Enzyme Inhibition and Regulation (5E) - Chem & Physics

1. In Lineweaver–Burk plots, which inhibition type leaves the $y$-intercept $\left(\frac{1}{V_{max}}\right)$ unchanged?

   Competitive inhibition. In Lineweaver-Burk plots, competitive inhibition increases the slope and x-intercept shift, but Vmax is unchanged, so y-intercept remains the same.

2. In Lineweaver–Burk plots, which inhibition type leaves the $x$-intercept $\left(-\frac{1}{K_m}\right)$ unchanged?

   Pure noncompetitive inhibition. Pure noncompetitive inhibition decreases Vmax (increases y-intercept) but does not affect Km, leaving the x-intercept unchanged in Lineweaver-Burk plots.

3. If a competitive inhibitor is present, how does increasing $[S]$ affect inhibition at high $[S]$?

   Inhibition is overcome; $V_{max}$ can still be reached. At high [S], substrate can outcompete the inhibitor for the active site, allowing the enzyme to reach its uninhibited maximum velocity.

4. What is the definition of $K_m$ in terms of $V_{max}$ on a Michaelis–Menten plot?

   $K_m$ is the $[S]$ at which $v_0=\frac{1}{2}V_{max}$. Km represents the substrate concentration at which the reaction rate is half-maximal, reflecting enzyme-substrate affinity in Michaelis-Menten kinetics.

5. Which inhibition type changes both $K_m$ and $V_{max}$ and binds both $E$ and $ES$ unequally?

   Mixed inhibition. Mixed inhibitors bind to both E and ES at a non-active site but with different affinities, altering both substrate binding (Km) and catalytic rate (Vmax).

6. Identify the inhibition type if both $K_m$ and $V_{max}$ decrease by the same factor in the presence of inhibitor.

   Uncompetitive inhibition. Proportional decrease in both parameters occurs when the inhibitor binds only ES, stabilizing it and altering kinetics uniformly.

7. What is feedback inhibition in a metabolic pathway?

   End product allosterically inhibits an earlier enzyme in the pathway. Feedback inhibition regulates pathways by having the final product bind allosterically to inhibit an upstream enzyme, preventing overproduction.

8. What is an allosteric inhibitor's typical effect on the apparent $K_{0.5}$ (or $K_m$-like term) in sigmoidal kinetics?

   Increases apparent $K_{0.5}$ (right-shifts the curve). Allosteric inhibitors reduce enzyme affinity for substrate, shifting the sigmoidal curve rightward and raising the [S] required for half-maximal velocity.

9. Which type of enzyme inhibition decreases $V_{max}$ but leaves $K_m$ unchanged?

   Pure noncompetitive inhibition. Pure noncompetitive inhibitors bind a site other than the active site on both E and ES equally, reducing effective enzyme concentration and thus Vmax, without affecting substrate affinity or Km.

10. In Lineweaver–Burk plots, which inhibition type produces parallel lines (same slope $\frac{K_m}{V_{max}}$)?

    Uncompetitive inhibition. Uncompetitive inhibition causes parallel lines in Lineweaver-Burk plots because it proportionally affects both Km and Vmax, preserving the slope Km/Vmax.

11. What is an allosteric activator's typical effect on the apparent $K_{0.5}$ (or $K_m$-like term) in sigmoidal kinetics?

    Decreases apparent $K_{0.5}$ (left-shifts the curve). Allosteric activators enhance enzyme affinity for substrate, shifting the sigmoidal curve leftward and lowering the [S] needed for half-maximal velocity.

12. Identify the inhibition type if $V_{max}$ decreases from $100$ to $50$ while $K_m$ remains unchanged.

    Pure noncompetitive inhibition. Halving of Vmax with unchanged Km suggests binding to a non-active site that reduces catalytic capacity without affecting substrate affinity.

13. What is the definition of catalytic efficiency in enzyme kinetics?

    $\frac{k_{cat}}{K_m}$. Catalytic efficiency measures how well an enzyme converts substrate to product at low [S], combining turnover rate kcat with substrate affinity via Km.

14. If a pure noncompetitive inhibitor is present, can increasing $[S]$ restore the original $V_{max}$?

    No; $V_{max}$ remains decreased. Pure noncompetitive inhibitors reduce the effective enzyme concentration, permanently lowering Vmax regardless of [S] increase.

15. Which type of inhibition decreases both $V_{max}$ and $K_m$ (typically by binding $ES$)?

    Uncompetitive inhibition. Uncompetitive inhibitors bind only to the ES complex, stabilizing it and reducing both Vmax and apparent Km by the same factor.

16. What is the Michaelis–Menten equation relating $v_0$, $V_{max}$, $[S]$, and $K_m$?

    $v_0=\frac{V_{max}[S]}{K_m+[S]}$. The equation describes hyperbolic enzyme kinetics, where initial velocity v0 approaches Vmax as [S] increases, with Km as the [S] yielding half Vmax.

17. What is the key distinction between allosteric enzymes and Michaelis–Menten enzymes in $v$ vs. $[S]$ shape?

    Allosteric enzymes show sigmoidal (not hyperbolic) kinetics. Allosteric enzymes exhibit cooperative binding, resulting in a sigmoidal v vs [S] curve, unlike the hyperbolic curve of non-cooperative Michaelis-Menten enzymes.

18. What is the relationship between $V_{max}$, $k_{cat}$, and total enzyme concentration $[E]_T$?

    $V_{max}=k_{cat}[E]_T$. Vmax is the maximum reaction rate, achieved when all enzyme is saturated, and equals the turnover number kcat times total enzyme concentration [E]T.

19. Identify the binding preference of an uncompetitive inhibitor: does it bind $E$, $ES$, or both?

    Binds only the $ES$ complex. Uncompetitive inhibitors preferentially bind the enzyme-substrate complex, often stabilizing it and reducing productive catalysis.

20. Identify the inhibition type if $K_m$ increases from $2\,\text{mM}$ to $6\,\text{mM}$ while $V_{max}$ stays constant.

    Competitive inhibition. The tripling of Km with unchanged Vmax indicates competition for the active site, characteristic of competitive inhibition.

21. What is the defining feature of irreversible enzyme inhibition on enzyme activity over time?

    Covalent or permanent inactivation reduces active $[E]_T$. Irreversible inhibitors form stable bonds with the enzyme, permanently decreasing the pool of functional enzyme and thus reducing activity persistently.

22. Identify the binding site for a competitive inhibitor relative to the substrate binding site.

    Active site (competes with substrate). Competitive inhibitors mimic substrate structure and bind directly to the enzyme's active site, preventing substrate access.

23. Which type of enzyme inhibition increases $K_m$ but leaves $V_{max}$ unchanged?

    Competitive inhibition. Competitive inhibitors bind the active site, competing with substrate and requiring higher [S] to reach half Vmax, thus increasing Km while Vmax is unchanged at saturating [S].

24. What is the Lineweaver–Burk (double reciprocal) form of Michaelis–Menten kinetics?

    $\frac{1}{v_0}=\frac{K_m}{V_{max}}\frac{1}{[S]}+\frac{1}{V_{max}}$. This linear transformation of the Michaelis-Menten equation allows plotting 1/v0 vs 1/[S] to determine Km and Vmax from slope and intercepts.

25. What does a lower $K_m$ generally indicate about an enzyme for its substrate?

    Higher apparent substrate affinity. A lower Km means the enzyme achieves half Vmax at lower [S], indicating stronger binding affinity for the substrate.

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