Aspirin is an irreversible inhibitor of PGH2 synthase, reacting with a Ser residue in the active site. You have incubated 5 µM enzyme with 1 mM aspirin and have found that 85% of the enzyme was inactivated in 30 minutes. A) Suppose you wanted to use 5 mM aspirin and inactivate 99% of the enzyme. For how long should you allow the reaction to proceed? B) Suppose that you incubate 1 µM enzyme with 1 mM aspirin for 30 minutes. Would you expect more or less than 85% of the enzyme to be inactivated? Explain your reasoning. C) Suppose that you incubate 1 mM enzyme with 1 mM aspirin for 30 minutes. Would you expect more or less than 85% of the enzyme to be inactivated? Explain your reasoning.
The Correct Answer and Explanation is:
We can use pseudo-first-order kinetics for irreversible inhibition when the inhibitor (aspirin) is in large excess over the enzyme.
The equation for first-order decay is:[E][E0]=e−kt\frac{[E]}{[E_0]} = e^{-kt}[E0][E]=e−kt
Given:
- 85% inactivated → 15% active enzyme remains
- [E]/[E0]=0.15[E]/[E_0] = 0.15[E]/[E0]=0.15
- [I]=1[I] = 1[I]=1 mM
- [E0]=5[E_0] = 5[E0]=5 µM
- Time = 30 min
First, solve for k:0.15=e−k⋅30⇒ln(0.15)=−30k⇒k=−ln(0.15)30≈1.897130≈0.06324 min−10.15 = e^{-k \cdot 30} \Rightarrow \ln(0.15) = -30k \Rightarrow k = \frac{-\ln(0.15)}{30} \approx \frac{1.8971}{30} \approx 0.06324\ \text{min}^{-1}0.15=e−k⋅30⇒ln(0.15)=−30k⇒k=30−ln(0.15)≈301.8971≈0.06324 min−1
Now, you want 99% inactivation → 1% enzyme remains active:0.01=e−kt0.01 = e^{-kt}0.01=e−kt
Solve for t with [I]=5[I] = 5[I]=5 mM (5× more inhibitor):
Rate increases proportionally:knew=5×0.06324=0.3162 min−1k_{\text{new}} = 5 \times 0.06324 = 0.3162\ \text{min}^{-1}knew=5×0.06324=0.3162 min−10.01=e−0.3162t⇒ln(0.01)=−0.3162t⇒t=4.60520.3162≈14.57 min0.01 = e^{-0.3162t} \Rightarrow \ln(0.01) = -0.3162t \Rightarrow t = \frac{4.6052}{0.3162} \approx 14.57\ \text{min}0.01=e−0.3162t⇒ln(0.01)=−0.3162t⇒t=0.31624.6052≈14.57 min
Answer A: Approximately 14.6 minutes are needed to inactivate 99% of the enzyme with 5 mM aspirin.
B)
If you incubate 1 µM enzyme with 1 mM aspirin (same inhibitor concentration), you now have a larger excess of inhibitor (1000-fold excess vs. 200-fold before).
This increases the likelihood of enzyme-inhibitor encounters. Because pseudo-first-order kinetics still applies and aspirin is in even greater excess, the inactivation will proceed faster.
Answer B: You would expect more than 85% of the enzyme to be inactivated in 30 minutes.
C)
Now you incubate 1 mM enzyme with 1 mM aspirin — the molar ratio is 1:1.
In this case, there is no excess of aspirin, so not all enzyme molecules can be inactivated. At most, only 1 mM of enzyme can be inactivated if all aspirin molecules react. However, this is under ideal conditions.
Also, with a 1:1 ratio, the reaction no longer follows pseudo-first-order kinetics — it’s a second-order reaction, and competition between enzyme molecules can slow the rate.
Answer C: You would expect less than 85% of the enzyme to be inactivated in 30 minutes.
