Michaelis Constant

[ranger4]

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The Michaelis constant Km is equal to the substrate concentration at half of Vmax. While Vmax varies based on enzyme concentration, Km does not. Why does Km not vary with enzyme concentration? So no matter how much enzyme is in a solution, the concentration of substrate at half the maximum reaction rate is always the same?

I understand you can show Km does not depend on enzyme concentration using the Michaelis-Menten equation (http://forums.studentdoctor.net/showthread.php?t=655704) but I'm trying to think of the answer conceptually.

 

[Chrome19]

The Michaelis constant Km is equal to the substrate concentration at half of Vmax. While Vmax varies based on enzyme concentration, Km does not. Why does Km not vary with enzyme concentration? So no matter how much enzyme is in a solution, the concentration of substrate at half the maximum reaction rate is always the same?

I understand you can show Km does not depend on enzyme concentration using the Michaelis-Menten equation (http://forums.studentdoctor.net/showthread.php?t=655704) but I'm trying to think of the answer conceptually.

Vmax is usually expressed on a "per mg" or "per mole" basis of the enzyme. It's therefore a defined quantity that's not affected by the number of enzyme molecules participating in the reaction. You can think of Vmax as an "intensive property".

Also recall that items such as Vmax and Km are calculated under limiting concentrations of enzyme. Therefore if a curve was generated under overwhelming amounts of enzyme, that curve would simply reflect the summation of the properties of the each individual enzyme molecule, each molecule having the same Vmax and Km for the respective substrate. Usually the apparent rate of the reaction would be so fast so that Km's and Vmax's cannot be inferred under this condition.

 

[MT Headed]

Km is the concentration of substrate at which the enzyme (no matter how much enzyme you have) will be running at "half speed".

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.

Conceptually, 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.

* I just ran an enzyme last week. My substrate concentration was 10^-5 through 10^-3. My enzyme concentration was less than 10^-8.

 

[Chrome19]

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.

I don't think this is correct. If you increase the number of enzyme participants, the RATE of the reaction will increase BUT Vmax is an intrinsic quality of the enzyme for the particular substrate, and will stay the same.

 

[MT Headed]

We disagree then. Could you be confusing Vmax and kcat?


http://en.wikipedia.org/wiki/Michaelis–Menten_kinetics
"It also follows that Vmax = kcat[E]0, where [E]0 is the enzyme concentration. kcat, the turnover number, is maximum number of substrate molecules converted to product per enzyme molecule per second."
--> kcat is intrinsic, but Vmax would double if the enzyme concentration doubles.


http://www.graphpad.com/curvefit/introduction63.htm
"Vmax (and V) are expressed in units of product formed per time."
--> If you double the amount of enzyme, you double the amount of product formed per unit time.


http://www.biology-online.org/dictionary/Vmax
"Since (Vmax) is described to be directly proportional to enzyme concentration, it can therefore be used to estimate enzyme concentration."

 

[Chrome19]

OK, I was wrong. kcat's and Km's are the constants not Vmax. I didn't think of that relation.

 

[Morsetlis]

The Michaelis constant Km is equal to the substrate concentration at half of Vmax. While Vmax varies based on enzyme concentration, Km does not. Why does Km not vary with enzyme concentration? So no matter how much enzyme is in a solution, the concentration of substrate at half the maximum reaction rate is always the same?

I understand you can show Km does not depend on enzyme concentration using the Michaelis-Menten equation (http://forums.studentdoctor.net/showthread.php?t=655704) but I'm trying to think of the answer conceptually.

Km demonstrates the affinity of an enzyme for a particular substrate, and Km is equal to the concentration of substrates at which the velocity of the reaction is equal to half of its maximum velocity possible for the particular enzyme-substrate interaction.

Vmax varies based on enzyme concentration before the more enzyme you have, the more substrates you can process maximally (the enzymes are the limiting factor).

Km does not vary with enzyme concentration because it is based on equilibrium constants, which as you know, does not vary with concentrations (in fact, they dictate the concentrations).

 

[ranger4]

Km is the concentration of substrate at which the enzyme (no matter how much enzyme you have) will be running at "half speed".

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.

Conceptually, 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.

* I just ran an enzyme last week. My substrate concentration was 10^-5 through 10^-3. My enzyme concentration was less than 10^-8.

Thanks for the help everyone! Your explanation makes sense, I hope you're right!

 

[Chrome19]

Thanks for the help everyone! Your explanation makes sense, I hope you're right!

He is right.