1 SDS PAGE = SDS polyacrylamide gel electrophoresis - PowerPoint Presentation

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SDS PAGE = SDS polyacrylamide gel electrophoresis

sodium dodecyl sulfate, SDS (or SLS): CH3-(CH2)11- SO4-- CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-SO4--

SDS

All the polypeptides are denatured and behave as random coils

All the polypeptides have the same charge per unit length

All are subject to the same electromotive force in the electric field

Separation based on the sieving effect of the polyacrylamide gel

Separation is by molecular weight only

SDS does not break covalent bonds (i.e., disulfides) (but can treat with mercaptoethanol for that) (and perhaps boil for a bit for good measure)

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Disulfides between 2 cysteines can be cleaved in the laboratory by reduction, i.e., adding 2 Hs (with their electrons) back across the disulfide bond.

One adds a reducing agent: mercaptoethanol (HO-CH2-CH2-SH). In the presence of this reagent, one gets exchange among the disulfides and the sulfhydryls:Protein-CH2-S-S-CH2-Protein  + 2 HO-CH2CH2-SH  ---> Protein-CH2-SH + HS-CH2-Protein   +  HO-CH2CH2-S-S-CH2CH2-OH

The protein's disulfide gets reduced (and the S-S bond cleaved), while the mercaptoethanol gets oxidized, losing electrons and protons and itself forming a disulfide bond. 

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Molecular weight

markers (proteins of known molecular weight)P.A.G.E.e.g., “p53”

12 18 48 80 110 130 160 140

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Sephadex bead

Molecular sieve chromatography

(= gel filtration, Sephadex chromatography)

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Sephadex bead

Molecular sieve chromatography

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Sephadex bead

Molecular sieve chromatography

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Sephadex bead

Molecular sieve chromatography

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Sephadex bead

Molecular sieve chromatography

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Plain

Fancy4oC (cold room)

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Non-spherical molecules get to the bottom faster

Larger molecules get to the bottom faster, and ….

Non-spherical molecules get to the bottom faster

~infrequent

orientation

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Handout 4-3: protein separations

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Most charged

and smallestLargest and most sphericalLowest MWLargest and least sphericalSimilar to handout 4-3, but Winners &native PAGE added

Winners:Winners:

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Enzymes = protein catalysts

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Each arrow = an ENZYME

Each arrow = an ENZYME

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H

2 + I22 HIH2 + I2

2 HI + energy“Spontaneous” reaction: Energy released Goes to the right H-I is more stable than H-H or I-I here i.e., the H-I bond is stronger, takes more energy to break it That’s why it “goes” to the right, i.e., it will end up with more products than reactants i.e., less tendency to go to the left, since the products are more stableChemical reaction between 2 reactants

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Change in Energy (Free Energy)

H2 + I22 HI{

-3 kcal/mole2H + 2Isay, 100 kcal/molesay, 103kcal/moleAtom pulled completely apart(a “thought” experiment)Reaction goes spontaneously to the rightIf energy change is negative: spontaneously to the right = exergonic: energy-releasingIf energy change is positive: spontaneously to the left = endergonic: energy-requiring

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H

2 + I22 HI

2 HIH2 + I22 HIH2 + I2Different ways of writing chemical reactionsH2 + I22 HI

2 HI

H

2 + I2

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Change in Energy (Free Energy)

H2 + I22 HI2H + 2I

{-3 kcal/molesay, 100 kcal/molesay, 103kcal/moleBut: it is not necessary to break molecule down to its atoms in order to rearrange them

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H

H+

IIHHII

I

I

HH

Transition state

(TS)

+

H

I

H

I

(2 HI)

H

H

+

I

I

(H

2

+ I

2

)

Products

Reactions proceed through a transition state

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Change in Energy

H2 + I22 HI2H + 2I

{-3 kcal/mole~100 kcal/moleH-H| |I-I(TS)ActivationenergySay,~20 kcal/mole

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Change in Energy

(new scale)H2 + I22 HI

{3 kcal/moleActivation energyHHII(TS)Allows it to happendetermines speed = VELOCITY = rate of a reactionEnergy neededto bring molecules together to forma TS complexNet energy change:Which way it will end up. the DIRECTIONof the reaction, independent of the rate2 separate concepts

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Concerns about the cell’s chemical reactions

DirectionWe need it to go in the direction we wantSpeedWe need it to go fast enough to have the cell double in one generation

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3 glucose’s

18-carbon fatty acidFree energy change: ~ 300 kcal per mole of glucose used is REQUIREDBiosynthesis of a fatty acidSo: 3 glucose

