CGS 5991 (2 Credits) Bioinformatics Tools
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1 CAP 5991 (3 Credits) Introduction to Bioinformatics CGS 5991 (2 Credits) Bioinformatics Tools Giri Narasimhan 8/26/03 CAP/CGS 5991: Lecture 1 1
2 Course Schedules CAP 5991 (3 credit) will meet every Tue from 11AM to 1:45PM. CGS 5991 (2 credit) will meet every Tue from 11AM to about 1PM. This course is not for CS students. Different exams and evaluation. 8/26/03 CAP/CGS 5991: Lecture 1 2
3 CAP/CGS 5991 Introduction to Bioinformatics Preliminaries Sequence Alignment Multiple Sequence Alignment Phylogenetic Analysis Molecular Structure Analysis Gene Recognition Genomics, Functional Genomics Proteomics Pattern Discovery Techniques Programming Environments: BioPerl Overview of Course Databases and Software Packages Statistics for Bioinformatics Sequencing and Mapping Computational Learning Methods HMM, NN, SOM, SVM, GA Computational Predictive Methods Microarray Data Analysis Digital Image Analysis Protein Structure Analysis: SPDBV Emerging Biotechnologies 8/26/03 CAP/CGS 5991: Lecture 1 3
4 CAP/CGS 5991 Software Packages Databases and Software Packages (GenBank, SWISS-PROT) Programming Environments (BioPerl) Sequence Alignment & Multiple Sequence Alignment (BLAST, CLUSTALW, CLUSTALX) Phylogenetic Analysis (CLUSTALW, Phylip, PAUP, PAML) Learning Methods (HMMPro, GeneCluster, ASOM) Pattern Discovery Techniques (GYM, TEIRESIAS, APRIORI) Molecular Structure Analysis (DALI, RASMOL, SPDBV) Microarray Analysis (CLUSTER, SAM, GeneCluster, TreeView) Statistical Software Packages (SAS, R) 8/26/03 CAP/CGS 5991: Lecture 1 4
5 Evaluation Semester Project (50 %) Homework Assignments (20 %) Exams (20 %) Class Participation (10 %) 8/26/03 CAP/CGS 5991: Lecture 1 5
6 Reading List and Schedule Watch that Course Home Page! 8/26/03 CAP/CGS 5991: Lecture 1 6
7 1. What is Bioinformatics? Introduction Analysis of biological data with computing & statistical tools. 2. The different aspects of Informatics: Data Management (Database Technology, Internet Programming) Analysis/Interpretation of Data (Data Mining, Modeling, Statistical Tools) Development of Algorithms/ Data Structures Visualization and Interface Design (HCI, Graphics) 3. How to assist biological research? propose new models or correlations based on data from experiments verify a proposed model using known data propose new experiments based on model or analysis use predicted information to narrow down search in a biological investigation 8/26/03 CAP/CGS 5991: Lecture 1 7
8 Overall Goals DNA Sequence Gene Protein Structure Gene Networks Function Metabolic Pathways 8/26/03 CAP/CGS 5991: Lecture 1 8
9 General Information Over 20 billion bases in the NCBI database: Human Genome has ~3 billion bp with 30,000+ genes. Viruses have 300bp to 300Kb (1 st one in 1978: Simian virus; 5Kb). 86 complete microbial genomes sequenced. Number of whole genomes sequenced is over 100 (not including over 1000 viruses), including: Caenorhabditis elegans, Arabidopsis thaliana, Drosophila melanogaster, Saccharomyces cerevisiae, Chromosomal maps for many organisms including: Mus musculus, Homo sapiens, Danio rerio, Zea mays, Oryza sativa Swiss-Prot has over protein sequences. 8/26/03 CAP/CGS 5991: Lecture 1 9
