Biochem Lec 11-Enzymes III: Inhibition

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26 Terms

1
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What are the 3 types of reversible inhibitors?

1) Competitive (most common)

2) Uncompetitive

3) Noncompetitive

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Where does a competitive inhibitor bind?

  • Binds reversibly to the active site of the enzyme → forms a nonproductive enzyme inhibitor complex (EI)

  • The substrate and inhibitor compete for binding at the active site

<ul><li><p>Binds reversibly to the active site of the enzyme → forms a nonproductive enzyme inhibitor complex (EI)</p></li><li><p>The substrate and inhibitor compete for binding at the active site</p></li></ul><p></p>
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What is Ki?

The equilibrium dissociation constant that defines the affinity of I for the enzyme. The higher the affinity, the more potent the inhibitor.

<p>The equilibrium dissociation constant that defines the affinity of I for the enzyme. The higher the affinity, the more potent the inhibitor.</p>
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What is the relationship between Ki and the affinity of the inhibitor for the enzyme?

Low Ki → high affinity

High Ki → low affinity 

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How does the unproductive EI complex fit into the enzyme/substrate to product reaction?

It adds to the beginning with E and S and is in equilibrium. It competes with the formation of ES.

<p>It adds to the beginning with E and S and is in equilibrium. It competes with the formation of ES.</p>
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What variables do competitive inhibitors impact?

  • Increase [S] needed to reach ½ Vmax → increase KM (KM<KMapp)

  • DO NOT affect Vmax

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What is the easiest way to identify competitive inhibition and calculate Ki?

Lineweaver-Burk plots because they make it easy to visualize how competitive inhibitors affect KM (change slope) but do not affect Vmax. Slope change helps calculate Ki.

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What is KMapp and how is it calculated?

M-M rate constant with inhibitor

<p>M-M rate constant with inhibitor</p>
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How do the Lineweaver-Burk plots of enzymes with and without competitive inhibitors differ? How are they the same?

With competitive inhibitor:

  • Steeper slope → larger KM (KMapp)

  • Less negative/larger X-int → larger KM (KMapp)

Y-int stays the SAME → no effect on Vmax

<p>With competitive inhibitor:</p><ul><li><p>Steeper slope → larger K<sub>M</sub> (K<sub>Mapp</sub>)</p></li><li><p>Less negative/larger X-int → larger K<sub>M</sub> (K<sub>Mapp</sub>)</p></li></ul><p>Y-int stays the SAME → no effect on V<sub>max</sub></p><p></p>
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How do drugs that are competitive inhibitors compare to substrates?

They are structurally similar:

  • Bind with similar chemical complementarity as the substrate to the enzyme active site

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Provide an example of a drug that is a competitive inhibitor. How does its structure compare to the natural substrate?

Methotrexate (anti-cancer) → target dihydrofolate reductase, an enzyme involved in nucleotide biosynthesis in tumor cells.

Very similar structure → only differs in 2 places

<p>Methotrexate (anti-cancer) →&nbsp;target dihydrofolate reductase, an enzyme involved in nucleotide biosynthesis in tumor cells. </p><p>Very similar structure → only differs in 2 places</p>
12
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What are competitive inhibitors often modeled after and why?

The transition state (X) because enzymes often have a higher affinity for the transition state.

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Why do enzymes often have a higher affinity for the transition state?

Enzymes often have higher affinity for the transition state than for the substrate because stabilizing this high-energy state lowers the activation energy and speeds up the reaction. If they bound the substrate more tightly, it would actually hinder catalysis by trapping it in the ground state.

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Provide an example of an enzyme that binds more strongly to the transition state and explain.

Proline racemase:

  • Ki for pyrrole 2-C (transition state analog) is 160x lower than the KM for proline → the transition state analog binds much more strongly to the enzyme

<p>Proline racemase:</p><ul><li><p>K<sub>i</sub>&nbsp;for pyrrole 2-C (transition state analog) is 160x lower than the K<sub>M</sub>&nbsp;for proline → the transition state analog binds much more strongly to the enzyme</p></li></ul><p></p><p></p>
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Where do uncompetitive inhibitors bind?

  • Uncompetitive inhibitors bind selectively to the ES complex, often adjacent to the active site → A ternary unproductive complex of ESI is formed

<ul><li><p>Uncompetitive inhibitors bind selectively to the ES complex, often adjacent to the active site → A ternary unproductive complex of ESI is formed</p></li></ul><p></p>
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How to uncompetitive inhibitors affect the reaction of substrate/enzyme to product?

They are added to the ES and form an equilibrium with ES and ESI

<p>They are added to the ES and form an equilibrium with ES and ESI</p>
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What variables to uncompetitive inhibitors affect?

KM AND Vmax → lower

KM → binding affinity appears greater for substrate because the substrate stays trapped in ESI complex

Vmax → prevents ES from completing the reaction, so fewer active complexes make product

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What do uncompetitive inhibitors NOT affect? How does this affect the Lineweaver-Burk plot?

The ratio of Km/Vmax stays the same → slope does not change

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How does the Lineweaver-Burk plot with/without an uncompetitive inhibitor compare?

With uncompetitive inhibitor:

  • Larger Y-int → lower Vmax

  • Smaller/more negative X-int → lower KM (KMapp)

Slope is the same for both

<p>With uncompetitive inhibitor:</p><ul><li><p>Larger Y-int → lower V<sub>max</sub></p></li><li><p>Smaller/more negative X-int → lower K<sub>M</sub> (K<sub>Mapp</sub>)</p></li></ul><p>Slope is the same for both</p>
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How is Kmapp calculated?

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How is Vmaxapp for uncompetitive inhibitors calculated?

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Where do noncompetitive inhibitors bind?

Noncompetitive inhibitors bind independently of the S to the E resulting in an unproductive complex (either EI or ESI)

<p>Noncompetitive inhibitors bind independently of the S to the E resulting in an unproductive complex (either EI or ESI)</p>
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What variables to noncompetitive inhibitors affect?

Lowers Vmax

No effect on KM

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How do noncompetitive inhibitors affect the reaction of enzyme/substrate to product?

Added to both S and E and ES. Forms equilibrium with ES to form ESI → in equilibrium with EI+S → in equilibrium with S+E

<p>Added to both S and E and ES. Forms equilibrium with ES to form ESI → in equilibrium with EI+S → in equilibrium with S+E</p>
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How do the Lineweaver-Burk plots of with/without a noncompetitive inhibitor compare?

With noncompetitive inhibitor:

  • Larger Y-int → lower Vmax

Same X-int → no effect on KM

<p>With noncompetitive inhibitor:</p><ul><li><p>Larger Y-int → lower V<sub>max</sub></p></li></ul><p>Same X-int → no effect on K<sub>M</sub></p><p></p>
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How is Vmaxapp calculated for noncompetitive inhibitiors?

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