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Taking the reciprocal of both sides of the Michaelis-Menten equation yields the Lineweaver-Burk Equation:

$ \dfrac{1}{V} = \dfrac{K_m}{V_{max}}\dfrac{1}{[S]}+ \dfrac{1}{V_{max}} $

Plotting a $ \dfrac{1}{V}$ vs. $\dfrac{1}{[S]}$ graph, I am told that:

y-int $= \dfrac{1}{V_{max}}$ and

x-int $= -\dfrac{1}{K_m}$

How are these relationships derived from the lineweaver-burk plot? I can see how the y-intercept can be equal to $\dfrac{1}{V_{max}}$ if $\dfrac{1}{[S]} = 0$, but I don't see how x-int $= -\dfrac{1}{K_m}$ by setting $\dfrac{1}{V} = 0$? Can someone demonstrate how these relationships were derived?

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  • $\begingroup$ Equation of a line: y=mx+c or y=m(x-d) where c is y-intercept and d is x-intercept. The general equation of a line is y-y₁=m(x-x₁) $\endgroup$ – WYSIWYG Oct 24 '16 at 7:03
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Set $ \dfrac{1}{V} = 0$ and solve for $\dfrac{1}{[S]}$:

$ 0 = \dfrac{K_m}{V_{max}}\dfrac{1}{[S]}+ \dfrac{1}{V_{max}} $

$ -\dfrac{1}{V_{max}} = \dfrac{K_m}{V_{max}}\dfrac{1}{[S]}$

$ -1 = {K_m}\dfrac{1}{[S]}$

$ -\dfrac{1}{K_m} = \dfrac{1}{[S]} = $ x-intercept

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  • $\begingroup$ (+1) ... and easy to remember if we analyze the situation from a dimensional analysis point of view. The x-intercept of the Lineweaver-Burk plot will have dimensions of 1/[S], so the answer cannot be -Km, -1/Vmax or - Km/Vmax as these are all dimensionally incorrect. Similarly, the y-axis intercept cannot be Vmax or 1/Km or Km/Vmax. It will have dimensions of 1/velocity ( and is 1/Vmax). The same applies to the Eadie-Hofstee plot, the Hanes plot, and all the rest of them (including Michaelis-Menten). $\endgroup$ – user1136 Oct 23 '16 at 13:32

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