Pseudo-kinetics

[G1SG2]

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Okay, so for saturation kinetics...lets say we have a solution with substrate and enzyme. The rate of the reaction is Rate=k[substrate][enzyme], and is 2nd order overall. Now, when our enzyme is saturated, adding more substrate won't increase the rate of the reaction, so the rate would APPEAR to be 0th order with respect to the substrate, but 1st order with respect to the enzyme, giving an overall apparent 1st order reaction with the rate law being k[enzyme].

However, the reaction is STILL considered a 2nd order reaction because after some time, the enzymes will free up and will take more substrates and increase the reaction rate, right? The 1st order kinetics are just temporary; we never say the reaction is ACTUALLY 1st order, but rather, its pseudo-first order at the point of saturation, right?

 

[sv3]

Okay, so for saturation kinetics...lets say we have a solution with substrate and enzyme. The rate of the reaction is Rate=k[substrate][enzyme], and is 2nd order overall. Now, when our enzyme is saturated, adding more substrate won't increase the rate of the reaction, so the rate would APPEAR to be 0th order with respect to the substrate, but 1st order with respect to the enzyme, giving an overall apparent 1st order reaction with the rate law being k[enzyme].

However, the reaction is STILL considered a 2nd order reaction because after some time, the enzymes will free up and will take more substrates and increase the reaction rate, right? The 1st order kinetics are just temporary; we never say the reaction is ACTUALLY 1st order, but rather, its pseudo-first order at the point of saturation, right?

Interesting stuff. I was with you until the bolded phrase. There can only be one max rate (unless you increase enzyme concentration). Once you hit a max rate, you can't go above it. I haven't come across an example where the rate order changes -isn't it a law? The appearance can change, the actual order never does (from what I've seen).

 

[G1SG2]

Interesting stuff. I was with you until the bolded phrase. There can only be one max rate (unless you increase enzyme concentration). Once you hit a max rate, you can't go above it. I haven't come across an example where the rate order changes -isn't it a law? The appearance can change, the actual order never does (from what I've seen).

Sorry, I should've been more clear-I meant when the enzymes take up the substrates waiting in line and the reaction once again appears to depend on the substrate as well. Like, after some time (after saturation, that is), if you add substrates, the reaction will seem to depend again on the substrate as well, since more enzymes will be free. Or will this never be apparent? Will the rate just level off (as we see in the graph) and thus, the reaction will continuously seem to be 0th order with respect to the substrate after saturation?

 

[sv3]

Sorry, I should've been more clear-I meant when the enzymes take up the substrates waiting in line and the reaction once again appears to depend on the substrate as well. Like, after some time (after saturation, that is), if you add substrates, the reaction will seem to depend again on the substrate as well, since more enzymes will be free. Or will this never be apparent? Will the rate just level off (as we see in the graph) and thus, the reaction will continuously seem to be 0th order with respect to the substrate after saturation?

Wow this can actually be tough. I was/am tempted to agree with you: If substrates keep turning into products, there won't be as much substrate, and then the enzymes won't be saturated. So I can see the rate dipping from it's max, and then if you add more substrate, the rate would again hit its max if its saturated again.

The thing messing me up now is equilibrium shifts. After using up so much substrate, could there be so much product that the reverse reaction occurs as frequently as the forward, permantely saturating the enzymes? (thus more substrates come from the reverse reaction, not you adding more substrate). In this case the rate stays at its max and doesn't dip.
I have no idea on this one........can you clear this part up for me?

 

[G1SG2]

Wow this can actually be tough. I was/am tempted to agree with you: If substrates keep turning into products, there won't be as much substrate, and then the enzymes won't be saturated. So I can see the rate dipping from it's max, and then if you add more substrate, the rate would again hit its max if its saturated again.

The thing messing me up now is equilibrium shifts. After using up so much substrate, could there be so much product that the reverse reaction occurs as frequently as the forward, permantely saturating the enzymes? (thus more substrates come from the reverse reaction, not you adding more substrate). In this case the rate stays at its max and doesn't dip.
I have no idea on this one........can you clear this part up for me?

Yeah, maybe....that would explain why the curve levels off in the saturation graph and doesn't change? Maybe the new substrates from the reactant side take turns with the products to get a chance with the enzyme? But yeah I agree, I think the rate law stays the way it is, after all, it is "pseudo" first order

 

[sv3]

Yeah, maybe....that would explain why the curve levels off in the saturation graph and doesn't change? Maybe the new substrates from the reactant side take turns with the products to get a chance with the enzyme? But yeah I agree, I think the rate law stays the way it is, after all, it is "pseudo" first order

Hey, can enzymes work like that? - like it doesn't matter if its the product or reactant? I was never sure of this because i thought enzymes were specific for one particular substrate. So once an enzyme catalyzes a reaction, the substrate now is in a different form - the product. I was never sure if the enzyme could recognize it and pull off the reverse reaction too. I think it does..........but looking for confirmation?

(i think this is why catalysts work both ways?)

the second guessing never ends............................!

 

[G1SG2]

Hey, can enzymes work like that? - like it doesn't matter if its the product or reactant? I was never sure of this because i thought enzymes were specific for one particular substrate. So once an enzyme catalyzes a reaction, the substrate now is in a different form - the product. I was never sure if the enzyme could recognize it and pull off the reverse reaction too. I think it does..........but looking for confirmation?

(i think this is why catalysts work both ways?)

the second guessing never ends............................!

Okay so we have E +S (eq arrows) ES---> E + P

No, I think some enzymes have an affinity for products (lol I think I even had a practice passage on this, something about the enzyme having more of an affinity for the product than the substrate). But if not, then once you have E + P, the E can take more substrates and continue converting them to products.

 

[sv3]

Okay so we have E +S (eq arrows) ES---> E + P

No, I think some enzymes have an affinity for products (lol I think I even had a practice passage on this, something about the enzyme having more of an affinity for the product than the substrate). But if not, then once you have E + P, the E can take more substrates and continue converting them to products.

Gotcha. Now, could E take P and turn it back into S? I think enzymes work both ways but not sure. I remember reading that a catalyzed reaction at equilibrium will have both forward and reverse reactions happening at the same rate so figured enzymes must be able to swing both ways?

 

[G1SG2]

Gotcha. Now, could E take P and turn it back into S? I think enzymes work both ways but not sure. I remember reading that a catalyzed reaction at equilibrium will have both forward and reverse reactions happening at the same rate so figured enzymes must be able to swing both ways?

Yup, since they increase the rate of both forward AND reverse reactions.

 

[sv3]

ahhh. Well ya, now i can see why the rate curve might not dip at all once it hits its max. Cause then you've got less substrate but more product and both will compete for the enzyme, still keeping it saturated. I'm sure this gets messy as hell once you get into numbers and stuff but at this point i understand it now. thanks!