Transcript Document

Goals of today’s lecture
1) Describe the basics of prokaryotic gene
regulation -operons, negative and positive
regulation
2) Illustrate the use of genetics in understanding
cellular processes
3) Cover some aspects of DNA-binding proteins
Gene Expression
• Regulating the amount of active gene
product (protein) by:
– Control of gene expression at the level of
transcription
– Control of at the level of translation
– Post-translational control
• Constitutive expression = gene product
made continuously
• Regulated expression = gene product made
on demand; expression can be induced or
repressed
Gene Expression & Information Flow
Replication
Transcription Translation Post-translation
RNA
Protein
Protein*
DNA
• Replication = produce exact copy of DNA for mitosis (cell
division) or reproduction (pass to the next generation)
• Transcription = transcribe DNA code into RNA (uses same
‘language’ of nucleic acids)
• Translation = translate nucleic acid code into a sequence of
amino acids (the primary structure of polypeptides)
• Post-translational modification = chemical modification to
activate a protein so it can function in the cell
Regulation of Gene Expression
in Prokaryotes
Catabolic Metabolism
• E. coli use many sugars for
metabolism
• Glucose is the preferred
source of carbon for E. coli
–Why?
• If glucose is not available,
bacteria can break down
lactose to generate glucose
Figure 17-2b-setup
Figure 17-2c-results
• E. coli produces high levels of b-galactosidase,
the enzyme that cleaves lactose to glucose +
galactose, only when lactose is present in the
environment. Thus, lactose (actually it is a
metabolite of lactose) acts as an inducer—a
molecule that stimulates the expression of a
specific gene.
• Jacques Monod found that b-galactosidase is not
expressed in E. coli cells grown in medium
containing glucose or glucose + lactose but only in
medium containing lactose and no glucose.
Genetic screening
unmutagenized cells
mutagenized cells
mutagen
105 cells
Mutant that can’t
utilize lactose as a
carbon source
each cell has
a different mutation
How to find the needle in the genetic haystack?
Which mutations effect lactose utilization?
Replica plating allows the identification of
genes that are essential to utilize lactose
~5000 colonies/plate
• Three classes of E. coli mutants defective in
lactose metabolism were isolated (Table 17.1):
lacZ, lacY, and lacI. lacI is a constitutive
mutant—one that has lost the ability to
regulate expression of a particular gene
because it produces a product at all times, not
just when an inducer is present.
Two proteins are critical
for E. coli to use lactose
and one is critical for
regulation of their
expression
Galactoside
permease
E. coli
b-Galactosidase
Glucose
Galactose
Lactose
Plasma membrane
lacl product
Section of E. coli
chromosome
lacl
b-Galactosidase Galactoside
permease
lacZ product lacY product
lacZ
Far away on the chromosome
lacY
Model for Operons in
Prokaryotes
• Portion of DNA including a set of genes involved in a
specific metabolic pathway
• Single regulatory region (operator + promoter)
• Generates single polycistronic RNA
• Repressor binds the operator and blocks RNA polymerase
• Repressor is the product of a regulatory gene
AUG
5’
UGA AUG
lacZ
UAA
lacY
3’
Figure 17-6a
Negative control: Regulatory protein shuts down
transcription
TRANSCRIPTION
No negative
control…
RNA polymerase
Gene sequence
Regulatory
protein
With negative
control…
No transcription
Figure 17-6b
Positive control: Regulatory protein triggers transcription
No positive
control…
No transcription
RNA polymerase
Regulatory
protein
With positive
control…
TRANSCRIPTION
Gene sequence
• Jacob and Monod proposed that the
lacI gene produces a repressor (the
LacI+ protein) that exerts negative
control over the lacZ and lacY genes.
The repressor was thought to bind
directly to DNA near or on the
promoter for the lacZ and lacY genes
(Figure 17.7).
Repressor present, lactose absent:
Repressor binds to DNA.
No transcription occurs.
The repressor blocks transcription
Repressor
synthesized
DNA
Normal
lacl gene
Repressor present, lactose present:
Lactose binds to repressor,
causing it to release from
DNA. Transcription occurs
(lactose acts as inducer).
Repressor
synthesized
Normal
lacl gene
lacZ
lacl+
RNA polymerase
bound to promoter
(blue DNA)
TRANSCRIPTION BEGINS
b-Galactosidase Permease
mRNA
lacZ
lacl+
RNA polymerase
bound to promoter
(blue DNA)
No repressor present, lactose present
or absent:
Transcription occurs.
No functional
repressor synthesized
Mutant
lacl gene
lacY
lacY
Lactose-repressor
complex
TRANSCRIPTION BEGINS
mRNA b-Galactosidase Permease
Lacl –
lacZ
RNA polymerase
bound to promoter
(blue DNA)
lacY
The Trp operon is also under negative control, but with
a twist
When tryptophan is present, transcription is blocked.
Repressor Tryptophan
No transcription
Operator
RNA polymerase
bound to promoter
When tryptophan is absent, transcription occurs.