18-carbon fatty acidSo getting a reaction to go in the direction you want is a major problem(to be discussed next time)Example

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Concerns about the cell’s chemical reactions

DirectionWe need it to go in the direction we wantSpeedWe need it to go fast enough to have the cell double in one generationCatalysts deal with this second problem, which we will now consider

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The catalyzed reaction

The velocity problem is solved by catalystsThe catalyst takes part in the reaction, but it itself emerges unchanged

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Change in Energy

H2 + I22 HI

Activation energywithoutcatalystHHII(TS)TS complexwith catalystActivation energyWITH thecatalyst

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Reactants in an enzyme-catalyzed reaction = “substrates”

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Reactants (substrates)

Not a substrateActive site or substrate binding site(not exactly synonymous,

could be just part of the active site)

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Substrate Binding

Unlike inorganic catalysts, enzymes are specific

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Small molecules bind with great specificity to pockets on ENZYME surfaces

Too far

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Unlike inorganic catalysts,

enzymes are specific                                      succinic dehydrogenaseHOOC-HC=CH-COOH <-------------------------------> HOOC-CH2-CH2-COOH +2H fumaric acid                                                     succinic acidNOT a substrate for the enzyme: 1-hydroxy-butenoate:    HO-CH=CH-COOH (simple OH instead of one of the carboxyls)Maleic acid

Platinum will work with all of these, indiscriminantly

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Enzymes work as catalysts for two reasons:

They bind the substrates putting them in close proximity.They participate in the reaction, weakening the covalent bonds of a substrate by its interaction with their amino acid residue side groups (e.g., by stretching).+

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Dihydrofolate reductase, the movie: FH2 + NADPH2

 FH4 + NADP or: DHF + NADPH + H+  THF + NADP+Enzyme-substrate interaction is oftendynamic.The enzyme protein changes its 3-D structureupon binding thesubstrate.http://chem-faculty.ucsd.edu/kraut/dhfr.mpg

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Chemical

kineticsSubstrate  Product (reactants in enzyme catalyzed reactions are called substrates)S  PVelocity = V = ΔP/ Δ tSo V also = -ΔS/ Δ

t (disappearance)From the laws of mass action:ΔP/ Δt = - ΔS/ Δt = k1[S] – k2[P]For the INITIAL reaction, [P] is small and can be neglected:ΔP/ Δt = - ΔS/ Δt = k1[S]So the INITIAL velocity Vo = k1[S]back reactionO signifies INITIAL velocity

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P vs. t

Slope = VoVo = ΔP/ Δ t

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P

t

[S1][S2][S3][S4]Effect of different initial substrate concentrations on P vs. t0.00.20.40.6

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P

Vo = the slope in each caset

[S1][S2][S3][S4]Effect of different initial substrate concentrations0.00.20.40.6Considering Vo as a function of [S](which will be our usual useful consideration):Slope = k1Vo = k1[S]Dependence of Vo on substrate concentraion

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We can ignore the rate of the

non-catalyzed reaction (exaggerated here to make it visible)Now, with an enzyme:

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Vo proportional to [S]

Vo independent of [S]Enzyme kinetics (as opposed to simple chemical kinetics)Can we understand this curve?

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Michaelis and Menten mechanism for the action of enzymes (1913)

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Michaelis-Menten mechanism

Assumption 1. E + S <--> ES: this is how enzymes work, via a complexAssumption 2. Reaction 4 is negligible, when considering INITIAL velocities (Vo, not V).Assumption 3. The ES complex is in a STEADY-STATE, with its concentration unchanged with time during this period of initial rates.  (Steady state is not an equilibrium condition, it means that a compound is being added at the same rate as it is being lost, so that its concentration remains constant.)X

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E + S

ESE + P

System is at equilibrium

Constant level

No net flow

System is at “steady state”

Constant level

Plenty of flow

Steady state is not the same as equilibrium

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Michaelis-Menten Equation(s)

[(k2+k3)/k1] +[S]k3[Eo][S]

Vo =If we let Km = (k2+k3)/k1, just gathering 3 constants into one, then:k3 [Eo] [S]Vo =Km + [S]See handout 5-1 at your leisure for the derivation (algebra, not complicated, neat)=

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k3 [Eo] [S]

Vo = Km + [S]Rate is proportional to the amount of enzyme

Otherwise, the rate is dependent only on SAt low S (compared to Km),rate is proportional to S:Vo ~ k3Eo[S]/KmAt high S (compared to Km),Rate is constantVo = k3EoAll the k‘s are constants for a particular enzyme

