How Does Motor Stepping Affect Spring Energy in a Processive Molecular System?

[andybham]

I am doing a project on tightly-coupled processive molecular motors. My 1D model consists of motors which literally step discrete step sizes of size U while executing a biased random walk along a filament which moves along its longitudinal axis with negligible drag. The motors are attached to a spring which is anchored at one end.

If you can imagine the motor has legs and is attached to a spring (~) of stiffness k.

~~k~~~~/\

I can't draw the springs but if you can imagine them attached to the motors like the one drawn above. Then a two motor system looks a bit like this...

-x dir ____________/\_____________/\___________ +x dir
filament

One property of this system is that the filament moves at a constant velocity and so it must have zero force on it at all times.

So if one motor steps along, the filament and the motors move in the -x direction by u/n ( where n is the number of motors),

So for 2 motors if any of the motors steps forward, then the filament has moves back u/2, the motor which has stepped has increased the extension in its spring by u/2 and the motor which has stayed still has compressed its spring by u/2 hence the total force on the filament is zero.

To generalise this to n motors, for any motor that takes a forward step,
1) the filament moves back u/n
2)the motor that has stepped has increased the extension in its spring by U(n-1)/n
3)the motor that didnt step decreased the extension in its spring by u/n

Now my question is...

What is the change in the energy in the spring when a motor steps?

The change in the energy of the motor that HASN'T stepped:
If the initial extension in its spring is xj then the final extension is xj-u/n
then the change in energy dE

Efinal - Einitial = sumj k/2(xj-U/n) - sumj k/2(xj)^2 right? (sumj = sum of all xj)

which simplifies down to sumj = k/2((U^2)/(n^2) - 2*xj*U/n)

The change in the energy of the motor that HAS stepped:

if initial extension is xi
then

dE = Efinal - Einitial = sumi k/2(xi+U-U/n)^2 - sumi k/2*(xi)^2

which I THINK simplifies down to

dE = sumi k/2*(2*xi*U*(n-1)/n+(U^2)/(n^2)(1-2n+n^2)

The trouble is I don't think this works!

AND

Do I need to sum the extensions of all the springs? or is it just the extensions of the springs of the motors that haven't stepped which goes into sumi and those which did step goes into sumj?

 

[Navin A S]

I think that the sumi should just include the extensions of the springs of the motors which haven't stepped and the sumj should include only the extensions of the springs of the motors which did step but I am not sure if this is correct...Any help would be greatly appreciated!

 

[ravenprp]

First of all, great job on your project! It seems like you have a good understanding of the mechanics involved in your 1D model.

To answer your question, the change in energy in the springs will depend on which motor steps. Let's break down the two cases:

1) When a motor steps and the filament moves back by u/n: In this case, the motor that stepped has increased the extension in its spring by u(n-1)/n, while the motor that didn't step has decreased the extension in its spring by u/n. So the change in energy in the springs would be:

dE = k/2 * ((U(n-1)/n)^2 - (u/n)^2) = k/2 * (U^2(n^2-2n+1)/n^2 - U^2/n^2) = k/2 * (U^2(n-1)^2/n^2)

2) When a motor doesn't step and the filament moves forward by u/n: In this case, the motor that didn't step has decreased the extension in its spring by u(n-1)/n, while the motor that stepped has increased the extension in its spring by u/n. So the change in energy in the springs would be:

dE = k/2 * ((-U(n-1)/n)^2 - (u/n)^2) = k/2 * (U^2(n^2-2n+1)/n^2 - U^2/n^2) = k/2 * (-U^2(n-1)^2/n^2)

As you can see, the change in energy for the two cases is just the same but with opposite signs. This makes sense because in both cases, the total change in energy in the springs should be the same.

To answer your second question, you only need to sum the extensions of the springs of the motors that have stepped (sumi) and the motors that haven't stepped (sumj). This is because the change in energy in the springs only depends on the extensions of the springs of the motors involved in the stepping process.

I hope this helps clarify your doubts. Keep up the good work on your project!