MCAT Biological and Biochemical Foundations of Living Systems Flashcards: 1a Enzyme Structure Catalysis

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This deck focuses on 1a Enzyme Structure Catalysis, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Biological and Biochemical Foundations of Living 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: What does a low $K_m$ indicate about an enzyme for its substrate?

Answer: High apparent affinity (less substrate needed to reach $\frac{V_{max}}{2}$). Lower $K_m$ means stronger binding, requiring less substrate for significant activity.

Flashcard 2: What is a cofactor in enzyme function?

Answer: A nonprotein helper required for activity (metal ion or organic molecule). Cofactors provide essential chemical functionalities that the protein component lacks for catalysis.

Flashcard 3: What is the transition state in an enzyme-catalyzed reaction?

Answer: The highest-energy, unstable intermediate along the reaction pathway. It represents the peak energy barrier that must be overcome for reactants to form products.

Flashcard 4: What are the effects of noncompetitive (pure) inhibition on $K_m$ and $V_{max}$?

Answer: $V_{max}$ decreases; $K_m$ unchanged. Inhibitor reduces functional enzyme concentration without impacting substrate binding affinity.

Flashcard 5: What is the definition of $V_{max}$ for an enzyme?

Answer: Maximum rate when enzyme active sites are saturated with substrate. $V_{max}$ quantifies the enzyme's full catalytic capacity when all sites are occupied.

Flashcard 6: What are the effects of uncompetitive inhibition on $K_m$ and $V_{max}$?

Answer: Both $K_m$ and $V_{max}$ decrease. Apparent affinity increases, but overall catalytic capacity is reduced due to trapped complexes.

Flashcard 7: What is an apoenzyme and what is a holoenzyme?

Answer: Apoenzyme: protein alone; holoenzyme: apoenzyme plus required cofactor(s). Apoenzyme lacks cofactors and is inactive; holoenzyme is the complete, functional form.

Flashcard 8: What is the active site of an enzyme?

Answer: The region that binds substrate and performs catalysis. This specific pocket on the enzyme facilitates substrate interaction and chemical transformation.

Flashcard 9: What are the effects of competitive inhibition on $K_m$ and $V_{max}$?

Answer: $K_m$ increases; $V_{max}$ unchanged. Inhibitor decreases apparent affinity but maximum velocity is achievable with excess substrate.

Flashcard 10: What is the induced-fit model of enzyme-substrate binding?

Answer: Substrate binding triggers a conformational change that improves fit. Enzyme flexibility allows adjustment to substrate shape, enhancing binding specificity and catalytic efficiency.

Flashcard 11: What parameter best represents catalytic efficiency for comparing enzymes?

Answer: $\frac{k_{cat}}{K_m}$. This ratio integrates turnover and affinity, enabling comparison of enzymes under non-saturating conditions.

Flashcard 12: Which statement is correct: enzymes change $K_{eq}$ or enzymes change $E_a$?

Answer: Enzymes change $E_a$ (decrease it); they do not change $K_{eq}$. Enzymes accelerate both forward and reverse reactions equally, preserving equilibrium but reducing energy barrier.

Flashcard 13: What is the definition of turnover number $k_{cat}$?

Answer: Catalytic cycles per enzyme per second at saturation: $k_{cat}=\frac{V_{max}}{[E]_T}$. $k_{cat}$ measures enzyme efficiency by quantifying substrate conversions per unit time at saturation.

Flashcard 14: What is the lock-and-key model of enzyme-substrate binding?

Answer: Active site is preformed and complementary to the substrate shape. Rigid enzyme structure ensures precise substrate recognition without conformational changes during binding.

Flashcard 15: Identify the type of inhibition: inhibitor binds only free enzyme at active site.

Answer: Competitive inhibition. Inhibitor competes with substrate for the active site, reducing available enzyme for catalysis.

Flashcard 16: What catalytic mechanism uses proton donation/abstraction by active-site residues?

Answer: General acid-base catalysis. Active-site amino acids act as acids or bases to stabilize intermediates via proton transfer.

Flashcard 17: What is the difference between a coenzyme and a prosthetic group?

Answer: Coenzyme binds transiently; prosthetic group is tightly or covalently bound. Binding duration distinguishes transient helpers from permanently integrated ones in enzyme structure.

Flashcard 18: Identify the type of inhibition: inhibitor binds only the $ES$ complex.

Answer: Uncompetitive inhibition. Inhibitor stabilizes the enzyme-substrate complex, preventing product release.

Flashcard 19: Identify the type of inhibition: inhibitor binds equally well to $E$ and $ES$ at an allosteric site.

Answer: Noncompetitive (pure) inhibition. Binding at a separate site affects enzyme function without altering substrate affinity.

Flashcard 20: What is the definition of an enzyme in biochemical terms?

Answer: A biological catalyst that increases reaction rate without being consumed. Enzymes accelerate biochemical reactions by lowering activation energy while remaining unaltered post-reaction.

Flashcard 21: What is the Michaelis constant $K_m$ in Michaelis-Menten kinetics?

Answer: Substrate concentration where $v_0 = \frac{V_{max}}{2}$. $K_m$ reflects enzyme-substrate affinity, indicating half-maximal velocity substrate level.

Flashcard 22: State the Michaelis-Menten equation for initial velocity $v_0$.

Answer: $v_0 = \frac{V_{max}[S]}{K_m + [S]}$. This hyperbolic equation models enzyme kinetics under steady-state assumptions for single-substrate reactions.

Flashcard 23: What is the primary way enzymes increase reaction rate?

Answer: They lower the activation energy $E_a$ by stabilizing the transition state. By reducing $E_a$, enzymes increase the fraction of molecules with sufficient energy to react.

Flashcard 24: Which thermodynamic quantity is unchanged by enzyme catalysis: $\Delta G$ or $E_a$?

Answer: $\Delta G$ is unchanged; $E_a$ is decreased. Enzymes affect reaction kinetics but not the overall free energy change or equilibrium position.

Flashcard 25: What catalytic mechanism uses a transient covalent enzyme-substrate intermediate?

Answer: Covalent catalysis. Temporary bond formation facilitates reaction by altering substrate reactivity.