michaelis-menten equationpldserver1.biochem.queensu.ca/~rlc/steve/313/campbell_bchm313... ·...

Michaelis-Menten equation

The ratio of kcat to Km can be used to describe an enzyme's

catalytic efficiency.

We also note that: kcatK m

=k1

k 2

k−1k 2

k1 is the on rate for binding. The efficiency of catalysis cannot be

greater than the “efficiency” of collisions

.

k2 / (k

-1 + k

2) describes the fraction of all encounters between E

and S that result in product formation. I.e. it is the efficiency of the conversion of ES to P.

Finally we can say: kcatK m

k coll

k 1

k coll

=efficiency of collisions

Catalytic efficiency

Why care about the kcat/Km ratio? We can use it to compare the

rates at which an enzyme catalyzes a reaction with different substrates. Suppose you have two different substrates S

1 and S

2 present at

concentrations [S1] and [S

2] along with the enzyme. The rates of

reactions will be given by:

v1=kcat 1[ES1] and v 2=kcat 2[ES2]

Recall that [ES] can be expressed in terms of free [E] and [S]:

[ES ]=[E ][S ]

K m

So then we can write the two rate equations as:

v1=kcat 1

K m1

[E ][S1] and v 2=kcat 2

K m2

[E ][S2]

Catalytic efficiency

Now we can divide one by the other and [E] drops out of the equation:

v1

v 2

=kcat 1/K m1[S1]

kcat 2/K m2[S2]

We can use this relationship to calculate the relative rates of any number of simultaneously occurring reactions with one enzyme.

Similarly, we can compare the forward and reverse reactions. If we define the forward and reverse reaction rates by the following:

v f=kcat f

K mf

[E ][S ] and v r=kcat r

K mr

[E ][P ]

If the two rates are equal, then S and P must be at their equilibrium concentrations. So we can write:

kcat f

K mf

[E ][S ]=kcat r

K mr

[E ][P ] and ∴kcat f /K mf

kcat r /K mr =K eq

Catalytic efficiency

kcat f /K mf

kcat r /K mr =K eq

From this we can see that the catalytic efficiencies of the forward and reverse reactions are related by the equilibrium constant.

Determination of kcat and Km

Just as for the analysis of binding equilibria our objective is to measure reaction rates at numerous concentrations of substrate and a fixed enzyme concentration.

Unless one is able to measure v at very low and very high [S], estimation of Vmax and K

m from a direct plot is difficult. (Actual

values for this data are Vmax=1 μM/s and Km = 7 μM-1.)

Determination of kcat and Km

Again, like the analysis of equilibrium binding, we can do a non-linear fit:

Non-linear least squares fit gives:Vmax = 0.96 +/- 0.05 μM/s and K

m = 6.4 +/- 0.8 μM-1

Determination of kcat and Km

We can use the Lineweaver-Burk plot (double reciprocal plot) to linearize the Michaelis-Menten equation:

1v=

K m

Vmax1[S ]

1

Vmax

A plot of 1/v versus 1/[S] should give a straight line with a slope of K

m/Vmax and a “y-intercept” of 1/Vmax. As in the case of

fitting equilibrium binding data with the double reciprocal plot, errors are distorted by the presence of the reciprocals.

Determination of kcat and Km

The Lineweaver-Burk plot (double reciprocal plot):

Linear least-squares fit gives:Vmax=0.88 +/- 0.05 μM/s and K

m = 5.0 +/- 0.5 μM-1

Determination of kcat and Km

An alternative way to linearize the Michaelis-Menten equation is analogous to the Scatchard plot for binding data:

v[S ]

=Vmax

K m

−1

K m

v

A plot of v/[S] versus v should give a straight line with a slope of -1/K

m and a “y-intercept” of Vmax/K

m. Errors in the data should

be less accentuated by the fact that the reciprocals of the velocity values (inherently more error-prone than [S]) are not used. This is known as the Eadie-Hofstee plot.

Determination of kcat and Km

The Eadie-Hofstee plot:

Linear least squares fit gives:Vmax = 0.94 +/- 0.05 μM/s and K

m = 5.9 +/- 0.7 μM-1

Determination of kcat and Km

Now running the experiment properly with data in triplicate:

Non-linear least squares fit gives:Vmax = 0.98 +/- 0.02 μM/s and K

m = 6.7 +/- 0.4 μM-1

Determination of kcat and Km

Now running the experiment properly with data in triplicate:

Vmax = 1.0 +/- 0.06 μM/s

Km = 7.1 +/- 0.7 μM-1

Vmax = 1.0 +/- 0.03 μM/s

Km = 7.0 +/- 0.5 μM-1

Lineweaver-Burk Eadie-Hofstee

Determination of kcat and Km

So if the linearization methods may be not so accurate (depending on the errors in the data), why would we continue to use them?

