I have found a diagram similar to the following one in my biology textbook. The diagram describes allosteric regulation.

enter image description here

However, I do not quite understand why the maximum reaction rate $V_{max}$ is still reached, even though I thought this is only possible for a competitive inhibition.

I think that the diagram should look something like this where the maximum reaction rate is not reached.

enter image description here

Is there simply an error in the diagram or have I misunderstood something here?

  • 1
    $\begingroup$ My textbook states clearly that $V_{max}$ remains the same in a case of competitive inhibition but is is lower in case of non-competitive inhibition under which, according to my textbook, you can count allosteric inhibition. Also I see diagrams like this everywhere where $V_{max}$ is lower for a non-competitive inhibition $\endgroup$ – HansMu158 May 6 '17 at 14:35
  • $\begingroup$ Your text-book is incorrect in saying that allosteric inhibition is not competitive. I do not know what your text book actually says (and do not care). The curves usually show the negative effector not reaching the maximum rate because a very high concentration would be needed to achieve this (and in any case the curves only reach this at infinite concentration). I have added my own answer to the question I cited (as I had been meaning to do for a while because of the confusion these terms cause). Look at my diagram carefully and you will see why allosteric negative effectors are competitive. $\endgroup$ – David May 6 '17 at 14:52
  • $\begingroup$ To clarify. The diagram in your textbook is probably correct, if idealized, but I cannot be sure as I presume you have redrawn in it from the spelling mistake "inhibtor". $\endgroup$ – David May 6 '17 at 14:55
  • $\begingroup$ Ok, so after watching this video on Khan Academy and reading your answer to the other question, I got it clear that allosteric inhibition can be competitive. So in the diagram in my textbook I have a competitive allosteric inhibition, right? $\endgroup$ – HansMu158 May 6 '17 at 14:56
  • $\begingroup$ Yes, I have redrawn it, but took caution to have it be exactly the same. I apparently messed up the spelling, though, sorry. $\endgroup$ – HansMu158 May 6 '17 at 14:57