Why doesn't $k_\text{cat}$ change in uncompetitive inhibition, given the fact that uncompetitive inhibition lowers the enzyme–substrate complex efficiency (which is the reason for lowering of $V_\text{max}$ as well)? Source of information: https://youtu.be/0ZiCqwtFMTs

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    $\begingroup$ Welcome to the Biology community! Aside from grammar and punctuation fixes, your question needs attention to the following issues: 1. Please mention the source of the information (kcat remaining unchanged in uncompetitive inhibition) and the context in which it was given. 2. 'enzyme–substrate work efficiency' is not a well defined term in biology. Can you word it better? Please edit your question to incorporate these changes. Thanks. $\endgroup$
    – Adhish
    Jul 30, 2020 at 17:39
  • $\begingroup$ The source: youtu.be/0ZiCqwtFMTs $\endgroup$ Jul 30, 2020 at 18:13
  • $\begingroup$ Enzyme- substrate complex. $\endgroup$ Jul 30, 2020 at 18:14
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    $\begingroup$ Please add these changes to the question itself (comments might not be read by everyone). $\endgroup$
    – Adhish
    Jul 30, 2020 at 18:15
  • $\begingroup$ Sorry, and thankyou $\endgroup$ Jul 30, 2020 at 18:19

1 Answer 1


Short answer

It all depends on how you look at $k_\text{cat}$ and enzyme concentrations

Long answer

The effect of an uncompetitive inhibitor on $k_\text{cat}$ can be interpreted in two ways.

Interpretation 1:

This is followed in the lecture series mentioned in the question.

See the video https://youtu.be/rNBEUGYu034 from 7:37 to 11:01 minutes. The instructor says:

... the turnover number basically describes the ability of that active site to actually transform the substrate molecules into the product molecules per unit time. Now because when the inhibitor is not bound to that particular enzyme-substrate complex, the active site's ability or efficiency to change that substrate to the product doesn't actually change, we see that the $k_\text{cat}$ value in uncompetitive inhibition also doesn't actually change...

Mathematically, $k_\text{cat}=\frac{V_\text{max}}{[\ce{E}]}$ (where $[\ce{E}]$ is the concentration of enzyme, excluding that which has been inactivated by inhibitor). Now, uncompetitive inhibition reduces both $V_\text{max}$ and $[\ce{E}]$ by the same factor. Thus $k_\text{cat}$ remains unchanged.

Interpretation 2:

Some people do things differently. They regard $k_\text{cat}$ as $\frac{V_\text{max}}{[\ce{E}]}$, where $[\ce{E}]$ is the total concentration of enzyme in the mixture (including inactivated forms). In this treatment, the uncompetitive inhibitor reduces $V_\text{max}$ but $[\ce{E}]$ stays the same. Thus $k_\text{cat}$ is reduced.

Qualitatively, the enzyme in the mixture is now a combination of active and inactivated forms, and thus has lower "efficiency" overall, i.e. lower $k_\text{cat}$.

  • $\begingroup$ Thanks for the answer. $\endgroup$ Jul 31, 2020 at 12:07

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