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At high S, Vo here = k3Eo, = Vmax

So the Michaelis-Menten equation can be written:Vmax [S]Vo =Km + [S]

k3 [Eo] [S]Vo =Km + [S]Simplest form=

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Understanding Vmax:

( the maximum intital velocity achievable with a given amount of enzyme )Now, Vmax = k3EoSo: k3 = Vmax/Eo= the maximum (dP/dt)/Eo, = the maximum (-dS/dt)/Eok3 = the TURNOVER NUMBERthe maximum number of moles of substrate converted to product per mole of enzyme per second; the maximum number of molecules of substrate converted to product per molecule of enzyme per secondTurnover number (k3) then is: a measure of  the enzyme's catalytic power.

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Some turnover numbers

(per second)Succinic dehydrogenase: 19 (below average)Most enzymes: 100 -1000The winner: Carbonic anhydrase (CO2 +H20 H2CO3)600,000That’s 600,000 molecules of substrate, per molecule of enzyme, per second.Picture it!You can’t.

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Km ?

Consider the Vo that is 50% of VmaxSo Km is numerically equal to the concentration of substrate required to drive the reaction at ½ the maximal velocityTry it: Set Vo = ½ Vmax in the M.M. equation and solve for S.

Vmax/2 is achieved at a [S] that turns out to be numerically equal to Km

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The equilibrium constant for this dissociation reaction is:

Consider the reverse of this reaction(the DISsociation of the ES complex):ES

k2k1E + SKd = [E][S] / [ES] = k2/k1(It’s the forward rate constant divided by the backward rate constant. See the Web lecture if you want to see this relationship derived)Another view of Km:==

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ES

k2k1E + SK

d = k2/k1Km = (k2+k3)/k1 (by definition)IF k3 << k2, then: Km ~ k2/k1But k2/k1 = Kd (from last graphic)so Km ~ Kd for the dissociation reaction (i.e. the equilibrium constant) (and 1/Km = ~ the association constant)So: the lower the Km, the more poorly it dissociates.That is, the more TIGHTLY it is held by the enzymeAnd the greater the Km, the more readily the substrate dissociates,so the enzyme is binding it poorly{Consider in reverse

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Km ranges

10-6M is good10-4M is mediocre10-3M is fairly poorSo Km and k3 quantitatively characterize how an enzyme does the job as a catalystk3, how good an enzyme is in facitiating the chemical change (given that the substrate is bound)Km, how well the enzyme can bind the substrate in the first place

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Got this far

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A

competitive inhibitor resembles the substrateEnzyme inhibition: competitive, non-competitive, and allostericCompetitive:

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A competitive inhibitor can be swamped out at high

substrate concentrationsHandout 5-3b

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-

Apparent (measured) Km increasesInhibitor looks like the substrate And, like the substrate, binds to the substrate binding site

Substrate concentrationVo+

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Biosynthetic pathway to cholesterol

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Zocor

(simvastatin)

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½ Vmax

w/o inhibitor½ Vmax withyet more inhibitorKm remains unchanged. Vmax decreases.

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Substrate

Non-competitive inhibitorExample: Hg ions (mercury) binding to –SH groups in the active site

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Non-competitive inhibitor exampleSubstrate still binds OK

But an essential participant in the reaction is blocked(here, by mercury binding a cysteine sulfhydryl)--CH2-SH

Hg

++

--CH2-SH

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= allosteric inhibitor

= substrate

+

Active

Inactive

Allosteric inhibitor binds to a different site than the substrate,

so it need bear no resemblance to the substrate

Active

The apparent

Km

OR the apparent

Vmax

or

both

may be affected.

The effects on the Vo vs. S curve are more complex and ignored here

Inhibitor

binding

site

Allosteric inhibition

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End product

End productEnd product

Feedback inhibition of enzyme activity, or “End product inhibition” First committed step is usually inhibitedAllosteric inhibitors are used by the cell for feedback inhibition of metabolic pathways

P

 Q  R  S  T  U  V

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              Thr deaminase

glucose  ......  --> --> threonine -----------------> alpha-ketobutyric acid SubstrateAllosteric inhibitorAlso here: Feedback inhibitor(is dissimilar from substrate)

protein

protein

isoleucine 

(and no other aa)ABC

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60 minutes, in a

minimal medium 20 minutes !, in a rich medium Rich medium = provide glucose + all 20 amino acids and all vitamins, etc.

