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Lectures 23-25Fall 2006
Prof. Daniel Kahne
Anti-Viral Agents
Fuzeon (blocks HIV fusion)
AZT (blocks reverse transcription)
Dupont-Merck Inhibitor (blocks HIV protease)
FUSION WITH CELL MEMBRANEAND ENTRY INTO CELL
UNCOATING OFRNA
RTRNA
DNA
TRANSCRIPTION
RNA
ASSEMBLY OF IMMATUREVIRIONS AND EXIT FROM CELL
TRANSLATION
RNA
VIRION MATURATION
2
General Chemotherapeutic Agents
O
O
O
O
OH
OO
O
HO
NH
OH
OO
O
O
Taxol (blocks mitosis)
Camptosar(blocks DNA replication)
N
N
O
O
OHO
ON
O
N
G0
Mechanism-specific Anti-Cancer Agents
Specific to Chronic MyelogenousLeukemia (CML)
N
N
NH
NH
N
N
N
O
Gleevec(blocks Bcr-Abl kinase
phosphorylation)
• Cell Signaling• Protein Kinases
3
OverviewRecognition
Signal transmissionand processing
Getting into the nucleusand transcription
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2
3
Lect 23-25. Cell Signaling and Regulation of Gene Expression:1. Regulation of gene expression.
a. Review of gene expressionb. Regulation of gene expression in prokaryotesc. Regulation of gene expression in eukaryotes (lecture notes)d. Inducers: Jacob and Monod diploid analysise. Le Chatelier's Principle
2. Growth factors and cell signaling.a. Growth factors and growth factor receptorsb. Tyrosine phosphorylation: a post-translational strategy for
modulating gene expressionc. Phosphates in biologyd. Ras, GTPases, and biological timescalese. MAP kinase cascade
Alberts pp. 532-539, 557-560, 267-281, 106-109, 150-154, 67McMurry pp 192-207, 335-338, 656-658
Lecture Readings
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Gene expression/transcription in prokaryotes
RNA polymerase binds to the -10 and -35 sequencesRNA polymerase is comprised of subunits:
Core RNA polymerase -- responsible for elongationHoloenyzme of transcript
σ factor -- guides RNA polymerase to the the start sight
Regulation of gene expression
• One σ factor (σ70) is responsible for transcribing 90% of genesin E. coli.
• When environmental conditions change, bacteria must turn onalternative genes to express proteins that allow them to thriveunder those new conditions.
• More specialized σ factors [(σH) and (σE)] are needed to activatedifferent genetic programs in the cell.
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How does one control subsets of genesthat use the same sigma factor?
Negative regulation -- repressors
RNA pol cannot bind promoter site when repressor is bound
Some sigma factors do not bind apromoter sequence without help.
Positive regulation -- activators
Activators help recruit sigma factors to increase gene expression
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Question
But how do repressors and activatorscontrol gene expression?
Answer
The activity of the repressors andactivators must change somehow.
A Demonstration
Regulation of gene expression changes inresponse to environmental signals.
Francois Jacob Jacques Monod
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All cells need energy
Glucose is the preferred carbon source.
OHOHO
OHHO
OH
Glucose
Central MetabolismCO2 + H2O + ATP (energy source)
Other sugarscan be converted to glucose
LactoseGalactose Glucose
OOHO
OHHO
OH
O
HO
HO
HO
OH
OH
O
HO
HO
HO
OH
OHOHO
OHHO
OH
!-Galactosidase
H20
+
enzymes
β-galactosidase only “recognizes” the galactose part of the lactose so…
β-linkage
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β-galactosidase can hydrolyze otherβ-linked galactosides
O-nitrophenyl-galactoside (ONPG)
galactose O-nitrophenol
ONPG hydrolysis produces a yellow product, ONP.A yellow color formed upon addition of ONPG indicates thepresence of active β-galactosidase.
!-Galactosidase
H20
O
O
HO
HO
HO
OH
NO2
OH
O
HO
HO
HO
OH O-
NO2+
Observation
OOHO
OHHO
OH
O
HO
HO
HO
OH
OHOHO
OHHO
OH
Glucose Lactose
E. coli E. coli
Only cells grown in the presence of lactose turn yellow when ONPG is added to each flask.
ONPG
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Explanation?
First Hypothesis:Cells grown in media containing lactose haveacquired the ability to use lactose throughsome beneficial mutation.
Test:Put cells grown in lactose media into glucosemedia. Now the addition of ONPG does notproduce a yellow color.
Another explanation
Hypothesis 2:β-galactosidase is always present but inactive unlesssubstrate is present (preprotein hypothesis).
Test:Determine whether all substrates for the enzymeinduce the activity of enzyme.
Result: Some substrates do not induce!
Where does the β-galactosidase come from?
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Jacob and Monod…found a mutant E. coli strain that produces a yellowcolor in the absence of lactose.
Hypothesis:All E. coli cells contain a gene for β-galactosidase (lacZ).The expression of lacZ is under the control of a regulatoryprotein. The activity of the regulatory protein is controlledby lactose.
lacI lacZ
E. coli cell
Diploid analysisprovides insight into gene regulation
Hypothesis:lacI DNA makes a diffusible molecule (a protein) that turnsoff β-galactosidase expression (negative control).
lacI lacZ
E. coli cell (wild type)
lacI lacZ
E. coli cell (mutant)
lacI lacZ
Diploid cell is clear!
