Enzymes are often described as following the two-step mechanism: E + S ES (fast)

Question

Enzymes are often described as following the two-step mechanism:

E + S ES (fast)

ES E + P (slow)

where E = enzyme, S = substrate, ES = enzyme-substrate complex, and P = product.

(a) If an enzyme follows this mechanism, what rate law is expected for the reaction?

(b) Suppose that, in the absence of the enzyme, a certain biochemical reaction occurs x times per second at normal body temperature (37 °C). In order to be physiologically useful, the reaction needs to occur 5000 times faster than when it is uncatalyzed. By how many kJ/mol must an enzyme lower the activation energy of the reaction to make it useful?

Explanation / Answer

a) The slow step will be the rate determining step,
   rate law -r = K [ES]
b) Use Arrhenius equation,
   ln (K2 / K1) = Ea / R [ 1/T1 - 1/T2 ]
K2 / K1 = 5000 ; T1 = 25+273.15 = 298.15 K ; T2 = 37+273.15 = 310.15K
   ln (5000) = Ea / 8.314 [ 1/298.15 - 1/310.15 ]
   8.5171 = Ea / 8.314 * [1.2977*10-4]
   65632.2207 = Ea / 8.314
   Ea = 545666.2834 J/mol = 545.6662 kJ/mol

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