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Lecture Comments (5)

1 answer

Last reply by: Professor Hovasapian
Mon Jun 27, 2016 7:44 PM

Post by Kaye Lim on June 14, 2016

Why enzyme Hexokinase is needed if the delG of the new coupled rxn is negative and spontaneous?

2 answers

Last reply by: Professor Hovasapian
Tue Nov 26, 2013 3:28 AM

Post by tiffany yang on November 13, 2013

Dear professor,

according to my study guide, delta G'= delta G' knot + 2.3 RT log Q
i think G'knot means:
  prime means: pH seven,  55M water,
  knot means one atm, 275 K       is this correct?

my confusion is that why is there a prime sign on the equation of my study guide; whereas yours doesn't. I feel that there shouldn't be a prime on the left side of the equation, because we 're looking for the delta G at that moment, at a new condition, rather than ph7, 55M water, one atm, I don't know why there's a prime sign on the left side. Does this mean that RT lnQ is being limited to ph=7,55M of water for the equation give by my teacher?   so delta G on the left side always reflect the most current condition in regard to pH, water concentration....etc.

second question is why can we find "delta" G'knot   by just using -RTlnK, I feel that since there's a delta, there should be a final - initial sort of thing. I mean how do we calculate delta if it's just at equilibrium, without knowing the condition it starts with.
Thank you so much. Your video cleared up so many of my confusion. You're incredible. I want to learn p chem from you next year too.

More on Thermodynamics & Free Energy

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • More on Thermodynamics & Free Energy 0:16
    • Calculating ∆G Under Standard Conditions
    • Calculating ∆G Under Physiological Conditions
    • ∆G < 0
    • ∆G = 0
    • Reaction Moving Forward Spontaneously
    • ∆G & The Maximum Theoretical Amount of Free Energy Available
    • Example Problem 1
    • Reactions That Have Species in Common
    • Example Problem 2: Part 1
    • Example Problem 2: Part 2- Enzyme Hexokinase & Coupling
    • Example Problem 2: Part 3
    • Recap

Transcription: More on Thermodynamics & Free Energy

Hello and welcome back to, and welcome back to Biochemistry.0000

In the last lesson, we started our discussion of thermodynamics; and today, we are going to finish off our discussion of thermodynamics, before we move on to another aspect of bioenergetics, which is going to be adenosine triphosphate.0003

OK, let's go ahead and continue on where we left off.0016

In the previous lesson, we ended with the following equation.0020

We said that the free energy change for a reaction is related to the equilibrium constant by this equation -RT, so ΔG = -RT ln Keq.0023

Now, this is under standard conditions.0040

Let me go ahead and put that there.0045

This is under standard conditions, and when I say standard conditions at this point, I am talking about the transformed biological standard because now, we are no longer dealing with chemistry standards.0047

We are dealing with biochemistry standards, so everything is pretty much the same except the pH is now 7, instead of 0.0061

Excuse me.0073

However, but most - well, not most, all, I will just write most - most physiological conditions, most physiological circumstances, do not have standard reactant concentrations.0076

Well, you know what, I am not going to write that; let me just say, let me just start again here, so we erase this.0112

OK, but under physiological conditions, under physio conditions, things are not standard.0126

In other words, your concentrations are not all the same - 1 molarity or 1 atmosphere.0144

You have different concentrations of different things, so you might have 3M adenosine triphosphate, and you might have 0.1M adenosine diphosphate.0151

When we talk about a standard, again, this is just so we have some point of reference and frame of reference by which to measure something.0162

But again, physiological conditions, concentrations are not going to be 1M, and they are certainly not going to be the same for all species that are involved.0170

OK, the actual free energy change, the actual free energy ΔG is calculated as follows.0178

We have a way of finding the actual free energy under a given set of physiological circumstances based on what we have calculated under standard conditions, deviation from that.0197

So, it is calculated as follows, and this is the important equation that we want you to know.0211

ΔG rxn, reaction, but you know what, I will just put ΔG equals the ΔG standard + RT ln Q.0220

Now, notice this is RT ln Q, not Keq because now, we are talking about a situation that is not at equilibrium.0234

The equation up here, this expresses a relationship between the equilibrium constant and the standard free energy change.0243

This says that these are just different ways of looking at the same thing, one is just a constant times the other.0250

