CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 1 -CSE 397-497:Computational Issues inMolecular BiologyLecture 2Spring 2004CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 2 -Important points to remember•Class attendance and participation are big part of grade.• Do assigned readings in advance – be prepared to discuss. (Lecture notes and readings will be posted on Blackboard.)• Each student enrolled must prepare and deliver one lecture.• Use of PowerPoint for slides is encouraged, but not required. (See http://www.openoffice.org for open-source version.)•Final project or paper due at end of course.• Final exam in event seminar format is not successful.CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 3 -Procedure for student lecturesBy 5:00 pm on Thursday, you give me ranked list of your top 3 topics in order of preference:•sequence comparison & alignment (pairwise & multiple),• sequencing and sequence assembly,•physical mapping of DNA,• phylogenetic trees,•genome rearrangements,• RNA and protein structure prediction,• DNA microarrays,• DNA computing.For each topic, there is a folder in Blackboard. When making your choice, look at these papers and also your textbook.order we will cover topics in courseCSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 4 -Procedure for student lecturesBy 5:00 pm on Friday, I will assign your topic and lecture date. (If you know you have a conflict for a specific date, tell me in advance – i.e., when you send me your ranked list of topics.)Two weeks before your lecture, you meet with me to discuss the material you plan to present (15 minutes).One week before your lecture, you show me your near-final lecture and a list of discussion topics ( 2 hours).One day before your lecture, we do a run-through ( 2 hours).CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 5 -Course gradingClass attendence / participation = 100 points.Lecture = 25 points (preparation) + 100 points (delivery).Final project or paper = 100 points.Final exam (if we need it) = 100 points.Scribe (if taking CSE 497)* = 25 points.* Note that CSE 397 and CSE 497 point totals will be different and each will be curved separately.CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 6 -The Central Dogma of Molecular Biology1. DNA copies its information in process involving many enzymes (replication).2. DNA codes for production of mRNA during transcription.3. mRNA migrates from nucleus to cytoplasm.4. mRNA carries coded information to ribosomes which "read" it and use it for protein synthesis (translation).http://allserv.rug.ac.be/~avierstr/principles/centraldogma.htmlCSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 7 -The Central Paradigm of BioinformaticsBy developing techniques for analyzing sequence data and the structures that result, we can attempt to understand the genetic nature of diseases.http://cmgm.stanford.edu/biochem218/CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 8 -“Junk” DNAhttp://www.accessexcellence.org/AB/GG/genes.htmlhttp://www.psrast.org/junkdna.htmAt this point in time, <10% of the DNA in the human genome can be associated with genes.The remainder is known as junk DNA because it has no apparent function.Recall that genes are contiguous stretches along a chromosome.However, recent studies are showing that non-coding DNA may play an important role in regulating gene expression (enhancing or suppressing expression of proximal genes). It's also used in forensic analysis as mutations are more likely in non-coding DNA regions than within genes (why?).CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 9 -Genetic inheritencehttp://www.accessexcellence.org/AB/GG/hapDIP.html1. Cells in mother and father both contain paired sets of chromosomes (diploid).2. Through meiosis, gametes (sex cells) contain only one chromosome from each pair (haploid).3. Fertilized egg cell (zygote) receives one chromosome from mother, one from father.4. Zygote splits and reproduces through mitosis to yield multi-cellular diploid organism.CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 10 -Crossing over (recombination)During meiosis, homologous chromosomes may cross over (recombine) forming chromosomes that mix genes from each parent.http://www.accessexcellence.org/AB/GG/comeiosis.htmlThe two chromosomes that form a pair are called homologous.Note that liklihood of recombination is function of distance between two genes. This observation is used in creating genetic linkage maps. Here we see recombination of gene c/C which appears in two forms (alleles). Genes ab (AB) are unlikely to recombine.CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 11 -GenomesComplete set of chromosomes that determines an organism is known as a genome.Note that each cell in an organism contains its entire genome!Sizes ofsome genomeshttp://www.cbs.dtu.dk/databases/DOGS/ http://www.nsrl.ttu.edu/tmot1/mus_musc.htm http://www.oardc.ohio-state.edu/seedid/single.asp?strID=324Mus musculusCSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 12 -We're more similar than you might thinkhttp://www.ornl.gov/sci/techresources/Human_Genome/graphics/slides/ttmousehuman.shtmlhttp://www.news.cornell.edu/releases/Dec03/chimp.life.hrs.html(The DNA of chimpanzees and humans is ~99% similar.)CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 13 -Genetic linkage map(107 – 108 base pairs)Studying a genomeMost genomes are enormous (e.g., 108 base pairs in case of human). Current sequencing technology, on the other hand, only allows biologists to determine ~103 base pairs at a time.This disparatey leads to some of the most interesting problems in computational biology.Physical map(105 – 106 base pairs)Sequencing(103 – 104 base pairs)ACTAGCTGATCGATTTAGCAGCAG...CSE 397-497: Computational Issues in Molecular BiologyLopresti · Spring 2004 · Lecture 2- 14 -Studying a