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Lecture Comments (3)

2 answers

Last reply by: Professor Hovasapian
Mon Feb 25, 2013 4:00 PM

Post by Nigel Hessing on February 25, 2013

Hmm, you didn't convert the gas constant into kilojoules first it should be 0.08315 as 8.315 is in joules.

Example Problems For Bioenergetics

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
  • Example 1: Calculate the ∆G°' For The Following Reaction 1:04
    • Example 1: Question
    • Example 1: Solution
  • Example 2: Calculate the Keq For the Following 4:20
    • Example 2: Question
    • Example 2: Solution
  • Example 3: Calculate the ∆G°' For The Hydrolysis of ATP At 25°C 8:52
    • Example 3: Question
    • Example 3: Solution
    • Example 3: Alternate Procedure
  • Example 4: Problems For Bioenergetics 16:46
    • Example 4: Questions
    • Example 4: Part A Solution
    • Example 4: Part B Solution
    • Example 4: Part C Solution
  • Example 5: Problems For Bioenergetics 29:27
    • Example 5: Questions
    • Example 5: Solution - Part 1
    • Example 5: Solution - Part 2

Transcription: Example Problems For Bioenergetics

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

We just finished our unit on bioenergetics, and today, we are going to start off with some example problems of bioenergetics.0004

I am actually going to be doing a couple of lessons on this; this is really, really, really important.0011

Bioenergetics is one of those topics that unfortunately, a lot of kids do not get a lot of practice in, and there is always going to be some sort of a hazy idea of what is it that is going on.0016

This idea of thermodynamics, this idea of ΔG, of spontaneity, of the coupling of an exergonic reaction with an endergonic reaction, the idea of oxidation-reduction, these are profoundly, profoundly important ideas.0027

And if you know what is happening both globally and in detail, so much of biochemistry and so much of everything in science is actually made so much easier.0044

I want to do a fair number of problems; some of them are going to be basic.0055

Some of them are going to be quite detailed, so let’s just jump right on in.0059

OK, example no. 1, let’s see.0064

Let’s start with black; how about that?0069

OK, so example 1, we would like to calculate the standard free energy change, the biochem’s standard free energy change for the following reaction.0074

We have aspartate + alpha-Ketoglutarate going to - actually let’s go ahead and leave it as an equilibrium here - an equilibrium with glutamate + oxaloacetate.0095

And we say that the Keq for this reaction is 0.147 at 25°C.0125

We would like you to calculate the free energy change for this given its Keq.0137

OK, well, we have a relation for this; it is really nice and straightforward.0142

The standard free energy change is equal to -R x T times the natural logarithm of the Keq.0146

We just put the numbers in- really, really great.0154

We have minus, so this is going to be 8.315, and I am going to write this out with its units so you see everything.0158

This is J/mol-K, let me make that a little bit more clear, and then we have the 298K; and then we have the natural logarithm of the Keq, which is 0.147, and then when we do this, we end up with, so notice the Kelvin cancels the Kelvin, we end up with -4,750J/mol, or if you prefer the kJ version, -4.75kJ/mol- there you go, that is it, really, really straight forward, definitely an exergonic reaction.0167

No problems there; OK, wait a minute.0217

No, I am sorry; this is not going to be negative because the logarithm of a number that is going to be less than 1 is going to be negative.0225

So, negative-negative makes it positive, so this is positive, which, of course, is confirmed by the fact that this is a really, really tiny Keq.0237

That means, it is not very favorable; that means, it favors not the products, but it actually favors the reactants.0247

This is +4.75kJ/mol, my apologies.0254

OK, let’s go to example no. 2.0260

We would like you to, this time, calculate the equilibrium constant for the following reaction.0266

It is just some basic stuff to get us going.0279

This time, we have L-malate, and let’s go ahead and write the reaction this way, in biochemical format.0284

NAD+ - oops, let me write this a little bit better - NAD+ comes in.0297

NADH + H+ goes out.0307

The enzyme that is catalyzing this is a dehydrogenase, right?0313

We just finished discussing how the NAD coenzymes work with dehydrogenases.0318

This is malate dehydrogenase, and it is going to convert this to oxaloacetate; and we are given a ΔG for this.0327

