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Chapter 6: Problem 6

Effect of Enzymes on Reactions Consider this simple reaction: \(\mathrm{S} \underset{\mathrm{k}_{2}}{\stackrel{\mathrm{k}_{1}}{\rightleftharpoons} \mathrm{P}} \quad\) where \(\quad K_{\mathrm{eq}}^{\prime}=\frac{[\mathrm{P}]}{[\mathrm{S}]}\) Which of the listed effects would be brought about by an enzyme catalyzing the simple reaction? a. increased \(k_{1}\) b. increased \(K_{\mathrm{eq}}^{\prime}\) c. decreased \(\Delta G^{\ddagger}\) d. more negative \(\Delta G^{\prime \circ}\) e. increased \(k_{2}\)

Short Answer

Expert verified
Increased \(k_1\), decreased \(\Delta G^{\ddagger}\), increased \(k_2\).

Step by step solution

01

Understanding Enzyme Catalysis

Enzymes act as catalysts in biochemical reactions. They speed up reactions without altering the equilibrium position or the thermodynamic properties of the reactants and products.
02

Effect on Reaction Rate Constants

Enzymes increase the reaction rate by lowering the activation energy, which specifically increases the rate constants, both forward (\(k_1\)) and backward (\(k_2\)). This means options 'a' (\(k_1\)) and 'e' (\(k_2\)) would increase.
03

Effect on Equilibrium Constant

The equilibrium constant (\(K_{eq}'\)) is only affected by changes in the relative energy of the products and reactants, which is not altered by enzymes. Hence, 'b' (\(K_{eq}'\)) is not affected by enzyme action.
04

Effect on Activation Energy

Enzymes lower the activation energy (\(\Delta G^{\ddagger}\)) for both forward and reverse reactions. This confirms option 'c' (\(\Delta G^{\ddagger}\)) would decrease.
05

Effect on Free Energy Change

The standard free energy change (\(\Delta G'^{\circ}\)) is a function of the initial and final states of the reaction, not the reaction pathway, and thus is unaffected by enzymes. Therefore, 'd' (\(\Delta G'^{\circ}\)) remains unchanged.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Rate Constants
Enzymes play a vital role in speeding up reactions by affecting the reaction rate constants. These constants, denoted as \(k_1\) for the forward reaction and \(k_2\) for the reverse reaction, determine the speed at which the reactants are converted into products and vice versa.

An enzyme achieves this by providing an alternative pathway for the reaction, which often requires less energy. Therefore, with enzymes in play, both \(k_1\) and \(k_2\) increase, leading to a faster reaction.
  • Increased \(k_1\) means the reaction moves more quickly from reactants to products.
  • Increased \(k_2\) means the conversion from products back to reactants is also enhanced.
This acceleration affects how quickly equilibrium is reached but not the equilibrium position itself.
Activation Energy
Activation energy, represented by \(\Delta G^{\ddagger}\), is the energy barrier that must be overcome for a reaction to proceed. Enzymes significantly lower this energy barrier, making it easier and quicker for a reaction to occur.

The lowering of activation energy is crucial because it directly influences the reaction rate.
  • A lower \(\Delta G^{\ddagger}\) means that less energy is required to initiate the reaction.
  • This effectively increases the reaction rates for both the forward and backward processes.
In essence, enzymes do not alter the starting or ending points of a reaction; instead, they make the journey between these points much faster and require less energy.
Equilibrium Constant
The equilibrium constant, \(K_{eq}'\), defines the ratio of the concentration of products to reactants when a reaction is at equilibrium. It is a representation of the balance between the forward and reverse reactions.

Importantly, enzymes do not affect \(K_{eq}'\). This constant is anchored by the energies of the reactants and products in their respective states and remains unaltered by the presence of a catalyst.
  • \(K_{eq}'\) remains constant regardless of enzymatic involvement.
  • Enzymes allow the equilibrium to be reached more quickly by increasing reaction rates.
Thus, while enzymes speed up how fast a system reaches equilibrium, they do not change the proportions between reactants and products at equilibrium.
Free Energy Change
The standard free energy change, \(\Delta G'^{\circ}\), represents the overall energy change from reactants to products under standard conditions. Unlike activation energy, which enzymes affect directly, \(\Delta G'^{\circ}\) is a measure of the potential of the reaction.

This value remains unaffected by the presence of enzymes because it is intrinsic to the chemical nature of the substances involved in the reaction.
  • \(\Delta G'^{\circ}\) is determined by the relative stability and energy levels of reactants and products.
  • Since enzymes do not shift the energies of the initial or final states, \(\Delta G'^{\circ}\) remains stable.
Enzymes simply provide a more efficient pathway for the reaction, making it faster without changing this fundamental energy difference.