What this equation tells us is that under a given set of circumstances, what is my temperature, what are my concentrations, given my base of a standard free energy change.0256

What is the free energy change under that circumstance; in other words, which way will the reaction actually go based on how things really are at that moment.0268

This is the equation that we want you to know; this is the one that is important.0277

Now, Q is exactly the same expression as Keq, except it has concentrations not at equilibrium; but it has concentrations at any given moment - now, 3 hours from now, 30 seconds from now - at the particular moment.0281

Q, the expression is still the same as the Keq, so it is going to be C raised to the C power, D raised to the D power, A raised to the A power and B raised to the B power, but it is at any given time, not at equilibrium.0298

The Keq is a measure of the concentrations at equilibrium.0307

It is actually characteristic for that reaction under a given set of circumstances, at a given temperature, but here, Q, at any given moment.0324

That is how we calculate the free energy for any given set of circumstances.0333

OK, now, when ΔG ends up being less than 0, this means the reaction is spontaneous.0340

In other words, it will proceed to the right as written until it reaches equilibrium.0356

That is what is interesting about this.0362

Remember what we said, the free energy change is a measure of the potential to want to reach equilibrium.0378

Once a reaction actually starts, as the reaction moves forward and moves towards equilibrium, the ΔG is actually going to rise, so -50, -40, -30, -20, -10, -5, 4, 3, 2, 1, 0.0383

When the system reaches equilibrium, now, the system is exactly where it wants to be.0402

It does not have a tendency to go this way or that way.0406

In other words, it has no more free energy to release, so now, the ΔG is 0.0410

So, as a reaction moves forward, ΔG changes and moves towards 0 until it reaches equilibrium.0415

Now, at equilibrium, the ΔG actually equals 0.0425

This is very, very important.0435

OK, now, at equilibrium, ΔG = 0; this means all of the free energy available to do work has been used up- that is it.0440

It is done; there is no more, once the system reaches equilibrium.0471

OK, this is the take-home lesson.0481

The measure of whether a given reaction under given conditions will move forward spontaneously as written is ΔG, not ΔG standard.0489

Under a given set of conditions, it is ΔG that we calculate from this equation right here, that tells us whether a certain reaction will move forward.0538

The ΔG standard is a point of reference; things deviate from that.0549

Sometimes, it will be more spontaneous than the standard conditions; sometimes it will be less spontaneous than the standard conditions.0554

It might switch altogether depending on what the conditions are; that is what is important here.0560

OK, now, let's rewrite the equation, ΔG = ΔG standard + RT ln Q.0565

Well, we said that at equilibrium, the free energy is equal to 0.0580

At equilibrium, that means this left side is equal to 0, that means ΔG + RT ln Q.0586

Well, when I move this over, what I end up getting is my original equation that I had from the last lesson, equals -RT, except now, because we are at equilibrium, instead of Q, we write Keq.0597

So, at equilibrium, you recover the relationship, and again, this relationship is a relationship between standard free energy change and the equilibrium constant; but at any given particular set of circumstances, it is ΔG that we are interested in.0613

That is what is going to tell us whether the reaction will move forward or will not move forward, and how much free energy we have at our disposal.0626

OK, now, ΔG is a measure of the maximum theoretical amount of free energy available.0635

The actual amount is much less.0675

ΔG is a measure of the maximum theoretical amount of free energy available.0679

So, let's say we have a particular reaction under a given set of conditions, and we calculated that the ΔG for this reaction is -50kJ/mol.0683

Well, that means that I have 50kJ/mol that I can do something with, whether it is build a macromolecule, or run some molecular motor or transport some solute across a membrane.0692

Whatever it is that the body needs to do it needs energy for, it has this 50kJ available to it, however, that is a theoretical number.0708

It is not going to use all of those 50kJ; the actual amount that ends up actually being used for useful work is actually a lot less than that, probably somewhere in the neighborhood of about 30 - 40% of that.0717

The rest of it is actually given off as heat.0727

OK, the actual amount is always less because some - actually most, I should say - free energy is always lost as heat entropy.0731

I have $50, but in order for me to actually go to a shop where I need to spend $50, I have to give $30 to the person at the door, so that leaves me with $20 to spend- that is what is going on.0772

I have to spend that $30 in order that I can do a little bit of shopping, if I want to do some shopping- that is all that is happening.0784