The free energy change is going to be 29.7kJ/mol, and it looks like this is positive, so we would like you to find the Keq for this.0338

Well, it is exactly the same as before, except now, you are just reversing the reaction.0347

You are finding the Keq instead of the ΔG.0351

Let’s go ahead and draw out some structures, just so we have a good sense of what is going on.0355

Again, this is biochemistry; we want to deal with structures as much as possible.0360

This much is just reasonable; we have C, C, C, C.0364

This is O-; we will put an H here.0369

This is L-malate; this is CH2, COO-.0373

There is that; there is this.0379

The NAD+ is there; the NADH plus the hydrogen ion.0383

And again, I will just write out the malate dehydrogenase, and what we get is our oxaloacetate, C, C, C, C, right?0390

What we have done, we have oxidized this; we have pulled off this hydrogen and this hydrogen.0408

We are left with that; we are left with COO- there, H2, COO- there.0412

Well, we know that the ΔG, our basic equation is -RT ln Keq.0418

We just rearrange this equation, and we get that the Keq is equal to E-ΔG / RT.0429

And then when we put the numbers in, we get E-29,700J/mol, right?0439

Now, notice, they gave the ΔG in kJ, but the R 8.315 is in J/mol-K, so we have to convert the ΔG to Joules.0457

That is why I have the 29,700 - again, that is going to be the biggest issue- is conversions - over 8.315.0469

It is going to be J/mol-K times...and it looks like this is going to be at 25°C, so we are there.0480

This cancels that- mole, Joule, Joule, mole; all the units cancel.0490

And what you end up is a Keq equal to 6.23 x 10-6; if I have done my arithmetic correct.0494

Again, arithmetic is ultimately...well, it is kind of irrelevant.0502

The idea is to understand what is going on, but 6.23 x 10-6- that is highly unfavorable.0506

That is confirmed by the ΔG, which was a +29.7kJ- highly unfavorable as written.0512

In other words, in this particular case, these are the things that are favored, not that- that is all this means.0520

OK, let’s see example no. 3.0528

Well, let’s go ahead and change colors here; I think I am going to go to blue for example no. 3.0535

Now, given the following information, given the following info, we would like you to calculate the standard free energy change for the hydrolysis of adenosine triphosphate, for the hydrolysis of ATP at 25°C.0543

If what we are given is the 2 following reactions, we are given glucose 6-phosphate + water, so its hydrolysis goes to glucose plus an inorganic phosphate, they tell us that the K1 of this one, the equilibrium constant, is equal to 270.0582

And they give us another reaction; they give us ATP + glucose goes to glucose 6-phosphate + ADP.0601

And they tell us that the equilibrium constant for this one equals 890.0614

What they want us to find using this information, is to find the free energy change for the hydrolysis of ATP.0619

OK, basically, what we are going to do is the following.0626

We are going to add the reactions, and then we are going to fiddle around, either with the Keqs or the ΔGs.0633

We are actually going to do this in 2 different ways; I am going to do it one way, then I am going to give you an alternate procedure.0642

I am going to add the reactions to get a net reaction.0648

Now, the equilibrium constant for the net reaction that I add is equal to K1 x K2.0655

Do you remember from chemistry, if you have 2 reactions with the particular equilibrium constant, if you add them, the equilibrium constant of the net reaction is the product of the equilibrium constants?0665

Let’s go ahead and add them; let's see.0679

Let’s go ahead and take...well, let me rewrite it.0683

Glucose 6-phosphate + H2O goes to glucose + inorganic phosphate, and then I have adenosine triphosphate + glucose goes to Glc, G6P, so glucose 6-phosphate + ADP.0689

Let me go ahead and add those; G6P cancels G6P.0713

Glucose cancels glucose, so what I am left with is my final reaction, which is the hydrolysis of ATP.0718

ATP + H2O goes to ADP + inorganic phosphate.0724

Now, the Keq...I will just write Knet.0732

I do not want to write eq; we know we are taking about an equilibrium constant.0739

So, the Knet is equal to K1 x K2, well, which is equal to 270 times the 890.0742

I just multiply the...right, I am going to get a Knet here.0751

It is going to be the product of the equilibrium constants of the individual reactions.0755

