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Special-topic Lecture Biosciences: Biological Sequence Analysis Leistungspunkte/Credit points: 5 (V2/Ü1) This course is taught in English language. Lecture form: The students will be required to work actively at home and during the tutorial in small groups to prepare about one third of the lecture content themselves. The material (from books and original literature) will be provided in the lecture. The lectures will be a mixture of ex-cathedra teaching, student presentations, and discussion. Topics to be covered: This course will enter into details of three selected topics in current genetics: (1) Chromatin and nuclear structure (2) Epigenetics (3) Cell programming: circadian rhythms and stem cells Biological Sequence Analysis SS 2009 – lecture 1 1 Aim of this lecture, „Lernziele“ The aim of this course is not to fully cover these three topics but to enter deeply into various details of these fields. This course should improve your ability to compile the necessary biological background that is relevant to your bioinformatics project from the original literature. During this course, you will have ample opportunity to explain biological details. In this way, you practise presentation skills and to use simple language for explaining difficult biology. Also, you should practise your english discussion skills. Biological Sequence Analysis SS 2009 – lecture 1 2 Content Chromatin and Nuclear structure 1 The DNA-encoded nucleosome organization of a eukaryotic genome, Nature 458, 362 (2009) 2 3 Nucleosome organization in the Drosophila genome, Nature 453, 358 (2008) Transcriptional regulation constains the organization of genes on eukaryotic chromosomes, PNAS 105, 15761 (2008) 4 De Novo Formation of a Subnuclear Body, Science 322, 1713 (2008) DNA processing and Epigenetics 5 A Global View of Gene Activity and Alternative Splicing by Deep Sequencing of the Human Transcriptome, Science 321, 956 (2008) 6 Chromatin signature reveals over a thousand highly conserved large non-coding RNAs in mammals, Nature 458, 223 (2009) 7 Shotgun bisulphite sequencing of the Arabidopsis genome reveals DNA methylation patterning, Nature 452, 215 (2008) 8 Chromatin dynamics during epigenetic reprogramming in the mouse germ line, Nature 452, 877 (2008) Cell programming 9 Oct-4 Induced Pluripotency in Adult Neural Stem Cells, Cell 136, 411 (2009) 10 A feedback regulatory loop involving microRNA-9 and nuclear receptor TLX in neural stem cell fate determination, Nature Struct. Mol. Biol. 16, 365 (2009) 11 Circadian Clocks I 12 The Arabidopsis Circadian Clock Incorporates a cADPR-Based Feedback Loop, Science 318, 1789 (2007) Biological Sequence Analysis SS 2009 – lecture 1 3 Tutorial Short weekly assignments are handed out for every paper. Up to two students can hand in a solved assignment. Send your solutions by e-mail to the tutor Barbara Hutter: [email protected] until Thursday 12 pm. The assignments will be discussed in the tutorial on the following Monday 1-2 pm. Schein condition 1 Only those students can get a „Schein“ who have obtained more than 50% of the points for all assignments, Schein condition 2 Up to two students are invited to present the content of the paper at the beginning of the next lecture. Every student has to present once during the semester. These presentations will not be graded. Biological Sequence Analysis SS 2009 – lecture 1 4 Schein = pass 3 written tests Schein condition 3 The successful participation in the lecture course („Schein“) will be certified upon fulfilling Schein conditions 1 and 2 and upon successful completion of 3 written 30 minute tests. All tests have to be passed. Each test covers the content of one lecture topic. Dates: at the beginning of lectures V5, V9 and V13. All students registered for the course may participate in the tests. The final mark on the Schein will be computed from the sum of the 3 test results. The tests will cover the lecture material (slides on the lecture website) and the papers and assignments for this topic. In case of illness please send E-mail to: [email protected] and provide a medical certificate. Those who miss or fail one test, will be given a second-chance oral exam. If you fail or miss more than one test (without excellent reason), you cannot get a Schein. Biological Sequence Analysis SS 2009 – lecture 1 5 Gene Transcription etc. Terms that you should remember from an introductory genetics lecture ... Genome Genes Introns, Exons Nucleus DNA-Polymerase mRNA Splicing Nuclear Pore Complex Ribosome tRNA Translation Biological Sequence Analysis SS 2009 – lecture 1 6 The cell nucleus HeLa cells stained for DNA with the Blue Hoechst dye. The central and rightmost cell are in interphase, thus their entire nuclei are labeled. On the left a cell is going through mitosis and its DNA has condensed ready for division. Schematic of typical animal cell, showing subcellular components. Organelles: (1) nucleolus (2) nucleus (3) ribosome (4) vesicle (5) rough endoplasmic reticulum (ER) (6) Golgi apparatus (7) Cytoskeleton (8) smooth ER (9) mitochondria (10) vacuole (11) cytoplasm (12) lysosome (13) centrioles wikipedia.org Biological Sequence Analysis SS 2009 – lecture 1 7 Chromatin structure How does all genomic DNA (length 2 m in human) fit into a tiny nucleus? Kornberg, 1974: Proposal that chromatin structure is based on a repeating unit of 8 histone Molecules and about 200 DNA base pairs. 