enzyme kinetics. catalyzed uncatalyzed formation of product is faster in the catalyzed reaction than...

ENZYME KINETICS

Consider reaction S (substrate) --> P (product)

catalyzed

uncatalyzed

•Formation of product is faster in the catalyzed reaction than inthe uncatalyzed reaction and initially is linear with time.

•We measure the enzyme activity by the rate of formation of product in the catalyzed reaction.

•Enzyme activity is proportional to the amount of enzyme and increases with an increase in the amount of substrate added.

What more can we say about the enzyme and the reaction from the dependence of the reaction rate on the conditions?

amount of enzyme

4x 2x 1x

amount of substrate

8x 4x 2x 1x

[P]

time time time

Analysis of an enzyme-catalyzed reaction E + S ES E + P, where k1, k-1, k2, k-2 are rate constants: the rate of E + S ES is k1[E][S], etc. Assume initial conditions: ES is formed quickly and [ES] is constant; [P] is small, so (k-2)[E][P] ~ 0 and the equations simplify to: E + S ES E+P,

k1 k-1

k2 k-2

k1 k-1

k2

Analysis of an enzyme-catalyzed reaction E + S ES E+P, where k1, k-1, k2, k-2 are rate constants Assume initial conditions: ES is formed quickly and [ES] is constant; [P] is small, so (k-2)[E][P] ~ 0 and the equations simplify to: E + S ES E+P,

k1 k-1

k2 k-2

k1 k-1

k2

Because [ES] is constant: the formation of ES must be balanced by the removal of ES: k1([Etotal] – [ES])[S] = (k2 + k-1)[ES]; and ([Etotal] – [ES])[S] = [(k2 + k-1)/k1][ES]

Analysis of an enzyme-catalyzed reaction (continued) E + S ES E+P, Here is where we are so far: ([Etotal] – [ES])[S] = [(k2 + k-1)/k1][ES] If we define Km ("Michaelis constant"), = (k2 + k-1)/k1 then ([Etotal] – [ES])[S] = Km[ES]; solving for [ES], we get [ES] = [Etotal][S]/(Km+[S]) Define the rate of reaction: V = k2[ES], so V = k2[Etotal][S]/(Km+[S]) Note that k2[Etotal] = Vmax (reaction is fastest when all enzyme is complexed with S) V = Vmax [S]/(Km+[S]) the Michaelis-Menten equation! This allows us to measure important characteristics of the enzyme, Km and Vmax, by varying substrate and measuring the reaction rate.

k1 k-1

k2

Graphing enzyme data Michaelis-Menten V = Vmax [S]/(Km+[S]) Graph V as a function of [S] As [S]<<Km, V = Vmax [S]/Km (At low [S], V approaches a straight line with respect to [S]) V = Vmax/(Km/[S] + 1) As [S] increases, V --> Vmax When [S] = Km, V = V max

V

[S]

1/2 Vmax

Km

Vmax

Lineweaver-Burke Take the inverse of Michaelis-Menten eqn: 1/V = (Km+[S])/(Vmax [S]) = (Km/Vmax)(1/[S]) + 1/ Vmax Plot 1/[V] vs 1/[S] Slope = Km/Vmax (allows linear interpolation) X-intercept = -1/Km Y-intercept = 1/Vmax

1/V

-1/Km 1/[S]

1/Vmax

Eadie-Hofstee V(Km + [S]) = Vmax [S] VKm + V[S] = Vmax [S] VKm/[S] + V = Vmax V = Vmax - Km(V/[S]) Plot V vs V/[S] Slope = - Km (allows a linear interpolation with a more even distribution of points.) Y intercept = Vmax X intercept = Vmax/Km

Vmax/Km V/[S]

V

Vmax

An example of the three plotting methods using some hypothetical dataand Excel to make the graphs:

Michaelis-Menten

S V 1 12 2 20 4 29 8 35 12 40

0

5

10

15

20

25

30

35

40

45

0 2 4 6 8 10 12 14

s

V

Vmax = 45 - 50 (?)Km = 2.5 - 3 (depending on Vmax)

1/[S] 1/V 1 0.083

0.5 0.05 0.25 0.034 0.125 0.029 0.083 0.025

Slope

‘Km/Vmax’ 0.063

y

inte rcep t ‘1/Vmax’ 0.01 95

Slope /y inte rcep t

Km 3.24

1/y

inte rcep t Vmax 51.3

Lineweaver-Burke

(I used Excel’s functions to get the slope and the y intercept and then calculatedKm and Vmax from those values.)

V/[S] V 12 12 10 20

7.25 29 4.37 35 3.33 40

Slope

"-Km" 3.07

y

intercept Vmax 49.9

0

5

10

15

20

25

30

35

40

45

0 5 10 15

v/s

v

Eadie-Hofstee

(Again, I used Excel’s functions to get the slope and the y intercept.)

Some implications of the enzyme measurements: 1. If k2 << k-1, then Km = (k2 + k-1)/k1 reduces to Km = k-1/k1 In this case, Km is close to a “dissociation constant”, which describes the tendency of a ligand to come off the protein (or other molecule) that binds to it. We get the same thing if [ES] << Etotal: ([Etotal] – [ES])[S] = Km[ES] reduces to Km = [Etotal][S]/[ES] Note that dissociation constant is the inverse of “affinity.” Kms for various enzymes range from 0.0004 to 122 mM. The low value shows very tight binding of substrate to enzyme; the high value, very loose binding.

Some more implications 2. The “Turnover number” (kcat) describes the rate of product formation per unit enzyme at saturating [S]: kcat = Vmax/[Etotal] (maximum catalytic activity). Turnover number is mols P formed per mol E per second: units in s-1 Turnover numbers for various enzymes range from 0.5 s-1 for lysozyme to 40,000,000 s-1 for catalase. 3. Catalytic efficiency is kcat/Km Efficiency is higher as turnover number rises and/or Km falls (affinity rises) For the most efficient enzymes, catalytic efficiency is limited by the diffusion of substrate to the enzyme. The maximum is 108 to 109 M-1s-1

Summary

•The reaction of an enzyme with its substrate to form product can be described quantitatively: the reaction rate (V) dependson the amount of enzyme (Vmax), the concentration of substrate ([S]), and the affinity of the enzyme for substrate (Km).

•Vmax and Km can be determined by measuring V as a functionof [S]. There are several methods for plotting the results to getVmax and Km.

•Under certain conditions, Km provides an indication of the affinityof an enzyme for its substrate. “Turnover number” and “Catalyticefficiency” provide even better measures of enzyme activity.