Question

when the reaction produces no intermediates, which of the following rate constant will be equal to $k_{cat}$?
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$e + s \xrightleftharpoonsk_{-1}{k_1} es \xrightarrow{k_2} e + p$
a) none of those
b) $k_2$

Explanation:

$k_{cat}$ (turnover number) represents the rate of the catalytic step where the enzyme-substrate complex (ES) converts to product (P) and free enzyme (E). In the given reaction scheme, this step is governed by the rate constant $k_2$, and there are no additional intermediates in the catalytic pathway, so $k_{cat}$ equals $k_2$.

Answer:

b) $k_2$