# Dynamics: Filament Polymerization

## Polymerization at Both Ends (cont.)

### Actin

Unlike microtubules, actin does not have the same critical concentration for both ends. The “$+$” end is the side where the filament grows and the “$-$” end is the side where the filament shrinks.

Plotting $\frac{dn_\pm}{dt}$ as a function of $c$ will illustrate a different phase space than that of microtubules. Assume that $% $, this gives us the following relationships between $c$ and $\frac{dn_\pm}{dt}$.

The three relationships above yield 3 different phases for actin.

$[1]$: Both ends shrink $[2]$: The $+$ end grows, $-$ end shrinks $[3]$: Both ends grow

For some value of $c$ within the regime of $[2]$ the rate at which the $+$ end grows equals the rate the $-$ end shrinks. This concentration is called the treadmilling concentration $c_T$.

For actin, $c_T\approx 0.16\mu M$. Both ends grows and shrink at the same rate resulting in the length of the actin remaining fixed. The system is in a steady state, however it is not in thermodynamic equilibrium because the filaments are constantly shifting. We can derive $c_T$ by noting that the total number of filaments $n$ remains constant.

## Problems with the Model

Our model, given by the rate equation has a couple of problems. $\text{Rate Equation: } \frac{dn}{dt} = \alpha-\mu$

• Nothing in this model says that $n$ cannot be negative.
• Physically, the number of filaments must be zero or positive.
• At small values of $n$ a continuous differential equation is no longer realistic. We can no longer approximate changes in $n$ as being continuous.

In order to try to solve these problems in our model we take a probabilistic approach. Let’s examine what may happen to a filament during small interval of time $\Delta t$ and their respective probabilities.

1. Add a filament: $\ \ \ \ \ \ \ \ p(n \rightarrow n+1) = \alpha\Delta t$
2. Subtract a filament: $\ p(n \rightarrow n-1) = \mu\Delta t$
3. Nothing happens: $\ \ \ \ p(n\rightarrow n) \ \ \ \ \ \ \ = 1 - (\alpha-\mu)\Delta t$

Using these relations we can formulate the probability that at some time $t+\Delta t$ the actin has $n$ filaments, $p_n(t+\Delta t)$. There are three different ways to get to a state with $n$ filaments: add 1 filament, subtract 1 filament, and the actin had $n$ filaments and nothing changed. Quantitatively this can be expressed as the following:

This can apply to all values of $n$ except for edge cases. For example, at $n=0$ the probability of having zero filaments is

We can re-write the two equations above to show how the probabilities vary as a function of time.

For $n>0: \ \ \dot{p}_n=\alpha \ p_{n-1}+ \mu \ p_{n+1} - (\alpha+\mu)p_n$

For $n=0: \ \ \dot{p}_0=\mu \ p_1-\alpha \ p_0$

The two above equations dictating the behavior of $\dot{p}_n$ are called the (stochastic) Master Equation.

## Expectation Value

We can calculate the expectation value (mean) of $n$ using the following formula:

The expression for $\frac{d\bar{n}}{dt}$ can be simplified by rewriting the different summation terms. By changing the summation index and by noting that for $n=0 \rightarrow n \ p_n=0$ we can make a few simplifications.

For the first summation term:

For the second summation term:

For the third dummation term:

This simplifies $\frac{d\bar{n}}{dt}$ to the following:

## If $% $, what happens at long times?

Let’s assume that as $t\rightarrow\infty$, the change in probabilities goes to zero, $\dot{p}_n\rightarrow0$. Using this assumption and the master equation we can find a recursion relation between the probabilities of different numbers of filaments.

For $n=0: p_1 = \frac{\alpha}{\mu}p_0$

For $n=1: p_2 = \frac{\alpha}{\mu}p_1 = \left(\frac{\alpha}{\mu}\right)^2p_0$

A pattern emerges, and we can therefore say that the probability of being in a state with $n$ filaments is given by

Let $\lambda \equiv \alpha/\mu$ so we can express the above statement as $p_n = \lambda^n \ p_0.$

In order for our probabilities to be normalized we enforce that all the probabilities must sum to one.

This is known as a geometric distribution, it is like a discrete version of an exponential distribution.

Now that we have a general form of $p_n$ we can calculate the mean of $n$.

This result is good because it shows that $\bar{n}$ does not become negative in this model.

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Written on February 21, 2015