Past exam of the biological physics course of the University of Oxford / 2017 / 2 / b

In spatially gated myosin attachment, for a freshly available head encountering an isolated site, take $P=0$ before the attachment window. Converting time derivatives to position derivatives with $dx/dt=v$ gives $v{P}^{\prime }=k_{on}(1-P)$ in the binding window, $P^{\prime }=0$ in the protected bound region, and $v{P}^{\prime }=-k_{off}P$ after detachment becomes allowed. Continuity givesFor $v<k_{on}\Delta x$, the plateau exceeds $1-e^{-1}$ and approaches one for much slower motion. For $v>k_{on}\Delta x$, the binding probability is lower, becoming $k_{on}\Delta x/v$ at very high speed. The spatial detachment tail has length $v/k_{off}$, so faster sliding makes a longer low-probability tail. The following is an original calculation of those profiles; the force panel is derived in c.

Reset condition. Geometric separation $D\gg x_0$ does not alone guarantee detachment before the next site: independently available encounters also need $(D-x_0)k_{off}/v\gg 1$. Without that extra timescale separation the head may still be attached to an earlier site.

For completeness, the periodic steady-state correction can be written explicitly. Put $a=k_{on}/v$, $b=k_{off}/v$, $A_0=1-e^{-a\Delta x}$ andHere $Q_0$ is the probability of entering the new window still attached to an earlier site. In the window the new-site probability is $P(x)=1-e^{-ax}-Q_{0}a(e^{-bx}-e^{-ax})/(a-b)$; after it, use its value $p_b=P(\Delta x)$ in the plateau and tail above. For $a=b$, replace $a(e^{-bx}-e^{-ax})/(a-b)$ by $ax{e}^{-ax}$. This follows from the old-site probability $Q(x)=Q_0{e}^{-bx}$ and new attachment rate $a[1-P-Q]$. The isolated-site result is recovered as $Q_0\to 0$.