Chapter 9: Reaction Mechanisms, Pathways, Bioreactions and Bioreactors

Derive the Rate Law for Competitive Inhibition

Competitive Inhibition

\( \text{E} + \text{S} \underset{k_2}{\overset{k_1}{\rightleftharpoons}} \text{E} \cdot \text{S} \xrightarrow{k_{\text{cat}}} \text{P} + \text{E} \)

\( \text{E} + \text{I} \underset{k_5}{\overset{k_4}{\rightleftharpoons}} \text{E} \cdot \text{I} \)

Developing the rate laws

\( r_P = -r_S = k_{\text{cat}} (E \cdot S) \)

\( \text{From } r_{E \cdot S} = 0 = k_1 (E)(S) - k_2 (E \cdot S) - k_{\text{cat}} (E \cdot S) \)

We obtain

\( (E \cdot S) = \frac{k_1 (E)(S)}{k_2 + k_{\text{cat}}} = \frac{(E)(S)}{K_M} \)

\( K_M = \frac{k_2 + k_{\text{cat}}}{k_1} \)

\( r_P = -r_S = k_{\text{cat}} (E \cdot S) = \frac{k_{\text{cat}} (E)(S)}{K_M} \)

From \( r_{E \cdot I} = 0 \)

\( r_{E \cdot I} = 0 = k_4 (E)(I) - k_5 (E \cdot I) \)

\( (E \cdot I) = \frac{k_4}{k_5} (E)(I) = \frac{(E)(I)}{K_I} \)

\( K_I = \frac{k_5}{k_4} \)

Total enzyme

\( E_t = (E) + (E \cdot S) + (E \cdot I) \)

\( E_t = (E) \left( 1 + \frac{(S)}{K_M} + \frac{(I)}{K_I} \right) \)

\( -r_S = r_p = \frac{k_{\text{cat}} E_t (S)}{K_M \left( 1 + \frac{(S)}{K_M} + \frac{(I)}{K_I} \right)} \)

\( \boxed{ -r_S = \frac{V_{\max} (S)}{K_M \left( 1 + \frac{(I)}{K_I} \right) + (S)} } \)

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