Identifying type of inhibitor from $K_m$ and $V_{max}$

Question

Apparently it is possible to identify whether an inhibitor is competitive or non-competitive from graphs of substrate concentration (x axis) and rate of reaction (y axis).

There needs to be a line for with inhibitor (constant concentration) and without inhibitor. The difference in $K_m$ and $V_{max}$ can then be interpreted to find out whether the inhibitor is competitive or not.

I know that if the effect of the inhibitor decreases with an increase in substrate concentration then inhibitor is competitive.

However I don't understand how to tell from the graphs mentioned earlier and $K_m$ and $V_{max}$. Could someone explain.

I have tried to google this but most of what comes up is scholarly articles which are beyond my level of understanding...

Answer 1

I think it is possible to identify the type of inhibition from (initial) velocity vs substrate-concentration curves, but it is difficult. The usual way this is done is by using a linear transformation of the Michaelis-Menten equation, such as the Lineweaver-Burk plot.

But you are right: for a reversible inhibitor, the way to identify the inhibition pattern (that is, to determine whether a reversible inhibitor is competitive, uncompetitive, mixed, or non-competitive) is to inspect the changes to the kinetic constants (usually K_m and V_max, but see below)

Before we get into how that is done, there are a few points we need to be aware of.

• The following only applies to reversible inhibitors. Irreversible inhibition, such as the inhibition of acetylcholinesterase by the nerve-gas sarin, is treated differently. (By 'reversible', it is simply meant that if the inhibitor is removed, by dilution for example, the inhibition goes away). In addition, tight-binding inhibitors are not considered.
• The following also only applies to single-substrate enzymes that obey the Michaelis-Menten equation. (But the analysis may be extended, without too much difficulty, to multi-substrate enzymes).
• We need to be very careful about the term 'non-competitive'. It means different things to different people. It is also often the least interesting pattern of inhibition.
• When analyzing inhibition patterns, it is often easier to analyzed the effect on V_max and V_max/K_m (the specificity constant), rather than on K_m and V_max. The reason for this is that Km is a complex kinetic constant and (as Dalziel and Fersht have shown), it should be thought of as the ratio of V_max and the specificity constant (V_max/K_m). In this respect, enzyme kineticists have 'messed things up' by considering the specificity constant to be the ratio of the maximum velocity and the Michaelis constant: it is the Michaelis constant that is the ratio. However, the paradigm of considering the specificity constant as V_max/K_m is now so ingrained, I'll stick with it. But the 'trick' in analyzing inhibition patters is to think in terms of V_max and V_max/K_m.
• Much of what follows may also be applied to reversible activation, but I am not going to go into that at all.

1. Reversible Inhibitor Patterns

We can now define our inhibition patterns, independent of any mechanism that gives rise to them, as follows:

• A competitive inhibitor has no effect on V_max but decreases the apparent value of V_max/K_m. We can also say, in 'old school terms', that a competitive inhibitor has no effect V_max but increases the apparent K_m value. Or, if we are going to 'visualize' things in terms of Lineweaver-Burk plots (see this wikipedia article), we can say that a competitive inhibitor has no effect on V_max but increases the apparent value of K_m/V_max

• An uncompetitive inhibitor decreases the apparent value of V_max but has no effect on V_max/K_m. Or, thinking in terms of reciprocals, an uncompetitive inhibitor increases the apparent value of 1/V_max but has no effect on K_m/V_max. In many ways, 'uncompetitive' is a a very poor term. Cornish-Bowden (2004) suggests the term 'catalytic inhibitor', and Laidler and Bunting use the term 'anti-competitive' to describe this type of inhibition.

• A mixed inhibitor decreases the apparent value of V_max and decreases the apparent value of V_max/ K_m. Or, thinking in terms of reciprocals, a 'mixed' inhibitor increases the apparent value of 1/V_max and increases the apparent value of K_m/V_max.
• A non-competitive inhibitor is best thought of as a special case of mixed inhibition where the apparent values of V_max and V_max/ K_m are decreased to the same extent. There is an interesting consequence of this: as the Michaelis constant may be thought of as the ratio of these two kinetic constants, it is unchanged in non-competitive inhibition. But it is difficult to envisage a realistic kinetic mechanism that results in this type of behavior. Cornish-Bowden (2004, pp 118-119) is very strong on this point (There is also a fourth edition of this great book).
• We now come to a 'tricky' bit. Some authorities, notably Cleland, do not distinguish between 'mixed' and 'non-competitive' inhibition, but instead call all cases where both V_max and V_max/ K_m are decreased 'non-competitive inhibition'. We need to be very careful on this one. As someone once said, enzyme kineticists would rather use each other's toothbrushes rather than use each other's nomenclature.