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Jacob and Monod’s model
lacI lacZ lacI lacZ
OOHO
OHHO
OH
O
HO
HO
HO
OH
lactose
lactose
OOHO
OHHO
OH
O
HO
HO
HO
OH
DNA
OOHO
OHHO
OH
O
HO
HO
HO
OH
E. coli cell (wild type)
Lactose’s effect on lac repressor is an example of Le Chatelier’sPrinciple. (A system at equilibrium, subjected to a stress, will adjust to
relieve the stress and restore the equilibrium.)
lac repressor
DNA-boundlac repressor
DNA-binding conformation
Ligand-binding conformation
Ligand-boundlac reprssor
Lect 23-25. Cell Signaling and Regulation of Gene Expression:1. Regulation of gene expression.
a. Review of gene expressionb. Regulation of gene expression in prokaryotesc. Regulation of gene expression in eukaryotes (lecture notes)d. Inducers: Jacob and Monod diploid analysise. Le Chatelier's Principle
2. Growth factors and cell signaling.a. Growth factors and growth factor receptorsb. Tyrosine phosphorylation: a post-translational strategy for
modulating gene expressionc. Phosphates in biologyd. Ras, GTPases, and biological timescalese. MAP kinase cascade
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Structure of a cell and its compartments
Cells also have eyes/ears/noses -- i.e., molecularmechanisms for sensing and responding to the environment
Problem: Not all signals get in
How do you transduce extracellular signals that cannotenter cells?
Lactose
EpidermalGrowth Factor(53 amino acids)
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A recognition event
The first step in signal transduction involves bindingof the growth factor to the receptor
Recognition 1
Growth factor receptors
Extracellular growthfactor binding domain
Transmembrane domain
Intracellular proteinkinase domain
In the absence of the growth factor, the growth factorreceptor kinase domain is inactive
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Activation of the receptor
inactive, monomeric active, dimeric
ligand
Ligand binding causes the receptor to dimerize, whichactivates the kinase
Protein kinases and phosphorylation
N
NN
N
NH2
O
OHOH
OPO
O-
O
HN
R O
OH
R PO
O-
O
P-O
O-
O
HN
R O
O
R
PO-O
O-
N
NN
N
NH2
O
OHOH
OPO
O-
O
P-O
OH
O+ +
ATP ADPTyrosine Residue Phosphorylated Tyrosine
Other Side Chains That Can Be Phosphorylated:
HN
R O
OHR
Serine Residue
HN
R O
OHR
Threonine Residue
Kinase
!"#
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Why do we need protein kinases?
ΔGoverall <<< 0
N
NN
N
NH2
O
OHOH
OPO
O-
O
PO
O-
O
P-O
O-
O
ATP
! " #
+
N
NN
N
NH2
O
OHOH
OPO
O-
O
P-O
O-
O
OH
OP
O-O
O-+H2O+
Phenol
ADP
Phosphate Ester
H3O++
S = substrate
P = product
TS‡ = transition state
E = enzyme
Uncatalyzedreaction
Enzyme-catalyzedreaction
E.S
E.TS‡
E-P
E + S
E + TS‡
E + P
ΔG‡
(cat)
Reaction coordinate
Free
ene
rgy
ΔG‡
(uncat)
ΔG°
Cellular processes depend on compounds that arekinetically stable but thermodynamically unstable.
Enzymes are requiredto use these compounds
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Kinases
Asp from kinase
Lys from kinase
3) proximity and orientation effects
1) acid catalysis
2) base catalysis
How does phosphorylation transduce a signal?
Kinases catalyze phosphate transfer using the principles of
N
NN
N
NH2
O
OHOH
OPO
O-
O
HN
R
O
OR
PO
O-
O
P-O
O-
O
H
O
O
HN
O
NH
O
Mn2+
H3N
N
O
O
OH
N
O
O
OH
N
O
O
OPO
OO
PO
OO
N
O
O
O
Phosphorylation alters the chemicalproperties of the side chains
Alcohol: small, polar group Phosphate: larger, charged group
serine
tyrosine
pKa1 = 2pKa2 = 7
Changing the size and polarity at specific aminoacids alters the structure and interactions of proteins
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General ways phosphorylation canaffect function:
• Regulation of enzymatic activity• Subcellular localization• Stability• Association with other molecules
Westheimer, F. H. Science, 1987, 235, 1173.
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ATP hydrolysis• The activation barrier (ΔG‡ ) to ATP hydrolysis is
large due to electrostatic repulsion.pKa1 = 2.15
pKa2 = 7.20
pKa3 = 12.15
pKa1 = 3.13
pKa2 = 4.76
pKa3 = 6.40
O
PHO
OH
OH
Phosphoric Acid
HO OH
OOOH
O OH
Citric Acid
Negative charges on the oxygens make ATP kinetically stable.
P
OO
O
+RO O
O
P
O
OO P
O
O
O
RO
!- !-
RO
Reaction Kinetics
Asp fromkinase
Lys fromkinase
Rate of reaction = Collision frequency
X Probability thatmolecules collide inthe right orientation
Probability thatmolecules collidewith enough energyto react
X
= [ SM ] (depends on P)
XProbability thatmolecules collide inthe right orientation
Probability thatmolecules collidewith enough energyto react
XMolecularvelocity
(depends onT)
X Reactioncross-section
(depends on sizeand shape)
X
= [ SM ] Rate constant(k)
XN
NN
N
NH2
O
OHOH
OPO
O-
O
HN
R
O
OR
PO
O-
O
P-O
O-
O
O
O
HN
O
H
NH
O
Mn2+
H3N