Binding Affinity Modelling
Law of Mass Action
When modelling chemical reactions, we used the law of mass action which explains and predicts behaviours of solutions in dynamic equilibrium. The basis for this law stems from the research conducted by Cato M. Guldberg and Peter Waage in the mid 18 hundreds. Guldberg and Waage recognized that chemical equilibrium is a dynamic process in which rates of reaction for the forward and backward reactions must be equal at chemical equilibrium. In order to find the equilibrium values for the chemicals involved, the differential equation governing the process must be found.
We have found said differential equation and then analytically solved it. After substituting in empirical data from the experiments, it is impossible to solve the equations to find the binding and unbinding constants (association and disassociation constants). Instead, we used a computational method to find these constants. Put simply this involved trying all the possible combinations of the constants (to the first significant figure) to find which best matches the empirical data. These results tell the biologists which zinc fingers have the best binding ability.
Another piece of code then uses these new found values to find out how much reactant to use, the related expenses, and how long the reaction will take.
We are currently working on a piece of code that will test the DNA sequence to see if they are viable options for our experiment.
The Equations
We will call the amount (concentration multiplied by volume) of the bacteria A, and the amount of binding site on the DNA (concentration multiplied by volume multiplied by binding sites per DNA sequence) as B.
The rate of change of A= the rate of unbinding-the rate of binding.
These rates are proportional to the amount of unbinded or binded molecules there are.
By letting
We can solve this differential equation using separation of variables.
Where χ is an unknown constant. To find χ we set A=A_0 at t=0.
Note that 4ac-b^2 is negative, so the square root cannot be calculated. We now rearrange to cancel out imaginary parts.
Therefore
Substituting the values for a, b, and c.
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Let
Then
