Regarding your first question, you are confusing the rate of association with the rate constant (kon). The rate is equal to kon[E][S]. Just a point about terminology.

In the single-turnover experiment, if the enzyme concentration is made high enough, the rate of association will not be rate-limiting, allowing measurement of the rate constant of the chemical step, which is not dependent on the enzyme concentration. Keep increasing the enzyme concentration until the rate constant no longer changes, and you will have measured the rate constant of chemical step.

This assumes that the phenomenon being measured is the chemical reaction. If, as you propose, the phenomenon being measured is not the chemical step, but is instead dissociation of the product, then the rate constant will be that of the dissociation step. If you aren't confident that you are measuring the rate constant of the chemical step by using a fluorescent probe, then consider using a quenched-flow method that allows you to extract the sample for later direct analysis of product formation.

Adam B Shapiro I'm curious why ST kinetics wouldn't be measuring the combination of the chemical reaction and product release. Intuitively, it seems to me that the ST kobs would be a product of both reaction steps' rate constants, especially in a situation where catalysis and release have similar rates.

Another suggestion I got from a professor is to perform the reaction in increasing viscosity on the idea that product release is diffusion limited, and cleavage is not. Thus, if increasing viscosity decreases the ST rate, I'll know release is rate limiting, whereas if it stays constant, the chemical step is rate-limiting.

I'm also not sure how resolving the reaction products will tell me which step I'm measuring in ST kinetics, because the product sizes would be the same either way, unless you're suggesting I isolate enzyme-substrate and enzyme-product complexes to see where the enzyme is bottlenecking, assuming they are stable enough for something like a pulldown assay. I could see this working, because ES complexes are in fact stable enough for such assays, though I don't know about the stability of the EP complex.

I also thought I could just resolve the products on denaturing PAGE to determine the size of the dominant nucleic acid product and then just synthesize it and directly measure its parameters, though it would not recapitulate any catalysis induced conformational changes that may stabilize the EP complex.

When you measure the overall rate of a process that has multiple steps, such as substrate cleavage followed by product release, by measuring a single rate constant, you can only measure the slowest (rate-limiting) step. If product release is fast relative to cleavage, then you will measure the rate of cleavage. But if product release is slower than cleavage, then product release is the rate you will measure. If you want to be sure that you are measuring the rate of cleavage even if the rate of product release is slower, then you have to measure cleavage directly. This could be done by a quench-flow method whereby you measure the concentration of product formed, but not necessarily released, at various very short times after starting the single-turnover reaction.

Imagine that cleavage and product release have identical rate constants (e.g. 10 s-1). What rate constant would you measure for the overall process of cleavage and product release by the method of monitoring release of the product by a fluorescence change? Would it be the product of the two rate constants (100 s-1), or would it be the same value as the rate constant of both processes (10 s-1)? From this you can see that it doesn't make sense for the overall rate constant to be the product of the two rate constants of the individual steps. The overall rate constant that is measured is that of the slowest step.

The chemical quench method allows you to stop the reaction by denaturing the enzyme. Then you can remove the sample and analyze it to see how much product has formed. Since the enzyme is denatured, the product will be separate from the enzyme, whether or not it dissociated prior to the quench. This allows you to measure the rate of the cleavage step without concern about whether it is faster or slower than the product dissociation step.

Since you are interested in a nuclease, the size of the products might be worth investigating to see if your enzyme is processive, i.e. it can remove nucleotides one at a time from the substrate without dissociating from it (relevant to an exonuclease). However, that was not the point I was trying to make.

Okay... basically, because the rate limiting step is slower, the enzyme will be saturated with either substrate or product, depending on whether cleavage or product release is faster, and then the rate observed is the rate limiting step.

Let's say the chemical step and product release are both 10 s-1, and release is required for the fluorescent signal. So in a reaction, would there be a 100 ms lag period before the rate reaches its maximum, during which time the enzymes are converting substrate to product?

Remember that this experiment was designed originally so that the rate of substrate binding would not be rate limiting, which was done by increasing the enzyme concentration to the point at which no further increase in the product release rate was observed. The reaction rate will be maximal at the very beginning (no detectable lag other than that due to mixing), and it will slow down exponentially as the enzyme-substrate complex is consumed. This is assuming a simple reaction scheme with only one step between substrate binding and product release.

Under these conditions, it is probable that the enzyme concentration will exceed the substrate concentration. If so, the enzyme will not be saturated with substrate. Also, under these chosen conditions, the observed rate would have to be either the rate of cleavage or the product release rate, whichever is slower.

If the slower rate constant is the rate of cleavage, the enzyme will contain substrate but no product, and the amount of free product will increase with time at the same rate as the rate of cleavage.

If the rate constant of product release is slower, the enzyme will, at some time after reaction initiation, contain product but no substrate, and the rate constant for appearance of free product will be equal to the product release rate constant.

It is generally assumed, that simple binding equilibria, where no covalent bonds are formed or broken, occur very fast compared to chemical reactions. Thus, usually the latter are rate-limiting. Also realise, that you have a mixing time in your equipment (say, 1 ms for a good stopped-flow instrument), and this is way longer than the time required to establish the binding equilibrium. Thus, binding times become important only if your equipment is very fast ("caged" compounds, temperature or pressure jump).

Sourav Roy