Talk:Competitive inhibition

From Wikipedia, the free encyclopedia
Jump to: navigation, search
WikiProject Molecular and Cellular Biology (Rated Start-class, High-importance)
WikiProject icon This article is within the scope of the WikiProject Molecular and Cellular Biology. To participate, visit the WikiProject for more information.
Start-Class article Start  This article has been rated as Start-Class on the project's quality scale.
 High  This article has been rated as High-importance on the project's importance scale.
 

When the article describes a detail about classical competitive inhibition in "the substrate's apparent affinity for the site is decreased," I disagree with the statement. I believe the apparent substrate affinity is actually increased and not decreased. In competitive inhibition, addition of more substrate will out compete the inhibitor and overcome the inhibition of the enzyme's catalytic rate; thus, the Vmax will be the same and only Km will be altered. For the true competitive inhibitor, the Vmax' (apparent Vmax for inhibited enzyme) will be the same as the real Vmax, while the Km' (apparent Km for the inhibited enzyme) will be greater than the real Km. Note that Km (Michaelis constant) represents (for enzyme reactions exhibiting simple Michaelis-Menten kinetics) the dissociation constant (affinity for substrate) of the enzyme-substrate (ES) complex.

I agree, this entry needs fixing. It is incorrect to say that maximum velocity will not be attained in competitive inhibition.

The affinity is the reciprocal dissociation constant.

Allosteric inhibition is something completely different than competitive inhibition, isn it? I thing that this article requires major revision... —Preceding unsigned comment added by MR Deidra (talkcontribs) 12:05, 3 September 2007 (UTC)

I was thinking the same thing. I'm removing the text about there being two types of competitive inhibition. If I'm mistaken here please re-insert this with more information that resolves more clearly the difference between "non-classical" competitive inhibition (I've not heard of this) and non-competitive/uncompetitive inhibition, and how this "non-classical" competitive inhibition is "competitive" even though it binds at an allosteric site.--Xris0 18:17, 12 September 2007 (UTC)

Getting V vx. [S] and Lineweaver-Burke plots up here would probably be useful if anyone has the time. --Xris0 18:23, 12 September 2007 (UTC)

With respect to the affinity of the substrate, I think what the second comments are trying to say is that Km is similar to the dissociation constant, and therefore an increased Km is indicative of weaker affinity for the enzyme, so the correct statement would be to say that the "apparent affinity of the substrate for the enzyme is decreased". I agree plots (and nicer equation style) would be nice, but I don't know how to add one.A2a2a2 01:58, 15 September 2007 (UTC)



Under the "mechanism" heading it first says "In competitive inhibition, the inhibitor binds to the same active site as the normal enzyme substrate" then a few lines down it says "Note that the inhibitor does not necessarily have to bind to the same active site that the substrate would bind to"

so which is correct? don't ask me, it just seems to contradict itself.124.178.136.93 (talk) 03:29, 24 April 2008 (UTC)

Do competitive inhibitors connect forever?[edit]

Curiously speaking: Does the competitive inhibitor stay on the active site "forever"? Or does it disconnect so that at some point a substrate can connect on to it? That's the only way I can understand that Vmax isn't lowered. An increase in substrate concentration shouldn't increase the rate if there's no more active sites available. The X (talk) 11:10, 9 May 2008 (UTC)


Some examples?[edit]

Was led to this article while doing some research on Caffeine. Could an editor with some expertise in this field maybe add to this article a few examples of Competitive inhibitors?

Thanks!Darrell Wheeler (talk) 11:05, 22 March 2009 (UTC)

Notation[edit]

What does the asterisk mean in things like the following?

E_T = ES*( \frac{K_m}{S} + 1 + \frac{I*K_m}{S*K_i} )= ES* \frac{K_m*K_i+S*K_i+I*K_m}{S*K_i}

And why are the parentheses not allowed their proper sizes, as follows?

E_T = ES*\left( \frac{K_m}{S} + 1 + \frac{I*K_m}{S*K_i} \right)= ES* \frac{K_m*K_i+S*K_i+I*K_m}{S*K_i}

Michael Hardy (talk) 15:44, 24 March 2009 (UTC)

allosteric site[edit]

a competitive inhibitor does not bind to an allosteric site on the enzyme, that's what non-competitive inhibitors do. competitive inhibitors "compete" with the substrate for the enzyme's active site. and depending on which one is in higher concentration, they can outcompete each other for the site. —Preceding unsigned comment added by 67.82.4.206 (talk) 16:55, 30 May 2010 (UTC)

Why make it more complicated than it needs to be?[edit]

In deriving the equation for the initial rate I don't understand why one needs to introduce the rate constants and then differentiate. The quasi-steady-state condition is assumed anyway. I've looked at Cornish-Bowden, Fersht and Voet & Voet, and as far as I can see they all simply manipulate the following equations (which are already used in the article here):
The total enzyme concentration E0 is given by:

 [E]_0 = [E] + [ES] + [EI]\,\!

 

 

 

 

(1)

The dissociation constant of the inhibitor from the enzyme is:

 K_I = \frac{[E][I]}{[EI]}

 

 

 

 

(2)

The Michaelis constant is (by definition):

 K_m = \frac{[E][S]}{[ES]}

 

 

 

 

(3)

The velocity (rate of production of product) is

 v_0 = k_2 [ES] \,\!

 

 

 

 

(4)

The maximum velocity is:

 V_\max = k_2 [E]_0 \,\!

 

 

 

 

(5)

Then one eliminates [EI], [ES], k2, [E]0 etc to get:
 v_0 = \frac{V_\max[S]}{K_m(1 + \frac{[I]}{K_i}) + [S]}
as in the article. What am I missing? Aa77zz (talk) 14:25, 4 October 2011 (UTC)

k3 vs k–3[edit]

[I just https://en.wikipedia.org/w/index.php?title=Competitive_inhibition&diff=518382814&oldid=518379865 undid a change] to the rate-constants in the reaction equation used in the Derivation section, but now I'm not so sure (or if I'm right technically, it's still confusingly written). Normally I've seen kx as the forward rate-constant and k–x as the corresponding backward rate-constant for an equilibrium, but what we have here is k–3 forward over k3 backward. The reason I undid it is because the Equation 2 for d[E]/dt appears to treat the term with k–3 as an increase in [E], which means that it's the reaction that has E as product (the forward reaction) and Equation 4 for d[EI]/dt treats k3 as a positive term and k–3 as a negative term, which seems consistent again with the equation as written. Could someone confirm that this is the intended way to write this equilibrium? DMacks (talk) 21:01, 17 October 2012 (UTC)