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

A skeletal-muscle fibre contains parallel myofibrils assembled from repeating sarcomeres between Z discs. Thin actin filaments extend from the Z discs; bipolar thick myosin II filaments occupy the central region. Overlap between them allows many myosin heads to interact with actin in parallel.

A rise in cytosolic calcium exposes actin binding sites through the troponin–tropomyosin regulatory system. ATP binding releases a strongly attached myosin head; hydrolysis primes its lever arm, attachment and product release permit a force-generating stroke, and another ATP allows detachment. The cycle converts chemical free energy into mechanical work rather than consuming actin or shortening the filaments themselves.

Relative filament sliding draws the Z discs together. The I band and H zone shorten as overlap increases, whereas the thick-filament A-band length is approximately unchanged. Calcium removal restores inhibition and permits relaxation. The many-head ensemble sustains force even though individual cross-bridges repeatedly attach and detach. This sliding-filament description concerns myosin II in muscle; the processive two-head walking of Myosin V in a is a related but distinct motor arrangement.

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 $Q_0\to 0$.

For the work of a spatially gated myosin head, the mechanical force exerted by an attached head in the increasing-$x$ direction is $f(x)=-dU/dx=\kappa (x_0-x)$. It assists motion before $x_0$ and resists it afterwards. For one site encounter the expected work isIn the narrow binding-window limit, retain the capture probability $p_b=1-e^{-k_{on}\Delta x/v}$ but neglect the attachment displacement compared with $x_0$. The positive work is $p_b\kappa x_0^2/2$. After $x_0$, use $s=x-x_0$ to obtain the negative contribution $-p_b\kappa {\int }_0^{\infty }s{e}^{-k_{off}s/v}ds=-p_b\kappa (v/k_{off}{)}^2$. HenceAn independent energy calculation gives the same factor: binding near zero loads energy $\kappa x_0^2/2$, while an exponentially distributed post-threshold attachment time has $⟨t^2⟩=2/k_{off}^2$, leaving mean elastic energy $\kappa v^2/k_{off}^2$ to be lost upon detachment. The factor of two in that second moment is essential.

One site passes every distance $D$, so average work per distance, or power divided by $v$, givesThe force tends to $\kappa x_0^2/(2D)$ as $v\downarrow 0$ and changes sign at $v=k_{off}{x}_0/\sqrt{2}$ in this idealized model. Above that speed forced sliding is resisted on average. The formula uses the reset condition in b, as well as $\Delta x\ll x_0$; if reset fails, replace $p_b$ by the periodic capture probability to leading order in $\Delta x/x_0$. For arbitrary window width, the isolated-site work is exactly ${\int }_0^{\Delta x}(k_{on}/v)e^{-k_{on}x/v}[\kappa (x_0-x{)}^2/2-\kappa v^2/k_{off}^2]dx$. The loaded spring energy must be supplied by the binding/power-stroke chemistry; this model does not specify an ATP efficiency. Real force-dependent transition rates and energy budgets limit extrapolation of the toy force-velocity law.