Sos-Binding

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Reaction

$\text{inactiveSos} + \text{activeEGFR} \rightleftharpoons \text{activeSos}$

Rate Equation

$v_{\text{Sos-binding}} = k_{\text{Sos-binding}} \cdot [\text{activeEGFR}][\text{inactiveSos}] - k_{\text{Sos-release}} \cdot [\text{activeSos}]$

General Information about the Mechanism

Sos is a guanine nucleotide release protein (GNRP), which binds via its C-terminal SH3 domain to Grb2 . Grb2 has two SH3 domains, but only the N-terminal one is included in the Sos binding process. The binding occures in a 2:1 stoichiometry. The complex then binds via the SH2 domain of Grb2 to the autophosphorylated EGFR. This binding might include a further protein, which is phosphorylated by the active EGFR, Shc. Through to the binding Sos is recruited to the membrane and can therefore activate Ras, the first enzyme of the MAP-kinase. Because of this process and further experiments ,which showed that the activation of Sos is rather through a recruitment to the membrane than to an enzymatic activation we decided to model this activation process as binding, whereas Xu et al modelled it as Michealis-Menten. One reason to model the activation as Michaelis-Menten kinetic might be the findings of Ichiba et al, that the negative influence of the termini on the catalytic domain of Sos is repressed by various mechanisms like, phosphorylation, calcium binding or diacylglycerol binding.

final Parameter

The errors of the final parameter are increased by 50% because only a similar reaction is measured.

$k_{\text{Sos-binding}}= 1.47 \cdot 10^8 \pm 4.0873 \cdot 10^7 \cdot 1.5\ \frac{1}{M \cdot s} = 1.47 \cdot 10^2 \pm 6.1310 \cdot 10^1\ \frac{1}{\mu M \cdot s}$

$k_{\text{Sos-release}} = 0.1488 \pm 2.643 \cdot 10^{-2} \cdot 1.5\ \frac{1}{s} = 0.1488 \pm 3.965 \cdot 10^{-2}\ \frac{1}{s}$

Parameter

Notes	References
$k_{\text{Sos-association}}= 3.34 * 10^6\ \frac{1}{Ms}$ $k_{\text{Sos-release}}= 0.14\ \frac{1}{s}$ $K_D = 42\ nM$ The binding of the p85αSH2-N domain to 17-mer phosphopeptide similar to a PDGFR region. (Hensmann et al. 1994)
The binding of the two different SH2 domains to phosphotyrosine containing peptides derived from an insuline receptor substrate.	Felder et al. (1993)

Parameter Calculation

In general it can be said that there is no direct binding between Sos and the EGFR. "Sos is constitutivly associated with the SH3 domain of Grb2", so the binding process modelled here resembles more the binding of the Sos-Grb2 complex to the EGFR. Because of the reason that "Grb2 consists of (only one)[...] SH2 domain", only the data derived from the binding of one SH2 domain to the peptide is used for the calculation.

$k_{\text{association}}=\frac{1}{5} \cdot (3.34 \cdot 10^6\ \frac{1}{Ms} + 0.48 \cdot 10^8\ \frac{1}{Ms} + 4.4 \cdot 10^8\ \frac{1}{Ms} + 0.33 \cdot 10^8\ \frac{1}{Ms} + 2.1 \cdot 10^8\ \frac{1}{Ms}) = 1.47 \cdot 10^8\ \frac{1}{Ms}$

$k_{\text{release}}=\frac{1}{5} \cdot (0.14\ \frac{1}{s} + 0.156\ \frac{1}{s} + 0.149\ \frac{1}{s} + 1.113\ \frac{1}{s} + 0.186\ \frac{1}{s}) = 0.1488\ \frac{1}{s}$

For the uncertainty estimation of the k_release the standard deviation of the five values (0.14 1/s, 0.156 1/s, 0.149 1/s, 0.113 1/s, 0.186 1/s) is calculated.

Then the relative error on the mean is calculated $\frac{2.643 \cdot 10^{-2}\ \frac{1}{s}}{0.1488\ \frac{1}{s}} = 1.776 \cdot 10^{-1}$. This value is assumed to be the relative error of each single measurement. Taking the measured K_D error into account the relative error on each single k_on value can be calculated. To finally estimate the uncertainty on k_on the average of these four values is taken and assumed as relative error on the calculated k_association value.

$\sigma_{k_{\text{association}}}\frac{0.1226 + 0.3224 + 0.1782 + 0.489}{4} \cdot 1.47 \cdot 10^8 \frac{1}{M \cdot s} = 4.0873 \frac{1}{M \cdot s}$

The value measured by (Hensmann et al. 1994) is not included in the uncertainty estimation.

KD value [nM]	error KD [nM]	relative error KD	error calculation kon	relative error kon
$3.27$	$0.18$	$\frac{0.18}{3.27} = 5.5 \cdot 10^{-2}$	$5.5 \cdot 10^{-2} - 1.776 \cdot 10^{-1}$	$1.226 \cdot 10^{-1}$
$0.34$	$0.17$	$\frac{0.17}{0.34} = 5.5$	$5.5 - 1.776 \cdot 10^{-1}$	$3.224 \cdot 10^{-1}$
$3.40$	$1.21$	$\frac{1.21}{3.40} = 3.56 \cdot 10^{-1}$	$3.56 \cdot 10^{-1} - 1.776 \cdot 10^{-1}$	$1.782 \cdot 10^{-1}$
$0.90$	$0.60$	$\frac{0.60}{0.90} = 6.67 \cdot 10^{-1}$	$6.67 \cdot 10^{-1} - 1.776 \cdot 10^{-1}$	$4.890 \cdot 10^{-1}$