Professor David Liu and Brian Tse, Life Sciences 1a page 1

Reaction Thermodynamics• Thermodynamics reflects the degree to which a reaction is favored

or disfavored

• Recall: G = Gibbs free energy = the energy available to do work

• ΔG = change in G of the system as state A becomes state B

EnergyΔG = negative“downhill”“favorable”“spontaneous”

A BEnergyΔG = positive “uphill”

“disfavored”A BΔG = zero A B at equilibrium

• ΔG can change over time and is dependent on many factorsincluding the reaction, concentrations, and temperature

Thermodynamics of a reaction and free energy

Two of the most important components of any chemical reaction are the reaction’s thermodynamicsand its kinetics. A reaction’s thermodynamics and its kinetics are distinct and there is no necessaryrelationship between the two.

A reaction’s thermodynamics is the degree to which the reaction is “favorable”— in other words, howmuch energy is released into the surroundings (or taken from the surroundings) as the substrates aretransformed into products. You recently learned the basic thermodynamics underlying protein folding,and many of the same concepts can be used to study chemical reactions. Recall that free energy (G) isthe energy available to do work and is expressed in units of energy per numbers of molecules (forexample, calories per mole). The quantity of energy released or absorbed by a system as the startingmaterials of a reaction become products is called the change in Gibbs free energy of the system (ΔG)..

A negative ΔG indicates that under a specified set of conditions, the system is releasing energy as thestarting state begins to transform into the ending state. In this case, the free energy (G) of the startingstate of the system is higher than that of the system some defined amount of time later. A process witha negative ΔG is “favorable”, or “downhill” under those specific set of conditions and takes placesspontaneously. Conversely, a positive ΔG describes a process that increases the system’s free energy.Such an “uphill” process is unfavorable under those specific set of conditions and does not take placespontaneously.

ΔG for starting materials becoming products in a chemical reaction is dependent on many factorsincluding the identity of the reactants, the concentrations of the substrates and products, and thetemperature of the reaction. G and ΔG are constantly changing for a system that is not at equilibrium.When starting materials are allowed to react, the relative concentrations of substrates and products beginto change. These changes, in turn, lower G such that the total free energy for the system decreases;therefore ΔG is negative as a system heads toward equilibrium. The free energy of the system willcontinue to decrease as the reaction progresses until ΔG = 0. When ΔG = 0, the system is at equilibriumand there is no additional net conversion of substrates to products (or vice-versa). At this point, the freeenergy of the system is now constant.

Professor David Liu and Brian Tse, Life Sciences 1a page 2

Thermodynamics: ΔG, ΔG°rxn, and Keq

• To study and compare reactions, need a standard state, definedas 1 M concentrations, 1 atm pressure, constant temperature

• ΔG° is defined as the change in G under standard state conditionsfor a hypothetical reaction that proceeds to 100% completion

• ΔG°, unlike ΔG, depends only on the identity of the reactants

A + B C + D

ΔG = ΔG°rxn + RT ln [C][D][A][B]

Therefore, ΔG°rxn = –RT ln Keq

At equilibrium, 0 = ΔG°rxn + RT ln Keq

Because ΔG can change constantly and depends on so many factors including reactant concentrations,it cannot be easily used to study or compare two different reactions. To compare and interpret thethermodynamics of various reactions, scientists have defined a “standard state” set of referenceconditions. This standard state is usually: substrate concentrations = 1 M, pressure = 1 atm, and aconstant temperature (which must be specified). Under these standard state conditions, ΔG= ΔG° bydefinition. ΔG° is the change in free energy that would result from the hypothetical complete conversionof all the starting materials into products under standard state conditions. Because in the standard stateall reaction conditions affecting free energy are pre-defined, the value of ΔG° depends solely on theidentities of the starting materials and products. Therefore, the concentrations of reactants or the extentto which a reaction is complete— two factors that influence ΔG— do not affect ΔG°.

Reactions of substrates high in free energy that generate products lower in free energy have negativevalues of ΔG°; conversely, reactions involving lower G° substrates that give rise to higher G° productshave positive values of ΔG°. When ΔG° for a reaction is < 0 (negative), that reaction is favorable or“downhill” under the standard state conditions and can take place spontaneously. Conversely, a reactionwith a ΔG° > 0 (positive) is unfavorable or “uphill” and does not occur spontaneously to completion understandard state conditions (although at equilibrium this reaction would still consist of some product whichwould represent less than 50% conversion; see below). Keep in mind that when considering biologicalprocesses, however, ΔG° is not a relevant value on its own because conditions in cells are very differentfrom standard state conditions and because reactions rarely proceed to completion. Nevertheless, ΔG° isvery useful for comparing the reaction thermodynamics of different processes.

