Mixed vs noncompetitive Inhibition

[hu2992]

I was looking at the list from AAMC for the list of biochemistry topics, and I noticed that they listed mixed inhibition and noncompetitive inhibition as two separate topics.
However, in the Khan Academy videos, it says that mixed and uncompetitive are the same.
Is there a difference between the two types of inhibition? If so, what are the differences?

EDIT: Sorry I meant that Khan Academy said that mixed and NONcompetitive are the same.

Thank you!

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[MrRed]

Uncompetitive Inhibition and Noncompetitive Inhibition are two separate entities. Noncompetitive Inhibition is a special case of mixed inhibition. Cue brief, but adequately detailed summary 😛

To start, the equation for the initial velocity of a Michaelis-Menten enzyme can be given by the equation:

Vo = Vmax/bKm + c(S)
Where b and c are both constants, specific for a given inhibitor/enzyme combo. We can use this equation for competitive, uncompetitive, and mixed inhibiton. (S) is the substrate concentration, Vmax is the maximum velocity the particular enzyme-catalyzed reaction can attain at very high substrate concentrations, Vo is the initial velocity of the reaction, and Km is the Michaelis constant.

An uncompetitive inhibitor will bind a site distinct from the active site of the enzyme (opposed to a competitive inhibitor which competes with the substrate for the active site) and will bind only to the enzyme-substrate (ES) complex.
In the presence of an uncompetitive inhibitor the typical Michaelis-Menten velocity equation: Vo = (Vmax/Km +c(S) ) becomes:

Vo = (Vmax/Km + c(S)) = Michaelis-Menten, uncompetitive inhibitor equation
The constant b is equal to 1 in uncompetitive inhibition, and thus it is essentially absent from the equation. The constant c has a non-zero value that is greater than 1, the constant depends on the inhibitor.

As you can see this means that at very high substrate concentrations Vo will approach Vmax/c; thus Vmax is decreased. Apparent Km will also decrease because, as we know, Km is (for practical purposes) the substrate concentration required to reach 1/2Vmax. In the equation above laid out for uncompetitive inhibition, we see that for Vo = Vmax/2 we must have 2 = Km + c(S). Thus the substrate concentration required to reach 1/2Vmax decreases by a factor of c, and the "apparent Km" decreases.

A mixed inhibitor is a separate entity from both the uncompetitive inhibitor and the competitive inhibitor.
A mixed inhibitor also binds at a site on the enzyme which is physically separate from the active site, however it can bind both to the enzyme itself and to the enzyme-substrate complex. As a brief aside the nomenclature for an enzyme-substrate complex is (ES), for the enzyme-inhibitor complex it is (EI) and for the enzyme-substrate-inhibitor complex it is (ESI).

Like uncompetitive inhibitors, a mixed inhibitor will usually affect both Km and Vmax.
For a mixed inhibitor the equation Vo = Vmax/bKm + c(S) holds true.
In this equation, for a mixed inhibitor, the constants b and c both have a non-zero value. In the case where b = c, the particular mixed inhibitor is called a noncompetitive inhibitor.

Noncompetitive inhibitors will decrease Vmax but leave Km unaffected. As substrate concentration increases such that it is much greater than Km, then Vo will approach Vmax/c, the same case we saw when discussing the uncompetitive inhibitor. However, because for a noncompetitive inhibitor b = c, Km will not change.
Again, the numerical value of Km is the substrate concentration required to reach 1/2Vmax. So for Vo=Vmax/2 we must have 2 = bKm + c(S). Because b and c are the same, both Km and (S) are multiplied by the same factor, and the apparent substrate concentration required to reach 1/2Vmax does not change.

In summary:

Michaelis-Menten velocity equation: Vo = Vmax/bKm + c(S)

Competitive Inhibitor: b = non-zero, greater than 1; c = 1
Apparent Km = increased; Vmax = decreased

Uncompetitive Inhibitor: b = 1, c = non-zero, greater than 1.
Km = decreased; Vmax = decreased

Mixed inhibitor: b = non-zero, greater than 1
c = non-zero, greater than 1
Vmax = decreased; Km = decreased


Noncompetitive Inhibitor
b = c = non-zero, greater than 1
Vmax = decreased, Km = unaffected

And for the record I learned all of this from reading Lehninger Principles of Biochemistry and would encourage you to read that book for any/all biochemistry concerns you might have. It literally has anything and everything you could ever want to know about biochemistry as it pertains to the MCAT, and much much more.

 

[RH8448]

They are different first of all.

Uncompetitive inhibition the Vmax decreases and the Km decreases. The inhibitor bind to the enzyme substrate complex only.

In mixed the Vmax decreases and Km increases. Inhibitor binds to enzyme and ESC.

It helps to draw graphs of all the different types.

 

[SkyJumper42]

They are different first of all.

Uncompetitive inhibition the Vmax decreases and the Km decreases. The inhibitor bind to the enzyme substrate complex only.

In mixed the Vmax decreases and Km increases. Inhibitor binds to enzyme and ESC.

It helps to draw graphs of all the different types.

He is asking for non-competitive vs mixed inhibition. I could use some clarification as well thanks.

 

[hu2992]

found this in the Kaplan book

 

[RH8448]

Oh sorry about that...

In non-competitive the Vmax decreases, however, the Km stays the same. This means the affinity of the enzyme with the substrate is not affected, however, the turn over rate is slowed down (kcat becomes smaller). The inhibitor binds allosterically to the E or ESC. It changes it in a way that only affect catalytic efficiency. You can prove this using the Lineweaver Burke plot which I think is the best way to remember the information. Know how to interpret the ([substrate], rate of reaction) graph too though.

Mixed inhibition is like I said not the same thing as non-competitive. They are similar to noncompetitive inhibitors because the mixed inhibitor binds to the E and ESC too. But, instead of just decreasing Vmax and thus decreasing kcat, the Km increases. In other words, the mixed inhibitor also affects the enzymes ability to bind its substrate.

So, these can be complex. I think it can be easy to confuse a lot of these as well. Confusing and wrong information is on the internet so you have to be careful what you learn. The best way to understand this is to - like I said - know the graphical info and how the curves look for all of them. Know how Km and Vmax are found on the graphs. Know the equations and how to solve them. Know what is actually happening in terms of what is binding to what. I recommend this video - actually I recommend all his videos. This guy is good and perhaps the best.

 

[hu2992]

Oh sorry about that...

In non-competitive the Vmax decreases, however, the Km stays the same. This means the affinity of the enzyme with the substrate is not affected, however, the turn over rate is slowed down (kcat becomes smaller). The inhibitor binds allosterically to the E or ESC. It changes it in a way that only affect catalytic efficiency. You can prove this using the Lineweaver Burke plot which I think is the best way to remember the information. Know how to interpret the ([substrate], rate of reaction) graph too though.

Mixed inhibition is like I said not the same thing as non-competitive. They are similar to noncompetitive inhibitors because the mixed inhibitor binds to the E and ESC too. But, instead of just decreasing Vmax and thus decreasing kcat, the Km increases. In other words, the mixed inhibitor also affects the enzymes ability to bind its substrate.

So, these can be complex. I think it can be easy to confuse a lot of these as well. Confusing and wrong information is on the internet so you have to be careful what you learn. The best way to understand this is to - like I said - know the graphical info and how the curves look for all of them. Know how Km and Vmax are found on the graphs. Know the equations and how to solve them. Know what is actually happening in terms of what is binding to what. I recommend this video - actually I recommend all his videos. This guy is good and perhaps the best.

thank you!