The slope of the linear plot of rate (V) versus substrate concentration ([S]), divided by the enzyme concentration ([E]), is kcat/Km. Here is why:

The Michaelis-Menten equation is V = Vmax[S]/(Km + [S]).

If Km >> [S], as in the case of a linear Michaelis plot, then V = Vmax[S]/Km.

Vmax is kcat[E], so V = kcat[E][S]/Km.

The slope of the V versus [S] plot is therefore kcat[E]/Km.

Divide the slope by [E] to get kcat/Km.

The slope of the linear plot of rate (V) versus substrate concentration ([S]), divided by the enzyme concentration ([E]), is kcat/Km. Here is why:

The Michaelis-Menten equation is V = Vmax[S]/(Km + [S]).

If Km >> [S], as in the case of a linear Michaelis plot, then V = Vmax[S]/Km.

Vmax is kcat[E], so V = kcat[E][S]/Km.

The slope of the V versus [S] plot is therefore kcat[E]/Km.

Divide the slope by [E] to get kcat/Km.

I agree with Adam answers ...good luck.

If it keeps increasing, the question is, whether it is catalyzed by an enzyme.

I agree with Adam if you desire to report kcat/Km. However, every enzyme cannot have an infinite Vmax and therefore must reach Km at a certain concentration of enzyme. Make sure you add all the substrate at once instead of adding incremental amounts and remember that the Vmax can be 'extrapolated' from several values of V0 vs [S] using the Lineweaver-Burke equation. If this equation does not give you a linear curve, then some modulator(s) may be adhering to your enzyme. Good luck.

Sammy,

1) Every enzyme reaction must have a Vmax value, if you can't reach it than you supposed to add more substrate. Nevertheless, you did not mention any technical or other difficulties with adding more substrate, so I suppose that you tried many substrate concentrations in a wide range. Therefore, I suppose the analysis mentioned by Adam does not necessarily yield a correct kcat/Km value (Km>>S does not hold), not to mention that you are probably interested in getting the Vmax and Km values. You might try to derive estimates based on linearizations of the Michaelis-Menten hyperbole, such as the Eadie-Hofstee or the Hanes plots. I do not recommend the double-reciprocal plot.

2) If you see some decline from linearity in the kinetic data, but, as you wrote, the activity keeps increasing, then it is possible to deal with a hyperbole kinetics associated with a non-specific increase duet o some non-enzymatic effect or, depending on the assay you use, an increase of the background in the assay. In this case you can try to remove this effect supposing that it depends linearly from substrate concentration. So I suggest to fit the data with a sum of a hyperbole a line, such as v = Vmax[S]/(Km + [S]) + const[S]

I hope this helps. Best: Karoly

Just to amplify the comments of others: Each enzyme molecule needs a certain amount of time to bind substrate, perform a reaction and release product. The first and third reaction depends on substrate and product concentration according to the law of mass action, but the speed of the second reaction does not, it is a property of the enzyme itself (given physical conditions). The number of molecules an enzyme molecule can turn over under optimal conditions (plenty of substrate, no product, that is, minimal required for the first and third reaction) is the turnover number kcat of that enzyme. Multiplied with [ES] you get the limiting velocity Vmax (according to IUPAC/IUBMB, "limiting" is used now instead of "maximal", as it is not a maximum in the mathematical sense).

If you see no saturation, then either your [S] is in the linear range of the Henri-Michaelis-Menten curve and you have to use higher substrate concentrations. Or you have a background reaction going on and are looking at the sum of the enzyme-catalysed hyperbolic plus another, linear reaction. The effects of that are eliminated by running a control reaction in parallel, which is exactly the same as the enzymatic, except that no enzyme is present. Product production in the control sample is subtracted from your experimental. In other word, instead of enzyme solution, add to your control an equal amoount of the buffer that the enzyme is dissolved in.