For the representative reaction of A + B ---> C + D, ΔG° and ΔG are related by a very usefulequation: ΔG = ΔG° + RT ln ([C][D] ÷ [A][B]). where T is temperature in degrees Kelvin and R is theideal gas constant. When the system is at equilibrium, ΔG = 0 and [C][D] ÷ [A][B] = Keq; therefore, atequilibrium this equation becomes ΔG° = –1.4 log10(Keq). A useful rule of thumb emerging from thisequation is that at room temperature, ~1.4 kcal/mol of free energy corresponds to a 10-fold change inthe equilibrium constant under standard state conditions.

Professor David Liu and Brian Tse, Life Sciences 1a page 3

Thermodynamics of Coupled Reactions

ATP ADP (P)Large

Energy

SmallEnergy A B ΔG° > 0

ΔG° < 0

ATP A B ADP (P) SmallEnergy ΔG° < 0

• Cells can enable uphill reactions to proceed efficiently bycoupling them with favorable reactions (e.g., ATP hydrolysis)

One must be careful interpreting ΔG° values. A positive ΔG° indicates that a reaction isunfavorable under the stated reaction conditions (which may or may not correspond to apositive ΔG°) and must be coupled to a more favorable process in order to take place to asignificant extent. In the reactions driving life processes, ATP hydrolysis, a highly favorableprocess (ΔG° = -7.4 kcal/mol) is often coupled to reactions having positive ΔG° values suchthat the overall process has a ΔG° < 0 and therefore can take place.

A negative ΔG° indicates that a reaction is favorable, but does not necessarily mean thatthe substrates will become the products on a reasonable time scale. For example, as wedemonstrated in class it is thermodynamically highly favorable (ΔG° << 0) if hydrogen gas andoxygen gas reacted to form water. However, in the absence of a flame or spark, this reactiondoes not take place at any appreciable rate at room temperature.

Professor David Liu and Brian Tse, Life Sciences 1a page 4

Reaction Thermodynamics Example

A BKeq = 10

ΔG° = –RT ln Keq = –1.4 kcal/mol at 298 K

time

conc

entra

tion start with 100% A

timeco

ncen

tratio

n

[B]/[A] atequilibrium

= 10

start with 100% B

time

ΔG 0

Atequilibrium,ΔG = 0& G is

constant

time

G decreasesas a systemapproachesequilibrium

free

ener

gy (G

)

Let’s integrate our understanding of reaction thermodynamics into an example. Imagine achemical reaction in which molecule A is transformed into molecule B. In this example, Keq forthe equilibrium between A and B as drawn above is 10. As you know, this Keq value meansthat at equilibrium, [B]/[A] = 10. Furthermore, based on the relationship between ΔG° andKeq described earlier, you also should be able to deduce the difference in free energy understandard state conditions between A and B— that is, the value of ΔG°. At room temperature(25 °C = 298 K), and using the value of R, the ideal gas constant, as approximately 2 cal/mol°K, we can calculate that ΔG° = –1.4 kcal/mol. This means that B is lower in free energy thanA under standard state conditions by 1.4 kcal/mol.

It’s important to realize that this value of ΔG° is constant— it doesn’t change regardless ofwhether the reaction has reached equilibrium, or even whether the reaction has taken placeyet. ΔG° depends only on the identity of A and B in this example.

If we start this reaction with a test tube containing 100% A, molecules of A will begin toconvert into molecules of B. The free energy of the system (G) decreases as the systemmoves toward equilibrium, and therefore ΔG starts out negative. Once the ratio of [B]:[A]reaches 10:1, the system is at equilibrium and no additional net changes in concentration takeplace. At this point, G is constant and ΔG is therefore zero.

We can also start the reaction with a test tube containing 100% B. In this case, moleculesof B will convert into molecules of A until the same equilibrium point is reached. Importantly,just as in the case of starting with 100% A, the free energy of the system decreases as thesystem moves toward equilibrium.

Professor David Liu and Brian Tse, Life Sciences 1a page 5

2 H2 + O2 (dilute)

The Thermodynamics Versus Kineticsof Blowing Stuff Up

2 H2 + O2 (dilute)heat

2 H2 + O2 (concentrated)heat

• All three of the above reactions are thermodynamically identical• Heat increases the rate of a reaction• Concentrating the substrates increases the rate of a reaction

very slow fast very fast

Why, and how do enzymes use these principles?

