Biology · Enzyme Catalysis and Inhibition
A transition state analog binds to an enzyme with dissociation constant Kd = 10-10 M, whereas the actual substrate S binds with Km = 10-3 M. Which conclusion is most justified based on thermodynamic principles of catalysis?
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 1The large difference in binding affinity indicates that the analog binds to an allosteric site rather than the active site
- 2The analog is a non-competitive inhibitor because its binding affinity is much higher than Km
- 3The enzyme binds the transition state 107 times more tightly than the substrate, reflecting the energy stabilization lowering activation energy✓
- 4The analog acts as a simple competitive inhibitor solely because of structural similarity to the substrate
Correct answer
The enzyme binds the transition state 107 times more tightly than the substrate, reflecting the energy stabilization lowering activation energy
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