MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 5e Enzyme Structure Catalysis

What this deck covers

This deck focuses on 5e Enzyme Structure Catalysis, 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: Which option best defines a prosthetic group in enzyme function?

Answer: A tightly bound cofactor that remains associated with the enzyme. Prosthetic groups enhance enzyme function by remaining covalently or tightly bound during catalysis.

Flashcard 2: Which option best defines competitive inhibition in enzyme kinetics?

Answer: Inhibitor binds active site; increases $K_m$; $V_{max}$ unchanged. Competitive inhibitors compete for the active site, requiring higher $[S]$ to achieve the same velocity.

Flashcard 3: What is the lock-and-key model of enzyme-substrate binding?

Answer: Active site is preformed and complementary to the substrate. This rigid model describes perfect geometric complementarity between enzyme and substrate without conformational changes.

Flashcard 4: Which option best defines the active site of an enzyme?

Answer: The specific region where substrate binds and catalysis occurs. The active site facilitates catalysis by providing a microenvironment that stabilizes the transition state through specific substrate interactions.

Flashcard 5: What is the catalytic triad commonly found in serine proteases?

Answer: Serine, histidine, and aspartate. These residues work together to facilitate nucleophilic attack and proton transfer in peptide bond hydrolysis.

Flashcard 6: If $[S] \gg K_m$, what is the approximate value of $v_0$?

Answer: $v_0\approx V_{max}$. High substrate concentrations saturate the enzyme, making the reaction rate independent of $[S]$.

Flashcard 7: What is the relationship between $\Delta G$ and enzyme catalysis for a given reaction?

Answer: Enzymes do not change $\Delta G$ of the reaction. Catalysts like enzymes accelerate reactions without altering the overall free energy change between reactants and products.

Flashcard 8: What is the induced-fit model of enzyme-substrate binding?

Answer: Substrate binding causes enzyme conformational change to fit better. This model explains how enzymes adapt their shape upon substrate binding to enhance catalytic efficiency.

Flashcard 9: What is the relationship between equilibrium constant $K_{eq}$ and enzyme catalysis?

Answer: Enzymes do not change $K_{eq}$. Since enzymes do not affect $\Delta G$, they leave the ratio of products to reactants at equilibrium unchanged.

Flashcard 10: What is the definition of $V_{max}$ in enzyme kinetics?

Answer: Maximum rate when enzyme is saturated with substrate. $V_{max}$ represents the enzyme's turnover rate when all active sites are occupied by substrate.

Flashcard 11: On a Lineweaver-Burk plot, what is the $y$-intercept equal to?

Answer: $\frac{1}{V_{max}}$. The y-intercept corresponds to the reciprocal of the maximum velocity in the Lineweaver-Burk equation.

Flashcard 12: Which option best describes how enzymes increase reaction rate?

Answer: They stabilize the transition state and provide a lower-$E_a$ pathway. Enzymes accelerate reactions by lowering the activation energy through transition state stabilization.

Flashcard 13: What is the definition of $K_m$ in Michaelis-Menten kinetics?

Answer: $[S]$ at which $v_0=\frac{V_{max}}{2}$. $K_m$ reflects the enzyme's affinity for the substrate, where half-maximal velocity is achieved.

Flashcard 14: What is the Lineweaver-Burk (double-reciprocal) form of Michaelis-Menten?

Answer: $\frac{1}{v_0}=\frac{K_m}{V_{max}}\frac{1}{[S]}+\frac{1}{V_{max}}$. This linear transformation allows determination of $K_m$ and $V_{max}$ from intercepts on a double-reciprocal plot.

Flashcard 15: Which option best defines a coenzyme in enzyme catalysis?

Answer: An organic cofactor, often vitamin-derived, required for activity. Coenzymes act as carriers of chemical groups or electrons during enzymatic reactions.

Flashcard 16: What is the primary effect of an enzyme on a reaction energy diagram?

Answer: It lowers the activation energy $E_a$. By stabilizing the transition state, enzymes reduce the energy barrier required for the reaction to proceed.

Flashcard 17: Which option best defines a cofactor in enzyme catalysis?

Answer: A nonprotein helper, often a metal ion, required for activity. Cofactors assist in catalysis by participating in redox reactions or stabilizing intermediates.

Flashcard 18: Which option best defines pure noncompetitive inhibition in enzyme kinetics?

Answer: Inhibitor binds allosteric site; decreases $V_{max}$; $K_m$ unchanged. Noncompetitive inhibitors reduce enzyme activity by binding elsewhere, without affecting substrate binding affinity.

Flashcard 19: If $[S] \ll K_m$, what is the approximate form of $v_0$?

Answer: $v_0\approx \frac{V_{max}}{K_m}[S]$. Low substrate levels make the reaction rate directly proportional to $[S]$, following first-order kinetics.

Flashcard 20: What is the difference between an apoenzyme and a holoenzyme?

Answer: Apoenzyme lacks cofactor; holoenzyme is the active complete enzyme. The apoenzyme requires the cofactor to form the functional holoenzyme for catalytic activity.

Flashcard 21: Which option best defines the transition state in enzyme-catalyzed reactions?

Answer: The highest-energy, unstable configuration between reactants and products. The transition state represents the peak of the energy barrier that must be overcome for the reaction to occur.

Flashcard 22: At $[S]=K_m$, what is $v_0$ expressed in terms of $V_{max}$?

Answer: $v_0=\frac{V_{max}}{2}$. At $[S]=K_m$, the enzyme operates at half its maximum capacity according to Michaelis-Menten kinetics.

Flashcard 23: What is the Michaelis-Menten equation for initial rate $v_0$?

Answer: $v_0=\frac{V_{max}[S]}{K_m+[S]}$. This equation models the hyperbolic relationship between initial velocity and substrate concentration in enzyme kinetics.

Flashcard 24: Which option best defines uncompetitive inhibition in enzyme kinetics?

Answer: Inhibitor binds $ES$; decreases $K_m$ and decreases $V_{max}$. Uncompetitive inhibitors bind only to the enzyme-substrate complex, altering both apparent affinity and maximum rate.

Flashcard 25: On a Lineweaver-Burk plot, what is the $x$-intercept equal to?

Answer: $-\frac{1}{K_m}$. The x-intercept is the negative reciprocal of $K_m$, indicating substrate affinity in the plot.