Raf-1-Removal

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Binding Reaction

$\text{Raf1} + \text{activePKA} \rightleftharpoons \text{removedRaf1} + \text{activePKA}$

Kinetic Equation

$\frac{k_{cat} \cdot [\text{activePKA}] \cdot [\text{Raf1}] \cdot (1-\frac{[\text{removedRaf1}]}{[\text{Raf1}] \cdot K_{eq}})}{Km_{\text{Raf1}}\cdot(1 + \frac{[\text{Raf1}]}{Km_{\text{Raf1}}} + \frac{[\text{removedRaf1}]}{Km_{\text{removedRaf1}}})}$

final Parameter

Because only a subunit was measured the standard deviation is increased by 1.5.

$k_{cat} = 1.92475 \pm 0.64133 \cdot 1.5\; \frac{1}{s} = 1.92475 \pm 0.962\; \frac{1}{s}$

$K_{m} = 42.7 \pm 8.984 \cdot 1.5\; \mu M = 42.7 \pm 13.4761\; \mu M$

Parameter

Calculation	Reference
The calculation of the vmax values that have the right unity is as follows: $1 Da = 1.66 \cdot 10^{-27} kg = 1.66 \cdot 10^{-21}\ mg$ $m(C \gamma) = 41000 Da = 6.806 \cdot 10^{-17}\ mg$ Failed to parse (lexing error): \text{#enzyme per mg} = \frac{1 mg}{m(C \gamma)} = 1.469 \cdot 10^{16} Failed to parse (lexing error): n(\text{enzyme per mg}) = \frac{\text{#enzyme/mg}}{6.023\cdot 10^{17}}= 0.024395\ \mu mol $v_{max}(1/min) = v_{max}(\frac{\mu mol}{min\cdot n(\text{enzyme})})$ $v_{max}(1/s)=\frac{v_{max}(1/min)}{60}$ $m(C \alpha) = 42000 Da = 6.972 \cdot 10^{-17}\ mg$ Failed to parse (lexing error): \text{#enzyme per mg} = \frac{1 mg}{m(C \gamma)} = 1.4343 \cdot 10^{16} Failed to parse (lexing error): n(\text{enzyme per mg}) = \frac{\text{#enzyme/mg}}{6.023\cdot 10^{17}}= 0.023813\ \mu mol $v_{max}(1/min) = v_{max}(\frac{\mu mol}{min\cdot n(\text{enzyme})})$ $v_{max}(1/s)=\frac{v_{max}(1/min)}{60}$ This calculation results in: Cα $v_{max}(\text{Kemptide}) = 4.829 \pm 1.120\ s^{-1}$ $v_{max}(\text{Histone}) = 2.1 \pm 0.84\ s^{-1}$ Cγ $v_{max}(\text{Kemptide}) = 0.20 \pm 0.07\ s^{-1}$ $v_{max}(\text{Histone}) = 0.57 \pm 0.20\ s^{-1}$ $k_{cat}\text{(average}) = \frac{4.829 + 2.1 + 0.20 + 0.57}{4}\; \frac{1}{s} = 1.92475\; \frac{1}{s}$ $\sigma_{k_{cat}(\text{average})} = \frac{\frac{1.120}{4.829} + \frac{0.84}{2.1} + \frac{0.07}{0.20} + \frac{0.20}{0.57}}{4} \cdot 1.92475 \frac{1}{s} = 0.64133\; \frac{1}{s}$	Zhang W. et al.(2004)

Km_P value and equilibrium constant

The K_m value for the product is assumed to be similar but slightly higher than the the K_m value of the substrate because of the similarity of the both species. Therefore the K_m value of the substrate is multiplied by 1.05 to gain the one of the product and the uncertainty is increased by increasing the error on Km_substrate by 50%.

For information about the equilibrium constant please see here.