1991, 1997: X-ray structures of the repeating unit, the nucleosome. There exist five histone types: H1, H2A, H2B, H3 and H4. One nucleosome contains 2 x H2A, 2 x H2B, 2 x H3, 2 x H4 Histones can be reversibly acetylated and de-acetylated. Kornberg & Lorch, Cell 98, 285 (1999) Biological Sequence Analysis SS 2009 – lecture 1 8 Chromatin structure The highly conserved nucleosome occurs essentially every 200 ± 40 bp throughout all eukaryotic genomes. The repeating nucleosome cores further assemble into higher-order structures which are stabilized by the linker histone H1. These compact linear DNA overall by a factor of 30-40. Packing of chromatin is generally repressive. Some transcription factors may invade at the terminal segments of the nucleosomal DNA. Binding of other TFs is promoted by the bending and supercoiling of DNA on a nucleosome. Biological Sequence Analysis SS 2009 – lecture 1 9 Nucleosome X-ray structure 146 bp of DNA are wrapped around the histone octamer in 1.65 turns of a flat, left-handed superhelix. DNA shows a double-helix twist value of 10.2 bp per turn. This is slightly different from the 10.5 bp value of free DNA. Luger et al. Nature 389, 251 (1997) Biological Sequence Analysis SS 2009 – lecture 1 10 Higher-order organization Tetranucleosome structure. Schalch et al. Nature 436, 138 (2005) Two different chromatin fiber models based on the tetranucleosome structure. http://www.structuralbiology.uzh.ch/research004h.asp Chromatin fiber models. Dorigo et al. Science 306, 1571 (2004) Biological Sequence Analysis SS 2009 – lecture 1 11 Transcription factor Reb1 RNA polymerase I enhancer binding protein; DNA binding protein which binds to genes transcribed by both RNA polymerase I and RNA polymerase II 176 Reb1 binding sites predicted in the genome. http://www.yeastgenome.org/cgi-bin/locus.fpl?dbid=S000000253 Biological Sequence Analysis SS 2009 – lecture 1 12 Effect of Reb1 on DNA Biological Sequence Analysis SS 2009 – lecture 1 13 Transcription factor Aft2 Iron-regulated transcriptional activator; Activates genes involved in intracellular iron use and required to iron homeostasis and resistance to oxidative stress. 80 AFT2 binding sites predicted in the genome. http://www.yeastgenome.org/cgi-bin/locus.fpl?locus=AFT2 Biological Sequence Analysis SS 2009 – lecture 1 14 Example: Monoallelic expression of odorant receptors The nose recognizes chemical information in the environment and converts it into meaningful neural signal, allowing the brain to discriminate among thousands of odorants and giving the animal its sense of smell. The mouse genome contains more than 1000 genes encoding olfactory receptors (ORs). This makes them the largest mammalian gene family. They are putative GPCRs and are located in clusters which are scattered throughout the genome. The large number of receptors suggests that each odor elicits a unique signature, defined by the interactions with a limited number of relatively specific olfactory receptors. From combinations of interactions, animals would then be able to sense more than 104–105 different odors. Shykind, Hum Mol Gen 14, R33 (2005) Biological Sequence Analysis SS 2009 – lecture 1 15 Monoallelic expression of odorant receptors Isolation of OR genes allowed studying the biology of olfaction. RNA in situ hybridization studies revealed two fundamental characteristics of OR expression. (1) neurons expressing a given receptor are restricted to one of 4 broad zones running across the olfactory epithelium. (2) within a zone, individual receptors are expressed sparsely and without apparent pattern. Quantitative analysis of these in situ hybridization experiments led to the suggestion that each neuron in the nose expresses only one or a few members of the gene family. Shykind, Hum Mol Gen 14, R33 (2005) Biological Sequence Analysis SS 2009 – lecture 1 16 Monoallelic expression of odorant receptors Subsequent analyses of cDNAs synthesized from single olfactory neurons just a single OR species could be isolated from each cell. This strengthened the ‘one neuron–one receptor’ hypothesis. Additionally it was found that ORs are transcribed from just one allele. Hypothesis by Buck and Axel in 1991: the olfactory sensory neuron selects a single receptor from just one allele of a spatially allowed subset of a widely dispersed gene family. Shykind, Hum Mol Gen 14, R33 (2005) Biological Sequence Analysis SS 2009 – lecture 1 17 Axonal Wiring in the Mouse Olfactory System The main olfactory epithelium of the mouse is a mosaic of 2000 populations of olfactory sensory neurons (OSNs). Each population expresses one allele of one of the 1000 intact odorant receptor (OR) genes. An OSN projects a single unbranched axon to a single glomerulus, from an array of 1600–1800 glomeruli in the main olfactory bulb. Within a glomerulus the OSN axon synapses with the dendrites of secondorder neurons and interneurons. Axons of OSNs that express the same OR project to the same glomeruli. Mombaerts, Ann Rev Cell Biol 22, 713 (2006) Biological Sequence Analysis SS 2009 – lecture 1 18 Monoallelic expression of odorant receptors The logic of the olfactory circuit rests upon this regulatory process as does the formation of the sensory map, which is dependent on receptor protein to guide the path-finding axon. Aberrant expression of multiple ORs per neuron may disrupt olfactory axon guidance and thus prevent accurate formation of the glomerular map. Once a neuron establishes its synapse in the olfactory bulb, it must remain committed to its OR. Any change in receptor would change the ligand specificity of the cell and confound the sensory map. Shykind, Hum Mol Gen 14, R33 (2005) Biological Sequence Analysis SS 2009 – lecture 1 19 Visualisation of monoallelic expression: Odorant receptor expression in axons (A) Whole mount view of a compound heterozygous mouse, age P30, genetically modified to express tau-lacZ and GFP from each allele of the P2 odorant receptor gene. Neurons express P2 monoallelically (green or red cells) in the olfactory epithelium (oe), and project their axons back into the olfactory bulb (ob) to form a glomerulus (gl, within white box). Nuclei are counterstained by Toto-3 (blue). (B) High power view of (boxed area in A) showing the convergence of P2 axons to a glomerulus (red and green fibers). Neighboring glomeruli are indicated by asterisks. How this mono-allelic expression works on a molecular level is apparently still unknown. Shykind, Hum Mol Gen 14, R33 (2005) Biological Sequence Analysis SS 2009 – lecture 1 20