I am studying the DNA binding and kinetic properties of a protein. I found that the Kd value for binding a DNA substrate are much lower than the respective Km value. Is this observation suggestive of the presence of a conformational change that precedes the chemical step? and how?
Hi there,
if your enzyme follows the simple scheme E+S ES--->E+P, KD
I might state the case a little differently from Dominque. We weren't told what the enzyme is, so it may not be an issue of processivity (such as by a polymerase or exonuclease). Assuming that this is a 1-substrate enzyme, such as an endonuclease, the observation that KD> kcat because KD=k-1/k+1 and KM=(k-1+kcat)/k+1. In words, the rate constant for dissociation of the ES complex is not greatly larger than the rate constant for catalysis.
Of course, if the enzyme has 2 or more substrates, the meaning of KM is more complicated and it can't be compared with KD so easily.
Hi Adam,
From a general point of view, I do agree with you : if KM>KD then kcat can't be considered 2 so at least n=3) so, as I already wrote it mathematically in my previous post, it's a matter of kcat being larger than k-1 (kcat=(n-1)*k-1and not "k-1 being not greatly larger than kcat") which actually means catalysis occurs faster than dissociation ((n-1) fold faster).
Hi Dominique, You're right. My answer could have been stated better.
If k-1 >> kcat (dissociation is much faster than catalysis), KM is approximately equal to KD. If this condition is not met, KM>KD. For example, if KM=3KD, then kcat=2k-1. In this example, catalysis is faster than dissociation.
However, more generally, KM>KD even if dissociation occurs somewhat faster than catalysis. For example, if k-1=3kcat, then KM=1.33KD.
Hi Adam and Dominique.
Thankyou for your responses.
I want to mention here that my protein is AP endonuclease which is a multifunctional protein. Apart from having AP endonuclease activity it also possesses 3'-5' exonuclease activity. The substrate that I have taken in the above experiment is an abasic site containing DNA. So the enzyme first cleaves at the abasic site and further employ its exonuclease activity to produce smaller sized products. How shall I deduce my results now from the above observation. i.e Km>>Kd in terms of indication of a CONFORMATIONAL CHANGE preceding the actual chemical step?
I don't think you can say anything definitive about conformational changes based solely on the above-mentioned observation. You are probably going to have to use biophysical techniques to investigate this phenomenon, such as pre-steady-state rapid kinetics to measure rates of individual steps.
One thing that concerns me is whether you can clearly delineate the chemical reaction for which the KM and KD were measured, given that the enzyme has 2 catalytic activities. Did you measure the initial rate of the endonuclease step, or the overall rate of both reactions?
I have also calculated rates for both reactions using separate oligonucleotide substrates. For endonuclease activity i used abasic site containing DNA while for exonuclease activity i used 3' recessed DNA as substrate. However, i have found out the dissociation constant using abasic DNA using and calculated initial rates for endonuclease step
I should be very careful about stating a precise value for KD because different techniques can give you a wide error range of its value. How did you measure the KD? I agree with Adam as regards the conformational change statements.
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