Biological Physics: Energy, Information, Life

368 Chapter 10. Enzymes and molecular machines[[Student version, January 17, 2003]]

Beyond equilibrium The Boltzmann distribution only applies to systems at equilibrium. To
tacklenonequilibrium situations, begin with the “perfect ratchet” case (very large energy step).
Wealready encountered the perfect ratchet when deriving the estimate Equation 10.2. Thus, as
soon as a ratchet arrives at one of the steps in the energy landscape, it immediately falls down the
step and cannot return; the probabilityP(x)isthusnearly zero just to the left of each step, as
shown in Figure 10.12.

Your Turn 10c

Verify that the functionP(x)=A(e−(x−L)f/kBT−1) vanishes atx=L,solves the Smoluchowski

equation with the potentialVtot(x)=fx,and resembles the curve sketched in Figure 10.12

between0andL.(HereAis some positive constant.) Substitute into Equation 10.3 to find that

j(1d)(x)iseverywhere constant and positive.

Youhavejust verified Sullivan’s third claim (the loaded S-ratchet can indeed make net rightward
progress), in the limiting case of a perfect ratchet. The constantAshould be chosen to make
P(x)dxaproperly normalized probability distribution, but we won’t need its actual value. Outside
the region between 0 andL,wemakeP(x)periodic by just copying it (see Figure 10.12).
Let us find the average speedvof the perfect S-ratchet. First we need to think about what
vmeans. Figure 10.12 shows the distribution of positions attained by a large collection ofM
ratchets. Even though the populations at each position are assumed to be constant, there can be
anet motion, just as we found when studying quasi-steady diffusion in a thin tube (Section 4.6.1
on page 121). To find this net motion, we count how many ratchets in the collection are initially
located in a single period (0,L), then find the average time ∆tit takes for all of them to cross the
pointLfrom left to right, using the fluxj(1d)found in Equation 10.3:

∆t=(number)/(number/time) =

(∫L

0

dxMP(x)

)/(

j(1d)

)

. (10.5)

Then the average speedvis given by

v=L/∆t=

(

Lj(1d)

)/(

M

∫L

0

dxMP(x)

)

. (10.6)

The normalization constantAdrops out of this result (and so doesM).
Substituting the expressions in Your Turn 10c into Equation 10.5 gives

∆t=

1

Df/kBT

∫L

0

dx

(

e−(x−L)f/kBT− 1

)

,

or

v=

(fL

kBT

) (^2) D
L

(

efL/kBT− 1 −fL/kBT

)− 1

. speed of loaded, perfect S-ratchet (10.7)

Although our answer is a bit complicated, it does have one simple qualitative feature: It’s finite.
That is, even though we took a very large energy step (the perfect ratchet case), the ratchet has a
finite limiting speed.

Your Turn 10d

a. Show that in the case ofzeroexternal force Equation 10.7 reduces to 2D/L,agreeing with our

rough analysis of the unloaded perfect ratchet (Equation 10.2).