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Oh! I used to love these kind of questions on enzyme kinetics back in college. We even did some lab experiments which involved extracting amylase and determining its activity. I don't work on enzymes anymore. However, let me see if I can be of any help. Note that my answer may not be perfect as I have never tried this experimentally. But, I am going to propose an approach which we can scrutinize later.

So, if I understand your question correctly, you have isolated a new enzyme. Hence, we have no way of knowing the standard values of constants for this enzyme. So we have to start from the scratch.

If I had to do this, I will work under following assumption -

• The enzyme molecule is a monomer, with each monomer having one catalytic site (If its a dimer of have more than one catalytic site, you molar concentration that you have already measured has to be multiplied by appropriate factor.
• It is not an allosteric enzyme or has substrate/product dependent activation/inhibition.

Alright then! we are good to go.

Now, to begin with we need to assume that we don't really know the concentration of this enzyme. And we have to determine the concentration using our experiment. In this case it won't matter how much of active fraction is present in your solution. We will get concentration and parameters of whatever is active based on our experiment.

From, what I can see based on the graph provided in the original question by Hive account@justtryme90, we already have a way to measure enzyme activity in vitro. I don't know what method was used, but I will assume that the substrate and product can be measured either using fluorescence or colorimetry in spectrometer with ease. Or even HPLC based fluorimetry, or in with NMR. Hence, it wont be difficult to set up two different experiments. One is enzyme kinetics based on substrate concentration; and another one would find amount of moles product formed in a given time, as a function of moles of substrate uused

Experiment 1:

In the first experiment (and it is important you do this first), you need to make a plot of enzyme activity based on substrate concentration. You start with very low substrate concentration ([S]<<[E]) and increase it gradually, in different tubes.

Let, [S0] be your initial concentration. Then you will have different tubes with concentration [S0], 2[S0], 4[S0] and so on. You, add the equal amount of enzymes (some few uL) to these tubes and measure the reaction speed. Just like you did to plot your graph. Hence, for each tube you you have V(S0), V(2S0), and so on catalysis rate. You have to take the linear range of your time to substrate/product concentration plot and calculate slope as y2-y1/x2-x1 for each graph.

Now, plot a graph based on your observation with X-axis as 1/[S] and Y-axis as 1/V. You should get a graph, as shown on your left. Find the X and the Y intercept of the graph. As, you can see the Y intercept is 1/Vmax and X intercept will help you calculate Km.

Since, Km is the concentration of substrate at where V = Vmax/2, it will help you calculating your saturation concentration. This is when all the active sites of your enzymes are occupied by the substrate, and hence increasing the substrate concentration further will not increase your rate of reaction further.

Experiment 2

Find the concentration of enzyme either by using its extinction coefficient and measuring the absorbance at 280 in a nanodrop (this will work best if your prep is very pure, otherwise use BCA or other method of estimation)

For this experiment you need to make multiple tubes with [S] >> [E], beyond 2*Km calculated in the previous experiment. You add the same amount of enzyme as in previous previous experiment in this tube and measure the time kinetics of product formation once again. Now, find the product formation rate , which should be equal to Vmax. Also, calculate concentration of both products and substrate at selected time points, in which your V = Vmax.

We know that -
d[P]/dt = K_cat[ES] = 0 (under steady state assumption)

As substrate gets converted to product the initial product concentration is
[S0] = [ES] + [S(t)] + [P(t)]
Where S(t) and P(t) are concentration of substrate and product measured at time t.
Hence,
[S0] - [S(t)] - [P(t)] = [ES]

d[P]/dt = K_cat( [S0] - [S(t)] - [P(t)] )

as long as [S] >> [E] , all of E is saturated by S. Hence, when you are measuring your [S] and [P] at high substrate concentration, you should be fine and d[P]/dt would be at maxima.

You substitute your measured product and substrate concentration in above equation, and calculate K_cat (see Uludag-Demirer et al., 2009).

Once you have the turnover number or K_cat, you find the total enzyme by using the following equation -

K_cat = Vmax/[E].

Note that it wont be different from [ES] you just calculated, since at Vmax, [ES] = [E].

Now, divide [E]/concentration of enzyme that you actually added based on 280nm absorbance in nanodrop or BCA estimation. This should yield active fraction of enzyme.

Some tips from a friend

So, I asked the same question to a friend who works in protein lab. He initially adviced to rather calculate the specific activity, instead of worrying about the active proportion. He suggested that you take different purified fractions. Find the fraction that has maximum activity. Then eventually, in the fraction with highest activity you measure kinetic parameters. You find the moles of substrate consumed per unit time. This gives you idea units of enzymes. Divide it by total amount of protein in your fraction (see specific activity). However, incase finding the active proportion, it is do able. Finding the turnover number and Vmax, would be the key.

References

Protein estimation

Uludag-Demirer et al., 2009. Estimating the turnover number in enzyme kinetic reactions using transient and stationary state data

Specific activity of enzymes

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