The mechanism of an enzymatic reaction may be described as: S is the substrate, E is the enzyme, ES — Physical Chemistry — Kinetics Chemistry Question
Kinetics of an Enzymatic Reaction
The mechanism of an enzymatic reaction may be described as:
S is the substrate, E is the enzyme, ES is the complex formed by S and E, and P is the product. k1, k–1 and k2 are the rate constants of the elementary reactions.
The rate of the enzymatic reaction, r, can be expressed as a function of the substrate concentration, c(S):
dc(P)/dt = r = k2 cT(E) c(S) / (KM + c(S))
t is the time,
c(P) is the product concentration,
cT(E) is the total enzyme concentration
and KM = (k–1 + k2)/k1.
Determine the variables x, y and z in the following rate equations:
-1 2
(S) (ES) – (S) (E) (ES) (S) (E) – ( ) (ES)z
xx y
dc dc k c c k c k c c k k c
dt dt
Model Answer
x = 1, y = –1, z = 1
Complete the following rate equation:
(E)d c
dt
Model Answer
d c(E) / dt = –k1 c(S)c(E) + k–1 c(ES) + k2 c(ES)
Penicillin (substrate) is hydrolyzed by β–lactamase (enzyme). The following data have been recorded when the total enzyme concentration was 10–9 mol dm–3.
x–axis: c–1(S) / (106 dm3 mol–1)
y–axis: r–1 / (106 dm3 min mol–1)
Determine the constants k2 and KM.
If c(S) = 0.01 KM, what is the concentration of the complex ES?
Model Answer
The reciprocal rate is plotted as a function of the reciprocal substrate concentration:
1/r = (KM / k2 cT(E)) × 1/c(S) + 1 / (k2 cT(E))
Intercept at 1/c(S) = 0 yields 1/r = 1/(k2 cT(E)) = 0.02×10^6 dm^3 min mol–1
With cT(E) = 1×10–9 mol dm–3 we obtain k2 = 50000 min–1
Intercept at 1/r = 0 yields 1/c(S) = – 1/KM = – 0.09×10^6
KM = 1.1×10–5
Alternatively, the slope is KM / (k2 cT(E)) = 0.22 min
KM = 1.1×10–5
The rate of the enzymatic reaction is given as
d c(P) / dt = k2 c(ES) = k2 cT(E) c(S) / (KM + c(S))
c(ES) = cT(E) c(S) / (KM + c(S))
c(ES) = cT(E) 0.01 KM / (KM + 0.01 KM)
c(ES) = 9.9×10–3 cT(E)
c(ES) = 9.9×10–12 mol dm–3
A competitive inhibitor I competes with the substrate and may block the active site of the enzyme:
I + E EI
If the dissociation constant of EI is 9.5×10–4 mol dm–3 and the total enzyme concentration is 8×10–4 mol dm–3, what total concentration of inhibitor will be needed to block 50 % of the enzyme molecules in the absence of substrate?
Model Answer
K = c(I) c(E) / c(EI) = c(I) 0.5 cT(E) / 0.5 cT(E) = c(I) = 9.5×10–4
The total inhibitor concentration is
cT(I) = c(I) + c(EI) = K + 0.5 cT(E) = 1.35×10–3 mol dm–3
Decide whether the following statements are true or false.
i) The rate of the enzymatic reaction, r, is reduced by the competitive inhibitor.
ii) The maximum value of the rate r is reduced by the competitive inhibitor.
iii) The concentration of the substrate S is unaffected by the competitive inhibitor.
iv) The activation energy of the enzymatic reaction is increased by the inhibitor.
Model Answer
i) true (the inhibitor reduces the free enzyme concentration and thus the rate of ES formation. A lower ES concentration results and leads to a smaller reaction rate)
ii) false (the maximum rate is reached for c(S) = where the inhibitor concentration can be ignored)
iii) false (the inhibitor reduces the free enzyme concentration and thus promotes the dissociation of the complex ES into E and S (Le Chatelier`s principle))
iv) false (the activation energy depends on the rate constants that are independent of concentrations)
A more detailed description of an enzymatic reaction includes the reverse reaction of the product back to the substrate. At the end of the enzymatic reaction, a chemical equilibrium is reached between the substrate and the product.
Decide whether the following statements are true or false.
i) The concentration of the product in the equilibrium is increasing with increasing concentration of the substrate.
ii) The concentration of the product in the equilibrium is increasing with increasing concentration of the enzyme.
iii) The concentration of the product in the equilibrium is higher, when the rate constant k2 is larger.
Model Answer
The enzyme is only a catalyst. The net reaction is S P
i) true (because K = ceq(P) / ceq(S))
ii) false (because K does not depend on the enzyme concentration)
iii) true (because K is the ratio of the rate constants for the forward and the reverse reaction)