1 Nanomotors

In this part of the course we will study nanomotors. First we will define what we mean by nanomotor. A nanomotor is a machine that is only nanometres or 10s of nanometres across. By machine I mean something that either moves in a controlled directed fashion (i.e., not randomly), or exerts forces. Here we will mainly be considering nanomotors that move things around. These nanomotors are like cars, they are machines for transporting things from A to B. But whereas a car is macroscopic, it is metres long, nanomotors are a billion times smaller. As atoms are typically a few tenths of a nm across, nanomotors are only tens or hundreds of atoms across. Scientists working on nanotechnology are working to make nanomotors, but so far they have only made very crude nanomotors. However, all living organisms rely totally on a large numbers of nanomotors. Our bodies are made of cells and each cells contain many nanomotors, a muscle cell can contain a billion nanomotors. Muscles can only exert forces because the cells of which they are composed can exert forces and the cells in turn can only exert forces because of the billion nanomotors they contain. As these nanomotors are (protein) molecules, they are also called molecular motors. Just as cars and trucks are there to transport people and goods around, say oranges from a Tesco’s depot to a Tesco’s superstore, many molecular motors in our cells are there to transport stuff (proteins etc) around a cell. Now, there is an obvious difference between a macroscopic object, such as an orange, and a molecule in solution. Oranges just sit there, whereas molecules diffuse around inside liquids. If a molecule is at the origin at time t = 0, we have found that at time t, the square root of the mean of the square of the distance of the molecule from the origin is

2 1/2 1/2 1/2 hr i
= (6D) t diffusion (1) because the molecule diffuses. Here D is the diffusion constant. So, an obvious question is: If molecules move via diffusion anyway, why do you need molecular motors to move them? The answer is that diffusion has no direction, you are as likely to go left as right, and so if a molecules needs to moved in a specific direction, to say a specific part of the cell, then diffusion is not adequate. Also, as we noted with diffusion in the atmosphere, the distance travelled increases only as the square root of time (as opposed to being linear in time as it is for motion in a straight line at constant speed) and so motion over large distances is very slow via diffusion.

1.1 Molecular motors and the 2nd Law of Thermodynamics To see how the 2nd Law of Thermodynamics (=the entropy can never decrease) applies to molecular motors, consider a single motor. The motors inside cells move along ‘railtracks’ inside cells, these are long (micrometres long or more) thin (∼ 10 nm) filaments, that criss-cross the cell. One type of filament, called a microtubule, is illustrated in Fig. 1. Note that it is made of a regular periodic array of molecules, it is essentially a one-dimensional crystal. Let us consider a motor moving in one dimension along one periodic row of pairs of the molecules, as shown in Fig. 1. The motor can bind to an array of positions along the row, one every 8 nm for this type of filament1. If a filament is, for example, 8 µm long this means the motor can bind to it at 1,000 different positions. If we just have a filament and a motor that is not consuming any fuel then the system will go to thermo- dynamic equilibrium where the entropy is a maximum (2nd Law). We know that the entropy is a maximum when all possible states, 1,000 of them here, are equally likely. Then the probability of being at any position is 1 −3 p(x)= = 10 (2) 103 1There are 13 rows in a filament and a motor can move from row to row quite easily, but for simplicity we neglect this. Including it just multiplies the numberof positions by 13.

1 The entropy S is then 103 103 −3 −3 3 S = − X pi ln pi = − X 10 ln 10 = ln 10 = 6.9. (3) i=1 x=1 where we label the binding sites from left to right as i = 1, 2,.... Now, molecular motors move in one direction. Let this motion be to the right. If one of the proteins is attached to a molecular motor, then it will tend to move to the right. Ultimately, it will end up as far right as it can be, i.e., it will be at the rightmost position for binding. This is a single binding site and so the probability of it being there, p = 1, while the probability of it being anywhere else, p = 0. 103 S = − X pi ln pi = −1ln1=0. (4) i=1 The entropy of the protein has, due to the directed motion of the molecular motor, decreased from 6.9 to 0. Of course, the total entropy must not decrease, and so the molecular motor must do work. This is just like a fridge. This pumps heat from a cold environment (inside the fridge) to a warmer environment (the surrounding room) and it cannot just do this as if it did so it would break the 2nd Law. Thus, all nanomotors must in addition to moving, generate entropy by producing heat. To produce this heat they will have to burn a chemical fuel of some sort. The molecular motors in our body burn a molecule called ATP, this powers a chemical reaction which allows them to move to one place, hence reducing the entropy due to the uncertainty in the position, In the above example, as the motion changed the entropy due to the position of the molecule, that of Eq. (4), by −6.9, the motor can compensate by producing δq = 6.9kT ≃ 2.8 × 10−20J of heat. This is for body temperature, for which kT ≃ 4 × 10−21J. This amount of heat produces δS = δq/kT = 9.2 of entropy