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Direction

of reactions in metabolism

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}

Energy differencedetermines the direction

of a chemical reagion Free

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For the model reaction

A + B C + D, written in the left-to-right direction indicated:Consider the quantity called the change in free energy associated with a chemical reaction, or: Δ G Such that:IF Δ G IS <0:THEN A AND B

WILL TEND TO PRODUCE C AND D(i.e., tends to go to the right).IF Δ G IS >0:THEN C AND D WILL TEND TO PRODUCE A AND B.(i.e., tends to go to the left)IF Δ G IS = 0:THEN THE REACTION WILL BE AT EQUILIBRIUM: NOT TENDING TO GO IN EITHER DIRECTION IN A NET WAY.

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Δ

G = Δ Go+ RTln([C][D]/[A][B])where A, B, C and D are the concentrations of the reactants and the products AT THE MOMENT BEING CONSIDERED.(i.e., these A, B, C, D’s here are not the equilibrium concentrations)R = UNIVERSAL GAS CONSTANT = 1.98 CAL / DEG K MOLE (R =~2)T = ABSOLUTE TEMP ( oK ) 0oC = 273oK; Room temp = 25o C = 298o K (T =~ 300)ln = NATURAL LOG

Δ Go = a CONSTANT: a quantity related to the INTRINSIC properties of A, B, C, and D

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Also abbreviated form:

Δ G = Δ Go+ RTlnQ (Q for “quotient”)Where Q = ([C][D]/[A][B])Qualitative term

Quantitative termJosiah Willard Gibbs(1839 - 1903)

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Δ

GoSTANDARD FREE ENERGY CHANGE of a reaction. If all the reactants and all the products are present at 1 unit concentration, then:Δ G = Δ Go + RTln(Q) = Δ Go + RTln([1][1] / [1][1])Δ

G = Δ Go + RTln(Q) = Δ Go + RTln(1)or Δ G = Δ Go +RT x 0,or Δ G = Δ Go,when all components are at 1….. a special case(when all components are at 1)“1” usually means 1 M

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So

Δ G and Δ Go are quite different,and not to be confused with each other.Δ Go allows us to compare all reactions under the same standard reaction conditions that we all agree to, independent of concentrations.So it allows a comparison of the stabilities of the bonds in the reactants vs. the products.It is useful.AND,It is easily measured.

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Because,

at equilibrium, Δ G = Δ Go + RTln(Q) = 0 and at equilibrium Q = Keq =(a second special case). So: at equilibrium, Δ G =

Δ Go + RTln(Keq) = 0And so: Δ Go = - RTln(Keq) So just measure the Keq,Plug in R and TGet: ΔGo, the standard free energy change[C]eq [D]eq[A]eq [B]eq

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E.g., let’s say for the reaction A + B C + D,

Keq happens to be:[C]eq[D]eq[A]eq[B]eqThen Δ Go = -RTlnKeq = -2 x 300 x ln(2.5 x 10-3) = -600 x -6 = +3600 3600

cal/mole (If we use R=2 we are dealing with calories)Or: 3.6 kcal/mole3.6 kcal/mole ABSORBED (positive number)So energy is required for the reaction in the left-to-right directionAnd indeed, very little product accumulates at equilibrium (Keq = 0.0025)= 2.5 x 10-3

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If

ΔGo = +3.6 for the reaction A + B < --- >C + DThen ΔGo = -3.6 for the reaction C + D <--- > A + B (Reverse the reaction: switch the sign)And:For reactions of more than simple 1 to 1 stoichiometries: aA + bB <--> cC + dD, ΔG = ΔG

o + RT ln [C]c[D]d                         [A]a[B]bNote:

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Some exceptions to the 1M standard condition:

1) Water: 55 M (pure water) is considered the “unit” concentration instead of 1MThe concentration of water rarely changes during the course of an aqueous reaction, since water is at such a high concentration. So when calulating Go, instead of writing in “55” when water participates in a reaction (e.g., a hydrolysis) we write “1.”This is not cheating; we are in charge of what is a “standard” condition, and we all agree to this: 55 M H20 is unit (“1”) concentration for the purpose of defining Go. Exception #1

:

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Exception #2:

In the same way, Hydrogen ion concentration, [H+]: 10-7 M is taken as unit concentration, by biochemists.since pH7 is maintained in most parts of the cell despite a reaction that may produce acid or base.This definition of the standard free energy change requires the designation ΔGo’

However, I will not bother. But it should be understood we are always talking about ΔGo’ in this course.

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Summary

ΔG = Go + RTln(Q)This combination of one qualitative and one quantitative (driving) term tell the direction of a chemical reaction in any particular circumstanceΔGo = - RTln(Keq)The ΔGo for any reaction is a constant that can be looked up in a book.