This is going to equal 240,300.0759

Now, I can go ahead and use my equation for ΔG to calculate the ΔG.0763

Now, my ΔG standard is equal to -RT ln Knet, which is equal to -8.315 x 298 times the logarithm of this number right here, 240,300.0769

When we run this calculation, we end up with -30.6kJ/mol.0794

You are going to get -30,600; I just converted it to kJ- that is it.0804

You have given 2 reactions; if you need to know something about another reaction that can come from the 2 reactions that you are given, the 2 or 3 or 4, when you add them up, I worked directly with the equilibrium constants because that was what was given to me.0810

I just multiplied the equilibrium constants and dealt with the equation.0825

Now, I will go ahead and do the alternate procedure, where I will take the individual reactions; I will use the Keqs to find the ΔGs, then I will just add the ΔGs.0829

Again, you are here; you want to go here.0837

You can go this way, or you can go this way; it does not really matter.0840

Let’s go ahead and do this; let’s do this one in red.0845

Again, this is just a question of personal taste; you know that is all it is- alternate procedure.0849

Some people’s minds work one way; some people’s minds work another.0855

As long as we end up in the same place and understand what is happening, that is what we are after.0858

We want you to understand what is happening, so alternate procedure.0862

Let me go ahead and calculate for reaction 1.0866

Our ΔG is equal to, well, -RT ln K.0874

So, it is -8.315 x 298 times the logarithm of the - let’s see, what have I got - logarithm of the 890.0879

OK, and I end up with -16.8kJ/mol.0897

That is one of the reactions; now, I will deal with the other reaction, reaction no. 20905

ΔG = -8.315 x 298 times the logarithm of 270, I believe, correct?0910

Yes, OK, 270, and for this one, I get -13.8kJ/mol.0918

Well, when I add those together, my ΔG net is equal to this, plus this, -30.6kJ/mol- that is it, very simple.0930

You can either work with the Keqs, and then use the ΔG equation, or you can use the Keqs that were given for each individual equation.0947

Get the ΔGs, and then add the ΔGs- that is all it is.0955

The only thing you have to watch out for is, again, with the Keq.0959

When you are adding one equation to another, you add their thermodynamic quantities, whether it is ΔG, ΔS, or ΔH.0962

You add the thermodynamic quantities to get the thermodynamic quantity of the net reaction.0968

When you are adding 2 equations together, the equilibrium constant for the net reaction is the product of the equilibrium constants for those 2 reactions- that is all you have to watch out for.0974

It is the thermodynamic quantities that are added, it is the Keqs that are multiplied.0987

OK, let’s see what we have got here.0991

OK, a little bit of a long problem here but a good one, nevertheless.0998

Let’s go ahead and stay with red, so example no. 4.1004

Now, given the following half reactions, we have crotonyl-coenzyme A + 2 electrons + 2 hydrogen ions goes to butyryl-coenzyme A, and we have a standard reduction potential of minus...let me write this a little bit better here.1014

The standard reduction potential on this is equal to -0.015V for that half reaction.1050

Now, we also have our NAD+ reaction + H+ + 2 electrons goes to NADH, and the standard reduction potential on this is -0.320.1059

Here are the questions that we would like you to deal with.1084

Now, let me make this M a little bit bigger, at 1M for each species and a pH equal to 7.0 - basically under standard conditions - we would like you to write the spontaneous reaction that will take place.1089

Write the spontaneous reaction that will occur.1116

That is going to be the first part, OK?1122

Now, part B, what we would like you to do is calculate what are the ΔG standard and the Keq for this reaction that you have written.1125

And part C, here is the interesting one.1146

Describe and demonstrate quantitatively what happens when we begin with the following concentrations, with a crotonyl-CoA concentration of 0.5, a but-CoA concentration of 1.0, an NAD+ concentration of 1.0 and an NADH concentration of 0.1.1151

We are going to switch the concentrations around; we would like you to describe what it is that is going to happen, and actually do it for us quantitatively, confirm what is going to happen quantitatively.1215

OK, well, let’s go ahead and take a look.1224

Part A, we notice that, again, we are looking at a reduction potential; so in a table of reduction potentials, they are all written as reductions.1229

You are going to take a look at the one that has the higher reduction potential.1237

