# Michaelis–Menten equation for multi-step reactions

I am trying to understand the following equation from the paper by Wang et al. [1, p. 5\ at the SI]:

### Michaelis–Menten Representation of the Kinesin Cycle

We assumed that there is a strong coupling between ATP turnover rate of kinesin and kinesin stepping on an MT. Each chemical step in the chemomechanical cycle of kinesin was assumed to be irreversible, except ATP binding and releasing (Fig. 1, step i). Consider the following scheme:

$$\large\ce{0 <->[k_\mathrm{ATP}\lbrack\ce{S}\rbrack][k_\mathrm{-ATP}] 1 ->[k_\mathrm{s}] 2 ->[k_\mathrm{attach}] 3 ->[k_\mathrm{ADP}] 4 ->[k_\mathrm{h}] 5 ->[k_\mathrm{-P}] 6}$$

Assume that $$F_i(t)$$ is the probability for the system to reach state $$6$$ if, at time $$t = 0$$, the system is in state $$i.$$ This function is governed by the backward master equations:

$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)\label{eqn:1}\tag{S1}$$

Why does equation \eqref{eqn:1} read as

$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)$$

and not

$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{-ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)?$$

I would like to change $$S(t)=\cos(t)$$ in time and see how it changes the states $$F_i(t)$$

### References

1. Wang, Q.; Diehl, M. R.; Jana, B.; Cheung, M. S.; Kolomeisky, A. B.; Onuchic, J. N. Molecular Origin of the Weak Susceptibility of Kinesin Velocity to Loads and Its Relation to the Collective Behavior of Kinesins. PNAS 2017, 114 (41), E8611–E8617. https://doi.org/10/gdj4h5.
• As pointed out in the comments, my answer was incorrect on some key points, so I've deleted it. I hope someone with better understanding will chime in. – Andrew 2 days ago