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Signaling
Ras - MAP kinase signaling
M
Receptor
tyrosine kinase
Shc
Grb2
Sos
Ras GAP
GDP
P
Ras
GTP
P
Raf P
MEK
P P
P P
ERK
P
Fos
P
P
P
Stat 3
P
Elk-1
Stimuli
Growth Factors (EGFR)
MAPK Kinase Kinase
Raf
MAPK kinase
MEK 1,2
MAPK
Response
Erk 1,2
Proliferation/growth/differentiation
Problem
ATP and GTP are structurally very similar, however Ras is
extremely selective in being activated by GTP and not ATP.
Biochemical analysis has determined that the Kd value for GTP
binding to Ras is 5.10-12 M and the Kon value is 2x10 6 M-1s-1. The
binding of ATP to Ras is much weaker but the Kon value is
identical.
Write the chemical equilibrium for the dissociation of Ras-GTP
to Ras and GTP. Label K on next to the appropriate arrow.
[ras] +[gtp]
[ras-gtp]
K on
Write the rate equations for both forward and reverse reactions.
At equilibrium, how do these rate constants relate to the equilibrium
constant for the dissociation of Ras-GTP to Ras and GTP (Kd)?
Forward rate = koff[Ras-gtp]
Reverse rate = k on [Ras] [gtp]
At equilibrium the rate of the forward and the reverse reactions are equal:
koff [Ras-gtp] = kon [Ras] [gtp]
Kd= koff/kon
Explain why Ras preferentially binds GTP over ATP in terms of specific
kinetic parameters.
The preferential binding of Ras to GTP versus ATP means that the Kd
value for Ras binding to GTP must be lower. Since the Kd=koff/kon and
the k on values are the same for GTP and ATP association with RAS
the difference must result from ATP having higher k off value than GTP.
Problem
When Ras is bound to GTP, it is capable to activate many protein
kinases, while when bound to GDP, Ras is inactive. For this reason Ras is
a molecular switch for many biochemical reactions. The figure below
shows GTP in the active site of Ras, where the numbers represent the
amino acid residues.
You make mutations in the
GTP binding pocket of Ras.
Give the most likely reason
why each mutation has the
stated effect:
1. Residue 16 is mutated to Arg
2. Residue 119 is mutated to
Leu
3. Residue 146 is Tyr
This slide highlights the interactions provided by kinases that accelerate
the phosphotransfer reaction. Asp deprotonates the Tyr hydroxyl,
increasing the nucleophilicity of the oxygen. Lys and Mg ion are essential to
stabilize the negative charges of the ATP.
Important take home messages
1. Cells have mechanisms to respond to environment.
2. Many signals are transduced through phosphorylation
cascades.
3. ATP is kinetically stable but thermodynamically labile
molecules. Enzymes that use APT as phosphate donor and
transfer the phopsphate onto an acceptor molecule are
called kinases.
4. Phosphorylation changes the behavior of proteins.
5. Phosphorylation cascades allow signal amplification.
6. Equilibrium dissociation constants are inversely correlated
with lifetime of complexes.
Problem
Upon binding of epidermal growth factor (EGF), two monomers of EGF
receptor dimerize to form an activated receptor. Each monomer then
phosphorylates multiple tyrosine residues on the other monomer. These
phosphorylated Tyr residues recruit signaling proteins involved in cell
proliferation.
1. If a Tyr that is normally phosphorylated upon EGFR dimerization is
mutated to phenylalanine, how will its ability to contribute to signaling
be affected? Explain.
2. Activation of EGFR promotes increased DNA synthesis. Mutations of
various EGFR tyrosine residues has the following effect on DNA synthesis.
Based on these results, which tyrosine residues are likely to play a role
in the signaling pathways that mediate DNA synthesis?
3. You create a mutant EGFR that lacks the kinase domain, but whose
ability to bind EGF and dimerize are not affected. If you express this
mutant in cells that already express wild type EGFR, what will be the
effect on EGF-activated signaling? Explain.
4. You examine the effect of the following mutant proteins on cell
proliferation in response to EGF. Indicate whether or not you would
expect cells expressing each mutant protein instead of wild type protein
to proliferate in the absence and presence of EGF.
5. Of the mutant protein in part 4., which would you expect to find in
cancer cells?
Problem
The Her2 receptor is a tyrosine kinase whose activation leads to cell growth
and proliferation. Many breast cancers, Her2 receptor mutations have
been isolated.
a. The ligand for the Her 2 receptor causes receptor dimerization. What is
the effect of receptor dimerization for many receptor tyrosine kinases?
b. A point mutation in the transmembrane domain of the Her2 receptor
converting a Val to Gln causes receptor to dimerize in the absence of
ligand. What effect would this mutation have upon the activation state of
the receptor?
c. In order to study receptor tyrosine kinase activity, scientists can
genetically modify the mutant Her2 receptor and eliminate the
intracellular domain. If this is the only one receptor expressed in a cell,
will the cell respond to the Her2 ligand?
d. If a heterodimer were to form containing one subunit of the receptor
described in part b and the other subunit of the receptor in part c, would
this complex be able to activate downstream targets?
e. Activation of Her2 leads to phophorylation of MAP kinase (MAPK). Why do
normal cells require either a phosphatase that removes the phosphate
group from MAP kinase or a MAP kinase that is rapidly degraded by
proteolysis.
Problems
One important ability of any signaling cascade is the ability to turn off,
consider the MAPK pathway where switches could be? Which one could be
the most critical?