The other reaction, that is the one that is going to be oxidized; you are going to switch that one.1241

In this case, the crotonyl-CoA reaction is -0.015, the NAD+ reduction is -0.320.1245

This is more negative, so this second reaction is the one that is actually going to switch.1257

This first reaction will stay as is; this will stay a reduction.1264

This will become an oxidation; that is what we are going to write.1269

We are going to write those 2, and we are just going to add them; and then we are going to add the reduction potentials to get a net reduction potential for that reaction.1272

Let’s write this out, so part A.1280

We have crotonyl-CoA + 2 electrons + 2 hydrogen ions goes to but-CoA.1284

And again, our standard reduction potential is -0.015.1298

And, of course, we switch the other one, so now it is going to be NADH.1305

Now, it is going to be on the left, and it is going to go to NAD+ + H+ + 2 electrons.1310

And, of course, when we switch the equation, we switch the sign, so it becomes now, +0.320V.1318

Now, we are going to add straight down; 2 electrons cancels 2 electrons.1327

One H cancels one of the Hs, and we are left with the following net reaction.1333

We have crotonyl-CoA plus the NADH; this is being reduced, right?1337

The crotonyl-CoA is being reduced to butyryl-CoA plus this thing, right there; and it is going to go but-CoA + NAD+ and its net.1347

I will write that, and I will write net; you see, just add those 2, and you get a +0.305.1365

This is the spontaneous reaction that is going to take place, if you put these together under the appropriate circumstances and with the right enzyme.1373

The crotonyl-CoA is going to be reduced by the NADH to but-CoA.1380

The NADH is going to be oxidized to NAD+- that is it.1385

It is always going to be like this; just take the one that has the higher reduction potential.1389

Leave it alone; take the one that has the lower reduction potential.1394

Switch it around, and then cancel the electrons; I mean, you might have to multiply by something to cancel electrons, but in general, it is always going to be 2 electrons 2, 4, 6- something like that.1397

Now, part B, well, part B is actually really, really easy; we wanted the ΔG, and we wanted the Keq.1407

Well, we have this, and we have a relation for that.1414

The standard free energy change is equal to -N x F times the E standard for the reaction.1419

We just put the numbers in; there are 2 electrons that are transferred in this reaction, so it is 2 there.1428

The Faraday constant is 96,485 C/mol of electrons transferred, and the E of the reaction is the 0.305.1433

When we do that, we end up with -58,856J or if you prefer -58.9kJ- that is it.1446

Here is your answer; that is our ΔG.1465

Well, we want the Keq; we want the K for the net reaction.1469

Well, that is easy; that is just the equation for ΔG rearranged.1473

It is going to be E to the -ΔG / RT.1478

When we put these values in E-58,856 divided by 8.315 - and it looks like the temperature is still 298, that is 25°C - we end up with a Keq.1485

The Knet is equal to a huge number: 2.06 x 1010- highly favorable reaction.1507

We know this already from the ΔG; the ΔG and the Keq, they are just alternate ways of describing where a reaction is, how far from equilibrium- that is it.1517

You have a coin; you have heads or tails.1529

This is just 2 sides of a coin; the relationship between the two is, it is just a constant RT- that is it.1531

Keq and ΔG, they tell us the same thing; they tell us how far from equilibrium something is.1539

In this particular case, it is really far from equilibrium; it wants to be over here, on the right-hand side- that is all it says.1544

OK, now, let’s see what we can do with part C.1552

Part C says, instead of starting with these 1M concentrations, the standard concentrations, we start with different concentrations.1558

So, we are going to have to use the Nernst equation and a reaction quotient, different numbers in the reaction quotient.1565

Let’s go ahead and do that; Part C, we are going to be using this.1572

E of the reaction is going to equal E standard - RT / nF x the ln of Q.1579

Now, Q here is going to be the following; It is going to be the concentration of the but-CoA, times the concentration of the NAD+, over the concentration of the crot-CoA, times the concentration of the NADH.1589

That is our reaction quotient; let’s go ahead and put the values in.1612

The E of the reaction is equal to...well, we found the E standard; sorry, not the chemical standard- the biochemical standard.1617