10 Genome Sizes Organism Size Date Est. # genes HIV type 1 10 Kb H. influenzae 1.8 Mb ,740 E. coli 4.7 Mb ,000 S. cerevisiae 12.1 Mb ,034 C. elegans 97 Mb ,099 A. thaliana 100 Mb ,000 D. melanogaster 180 Mb ,061 M. musculus 3 Gb 2002 ~30,000 H. sapiens 3 Gb ,000+ 8/26/03 CAP/CGS 5991: Lecture 1 10
11 Caenorhabditis Elegans Entire genome st animal; 26 th organism 8 year effort 97 million bases 19,099 genes 402 gene clusters 12,000 genes with known function thousands of mutants 7000 families of repeats Multicellular organism Nematode (phylum) Easy to experiment with Easily observable 959 cells 302 nerve cells 36% of proteins common w/ human 8/26/03 CAP/CGS 5991: Lecture 1 11
12 Homo sapiens 15 year effort, 3 billion bases, 100,000 gaps Variable density of: Genes, SNPs, CpG islands, recombination rates ~1.1% of the genome codes for proteins ~ % of the genome consists of repeat sequences ~10 % of the genome consists of repeats called ALUs ~5 % of the genome consists of long repeats (>1 Kb) ~ 50 transposon-derived genes 223 genes common with bacteria that are missing from worm, fly or yeast. 8/26/03 CAP/CGS 5991: Lecture 1 12
13 (Approximate) String Matching Input: Text T, Pattern P Question(s): Does P occur in T? Find one occurrence of P in T. Find all occurrences of P in T. Count # of occurrences of P in T. Find longest substring of P in T. Find closest substring of P in T. Locate direct repeats of P in T. Many More variants Applications: Is P already in the database T? Locate P in T. Can P be used as a primer for T? Is P homologous to anything in T? Has P been contaminated by T? Is prefix(p) = suffix(t)? Locate tandem repeats of P in T. 8/26/03 CAP/CGS 5991: Lecture 1 13
14 The Suffix Tree Data Structure Borrelia burgdorferi: 1 million bases Shotgun Sequencing: 4612 fragments 2 million bases long totally Using suffix trees - 15 min for Fragment Assembly Using Dynamic Programming - 10 days 8/26/03 CAP/CGS 5991: Lecture 1 14
15 Repeats in DNA Sequences Genomic Imprinting: Some genes are expressed only when inherited from one specific parent. 16 such genes are known; 5 inherited from mother; rest from father. These 16 genes have a lot of repeats. Repeats are of size 25 to 120 bp and of total length The repeats are unique to these imprinted regions. They have no obvious homolgy to each other or to other highly repetitive mammalian sequences. Repeats are also known to be responsible for several genetic diseases: Fragile X, Huntington s disease, Kennedy s disease, myotonic dystrophy, ataxia. 8/26/03 CAP/CGS 5991: Lecture 1 15
16 Drosophila Eyeless vs. Human Aniridia 24 IERLPSLEDMAHKGHSGVNQLGGVFVGGRPLPDSTRQKIVELAHSGARPCDISRILQVSN 83 I R P+ M + HSGVNQLGGVFV GRPLPDSTRQKIVELAHSGARPCDISRILQVSN 17 IPRPPARASMQNS-HSGVNQLGGVFVNGRPLPDSTRQKIVELAHSGARPCDISRILQVSN GCVSKILGRYYETGSIRPRAIGGSKPRVATAEVVSKISQYKRECPSIFAWEIRDRLLQEN 143 GCVSKILGRYYETGSIRPRAIGGSKPRVAT EVVSKI+QYKRECPSIFAWEIRDRLL E 76 GCVSKILGRYYETGSIRPRAIGGSKPRVATPEVVSKIAQYKRECPSIFAWEIRDRLLSEG VCTNDNIPSVSSINRVLRNLAAQKEQ 169 VCTNDNIPSVSSINRVLRNLA++K+Q 136 VCTNDNIPSVSSINRVLRNLASEKQQ TEDDQARLILKRKLQRNRTSFTNDQIDSLEKEFERTHYPDVFARERLAGKIGLPEARIQV Q RL LKRKLQRNRTSFT +QI++LEKEFERTHYPDVFARERLA KI LPEARIQV 222 SDEAQMRLQLKRKLQRNRTSFTQEQIEALEKEFERTHYPDVFARERLAAKIDLPEARIQV WFSNRRAKWRREEKLRNQRR 477 WFSNRRAKWRREEKLRNQRR 282 WFSNRRAKWRREEKLRNQRR 301 8/26/03 CAP/CGS 5991: Lecture 1 16
