7. MicroRNA. (a) P(X = 14) = dbinom(14; 20; 0:7) = 0:191639. (b) P(X · 14) = pbinom(14; 20; 0:7) = 0:5836292. (c) P(X > 10) = 1 ¡ P(X · 10) = 1 ¡ pbinom(10; 20; 0:7) = 0:9520381. (d) P(10 · X · 15) = P(X · 15)¡P(X · 9) = pbinom(15; 20; 0:7) ¡ pbinom(9; 20; 0:7) = 0:7453474. (e) 20 ¢ 0:7 = 14. (f) sqrt(20* 0.7 * 0.3)=2.04939. 8. Zyxin. (a) P(X · 1:2) =pnorm(1.2,1.6,0.4)=0.1586553. (b) P(1:2 · X · 2:0) =pnorm(2.0,1.6,0.4) - pnorm(1.2,1.6,0.4)=0.6826895. (c) P(2:4 · X · 0:8) =pnorm(2.4,1.6,0.4) - pnorm(0.8,1.6,0.4)=0.9544997.228 APPENDIX A. ANSWERS TO EXERCISES (d) x0:025 =qnorm(0.025,1.6,0.4)=0.8160144. Similarly, x0:975 = 2:383986. (e) x <- rnorm(1000,1.6,0.4) gives mean(x) = 1.608401 and sd(x)=0.4022082. Both are close to the values in the population. 9. Some computations on Golub et al. (1999) data. (a) The tree larges t-value 57.8, 55.2, and 47.5 are extremely large. data(golub, package="multtest") gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) tval <- apply(golub[,gol.fac=="ALL"],1,function(x) sqrt(27) * mean(x)o <- order(tval,decreasing=TRUE) tval[o[1:3]] golub.gnames[o[1:3],2] (b) The scrip below gives 2185 ratios between 0.5 and 1.5. sdall <- apply(golub[,gol.fac=="ALL"],1, sd) sdaml <- apply(golub[,gol.fac=="AML"],1, sd) sdratio <- sdall/sdaml sum( sdratio > 0.5 & sdratio < 1.5) 10. Extreme value investigation. The blue line (extreme value) ¯ts to the black line (density of generated extreme data) much better than the red line (normal distribution). an <- sqrt(2*log(n)) - 0.5*(log(log(n))+log(4*pi))*(2*log(n))^(-1/2) bn <- (2*log(n))^(-1/2) e <- double(); n <- 10000 # Serfling p.90 for (i in 1:1000) e[i] <- (max(rnorm(n))-an)/bn plot(density(e),ylim=c(0,0.5)) f<-function(x){exp(-x)*exp(-exp(-x))} curve(f,range(density(e)$x),add=TRUE,col = "blue") curve(dnorm,add=TRUE,col = "red") Answers exercise chapter 4: Estimation and Inference 1. Gene CD33. Use agrep("^CD33",golub.gnames[,2]) to ¯nd 808. (a) The code229 library(multtest);data(golub) gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) shapiro.test(golub[i,gol.fac=="ALL"]) gives p-value = 0.592 and changing ALL into AML gives p-value = 0.2583. Hence, for normality is accepted. (b) var.test(golub[i,] ~ gol.fac) gives p-value = 0.1095 so equal- ity of variances is accepted. (c) t.test(golub[i,] ~ gol.fac, var.equal = TRUE) gives p-value = 1.773e-09, so equality of means is rejected. (d) Yes, t = -7.9813 is quite extreme. 2. Gene MYBL2 V-myb avian myeloblastosis viral oncogene homolog-like 2. Take i <- 1788. (a) Use boxplot(golub[i,] ~ gol.fac) to observe from the box- plot that one is quite certain that the null-hypothesis of no exper- imental e®ect holds. (b) t.test(golub[i,] ~ gol.fac, var.equal = TRUE) gives p-value = 0.8597, so that the null hypothesis of equal means is accepted. 3. HOXA9. Use i <- 1391. (a) shapiro.test(golub[i,gol.fac=="ALL"]) gives p-value = 1.318e-07, so that normality is rejected. (b) wilcox.test(golub[i,] ~ gol.fac) gives p-value = 7.923e-05, so that equality of means is rejected. Note that the p-value from Grubbs test of the ALL expression values is 0.00519, so the null hypothesis of no outliers is rejected. Nevertheless the Welch two- sample T-test is also rejects the null-hypothesis of equal means. Its t-value equals -4.3026 and is quite large. 4. Zyxin. (a) Searching NCBI UniGene on zyxin gives BC002323.2. (b) Use chisq.test(as.data.frame(table(read.GenBank(c("BC002323.2"))))$Freq) to ¯nd p-value < 2.2e-16, so that the null-hypothesis of equal frequencies is rejected.230 APPENDIX A. ANSWERS TO EXERCISES (c) We download and store the frequencies of the sequences in x and y. Next the empirical probabilities from y are use to predict the frequencies from y. x <- as.data.frame(table(read.GenBank(c("X94991.1"))))$Freq y <- as.data.frame(table(read.GenBank(c("BC002323.2"))))$Freq >chisq.test(x, p=y/sum(y)) Chi-squared test for given probabilities data: x X-squared = 0.0277, df = 3, p-value = 0.9988 5. Gene selection. ptg <- apply(golub, 1, function(x) t.test(x ~ gol.fac, alternative = c("greater"))$p.value) golub.gnames[order(ptg)[1:10],2] 6. Antigenes. library(multtest); data(golub) gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) pt <- apply(golub, 1, function(x) t.test(x ~ gol.fac)$p.value) index <-agrep("^antigen",golub.gnames[,2]) golub.index<-golub[index,] pt.index<-pt[index] golub.gnames.index<-golub.gnames[index,] golub.gnames.index[order(pt.index)[1:length(index)],2] 7. Genetic Model. From the output below the null hypothesis that the probabilities are as speci¯ed is accepted. > chisq.test(x=c(930,330,290,90),p=c(9/16,3/16,3/16,1/16)) Chi-squared test for given probabilities data: c(930, 330, 290, 90) X-squared = 4.2276, df = 3, p-value = 0.2379 8. Comparing two genes.231 all66 <- golub[66,gol.fac=="ALL"] all790 <- golub[790,gol.fac=="ALL"] boxplot(all66,all790) mean(all66);mean(all790) median(all66);median(all790) sd(all66);sd(all790) IQR(all66)/1.349 ;IQR(all790)/1.349 mean(all66);mean(all790) mad(all66);mad(all790) shapiro.test(all66);shapiro.test(all790) 9. Normality tests. library(multtest);data(golub) gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) allsh <- apply(golub[,gol.fac=="ALL"], 1, function(x) shapiro.test(x)$p.value) amlsh <- apply(golub[,gol.fac=="AML"], 1, function(x) shapiro.test(x)$p.value) > 100 * sum(allsh>0.05)/length(allsh) [1] 58.27598 > 100 * sum(amlsh>0.05)/length(amlsh) [1] 78.5644 > 100 * sum(allsh>0.05 & allsh>0.05)/length(allsh) [1] 58.27598 10. Two-sample tests on gene expression values of the Golub et al. (1999) data. (a) data(golub, package = "multtest"); gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) pt <- apply(golub, 1, function(x) t.test(x ~ gol.fac)$p.value) pw <- apply(golub, 1, function(x) wilcox.test(x ~ gol.fac)$p.value) o <- order(pt,decreasing=FALSE) > golub.gnames[o[1:10],2] [1] "Zyxin" [2] "FAH Fumarylacetoacetate" [3] "APLP2 Amyloid beta (A4) precursor-like protein 2" [4] "LYN V-yes-1 Yamaguchi sarcoma viral related oncogene homolog" [5] "CST3 Cystatin C (amyloid angiopathy and cerebral hemorrhage)"232 APPENDIX A. ANSWERS TO EXERCISES [6] "X-LINKED HELICASE II" [7] "RB1 Retinoblastoma 1 (including osteosarcoma)" [8] "TOP2B Topoisomerase (DNA) II beta (180kD)" [9] "TCRA T cell receptor alpha-chain" [10] "T-COMPLEX PROTEIN 1, GAMMA SUBUNIT" (b) > o <- order(pw,decreasing=FALSE) > golub.gnames[o[1:10],2] [1] "FAH Fumarylacetoacetate" [2] "Zyxin" [3] "CST3 Cystatin C (amyloid angiopathy and cerebral hemorrhage)" [4] "ELA2 Elastatse 2, neutrophil" [5] "TCF3 Transcription factor 3 (E2A immunoglobulin enhancer bindin[6] "Macmarcks" [7] "LYN V-yes-1 Yamaguchi sarcoma viral related oncogene homolog" [8] "CD33 CD33 antigen (differentiation antigen)" [9] "VIL2 Villin 2 (ezrin)" [10] "APLP2 Amyloid beta (A4) precursor-like protein 2" 11. Biological hypotheses. (a) n = 1000, p = 0:05 so np = 50 (b) pbinom(9,1000,.05)=5:24 ¢ 10¡13. (c) sum(dbinom(6:1000,1000,.05))=1. (d) sum(dbinom(2:8,1000,.05))= 8:8 ¢ 10¡14. 12. Programming some tests. (a) data(golub,package="multtest") gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) x <- golub[1042,gol.fac=="ALL"] n <- length(x) y <- golub[1042,gol.fac=="AML"] m <- length(y) t <- (mean(x)-mean(y))/sqrt(var(x)/n + var(y)/m) v <- (var(x)/n + var(y)/m)^2/( (var(x)/n)^2/(n-1) + (var(y)/m)^2/(m-12*pt(-abs(t),v) mean(x) - mean(y) + qt(0.025,v)* sqrt(var(x)/n + var(y)/m) mean(x) - mean(y) + qt(0.975,v)* sqrt(var(x)/n + var(y)/m)233 (b) z <- golub[1042,] > sum(rank(z)[1:27]) - 0.5*27*(27+1) [1] 284 (c) x <- golub[1042,gol.fac=="ALL"] y <- golub[1042,gol.fac=="AML"] w <- 0 for (i in 1:27) w <- w + sum(x[i]>y) > w [1] 284 Answers to exercises of Chapter 5 Linear Models 1. Analysis of gene expressions of B-cell ALL patients. library(ALL); data(ALL) ALLB <- ALL[,ALL$BT %in% c("B","B1","B2","B3","B4")] > table(ALLB$BT) B B1 B2 B3 B4 T T1 T2 T3 T4 5 19 36 23 12 0 0 0 0 0 psw <- apply(exprs(ALLB), 1, function(x) shapiro.test(residuals(lm(x ~ ALLB$BT)))library(lmtest) pbp <-apply(exprs(ALLB), 1, function(x) as.numeric(bptest(lm(x ~ ALLB$BT),studentize = FALSE)$p.value)) > sum(psw > 0.05) [1] 6847 > sum(pbp > 0.05) [1] 10057 > sum(psw > 0.05 & pbp > 0.05) [1] 6262 2. Further analysis of gene expressions of B-cell ALL patients. > panova <- apply(exprs(ALLB), 1, function(x) anova(lm(x ~ ALLB$BT))$Pr[1]) > featureNames(ALLB)[panova<0.000001] [1] "1125_s_at" "1126_s_at" "1134_at" "1389_at" "1500_at"234 APPENDIX A. ANSWERS TO EXERCISES [6] "1866_g_at" "1914_at" "205_g_at" "31472_s_at" "31615_i_at" [11] "31616_r_at" "33358_at" "35614_at" "35991_at" "36873_at" [16] "37809_at" "37902_at" "38032_at" "38555_at" "39716_at" [21] "40155_at" "40268_at" "40493_at" "40661_at" "40763_at" [26] "41071_at" "41139_at" "41448_at" "873_at" > pkw <- apply(exprs(ALLB), 1, function(x) kruskal.test(x ~ ALLB$BT)$p.va> featureNames(ALLB)[pkw<0.000001] [1] "1389_at" "1866_g_at" "38555_at" "40155_at" "40268_at" > panovasmall <- panova < 0.001 > pkwsmall <- pkw < 0.001 > table(panovasmall,pkwsmall) pkwsmall panovasmall FALSE TRUE FALSE 12172 38 TRUE 124 291 There are 124 signi¯cant gene expressions from ANOVA which are not signi¯cant on Kruskal-Wallis. There are only 38 signi¯cant gene ex- pressions from Kruskal-Wallis which are non-signi¯cant according to ANOVA. The tests agree on the majority of gene expressions. 