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E. coli = E. lephant ? 1965 1966 F. Jacob J. Monod A. Pardee D. Hawthorne H. Douglas Y. Oshima MCB 140 11/27/06 1 Analogy and homology as tools in genetic investigation Animal Mandibular Arch (ventral) Mandibular Arch (dorsal) Hyoid Arch (dorsal) Shark Meckel's cartilage Palatoquadrate cartilage Hyomandibular cartiliage Amphibian Articular (bone) Quadrate (bone) Stapes Mammal Malleus Incus Stapes MCB 140 11/27/06 2 MCB 140 11/27/06 3 a cells produce a pheromone and a receptor a cells produce a pheromone and a receptor diploid (a/a) cells produce none of the above MCB 140 11/27/06 4 Shmoo Al Capp (1948) – Li’l Abner MCB 140 11/27/06 5 Marsh and Rose diagram MCB 140 11/27/06 6 The phenotype of a haploid yeast cell with respect to mating is determined by transcription factors An a cell produces two transcription factors, Mata1p and Mata2p, that ensure expression of a specific genes, including the pheromone and receptor, and repress expression of a specific genes. In an a cell, Mata1p and Mata2p are not expressed, and a different transcription factor is expressed, Mata1p. The a genes are off, and the a genes (pheromone and receptor) are on. MCB 140 11/27/06 7 A.9 MCB 140 11/27/06 8 Amazing but true A wild-type haploid yeast cell contains THREE copies of mating type-determining genes: • Copy #1: the a1 and a2 genes (silent). • Copy #2: the a1 and a2 genes (also silent). • Copy #3: An additional copy of genes in item 1, or of the genes in item 2, but active. Whichever genes are contained in copy #3 determines the mating type. MCB 140 11/27/06 9 A.11 A.12 MCB 140 11/27/06 10 “An easily understood, workable falsehood is more useful than an incomprehensible truth.” MCB 140 11/27/06 11 a2 a1 a cell HMLa silent cen a2 a1 a2 a1 MAT HMRa active silent MCB 140 11/27/06 12 Loss of silencing at the silent mating type cassettes creates a “nonmater” – a haploid that is a/a and that thinks it’s a diploid. a2 a1 a cell HMLa active cen a2 a1 a2 a1 MAT HMRa active active MCB 140 11/27/06 13 Screen for silencing mutants A sample “screen”: 1. Take haploid cells. 2. Mutate them. 3. Screen for those that don’t mate. Problem: mating is so much more than proper silencing of mating type loci!! MCB 140 11/27/06 14 The mating pheromone response Also see Fig. A.13. Jeremy Thorner Thorner diagram MCB 140 11/27/06 15 How to screen for silencing mutants a2 a1 a cell HMLa silent a2 a1 a2 a1 MAT HMRa active silent cen Jasper Rine and Ira Herskowitz (1987) Genetics 116: 9-22. MCB 140 11/27/06 16 How to screen for silencing mutants a2 a1 HMLa silent cen a2 a1 a2 a1 mata1-1 HMLa active silent Note: mata1-1 is a special allele of the a gene – it is recessive to a Jasper Rine and Ira Herskowitz (1987) Genetics 116: 9-22. MCB 140 11/27/06 17 Rine schematic mate to a cells Jasper Rine and Ira Herskowitz (1987) Genetics 116: 9-22. MCB 140 11/27/06 18 The data • Colonies screened: 675,000 • Colonies that mated to a: 295 • Major complementation groups: 4 silent information regulators: SIR1, SIR2, SIR3, SIR4 Jasper Rine and Ira Herskowitz (1987) Genetics 116: 9-22. MCB 140 11/27/06 19 Question What molecular mechanisms are responsible for silencing at the mating type loci? heterochromatin formation in metazoa prostate cancer breast cancer ageing “normal” gene regulation in mammals MCB 140 11/27/06 20 Homework MCB 140 11/27/06 21 How can one explain the evolution of two distinct mating types in budding yeast? Surely a pathway