MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 5e Enzyme Inhibition Regulation

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This deck focuses on 5e Enzyme Inhibition Regulation, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

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

Answer: 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.

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

Answer: 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.

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

Answer: 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.

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

Answer: $K_m$ 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.

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

Answer: 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).

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

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

Flashcard 7: What is feedback inhibition in a metabolic pathway?

Answer: 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.

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

Answer: 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.

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

Answer: 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.

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

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

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

Answer: 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.

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

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

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

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

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

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

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

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

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

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

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

Answer: 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.

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

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

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

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

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

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

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

Answer: 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.

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

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

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

Answer: 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].

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

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

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

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