Adam Shapiro gives the classical theoretical steady-state MM kinetics answer - the derivation he gives is textbook correct. However, I  also agree with Gustavo and Karoly - it is impossible for an enzyme not to reach reach a maximal rate as Vmax always dependent on [E]total - so unless you are proposing to break the 1st law of thermodynamics by proposing the creation of more enzyme during the assay, taking the purely theoretical approach is flawed. This can also be observed by making the simple point that Adam's theoretical explanation requires Km >> [S] (i.e in the pseudo-1st order phase of the MM function, well below the Km)   and this cannot be assumed. As elegantly pointed out by Karoly, there are too may possible confounders in your experimental design to be sure what is occurring. But, as he states, be sure you really are using a very wide range of substrate and also ensure (as stated by Gustavo) that the reaction really is enzyme catalysed. There are many things that could be experimentally wrong that you need to check before you assume you can't reach Vmax. If you are seeing non-specific rate increase(enzyme-independent) than Karoly's modified MM equation v = Vmax[S]/(Km + [S]) + const[S] is not a bad approach as most non-specific effects will increase rate linearly with [S].

The occasions I have seen Adam's theoretical approach used is not because Vmax doesn't exist (it always does) but because the reagent may too horrendously expensive to use in such high concentrations, or the instrumentation used for measuring the reaction rate signal is being saturated before reaching Vmax. If, after checking and thinking logically about possible errors in your experimental set-up, you still can't get a direct Vmax value, then Adam's methods can be used. Follow Adam's linear plot and derive the Km value. If this value is many times greater than the upper limits of the [S] you were able to use (and, hence, so will the Vmax be) than Km >> [S] is obviously correct and this approach used with confidence. 

Nice problem to post. Good luck

I have personally encountered the situation where a linear Michaelis plot was obtained within the range of substrate concentrations that were technically feasible. Nevertheless, by using progress curve analysis, I was able to obtain Km and Vmax values. In this method, instead of measuring the initial rate, you run the reaction to completion, or at least well beyond the initial rate, so that the rate slows down due to substrate depletion. You can fit this curve to the integrated form of the M-M equation, if you are sure there is no product inhibition. Alternately, you can use numerical integration to fit the curve to a kinetic model, which can include product inhibition. It takes some trial and error to find the sweet spot of substrate concentration and enzyme concentration for these methods.

You can try with several options:

I recommend you to increase the substrate concentration first. About calculation and reporting you may use either Michaelis-Menten or the double-reciprocal (Lineweaver–Burk). Although the double-reciprocal was derived from Michaelis-Menten, it is sometimes easier to interpret.

I have used high cost substrate BAQ at low concentrations.

I always am recommending the reliable and sensive HPLC-photometric enzyme assay (please see file; J Chrom B BIN LIP Km).   HPLC method requires the substrate in very small amount; i.e., only 0.1 mL of reaction mixture is required to perform the enzymatic reaction.   By the way, Sigma has previously stopped selling our important substrate of BAQ (biotinyl-6-aminoquinoline), but HPLC method can be performed to publish the above article due to minimum use of substrate (measurement of initial velocity at low substrate concentrations).   Further, enzyme kinetics are measured around the concentrations of Km (5, 7.5, 10, 15, and 20 μM for BAQ, and 5, 10, 15, 20, and 25 μM for LAQ), and the sensitive HPLC method is indispensable.  

We have previously developed an HPLC-photometric-carboxylase assay in order to get rid of the use of dangerous radio-isotope (please see file; 4-kinds of carboxylase). 

By the way, such a characteristics is observable in hydrophobic membrane glycoprotein enzymes such as thiol-type rat brain and rat kidney biotinidases.   Then, I have diluted the homogenate, soluble, and membrane fractions containing enzymes by 0.1 M sodium-phosphate buffer (pH 6.8; with 1 mM EDTA) in series and biotinidase activity have been determined (10 mM 2-mercaptoethanol (2-ME) has been added immediately before assay).   The infinitely-diluted enzyme activity has been estimated graphically and expressed as Vmax or V.   It is noteworthy that such an assay with dilution can uniquely performable only by using sensitive HPLC-photometric enzyme-assay method.   Safe dilution of homogenate (membrane and soluble fractions) can uniquely achievable by using the HPLC-protein assay (please see file; HPLC-Surf-SEC protein determination method).