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Why is the Km of an enzyme half of Vmax?

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closed as off-topic by Bryan Krause, theforestecologist, WYSIWYG Jan 9 at 10:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework questions are off-topic on Biology unless you have shown your attempt at an answer. For more information see our homework policy." – Bryan Krause, theforestecologist, WYSIWYG
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ We welcome new users to SE Biology, but we expect them to read the help on asking questions before posting, and to look at examples of well-presented questions. Your question, I am afraid does not come up to the mark. You have put half your question in the title, and the rest as a single line. The premise is incorrect in actual fact — Km is the substrate concentration at Vmax/2, and I have no idea what you mean by "any other specific reason". This is why your question has been downvoted. $\endgroup$ – David Dec 14 '18 at 16:25
  • $\begingroup$ As Jam has pointed out, for a single-substrate enzyme that obeys the Michaelis-Menten equation, Km is always the substrate concentration at Vmax/2. But there is another way of thinking about Km after Fersht and Dalziel. It is the ratio of the apparent first-order rate constant (kcat) and the apparent second order rate constant (the specificity constant or kcat/Km). Unlike kcat and the specificity constant, Km is independent of enzyme concentration $\endgroup$ – user1136 Dec 14 '18 at 17:23
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$K_M$ is a constant and is not $V_{\mathrm{max}}/2$ as they have different dimensions. But by the Michaelis Menten equation, $V=\frac{V_{\mathrm{max}}[S]}{K_M+[S]}$. So, when $[S]=K_M$, we have $V=\frac{V_{\mathrm{max}}[S]}{[S]+[S]}=\frac{V_{\mathrm{max}}}{2}$. Hence, $K_M$ is the substrate concentration, at which we would have $V=V_{\mathrm{max}}/2$.

Reference: Worthington Biochemical Corporation

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  • $\begingroup$ We can also see that $K_M$ can't be identically $V_{\mathrm{max}}/2$ since $V_{\mathrm{max}}$ is dependent on the initial enzyme concentration while $K_M$ isnt. $\endgroup$ – Jam Dec 14 '18 at 15:23
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    $\begingroup$ Km is never Vmax/2. It is the substrate concentration at Vmax/2. I would sort out the question before giving an answer. In this case I would not answer the question because it is so badly presented and shows so little effort on the poster's part. See "Answer well-asked questions" in the Help on answering questions. And the suggestion that a naive user reads a "credible textbook on enzyme kinetics" is, well, incredible. Again, point him to a page online. And if you answer take the trouble to set it out in a readable manner. $\endgroup$ – David Dec 14 '18 at 16:20
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    $\begingroup$ [Edited] I believe your comment and my answer say the exact same thing. I've told the OP that what they've said is false. In my opinion, even if it's a low effort question, the OP deserves to know that they're wrong. If you would like the OP's question to be deleted, then please vote for that. If you take any particular issue with my formatting that suggests my answer isn't readable, please point it out. $\endgroup$ – Jam Dec 14 '18 at 17:11

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