So there we have it: the two cases that 'demarcate' reversible inhibition are where only the apparent V_max is changed (uncompetitive inhibition) and where only the apparent V_max/K_m (the specificity constant) is changed (competitive inhibition). 'Mixed' inhibition is where the apparent values of both these kinetic constants are affected, and a special case of 'mixed' inhibition is where the the apparent values of both kinetic constants are decreased to the same extent, resulting in no change to the Michaelis constant.

2. Reversible Inhibitor Mechanisms

So far, I have said nothing about the mechanisms that might give rise to these inhibition patterns. Preliminaries:

• In what follows, $K_m$ is the Michaelis constant, $V_{max}$ is the maximum velocity, $s$ is substrate concentration, $i$ is inhibitor concentration, and $K_{i}$ and $K_{ii}$ are inhibition constants. Both $v$ and $v_i$ refer to the initial velocity.
• Inhibition patterns are analyzed using the Lineweaver-Burk plot. This is convenient as in such a plot (see this wikipedia article), the y-axis intercept equals $1$/$V_{max}$ and the slope equals $K_m$/$V_{max}$. Changes to the apparent value of $V_{max}$ are manifested as a y-axis-intercept effect, and changes to the specificity constant ($V_{max}$/$K_m$) are manifested as a slope effect. In addition, the x-axis intercept equals $-1$/$K_m$, so that changes to the apparent $K_m$ value, or lack of such a change, is easily recognized.
• The Lineweaver-Burk plot is not the only linear transformation of the Michelis-Menten equation, or even the best one (see here). Other plots are the Hanes-Woolf plot and the Eadie-Hofstee plot. As someone else once said, biochemists worship at the alter of the straight line. I use the Lineweaver-Burk plot because, IMO at least, it is the most intuitive.
• It needs to be borne is mind that many kinetic mechanisms may give rise to an inhibition pattern. Many kinetic mechanisms may give rise to competitive inhibition, for example. What follows are illustrative examples.
• All of the plots below (generated using Mathematica) were created with $V_{max}$ = 100 and $K_m$ = 10. When only one inhibition constant was required ($K_i$), it was set to 100. When two were required, the second ($K_{ii}$) was set to 20 (All 'arbitrary units'). The plots, of course, are very easy to generate and may be done with many software applications.

(a) Competitive Inhibition

Let's consider reversible inhibition in single-substrate enzyme described by the mechanism shown above, where both the inhibitor and substrate compete for the 'free' enzyme.

Derivation of the rate law for this mechanism using either the equilibrium or steady-state assumption, leads to an equation of the following form (nice derivations are given in Segel, 1975):

$$ v_{i} = { {{{V_{max}}}\ s\ }\over{K_{m}(1 + {i\over{K_{i}}})} + s}\ \ \ \ \ (1)$$

Representative plots of Eqn (1), showing the effect of increasing inhibitor concentrations:

Taking reciprocals of Eqn (1) followed by rearrangement leads to the Lineweaver-Burk linear transformation:

$${1\over{v_i}}=\frac{K_m}{V_{max}}(1 + {i\over{K_i}})( {1\over{s}}) + {1\over{V_{max}}}\ \ \ \ \ (2)$$

It is immediately obvious that the inhibitor increases the apparent value of $K_{m}$ / $V_{max}$ but does not effect 1 / $V_{max}$.

That is, it effects the specificity constant ($V_{max}$ / $K_{m}$) but not $V_{max}$. Inhibition is therefore competitive

Plots of Eqn (2) predict a family of straight lines intersecting on the y-axis at $1$/$V_{max}$, and the competitive inhibition pattern is easily recognized:

In terms of the Lineweaver-Burk transformation, a competitive inhibitor causes the slope to increase but does not change the y-axis intercept.

