Problem Statement: Consider the following enzyme reaction mechanism:
k1 k2
E + S <------> ES ------> E + P
k-1
where the rate constants are: k1=10,
k-1=10, and k2=1. Initially charge the
bioreactor with an equal amount of enzyme and substrate of 1 unit
each. Find the time at which the substrate concentration in the
reactor drops to one half of the initial charge. Compare a) the
result obtained by directly integrating the dynamic equations
derived from elemental reaction mechanism against b) that from
the Michaelis-Menten rate expression.
Solution:
The following Mathcad worksheet solves the specific problem
with the given set of parameters.
Comparison of Michaelis-Menten Expression to the Full Mechanism
Integrate ODEs Arising from Enzyme Kinetic Mechanisms
The following files, albeit written slightly differently or
having a different set of model parameters, solve an identical
set of ODEs. They demonstrate different ways of using Mathcad.
The simulations on a closed batch reactor clearly shows that the Michaelis-Menten expression is valid when E0<<S0 and that it is invalid when E0~S0. However, it is important to realize that we are simulating a batch reactor in the next exercise, not a biological cell. The Michaelis-Menten expression is routinely used to describe enzymatic reactions inside cells despite that the relationship E0<<S0 does not always hold. This is because a cell is not a closed batch system. There is constant exchange of materials with the surrounding, and we do not start up a cell at t=0 with no ES complex. If you provide more appropriate conditions and simulate a biological cell rather than a batch reactor, you should see that the Michaelis-Menten expression remains a good approximation even when E0~S0. (Question: how would you modify the equations to simulate a cell?)
See Michaelis-Menten Kinetics with enzyme deactivation, when the enzyme activity is time-dependent due to deactivation.
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