Enzymes play a key role in many chemical reactions in living systems. Some enzyme-catalysed reaction — Physical Chemistry — Kinetics Chemistry Question
Enzyme catalysis
Enzymes play a key role in many chemical reactions in living systems. Some enzyme-catalysed reactions are described in a simple way by the Michaelis-Menten mechanism, as given below.
E + S ⇌ ES (rate constants forward, backward)
ES ⇌ E + P (rate constant forward)
where E stands for the enzyme, S stands for the substrate on which it acts and P, the end product of the reaction. and are the forward and backward rate constants for the first step, respectively, and is the forward rate constant for the second step.
Ignore the backward rate for the second step and assume that the enzyme equilibrates with its substrate very quickly.
In an experiment, the initial rate (of formation of P) is determined for different concentrations of the substrate, keeping the total concentration of enzyme fixed at mol dm. The following graph is obtained.
The graph is linear for small [S] and it approaches a constant value for large [S]. Show that these features are consistent with the Michaelis-Menten mechanism. (Use steady state approximation for the intermediate step.)
Model Answer
i. The differential rate equations for the Michaelis-Menten mechanism are
d[ES]/dt = k1 [E][S] - k1' [ES] - k2 [ES] (1)
d[P]/dt = k2 [ES] (2)
In the steady-state approximation, d[ES]/dt = 0 (3)
Eq. (1) then gives [ES] = k1 [E][S] / (k1' + k2) (4)
Now [E]o = [E] + [ES] (5) where [E]o is the total enzyme concentration.
Equations (4) and (5) give
[ES] = [E]o [S] / ([S] + Km) (6)
where Km = (k1' + k2) / k1 is the Michaelis-Menten constant.
From eq. 2: d[P]/dt = k2 [E]o [S] / ([S] + Km) (7)
Since the backward rate is ignored, our analysis applies to the initial rate of formation of P and not close to equilibrium. Further, since the enzyme concentration is generally much smaller than the substrate concentration, [S] is nearly equal to [S]0 in the initial stage of the reaction. Thus, according to the Michaelis-Menten mechanism, the initial rate versus substrate concentration is described by eq. (7), where [S] is replaced by [S]0.
For [S] << Km,
Initial rate = k2 [E]o [S] / Km (8)
i.e., initial rate varies linearly with [S].
For [S] >> Km,
Initial rate = k2 [E]0 (9)
i.e. for a large substrate concentration, initial rate approaches a constant value k2[E]0.
Thus the indicated features of the graph are consistent with Michaelis-Menten mechanism.
Determine the rate constant k2 for the second step.
Model Answer
The asymptotic value of initial rate is k2 [E]0. From the graph,
k2 [E]0 = 3.0×10^{–6} mol dm^{–3} s^{–1}
With [E]0 = 1.5×10^{–9} mol dm^{–3}
We get k2 = 2.0×10^3 s^{–1}
Predict the initial rate on the basis of the Michaelis-Menten mechanism for the substrate concentration [S] = 1.0×10^{–4} mol dm^{–3}.
Model Answer
From eq. (7), for [S] = Km, the initial rate is half the asymptotic value. From the graph, therefore,
Km = 5.0×10^{–5} M
For [S] = 1.0×10^{–4} mol dm^{-3} using eq. (7) again,
Initial rate = (2.0×10^3 s^{–1}) (1.5×10^{–9} mol dm^{–3}) (1.0×10^{–4} mol dm^{–3}) / (5.0×10^{–5} mol dm^{–3} + 1.0×10^{–4} mol dm^{–3})
= 2.0×10^{-6} mol dm^{–3} s^{–1}
Determine the equilibrium constant for the formation of the enzyme – substrate complex ES.
Model Answer
We have Km = (k1' + k2) / k1 = 5.0×10^{–5} mol dm^{–3}
The enzyme equilibrates with the substrate quickly, that is the first step of equilibration between E, S and [ES] is very fast. This means that k1' is much greater than k2. Therefore, neglecting k2 above,
k1' / k1 = 5.0×10^{–5} mol dm^{–3}
The equilibrium constant K for the formation of ES from E and S is
K = k1 / k1' = 2.0×10^4 M^{-1}
The experiment above studied at 285 K is repeated for the same total enzyme concentration at a different temperature (310 K), and a similar graph is obtained, as shown below.
Determine the activation energy for the conversion of ES to E and P.
