In the simplest case of a monomeric enzyme with a single active site, the Km is independent of the enzyme concentration, in principle. However, if the measurement is not done under the right conditions for Michaelis-Menten kinetics, the Km may appear to vary with the enzyme concentration. The enzyme concentration must be much lower then the substrate concentration, and you must measure the initial rate of the reaction. If the enzyme concentration is too high, these conditions may be violated.

Hi,

Vmax is maximum for a particular enzyme in a defined set of conditions. You cannot increase it further, it is the maximum. That is the last part of your question. So, for a particular Vmax the Km is always the same.

Does Km change with enzyme concentration? Yes, it can. But that is not related to Vmax. Km can change with enzyme concentration if say the enzyme is oligomeric and the subunits are talking to each other (I am over simplifying here). When there are more than one subunit and more that one substrate, things gets slightly more complex.

Hopefully this will help. 

In the simplest case of a monomeric enzyme with a single active site, the Km is independent of the enzyme concentration, in principle. However, if the measurement is not done under the right conditions for Michaelis-Menten kinetics, the Km may appear to vary with the enzyme concentration. The enzyme concentration must be much lower then the substrate concentration, and you must measure the initial rate of the reaction. If the enzyme concentration is too high, these conditions may be violated.

Km is the concentration of substrate at which the enzyme will be running at "half speed". If you doubled the amount of enzyme, sure the Vmax is going to increase. If you doubled the amount of enzyme, sure the Vmax is going to increase. You have twice as many workers. 1/2 Vmax will increase too, obviously. But Km, the amount of substrate at which half of the enzymes are working and half of the enzymes are bored and txting on their iphones, will remain the same. These problems are typically set up such that there is an overwhelming amount of substrate, relatively little enzyme, and the empty enzyme starts processing the substrate by randomly bumping into it and sticking to it. So doubling the amount of enzyme simply doubles the number of workers who still randomly bump into the enormous amounts of substrate at half of their capacity.

The Km is only related with the enzyme,when the enzyme is given,its Km will not change no matter how or what  the condition changes.

Hi there,

Why do you write Vm=2Km (this is false)? Is it because v0=Vm/2 is observed when [S]0=Km? Whatever forget this forged equation!

The answer is simply no : Km is independent on the concentration of enzyme if you are doing the proper experiment ([S]0 being always much greater than [E]0). And yes Vm does vary with [E]0 that's why enzymologists prefer expressing enzyme parameters as kcat and Km which are both independent of concentrations used during experiments.

While Vmax is the product of the catalytic constant and of the total enzyme concentration as mentioned above by A Shapiro, Km should be independent. If it is not the case, could be interesting to see the polymeriztaion state of the enzyme and to analyze in details the appearance of product as a function of time and see if linear (should be...)....If there is a deviation (lag or burst).

The KM does not vary with enzyme concentration (tit's a "constant", right?). What can vary is the Vmax, therefore efficiency of an enzyme is usually expressed as the ratio Vmax/E. 

If you need more details, please refer to the "oldy" but "goldy" book written by Segel IH: Enzyme kinetics.

there is also a very good book written by A Cornish-Bowden I recommend to students, entitled Fundamental in Enzyme kinetics.

I think the question has been answered already but it is worth repeating that the questioner is obviously very confused about Michaelis-Menten kinetics. Km has dimensions of CONCENTRATION. It is the concentration at which the rate is half-maximum, i.e., the rate is Vmax/2. Vmax is a rate and has dimensions of CONCENTRATION/TIME. Vmax is directly proportional to enzyme concentration [E] so if [E] is varied Vmax will vary in tandem. This will not affect Km which remains the substrate concentration at which the rate is exactly Vmax/2

Is there any reference for the statement "Km is independent of Enzyme Titre". Please share if there is any?

Hi Lohit

have a look in this article may be it will help

it explain how Km does not vary with enzyme concentration  

if you add more enzyme it will increase the rate so the slope for new condition will be higher but the vmax will reach at same concentration as before 

you can also check video on youtube 

"Why Km does not change with the enzyme concentration"

t-10:00 

http://ocw.mit.edu/courses/biology/7-014-introductory-biology-spring-2005/readings/l5_bioche_deltag.pdf

Km is the name given to this term (Briggs-Haldane): Km  = (kr+kcat)/kd, where kd is the rate constant for ES formation, kr is the rate constant for ES breakage into E and S, and kcat is the rate constant for ES breakage into product and enzyme. Clearly, all three rate constants are constant (at constant temperature etc.), ie. they're independent of enzyme concentration, so the combination of constants must be another constant, Km. 

E + S ES -> E + P

https://depts.washington.edu/wmatkins/kinetics/michaelis-menten.html

I have the same confusion, and it seemed to me that the Km should vary with the concentration of enzyme. Conceptually Km refers to the concentration of substrate at which the reaction rate is half of the Vmax, and in other words, it refers to the concentration of substrates at which the enzyme was half converted to ES. Right? While enzyme concentration increases, the amount of substrate was also increased to half saturate the enzyme, right? Km therefore should increase as the enzyme concentration increase, and this seems very perfectly logic.

According to the requirements of Michaelis-Menten kinetics, [S]total>>[E]total so that [S]free is approximately equal to [S]total. A small increase in [E]total therefore has no significant effect on [S]free. (In other words, increasing the enzyme concentration does not significantly deplete the total substrate concentration.) Since [S]free is not significantly affected by an increase is [E]total, there is no significant effect on Km either.