Figure 1: Schematic showing two types of molecular motor, dynein and kinesin, and one type of filament, called a microtubule (i.e., one type of the ‘railtrack’ inside cells that motors run along). I don’t expect you to remember the names of these motors and filaments. However note that the filament is made of rows of molecules arranged in a helical fashion and that each row is made of a repeated (i.e., periodic) sequence of pairs of molecules. The two parts of the pair are coloured in different colours (green and blue, or in B&W photocopy light and dark grey). The microtubule filament is made of a helix 13 rows of these pairs of molecules. The period is about 8 nm and so a motor can bind at a whole sequence of positions along a microtubule that are 8 nm apart.

2 u(x) 3) diffusion 7) M

M to M* 2) x 4) 6) u(x) M* to M 8) M*

5) 1)

l x

Figure 2: Schematic of the potentials (dashed lines) as a function of x, u(x), for both states of the molecular motor: M (top) and M ∗ (bottom). The motor itself is indicated by the black circle. The dotted lines indicate the motor going from state M to M ∗ or vice versa, and it either diffusing, in state M, or moving to the bottom of the potential well, in state M ∗. The progression of the motor shown is: 1) starts in state M ∗ in leftmost potential well, 2) M ∗ → M, 3) diffuses (by chance to right), 4) M → M ∗, 5) moves to the bottom of the potential that is second from the left, 6) M ∗ → M, 7) diffuses to the left, 8) moves back to the bottom of the potential that is second from the left. and so the total entropy is then zero — which is as low as it is possible to go. Any lower and the total entropy change is negative. In molecular motors this δq comes from burning molecules like ATP, which produces heat.

1.2 A toy model for nanomotors: The Brownian Ratchet Real molecular motors are complex and poorly understood. It is difficult to experimentally work out how they function. Light microscopy, for example, is useless as the motors are much less than the wavelength of light across. However, there is a simple model that illustrates how motors can exploit diffusion to move in a directed way, by burning a fuel. This is the Brownian ratchet, which goes back to an idea of Feynman’s in the 1960s. It is also called a diffusive ratchet. The idea is to use a chemical reaction (molecular motors burn ATP as a fuel), that takes the motor from one state, call it M, to another, call it M ∗, and back in order to rectify diffusion. By rectify I mean allow motion in one direction, we will take this to be to the right, while preventing it in the other direction, to the left. Our Brownian ratchet only moves in 1 dimension, which is realistic as the molecular motors move up and down filaments inside cell. This rectification is like a rectifier in an electric circuit, which only allows a current to flow in 1 direction. So, the motor is always restricted to move along the x axis (in practice because it is bound to a filament). However, in state M it can freely diffuse along the x axis, whereas in state M ∗ it feels a sawtooth potential

3 u(x). See Fig. 2 for schematics of the 2 potentials. By freely diffuse we mean that the potential u(x) the motor feels in state M is a constant so there are no forces on it. In state M it has a diffusion constant D. Also, it is important to note that the sawtooth potential the molecular motor feels in state M ∗ is highly asymmetric, as you go from left to right the potential gradually drops, over a distance l, then suddenly increases again. It is this asymmetry that is going to rectify the motion of the motor. We start with the motor in state M ∗ and at the bottom of the potential. The potential well is assumed to be deep, i.e., deeper than the thermal energy kT . Therefore, in state M ∗ the molecule quickly heads towards the minimum and stays there. Then, if we start in state M ∗, the sequence of events that occurs is:

1. A motor in state M ∗ stays in that state only for a time τ ∗, on average, before converting to state M, i.e., M ∗ → M. So after a time τ ∗ the motor is in state M.

2. In state M the potential is flat, so it freely diffuses with diffusion constant D. Thus, t seconds after it flipped to state M, it will have diffused a distance of about (Dt)1/2 — it is equally likely to be in either direction.

3. The motor stays in state M for only a short period of time, on average in stays only for a time τ. Now, we assume that τ is so small that (Dτ)1/2 ≪ l, i.e., in the time the motor is in the state M in which it can freely diffuse it can only diffuse a distance much less than the period of the sawtooth potential in state M ∗.