Professor David Liu and Brian Tse, Life Sciences 1a page 6

Reaction Kinetics: Ratesand Rate Constants

• Kinetics determines the rate at which reactions occur

transition state

S [TS]‡ PLow energy TS = fast reactionHigh energy TS = slow reaction

• For the above simple reaction, rate = k[S1][S2]

S1 S2 [TS]‡ P

• k = the rate constant, a function of ΔG‡, the transition statefree energy (larger ΔG‡ = smaller k = slower reaction)

Kinetics of a reaction and ΔG‡

The reason why hydrogen and oxygen gas won’t react (burn) at any significant rate at roomtemperature is explained by the second crucial component of a chemical reaction: thereaction’s kinetics. While thermodynamics are most informative about the outcome of areaction once it reaches equilibrium, the kinetics of a reaction describe the rate at which thereaction can take place under a given set of reaction conditions. Because living systems arenever at equilibrium (indeed, a defining characteristic of life is that it manages to sustain itselffar from equilibrium), an understanding of reaction kinetics is crucial to understanding thechemistry of life.

In order for the substrates of a chemical reaction to become products, the free energy ofthe starting materials must be increased until it exceeds a certain threshold value. When theenergy of a system reaches this threshold, the system is said to exist in the transition state,the state of the highest free energy during the course of a chemical reaction. The ‡ symbol isused to indicate that a descriptor refers to a transition state. Because the energy of thetransition state is at a maximum, the system can spontaneously become products (or revert tostarting materials) without requiring the infusion of any additional energy. The change in freeenergy between the substrates and the transition state is called ΔG‡, also known as theactivation energy of a reaction.

Professor David Liu and Brian Tse, Life Sciences 1a page 7

Reaction Kinetics: Ratesand Rate Constants

• Kinetics determines the rate at which reactions occur

transition state

S [TS]‡ PLow energy TS = fast reactionHigh energy TS = slow reaction

• For the above simple reaction, rate = k[S1][S2]

S1 S2 [TS]‡ P

• k = the rate constant, a function of ΔG‡, the transition statefree energy (larger ΔG‡ = smaller k = slower reaction)

(notes continued from previous page)

The difference between a fast reaction and one that occurs slowly is simply the value of ΔG‡. The larger the activation energy barrier, the more energy must be infused into thesubstrates (most commonly, by heating the reaction) before they can become products, andtherefore the slower the reaction. Conversely, reactions with almost no ΔG‡ proceedextremely quickly.

The “rate” of a chemical reaction is dependent on k, the rate constant, as well as on theconcentrations of each of the reactants. The rate constant (k) is related to e(-ΔG‡/RT) and asyou probably have deduced, larger values of ΔG‡ result in smaller rate constants. The rate of asimple one-step reaction is equal to k multiplied by the concentrations of each of thesubstrates. Doubling the concentration of each of two substrates in the simple reaction shownabove will therefore increase the rate of the reaction by four-fold. Rates of reactions alwayshave units of molecules per time or concentrations per time (for example, moles of productformed divided by seconds, or moles of substrate consumed per liter divided by seconds).

It is crucial that you understand the distinction between Keq and k. The former is anequilibrium constant, a ratio of concentrations at equilibrium. The latter is a rate constant, ameasure of how quickly a given substrate becomes a given product. Confusingly, researchersoften abbreviate Keq using K, emphasizing the extreme importance of clearly notating upper-case K and lower-case k when describing reactions!

Professor David Liu and Brian Tse, Life Sciences 1a page 8

Relationship Between Keq, kfor, and krev

kforward

kreverseA + B C + D

Therefore, kforward/kreverse = Keq

Forward rate = kforward [A][B]Reverse rate = kreverse [C][D]

At equilibrium, forward rate = reverse rate

kforward [A][B] = kreverse [C][D]

kforward/kreverse = [C][D] [A][B]

Although there is no necessary relationship between thermodynamics and kinetics— i.e.,you cannot infer Keq from k alone or vice-versa, there is a useful relationship between theequilibrium constant, the rate constant of the forward reaction (kforward), and the rate constantof the reverse reaction (kreverse). Recall that at equilibrium, the rate (not rate constant!) of theforward reaction equals the rate of the reverse reaction and no net transformation ofsubstrates into products takes place. Since each rate equals the rate constant multiplied bythe concentration of each substrate, at equilibrium you can equate in the example abovekforward [A][B] = kreverse [C][D]. At equilibrium, [C][D]/([A][B]) is also equal to Keq. Thereforeyou can deduce that kforward/kreverse = Keq.