17 Sequence Alignment HBA_HUMAN HBB_HUMAN GSAQVKGHGKKVADALTNAVAHVDDMPNALSALSDLHAHKL G+ +VK+HGKKV A+++++AH+D LS+LH KL GNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKL HBA_HUMAN LGB2_LUPLU GSAQVKGHGKKVADALTNAVAHV---D--DMPNALSALSDLHAHKL H+ KV + +A ++ +L+ L+++H+ K NNPELQAHAGKVFKLVYEAAIQLQVTGVVVTDATLKNLGSVHVSKG HBA_HUMAN F11G11.2 GSAQVKGHGKKVADALTNAVAHVDDMPNALSALSD----LHAHKL GS+ + G + +D L ++ H+ D+ A +AL D ++AH+ GSGYLVGDSLTFVDLL--VAQHTADLLAANAALLDEFPQFKAHQE HBA_HUMAN: Human Alpha Globin HBB_HUMAN: Human Beta Globin F11G11.2 : Leghaemoglobin from yellow lupin Needleman-Wunsch Smith-Waterman 8/26/03 CAP/CGS 5991: Lecture 1 17
18 Sequence Alignment Input: Sequence A, Sequence B, Database D Question(s): AlignA and B. Determine similarity (A, B). AlignA and D. Find sequence in D with maximum similarity to A. Many More variants 8/26/03 CAP/CGS 5991: Lecture 1 18
19 Molecular Biology Background 8/26/03 CAP/CGS 5991: Lecture 1 19
20 The 2 star players Protein DNA 8/26/03 CAP/CGS 5991: Lecture 1 20
21 The Players DNA String with alphabet {A, C, G, T} Nucleotides/Bases RNA String with alphabet {A, C, G, U} Bases Protein String with 20-letter alphabet Amino acids/residues 8/26/03 CAP/CGS 5991: Lecture 1 21
22 Central Dogma DNA acts as a template to replicate itself. DNA is transscribed into RNA. RNA is translated into Protein. DNA RNA Protein 8/26/03 CAP/CGS 5991: Lecture 1 22
23 Chromosomes 8/26/03 CAP/CGS 5991: Lecture 1 23
24 Chromosomes 8/26/03 CAP/CGS 5991: Lecture 1 24
25 DNA Molecule 8/26/03 CAP/CGS 5991: Lecture 1 25
26 DNA Complementary Bases 8/26/03 CAP/CGS 5991: Lecture 1 26
27 Proteins Amino acids 8/26/03 CAP/CGS 5991: Lecture 1 27
28 RNA 8/26/03 CAP/CGS 5991: Lecture 1 28
29 The Genetic Code 8/26/03 CAP/CGS 5991: Lecture 1 29
30 8/26/03 CAP/CGS 5991: Lecture 1 30
31 DNA RNA Protein 8/26/03 CAP/CGS 5991: Lecture 1 31
32 8/26/03 CAP/CGS 5991: Lecture 1 32
33 8/26/03 CAP/CGS 5991: Lecture 1 33
34 DNA Transcription 8/26/03 CAP/CGS 5991: Lecture 1 34
35 Transcription Initiation 8/26/03 CAP/CGS 5991: Lecture 1 35
36 Transcription 8/26/03 CAP/CGS 5991: Lecture 1 36
37 Transcription Steps RNA polymerase needs many transcription factors (TFIIA,TFIIB, etc.) (A) The promoter sequence (TATA box) is located 25 nucleotides away from transcription initiation site. (B) The TATA box is recognized and bound by transcription factor TFIID, which then enables the adjacent binding of TFIIB. DNA is somewhat distorted in the process. (D) The rest of the general transcription factors as well as the RNA polymerase itself assemble at the promoter. What order? (E) TFIIH then uses ATP to phosphorylate RNA polymerase II, changing its conformation so that the polymerase is released from the complex and is able to start transcribing. As shown, the site of phosphorylation is a long polypeptide tail that extends from the polymerase molecule. 8/26/03 CAP/CGS 5991: Lecture 1 37
38 Transcription Factors The general transcription factors have been highly conserved in evolution; some of those from human cells can be replaced in biochemical experiments by the corresponding factors from simple yeasts. 8/26/03 CAP/CGS 5991: Lecture 1 38
39 Protein Synthesis 8/26/03 CAP/CGS 5991: Lecture 1 39
40 Protein Synthesis: Incorporation of amino acid into protein 8/26/03 CAP/CGS 5991: Lecture 1 40
41 8/26/03 CAP/CGS 5991: Lecture 1 41
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