Delays and hysteresis in transient kinetics

Realistically, however, this lag phase can be too small to be observed. To see how significant the lag phase is, we compute the inflection point at which the curvature turns from positive to negative. Setting d2c(t )/dt2 = 0, we solve the inflection point t = —ln()4). (Whether one can observe a lag phase depends on whether the time resolution of the measurement is greater than t. ) 

Here we can see a pattern in the mathematics. Note that the initial slope of c(t) is = yb 0). Hence, if b(0) = 0, then = 0, and = y = yaa(0). By the same argument, if a kinetic process starts with a species that is more than two steps away from the species C, then even the initial curvature in c(l) will be zero. 

An example of a kinetic scheme that displays hysteresis occurs when a slow conformation change between two enzyme-substrate complexes is required before the product can be released  

Invoking the quasi-steady approximation, the steady state flux expression for this system can be shown to be 3 

Enzyme binding and conformational change cause a delay of approximately ten seconds in the appearance of product in the simulations based on the full kinetics of Equation (4.37) compared to the simplified kinetics of Equation (4.40). This delay results in a positive second derivative for [P] at early times. 

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