student

[faizan_ahmad]

An enzyme in the native buffer solution exists in the equilibrium, native (N) state <--> denatured (D) state. So it has its KM and Vmax in this buffer. If one adds a cosolute to the enzyme solution, and if the cosolute shifts the equilibrium from D state to N state (i.e., the enzyme gets stabilized), then will the cosolute also effect the KM and Vmax?

[Gyrase]

Km is constant but Vmax will increase. Of course if the enzyme has michaelis-menten kinetics at all!

cheers,

DG

[faizan_ahmad]

Km is constant but Vmax will increase. Of course if the enzyme has michaelis-menten kinetics at all!

cheers,

DG

 

Thanks!

I am sure your argument is that Km is not a function of [E]. We have measured Km of the enzyme in the presence and absence of urea which shifts the N state <--> D state equilibrium from left to the right. We have observed that Km is increased in the presence of urea. Then this argument fails to explain this observation.

FA

[Hobbz]

Gyrase and I had an extensive discussion on Km recently and we should be on our "A" game for this. So that's a very thought provoking system you have there with the equilibrium between the native and denatured protein in the urea solution. First, your equilibrium between the native and denatured protein would have to equilibriate on a time scale similar to the half-life of your substrate in the system in order to consider it is affecting anything. Second, it still should not change the Km if the equilibrium of the denatured and native enzyme shifts fast enough. It may appear that way when plotting your data on a Michaelis-Menten plot but your system is not following Michaelis-Menten assumptions because the total concentration of active enzyme is changing in your sample. You would need to derive new equations in order to properly calculate the Km of your enzyme in this environment and I that's beyond my ability. Third, your Km may actually be changing because urea may be altering the structure of the native enzyme which could allow for the substrate to get to the enzymes active site faster or slower and leave faster or slower.

 

I hope that helps and I may be wrong.

[faizan_ahmad]

I thank both of you! We equilibrate the enzyme in urea solution. The enzyme is, say 25%, denatured as measured by several conformational properties. Our buffer and substrate solutions contain the same amount of urea. So all activity measurements are done at constant urea solution. I am sure you will now come up with different answers for Km and Vmax now.

[Hobbz]

So, you never calculated Vmax, kcat, or Km without Urea in the sample?

[Gyrase]

What technique have you used for determining native denature ratio? CD?

please notice when you say 25% denatured, it dose not mean 25% totally denatured and 75% are totally native. urea denatures all of your enzyme molecules! so we are not talking about your original enzyme we are talking about a new 3d structure!