lecture 12 - control of enzyme activity Flashcards
What are enzyme inhibitors?
Compounds that bind to enzymes and reduce their activity
What are the 2 classes of enzyme inhibitors?
Reversible and irreversible
What are irreversible inhibitors?
Inhibitors that permanently bind covalently to enzymes, usually in the active site
What type of bonds are formed between irreversible inhibitors and amino acids in the active site of enzymes?
Covalent bonds
What are the 3 classes of reversible inhibitor?
Competitive, non-competitive, mixed
How do reversible inhibitors bind to enzymes?
Non-covalently
What is a competitive inhibitor?
A reversible inhibitor that’ competes directly with the substrate for the active site, creating two possible mutually exclusive binding events
How does a Vo vs [S] graph change when a competitive inhibitor is added?
No change in Vmax, because infinite [S] eventually outcompetes the inhibitor. However, the curve is flatter and moved to the right because a higher [S] is needed to reach Vmax - therefore higher Km
How is the Km value affected when a competitive inhibitor is added?
Increased Km - because a greater [S] is needed to reach Vmax/2
How is the Vmax value affected when a competitive inhibitor is added?
Unaffected
How does a Lineweaver-Burke Plot change when a competitive inhibitor is added?
Steeper line, same y intercept, higher x-intercept (less negative, smaller number)
What is a non-competitive inhibitor?
An inhibitor that binds a different site to the substrate (not the active site), so has no affect on the binding of S (in the case of pure non-competitive inhibition) . However, binding of I changes the structure of the active site such that transition state stabilisation of S is no longer optimal
How does a Vo vs [S] change when a non-competitive inhibitor is added?
Vmax is less, so curve reaches the asymptote at a lower point. Km remains the same because the binding affinity is the same
How is the Km value affected when a non-competitive inhibitor is added?
Km value is unchanged because the binding affinity of the substrate is not affected by a non-competitive inhibitor
How is the Vmax value affected when a non-competitive inhibitor is added?
Decreases because the product is made more slowly as transition state stabilisation is impaired by the inhibitor