TRANSCRIPTION
5 genes coding for enzymes involved
in tryptophan synthesis
RNA polymerase
bound to promoter
Figure 17-10
lac operon
trp operon
Catabolism
(breakdown of lactose)
Anabolism
(synthesis of tryptophan)
Repressor
Lactose
Repressor
Tryptophan
Tryptophan
binds to
repressor
Lactose binds
to repressor
Lactoserepressor
complex
releases from
operator
Operator
Tryptophanrepressor
complex binds
to operator
Operator
Transcription
of lac operon
TRANSCRIPTION
No more
transcription
of trp operon
Why doesn’t beta-galactosidase get induced in media
containing Glucose?
Catabolite Repression
CAP regulates lac operon positively and
requires cAMP for DNA binding
When cAMP is present:
cAMP binds to CAP and the
cAMP-CAP complex binds
to DNA at the CAP site.
RNA polymerase binds
the promoter efficiently.
Transcription occurs frequently.
When cAMP is absent:
CAP does not bind to DNA.
RNA polymerase binds
the promoter inefficiently.
Transcription occurs rarely.
cAMP
CAP
FREQUENT TRANSCRIPTION
CAP site
Operator lacZ
lacY
lacA
RNA polymerase bound
tightly to promoter (blue DNA)
CAP
INFREQUENT TRANSCRIPTION
CAP site
Operator lacZ
RNA polymerase bound
loosely to promoter (blue DNA)
lacY
lacA
Cyclic AMP (cAMP) is synthesized when glucose levels are low
Glucose inhibits the activity of the enzyme adenylyl cyclase, which catalyzes production of cAMP from ATP.
Glucose inhibits
this enzyme
ATP
Adenylyl cyclase
cAMP
Two phosphate
groups
The amount of cAMP and the rate of transcription of the lac operon are inversely related to the concentration
of glucose.
CAP
HIGH
glucose
concentration
INACTIVE
adenylyl cyclase
LOW
cAMP
CAP does not
bind to DNA
Infrequent transcription
of lac operon
(Cell continues to use
glucose as energy source.)
cAMP
CAP
LOW
glucose
concentration
ACTIVE
adenylyl cyclase
HIGH
cAMP
CAP-cAMP complex
binds to DNA
Frequent transcription
of lac operon
(Cell uses lactose
if lactose is present.)
Dual Regulation of lac operon
• Negative control by lac repressor >> needs the inducer
(lactose) to inactivate the lac repressor
• Positive control by CAP (activated by high [cAMP]
resulting from low [glucose]) >> determines rate of
transcription if the operator is NOT blocked by the
repressor
Figure 17-15
lac operon
Repressor
Promoter
INFREQUENT TRANSCRIPTION
Glucose HIGH
Lactose LOW
CAP
site
Operator
lacZ
lacY
lacA
RNA polymerase bound
loosely to promoter
INFREQUENT TRANSCRIPTION
Glucose HIGH
Lactose HIGH
CAP
site
Operator
RNA polymerase bound
loosely to promoter
lacY
lacA
Inducer-repressor complex
FREQUENT TRANSCRIPTION
Glucose LOW
Lactose HIGH
lacZ
CAP
site
Operator
RNA polymerase bound
tightly to promoter
lacZ
lacY
lacA
Figure 17-11-1
DNA FOOTPRINTING
Radioactive
atom
DNA
Repressor
protein
No repressor
1. Generate fragments
2. Divide fragments
3. Cut fragments with nuclease
from the DNA region
of interest, such as
the lac operon of
E. coli. Attach a label
to end of fragments.
into two samples.
Add repressor protein
to one sample. The
repressor will bind to
the operator.
to produce fragments of different
lengths. Repressor protects
operator DNA from nuclease
cleavage.
Figure 17-11-2
DNA FOOTPRINTING
“Footprint”
No cuts occurred in the DNA
region protected by the repressor.
This region must be the operator.
Largest fragments
(cut far from label)
4. Load fragments
into two lanes in a
gel. Sort by size via
electrophoresis. (The
fragments with a
label will be visible.)
Smallest fragments
(cut close to label)
A DNA sequencing
reaction can be
used to determine
the sequence of the
“footprint.”
You isolate an E. coli mutant that has high expression of the
lacZ gene even in the absence of lactose (media has no glucose).
You think you have a mutation in the lacI repressor, but this turns
out not to be the case. Which of the following do you think best
explains your mutant?
A)
B)
C)
D)
Mutation in the lacZ gene that increases its activity.
Mutation in the O1 binding site that reduces lacI binding.
Mutation in the O1 binding site that increases lacI binding.
Mutation inactivating the lacY gene.
Summary of Prokaryotic
Gene Regulation
• Prokaryotic genes that code for enzymes in a specific
metabolic pathway are clustered in groups and
regulated together = operon
• Lac operon discovered by Jacob & Monod
• Key advantage: single ‘on-off’ switch to coordinate
gene expression
• Switch = operator that controls access of RNA
polymerase to the promoter
• Repressor protein binds to the operator
• Repressor proteins are subject to regulation (positive or
negative) by the metabolic substrates and products of
the pathway.