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Question

Which one of the following statements is CORRECT for enzyme catalyzed reactions? ($\Delta G$ is Gibbs free energy change, $K_{eq}$ is equilibrium constant)

The correct answer is
Enzymes do not affect $\Delta G$ or $K_{eq}$

Enzyme Catalysis Fundamentals

Enzymes are biological catalysts that accelerate the rate of biochemical reactions. They achieve this by lowering the activation energy required for the reaction to proceed.

Thermodynamics of Enzyme Action

Key thermodynamic parameters determine the feasibility and extent of a reaction:

  • Gibbs Free Energy Change ($\Delta G$): This parameter indicates the spontaneity of a reaction. A negative $\Delta G$ signifies a spontaneous reaction, while a positive $\Delta G$ indicates a non-spontaneous reaction. At equilibrium, $\Delta G = 0$.
  • Equilibrium Constant ($K_{eq}$): This represents the ratio of products to reactants at chemical equilibrium. It reflects the position of the equilibrium.

Enzymes' Effect on $\Delta G$ and $K_{eq}$

Crucially, enzymes do not alter the overall thermodynamics of a reaction:

  • $\Delta G$: Enzymes do not change the free energy difference between the reactants and the products. The initial and final states of the reaction remain energetically the same, regardless of whether an enzyme is present. Therefore, enzymes do not affect $\Delta G$.
  • $K_{eq}$: The equilibrium constant ($K_{eq}$) is directly related to $\Delta G$ via the equation $\Delta G = -RT \ln K_{eq}$. Since enzymes do not change $\Delta G$, they also cannot change $K_{eq}$. Enzymes help the reaction reach equilibrium faster but do not shift the equilibrium position itself.

Conclusion: Enzymes increase reaction rates by lowering activation energy but do not affect the overall free energy change ($\Delta G$) or the equilibrium constant ($K_{eq}$). They only help the system reach the existing equilibrium state more quickly.

Therefore, the correct statement is that enzymes do not affect $\Delta G$ or $K_{eq}$.

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Important Questions from Enzyme Kinetics and Michaelis Menten Equation

  1. The catalytic efficiency of an enzyme following Michaelis-Menten kinetics is defined by
  2. You are characterizing a new enzyme isolated and purified in the laboratory. If the maximum velocity of the enzyme is $1800 \text{ } \mu moles \text{ L}^{-1}  \text{min}^{-1}$ and the total concentration of the enzyme in the reaction mixture is $1.5 \mu \text{M}$, then the turnover number of the enzyme is _______ $\text{s}^{-1}$. (answer in integer)

  3. You have purified an enzyme using a series of chromatographic methods. It was observed that a $10 \mu \text{  g mL}^{-1}$ of this purified enzyme converted $10 \text{ mM}$ substrate per hour at 25$^{\circ}$C and pH 7. Its specific activity is _______ $\text{IU  } \mu\text{g}^{-1}$. (rounded off to three decimal places)

  4. Within the Michaelis-Menten framework, the ratio of $v_0/V_{max}$ 

    when $[S] = 20 \times K_m$ is _________. 

    (Round off to two decimal places)

  5. The activity of lactate dehydrogenase can be measured by monitoring the following reaction: 

    Pyruvate + NADH $ \longrightarrow $ Lactate + $NAD^+$ 

    The molar extinction coefficient of NADH at 340 nm is $6220 \ M^{-1}.cm^{-1}$. $NAD^+$ does not absorb at this wavelength. In an assay, $25 \ \mu L$ of a sample of enzyme (containing $5 \ \mu g$ protein per mL) was added to a mixture of pyruvate and NADH to give a total volume of 3 mL in a cuvette of 1 cm pathlength. The rate of decrease in absorbance at 340 nm was $0.14 \ min^{-1}$. The specific activity of the enzyme will be ____________________ $ \mu mol.min^{-1}.mg^{-1}$.

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