I will try to respond. Normally when you use enzymes you apply Michaelis Menten formula, however, you need to apply it depending on the context. There might be at times where you don't use it. It will be the times where you don't follow the model. And there are several. Two substrates, enzymes with particularities:inhibition, active site "turning off", maybe another modulatory site.

You are right when there is an exponential phase and a plateau. But enzymes may be degraded or broken down or denatured. In an natural environment, enzymes may have a turn over or half life as a protein, although it is considered perfect as they are recycled. They might work for a time, although it depends how they work. You might use an enzymatic assay where you use at least two enzymes, and then (normally) measure it at a certain wave length at a spectophotometer. Cofactors may be used for enzymes to improve stability and their function. This means a small range of pH is where they work (if you say cytosolic range).

To measure you might use any, nmol, U/mg, activity (units), maybe by quantity of substrate. Normally U (units).

"enzymes can become saturated and won’t work once they reach this point"

I think you may be referring to the concept of saturable binding. The active site of the enzyme, where the substrate binds, has a certain affinity for the substrate. The proportion of the enzyme that is performing catalysis depends on the proportion that has substrate bound at any instant. This depends on the concentration of the substrate (it can also depend on the concentration of the enzyme if it is extremely high) relative to its affinity, as measured by its Km*. If the substrate concentration is much higher than its Km, the active site is said to be saturated, meaning that all the enzyme is performing catalysis. It can't go any faster, unless the enzyme concentration is increased. The enzyme doesn't stop working because of saturation; it is working at top speed. To increase the rate at that point, you have to increase the enzyme concentration. The initial rate of the reaction (before the substrate starts to become depleted and the products build up) is directly proportional to the enzyme concentration. Some enzyme reactions will never go to completion because they are reversible; they just reach an equilibrium. The equilibrium can be forced in the direction of the products, however, by removing at least one of the products as they form. This can sometimes be done by including a second enzyme that uses the product as its substrate.

*The Michaelis constant, or Km, which is not necessarily a dissociation constant, is the concentration of substrate at which the catalytic rate is one-half the maximal rate.

You have a misconception, which I totally understand due to your background. When the enzymes become "saturated" it doesn't mean they "don't work". They will keep working, the reactions will still occur, it's just that the speed of the reaction won't increase any further even if you add more substrate.

Enzymes are not consumed in reactions. They're like workers in a factory, they process the material (substrate) into product, but they themselves are not consumed to become product. So the amount of workers will remain the same over time.

Now, if you have a lot of workers and little material, many of them will just lazy around doing nothing. In this situation, if you add more material, those free workers will get to work, and you produce the product faster.

But, if you keep adding material faster, at some point all the workers will be busy. At that point, even if you pile another mountain of material at the gate every minute, the speed that you produce the product will not increase any further. We refer to this situation as the enzyme is "saturated". Make no mistake, even when it's "saturated", the production is still occurring. And if you stop adding material (substrate), the backlog of substrate you have already added will slowly be processed by the enzyme until all of it becomes product.

So in the end, you'll get your reaction and your product no matter how much enzyme you add. Even if it's just 1 single molecule of enzyme, it will get the job done, eventually. The only difference is time, i.e. how long you have to wait until that amount of substrate you added turns into product. We usually present this speed through "enzyme unit", or U (which you can find on the label of all enzyme bottles, e.g. 1 U = 12 g). 1 U of an enzyme is the amount of enzyme that if all those enzyme molecules get to work (equivalent to enzyme saturation), they will convert 1 micromole of substrate to product every minute.

So the amount of enzyme you need to use all comes down to how long you prefer to wait. If, for example, you have 300 micromoles of substrate, and you only want to wait 20 minutes for the reaction to finish, you will need to add:

Amount of enzyme = 300/20 = 15 (U)

Please understand that this is just a grossly simplified explanation of the most basic. When you get to the actual enzyme speed, there are a bunch of other factors you need to mind, such as the chance of collision between enzyme and substrate molecules (which depends on temperature and dilution, similar to normal chemical reaction), the nature of the enzyme (which determines how stickily it catch the substrate molecule when they collide or whether they'll just bounce off each other, how stickily the product bounds to the enzyme and delays further substrate coming in, whether the enzyme needs to change its structure and therefore requires a recovery time, etc.), how many substrates take part in the reaction, the presence of inhibitor/inducer, etc. It can get complicated super fast and there's a whole field of study for it. If you're interested in it, you can start by learning about Michaelis–Menten kinetics, which is the most fundamental of them all.