3. Finding the ten best best genes among gene expressions of B-cell ALL patients. > sort(panova)[1:10] 1914_at 1389_at 38555_at 33358_at 40268_at 39711.466523e-14 5.891702e-14 4.873245e-10 1.117406e-09 1.145502e-09 4.74861540763_at 37809_at 36873_at 1866_g_at 5.256410e-09 2.155457e-08 2.402379e-08 3.997065e-08 > sort(pkw)[1:10] 1389_at 40268_at 38555_at 1866_g_at 40155_at 1912.348192e-09 7.764046e-08 1.123068e-07 2.335279e-07 6.595926e-07 1.0745251125_s_at 40662_g_at 38032_at 40661_at 1.346907e-06 1.384281e-06 1.475170e-06 1.719456e-06 npanova <- names(sort(panova)[1:10]) npkw <- names(sort(pkw)[1:10])235 > intersect(npanova,npkw) [1] "1914_at" "1389_at" "38555_at" "40268_at" "1866_g_at" 4. A simulation study for ANOVA. > x <- matrix(rnorm(90000),nrow = 10000, ncol = 9) > a <- gl(3,3) > panova <- apply(x, 1, function(x) anova(lm(x ~ a))$Pr[1]) > sum(panova<0.05) [1] 514 The number of false positives is 514. The expected number is ® ¢ n = 0:05 ¢ 10; 000 = 500, which is quite close to the observed. A matrix with di®erences between three groups of gene expression val- ues. sigma <- 1; n <- 10000 data <- cbind(matrix(rnorm(n*3,0,sigma),ncol=3), matrix(rnorm(n*3,1,sigma), ncol = 3),matrix(rnorm(n*3,2,sigma), ncol = 3)) a <- gl(3,3) panova <- apply(data, 1, function(x) anova(lm(x ~ a))$Pr[1]) > sum(panova<0.05) [1] 3757 > pkw <- apply(data, 1, function(x) kruskal.test(x ~ a)$p.value) > sum(pkw<0.05) [1] 1143 Thus the number of true positives from ANOVA is 3757 and the num- ber of false negatives is 6243. For the Kruskal-Wallis test there are 1143 true positives and 8857 false negatives. This can be impoved by increasing the number of gene expressions per group. Answers to exercises of Chapter 6: Micro Array Analysis. 1. Gene ¯ltering on normality per group of B-cell ALL patients.236 APPENDIX A. ANSWERS TO EXERCISES library("genefilter") data(ALL, package = "ALL") ALLB <- ALL[,ALL$BT %in% c("B1","B2","B3","B4")] f1 <- function(x) (shapiro.test(x)$p.value > 0.05) sel1 <- genefilter(exprs(ALL[,ALLB$BT=="B1"]), filterfun(f1)) sel2 <- genefilter(exprs(ALL[,ALLB$BT=="B2"]), filterfun(f1)) sel3 <- genefilter(exprs(ALL[,ALLB$BT=="B3"]), filterfun(f1)) sel4 <- genefilter(exprs(ALL[,ALLB$BT=="B4"]), filterfun(f1)) selected <- sel1 & sel2 & sel3 & sel4 library(limma) x <- matrix(as.integer(c(sel2,sel3,sel4)),ncol = 3,byrow=FALSE) colnames(x) <- c("sel2","sel3","sel4") vc <- vennCounts(x, include="both") vennDiagram(vc) 137 pass filter 2 but not the other 510 pass filter 2 and 3 but not 4 1019 pas filter 2 and 4 but not 3 5598 pass filter 2, 3 and 4. etc. 2. Analysis of gene expressions of B-cell ALL patients using Limma. library("ALL"); library("limma");library("annaffy");library(hgu95av2.db) data(ALL) ALLB <- ALL[,ALL$BT %in% c("B1","B2","B3","B4")] design.ma <- model.matrix(~0 + factor(ALLB$BT)) colnames(design.ma) <- c("B1","B2","B3","B4") cont.ma <- makeContrasts(B2-B1,B3-B2,B4-B3,levels=factor(ALLB$BT)) fit <- lmFit(ALLB, design.ma) fit1 <- contrasts.fit(fit, cont.ma) fit1 <- eBayes(fit1) topTable(fit1, coef=2,5,adjust.method="fdr") tab <- topTable(fit1, coef=2, number=20, adjust.method="fdr") anntable <- aafTableAnn(as.character(tab$ID), "hgu95av2", aaf.handler()) saveHTML(anntable, "ALLB1234.html", title = "B-cell ALL of stage 1,2,3,4"3. Finding a row number: grep("1389_at",row.names(exprs(ALL))). 4. Remission (genezing) from acute lymphocytic leukemia (ALL).237 library(ALL); data(ALL) table(pData(ALL)$remission) remis <- which(pData(ALL)$remission %in% c("CR","REF")) ALLrem <- ALL[,remis] remfac <-factor(pData(ALLrem)$remission) pano <- apply(exprs(ALLrem),1,function(x) t.test(x ~ remfac)$p.value) sum(pano<0.001) > sum(pano<0.001) [1] 45 library(hgu95av2.db) names <- featureNames(ALLrem)[pano<.001] ALLremsel<- ALLrem[names,] symb <- mget(names, env = hgu95av2SYMBOL) genenames <- mget(names,hgu95av2GENENAME) listofgenenames <- as.list(hgu95av2GENENAME) unlistednames <- unlist(listofgenenames[names],use.names=F) > grep("p53",unlistednames) [1] 12 21 > length(unique(unlistednames)) [1] 36 5. Remission achieved. library(ALL); data(ALL) ALLCRREF <- ALL[,which(ALL$CR %in% c("CR","REF"))] pano <- apply(exprs(ALLCRREF),1,function(x) t.test(x ~ ALLCRREF$CR)$p.value) > sum(pano<0.0001) [1] 11 > featureNames(ALLCRREF)[pano<.0001] [1] "1472_g_at" "1473_s_at" "1475_s_at" "1863_s_at" "34098_f_at" "36574_at" library("hgu95av2.db") affynames <- featureNames(ALLCRREF)[pano<.0001] genenames <- mget(affynames, env = hgu95av2GENENAME)238 APPENDIX A. ANSWERS TO EXERCISES > grep("oncogene",genenames) [1] 1 2 3 affytot <- unique(featureNames(ALLCRREF)) genenamestot <- mget(affytot, env = hgu95av2GENENAME) > length(grep("oncogene",genenamestot)) [1] 239 > length(genenamestot) [1] 12625 > dat <- matrix(c(12625,239,11,3),2,byrow=TRUE) > fisher.test(dat) Fisher's Exact Test for Count Data data: dat p-value = 0.002047 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 2.562237 54.915642 sample estimates: odds ratio 14.39959 6. Gene ¯ltering of ALL data. library("ALL") data("ALL") table(ALL$BT) ALLT23 <- ALL[,which(ALL$BT %in% c("T2","T3"))] library(genefilter) f1 <- function(x) (shapiro.test(x)$p.value > 0.05) f2 <- function(x) (t.test(x ~ ALLT23$BT)$p.value < 0.05) sel1 <- genefilter(exprs(ALLT23[,ALLT23$BT=="T2"]), filterfun(f1)) sel2 <- genefilter(exprs(ALLT23[,ALLT23$BT=="T3"]), filterfun(f1)) sel3 <- genefilter(exprs(ALLT23), filterfun(f2)) > sum(sel1 & sel2 & sel3)239 [1] 905 > sum(sel1 & sel2) [1] 9388 > sum(sel3) [1] 1204 7. Stages of B-cell ALL in the ALL data. library("ALL") library("limma"); allB <- ALL[,which(ALL$BT %in% c("B1","B2","B3","B4"))] facB123 <- factor(allB$BT) cont.ma <- makeContrasts(B2-B1,B3-B2,B4-B3, levels=facB123) design.ma <- model.matrix(~ 0 + facB123) colnames(design.ma) <- c("B1","B2","B3","B4") fit <- lmFit(allB, design.ma) fit1 <- contrasts.fit(fit, cont.ma) fit1 <- eBayes(fit1) > topTable(fit1, coef=2,5,adjust.method="BH") ID logFC AveExpr t P.Value adj.P.Val B 6048 35991_at 0.5964481 4.144598 6.624128 2.578836e-09 0.0000325578 10.842989 3909 33873_at 0.5707770 7.217570 6.083524 2.891823e-08 0.0001825464 8.625253 5668 35614_at 1.7248509 5.663477 5.961231 4.946078e-08 0.0002081474 8.132884 6776 36711_at -2.3664712 7.576108 -5.759565 1.187487e-07 0.0003054110 7.329631 7978 37902_at 0.8470235 4.258491 5.742783 1.276579e-07 0.0003054110 7.263298 > sum(fit1$p.value<0.05) [1] 4328 8. Analysis of public micro array data. library(GEOquery); library(limma); library(hgu95av2.db);library(annaffy) gds486 <- getGEO("GDS486"); eset486 <- GDS2eSet(gds486,do.log2=T) nrmissing <- apply(exprs(eset486), 1, function(x) sum(is.na(x)) ) eset486sel <- eset486[nrmissing<1,] pval486sel <- apply(exprs(eset486sel), 1, function(x) t.test(x ~ eset486sel$cellpval486 <- nrmissing pval486[pval486==0]<-pval486sel240 APPENDIX A. ANSWERS TO EXERCISES pval486[pval486>1]<-1 gds711 <- getGEO("GDS711"); eset711 <- GDS2eSet(gds711,do.log2=T) nrmissing <- apply(exprs(eset711), 1, function(x) sum(is.na(x)) ) eset711sel <- eset711[nrmissing<1,] panova711sel <- apply(exprs(eset711sel), 1, function(x) anova(lm(x ~ esetpval711sel <- panova711sel pval711 <- nrmissing pval711[pval711==0]<-pval711sel pval711[pval711>1]<-1 gds2126 <- getGEO("GDS2126"); eset2126 <- GDS2eSet(gds2126,do.log2=T) nrmissing <- apply(exprs(eset2126), 1, function(x) sum(is.na(x)) ) eset2126sel <- eset2126[nrmissing<1,] pval2126sel <- apply(exprs(eset2126sel), 1, function(x) anova(lm(x ~ esetpval2126 <- nrmissing pval2126[pval2126==0]<-pval2126sel pval2126[pval2126>1]<-1 sumpval <- pval486 + pval711 + pval2126 o <- order(sumpval,decreasing=FALSE) genenames <- names(sumpval[o[1:20]]) symb <- "aap" for (i in 1:20) symb[i] <- get(genenames[i], env = hgu95av2SYMBOL) > symb [1] "GADD45A" "DUSP4" "OAS1" "STAT1" "STAT1" "AKR1C3" "PSMB9" [16] "TKT" "NFKB1" "SLC7A5" "CXCL2" "DLG5" library("KEGG");library("GO");library("annaffy") atab <- aafTableAnn(genenames, "hgu95av2", aaf.handler() ) saveHTML(atab, file="ThreeExperiments.html") # p53 plays a role. 9. Analysis of genes from a GO search. library(ALL) data(ALL,package="ALL")241 ALLP <- ALL[,ALL$mol.biol %in% c("ALL1/AF4","BCR/ABL","NEG")] neg <- which(ALLP$mol.biol=="NEG") aal1 <- which(ALLP$mol.biol=="ALL1/AF4") bcr <- which(ALLP$mol.biol=="BCR/ABL") orderpat <- c(neg,aal1,bcr) ALLP <- ALL[,ALL$mol.biol %in% c("ALL1/AF4","BCR/ABL","NEG")] ALLPo <- ALLP[,c(neg,aal1,bcr)] facnr <- c(rep(1,74),rep(2,10),rep(3,37)) nab.fac <- factor(facnr,levels=1:3, labels= c("NEG","ALL1/AF4","BCR/ABL")) panova <- apply(exprs(ALLPo), 1, function(x) anova(lm(x ~ nab.fac))$Pr[1]) library("GO"); library("annotate"); library("hgu95av2") GOTerm2Tag <- function(term) { GTL <- eapply(GOTERM, function(x) {grep(term, x@Term, value=TRUE)}) Gl <- sapply(GTL, length) names(GTL[Gl>0]) }> GOTerm2Tag("protein-tyrosine kinase") [1] "GO:0004713" probes <- hgu95av2GO2ALLPROBES$"GO:0004713" > sum(panova[probes]<0.05) [1] 86 > sum(panova[probes]<1) [1] 320 > sum(panova<0.05) [1] 2581 > sum(panova<1) [1] 12625 > fisher.test(matrix(c(12625, 2581,320,86),2,byrow=TRUE)) Fisher's Exact Test for Count Data242 APPENDIX A. ANSWERS TO EXERCISES data: matrix(c(12625, 2581, 320, 86), 2, byrow = TRUE) p-value = 0.03222 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 1.019848 1.679625 sample estimates: odds ratio 1.314569 the odds ratio differs significantly from zero; there are more significanAnswers to exercises of Chapter 7: Cluster Analysis and Trees. 