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Author Topic: Enzyme Kinetics: Units of kcat?  (Read 42968 times)

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kristinase

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Enzyme Kinetics: Units of kcat?
« on: October 07, 2009, 12:19:31 PM »

I have my Michaelis-Menten plot of rate (units of mols product/hr/mol enzyme) vs. substrate concentration.

From here, I have used a Hanes-Woolf plot to determine Km and Vmax. My Vmax is in units of mols product/hr/mol enzyme. I am getting confused by units as I try to use the equation kcat = Vmax/E, where E is the total enzyme concentration. Why don't the Units of Vmax/E cancel to give units of inverse time, like I know the units of kcat should be?
What units should E be in? If I use concentration units such as mols of enzyme/L, then when trying to cancel the units, I'm left with a mess of: (mols product/(hr* mols enzyme)) / (mol/L enzyme).

Thanks
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renge ishyo

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Re: Enzyme Kinetics: Units of kcat?
« Reply #1 on: October 07, 2009, 02:59:16 PM »

There is some problem with your graph in the sense that your units of Vmax to start with contain an extra enzyme term in the denominator that shouldn't be there. Because I can't see your data, I will settle for briefly describing what the situation should be.

You equation should be Vmax = kcat[E]

In this, Vmax = the rate of substrate converted to product per unit time which I will write as "[R]/t"

The units of kcat should be [R]/t[E] such that when you multiply kcat by the total enzyme concentration the units of vmax come out to [R]/t which is what you measured using the spectrometer or whatever. The total enzyme concentration must cancel with the units of kcat to give the proper units for vmax.

The interpretation of kcat (in spite of its funky units) is that it gives the amount of substrate converted to product per unit time per enzyme. Here is the idea with some easy numbers to see what I mean. Let Vmax = 10 molecules of substrate converted to product per second. If two enzyme molecules were added to the flask before this experiment began then kcat would be:

kcat = Vmax/[E] = 10/2 = 5 molecules of substate/1second * 1 enzyme molecule = 5 s-1

So a kcat of 5 means that each individual enzyme "turned over" 5 molecules of substrate every second (and kcat is frequently called the "turnover number" as a result). Some books drop the units for the substrate over enzyme concentrations and just list the units for kcat as s-1 as shown above. This is because the interpretation given above is considered to be "understood" (yay for ambiguity!).

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kristinase

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Re: Enzyme Kinetics: Units of kcat?
« Reply #2 on: October 08, 2009, 07:59:07 AM »

Thank you, that helps a lot. The kcat units were very ambiguous and not well explained in any of my textbooks.

I used those originally stated Vmax units because I had seen some examples in which the units of Vmax was in mol/(time*mg of enzyme). I thought that it was a good point to account for the amount of enzyme in the reaction. This still leaves me wondering - don't the Km and Vmax depend on enzyme concentration? Is there anyplace in these calculations that accounts for this?

Lets say my Vmax is 5 nmol product per minute, but I would think that this will also depend on the amount of enzyme present when I did my assay. If I had used 1 nmol enzyme in 1mL, and next time I used 2nmol enzyme in 1L, won't I now see a Vmax of 10 nmol of product per minute? There's twice as much enzyme in that reaction to make twice as much product, right?

Thanks again for sharing your enzyme kinetics knowledge!
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renge ishyo

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Re: Enzyme Kinetics: Units of kcat?
« Reply #3 on: October 08, 2009, 08:28:37 AM »

Vmax is dependent on the enzyme concentration in just the way you say. You can see this clearly from the formula:

Vmax = kcat[E]

As the total enzyme concentration "[E]" increases, so does Vmax as given by the above formula. However, Km does *not* vary as the total enzyme concentration varies. You can see this from the Michaelis-Menten equation:

V = Vmax[R]/(Km + R)

The value of the Km can be solved for at 1/2 Vmax. Making the substitution 1/2 Vmax = V in the equation above gives:

1/2Vmax = Vmax[R]/(Km+[R])

You can see that Vmax cancels out of the equation at this point:

1/2Vmax = Vmax[R]/(Km+[R])

If you go forward and solve for [Km], you will get that Km = [R], where [R] is the substrate concentration at 1/2 Vmax.

This result may seem weird at first, but it makes sense when you think about it. Increasing the amount of enzyme increases the number of enzyme active sites, but it does not increase the *affinity* of the enzyme active site for binding its substrate. If you have an enzyme with a high Km (meaning that the substrate binds poorly to the enzyme), the substrate will still bind poorly to the enzyme even if you add in a ton of extra enzymes...the reaction will go faster (higher vmax) because there are more enzymes for the substrate to bind to, but the Km which describes how well the substrate binds will not change.
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Yggdrasil

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Re: Enzyme Kinetics: Units of kcat?
« Reply #4 on: October 08, 2009, 08:32:54 AM »

Yes, Vmax is dependent on enzyme concentration from the equation you provided:  Vmax = kcat[E], where kcat is independent of the enzyme concentration.  So, you are correct, if you double the enzyme concentration, your Vmax should double.

Since the rates you report have already been divided by the amount of enzyme, you do not need to divide by the enzyme concentration again to get kcat.  Given that you are plotting V/[E] versus [S ], The value at which your Michaelis-Menten curve saturates will be your kcat (note that the moles of product and moles of enzyme can be thought to cancel assuming one active site per enzyme, giving units of inverse time).

Km, however, is not dependent on the concentration of enzyme.

(edit: renge ishyo beat me to, but I'll leave my explanation anyway in case it helps)
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renge ishyo

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Re: Enzyme Kinetics: Units of kcat?
« Reply #5 on: October 08, 2009, 08:34:53 AM »

(edit: I'm sure the TC won't complain, the more the merrier  ;) )
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