How do I measure activity of an enzyme if it catalyses a two-step reaction? The final product is what I want.
To get a satisfying answer, you'll need to provide a bit more detail about the enzyme. Generally speaking, there are two classes of "two-step" reaction mechanisms. The sequential mechanism describes the binding of two substrates to the same form of the enzyme, typically in discrete binding sites. The ping-pong mechanism describes the binding of each substrate to a different form of the enzyme and forming a covalent intermediate.
https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/3%3A_Enzyme_Kinetics/3.4%3A_Multisubstrate_Systems
There are two rates, k1 and k2. Depending on the nature of your enzyme, it may be possible to measure the total rate (k1+k2) by adding both substrates and quantifying product release and/or substrate depletion the same way you would for any one-step enzyme. For ping-pong enzymes this can be more complicated and depends on your approach but is still possible.
For steady-state kinetics, you measure the rate of an enzyme reaction that occurs in 2 steps the same way you measure the rate of a reaction that occurs in one step: Put all the substrates in at the beginning and measure the initial rate of formation of the final product.
I worked on NAD-dependent DNA ligase, which goes by the Ping Pong mechanism mentioned by Alexander C. Anderson for a reaction with 2 substrates and 3 products, in 3 steps, as follows:
E + NAD+ E-AMP + NMN
E-AMP + nDNA E.AMP-nDNA E + AMP + DNA
where
E is the NAD-dependent DNA ligase
NAD+ is nicotinamide adenine dinucleotide (which can be viewed as AMP-NMN)
NMN is nicotinamide mononucleotide
AMP is adenosine monophosphate
nDNA is nicked DNA
DNA is unnicked DNA
One of the intermediate stages in the reaction is a covalent adduct of AMP with the enzyme, and the other is a covalent adduct of AMP with nDNA.
Overall, the reaction is NAD+ + nDNA AMP + NMN + DNA
If NAD+ is substrate A and nDNA is substrate B, the rate of product (in this case both AMP and unnicked DNA) formation (which applies to any Ping Pong mechanism with 2 substrates) is
V = (Vmax[A][B])/(KmA[B] + KmB[A] + [A][B])
The rate equation may be a bit more complicated if there is substrate inhibition caused by the second substrate binding before the first one. That will show up as a Michaelis plot of rate versus substrate concentration that rises as the substrate B concentration rises at first, then falls at higher substrate B concentrations:
V = (Vmax[A][B])/{KmA[B](1 + [B]/KiB) + KmB[A] + [A][B]}
where KiB is the substrate inhibition constant.
Adam B Shapiro Thank you for your advice! Actually, the reaction of my enzyme is E + A + B C + B D.
Where A is the limited substrate, B is the excess substrate, C and D are the intermediate product (which is not expected to accumulate) and the final product (which is what I wanna synthesize), respectively. The enzyme catalyzes the formation of the product via a sequential mechanism.
Overall, the reaction is A + 2B C (which in the reaction condition couldn't be detected) + D
In the reaction condition, product C would be transformed very soon into product D. Could I put all substrates and measure the activity by monitoring the release of product D in this case?
Alexander C. Anderson Thanks for the proposal! Actually, the reaction of my enzyme is E + A + B C + B D.
Where A is the limited substrate, B is the excess substrate, C and D are the intermediate product (which is not expected to accumulate) and the final product (which is what I wanna synthesize), respectively. The enzyme catalyzes the formation of the product via a sequential mechanism.
Overall, the reaction is A + 2B C (which in the reaction condition couldn't be detected) + D
In the reaction condition, product C would be transformed very soon into product D. Could I put all substrates and measure the activity by monitoring the release of product D in this case?
Well then, you're in luck! If it's straightforward to detect and quantify D, and you can use the natural substrates A and B at appropriate/physiologically relevant concentrations, then you can certainly report the activity of the enzyme in terms of units of D turned over per unit of time.
This isn't the case for all enzymes, which is where the assay tends to get more complicated, but it seems like your case is pretty manageable. Best of luck with the experiment!
In general, I agree with Alexander C. Anderson.The only concern I would have is whether there is any significant product inhibition by C.
Because of the involvement of 2 molecules of substrate B along with one of A the rate equation I wrote is not the right one for the overall reaction. This is really a 3-substrate reaction in which two are the same molecule. The numerator would be Vmax[A][B]2. The denominator depends on the precise details of the kinetic mechanism. There would probably be two different Kms for the first and second molecules of B.
An example of an enzyme catalyzing such a reaction with 2 molecules of the same substrate is D-alanine-D-alanine ligase. The overall reaction is:
2 D-Ala + ATP D-Ala-D-Ala + ADP + inorganic phosphate
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Adam B Shapiro Alexander C. Anderson Thank you for both of your answers!
I think there was no considerable evidence for the worry of product inhibition by C. In fact, I think it's better that C could bind to the enzyme because C is the actual substrate to form D.
Nonetheless, I would try to do the experiment and see if the curve would be fine.
Thanks a lot! Best wishes for your research!
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