Why is the Km of an enzyme half of Vmax? [closed]

Why is the Km of an enzyme half of Vmax?

closed as off-topic by Bryan Krause♦, theforestecologist♦, WYSIWYGJan 9 at 10:57

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• 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. – David Dec 14 '18 at 16:25
• 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 – user1136 Dec 14 '18 at 17:23

$$K_M$$ 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 $$V=V_{\mathrm{max}}/2$$.
• 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. – Jam Dec 14 '18 at 15:23