Question
What is the chemical nature of the
repressor?
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MCB 140 – 11/8/2006
If you think about it …
•
•
The repressor has to directly bind to a
specific DNA sequence (the operator) in
the E. coli genome.
From a reverse-engineering perspective,
the simplest way to design something
that interacts with DNA sequencespecifically is to use a nucleic acid that is
complementary to that DNA!
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MCB 140 – 11/8/2006
• In 1959, a biochemical experiment was
done “proving” that the lac repressor is not
a protein, and is most likely an RNA
molecule
• In 1965, a genetic experiment was done
proving that biochemical experiment
entirely wrong
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1959:
(When the z+ gene and the i+
gene arrive in an i- cytoplasm
(no repressor), the z+ gene
becomes active, and stays
active for about 2 hours. At that
point, the repressor is made,
and shuts the z+ gene off.
This offers an elegant
opportunity to determine the
biochemical nature of the
repressor: add an inhibitor of
protein synthesis, and let the
cells spend their first two hours
post-mating in that good stuff.
A.B. Pardee and L.S. Prestidge
BBA 36: 545.
If the repressor is a protein, then inhibiting its synthesis following
transfer of z+ into an i- cytoplasm should allow for
constitutive synthesis of b-galactosidase!
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The result
• Inhibiting protein synthesis does not destroy the
“plateau” effect: b-gal still goes off!
• Conclusion: “… the repressor probably is not a
protein, since it was made when [protein]
synthesis was inhibited. Ribonucleic acid would
seem a likely candidate for the role of the
repressor.”
A.B. Pardee and L.S. Prestidge BBA 36: 545.
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A distinction between
genetics and biochemistry
• The validity of Pardee’s conclusion is
unequivocally dependent on a biochemical
phenomenon (total inhibition of protein
synthesis by 5-methyltryptophan)
• It turns out that not all protein synthesis in
E. coli is inhibited by 5-me-T…
• A genetic experiment (=crosses between
strains of different genotype) was
performed to prove Pardee wrong
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8.28
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Let’s mate
1. Make an i- strain of E. coli (constitutive)
2. Make several such strains (=different
kinds of mutations)
3. Mate those i- E. coli to other E. coli
carrying nonsense suppressor tRNA
genes
4. See what happens.
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“Suppression of and complementation among mutants of the regulatory
gene of the lactose operon of Escherichia coli.”
Bourgeois S, Cohn M, Orgel LE.
JMB 14: 300 (1965)
Cross an i- strain with a tRNAsup strain
Measure activity of b-gal (+ or - lactose in the medium).
1.
2.
Suppressor:
lactose
Strain
i+
i-
none (wt)
no
yes
8
4400
su1
no
6
yes
4400
su3
no
14
yes
3900
7600 7000 1200 1550 3400 7100
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MCB 140 – 11/8/2006
In their own words
“We know … that the suppressor strains
used here act at the level of translation by
allowing the chain-terminating codon to be
read as an amino acid. We have shown
suppression of i- mutations by these
suppressors and conclude therefore that
the i gene of the lactose operon codes for
a protein.”
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stimulus
+
+
Regulation of genes occurs via the interaction of transacting factors (proteins) with cis-acting sequences
near the genes themselves.
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François Jacob:
“If it’s true for E. coli, it must be true for E. lephant.”
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Budding (brewer’s and baker’s) yeast, Saccharomyces cerevisiae
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Yeast ferment all available sugar
even in the presence of oxygen
Mammals:
sugar
glycolysis
respiration
pyruvate
TCA+O.P.
CO2+H2O
S.c.:
sugar
glycolysis
pyruvate
fermentation
C2H5OH+CO2
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A major evolutionary conservation of cellular
response to sugar in the medium
In Saccharomyces cerevisiae (budding yeast = a
fungus = a eukaryote):
1. Enzymes for utilization of the sugar galactose
are induced ~1000-fold by galactose.
2. These enzymes are also severely repressed
by glucose in the medium.
3. Thus, for these genes to be induced fully, the
medium must contain galactose and no
glucose.
Just like E. coli.
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MCB 140 – 11/8/2006
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
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First experiment
(Howard Douglas, 1963)
Goal: make i- yeast – that is, yeast that
synthesize galactose-metabolizing
enzymes constitutively.
1. Take mutant strain (gal3) that has a
markedly delayed response to galactose
and does not grow on it very well at all.
2. Grow on galactose – see what grows.
Douglas HC, Penroy G (1963) A gene controlling inducibility of the galactose pathway enzymes in Saccharomyces. Biochim Biophys Acta 68: 155.
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What grew
1. One would expect revertants of the gal3
mutation. Those didn’t show up.