Professor David Liu and Brian Tse, Life Sciences 1a page 9

Reaction Energy Diagrams

S

P

reactants(substrates) products

transition state

[TS]‡

ΔG°‡

ΔG°rxn

S [TS]‡ P• Reaction energy diagrams

depict changes in ΔG° asone substrate moleculebecomes one productmolecule

• X-axis: the reactioncoordinate

• Y-axis: free energy underdefined conditions (doesnot change withconcentration of S or P)

G

Reaction energy diagrams

Concepts in reaction thermodynamics and kinetics are best illustrated by drawing a reaction energydiagram. Such a diagram plots the free energy of a particular molecule undergoing a reaction on the Y-axis. The X-axis is the reaction coordinate, a measure of the degree of progress in the reaction (i.e., theextent to which a starting material molecule has changed into a product molecule. Different molecules(or collections of molecules) are represented in reaction energy diagrams as points in the diagram.These points are positioned vertically at their estimated free-energy levels, with more stable (lowerenergy) species closer to the bottom of the diagram. The points are placed horizontally to indicate whereon the reaction coordinate they are thought to form, with earlier-forming species positioned to the leftside of the diagram, and later-forming species to the right side. The transition state of a reaction isrepresented by a point reflecting its free energy and location along the reaction coordinate. By definitionthe highest point on the reaction energy diagram must be the transition state of the reaction.

Because the products of a favorable reaction are lower in free energy than that of the substrates (ΔG< 0), the overall decrease in free energy for a favorable reaction is represented by an overall downwardslope in the reaction energy diagram going from the starting materials to the products. Of course theenergy of the system during a chemical reaction does not simply change as a straight line connecting thestarting materials and products— instead, the reaction energy diagram must also reflect the higherenergy of the transition state. Typically a smooth curve (occasionally called the energy landscape of areaction) is used to connect the starting materials to the transition state to the products of a reaction in areaction energy diagram. Of course the exact free-energy values of all points on the curve other than thestarting material and product are virtually impossible to ascertain. Nevertheless, the reaction energydiagram is a powerful tool for analyzing chemical reactions.

It is important to realize that the Y-axis of a reaction energy diagram reflects the free energy under aspecific set of conditions, for example, under standard state conditions. Even if those specific conditionsare not standard state conditions (1 M, 1 atm, etc.), the Y-axis free energy values in a reaction energydiagram do not change as the concentrations of S or P change.

Professor David Liu and Brian Tse, Life Sciences 1a page 10

A B C D E

Transition States and Intermediates

‡ ‡intermediate

A

E

C

[B]‡

[D]ॠTransition states are

energy maxima;intermediates are localenergy minima

• Note: only moleculeswith the same combinedmolecular formula canbe placed on the sameenergy diagram!

G

Although the reactions we’ve depicted above represent the simplest possible cases, morecomplex variants are quite common in the reactions of life. A reaction can have multipletransition states, although the transition state with the highest ΔG‡ is the one that plays thelargest role in determining the overall kinetics of the reaction. Likewise, many reactionsproceed through intermediates that are neither starting materials nor products nor transitionstates. Instead, intermediates are “resting points” along a chemical reaction in which partiallyconverted starting materials might exist for a period of time. In contrast, transition states bydefinition are fleeting. In a reaction energy diagram, the difference between an intermediateand a transition state is very clear: intermediates lie in minima (wells), while transition statesare represented as maxima (hills).

How do scientists know what transition states look like? Because a transition state is, bydefinition, a fleeting structure with no significant lifetime, it is extremely difficult for scientiststo directly observe a transition state. Instead, scientists mostly use the outcomes of chemicalreactions to infer the structure of the transition states. In many cases, key features such asthe approximate shape, charge, and atom connectivity of a transition state can be deduced byexamining reaction outcomes.

A final, but very important note about reaction energy diagrams: it is only meaningful to putmolecules on the same free energy diagram that have the same combined chemical formula.Since the starting materials and products of a simple chemical reaction must (under theprinciple of the conservation of matter) contain the same atoms, they can always be placed onthe same energy diagram, assuming that the scientist has correctly depicted them! If youforget one of the starting materials or products or parts of a molecule when you depict achemical reaction, however, creating a reaction energy diagram will be a confusing (andultimately futile) exercise. Therefore it is crucial that all species depicted on the same energydiagram share the same combined chemical formula.