We can also go a step further. The slopes of the above lines are given by the following Eqn:

$$slope = \frac{K_m}{V_{max}} + \frac{K_m}{V_{max} K_{i}}{i}\ \ \ \ \ (3)$$

Thus a plot of slope vs inhibitor-concentration is predicted to be a straight line which intersects the x-axis at -$K_i$

Such replots serve two functions. Firstly, they allow determination of the $K_{i}$ value. In this case, the x-axis intercept is -100, which is not too surprising as 100 was the value of $K_i$ chosen in the simulation. Secondly, they check for unexpected kinetic complexity. A curved slope replot, for example, might be indicative of partial competitive inhibition, where the EI complex can perhaps breakdown to give product. Such kinetic complexity is probably rare with single-substrate enzymes, but may occur in multi-substrate enzymes (and may require the rejection of a simple kinetic mechanism as an explanation of kinetic data). Segel (1975) is very strong on partial inhibition, and the mechanisms that may give rise to it. When the slope replot is linear we may speak of linear competitive inhibition (see Cornish-Bowden, 2004).

A number of points may be made about competitive inhibition:

• One of the 'hallmarks' of competitive inhibition is that the inhibitory effect may be overcome by adding excess substrate.
• A competitive inhibitor need not bind the the active site. All that is required is that it binds to the free enzyme in a manner that prevents substrate binding. An allosteric inhibitor, for example, may be competitive. But, of course, competition between the substrate and inhibitor for the same active site is one way that competitive inhibition may arise (assuming that binding of inhibitor prevents substrate binding). Segel (1975) is very strong on this point.
• A good example of a competitive inhibitor is malonic acid, which inhibits succinate dehydrogenase (see Segel, 1975). In the world of two-substrate kinetics, pyrazole is a competitive inhibitor, with respect to ethanol, of horse liver alcohol dehydrogenase, and a classic paper (Li and Theorell) showing this is available here

Finally, let's reiterate this point: Eqn (1) describes one mechanism that gives rise to a competitive inhibition pattern. It is certainly not the only one.

(b) Uncompetitive Inhibition

Now let's consider a mechanism, described by the diagram above, where the inhibitor cannot bind to the 'free' enzyme, but instead bind to the enzyme-substrate complex (to give an abortive EAI ternary complex)

Derivation of the rate law his mechanism (again by making either the steady-state or equilibrium assumption; see Segel, 1975) leads to an equation of the following form:

$$ v_{i} = { {{{V_{max}}}\ s\ }\over{K_{m}} + (1 + {i\over{K_{i}}}) s}\ \ \ \ \ (3)$$

Transformation to the Lineweaver-Burk form:

$${1\over{v_i}}=\frac{K_m}{V_{max}}( {1\over{s}}) + {1\over{V_{max}}}(1 + {i\over{K_i}})\ \ \ \ \ (4)$$

In this case, the inhibitor increases the apparent value of 1 / $V_{max}$ but does not effect $K_{m}$ / $V_{max}$

In other words, it effects the apparent value of $V_{max}$, but has no effect on the specificity constant ($V_{max}$ / $K_{m}$). Inhibition is therefore uncompetitive.

Furthermore, unlike the case of competitive inhibition, increasing the substrate concentration does not abolish inhibition.

An uncompetitive inhibitor causes the slope of a Lineweaver-Burk plot to increase, but does not change the y-axis intercept of such a plot.

Therefore double-reciprocal plots of $1$/$v_i$ vs $1$/$s$ at different $i$ form a family of parallel lines.

In this case, the $K_{i}$ may be determined from an intercept-replot, where the x-axis intercept is -$K_i$.

$$intercept={1\over{V_{max}}} +{i\over{V_{max} {K_i}}}\ \ \ \ \ (5)$$

(c) Mixed Inhibition

Now let's consider a mechanism where the inhibitor may bind to either the 'free' enzyme or the enzyme-substrate complex and (to keep things somewhat realistic) where the substrate may bind to either the free enzyme or the enzyme-inhibitor complex.

Under certain simplifying assumptions (see Segel, 1975) the mechanism shown above may give rise to the following rate law:

$$ v_{i} = { {{{V_{max}}}\ s\ }\over{K_{m}(1 + {i\over{K_{i}}})} + (1 + {i\over{K_{ii}}}) s}\ \ \ \ \ (6)$$

In this case, there are two inhibition constants, one 'governing' the binding of inhibitor to the 'free' enzyme ($K_{i}$) and one 'governing' the binding of inhibitor to the enzyme-substrate complex ( $K{ii}$).

