J.R. S. answered • 12/29/17
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Ph.D. in Biochemistry--University Professor--Chemistry Tutor
The Kd is the dissociation constant and is for the reaction ES ==> E + S, not E + S ==> ES. This latter reaction is for the association constant, Ka. So, you are correct that the expression for K is products/reactants. This is also true for Kd, which is actually [E][S]/[ES]. So, the Kd is a ratio of koff/kon (note these are lower case "k" values which are rate constants). Also recall that in enzyme kinetics, the conversion of ES to E + P is assumed to be negligible because we use initial rates before much P is formed and thus k3 is not included. Also, in enzyme kinetics, the Ka is given the term Km for the Michaelis constant.
J.R. S.
tutor
I’ll take your questions one at a time. The Kd and Km are NOT the same, but yes, they are sort of a measure of binding of the substrate to the enzyme, but not the frequency. The Km is sort of an affinity constant which is the [S] to reach 1/2 max. velocity.
The reason we measure product formation early (initial rates) is because the Michaelis Menten equation is based on the assumption that k3 is negligible and no product is formed. This is essential the case during the ear stages of the enzyme reaction. If you measure the kinetic constants after significant product formation, the equations become much more complex because you need to factor in k3 as the rate constant for ES => product. Not to mention the possible complicating factor of product inhibition.
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12/29/17
Chris Y.
Ok, so just breaking that down:
So Km and Kd are not the same K, but the approx. measure the same concept, correct?
If an enzyme has a high Km, then the enzyme is more likely to dissociate from the substrate and at any instant in time (not over a period of time) in order for there to be a high enough concentration of ES to allow for the enzymatic activity to equal 1/2 of Vmax?
I'm thinking that the reason why its not a frequency is because Km would be the [S] to equal 1/2 of Vmax at any instant in time, correct?
So then Km does not measure the productivity of the enzyme at all? Just the [S] necessary to create a high enough [ES] so that the enzyme will function at 1/2 Vmax? So its the likelihood of that enzyme to bind to the substrate without having to increase the [S] and reach 1/2 Vmax? Sorry thats a few ways of saying almost the same thing. I just want to make sure that the way I'm understanding it is correct.
As well, kcat is a measure of ES <--> E +P, correct? is kcat essentially k3?
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12/31/17
J.R. S.
tutor
So Km and Kd are not the same K, but the approx. measure the same concept, correct? -
Yes, correct. They are both sort of measures of the affinity of S for E.
If an enzyme has a high Km, then the enzyme is more likely to dissociate from the substrate and at any instant in time (not over a period of time) in order for there to be a high enough concentration of ES to allow for the enzymatic activity to equal 1/2 of Vmax? - Not so much that the enzyme is more likely to dissociate from the substrate (that would be Kd), but rather a high Km suggests that the substrate is less likely to bind to the E in the first place.
I'm thinking that the reason why its not a frequency is because Km would be the [S] to equal 1/2 of Vmax at any instant in time, correct? - Yes, correct
So then Km does not measure the productivity of the enzyme at all? Just the [S] necessary to create a high enough [ES] so that the enzyme will function at 1/2 Vmax? - No, not exactly. Km is a combination of factors (rate constants - see below) so it does incorporate what you call productivity. Since the Km expression has in it the term for Vmax, this includes "productivity". Just so we are using the same terms, E + S <==> ES ==> P + E where k1 is forward reaction for E+S =>ES and k2 is reverse reaction; k3 is forward reaction for ES => P. Km is (k2+k3)/k1, so it does contain "productivity). Since v = Vmax [S]/Km + [S] again Km contains the Vmax term so inherently has "productivity" within it.
As well, kcat is a measure of ES <--> E +P, correct? is kcat essentially k3? Yes, that is correct. kcat can be viewed as the turnover number or the number of substrate molecules converted to product per second. See ES => P + E. It is calculated also as Vmax/[E].
I hope this all makes some sense. I find it difficult to explain completely enzyme kinetics in this format of "email", without a white board, and other supplemental material. I've done my best.
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12/31/17
Chris Y.
12/29/17