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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