1. Cluster analysis on the "Zyxin" expression values of the Golub et al. (1999) data. data(golub, package="multtest") data <- data.frame(golub[2124,]) gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) stripchart(golub[2124,]~gol.fac, pch=as.numeric(gol.fac)) plot(hclust(dist(clusdata,method="euclidian"),method="single")) initial <- as.matrix(tapply(golub[2124,],gol.fac,mean), nrow = 2, ncol=1, cl<- kmeans(data, initial, nstart = 10) table(cl$cluster,gol.fac) n <- length(data); nboot<-1000 boot.cl <- matrix(0,nrow=nboot,ncol = 2) for (i in 1:nboot){ dat.star <- data[sample(1:n,replace=TRUE)] cl <- kmeans(dat.star, initial, nstart = 10) boot.cl[i,] <- c(cl$centers[1,],cl$centers[2,]) }> quantile(boot.cl[,1],c(0.025,0.975)) 2.5% 97.5% -1.07569310 -0.03344292243 > quantile(boot.cl[,2],c(0.025,0.975)) 2.5% 97.5% 0.731493 1.784468 2. Close to CCND3 Cyclin D3. library("genefilter"); data(golub, package = "multtest") closeg <- genefinder(golub, 1042, 10, method = "euc", scale = "none") golub.gnames[closeg[[1]][[1]],2] boxplot(golub[394,] ~gol.fac) 3. MCM3. data(golub, package = "multtest") x <- golub[2289,]; y <- golub[2430,] plot(x,y) which.min(y) # the plot suggests the smallest y as the outlier > cor.test(x[-21],y[-21]) Pearson's product-moment correlation data: x[-21] and y[-21] t = 10.6949, df = 35, p-value = 1.42e-12 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.7690824 0.9341905 # much smaller sample estimates: cor 0.875043 # much larger than 0.6376217 nboot <- 1000; boot.cor <- matrix(0,nrow=nboot,ncol = 1) data <- matrix(c(x[-21],y[-21]),ncol=2,byrow=FALSE) for (i in 1:nboot){ dat.star <- data[sample(1:nrow(data),replace=TRUE),] boot.cor[i,] <- cor(dat.star)[2,1]} > mean(boot.cor) [1] 0.8725835 # very similar to cor.test > quantile(boot.cor[,1],c(0.025,0.975)) 2.5% 97.5% 0.7755743 0.9324625 # very similar to cor.test244 APPENDIX A. ANSWERS TO EXERCISES 4. Cluster analysis on part of Golub data. library(multtest);data(golub); gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) o1 <- grep("oncogene",golub.gnames[,2]) plot(hclust(dist(golub[o1,],method="euclidian"),method="single")) o2 <- grep("antigene",golub.gnames[,2]) plot(hclust(dist(golub[o2,],method="euclidian"),method="single")) o3 <- grep("receptor",golub.gnames[,2]) plot(hclust(dist(golub[o3,],method="euclidian"),method="single")) 5. Principal Components Analysis on part of the ALL data. library(ALL); data(ALL) ALLB <- ALL[,ALL$BT %in% c("B1","B2","B3")] panova <- apply(exprs(ALLB), 1, function(x) anova(lm(x ~ ALLB$BT))$Pr[1]) ALLBsp <- ALLB[panova<0.001,] > dim(exprs(ALLBsp)) [1] 499 78 > min(cor(exprs(ALLBsp))) [1] 0.5805595 > eigen(cor(exprs(ALLBsp)))$values[1:5] [1] 65.2016203 2.9652965 2.4781567 0.7556439 0.6040647 data <- exprs(ALLBsp); p <- ncol(data); n <- nrow(data) ; nboot<-1000 eigenvalues <- array(dim=c(nboot,p)) for (i in 1:nboot){dat.star <- data[sample(1:n,replace=TRUE),] eigenvalues[i,] <- eigen(cor(dat.star))$values} > for (j in 1:p) print(quantile(eigenvalues[,j],c(0.025,0.975))) 2.5% 97.5% 63.43550 66.77785 2.5% 97.5% 2.575413 3.530350 2.5% 97.5% 2.081573 2.889933 2.5% 97.5%245 0.6475809 0.9942871 #Hence, the first three are significant! 2.5% 97.5% 0.5067404 0.7482680 2.5% 97.5% biplot(princomp(data,cor=TRUE),pc.biplot=T,cex=0.5,expand=0.8) 6. Some correlation matrices. eigen(matrix(c(1,-0.8,-0.8,1),nrow=2)) eigen(matrix(c(1,0.8,0.8,0.8,1,0.8,0.8,0.8,1),nrow=3)) eigen(matrix(c(1,-0.5,-0.5,-0.5,1,-0.5,-0.5,-0.5,1),nrow=3)) > 2.6/3 * 100 [1] 86.66667 > eigen(matrix(c(1,0.8,0.8,0.8,1,0.8,0.8,0.8,1),nrow=3))$vectors [,1] [,2] [,3] [1,] -0.5773503 0.8164966 0.0000000 [2,] -0.5773503 -0.4082483 -0.7071068 [3,] -0.5773503 -0.4082483 0.7071068 Answers to exercises of Chapter 8: Classi¯cation Methods. 1. Classi¯cation tree of Golub data. Use recursive partitioning in rpart library(multtest);data(golub); gol.fac <- factor(golub.cl,levels=0:1, labels= c("ALL","AML")) maxgol <- apply(golub[,gol.fac=="ALL"], 1, function(x) max(x)) mingol <- apply(golub[,gol.fac=="AML"], 1, function(x) min(x)) sum(maxgol < mingol) > which.min(maxgol - mingol) [1] 2124 > golub.gnames[2124,] [1] "4847" "Zyxin" "X95735_at" > boxplot(golub[2124,] ~gol.fac) gol.rp <- rpart(gol.fac ~ golub[2124,], method="class", cp=0.001) plot(gol.rp, branch=0,margin=0.1); text(gol.rp, digits=3, use.n=TRUE)246 APPENDIX A. ANSWERS TO EXERCISES grep("Gdf5",golub.gnames[,2]) > grep("Gdf5",golub.gnames[,2]) [1] 2058 gol.rp <- rpart(gol.fac ~ golub[2058,], method="class", cp=0.001) plot(gol.rp, branch=0,margin=0.1); text(gol.rp, digits=3, use.n=TRUE) gol.rp <- rpart(gol.fac ~., data.frame(t(golub)), method="class", cp=0.plot(gol.rp, branch=0,margin=0.1); text(gol.rp, digits=3, use.n=TRUE) 2. Sensitivity versus speci¯city. (a) library(multtest);library(ROCR);data(golub) golub.clchanged <- -golub.cl +1 pred <- prediction(golub[1042,], golub.clchanged) perf <- performance(pred, "sens", "spec") plot(perf) (b) The function is essentially the same. (c) Use auc as before. 