OK, let's do an example problem here; let's go ahead and do this in blue.0792

Oops, let's make sure we get it in blue here, so example problem no. 1.0797

OK, let's consider again, our glucose 1-phosphate reaction to glucose 6-phosphate and our ΔG standard for that, -7.3kJ/mol.0809

That is actually a little bit more than that; I must have gotten this number from a different...OK, so, -7.3kJ/mol.0828

Now, if the concentration of G1P equals 0.3M, and the concentration of G6P equals, let's say, 1.7M, OK, is this reaction spontaneous as written at 25°C.0835

OK, well, we have 0.3M; we have 1.7M, so these are not standard conditions.0875

The concentrations are different, but we have an equation that deals with this.0880

We have the ΔG under a given set of conditions equals the ΔG under standard conditions plus RT ln and Q.0885

Now, we just go ahead and put these in.0895

And, of course, Q is going to be the products, the G6P concentration over the G1P concentration, which is the reactants.0899

And again, we have to account for the units, so we have got, well, the ΔG standard is -7.3kJ/mol, so let's use -7300J/mol + 8.315 is our T is 298, and we have the logarithm of Q, which is going to be 1.7/0.3.0911

Concentration of G6-phosphate is 1.7; concentration of G1-phosphate is 0.3, and when we run this calculation, we end up with -7300 + 4298 for a total of -3002J/mol.0946

OK, now, the ΔG is -3002J/mol, which is -3.0kJ/mol.0978

Well, it is still spontaneous; it is negative, and it is -3kJ/mol.0991

But notice, under standard conditions, it is more than twice as spontaneous.0996

That is -7.3kJ/mol, so under these circumstances, I only have about 3000J of free energy available; and let's say we only end up using about 50% of that, so I am really going to only end up with about 1500J that is going to actually do useful work.1001

In this particular case, if the conditions were standard, I have 7.3kJ/mol, so I would have a lot more available.1021

I would have about 3.5, 3.6kJ actually available for real useful work.1029

Under these conditions, it is the ΔG that is going to tell us how much free energy we actually have available, if the reaction is still spontaneous.1037

Obviously, if these numbers change, we can actually change the spontaneity.1046

We can make it so that it actually does not want to move forward under standard conditions.1051

Perhaps, it wants to move backward depending on what these numbers are.1056

OK, now, let's see what we can do; now, let's leave it as blue.1062

Individual reactions that have species in common can be added, and common species can be cancelled on either side of the arrows to yield a net equation.1072

Do you remember back in general chemistry when you guys did Hess's law?1128

You wanted to find the enthalpy of a particular reaction.1132

Well, you knew that if you had the enthalpy of like 2 or 3 or 4 reactions, that when you added those reactions together, and manipulated them in such a way, so that when you actually added them straight down and species cancelled, if you ended up with your net reaction, what you can do is just add the enthalpies; and that will give you the enthalpy of the reaction that you want.1135

This is the same thing we are doing here; we can do the same for free energy, the same thing that we do for enthalpy.1154

We can add equations, cancel species, and the final equation that we get is perhaps the equation that we are actually interested in- that is what is going on here.1161

Now, the actual free energy of the net equation equals to the free energy of the first equation plus the free energy of the second equation, so the free energies are additive.1171

However, the equilibrium constant of the net equation is equal to the equilibrium constant times the second equilibrium constant.1185

So, the equilibrium constant of the first equation times the equilibrium constant of the second.1199

Keq is multiplicative; the free energy is additive- very, very important.1203

Let's do an example here.1210

OK, now, the first step of glycolysis, which we will be getting to very soon, which is the breakdown of glucose, is glucose plus an inorganic phosphate.1216

Well, I will just write it as a single arrow - goes to glucose 6-phosphate + H2O, and the ΔG for this is equal to +13.8kJ/mol.1241

This is positive; it is a positive ΔG.1259

What this tell me is that this reaction does not want to go this way under normal circumstances.1263

In fact, it wants to go the other way under normal circumstances because this is positive.1267

It is spontaneous in the reverse direction.1272

This reaction is not thermodynamically favorable; let's write it in red.1276

This reaction is not thermodynamically favorable, and if the reaction takes place all the time, so how does it proceed?1283

Very, very interesting, this is very, very profound- what we are about to do here.1313

