Signaling

Ras - MAP kinase signaling

MEK

GAP RasGTP

RasGDP

ERK

Raf

PP

PP P

P P

Shc

SosGrb2

Receptor tyrosine kinase

M

FosP P

Elk-1PP

Stat 3P

Growth Factors (EGFR)Stimuli

RafMAPK Kinase Kinase

MEK 1,2MAPK kinase

Erk 1,2MAPK

Proliferation/growth/differentiation Response

ProblemATP 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 Kdvalue 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.

ProblemWhen 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 Arg2. Residue 119 is mutated to

Leu3. 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 phosphorylationcascades.

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.

ProblemUpon 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?

ProblemThe 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.

ProblemsOne 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

Raf

Mek 1,2

Erk 1,2P

P

P

Erk 1,2 P

Problem1. Bacterial pathogens modulate activation of MAPK in a way to escape anti-bacterial 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, anti-phospho-erk, anti-erk and anti-mek.

While the western blotting analysis demonstrates activation of MEK by detecting the presence of phospho-MEK, the downstream activation of erkis absent in infected cells. Thus, bacterial infection stops erk activation.

erk P-erk2. 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 non-stimulated 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 Ablhave homology in the kinase domain, and also in two domains, called Srchomology 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.

Problem1. 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 adenine-binding pocket between the large and small lobes of the kinasedomain. 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 Glu670 Thr673 Cys810 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 form6. Combination therapy may be a promising new therapeutic approach

for treatment of CML.

GOOD LUCK !!!!!