3. Comparing Classi¯cation Methods. library(rpart) predictors <- matrix(rnorm(100*4,0,1),100,4) colnames(predictors) <- letters[1:4] groups <- gl(2,50) simdata <- data.frame(groups,predictors) rp<-rpart(groups ~ a + b + c + d,method="class",data=simdata) predicted <- predict(rp,type="class") table(predicted,groups) plot(rp, branch=0,margin=0.1); text(rp, digits=3, use.n=TRUE) > table(predicted,groups) groups predicted 1 2 1 41 12247 2 9 38 library(e1071) svmest <- svm(predictors, groups, data=df, type = "C-classification", kernel = "lsvmpred <- predict(svmest, predictors, probability=TRUE) > table(svmpred, groups) groups svmpred 1 2 1 31 25 2 19 25 library(nnet) nnest <- nnet(groups ~ ., data = simdata, size = 5,maxit = 500, decay = 0.01, Maxpred <- predict(nnest, type = "class") > table(pred, groups) # prints confusion ma groups pred 1 2 1 45 10 2 5 40 The misclassi¯cation rate of rpart, svm, and nnet is, respectively, 21/100, 44/100, and 15/100. If we increase the number of predictors, then the misclassi¯cation rate decreases. 4. Prediction of achieved remission. library(ALL); library(hgu95av2.db); library(rpart); data(ALL) ALLrem <- ALL[,which(pData(ALL)$remission %in% c("CR","REF"))] remfac <-factor(pData(ALLrem)$remission) pano <- apply(exprs(ALLrem),1,function(x) t.test(x ~ remfac)$p.value) names <- featureNames(ALLrem)[pano<.001] ALLremsel<- ALLrem[names,] data <- data.frame(t(exprs(ALLremsel))) all.rp <- rpart(remfac ~., data, method="class", cp=0.001) plot(all.rp, branch=0,margin=0.1); text(all.rp, digits=3, use.n=TRUE) rpart.pred <- predict(all.rp, type="class") > table(rpart.pred,remfac) remfac248 APPENDIX A. ANSWERS TO EXERCISES rpart.pred CR REF CR 93 1 REF 6 14 > 7/(93+1+6+14) [1] 0.06140351 > mget(c("1840_g_at","36769_at","1472_g_at","854_at"), env = hgu95av2GENE$`1840_g_at` [1] NA $`36769_at` [1] "retinoblastoma binding protein 5" $`1472_g_at` [1] "v-myb myeloblastosis viral oncogene homolog (avian)" $`854_at` [1] "B lymphoid tyrosine kinase" 5. Gene selection by area under the curve. library(ROCR); data(golub, package = "multtest") gol.true <- factor(golub.cl,levels=0:1,labels= c("TRUE","FALSE")) auc.values <- apply(golub,1, function(x) performance(prediction(x, gol.true),"auc")@y.values[[1]]) o <- order(auc.values,decreasing=TRUE) golub.gnames[o[1:25],2] 6. Classi¯cation Tree for Ecoli. ecoli <- read.table("http://www.grappa.univ-lille3.fr/~torre/Recherche/Dadownloads/ecoli/ecoli.data",sep=",",header = TRUE) colnames(ecoli) <- c("SequenceName","mcg","gvh","lip","chg","aac","alm1",ecolisel<- ecoli[which(ecoli$ecclass %in% c("cp","im","pp")),] ecolisel$ecclass <- factor(ecolisel$ecclass, levels=c("cp","im","pp")) library(rpart) rpfit <- rpart(ecolisel$ecclass ~ mcg + gvh + lip + aac + alm1 + alm2,datplot(rpfit, branch=1,margin=0.1); text(rpfit, digits=3, use.n=TRUE) title(main = "rpartfit ecoli classes cp im and pp")249 predictedclass <- predict(rpfit, type="class") table(predictedclass,ecolisel$ecclass) #predictors are alm1, gvh and im > (1+2+7+4)/length(ecolisel$ecclass) [1] 0.05166052 Answers to exercises of Chapter 9: Analyzing Sequences 1. Writing to a FASTA ¯le. choosebank("genbank"); library(seqinr) query("ccnd3hs","sp=homo sapiens AND k=ccnd3@") ccnd3 <- sapply(ccnd3hs$req, getSequence) x1 <- DNAStringSet(c2s(ccnd3[[1]])) write.XStringSet(x1, file="ccnd3.fa", format="fasta", width=80) ccnd3c2sn <- sapply(ccnd3, c2s) x1 <- DNAStringSet(ccnd3c2sn) write.XStringSet(x1, file="ccnd3n.fa", format="fasta", width=80) An alternative would be to use the write.dna function of the ape package. 2. Dotplot of sequences. seq1 <- sample(c("A","G","C","T"),100,rep=TRUE,prob=c(0.1,0.4,0.4,0.1)) seq2 <- sample(c("A","G","C","T"),100,rep=TRUE,prob=c(0.1,0.4,0.4,0.1)) par(mfrow=c(1,2)) dotPlot(seq1, seq2, main = "Dot plot of different random sequences\nwsize = 1, wsdotPlot(seq1, seq1, main = "Dot plot of equal random sequnces\nwsize = 1, wstep = par(mfrow=c(1,1)) par(mfrow=c(1,2)) dotPlot(seq1, seq2, main = "Dot plot of different random sequences\nwsize = 3, wsdotPlot(seq1, seq1, main = "Dot plot of equal random sequnces\nwsize = 3, wstep = par(mfrow=c(1,1)) par(mfrow=c(1,2)) dotPlot(seq1, seq1, main = "Dot plot of different random sequences\nwsize = 3, ws250 APPENDIX A. ANSWERS TO EXERCISES dotPlot(seq1, seq1[100:1], main = "Dot plot of equal random sequnces\nwsipar(mfrow=c(1,1)) x <- c("RPLWVAPDGHIFLEAFSPVYK") y <- c("RPLWVAPDGHIFLEAFSPVYK") z <- c("PLWISPSDGRIILESFSPLAE") choosebank("genbank"); library(seqinr) query("ccnd3hs","sp=homo sapiens AND k=ccnd3@") ccnd3 <- sapply(ccnd3hs$req, getSequence) sapply(ccnd3hs$req, getName) ccnd3prot <- sapply(ccnd3hs$req, getTrans) dotPlot(ccnd3prot[[1]], s2c("EEEVFPLAMN"), main = "Dot plot of two proteidotPlot(ccnd3prot[[7]], ccnd3prot[[8]], main = "Dot plot of two protein\ndotPlot(s2c(x), s2c(z), main = "Dot plot of two protein\nwsize = 1, wstep 3. Local alignment. library(seqinr);library(Biostrings);data(BLOSUM50) x <- s2c("HEAGAWGHEE"); y <- s2c("PAWHEAE") s <- BLOSUM50[y,x]; d <- 8 F <- matrix(data=NA,nrow=(length(y)+1),ncol=(length(x)+1)) F[1,] <- 