I know definition of Km and Kcat. when affinity of enzyme to substrate in the two enzyme ( wild_type and without small domain) haven't had any change, how does the rate of catalysis increase? My enzyme is cellulase, I delete small domain from it. There is no change In km, but Kcat of truncated enzyme increases. This domain has some hydrophobic interaction with catalytic domain .
Hi there,
Your results show that the small domain acts as a non competitive inhibitor in the full length enzyme. The contact between the two domains leads to a less efficient catalysis without affecting the binding of substrate. Have you tried to co express the small domain and the truncated form to see if they are able to interact in a stable manner without covalent linkage?
Hi dear Dominique
TNX for answer
no, I just express truncated domain.
Dear Arshad
thank you for complete answer . reaction mix is the same for both of them
Regards
Hi, the effect can be explained by increased dynamics of the catalytic domain after the removal of the small domain. If the enzyme undergoes a conformational change related to binding the substrate and this conformational step is rate-limiting for catalytic turnover, then after its acceleration (e.g. by increasing flexibility) kcat will increase while Km remains the same. This will be true to the extent that the rate constant for this conformational change is rate-limiting or equal to the next catalytic step(s). If the question is still relevant, I can send kinetic calculations for 3 and 4 step models, reversible and irreversible, and related simulations that show this effect quantitatively. Best wishes, Zbynek
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