Would you expect to activate G protein coupled receptors or receptor
tyrosine kinases by exposing cells to antibodies that bind to the receptive
protein?
If the cell surface receptors can rapidly signal to the nucleus by activating
latent gene regulatory proteins like stats at the plasma membrane, why do
most cell surface receptors use long, indirect signaling cascades to
influence gene transcription?
Ras
P
Raf
P Mek 1,2
P Erk 1,2
Erk 1,2 P
Problem
1. Bacterial pathogens modulate activation of MAPK in a way to escape antibacterial immune responses. Based on the provided data explain how
bacterial infection affects erk activation.
This image represents
immunoblot data from total
protein extracts from HeLa cells
infected with wild type bacteria
(WT) or activated by chemical
treatment with PMA. The MAPK
activation is detected by blotting
with anti-phospho MEK, antiphospho-erk, anti-erk and antimek.
While the western blotting analysis demonstrates activation of MEK by
detecting the presence of phospho-MEK, the downstream activation of erk
is absent in infected cells. Thus, bacterial infection stops erk activation.
erk
P-erk
2. MAPK pathway activation leads
to phosphorylation of MAPK (erk)
and translocation of the activated
erk into the nucleus. The presence
of the activated MAPK – erk in the
cytosol and nucleus can be
visualized by staining of infected
HeLa cells with monoclonal
antibodies specific for the
phosphorylated forms of erk ( in
red ) or non-phosphorylated erk (in
green). NS stands for nonstimulated cells, PMA – stands for
PMA treated cells (PMA is an
inducer of erk activation) and WT
stands for infected HeLa cells.
What is the consequence of
bacterial infection on localization
of erk?
3. Scientists have suspected that the effects of bacterial infection
on erk activation depends on an activity of a protein, called OspF.
Design an experiment to test this hypothesis.
Create bacterial strains that express wild type and mutant
(non-functional ) OspF protein. Infect HeLa cells with the
two different bacteria and analyze erk activation.
Abl is a member of large family of protein kinases, called Src. Src and Abl
have homology in the kinase domain, and also in two domains, called Src
homology 2 (SH2) and Src homology 3 (SH3). Both Src and Abl are
modified at the amino terminus with fatty acid called myristate.
Src can exist in inactive and active forms. The regulatory apparatus that
governs the state consists of: a “latch”, a “clamp” and a “switch”. When the
Tyr 527 is phosphorylated, the latch interacts with the SH2 domain. The
linker interacts with SH3 domain. Both these interactions allow the clamp to
maintain the kinase in inactive, closed conformation.
To acquire active open conformation, Tyr 527 needs to be dephosphorylated,
which leads to unclamping. The Sh2 and Sh3 domains are released and able to
interact with other molecules. Tyr 416 is accessible for phosphorylation.
In abl, the latch involves an interaction between the N-terminal domain of
the peptide with a hydrophobic pocket in the C-lobe of the kinase domain.
This interaction is mediated by a long lipid chain, that is attached to the
N terminus. Unlatching is accomplished by the release of myristate.
Bcr-Abl, the fusion protein encoded by the Phyladelphia chromosome, is a
chimeric protein in which the cap portion of abl is substituted with a
portion from the bcr molecule. This change affects the latching
mechanism. The protein is constitutively active.
Because the latch is missing the inactive closed conformation is less stable.
The equilibrium is shifted towards the inactive “ open” form of the protein.
This form can be activated by a single autophosphorylation event.
The protein spends far more time in active open conformation.
Problem
1. Gleevec inhibits CML by binding to an stabilizing one of three forms
of Bcr-Abl, the closed inactive form. How does this stabilization
reduce Bcr-Abl activity with respect to LeChatelier’s principle?
Since the concentration of Bcr-Abl in the cell is relatively
constant, Gleevec binding results in decreasing the concentration
of active Bcr-Abl.
2. It has been discovered that Gleevec can inactivate another tyrosine
kinase, called c-kit. Gleevec mediated inactivation of c-kit differs
from that of bcr-abl. In this case Gleevec binds to the adeninebinding pocket between the large and small lobes of the kinase
domain. How would that binding affect c-kit function?
Since the adenine binding pocket is occupied by Gleevec, c-Kit can
no longer bind ATP and phosphorylate other molecules.
3. Please identify the four amino acids and the type of molecular
interactions between Gleevec and c-Kit.
640 Glu
670 Thr
673 Cys
810 Asp
4. Over time, cancerous cells evolve to become resistant to Gleevec. A
common Gleevec resistant mutant occurs where residue 670 mutates
to isoleucine. Please explain two ways in which this mutation may
reduce Gleevec activity?
Steric hindrance- mutation of Thr into Ile results ina larger
residue being present at that location. The larger size will reduce
Gleevec binding.
Loss of hydrogen bonding – the mutation will replace an amino acid
with polar sidechain for one with hydrophobic sidechain. This will
result in a loss of hydrogen bonding and increase in ethalpy.
Gleevec-resistant versions of Abl are still inhibited by Desatinib,
because the drug works by binding to the active conformation of Abl.
Conclusions
1. Protein kinases contain domains that fold in to conserved
structures.
2. Protein kinases catalyze phosphorylation through proximity,
orientation, and electronic effects.
3. Bcr-Abl is missing a critical component that controls activation of
the kinase, rendering the molecule constitutively active.
4. Gleevec binds the inactive, closed abl, shifting the equilibrium
towards this form.
5. Dasatinib binds the active form
6. Combination therapy may be a promising new therapeutic approach
for treatment of CML.
GOOD LUCK !!!!!