0 ; F[,1] <- 0 rownames(F) <- c("",y); colnames(F) <- c("",x) for (i in 2:(nrow(F))) for (j in 2:(ncol(F))) {F[i,j] <- max(c(0,F[i-1,j-1]+s[i-1,j-1],F[i-1,j]-d,F[i,j-1]-d))} > max(F) [1] 28 4. Probability of more extreme alignment score. library(seqinr);library(Biostrings);data(BLOSUM50) randallscore <- c(1,1) for (i in 1:1000) { x <- c2s(sample(rownames(BLOSUM50),7, replace=TRUE)) y <- c2s(sample(rownames(BLOSUM50),10, replace=TRUE)) randallscore[i] <- pairwiseAlignment(AAString(x), AAString(y), substitut251 gapOpening = 0, gapExtension = -8, scoreOnly = TRUE) }> sum(randallscore>1)/1000 [1] 0.003 > plot(density(randallscore)) 5. Prochlorococcus marinus. library(seqinr) choosebank("genbank") query("ccmp","AC=AE017126 OR AC=BX548174 OR AC=BX548175") ccmpseq <- sapply(ccmp$req,getSequence) gc <- sapply(ccmpseq, GC) > wilcox.test(gc[1:2],gc[3:9]) Wilcoxon rank sum test data: gc[1:2] and gc[3:9] W = 0, p-value = 0.05556 alternative hypothesis: true location shift is not equal to 0 > t.test(gc[1:2],gc[3:9]) Welch Two Sample t-test data: gc[1:2] and gc[3:9] t = -5.8793, df = 1.138, p-value = 0.08649 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.4507417 0.1079848 sample estimates: mean of x mean of y 0.3362065 0.5075849 gc in the left group is lower, the tests are not significant. 6. Sequence equality. \begin{verbatim} library(seqinr) choosebank("genbank") query("ccnd3hs","sp=homo sapiens AND k=ccnd3@")252 APPENDIX A. ANSWERS TO EXERCISES sapply(ccnd3hs$req,getLength) > ccnd3prot <- sapply(ccnd3hs$req, getTrans) > table(ccnd3prot[[1]]) * A C D E F G H I K L M N P Q R S T V W Y 1 31 12 12 21 6 14 7 10 10 41 9 1 17 16 22 19 18 15 3 8 > table(ccnd3prot[[2]]) * A C D E F G H I K L M N P Q R S T V W Y 1 30 12 12 21 6 14 7 10 10 41 9 1 17 16 22 20 18 15 3 8 # Hence, there is only one difference! > which(!ccnd3prot[[1]]==ccnd3prot[[2]]) [1] 259 7. Conserved region. ID XRODRMPGMNTB; BLOCK AC PR00851A; distance from previous block=(52,131) DE Xeroderma pigmentosum group B protein signature BL adapted; width=21; seqs=8; 99.5%=985; strength=1287 XPB_HUMAN|P19447 ( 74) RPLWVAPDGHIFLEAFSPVYK 54 XPB_MOUSE|P49135 ( 74) RPLWVAPDGHIFLEAFSPVYK 54 P91579 ( 80) RPLYLAPDGHIFLESFSPVYK 67 XPB_DROME|Q02870 ( 84) RPLWVAPNGHVFLESFSPVYK 79 RA25_YEAST|Q00578 ( 131) PLWISPSDGRIILESFSPLAE 100 Q38861 ( 52) RPLWACADGRIFLETFSPLYK 71 O13768 ( 90) PLWINPIDGRIILEAFSPLAE 100 O00835 ( 79) RPIWVCPDGHIFLETFSAIYK 86 library(Biostrings);data(BLOSUM50) x <- c("RPLWVAPDGHIFLEAFSPVYK") y <- c("RPLWVAPDGHIFLEAFSPVYK") z <- c("PLWISPSDGRIILESFSPLAE") x == y pairwiseAlignment(AAString(x), AAString(z), substitutionMatrix = "BLOSUM5> pairwiseAlignment(AAString(x), AAString(y), substitutionMatrix = "BLOSUGlobal Pairwise Alignment253 1: RPLWVAPDGHIFLEAFSPVYK 2: RPLWVAPDGHIFLEAFSPVYK Score: 154 >> z <- c("PLWISPSDGRIILESFSPLAE") >> x == y [1] TRUE > pairwiseAlignment(AAString(x), AAString(z), substitutionMatrix = "BLOSUM50",gapGlobal Pairwise Alignment 1: RPLWVAP-DGHIFLEAFSPVYK 2: -PLWISPSDGRIILESFSPLAE Score: 85 8. Plot of CG proportion from Celegans. (a) Produce a plot of the CG proportion of the chromosome I of Cel- egans (Celegans.UCSC.ce2) along a window of 100 nucleotides. Take the ¯rst 10,000 nucleotides. library(seqinr) source("http://bioconductor.org/biocLite.R") biocLite("BSgenome.Celegans.UCSC.ce2") library(BSgenome.Celegans.UCSC.ce2) GCperc <- double() for (i in 1:10000) GCperc[i] <- GC(s2c(as.character(Celegans$chrI[i:(i+100)])plot(GCperc,type="l") (b) A binding sequence of the enzyme EcoRV is the subsequence GATATC. How many exact matches has Chromosome I of Cel- egans. > subseq <- "gatatc" > countPattern(subseq, Celegans$chrI, max.mismatch = 0) [1] 3276 > length(s2c(as.character(Celegans$chrI))) * (1/4)^6 [1] 3681.759 9. Plot of codon usage.254 APPENDIX A. ANSWERS TO EXERCISES data(ec999) ec999.uco <- lapply(ec999, uco, index="eff") df <- as.data.frame(lapply(ec999.uco, as.vector)) row.names(df) <- names(ec999.uco[[1]]) global <- rowSums(df) title <- "Codon usage in 999 E. coli coding sequences" dotchart.uco(global, main = title) choosebank("genbank"); library(seqinr) query("ccndhs","sp=homo sapiens AND k=ccnd@") ccnd <- sapply(ccndhs$req, getSequence) ccnd.uco <- lapply(ccnd3, uco, index="eff") df <- as.data.frame(lapply(ccnd.uco, as.vector)) row.names(df) <- names(ccnd.uco[[1]]) global <- rowSums(df) title <- "Codon usage in ccnd3 homo sapiens coding sequences" dotchart.uco(global, main = title) Answers to exercises of Chapter 10: Markov Models. 1. Visualize by a transition graph the following transition matrices. Con- sult your teacher. 