Well, let's look at another reaction; I will do this in blue.1321

Let's look at adenosine triphosphate plus water; so, the hydrolysis of ATP to ADP plus inorganic phosphate.1331

Well, we have seen this one several times; the ΔG for this is -30.5kJ/mol- highly exergonic, definitely wants to move forward.1340

What I am going to do is let me see if I can put this reaction and this reaction together, and let me see what I come up with.1352

OK, I have reaction no. 1, which is Glc + PI goes to G6P + H2O, and we said that the ΔG for this equals - what did we say - 13.8.1361

I will not write the units; this is kJ/mol.1386

And our second reaction, we have ATP + H2O goes to ADP plus an inorganic phosphate, and the ΔG for this one, we said was -30.5 kJ/mol, so watch what happens.1388

H2O cancels H2O; PI cancels PI.1407

What I am left with is glucose + ATP goes to G6P, glucose 6-phosphate plus ADP.1412

We said that the ΔGs are additive, so 13. 8 - 30.5; I end up, and I am hoping you will confirm that I have done my arithmetic here.1423

So, the ΔG of the net reaction is equal to -16.7kJ/mol.1435

OK, look at what we have done.1448

We have taken glucose 6-phosphate; we want to do this reaction.1452

We want to take glucose to glucose 6-phosphate, but under normal circumstances, we cannot just put it in the vicinity of some inorganic phosphate and hope to God that it reacts.1455

It is not going to react; 13.8, it is not thermodynamically favorable.1463

However, if I couple it to a highly exergonic reaction like the hydrolysis of ATP, which is highly exergonic, by adding them together, I end up with a reaction that actually goes from glucose to glucose 6-phosphate.1467

It is thermodynamically favorable; now, the reaction has become viable.1487

What I have done is I have coupled a non-thermodynamically viable reaction with one that is very thermodynamically viable, and in the process, I have actually created a way for the process to happen that is thermodynamically favorable.1491

This is called coupling.1506

OK, now, let me do this in black.1509

Now, the conversion of Glc to G6P becomes viable.1516

The enzyme hexokinase, which catalyzes this particular reaction, that reaction, provided an alternate pathway by coupling - that is the important word here, you will see it a lot - an endergonic reaction, reaction no. 1 to an exergonic reaction, reaction no. 2, through common intermediates.1536

We are going to use the energy available here to compensate for the fact that there is not enough energy there.1593

OK, the energy available in or from - I should say - ATP hydrolysis is used to drive an otherwise endergonic reaction.1604

This is a huge theme in biochemistry; this is how the body does it.1641

It couples fundamentally endergonic reactions; again, doing things that the body needs to do takes a hell of a lot of energy.1646

So, these reactions do not just happen spontaneously; they need an energy source.1653

Adenosine triphosphate is your energy source; it couples endergonic reactions with the exergonic reaction of ATP to run these reactions forward.1658

That is what is going on, is to drive, otherwise, endergonic reaction but by a different pathway; and that is what is important, this idea of a different pathway.1667

OK, now, we have our first reaction: glucose + phosphate.1686

OK, and here is what is important to remember: when we couple this reaction, the reaction that is actually taking place in the enzyme is this reaction right here.1693

Glucose is going to be brought into contact with ATP; a phosphate group is going to be transferred to glucose.1702

It is going to turn into glucose 6-phosphate, and it is going to release the ADP.1708

So, when the enzyme opens up again, it is going to release the new product, and it is going to release the ADP.1712

Now, what we say, what you will often hear biochemists say, what you will hear your teacher say, what you will read in books, is that the reaction of glucose to glucose 6-phosphate is driven by the hydrolysis of ATP.1718

However, it is very, very important to remember that ATP is not actually hydrolyzed.1735

Water does not actually come in and break it up to produce a phosphate, and then this free phosphate reacts with this.1740

Reaction no. 1 and reaction no. 2 do not actually take place; neither one of these takes place.1745

What takes place is the coupled reaction; this one right here.1751

This is very, very important; I think it is very unfortunate.1755

Biochemistry is filled with description of things that, I think, leaves students astray.1760

So, when we say a reaction is driven by the hydrolysis of ATP, we are not saying that ATP actually undergoes hydrolysis.1765

What we are saying is that we have coupled that reaction to the endergonic reaction, some species have cancelled, and what you end up with is a net reaction.1773

