Vmax and absolute activity

[bluequestions]

Hi! The following is a question from C/P (#55) from EK FL 1. You don't need any info from the passage to answer this question, so I'm just posting the question below:
During one experiment, a fixed amount of D-trehalose (substrate) was added and the activity was 60% of its maximum, an absolute value of 35 nmol/min. The trypsin (enzyme) concentration was then doubled while the concentration of D-trehalose was held constant. How did the activity of the trypsin change during this time?

Answer: It increased from 35 nmol/min to 50 nmol/min.

I picked the wrong answer, which said that the activity remains at 35nmol/min. When I looked at the solution, I understand the explanation which says that increasing the enzyme means more chances of ES complex forms, which means greater activity of the reaction.
However, I think I'm still a bit confused. It seems from the question that increasing concentration of enzyme allowed greater activity, does this imply that originally there's not enough enzyme to take up all the enzymes around, therefore the activity was only 60% of max? but doesn't this also imply that all the enzymes originally were saturated already? If this is the case, according to M-M kinetics, shouldn't the activity reach the maximum rate or activity at this point, which is the definition of Vmax (rate of reaction when all enzymes are saturated)?

Thank you!

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[aldol16]

The Vmax is not the same in both cases. Imagine this scenario. You have a printer that can print at a rate of 10 pages/minute. The printer is the enzyme. Paper is the substrate. The Vmax here is 10 pages per minute. Now, you buy four other printers just like it. So now, wouldn't you say your rate of printing has changed? Now, you can print 50 pages per minute. In other words, your Vmax has changed because you have more printers (enzymes).

 

[Neuroplasticity]

However, I think I'm still a bit confused. It seems from the question that increasing concentration of enzyme allowed greater activity, does this imply that originally there's not enough enzyme to take up all the enzymes around, therefore the activity was only 60% of max? but doesn't this also imply that all the enzymes originally were saturated already? If this is the case, according to M-M kinetics, shouldn't the activity reach the maximum rate or activity at this point, which is the definition of Vmax (rate of reaction when all enzymes are saturated)?

Thank you!

Yes, Vmax changes, but I think he is asking why is the original case not at Vmax.

Increasing concentration of enzyme causing greater activity does not imply that the original enzyme was saturated (not at Vmax).
1. It is like initially you have 1 printer that has to wait an average of 5 sec. for a paper to randomly fall in its tray. This printer is not saturated with paper, and not at its max rate.
2. Then you increase to 10 printers, and now there is a higher chance that paper will randomly fall into one of their trays. So the rate increases. This is like you said the solution says "increasing the enzyme means more chances of ES complex forms."

 

[aldol16]

Yes, Vmax changes, but I think he is asking why is the original case not at Vmax.

I see. When doing experiments like these, we usually do it at large excess substrate concentration. It's bad practice to do M-M kinetics at non-enzyme-limiting conditions. You want to measure Vmax so you want your enzyme to be in a solution filled with substrate that it binds.

 

[Neuroplasticity]

I see. When doing experiments like these, we usually do it at large excess substrate concentration. It's bad practice to do M-M kinetics at non-enzyme-limiting conditions. You want to measure Vmax so you want your enzyme to be in a solution filled with substrate that it binds.

I have never done any experiment like this, but that certainly makes sense. And that you for providing so many detailed responses here, I'm sure I would have appreciated it when I was studying for the MCAT.

 

[bluequestions]

Hi guys, sorry for the late reply! First of all, thank to both of you for the answers! I still have some questions I want to clear up:

Increasing concentration of enzyme causing greater activity does not imply that the original enzyme was saturated (not at Vmax)

Sorry, did you mean "does not imply that the original enzyme was UNsaturated"?

So in the original scenario in the question, is the experiment done under non-excess substrate substrate condition then? hence why the enzyme is only reaching at 60% of its Vmax, instead of full Vmax?

And just to confirm, depending on the amount of enzymes that's in the sample ran, we get different Vmax if we vary enzyme concentration in each trial?

Thank you again!

 

[Neuroplasticity]

Sorry, did you mean "does not imply that the original enzyme was UNsaturated"?

I did not mean that, you said it implied that the enzyme was saturated, and I am saying that it does not imply that the enzyme was saturated.

So in the original scenario in the question, is the experiment done under non-excess substrate substrate condition then? hence why the enzyme is only reaching at 60% of its Vmax, instead of full Vmax?

And just to confirm, depending on the amount of enzymes that's in the sample ran, we get different Vmax if we vary enzyme concentration in each trial?

Yes and yes.

I have a question though, why did doubling the total enzyme not double the rate? M-M equation shows that doubling total enzyme doubles rate?

 

[aldol16]

I have a question though, why did doubling the total enzyme not double the rate? M-M equation shows that doubling total enzyme doubles rate?

My gut tells me that it's simply a bad question from EK, as per usual with these companies. That's the simplest explanation. But we could come up with specific circumstances where doubling enzyme concentration wouldn't double rate. One obvious one is that we're not at large excess of substrate conditions. Remember that doubling enzyme doubles Vmax. Vmax only applies when substrate is in excess - in this way, Vmax is actually a measure of the maximum efficiency of the enzyme. The enzyme doesn't work at maximum efficiency all the time - that is, not at non-excess substrate conditions.

 

[bluequestions]

Thank you!