4. After a time τ the motor returns to state M ∗. It has either diffused to the left or to the right, each is equally likely, i.e., each has a probability of 1/2. If it has diffused to the left it has only moved a little way, ≪ l, to the left, and so on its return to the M ∗ state it will just slide down to the bottom of the same well it was at the beginning: it has not gone backwards. However, if it has diffused to the right, it will have gone over the top of the sawtooth potential in the potential and so when it returns to the state M ∗ it is in the next valley along. It will then slide down to the next minimum in the potential to the right, and so will have moved forward a distance l to the right.

5. The motor is now in state M ∗ at the bottom of the potential well (either the same one as before or the next one to the right). The motor is now ready to start another sequence 1) to 4).

After sufficient time has elapsed that the sequence of M to M ∗ and back to M has been repeated many times, the net effect is that during each M ∗ → M → M ∗ cycle which takes on average a time τ ∗ + τ, there is a 50% probability that the motor has moved l to the right and a 50% probability that the motor has stayed where it was. It will not have moved to the left: the sawtooth potential has rectified the diffusional motion. As it moves a distance l with 50% probability every τ ∗ + τ seconds the average velocity is 1 l average velocity = . (5) 2 τ ∗ + τ Our expression for the velocity is half the period of the potential, divided by the sums of the times in each state, τ ∗ and τ. In cells, the motors run along two types of filament, one of these types is called a microtubule. It has a rough surface with a periodicity of 8nm. Thus we know l, it is of order 10nm, to within the approximations we are currently using. Also, motor speeds have been measured and they are typically about 1000 nm s−1 or a little less. The final fact is that motors burn one molecule of ATP per 10 nm step and that burning a single ATP molecule releases about 10−19J of energy. Let us see whether these observations make sense in terms of our model. A velocity of 1000 nm s−1 implies that 1 10 nm step takes 10−2s. Thus τ + τ ∗ cannot be more that 10−2s. Let us consider these 2 times separately, first τ. This is the time taken to diffuse over a peak in the potential but not too far (to avoid it

4 going backwards). We expect the peak to be 1 to 2 nm wide, i.e., larger than an atom but less than 10 nm of course. The diffusion constant for a protein is given approximately by the Stokes-Einstein expression kT D = (6) 6πaη where a is the radius of the protein and η is the viscosity of the inside of a cell. Motors are quite big proteins so we take a = 10nm. The viscosity of water η ≃ 10−3Pa s. However, cells are full of proteins and other molecules so to a protein they feel around 100 times thicker than water, so we take a viscosity η = 0.1Pa s, inside a cell. Thus we have a diffusion constant D ≃ 10−13m2s−1 = 105nm2s−1. Also, note that the term on the bottom, 6πaη is the drag coefficient, i.e., if a protein molecule is dragged through the cell at a velocity v, the drag on the particle is

−8 drag force = 6πaηv ≃ 10 v N. (7)

Now, we return to τ the time the motor takes to diffuse approximately 1 nm when it is in the freely diffusing state M. The distance diffused is approximately (Dt)1/2 ≃ (105t)1/2nm for a protein. This distance equals 1nm when t = 10−5s. We know that on average 1 step takes about 10−2s. Therefore, diffusion over the required short distance is easily fast enough to give us the velocities measured in experiment. Now, let us consider the other time, τ ∗. This is the time it takes to fall down the potential well in state M ∗. Going to the bottom of the well involves being pulled a distance of order l, i.e., 10 nm. This is in the thick viscous environment of the cell and so it involves doing work against friction. As we have only burned one ATP molecule we can do no more than 10−19J of work. The work done is just force times distance, and we know the force as a function of the velocity v that we move to the bottom of the potential well. So, we have that −16 work done against friction = force × l = 6πaηvl ≃ 10 v J. (8) This increases with v of course, for a v = 1000nm s−1, the work done is 10−22J. This is much less than the work that can be done with a single ATP molecule. So, we have found that our model is consistent with the experimental data. When we put experimentally measured values for the parameters such as protein size, ATP energy etc., then we found that the maximum velocity within the model was more than the velocity measured in experiment. If v were 1000 times larger then the drag would be 1000 times larger, as it increases linearly with v, and so 10−19 not 10−22J would be consumed. This would be the limit to how fast the motor could move. Motors are not observed to move this fast, however, they are observed to move cargoes, pulling cargoes along will increase the drag. The drag increases with the radius, so if a motor pulls a cargo about 100 times its size, i.e., 1000 nm, at about 1000 nm s−1, then it burns 10−21. This is about 10% of the maximum enegrgy in ATP. So, then it is working at about 10% efficiency, which is comparable to the efficiency of an internal combustion engine. (Of course, as the cargo is so small, its weight is not an issue.)

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