Mª Angeles Zorrilla Lopez-Perea Adam B Shapiro Duy Minh Pham Thank you very much for helping me understand looks like I’ll have to learn:

michaelis-menten kinetics

This ‘saturation’ behaviour is similar to how a weak acid/base works how you can shift the equilibrium to the direction of the products by raising the concentration of the reactant or removing a product, except you can increase the rate by adding more enzyme which I’m not sure is posdible with a weak acid/base.

Apart from the kinetics is there anything else I’d need to learn?

Seems like applying the michaelis-menten formula will give me the concentration of substrate I need to avoid based on the concentration of enzyme so it doesn’t become ‘saturated’

As a hard headed approach if went with an excess of enzymes (just dump a load in) and some substrate would thst guarantee the fastest rate of reaction?

This saturation is above all the plateau phase. It could be under certain circumstances although essentially is when it reaches this point. Maybe the build up of product and make sure the enzymes have the right substrate to react. All you mean by substrate makes sure it is not an inhibitor. It binds strongly to the active site. Maybe so tightly they will not come loose. There are irreversible and reversible inhibitors. The environment too. Temperature, fI, pH, substrate.

Lareab Ashraf In theory, enzyme saturation is not necessarily something bad that should be avoided. In fact, it's the state that give you the maximum reaction speed for a set amount of enzyme. Of course, you can always add more enzyme to increase the speed further, but if you set the substrate concentration below the saturation concentration, you're actually not utilizing all the enzyme molecules you have. In other words, you're having some workers lazying around doing nothing, and therefore, you're not using the enzyme with the best efficiency.

In practice, however, there are a few things to mind before you throw too much enzyme and/or substrate into the mix. Enzymes are proteins, and their substrates most of the time are large mass organic compounds as well. Therefore, if you add too much of these compounds into the solution, they tend to increase viscosity, which in turn slows down the movement of these molecules themselves, making it harder for them to meet each other. As a result, it may decrease the speed of the reaction.

In some rare cases, substrate at too high a concentration may become inhibitor of the enzyme itself. The molecular explanation of this phenomenon is that an enzyme can bind a substrate at more than one position, or "site". The site that perform the reaction (the catalytic site) usually has much higher affinity to the substrate than other sites, therefore the substrate prioritizes it. However, at too high substrate concentration, some substrate molecules are bound to bind to the other sites, and when they do, they may change the structure/function of the enzyme, preventing it from performing the catalysis. This is called "nonproductive binding".

Therefore, in practice, the substrate and enzyme concentration should be maintained at a sweet spot, above the saturation ratio to get the maximum efficiency with the enzyme, but below that which may cause substrate inhibition/viscosity (this depends on each enzyme/substrate). This range is usually quite large though, so it's pretty easy to achieve.

The biggest issue most of the time is not enzyme or substrate, but rather product. As I mentioned, after each catalysis, the enzyme molecule needs a certain amount of time to detach the product from itself. This detachment will become harder if there are a lot of product floating around (Fick's law of diffusion). And this will get worse as product accumulates over time if you don't remove them. Therefore, ideally, you would want to continuously remove as much product as you can from the reaction solution.

Duy Minh Pham Mª Angeles Zorrilla Lopez-Perea in my case the reaction is much more chemical than biological in nature and so much less complex, the substrate being a lactone, here is how it goes:

chemical version:

lactone + NaOH -> NaSaltWeakAcid

biological version:

enzyme = paraoxanase

I’m not sure if I could use any lactonase enzyme, paraoxanase is the one in human blood that drives this reaction.

......... enzyme

lactone weak acid

as you can see the biological version is more efficient, and makes the weak acid (the product I want) directly rather than a sodium salt, so I could skip having to remove sodium, and also skip having to use a reagent, I could just recycle the enzyme instead.

Enzyme reactions should usually be performed in a buffered aqueous medium to maintain the pH, since enzymes have optimal pH for their activity. If it is an enzyme in blood, the pH optimum is probably going to be near neutral (the pH of blood plasma is typically around 7.4). Since buffers are usually salts, there will probably be a cation of some sort around. Moreover, the human enzyme seems to require Ca2+ for its activity.

Thank you for the info Adam B Shapiro I was thinking of using baking soda but the salt of a calcium base rather than sodium base seems to be a better option from your answer, can I ask how you came about this detailed info?

I checked the BRENDA database: brenda-enzymes.org. There are 3 enzymes called paraoxonase in that database, but I didn't know which one was yours. Check the literature on the enzyme you are using to find the optimal conditions for its activity.

I advise you to consult with a local specialist in enzyme biochemistry to learn the proper handling of enzymes to preserve their activity.