● Ease of plotting when you have no computer to do a non-linear fit.

● Identification of kinetics that do not fit the Michaelis-Menten model.

● Analysis of kinetics of enzyme inhibition.

Reversible Inhibition of Enzyme Activity

Consider the general scheme of inhibition:

E + S ES E + P + + I I

EI ESI

Km

kcat

KI1

KI2

Four possible types of inhibition can exist, depending on the values of K

I1 and K

I2.

Reversible Inhibition of Enzyme Activity

We want to find and equation for the rate of reaction in terms of [E]

T, [S], [I], and the parameters, k

cat, K

m, K

I1, and K

I2

v=k cat [ES ]

[ES ]=[E ][S ]

K m

[EI ]=[E ][I ]

K I1

[ESI ]=[ES ][I ]

K I2

=[E ][S ][I ]

K m K I2

[E ]T=[E ][EI ][ES][ESI ]

Reversible Inhibition of Enzyme Activity

Substituting the expressions for [ES], [EI] and [ESI] into the last equation gives:

v=k cat [E ][S ]

K m

[E ]T=[E ][E ][I ]

K I1

[E ][S ]

K m

[E ][S ][ I ]

K mK I2

If we divide that into:

we get:

v[E ]T

=

k cat [S ]

K m

1[I ]K I1

[S ]K m

[S ][I ]K m K I2

Reversible Inhibition of Enzyme Activity

Which can be rearranged to:

v=k cat [E ]T [S ]

K m 1[I ]K I1

[S ]1 [S ][I ]K m K I2

Dividing numerator and denominator by (1 + [I]/K

I2) gives:

v=

k cat

1 [ I ]K I2

[E ]T [S ]

K m

1 [I ]K I1

1 [I ]

K I2[S ]

Reversible Inhibition of Enzyme Activity

Now we can say, that for a given [I], we can express that equation in a form with apparent k

cat and K

m values:

v=kcat app [E ]T [S ]

Kmapp[S ]

kcat app=kcat

1 [I ]K I2

Kmapp=Km

1 [I ]K I1

1 [I ]

K I2

where:

Types of Reversible Inhibition

● Competitive

● Non-competitive

● Uncompetitive

● Mixed

Competitive Inhibition

kcatKm

+

+

kcat app=kcat

Kmapp=Km 1 [I ]K I1

+

KI1

Competitive Inhibition

Characteristic plots

v

[S]

Vmax

+I

- I

1/[S]

1/v

+I

- I

1/Vmax

v/[S]

v

- I

+I

Non-competitive Inhibition

kcatKm

+

+

kcat app=kcat

1 [I ]K I2

Kmapp=Km

+

+

KI1

KI2

KI1 = K

I2

Non-competitive Inhibition

Characteristic plots:

v

[S]

Vmax

+I

- I

1/[S]

1/v

+I

- I v/[S]

v

- I

+I

Uncompetitive Inhibition

kcatKm+

kcat app=kcat

1 [I ]K I2

Kmapp=

Km

1 [I ]K I2

+

+

KI2

KI1 → ∞

Non-competitive Inhibition

Characteristic plots:

v

[S]

Vmax

+I- I

1/[S]

1/v

+I

- I v/[S]

v

- I

+I

Mixed Inhibition

kcatKm

+

+

kcat app=kcat

1 [I ]K I2

+

+

KI1

KI2

KI1 ≠ K

I2

Kmapp=Km1 [I ]

K I1

1 [I ]K I2

Mixed Inhibition

Characteristic plots will vary, depending on the relative values of K

I1 and K

I2.

v

[S]

Vmax

+I- I

v

[S]

Vmax

+I- I

K I1K I2K I1K I2

1/[S]

1/v

+I

- I

1/[S]

1/v

+I

- I

Mixed Inhibition

The intersection point between the lines in a Lineweaver-Burk plot can be found from the following two equations:

1[S ]

=−1K m

K I1

K I2

1v=

1Vmax 1−K I1

K I2

So if KI1 > K

I2 the intersection point will be below the horizontal axis

and if KI1 < K

I2, then it will be above the horizontal axis.

Something to think about

We said that a plot of v versus [S] exhibited saturation

v

[S]

whereas v versus [E] did not. Why would that be?

v

[E]