Taking reciprocals, the corresponding Lineweaver-Burk transformation may be expressed as follows: $${1\over{v_i}}=\frac{K_m}{V_{max}}(1 + {i\over{K_i}})( {1\over{s}}) + {1\over{V_{max}}}(1 + {i\over{K_{ii}}})\ \ \ \ \ (7)$$

The inhibitor increases both apparent the value of 1 / $V_{max}$ and (by not necessarily the same factor) and the apparent value of $K_{m}$ / $V_{max}$. Inhibition is therefore mixed.

Eqn (7) predicts a family of straight lines that intersect at a single point:

$$({x, y})= (-\frac{K_{ii}}{K_i K_m}, \frac{K_{i} - K_{ii}}{K_i V_{max}}) = (-0.02,0.008) \ \ \ \ \ (8) $$

In this case, slope and intercept replots may be used to determine the values of $K_i$ and $K_{ii}$. A detailed analysis of such plots is given in Segel (1975).

(d) Non-Competitive Inhibition

We now come to the case of non-competitive inhibition, which (as stated above) is best considered a special case of mixed inhibition. When $K_i$ = $K_{ii}$, Eqn (6) simplifies to the following:

$$ v_{i} = { {{{V_{max}}}\ s\ }\over(1 + {i\over{K_{i}}})(K_{m} +s)}\ \ \ \ \ (9)$$

The Lineweaver-Burk transformation:

$${1\over{v_i}}=\frac{K_m}{V_{max}}(1 + {i\over{K_i}})( {1\over{s}}) + {1\over{V_{max}}}(1 + {i\over{K_{i}}})\ \ \ \ \ (10)$$

Eqn (10) predicts a family of lines where increasing $i$ affects both the slope and intercept to the same extent, and which intersect on the x-axis at $-1$/$K_m$

But why should $K_i$ equal $K_{ii}$ in any realistic case?

Notes

Fersht now owns the copyright to his book, and is distributing it free of charge

All issues of Acta Chem Scand (1947 - 1999), including many classic papers, are available on-line

An example of a steady-state rate law derivation is given in this Biology Stack Exchange answer

References

Cook, P. F. & Cleland, W. W. (2007). Enzyme Kinetics and Mechanism. Garland Science Publishing (Taylor & Francis Group). London & New York.

Cornish-Bowden, A. (2004). Fundamentals of Enzyme Kinetics. 3rd edn. Portland Press Ltd, London.

Dalziel, K. (1957). Initial steady state velocities in the evaluation of enzyme-coenzyme-substrate reaction mechanisms. Acta Chem. Scand. 11, 1706 - 1723. [pdf]

Dalziel, K. (1975). Kinetics and mechanism of nicotinamide-nucleotide-linked dehydrogenases. In The Enzymes, 3rd edn., Vol. 11. Boyer, P. D., Ed. pp 1 - 60. Academic Press, New York.

Fersht, A. (1999). Structure and Mechanism in Protein Science. A Guide to Enzyme Catalysis and Protein Folding. W. H. Freeman, New York. [pdf]

Li, T. - K. & Theorell, H. (1969). Human liver alcohol dehydrogenase: Inhibition by pyrazole and pyrazole analogs. Acta Chem. Scand. 23, 892 - 902. [pdf]

Lineweaver, H. & Burk, D. (1934). The determination of enzyme dissociation constants. J. Am. Chem. Soc. 56, 658 - 666.

Segel, I. H. (1975). Enzyme Kinetics. Behavior and Analysis of Rapid Equilibrium and Steady-State Enzyme Systems. John Wiley & Sons, Inc., New York.

Answer 2

Competitive inhibitor competes for the active site. Therefore it will interfere with the binding of the substrate thereby increasing the apparent K_M.

A strictly non-competitive inhibitor does not compete for the active site. It however inhibits the catalysis by reducing the available molecules of active enzyme, E₀ (if it is a perfect inhibitor), thereby lowering the V_max.

There can be mixed inhibitors too and there can be different kinds of mixed inhibition. For example an inhibitor that interferes with catalysis and can also compete for the active site.

Depending on how you define K_M, a non-competitive inhibitor may or may not change it. If such inhibitor can bind to the enzyme at non-active site and reduce its k_cat, it changes the K_M as well as V_max in case of Briggs-Haldane kinetics.