It is 0.305 - 8.315 x 298 divided by their 2 electrons that are transferred.1625

96,485 is the Faraday constant; the logarithm of...well, the but-CoA concentration was 1.1639

The NAD+ concentration was 1; the crot-CoA was going to be 0.5, and the NADH is going to be 0.51651

When we put that in, we get 0.305 - 0.0385 = 0.267- there we go.1663

Now, we had an initial, under standard conditions, the reduction potential for the net reaction was 0.0305.1682

It is going down a little bit; it is still spontaneous.1693

This is still positive; it is still a spontaneous reaction, but now, it is less spontaneous.1696

Under these conditions, there is less free energy available to do useful work.1702

There is still plenty of free energy available; I mean, 0.267 is a huge number relatively speaking, but there is less than there was under standard conditions.1707

Now, our qualitative response is the reaction is still spontaneous.1717

It will still go in the direction that we have written it but less so- that is it; that is all that is going on here.1724

OK, let’s go ahead and finish off with a very interesting kind of example.1737

Let’s see if we can, sort of, make sense of this.1744

Yes, Alright; I think I am going to do this one in red, and I think I am going to actually start on another page.1748

Let me go ahead and move to the next page here; OK, so example 5.1763

Let me see if I have enough, alright, I do, so example 5.1768

Now, consider the following reaction.1775

It is going to be ATP + H2O, the hydrolysis of the adenosine triphosphate going to ADP, + inorganic phosphate.1787

We have the ΔG standard for this, is equal to -30.5kJ/mol of ATP.1801

Now, ΔG is under biochem standards - right - of pH equal to 7.0 or a hydrogen ion concentration equal to 10-7.1810

You know, I have never liked the notion of pH; it just really, really bothers me.1837

It has always bothered me; we are dealing with concentrations.1840

Let’s just deal with concentrations, but pH is everywhere, so that is fine.1844

So, pH7, that just means that the hydrogen concentration is 10-7.1849

OK, now, here is our question.1853

If we drop the pH to 4.0 from 7.0, in other words, if we make this environment more acidic, pH to 4.0, our question is: will the reaction above become more exergonic or less exergonic.1859

In other words, if we drop the pH and make it more acidic, is there going to be more free energy available to do work, or is there going to be less free energy able to do work.1896

Is this going to get negative, more free energy; or is it going to get positive, less free energy- that is our question.1905

And, of course, the quantitative version: what will be - there is the qualitative and the quantitative - the new ΔG standard under conditions of pH4 instead of pH7- that is the real question.1913

OK, well, qualitatively, we can answer the first part as follows.1935

Let me go ahead and draw a little bit of a line here.1940

Qualitatively, we can answer the first part as follows.1944

Here is where we are getting into the details of what it is that is going on; often in biochemistry or in chemistry, there are going to be different things within a given reaction.1963

There are going to be different aspects that we are interested in; we do not want all of the details all the time.1970

It is just going to clutter things up; in this particular case, we want the details, and what I mean by details is we want to talk about what is going on where- where is each particle going, and what is the charge on each particle.1975

Well, the hydrolysis of ATP actually takes place like this.1985

ATP is carrying a 4- charge.1991

When it is hydrolyzed by a water molecule, it is going to form an ADP molecule, which is carrying a 3- charge plus an inorganic phosphate, which in general, is carrying a 2- charge; and it is going to have this.1996

It is going to release some hydrogen ion into the aqueous medium- this is what is really going on.2013

We did not write that up here; notice, we did not have this up there.2018

Now, dropping the pH means raising the hydrogen ion concentration2022

Well, if you raise the hydrogen ion concentration, you know by le Chatelier's principle that you are going to end up pushing the reaction that way.2028

That is what is going to happen here; qualitatively, you can answer this question.2035

It is actually going to become less exergonic; there is going to be less free energy available.2039

This -30.5 is going to get less negative; it is going to go up to maybe a -20, -15, -10, who knows- that is what happens.2045

It is good to know exactly what is going on in a particular reaction.2052

In general, we do not really concern ourselves with stuff like that, but it really is a great idea to understand the details.2056

So, yes, we want you to have a big picture, but there are certain things, certain reactions, that are so ubiquitous, we need you to know exactly what is happening- this is what is going on.2063

