Hi there,

If working with crude extracts obtained using exactly the same extraction protocol and the same enzyme assay, Km should be unchanged as it's not dependent on enzyme concentration and Vm should vary according to the enzyme content of the sample used for the assay. As Vm is directly dependent on the enzyme concentration, the richer sample will give the higher Vm.

Usually Vmax is used in studies with varying amounts of substrates or inhibitors. I think that an alternate question would be to ask if kcat (or turnover number) would change, and there the answer would very likely be "no".

In looking at amount of enzyme expressed in tissue, or in vitro, it's the total catalytic activity that's measured in units something like umoles/min/mg. So while catalytic activity [or catalytic concentration] will certainly increase with increased expression, the kcat  will, along with Km, almost certainly not. With increased amount of enzyme with a constant kcat, measured velocities would certainly increase.

The classic book by Cornish-Bowden (now in its 4th edition) is  a good reference:

https://books.google.com/books?id=_3fqaYnrdosC&dq=kcat+enzyme+kinetics&lr=&source=gbs_navlinks_s

None!

Hi All, thanks for your responses. 

@Rob: may I please request you to provide some insight?

The substrate affinity (defined roughly as KM) and the catalytic efficiency (Vmax) are properties of the enzyme, irrespective of the amount. More properly, as Robert has stated, we should consider kcat. Now, if you express more enzyme, the measured activity will be higher, because v is proportional to Vmax * [E]. Double the enzyme, double the catalytic activity. But KM and Vmax do not change. However, I seem to have lost my caveat whe I posted my reply if over-expressing an enzyme makes it aggregate, bind to other proteins etc, then the catalytic properties will change. Do not confuse v, the measured velocity, with Vmax, the maximal velocity attainable by that enzyme.

I think your question demonstrates that we do need to maintain our teaching of fundamentals of enzymology and metabolism - of course I speak as a biochemist from the last millennium, so "I would say that, wouldn't I?' 

Rob

[EDITED to correct silly typos]

 Thanks for the insight Rob! I really appreciate it. I also think that we need to maintain our teachings of fundamentals of enzymology and metabolism. While I agree that Km is the true fundamental property of an enzyme substrate pair, I think you might want to double check about Vmax being a 'true' fundamental property. In my opinion, Vmax in its literal meaning is a product of turnover number and enzyme concentration. Because of the dependence on enzyme concentration, Vmax will be impacted by the number of total active sites present in the enzyme solution, which in turn will be increased by increased enzyme expression. Please correct me if I am wrong. I do not have a biochemistry background and have only preliminary knowledge about enzyme kinetics.  

 Vmax being defined as kcat*[E]0 (even in last century enzymology lectures), it does actually vary with [E]0 which represents the total concentration of enzyme used in the enzyme activity assay. So it can't be said that Vmax is a constant as [E]0 may vary . Only kcat and KM are intrinsic constants of the enzyme towards the substrate considered. In summary Kamran is right and Rob is wrong about Vmax.

Dominque: I'm going to disagree a bit:

When I cite Vmax in my work, I would say (for example)

Vmax = 123 umol/min/mg protein

THis then allowed me to calculate velocity according to the amount of enzyme added. It follows that more enzyme - no change in Vmax, but higher v.

I think this is mostly semantics.

In my mind (also from the last century), speaking about Vmax makes sense only at a constant amount of enzyme and (usually) with varying amounts of substrate and or inhibitor.

If I can take the liberty of rephrasing Kamran's original question it would be: What will be the impact of increased enzyme expression on its kcat, v, and Km?

And my answer would be (within some constraints): none, increase, and none

Thanks Rob, Dominique, Robert for nice explanations! My knowledge of enzyme kinetics has really increased after this discussion. And this is what matters. 

Thanks,

Dear Rob,

µmol/min/mg of protein is not a velocity unit. What you define as Vmax is actually specific activity which indeed is unique for a enzyme-substrate couple. 

Hi Kamran, Rob and Dominique,

µmol/min/mg of protein is a way of expressing specific activity (for a purified enzyme preparation or even for a crude extract). If concentration of the substratr(s) is saturating (optimally, it should be) this would approximate the 'specific' Vmax.

For pure enzyme preparations, this specific Vmax can  easily be translated into kcat - you just need to know the enzyme's molecular mass and make the simple, but crucial,, assumption that all of the enzyme molecules are active.

kcat , we all agree on this, is constant under a given set of conditions (pH, temperature, tipe of substrate etc.), so the 'specific' Vmax is not expected to change with the yield of the initial enzyme expression. If it does,, it suggests that some complicating factors are at play (for example, a variable fraction of enzyme molecules may be expressed or purified in an inactive form).

On the other hand, if we are dealing with a crude extract, or with an only partially purified enzyme prep, the specific maximum activity  (µmol/min/mg of protein) is NOT directly convertible to kcat and may well vary with preparation. In particular it is expected to increase proportionally with the yield of the expression.

(I am writing all this to summarize the facts; the answers provided by Rob, Dominique and others were all substantially correct, but one needs to put everything together to get a comprehensive picture)