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How to satisfy Cycle (im)balance at all driving forces’ values and at all concentrations:

Lorin Milescu and Fred Sachs

If you don’t care about theory

  1. Make all the preexponential terms equal (or if you need to start with them unequal, make the product in one direction around the loop equal to the product in the reverse direction. In general, I find it much simpler to set each rate equal by default. This assures that at zero stimulus the loop is in balance
  2. If there is a concentration driven rate, then the concentration dependent terms must be equal in the forward and back directions.
  3. If the reaction has a stimulus driven rates, such as voltage, make the sum of the exponential dependencies of the rates in one direction (the Q terms of the rate constants) equal to the sum in the other direction.


More mathematically

First, we modify Eq. 25 in [1] to include an imbalance factor m (useful for some applications involving energy consumption such as an ATPase):

$$\prod q_{ij} = m \cdot \prod q_{ji} \rightarrow \;$$ $$\sum \ln q_{ij} = \ln m + \sum q_{ji} \rightarrow \;$$ $$\sum \left [ v_{ij} + \ln C_{ij} + k^1_{ij} \cdot V_{ij} \right ] = \ln m + \sum \left [ v_{ji} + \ln C_{ji} + k^1_{ji} \cdot V_{ji} \right ] \rightarrow \;$$ $$\sum v_{ij} + \sum k^1_{ij}\cdot V_{ij} - \sum v_{ji} - \sum k^1_{ji}\cdot V_{ji} = \ln m + \sum \ln C_{ji} - \sum \ln C_{ij}$$

Suppose there is a single driving force V, and a single ligand L (hence \(C_{ij}\) is either 1 if rate is not concentration dependent or [L] if concentration dependent):

$$\sum v_{ij} - \sum v_{ji} + V \cdot \left ( \sum k^1_{ij} - \sum k^1_{ji} \right ) = \ln m + \sum \ln C_{ji} - \sum \ln C_{ij}$$

Write it as (eliminate ln m for clarity):

$$x + V \cdot y = C$$

The above equation should be satisfied by all values of V, and by all values of C. Say we add another equation for a different value of V:

$$x + V_1 \cdot y = C, x + V_2 \cdot y = C$$

The above system is solved for \(y = \frac{C-C}{V_2-V_1} = 0\). Hence:

$$\sum k^1_{ij} - \sum k^1_{ji} = 0$$

In English: the charge moved clockwise should equal the charge moved counterclockwise. This constraint can be automatically enforced by the algorithm. The same can be done for C:

$$x + V \cdot y = C_1, x + V \cdot y = C_2$$

The system is solved for \(C_1 - C_2 = 0\). The only possible solution is \(C_1 = 0\) and \(C_2 = 0\). Hence:

$$\sum \ln C_{ji} = \sum \ln C_{ij}$$

In English: for each ligand, the number of clockwise concentration dependent rates should equal the number of counterclockwise concentration dependent rates. This constraint can be enforced by the QuB user, by specifying appropriately the concentration dependent rates.

References

Milescu,L., Akk,G., & Sachs,F. Maximum likelihood estimation of ion channel kinetics from macroscopic currents. Biophysical J. 88, 2494-2515 (2005).


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