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Enzyme saturation curve

From the above picture it can be seen that, in the region "B" the activity of enzyme is not proportional to the substrate concentration.

Why don't we achieve enzyme saturation linearly? Why do we go through that curved phase( region B )?

Why don't all enzyme gets saturated at a point?

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    $\begingroup$ The binding of substrate S to enzyme E is an equilibrium which means that there will always be some free enzyme (V is proportional to ES). Section A of the line isn't really linear at all. $\endgroup$
    – Alan Boyd
    Commented Jul 13, 2017 at 15:02
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    $\begingroup$ At very high substrate concentration, the process will become diffusion limited. At that point adding additional substrate will no longer improve enzyme effectiveness, thus we get a curve and not a straight line as the enzyme concentration is constant. $\endgroup$
    – Jeppe Nielsen
    Commented Jul 13, 2017 at 23:56

1 Answer 1

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Your image is not correct.

According to Lehninger (2000):

At relatively low concentrations of substrate, V0 increases almost linearly with an increase in [S]. At higher substrate concentrations, V0 increases by smaller and smaller amounts in response to increases in [S]. Finally, a point is reached beyond which increases in V0 are vanishingly small as [S] increases. This plateau-like V0 region is close to the maximum velocity, Vmax. (emphasis mine)

This is the image that comes after that paragraph:

enter image description here

FIGURE 6–11 Effect of substrate concentration on the initial velocity of an enzyme-catalyzed reaction. Vmax is extrapolated from the plot, because V0 approaches but never quite reaches Vmax. The substrate concentration at which V0 is half maximal is Km, the Michaelis constant.

Thus, regarding your questions ("Why don't we achieve enzyme saturation linearly? Why do we go through that curved phase( region B )?"), the answer is: there is no line, it's all curved.

Source:

  • Lehninger, A., Nelson, D. and Cox, M. (2000). Principles of biochemistry. New York: Worth Publishers.
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  • $\begingroup$ But is this really an answer to the question? Despite his mistaken diagram, the basic point he is asking is why the curve is hyperbolic rather than linear. This, of course is trivial and answered in all texts, but an unhelpful answer really isn't a lot of use. Surely better to scrub the question. $\endgroup$
    – David
    Commented Jul 13, 2017 at 17:46
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    $\begingroup$ OP accepted the answer. So, I guess the answer is fine. $\endgroup$
    – Roni Saiba
    Commented Jul 14, 2017 at 0:36
  • $\begingroup$ @David I got my answer from another question's answer ( Mattdo's answer to biology.stackexchange.com/questions/9024/…) Which put emphasis on diffusion. But I accepted Gerardo answer because it highlighted mistake and there was no other answer. $\endgroup$
    – JM97
    Commented Jul 14, 2017 at 1:31
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    $\begingroup$ @JM97 Please unmark my answer as the accepted one, specially if you think that it was unhelpful. Normally, nobody writes another answer if there is already an accepted one. So, by unmarking my answer, you get more chances of receiving the answer you wish, whatever it is. $\endgroup$
    – user24284
    Commented Jul 14, 2017 at 1:35

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