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.

\[c^*_+ \neq c^*_-\]

Plotting \(\frac{dn_\pm}{dt}\) as a function of \(c\) will illustrate a different phase space than that of microtubules. Assume that \(c^*_+ < c^*_-\), this gives us the following relationships between \(c\) and \(\frac{dn_\pm}{dt}\).

\[\ \ \ \ \ \ \ \ \ \ c < c^*_+ \ \Rightarrow \ \frac{dn_+}{dt}<0 \text{ and } \frac{dn_-}{dt} < 0 \ \ [1]\\ c^*_+< c < c^*_- \ \Rightarrow \ \frac{dn_+}{dt}>0 \text{ and } \frac{dn_-}{dt} < 0 \ \ [2]\\ \ \ \ \ \ \ \ \ \ \ c < c^*_+ \ \Rightarrow \ \frac{dn_+}{dt} > 0 \text{ and } \frac{dn_-}{dt} > 0 \ \ [3]\]

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 \(c_T\).

\[c=c_T \Rightarrow \frac{dn_+}{dt} = \frac{dn_-}{dt}\]

Simple treadmilling animation

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.

\[\frac{dn}{dt} = 0 \\ \frac{dn}{dt} = \frac{d(n_++n_-)}{dt} = \frac{dn_+}{dt} + \frac{dn_-}{dt} \\ \Rightarrow \frac{dn_+}{dt} = - \frac{dn_-}{dt} \neq 0 \\ k_+ c_T - \mu_+ = -k_- c_T + \mu_- \\ \Rightarrow c_T = \frac{\mu_+ + \mu_-}{k_+ + k_-}\]

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:

\[p_n(t+\Delta t) = p_{n-1}(t)\alpha\Delta t+ p_{n-1}(t)\mu\Delta t + p_n(t)(1 -\alpha\Delta t-\mu\Delta t)\]

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

\[p_0(t+\Delta t) = p_1(t)\mu\Delta t + p_0(t)(1-\alpha\Delta t).\]

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

\[\frac{p_n(t+\Delta t) - p_n(t)}{\Delta t} = \dot{p}_n\\\]

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:

\[\left<n\right> \equiv \bar{n} = \sum_{n=0}^\infty n \ p_n \\ \frac{d\bar{n}}{dt}=\frac{d}{dt} \sum_{n=0}^\infty n \ p_n = 0 \cdot\frac{dp_0}{dt}+\sum_{n=1}^\infty n \ \frac{p_n}{dt} \\ \frac{d\bar{n}}{dt}=\alpha\sum_{n=1}^\infty n\ p_{n-1} + \mu\sum_{n=1}^\infty n\ p_{n+1} -(\alpha+\mu)\sum_{n=1}^\infty n \ p_n\]

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:

\[\sum_{n=1}^\infty n\ p_{n-1} \rightarrow \sum_{n'=0}^\infty(n'+1)p_{n'} = \sum_{n'=0}^\infty n' \ p_{n'} + \sum_{n'=0}^\infty p_{n'} \\ = \bar{n} + 1\]

For the second summation term:

\[\sum_{n=1}^\infty n\ p_{n+1} \rightarrow \sum_{n=0}^\infty n\ p_{n+1} \rightarrow \sum_{n'=1}^\infty (n'-1)p_{n'} \\ = \sum_{n'=1}^\infty n'\ p_{n'} - \sum_{n'=1}^\infty p_{n'} \\ = \bar{n} - (1-p_0)\]

For the third dummation term:

\[\sum_{n=1}^\infty n \ p_n \rightarrow \sum_{n=0}^\infty n \ p_n \\ = \bar{n}\]

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

\[\frac{d\bar{n}}{dt} = \alpha\ \bar{n}+\alpha+\mu\ \bar{n}-\mu+\mu \ p_0 -\alpha \ \bar{n}- \mu \ \bar{n} \\ \Rightarrow \alpha-\mu+\mu \ p_0\]

If \(\alpha<\mu\), 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

\[p_n = \left(\frac{\alpha}{\mu}\right)^n \ p_0.\]

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.

\[1 = \sum_{n=0}^\infty \lambda^n \ p_n = p_0 \ \sum_{n=0}^\infty \lambda^n = p_0 \frac{1}{1-\lambda} \\ \Rightarrow p_0 = 1-\lambda \\ p_n = (1-\lambda) \ \lambda^n\]

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 \(n\).

\[\bar{n} = \sum_{n=0}^\infty n(1-\lambda)\lambda^n \\ \bar{n} = (1-\lambda)\sum_{n=0}^\infty n \ \lambda^n = (1-\lambda)\lambda \ \partial_\lambda \sum_{n=0}^\infty \lambda^n \\ \bar{n} = (1-\lambda)\lambda \ \partial_\lambda \left(\frac{1}{1-\lambda}\right) \\ \Rightarrow \bar{n}(t\rightarrow\infty) = \frac{\lambda}{1-\lambda} \\ \bar{n}(t\rightarrow\infty) = \frac{\alpha}{\mu-\alpha} > 0\]

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

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