Adaptation

We will continue our discussion on adaptation from the last lecture. We want to answer the question of what causes the sharp decrease (response) in activity \(A\) due to an increase in ligand concentration \(l\).

Response Curve

The response curve plots the activity as a function of ligand concentration. In a system with adaptation the response curve will change as the ligand concentration in the environment changes. Can we describe the response function \(A(l)\)?

Statistical Mechanics Approach

Let’s consider a system of receptors at a fixed methylation level \(m\).

A single receptor can be in one of four states characterized by being (un)bound and being (in)active. Let \(a\) represent the activity of a receptor and \(b\) represent its boundedness.

\[\begin{align*} \{ a = 0, \ b = 0 \} &\equiv \text{inactive and unbound} \\ \{ a = 1, \ b = 0 \} &\equiv \text{active and unbound} \\ \{ a = 0, \ b = 1 \} &\equiv \text{inactive and bound} \\ \{ a = 1, \ b = 1 \} &\equiv \text{active and bound} \end{align*}\]

The reactions in our system look like the following

\[\text{R}+\text{L} \rightleftharpoons^{k_+}_{k_-} \text{C} \ \ .\] \[\begin{align*} \text{R} &\equiv \text{concentration of unbound receptors} \\ \text{L} &\equiv \text{concentration of ligand} \\ \text{C} &\equiv \text{bound receptor-ligand complex} \end{align*}\]

This yields the rate equation

\[\dot{c} = k_+ rl - k_- c \ \ .\]

Which has the steady-state solution

\[0 = k_+ rl - k_- c \ \ .\]

The concentration of receptors is assumed to be constant.

\[c + r = 1/V = \text{constant} \\ \Rightarrow k_- c = k_+ ( \tfrac{1}{V} - c)l\] \[cV = \frac{k_+ l}{k_+ l + k_-} = \frac{l}{l+K_i} \\ K_i \equiv \tfrac{k_-}{k_+} \\ \Rightarrow rV = \frac{K_i}{l+K_i}\]

What are the probabilities \(P_{ab}\) of the receptors being in the different states?

\[\frac{P_\text{bound}}{P_\text{unbound}} = \frac{P_{01}}{P_{00}}=\frac{c}{r} = \frac{l}{K_i}\]

Through an analogous argument we can get the ratio of probabilities for the active case (\(a=1\)).

\[\frac{P_{11}}{P_{10}} = \frac{l}{K_a}\]

In order to relate the probabilities for active and inactive receptors let

\[\frac{P_{10}}{P_{00}} = e^{f(m)}\]

describe the ratio between active and inactive states with and without ligand. The ratio is a function of the methylation level as indicated by the term \(f(m)\).

These ratios along with the normalization condition gives us four equations with four unknown variables.

\[\Rightarrow P_{00} = \left[ 1 + \frac{l}{K_i} + e^f \left( 1 + \frac{l}{K_a} \right) \right]^{-1}\]

We are interested in the marginal probability distribution \(P_a\) which is the probability of being in an active state regardless of boundedness.

\[P_a = \sum_b P_{ab}\] \[\text{let } \ \frac{P_{a=1}}{P_{a=0}} = \frac{P_{10}+P_{11}}{P_{00}+P_{01}} \equiv e^F\\ \Rightarrow F = F(m,l) = f(m) + \ln \left( \frac{1 + l/K_a}{1 + l/K_i} \right)\]

Let’s examine some limiting cases for \(F\).

\[\begin{align*} l = 0 &: F(m,l) \to f(m) > 0 \\ l = \infty &: F(m,l) \to f(m) + \ln \left(\tfrac{K_i}{K_f} \right) \\ &\ \ \ F(m,l) < 0 \end{align*}\]

Unfortunately, this simple model of activty does not reproduce experimental results. Bacteria receptors are not this simple so the model breaks down. One hypothesis for a more realistic model is that receptors are coupled to one another.

Monod, Wyman, Changeux (MWC) Model

In the MWC model receptors that are near one another can affect each other.

• Receptors are grouped into subclusters.
  • \(N\) is the number of receptors in one subcluster.
• All receptors in one subcluster must have the same activity state.

\[P_{A=0} + P_{A=1} = 1\] \[\frac{P_{A=1}}{P_{A=0}} = e^{NF} \\ \Rightarrow P_{A=1} = \frac{e^{NF}}{1+e^{NF}} \ , \ \ P_{A=0} = \frac{1}{1+e^{NF}}\]

We can measure the average activity.

\[\bar{A} = \sum_{A=0}^1 A P_A = P_1 = \frac{e^{NF}}{1+e^{NF}} = \left( 1+e^{NF} \right)^{-1}\] \[\text{recall, } \ F = f(m) + \ln \left( \frac{1 + l/K_a}{1 + l/K_i} \right) \\ \Rightarrow e^{NF} = \underbrace{e^{Nf(m)}}_{c(m)} \left( \frac{1 + l/K_a}{1 + l/K_i} \right)^N \\ \bar{A} = \frac{c (1+l/K_a)^N}{c(1+l/K_a)^N+(1+l/K_i)^N}\]

As we have done before, let’s examine the average activity in some limiting cases.

\[\begin{align*} l>>K_a>>K_i \ &: \ \bar{A} \to \frac{c}{c+(K_a/K_i)^N} \xrightarrow{c<<(K_a/K_i)^N} 0 \\ l << K_i << K_a \ &: \ \bar{A} \to \frac{c}{c+1} \xrightarrow{c>>1} 1 \end{align*}\]

Notice that the term \(c\) is a function of the methylation levels, and so the methylation level changes the mean activation behavior. \(\{K_a,K_i\}\) sets the range of \(l\) for the response in activity can vary from 0 to 1.

\[\frac{d\bar{A}}{dl} = - \left[1+e^{-NF}\right]^{-2}(-N)e^{-NF}\frac{dF}{dl} \\ \Rightarrow \frac{d\bar{A}}{dl} = -N\bar{A}(1-\bar{A})\frac{ K_a-K_i}{(l+K_a)(l+K_i)}\]

So the change in the mean activity depends on cooperativity through the number of receptors per subcluster. The term \(\bar{A}(1-\bar{A})\) is maximized at \(\bar{A} = \tfrac{1}{2}\), and \(K_a>>K_i\) maximizes the last term in the derivative. Therefore at those parameter regimes the activity is very sensitive to changes in the ligand concentration.

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