It is that net reaction that a particular enzyme is catalyzing, but we use the terminology "driven by the hydrolysis of ATP", so neither one of these reactions is taking place.1781

ATP is not being hydrolyzed; please understand that.1791

It is very, very important; I am sorry that it has turned out like this, that people speak so loosely.1793

They should definitely speak a little bit more clearly, but please know that ATP is not being hydrolyzed.1801

An alternate pathway is being provided; it is being driven by the energy available from ATP hydrolysis, this 30.5kJ/mol- that is what is happening here.1808

Let me go ahead and write that down in red.1820

Now, neither - actually, you know what, let me do this on the next page here, OK, next page, OK - reaction 1, nor reaction 2 is actually taking place.1825

What takes place is glucose + ATP going to glucose 6-phosphate + ADP.1860

That is the reaction that is taking place in the hexokinase.1874

Let me go ahead and draw out what is actually happening here; let me draw out my glucose here.1879

You know what, let me make this a little bit lower; no, it actually does not matter.1888

That is OK, boom, boom, boom, boom, boom; let's go ahead and make this the beta form.1894

This OH; this is down.1902

This is up, and this is down.1904

Now, we have CH2OH; this is our no. 6 carbon.1905

Over here, this is our no. 6 carbon; we have adenosine triphosphate.1911

Let me go ahead and write this one over here.1916

We have O, P, O, P, O, P, O, ribose, and adenine, boom, minus boom, minus boom, minus; and this is minus.1920

So, what is happening is the following; these electrons going that way, kicking up that way, this is glucose.1941

This is adenosine triphosphate; what you end up with is glucose 6-phosphate, which looks...actually, do I have another page here?1952

Yes, I do, good; and what I have now, is this CH2.1961

I have O; I have P.1972

I have O; I have O, OH, OH, OH, OH.1975

What I have is glucose 6-phosphate + adenosine diphosphate, which is going to be O, P, O, P, O, ribose and adenine, just so you can actually see what is happening here.1985

This is ADP; this is the reaction that is taking place in the enzyme.2007

The phosphorylation of glucose to glucose 6-phosphate is driven by the hydrolysis of HTP.2013

It is driven by the energy that comes from the hydrolysis of HTP, but the reactions are coupled, species cancel, so the net reaction that takes place is this one right here.2020

The glucose reacts directly with ATP to form glucose 6-phosphate and ADP.2031

You will see it written like this.2038

Glc, glucose 6-phosphate, ATP comes in; ADP goes out.2042

This is the biochemical shorthand for it; this is how you will see it, and hexokinase.2051

They will usually write the enzyme that catalyzes it there and anything else.2058

In this particular case, there is going to be magnesium involved, magnesium ion; but we will get to that when we talk about glycolysis.2062

But, very, very important to remember: do not mistake what they say with what is going on.2068

It is the net reaction that is taking place; that is the reaction that is being catalyzed by the enzyme.2072

OK, now, let's go ahead and finish up here.2079

OK, let me see, alright.2083

Let's go ahead and do a quick recap before we leave our thermodynamics and spend more time talking about this idea of reaction coupling.2088

Quick recap: we said that the ΔG standard, there is a relationship between that and the equilibrium constant for a given reaction.2102

That relationship is ΔG is equal to -RT ln Keq.2112

So, ΔG and Keq are different ways of representing the potential for a reaction to want to go to equilibrium as written under standard conditions.2120

OK, we had ΔG = ΔH -T ΔS.2130

It says that the free energy...yes, I will go ahead and just write it for this one.2140

That is fine; that is fine.2146

This says that the free energy is also related to two other thermodynamic properties, which is the enthalpy, the heated reaction, and the entropy, the degree of disorder of the reaction.2149

OK, this, this, if you want, you can set this equal to this, in case you need to derive a relationship between this and this or this and this, so it is good.2159

And, of course, the final one and the one that is probably most important, well, this one is definitely, profoundly important, so the ΔG of a given reaction under a given set of conditions that could be any temperature, any concentrations for the reactants and products, that is equal to the ΔG standard + RT ln Q, where Q is the reaction quotient, equals C, D, over A, and over B.2169

You know what, that is fine; I will go ahead and put these D and A, and this is at any given moment.2207

Thermodynamics- this is pretty much what we are concerned with.2218

OK, thank you for joining us here at Educator.com2223

We will see you next time, bye-bye.2226