could have just evolved for the fusion of two identical haploid cells? MCB 140 11/27/06 22 Two mating types have evolved under selective pressure to avoid inbreeding One evolutionary advantage of mating is the production of novel genotypic combinations via the fusion of two genomes with different life histories. D1 D1 x M D2 D2 MCB 140 11/27/06 23 Granddaughters of any given mother can switch mating type MCB 140 11/27/06 24 MCB 140 11/27/06 25 [email protected] MCB 140 11/27/06 26 a2 a1 a cell a cell a2 a1 a2 a1 MAT HMRa silent active silent a2 a1 a2 a1 a2 a1 MAT HMRa HMLa HMLa cen cen MCB 140 11/27/06 27 Epigenetic inheritance • In an a strain, the genetic information at MAT and at HMLa is identical. • The one at MAT is expressed, but the one at HML is not – it is epigenetically silenced. Epigenetic: mitotically stable (persists through cell division) change in gene expression state that is not associated with a change in DNA sequence. Examples: X chromosome inactivation; imprinted genes; transgene silencing in gene therapy. MCB 140 11/27/06 28 Compaction into chromatin brings the eukaryotic genome to life 15,000x compaction < 10-5 metres > 1 metre MCB 140 11/27/06 29 “Beads on a string” MCB 140 11/27/06 30 The Nucleosome Core Particle: 8 histones, 146 bp of DNA MCB 140 11/27/06 31 Histones: Conserved and Charged H.s. = Lycopersicon esculentum MCB 140 11/27/06 32 MCB 140 11/27/06 33 “Extremely conserved histone H4 N terminus is dispensable for growth but essential for repressing the silent mating loci in yeast” (M. Grunstein) Fig. 3 kayne Kayne et al. (1988) Cell 55: 27-39. MCB 140 11/27/06 34 Fig. 6 and 7 of Kayne. Kayne et al. (1988) Cell 55: 27-39. MCB 140 11/27/06 35 Kayne et al. (1988) Cell 55: 27-39. MCB 140 11/27/06 36 Acetylation of lysine in histone tail neutralizes its charge (1964) MCB 140 11/27/06 37 “Genetic evidence for an interaction between SIR3 and histone H4 in the repression of the silent mating loci in Saccharomyces cerevisiae” Reverse genetics: introduce point mutations in H4 tail!! Johnson et al. (1990) PNAS 87: 6286-6290. MCB 140 11/27/06 38 Table 2 Johnson et al. (1990) PNAS 87: 6286-6290. MCB 140 11/27/06 39 MCB 140 11/27/06 40 And 5 years later … Sir3p and Sir4p bind H3 and H4 tails Hecht et al. (1995) Cell 80: 583. MCB 140 11/27/06 41 Houston, we have a … Every nucleosome in the cell has an H3 and H4 tail (two of each, actually). Why do the SIRs bind only where they bind? MCB 140 11/27/06 42 The silencers “Hawthorne deletion” (1963) and onwards: two silencers flank the mating type loci: MCB 140 11/27/06 43 The key question How do the SIRs spread from the silencer and over the mating type loci genes? = how do the SIRs actually silence txn? MCB 140 11/27/06 44 Roy Frye (Pitt) “Characterization of five human cDNAs with homology to the yeast SIR2 gene: Sir2-like proteins (sirtuins) metabolize NAD and may have protein ADPribosyltransferase activity” BBRC 260: 273 (1999). 1. Bacteria have proteins homologous to Sir2. 2. So do humans (>5). 3. The bacterial proteins are enzymes, and use NAD to ADP-ribosylate other proteins. MCB 140 11/27/06 45 J. Denu: Sir2p is a NAD-dependent histone deacetylase (HDAC) Sir2p Tanner et al., PNAS 97: 14178 (2000) MCB 140 11/27/06 46 Rusche L, Kirchmaier A, Rine J (2002) Mol. Biol. Cell 13: 2207. MCB 140 11/27/06 47 Histone tail acetylation promotes chromatin unfolding (somehow) acetylation MCB 140 11/27/06 48 Next time: the genetics of heterochromatin in metazoa MCB 140 11/27/06 49