Model Answer
From the graph at the new temperature, k2 [E]0 = 6.0×10^{–6} mol dm^{–3} s^{–1}
i.e. k2 = 6.0×10^{–6} mol dm^{–3} s^{–1} / 1.5×10^{–9} mol dm^{–3} = 4.0×10^3 s^{–1}
Using Arrhenius relation for temperature dependence of rate constant: k = A e^{-Ea/RT}
where Ea is the molar activation energy.
ln (k(T2)/k(T1)) = (Ea / R) * (1/T1 - 1/T2)
Now k(310) / k(285) = 2.0
R = 8.314 J mol^{–1} K^{–1}
Ea = 20.4 kJ mol^{–1}
One interesting application of the ideas above is the way enzyme catalysed reactions inactivate antibiotics. The antibiotic penicillin is, for example, inactivated by the enzyme penicillinase secreted by certain bacteria. This enzyme has a single active site. Suppose, for simplicity, that the rate constants obtained in a above apply to this reaction. Suppose further that a dose of 3.0 μmol of the antibiotic triggers the release of 2.0×10^{–6} μmol of the enzyme in a 1.00 cm3 bacterial suspension.
Determine the fraction of the enzyme that binds with the substrate (penicillin) in the early stage of the reaction.
Model Answer
The fraction of the enzyme that binds with the substrate is, from eq. (6):
[ES] / [E]o = [S] / ([S] + Km)
where [S] is nearly equal to [S]o in the initial stage of the reaction.
Now [S]o = 3.0×10^{-6} mol / 1×10^{-3} dm^3 = 3.0×10^{-3} mol dm^{-3}
and Km = 5.0×10^{–5} mol dm^{–3}
[ES]/[E]o = 3.0×10^{-3} / (5.0×10^{-5} + 3.0×10^{-3}) = 0.98
Nearly the whole of the enzyme is bound with the substrate.
Determine the time required to inactivate 50 % of the antibiotic dose.
Model Answer
From Eq. 7, integrating the equation gives:
-d[S]/dt = k2 [E]o [S] / ([S] + Km)
Km ln([S]o/[S]) + [S]o - [S] = k2 [E]o t (13)
If at t = T, [S] = 1/2 [S]o
Km ln 2 + 1/2 [S]o = k2 [E]o T (14)
Here [E]o = 2.0×10^{-12} mol / 1.0×10^{-3} dm^3 = 2.0×10^{-9} mol dm^{-3}
k2 = 2.0×10^3 s^{-1}
Km = 5.0×10^{-5} mol dm^{-3}
[S]o = 3.0×10^{-3} mol dm^{-3}
Substituting these values in eq. (14) gives
T = 384 s
Thus 50 % of the antibiotic dose is inactivated in 384 s.
In order to control the inactivation of penicillin a substance is introduced which has a similar structure to penicillin and is able to occupy the enzyme site but otherwise it is completely unreactive. This naturally inhibits the enzyme-catalysed reaction. The degree of inhibition, i, is defined by the relation:
i = 1 - r/r0
where r and r0 are the initial rates of reactions with and without the inhibitor, respectively.
Consider again the Michaelis-Menten type of mechanism to describe the situation:
E + S ⇌ ES (rate constants k1, k1')
E + I ⇌ EI (rate constants k3, k3')
ES → E + P (rate constant k2)
Show that the degree of inhibition decreases with increase in concentration of the substrate (for constant concentration of the inhibitor), and the inhibitor ceases to be effective for large substrate concentrations. (This is known as competitive inhibition.)
Model Answer
The differential rate equations for the situation are:
d[ES]/dt = k1[E][S] - k1'[ES] - k2[ES] (15)
d[EI]/dt = k3[E][I] - k3'[EI] (16)
d[P]/dt = k2[ES] (17)
where k3 and k3' are the forward and backward rate constants for the enzyme-inhibitor reaction.
Applying steady-state approximation to [ES] and [EI],
[E][S] / [ES] = (k1' + k2) / k1 = Km (18)
And [E][I] / [EI] = k3' / k3 = KI (19)
Now [E]0 = [E] + [ES] + [EI] (20)
[E]0 = [ES] (1 + Km/[S] + Km[I]/(KI[S])) (21)
d[P]/dt = k2[E]0 / (1 + Km/[S] + Km[I]/(KI[S])) (22)
Here, KI = k3' / k3 is the equilibrium constant for the dissociation of EI to E and I.
The degree of inhibition is i = 1 - r/r0
Using eq. 22
i = ([I]/KI) / (1 + [S]/Km + [I]/KI) (23)
For fixed [I], i decreases with increase in [S] (competitive inhibition), and for large [S], i → 0, i.e. the inhibitor ceases to play any role.
For low substrate concentration of penicillin, determine the concentration of the inhibitor that reduces the rate of the inactivation of penicillin by a factor of 4. The dissociation constant of enzyme-inhibitor complex is given to be 5.0×10^{–5}.
Model Answer
For small [S],
i = ([I]/KI) / (1 + [I]/KI)
If r/r0 = 1/4, i = 3/4
i.e. [I] = 3 KI = 3 x 5.0×10^{-5} = 1.5×10^{-4}
The inhibitor concentration required to reduce the rate of inactivation by a factor of 4 is 1.5×10^{-4} mol dm^{-3}; i.e., 0.15 μmol in a volume of 1.00 cm3.