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

Study 5e Enzyme Structure Catalysis in MCAT Chemical and Physical Foundations of Biological Systems with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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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.

How to use these flashcards

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.

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

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QUESTION

What is the relationship between equilibrium constant KeqK_{eq}Keq​ and enzyme catalysis?

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ANSWER

Enzymes do not change KeqK_{eq}Keq​. Since enzymes do not affect ΔG\Delta GΔG, they leave the ratio of products to reactants at equilibrium unchanged.

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Flashcard 1: What is the relationship between equilibrium constant KeqK_{eq}Keq​ and enzyme catalysis?

Answer: Enzymes do not change KeqK_{eq}Keq​. Since enzymes do not affect ΔG\Delta GΔG, they leave the ratio of products to reactants at equilibrium unchanged.

Flashcard 2: 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 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: What is the definition of VmaxV_{max}Vmax​ in enzyme kinetics?

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

Flashcard 5: What is the primary effect of an enzyme on a reaction energy diagram?

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

Flashcard 6: What is the relationship between ΔG\Delta GΔG and enzyme catalysis for a given reaction?

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

Flashcard 7: Which option best describes how enzymes increase reaction rate?

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

Flashcard 8: What is the Michaelis-Menten equation for initial rate v0v_0v0​?

Answer: v0=Vmax[S]Km+[S]v_0=\frac{V_{max}[S]}{K_m+[S]}v0​=Km​+[S]Vmax​[S]​. This equation models the hyperbolic relationship between initial velocity and substrate concentration in enzyme kinetics.

Flashcard 9: 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 10: 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 11: 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 12: 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 13: What is the definition of KmK_mKm​ in Michaelis-Menten kinetics?

Answer: [S][S][S] at which v0=Vmax2v_0=\frac{V_{max}}{2}v0​=2Vmax​​. KmK_mKm​ reflects the enzyme's affinity for the substrate, where half-maximal velocity is achieved.

Flashcard 14: 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 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: At [S]=Km[S]=K_m[S]=Km​, what is v0v_0v0​ expressed in terms of VmaxV_{max}Vmax​?

Answer: v0=Vmax2v_0=\frac{V_{max}}{2}v0​=2Vmax​​. At [S]=Km[S]=K_m[S]=Km​, the enzyme operates at half its maximum capacity according to Michaelis-Menten kinetics.

Flashcard 17: If [S]≫Km[S] \gg K_m[S]≫Km​, what is the approximate value of v0v_0v0​?

Answer: v0≈Vmaxv_0\approx V_{max}v0​≈Vmax​. High substrate concentrations saturate the enzyme, making the reaction rate independent of [S][S][S].

Flashcard 18: If [S]≪Km[S] \ll K_m[S]≪Km​, what is the approximate form of v0v_0v0​?

Answer: v0≈VmaxKm[S]v_0\approx \frac{V_{max}}{K_m}[S]v0​≈Km​Vmax​​[S]. Low substrate levels make the reaction rate directly proportional to [S][S][S], following first-order kinetics.

Flashcard 19: What is the Lineweaver-Burk (double-reciprocal) form of Michaelis-Menten?

Answer: 1v0=KmVmax1[S]+1Vmax\frac{1}{v_0}=\frac{K_m}{V_{max}}\frac{1}{[S]}+\frac{1}{V_{max}}v0​1​=Vmax​Km​​[S]1​+Vmax​1​. This linear transformation allows determination of KmK_mKm​ and VmaxV_{max}Vmax​ from intercepts on a double-reciprocal plot.

Flashcard 20: On a Lineweaver-Burk plot, what is the yyy-intercept equal to?

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

Flashcard 21: On a Lineweaver-Burk plot, what is the xxx-intercept equal to?

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

Flashcard 22: Which option best defines competitive inhibition in enzyme kinetics?

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

Flashcard 23: Which option best defines pure noncompetitive inhibition in enzyme kinetics?

Answer: Inhibitor binds allosteric site; decreases VmaxV_{max}Vmax​; KmK_mKm​ unchanged. Noncompetitive inhibitors reduce enzyme activity by binding elsewhere, without affecting substrate binding affinity.

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

Answer: Inhibitor binds ESESES; decreases KmK_mKm​ and decreases VmaxV_{max}Vmax​. Uncompetitive inhibitors bind only to the enzyme-substrate complex, altering both apparent affinity and maximum rate.

Flashcard 25: 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.