2. What did show up was true i- cells –
yeast that synthesized the GAL enzymes
constitutively (that’s why they grew).
3. They made i+ / i- cells – they were
inducible.
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Conclusion (ta-daaa!)
“The inducibility in gene in yeast fits the
description proposed by Monod and Jacob
for regulator genes. … By analogy with the
lactose system in E. coli, the galactoseinducible state in yeast corresponds to the
production of a repressor, due to i+, while
the constitutive state, due to i-, represents
a failure to repression.”
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MCB 140 – 11/8/2006
Nothin’ but net
Indeed, the i+ gene is now called GAL80.
Its product, Gal80p, is a repressor of GAL
genes.
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But
“We might have expected, by further
analogy with the systems in E. coli, to find
mutations linked to the structural galactose
genes and to be expressed as cis
dominant constitutives, but these have not
yet been detected in our material.”
In other words, they wanted to find “operator” mutations, and didn’t find them.
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Screen for gal cells
H. Douglas, D. Hawthorne (1964):
1. Take wt haploid yeast.
2. Zap them with UV light.
3. Replica-plate to find those that are gal.
4. 1:1000 are mutant.
Douglas HC, Hawthorne D (1964) Enzymatic expression and genetic linkage of genes controlling lactose utilization in Saccharomyces. Genetics 49: 837.
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7.5a
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A.8
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Initial analysis of mutants
1.
2.
3.
4.
Cross mutant to wt.
Sporulate.
Dissect the tetrad (ascus)
Confirm that 2 spores are GAL and 2 are
gal.
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What came out
Large number of mutants in different genes:
• GAL1 six mutants
• GAL2 five mutants
• GAL3 six mutants
• GAL4 two mutants
• GAL5 nine mutants
• GAL7 four mutants
• GAL10 one mutant
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“When genes are linked,
PDs exceed NPDs”
5.18
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Linkage analysis
Gene pair
gal1/gal7
gal1/gal10
gal7/gal10
gal1/gal4
gal3/gal4
PD
313
59
72
21
20
# of tetrads
NPD
0
0
0
23
13
T
0
0
0
56
48
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MCB 140 – 11/8/2006
Conclusions
• The following genes are very closely
linked: GAL1, GAL7, and GAL10
• The GAL4 gene is not linked to those
three
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MCB 140 – 11/8/2006
What does what
Genotype
wild type
gal1
gal7
gal10
gal10/igal1/gal7
gal1/gal10
gal4
gal4/i-
Kinase
13-24
0
6
2.7
13
0
0
0
0
Enzymatic activities
Transferase Epimerase
9-14
8-34
14.9
87.2
0
33.8
2.2
0
15.5
0
0
20.3
6.1
0
0
0
0
0
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MCB 140 – 11/8/2006
The fun part of this complete
breakfast
The gal4 mutation is unlinked to the enzyme
genes, and yet the GAL4 gene product is
required for their synthesis.
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MCB 140 – 11/8/2006
Ta-daaa!!
“The closely linked genes GAL1, GAL7, and
GAL10 [code for] the galactose pathway
enzymes, galactokinase, transferase, and
epimerase. The mutation gal4 blocks the
synthesis of these enzymes, but unlike the
phenotypically similar mutation Oo in E.
coli is complementable and is not linked to
the genes whose expression it controls.”
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MCB 140 – 11/8/2006
Summary ctd.
“GAL4 apparently produces a cytoplasmic
product required for the expression of
GAL1, GAL7, and GAL10.
The role of the repressor might be to prevent
the synthesis or the activity of this
product.”
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MCB 140 – 11/8/2006
But still
“In considering a problem of an operator
locus in yeast, it is of interest that [we]
have been unable to find a constitutive
mutant analogous to an OC type…”
= cis dominant mutation that leads to constitutive galactose enzyme synthesis
“find mutation in operator that does not bind repressor…”
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MCB 140 – 11/8/2006
1966: got it!
Complex mutagenesis screen in a diploid
strain for constitutive mutants – exactly
analogous to such screen in E. coli, where
a diploid was used to prevent isolation of
mutants in the repressor gene itself.
Got one dominant mutation that was very
tightly linked to the GAL4 gene.
Douglas HC, Hawthorne D (1966) Regulation of genes controlling synthesis of the galactose pathway enzymes in yeast. Genetics 54: 911.