Professor David Liu and Brian Tse, Life Sciences 1a page 11

Drawing (Simple) Transition States

1) Inventory the bondsbeing made and brokenin one chemical step

O

NH

OH

O

NH

OHbond broken

bond made

3) Assign formal charges(e.g., assume partiallyformed bonds representhalf a bond)

O

NH

OH

O making 2bonds is neutral;O making 1 bond(with 3 lone pairs)is –1; O thatmakes ~1.5bonds is “δ–”

δ— δ—

O thatmakes~1.5 bondsis “δ–”

Note overall charge conservation!

O

NH

OH2) Draw a single structure

in which each of thesechanging bonds ispartially formed (dashed)

Drawing Transition States

The difficulty of their direct observation can make transition states challenging to draw in detail.Nevertheless, the essential features of transition states, including those features that are affected byenzymes, can often be depicted using the strategy shown here.

To draw a simple transition state, start by taking inventory of the bonds that are made and brokenover the course of the single chemical step of interest. When we draw starting materials, intermediates,and products of chemical reactions, these bonds are either 100% formed or 100% broken. In contrast, atransition state contains partially formed covalent bonds. Therefore, after identifying the bonds that aremade and broken, draft a single structure— beginning with the starting materials— in which thesechanging bonds are partially formed. Partially formed bonds are drawn using dashed lines (not to beconfused with hydrogen bonds, which are also often drawn using dashed lines).

In the final step of drawing a transition state, examine each atom of your proposed structure andassign appropriate formal charges. Use the formal charge principles we previously discussed to assignpartial charges (indicated by the symbol δ) to those atoms that participate in partial covalent bonds. Forexample, recall that an O atom that makes two bonds (and has two lone pairs) is neutral, while an Oatom that makes one bond (and has three lone pairs) has a –1 charge. An O atom that makes one fullbond and one partial bond (roughly 1.5 bonds) will therefore have a δ– charge.

The process of drawing a transition state using the above strategy is applied to the first step of amidebond hydrolysis in this slide. Carefully go through this example, and then pick a different step from anyreaction shown earlier in this course and try your own hand at drawing its transition state. Remember tobracket and place a ‡ around your finished structure to specify its fleeting, transition state status.

Professor David Liu and Brian Tse, Life Sciences 1a page 12

Lectures 15-16: The Molecular Basis of Enzyme Catalysis: HIV Protease

1. The function and structure of HIV proteasea. Introduction to proteasesb. Discovery of HIV proteasec. Overview of the three-dimensional structure of HIV protease

2. Chemical reactions and the energies driving thema. Amide bond cleavage: the reaction catalyzed by HIV proteaseb. Thermodynamics of a reaction and free energyc. Kinetics of a reaction and ΔG‡

d. Reaction energy diagramse. Transition states, intermediates, and how to draw them

3. How enzymes accelerate chemical reactions: the case of HIV proteasea. Catalysts alter a reaction’s kinetics, but not its thermodynamicsb. Chemical strategies behind enzyme catalysis

i. Proximity and orientation effectsii. Nucleophilicity and electrophilicityiii. Acid and base catalysis

4. The molecular basis of substrate specificitya. Trypsin substrate specificityb. HIV protease substrate specificity

Let’s now apply what we’ve learned about reaction thermodynamics and kinetics to theamide bond hydrolysis reaction catalyzed by HIV protease.

While it is sometimes possible to predict if a reaction under standard state conditions isfavorable (ΔG° < 0) or unfavorable (ΔG° > 0) with the assistance of several tables ofempirical and theoretical data, it is usually not possible to predict ΔG of a reaction simply byinspection, especially when the reaction does not take place under standard state conditions.The cleavage of an amide bond in water under typical physiological conditions, however, hasbeen empirically determined to be a favorable reaction.

Despite the fact that amide hydrolysis is a favorable reaction, proteins are notoriouslystable biological macromolecules, with very slow rates of spontaneous hydrolysis arising fromhigh activation energy barriers along the amide hydrolysis reaction coordinate. In fact thehalf-life of a typical peptide bond in the absence of proteases is estimated to be ~1 to 100years. This stability is crucial to life, as many of the structural and functional proteins in yourbody play roles that require long lifetimes. Yet on the other hand, many biological processes(including digestion of proteins and the processing of HIV’s polyproteins into mature viralproteins) require the rapid hydrolysis of proteins. This apparent dilemma between thestability of proteins and the need for rapid protein hydrolysis is resolved by the ability ofenzymes such as HIV protease to accelerate amide bond hydrolysis when needed.

