In the classic enzyme-catalyzed mechanism, the Michaelis-Menten rate law is v = (Vmax [S])/(Km + [S]); what are Vmax and Km in terms of the rate constants and enzyme concentration?

• A. v = (Vmax [S])/(Km + [S]); Vmax = k2[E]T; Km = (k-1 + k2)/k1
• B. v = kcat [E][S]/Km
• C. v = k2 [E][S]/(Km + [S])
• D. v = k1[E][S]/(k-1 + k2)

Answer: (A)

In the Michaelis–Menten mechanism, the rate at which product forms is tied to how fast the enzyme–substrate complex ES turns into product and how much total enzyme is present. The maximum rate occurs when all enzyme molecules are engaged in turnover, so Vmax equals k2 multiplied by the total enzyme concentration [E]T (often written as Vmax = kcat [E]T, with kcat synonymous with k2 in this mechanism). The substrate concentration needed to reach half of that maximum rate is defined by Km, which combines the rates of forming ES and breaking it apart or turning it into product: Km = (k−1 + k2)/k1. This expresses Km as a balance among binding (k1) and dissociation (k−1) plus conversion to product (k2). Therefore the correct relations are Vmax = k2 [E]T and Km = (k−1 + k2)/k1, leading to the rate law v = (Vmax [S])/(Km + [S]). Other expressions that omit [E]T, or use [E] instead of [E]T, or place the constants differently, do not match the standard derivation of the Michaelis–Menten form.