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- galactose
Gal80p
GAL7
GAL4
GAL10
GAL1
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+ galactose
Gal80p
GAL4
Gal4p
GAL7
GAL10
GAL1
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- galactose in gal81 strain
Gal80p
gal81
GAL4
Gal4p
GAL7
GAL10
GAL1
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wrong
(correct between 1966-1978)
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Gal4p is synthesized at all times,
irrespective of the presence of galactose
Yasuji Oshima (Osaka)
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 Tokyo Kokuritsu Hakubutsukan (Ueno)
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sumi-e
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gal4-4 – a ts allele of GAL4
GAL at 25 and gal at 35
1. Take wt and gal4-4 cells.
2. Grow at 35 in glucose
3. Move them to 25
4. Immediately add galactose
5. Measure galactokinase.
6. Compare wt and gal4-4.
gal4-4 (3525)
gal4-4 (25)
wild-type
44
Matsumoto et al. J. Bacteriol. 134: 446 (1978).
MCB 140 – 11/8/2006
Epistasis (1982-84)
A gal80 cell synthesizes galactose utilization enzymes
constitutively. A gal4 cell does not synthesize those
enzymes under any condition. A double-mutant gal80
gal4 cell has the same phenotype as a gal4 cell – no
enzyme synthesis.
GAL4 is epistatic to GAL80.
Gal4p acts downstream of Gal80p.
A “superrepressor” allelic form of GAL80 (GAL80S) does
not respond to galactose. Overexpression of Gal4p
can overcome its effect! That is, GAL4 GAL80S is
uninducible, but GAL4high copy GAL80S is inducible.
Johnston and Hopper PNAS 79: 6971 (1982).
Torchia et al. MCB 4: 1521 (1984).
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SUPPRESSION: A given mutation (A) has a discrete phenotype that is not normal, i.e. not
wild-type. The presence of the second mutation (B, the suppressor mutation) causes the AB
double mutant to display a phenotype that is normal or near-normal. Thus, a suppressor
mutation rescues or restores or repairs, in whole or in part, the defect caused by the first
mutation. Examples: A nonsense mutation in a gene can be suppressed by a mutation in a
tRNA gene in which the anticodon has been mutated to read the nonsense codon
(informational suppressor). A temperature-sensitive mutation that destabilizes a protein can
be corrected by a compensating mutation in another gene product that binds to and acts in a
complex with the first gene product (extragenic suppressor). Overproduction of a
transcription factor can overcome the need for a protein kinase in the pathway that is
normally needed to activate that transcription factor (dosage suppressor). Loss of a
repressor of a gene can compensate for absence of the positive regulator normally needed
to turn on that gene (bypass suppressor).
EPISTASIS: Two mutations (A and B) each have discrete phenotypes that are readily
distinguishable from each other, and neither are normal (both are non-wild-type). If the
phenotype of the AB double mutant resembles that of one of the two single mutants and not
the other, that mutation is said to be epistatic over the other. Thus, if the AB double mutant
looks like the A mutant alone, mutation A is said to be epistatic over mutation B; conversely,
if the AB double mutant looks like the B mutant alone, mutation B is said to be epistatic over
mutation A. Examples: In yeast, the ade3 mutation blocks purine biosynthesis early in the
pathway, whereas the ade2 mutation blocks the pathway later. Both require Ade in the
medium to grow. The ade2 mutant has a white colony color, but the ade2 mutant has a pinkto-red colony color (because the metabolic intermediate that accumulates in the ade2 mutant
polymerizes to form a pigment). The ade2 ade3 double mutant still requires Ade to grow, but
displays a white colony color. Thus, the ade3 mutation is said to be epistatic to the ade2
mutation (which make sense because ade3 blocks production of the intermediate that would
otherwise accumulate in the ade2 mutant). In flies, apterous (ap) mutations block wing
formation, whereas curled wing (cw) causes a dramatic change in wing morphology. Neither
have normal wings. A ap cw double mutant has a wingless phenotype. Thus, ap is epistatic
to cw. Thus, in the pathway for construction of a wing, ap functions before cw.
Prof. Jeremy Thorner, UCB
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MCB 140 – 11/8/2006
Also “described” on pp. 592-593, under the amusing heading
“The yeast GAL system is another complex regulatory mechanism.”
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How genes respond to environmental stimuli
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More from Dr. Jacob
“I have always been convinced that the
same principles operating in bacteria are
also operating in higher organisms with
added complexity. The question therefore
is to understand what kind of complexity is
involved and how it is generated.”
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MCB 140 – 11/8/2006
Next time
The answer to Dr. Jacob’s question.
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Further reading
1. T. Brock The Emergence of Bacterial
Genetics (CSHL Press 1990)
2. M. Ptashne, A. Gann Genes and Signals
(ibid)
3. H. Judson The eighth day of creation
(Simon and Schuster 1979)
4. S. Brenner My Life in Science (BMC
Press 2001)
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