Dropping the pH will push the reaction that way.2075

OK, now, let’s do the quantitative; now, let’s do the math.2080

OK, let’s see what we can do here; I think I am going to start that on the next page.2085

Now, the biochem - so, let’s write this all out - standard for ΔG, it already accounts for a hydrogen ion concentration equal to 10...let me just write it here.2091

It already accounts for the fact that the hydrogen ion concentration is equal to 10-7M.2120

Now, what we have to do is...well, let me write it out.2130

Well, I will say it; what we have to do is recover not the biochem standard but the normal chemical standard that does not account for this, where we say 1M concentration of every species.2135

Well, that 1M concentration also includes the hydrogen ion.2151

A 1M concentration, hydrogen ion, -log of 1, you are going to get a pH of 0; but we cannot do pH of 0.2155

This is why the biochemical standard uses a pH of 7; that is why we have this thing right here, the standard plus that little mark that lets us know that we are at pH7.2165

However, in order to actually find what the new ΔG standard is at a different pH - pH4 in this case - we have to recover the original chemical standard and then work from there forward, and find a new biochemical standard.2174

That is what we are going to do; we are going to run 2 calculations.2190

What we have to do, and this is the kind of analysis that you want to think about.2193

You want to understand what it is that we mean when we say biochemical standard.2198

When you see some number like a reduction potential that says that we have accounted for a pH of 7, what does that mean?2203

It means that we have taken the chemical standard; we have switched the number around.2211

We have recalculated, and we have entered new numbers in this table.2215

Well, in order to find a new standard, we have to go back to the original that we came from and then work forward from there.2219

What we have to do is recover that ΔG standard, then calculate a new ΔG standard biochem at a pH equal to 4.2226

That is what we are doing; this pH7, this pH4, it is a new biochemical standard.2252

Let’s go ahead and actually recover our ΔG; well, here is the equation.2260

The ΔG biochemical standard is equal to the ΔG chemical standard + RT ln.2267

Everything is the same, ADP, PI; but at this time, the H is put in there.2277

Now, we are using the entire equation; see, this is where it comes from.2288

We take the chemical standard; we recalculate using this equation for finding the new ΔG by including the H.2293

When we get this number, that is the number we put into the tables put into biochemistry texts.2299

It is the ΔG chemical standard + RT ln with the actual hydrogen ion in the reaction quotient- there, over ATP.2304

OK, now, let’s, this is our variable.2318

We want to find this so that we can go backward from that; well, we want to know the ΔG standard.2322

It is the -30.5; our equation is -30,500 - remember, we are working in Joules - equals the chemical standard + 8.315 x 298 times the logarithm of...well, the concentration of ADP is 1.2329

The concentration of PI is 1; the hydrogen ion concentration at pH7 is 10-7.2351

The concentration of ATP is 1; when we do this and solve for ΔG, what we get is the following: ΔG chemical standard = 9,438J/mol.2360

Notice, this is positive; this is our chemical standard.2380

We have recovered it from the biochem standard; we have just worked this equation backwards.2384

Instead of looking for something on the left side of the equation, we are looking for this thing.2388

Now, we take this number, and we readjust using a pH of 4 or a hydrogen ion concentration of 10-4.2393

We go, ΔG, biochem standard for pH equal to 4, is going to equal 9,438 plus same thing, 8.315, the RT, + 8.315, times our 298, times the logarithm of, again, this time it is 1 for ADP.2407

It is 1 for PI, except now, it is going to be 10-4 / 1 for ATP.2444

OK, when we run this calculation, we end up with the following number.2454

ΔG standard for a pH equals 4 is equal to -13,384J/mol or -13.4kJ/mol.2461

Notice, this has confirmed the fact that it is less exergonic than the -30.5.2483

We knew what is going to happen qualitatively; now, we took care of it quantitatively.2490

This confirms it, and that is it.2496

Again, the biochem standard accounts for the pH equalling 7, the hydrogen ion concentration being 10-7.2499

I have to use that equation to actually recover the chemical standard, and from there, calculate a new biochemical standard by including the pH now, which is 4 or the hydrogen ion concentration being 10-4.2509

I hope that makes sense; thank you for joining us here at

We will see you next time for more problems on bioenergetics.2526

Take care, bye-bye.2531