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Most popular questions from this chapter

Intracellular Concentration of Enzymes To approximate the concentration of enzymes in a bacterial cell, assume that the cell contains equal concentrations of 1,000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100,000 . Assume also that the bacterial cell is a cylinder (diameter \(1.0 \mu \mathrm{m}\), height \(2.0 \mu \mathrm{m}\) ), that the cytosol (specific gravity \(1.20\) ) is \(20 \%\) soluble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the average molar concentration of each enzyme in this hypothetical cell.

Irreversible Inhibition of an Enzyme Many enzymes are inhibited irreversibly by heavy metal ions such as \(\mathrm{Hg}^{2+}, \mathrm{Cu}^{2+}\), or \(\mathrm{Ag}^{+}\), which can react with essential sulfhydryl groups to form mercaptides: $$ \text { Enz-SH }+\mathrm{Ag}^{+} \rightarrow \text { Enz-S-Ag }+\mathrm{H}^{+} $$ The affinity of \(\mathrm{Ag}^{+}\)for sulfhydryl groups is so great that \(\mathrm{Ag}^{+}\) can be used to titrate - SH groups quantitatively. An investigator added just enough \(\mathrm{AgNO}_{3}\) to completely inactivate a \(10.0 \mathrm{~mL}\) solution containing \(1.0 \mathrm{mg} / \mathrm{mL}\) enzyme. A total of \(0.342 \mu \mathrm{mol}\) of \(\mathrm{AgNO}_{3}\) was required. Calculate the minimum molecular weight of the enzyme. Why does the value obtained in this way give only the minimum molecular weight?

The Effects of Reversible Inhibitors The MichaelisMenten rate equation for reversible mixed inhibition is written as $$ V_{0}=\frac{V_{\max }[\mathrm{S}]}{\alpha K_{\mathrm{m}}+\alpha^{\prime}[\mathrm{S}]} $$ Apparent, or observed, \(K_{\mathrm{m}}\) is equivalent to the [S] at which $$ V_{0}=\frac{V_{\max }}{2 \alpha^{\prime}} $$ Derive an expression for the effect of a reversible inhibitor on apparent \(K_{\mathrm{m}}\) from the previous equation.

Kinetic Inhibition Patterns Indicate how the observed \(K_{\mathrm{m}}\) of an enzyme would change in the presence of inhibitors having the given effect on \(a\) and \(\alpha^{\prime}\) : a. \(\alpha>\alpha^{\prime} ; \alpha^{\prime}=1.0\) b. \(\alpha^{\prime}>\alpha\) c. \(\alpha=\alpha^{\prime} ; \alpha^{\prime}>1.0\) d. \(\alpha=\alpha^{\prime} ; \alpha^{\prime}=1.0\)

Properties of an Enzyme of Prostaglandin Synthesis Prostaglandins are one class of the fatty acid derivatives called eicosanoids. Prostaglandins produce fever and inflammation, as well as the pain associated with inflammation. The enzyme prostaglandin endoperoxide synthase, a cyclooxygenase, uses oxygen to convert arachidonic acid to \(\mathrm{PGG}_{2}\), the immediate precursor of many different prostaglandins (prostaglandin synthesis is described in Chapter 21 . Ibuprofen inhibits prostaglandin endoperoxide synthase, thereby reducing inflammation and pain. The kinetic data given in the table are for the reaction catalyzed by prostaglandin endoperoxide synthase in the absence and presence of ibuprofen. a. Based on the data, determine the \(V_{\max }\) and \(K_{\mathrm{m}}\) of the enzyme. \(\begin{array}{ccc}\begin{array}{c}\text { [Arachidonic } \\ \text { acid] }(\mathrm{mM})\end{array} & \begin{array}{c}\text { Rate of formation of } \\\ \mathrm{PGG}_{2}\left(\mathrm{mM} \mathrm{min}^{-1}\right)\end{array} & \begin{array}{c}\text { Rate of formation of } \\ \mathrm{PGG}_{2} \text { with } 10 \mathrm{mg} / \mathrm{mL}\end{array}\end{array}\) \begin{tabular}{ccc} ibuprofen & \(\left(\mathrm{mM}^{-1} \mathrm{~min}^{-1}\right)\) \\ \hline \(0.5\) & \(23.5\) & \(16.67\) \\ \(1.0\) & \(32.2\) & \(30.49\) \\ \(1.5\) & \(36.9\) & \(37.04\) \\ \(2.5\) & \(41.8\) & \(38.91\) \\ \(3.5\) & \(44.0\) & 25 \\ \hline \end{tabular} b. Based on the data, determine the type of inhibition that ibuprofen exerts on prostaglandin endoperoxide synthase.
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