MCB 140 11/27/06 1 E. coli = E. lephant ? F. Jacob J. Monod A. Pardee D. Hawthorne H. Douglas Y....
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Transcript of MCB 140 11/27/06 1 E. coli = E. lephant ? F. Jacob J. Monod A. Pardee D. Hawthorne H. Douglas Y....
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1MCB 140 11/27/06
E. coli = E. lephant ?
F. JacobJ. MonodA. Pardee
D. HawthorneH. DouglasY. Oshima
1965 1966
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2MCB 140 11/27/06
Analogy and homology as tools in genetic investigation
AnimalMandibular Arch (ventral)
Mandibular Arch (dorsal)
Hyoid Arch(dorsal)
Shark Meckel's cartilagePalatoquadrate cartilage
Hyomandibular cartiliage
Amphibian Articular (bone) Quadrate (bone) Stapes
Mammal Malleus Incus Stapes
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3MCB 140 11/27/06
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4MCB 140 11/27/06
a cells produce a pheromone and receptor
cells produce pheromone and a receptor
diploid (a/) cells produce none of the above
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5MCB 140 11/27/06
ShmooAl Capp (1948) – Li’l Abner
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6MCB 140 11/27/06
Marsh and Rose diagram
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7MCB 140 11/27/06
The phenotype of a haploid yeast cell with respect to mating is determined by
transcription factors
An cell produces two transcription factors, Mat1p and Mat2p, that ensure expression of specific genes, including the pheromone and receptor, and repress expression of a specific genes.
In an a cell, Mat1p and Mat2p are not expressed, and a different transcription factor is expressed, Mata1p. The genes are off, and the a genes (pheromone and receptor) are on.
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8MCB 140 11/27/06
A.9
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9MCB 140 11/27/06
Amazing but true
A wild-type haploid yeast cell contains THREE copies of mating type-determining genes:
• Copy #1: the 1 and 2 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.
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10MCB 140 11/27/06
A.11
A.12
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11MCB 140 11/27/06
“An easily understood, workable falsehood is more useful than an incomprehensible truth.”
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12MCB 140 11/27/06
cen MATHML HMRa
a1a2
cell
active silentsilent
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13MCB 140 11/27/06
Loss of silencing at the silent mating type cassettes creates a “nonmater” – a haploid
that is a/ and that thinks it’s a diploid.
cen MATHML HMRa
a1a2
cell
active activeactive
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14MCB 140 11/27/06
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!!
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15MCB 140 11/27/06
The mating pheromone response
Jeremy Thorner
Thorner diagramAlso see Fig. A.13.
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16MCB 140 11/27/06
How to screen for silencing mutants
cen MATHML HMRa
a1a2
a cell
a1a2
active silentsilent
Jasper Rine and Ira Herskowitz (1987) Genetics 116: 9-22.
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17MCB 140 11/27/06
How to screen for silencing mutants
cen mata1-1HML
a1a2
active silentsilent
Jasper Rine and Ira Herskowitz (1987) Genetics 116: 9-22.
HML
Note: mata1-1 is a special allele of the a gene – it is recessive to
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18MCB 140 11/27/06Jasper Rine and Ira Herskowitz (1987) Genetics 116: 9-22.
Rine schematic
mate to a cells
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19MCB 140 11/27/06
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.
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20MCB 140 11/27/06
Question
What molecular mechanisms are responsible for silencing at the mating type loci?
heterochromatin formation in metazoaprostate cancer breast cancer ageing “normal” gene regulation in mammals
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21MCB 140 11/27/06
Homework
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22MCB 140 11/27/06
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?
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23MCB 140 11/27/06
Two mating types have evolved under selective pressure to avoid inbreeding
M
D1
D2
D1
D2
One evolutionary advantage of mating is the production of novel genotypic combinations via the fusion of two
genomes with different life histories.
x
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24MCB 140 11/27/06
Granddaughters of any given mother can switch mating type
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25MCB 140 11/27/06
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27MCB 140 11/27/06
cen MATHML HMRa
a1a2
cell
cen MATHML HMRa
a1a2
a cell
a1a2
active silentsilent
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28MCB 140 11/27/06
Epigenetic inheritance
• In an strain, the genetic information at MAT and at HML 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.
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29MCB 140 11/27/06
> 1 metre< 10-5 metres
15,000x compaction
Compaction into chromatin brings the eukaryotic genome to life
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30MCB 140 11/27/06
“Beads on a string”
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31MCB 140 11/27/06
The Nucleosome Core Particle:8 histones, 146 bp of DNA
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32MCB 140 11/27/06
Histones: Conserved and Charged
H.s. = Lycopersicon esculentum
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33MCB 140 11/27/06
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34MCB 140 11/27/06
“Extremely conserved histone H4 N terminus is dispensable for growth but essential for repressing
the silent mating loci in yeast” (M. Grunstein)
Kayne et al. (1988) Cell 55: 27-39.
Fig. 3 kayne
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35MCB 140 11/27/06Kayne et al. (1988) Cell 55: 27-39.
Fig. 6 and 7 of Kayne.
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36MCB 140 11/27/06Kayne et al. (1988) Cell 55: 27-39.
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37MCB 140 11/27/06
Acetylation of lysine in histone tail neutralizes its charge (1964)
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38MCB 140 11/27/06
“Genetic evidence for an interaction between SIR3 and histone H4 in the repression of the silent
mating loci in Saccharomyces cerevisiae”
Johnson et al. (1990) PNAS 87: 6286-6290.
Reverse genetics: introduce point mutations in H4 tail!!
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39MCB 140 11/27/06Johnson et al. (1990) PNAS 87: 6286-6290.
Table 2
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40MCB 140 11/27/06
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41MCB 140 11/27/06
And 5 years later …
Sir3p and Sir4p bind H3 and H4 tails
Hecht et al. (1995) Cell 80: 583.
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42MCB 140 11/27/06
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?
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43MCB 140 11/27/06
The silencers
“Hawthorne deletion” (1963) and onwards:
two silencers flank the mating type loci:
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44MCB 140 11/27/06
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?
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45MCB 140 11/27/06
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 ADP-ribosyltransferase 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.
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46MCB 140 11/27/06
J. Denu: Sir2p is a NAD-dependenthistone deacetylase (HDAC)
Tanner et al., PNAS 97: 14178 (2000)
Sir2p
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47MCB 140 11/27/06Rusche L, Kirchmaier A, Rine J (2002) Mol. Biol. Cell 13: 2207.
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48MCB 140 11/27/06
acetylation
Histone tail acetylation promotes chromatin unfolding (somehow)
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49MCB 140 11/27/06
Next time: the genetics of heterochromatin
in metazoa