2. Computing probabilities. The answers are provided by the following. > P <- matrix(c(3/4,1/4,1/2,1/2),2,2,byrow=T) > pi0 <- c(1/2,1/2) > pi0 %*% P [,1] [,2] [1,] 0.625 0.375 > P %*% P [,1] [,2] [1,] 0.6875 0.3125 [2,] 0.6250 0.3750 > P [,1] [,2] [1,] 0.75 0.25 [2,] 0.50 0.50255 3. Programming GTR. Use ¼A = 0:15, ¼G = 0:35, ¼C = 0:35, ¼T = 0:15, ® = 4, ¯ = 0:5, ° = 0:4, ± = 0:3, ² = 0:2, and ³ = 4. (a) Program the rate matrix in such a manner that it is simple to adapt for other values of the parameters. library(Matrix) piA <- 0.15; piG <- 0.35; piC <- 0.35; piT <- 0.15 alpha <- 4; beta <- 0.5; gamma <- 0.4; delta <- 0.3 epsilon <- 0.2; zeta <- 4 Q <- matrix(data=NA,4,4)Q[1,2] <- alpha * piG; Q[1,3] <- beta * piC; Q[1,4] <- gamma * piT Q[2,1] <- alpha * piA; Q[2,3] <- delta * piC; Q[2,4] <- epsilon * piT Q[3,1] <- beta * piA; Q[3,2] <- delta * piG; Q[3,4] <- delta* piQ[4,1] <- gamma * piA; Q[4,2] <- epsilon* piG; Q[4,3] <- zeta * piC diag(Q) <- 0 diag(Q) <- -apply(Q,1,sum) Q <- Matrix(Q) > Q 4 x 4 Matrix of class "dgeMatrix" [,1] [,2] [,3] [,4] [1,] -1.635 1.400 0.175 0.060 [2,] 0.600 -0.735 0.105 0.030 [3,] 0.075 0.105 -0.225 0.045 [4,] 0.060 0.070 1.400 -1.530 (b) The transversion rate is larger then the transition rate because the blocks outside the main diagonal have lower values. (c) The probability transition matrix is > P <- as.matrix(expm(Q)) > P [,1] [,2] [,3] [,4] [1,] 0.32199057 0.51569256 0.1392058 0.02311107 [2,] 0.22097363 0.64908639 0.1115233 0.01841667256 APPENDIX A. ANSWERS TO EXERCISES [3,] 0.05203969 0.09913633 0.8263804 0.02244359 [4,] 0.04621015 0.08457814 0.6397090 0.22950271 rownames(P) <- colnames(P) <- StateSpace <- c("a","g","c","t") pi0 <- c(1/4,1/4,1/4,1/4) markov2 <- function(StateSpace,P,n){ seq <- matrix(0,nr=n,nc=1) seq[1] <- sample(StateSpace,1,replace=T,pi0) for(k in 1:(n-1)){ seq[k+1] <- sample(StateSpace,1,replace=T,P[seq[kreturn(seq) } seq <- markov2(StateSpace,P,99) 4. Distance according to JC69. (a) accnr <- paste("AJ5345",26:27,sep="") seqbin <- read.GenBank(accnr, species.names = TRUE, as.character = FADown load the sequences AJ534526 and AJ534527. Hint: Use as.character = TRUE in the read.GenBank function. (b) Two solution of computing the proportion of di®erent nucleotides are dist.dna(seqbin, model = "raw") p <- sum(seq$AJ534526 != seq$AJ534527)/1143 (c) Simply insert the obtained p in the formula d <- -log(1-4*p/3)*3/4.Appendix B References Dalgaard, P. (2002). Introductory statistics with R. New York: Springer. Bain, L.J. & Engelhardt, M. (1992). Introduction to probability and mathe- matical statistics. Paci¯c Grove: Duxbury. Becker, R.A., Chambers, J.M. & Wilks, A.R. (1988). The new S language. New Jersey: Bell Telephone Laboratories. Beran, B. & Srivastava, M.S. (1985). Bootstrap tests and con¯dence regions for functions of a covariance matrix. The Annals of Statistics, 13, 95- 115. Beran, R. & Ducharme, G.R. (1991). Asymptotic theory for bootstrap meth- ods in statistics. Montreal: Centre de recherche math¶ematique. Breiman, L., J. H. Friedman, R. A. Olshen, and C. J. Stone. (1984) Classi- ¯cation and Regression Trees. Monterey: Wadsworth. Breusch, T.S. & Pagan A.R. (1979). 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REFERENCESIndex aggregation, 95 Anderson-Darling test, 64 annotation, 104 background correction, 94 Binomial test, 58 BLOSUM50, 185 bootstrap, 127, 132 box-and-wiskers-plot, 20 calculator, 4 chi-squared distribution, 37 chi-squared test, 59 classi¯cation tree, 150 confusion table, 158 construct a sequence, 4 correlation coe±cient, 130 data matrix, 6 data vector, 5 density, 41 design matrix, 101 dinucleotide, 176 distance, 118 downloading sequences, 174 F-distribution, 40 F-test, 57 Fisher test, 62 frequency table, 17 genBank, 18 gene ¯ltering, 97 gene ontology, 107 GO, 107 gol.fac, 11 Golub et al. (1999) data, 10 grep, 12 help, 3 histogram, 19 homoscedasticity, 85 install R, 1 installing Bioconductor, 2 installing R, 2 interquartile range, 25 k-means cluster analysis, 125 Kruskal-Wallis test, 87 linear model, 74 matrix computations, 8 mean, 24 median, 24 median absolute deviation, 25 misclassi¯cation rate, 158 mismatch, 91 model matrix, 101 Needleman-Wunsch, 184 neural network, 162 normal distribution, 35 263264 INDEX normality of residuals, 85 normality test, 63 normalization, 94 one sample t-test, 51 one sided hypothesis, 48 one-way analysis of variance, 77 packages, 2 perfect match, 91 Phylogenetic tree, 203 predictive power, 147 principal components analysis, 133 Quantile-Quantile plot, 22 quartile, 20 query language, 173 receiver operator curve, 148 rma, 95 running scripts, 13 sample variance, 25 sensitivity, 147 Shapiro-Wilk test, 63 signi¯cance level, 48 single linkage cluster analysis, 121 speci¯city, 147 standard deviation, 25 stripchart, 19 support vector machine, 161 T-distribution, 39 training set, 159 triangle inequality, 118 two sided hypothesis, 48 two-sample t-test, 54 validation set, 159 Wilcoxon rank test, 65 Z-test, 48