Professor David Liu and Brian Tse, Life Sciences 1a page 13

Enzyme-product complexFree product(s)

Productrelease

Free enzyme

SubstrateAn Enzymein Action

Enzyme-substratecomplex

Catalyzedreaction

[TS]‡

Substratebinding

3. How enzymes accelerate chemical reactions: the case of HIV protease

We are ready to analyze the specific chemical ways in which enzymes can acceleratereactions. At the molecular level, an enzyme-catalyzed reaction unfolds in the following way:the enzyme randomly encounters the substrate in solution. Occasionally such an encounterwill take place in a manner that allows the enzyme to bind to the substrate. When bound tothe enzyme, the substrate experiences a precisely tailored environment that facilitates thesubstrate’s transformation into the transition state of the reaction. The enzyme-stabilizedtransition state can then undergo additional changes to become enzyme-bound product. Theenzyme-product complex then dissociates into free product and free enzyme. The releasedenzyme is ready to catalyze the conversion of another molecule of substrate into product.

Professor David Liu and Brian Tse, Life Sciences 1a page 14

Catalysts Lower ΔG‡

Lower ΔG‡ = Faster reaction

• Enzymes catalyze reactions primarily by decreasing ΔG‡

Catalysts emergeunchanged andtherefore never

alter the ΔG° for thereaction

• Catalysis affects a reaction’s kinetics, not its thermodynamics

(Fisher-Price version)

Catalysts alter a reaction’s kinetics, but not its thermodynamics

Catalysts including enzymes accelerate chemical reactions by lowering the activation barrierof the reaction (i.e., by decreasing ΔG‡) and/or by increasing the effective concentration of thesubstrates. Even though a catalyst may participate in a chemical reaction in a variety of ways,by the time the reaction is complete, a true catalyst by definition must be regenerated inexactly the same form that existed at the start of the reaction.

Because a catalyst is not changed by a chemical reaction, it cannot alter thethermodynamics of the reaction. Therefore, a catalyst cannot make an unfavorable reactionfavorable, or even make a favorable reaction more favorable— the ΔG of a reaction is neverchanged by a catalyst. You can understand this very important point by appreciating that thefree energy of the catalyst is identical in the substrate side of a reaction and in the productside because the catalyst emerges in the same form before and after a reaction; therefore thecatalyst does not contribute to any change in free energy (ΔG) that occurs when the reactiontakes place.

Professor David Liu and Brian Tse, Life Sciences 1a page 15

Enzyme Catalysis(More Accurate)

Uncatalyzedreaction

Enzyme-catalyzedreaction

E-S

E-TS‡

E-P

E + S

E + TS‡

E + P

S = substrateP = productTS‡ = transition stateE = enzyme

• The enzyme muststabilize the transitionstate more than theenzyme stabilizes thesubstrates in order toaccelerate the reaction(so ΔG‡

cat < ΔG‡uncat)

ΔG‡

(enzymecatalyzed)

Reaction coordinate

Free

ene

rgy

ΔG‡

(uncatalyzed)

(More complete version)

An enzyme most frequently lowers ΔG‡ by stabilizing the transition state of a chemicalreaction (in other words, by lowering G‡) more than it stabilizes the substrates. Returning toour reaction energy diagram of an uncatalyzed and catalyzed chemical reaction, you canappreciate why an effective catalyst cannot stabilize both the substrate and the transition stateequally— if this were the case, the ΔG‡ of the catalyzed reaction would be the same as the ΔG‡ of the uncatalyzed reaction, and no rate acceleration of the reaction would result. Aneffective catalyst must preferentially stabilize the transition state.

Professor David Liu and Brian Tse, Life Sciences 1a page 16

Breakout QuestionChorismate mutase is an enzyme that catalyzes the conversion ofchorismate to prephenate. Which of the following predicts the behaviorof two identical solutions that start with 0.1 M chorismate, to which a tinyamount of chorismate mutase is added to only one solution?

A) The [chorismate] in the solution with enzyme is always at least as high as the[chorismate] in the solution without enzyme.

B) The solution with the enzyme will generate more free energy after anyamount of time (short or long) than the solution without the enzyme.

C) At equilibrium, [prephenate] is equal in the two solutions.D) The value of ΔG° approaches zero as both reactions approach equilibrium

chorismate prephenateO

O

O

OH

O

O

HO

O

CO2

O

O

ΔG° < 0