 Hello. Hello. I think we'll get started. So today's material unfortunately will be on the exam. So it's entirely possible. We're not going to get through this whole lecture. So wherever we get done, actually if you ask us questions, they'll be left on this test and more on the harder test. So the first test is a little easier. So factor that in as you prioritize questions. So wherever we stop, that's what you need to know for the exam. And if this wants to work, this being sporadic. There we go. Okay. So we have a lot of different various dates to keep in mind. The TA-led review sessions coming up on Sunday. Sarah is orchestrating that event. That's at 8 p.m. in the Science Center in the main gathering area. I have office hours today for two hours right after class at Sydney Frank Hall. TAs are available by appointment. Your sapling problem set, third one is due on Monday. So this is another way to get ready for the exam. In general, if I could have your attention. Hello. The other students are getting frustrated. The difficulty of the sapling problem sets is it's more difficult than what you're going to see on the exam. So if you look at the type of questions that are getting on sapling, look at some of those old exams and say it's sort of a little bit unbalanced. So we're not going to ask as difficult questions as the saplings. So in the midterm, we need to split by last name. This is just not a way for us to take an exam. We're all on top of each other. There's nowhere for your coffee and your get yourself nice and calm and a zen moment for the exam. So we're going to spread us out. So we almost have every other seat seating for the exam. And I need your help. You need to go to the right place. If you go to the wrong place, we might not be able to grade your exam. So if you turn your exam in, we'll know what your last name is. Just so you know. So I'll have a nice thing on the front of the board. And to remember, last names A through M are in Macmillan. Last names to intercede to bird. Let's settle down, please. Okay. So, and that's Ernie and Burtville. That's the thing with the glass greenhouse on top. All right. So we're going to dive right in. And so last time we left off with a few slides remaining, we're thinking about the standard change in free energy for a variety of reactions. We talked about a lot of different hydrolysis reactions. Today we're going to be thinking, there's other things you can do with molecules like rearrangements. We'll see some examples today of rearrangements. Generally, they have a smaller change, standard changes for energy. But what we're going to be doing today is the partial oxidation of glucose. And so the full oxidation of glucose releases tremendous amounts of energy. Every two electrons removed from a carbon atom of glucose, or every two electrons removed, it's about 200 kilojoules per mole. And when you think about, you know, the magnitude of that, ATP hydrolysis is only 60. So that's a tremendous amount of energy. And today we're in a step-wise process. We're going to be removing just two electrons from glucose. And that alone is going to drive glycolysis. And so we can have what's referred to as substrate level phosphorylation. So in other words, ATP hydrolysis is less exergonic than hydrolysis of other molecules. And we'll be learning today some reasons why these other molecules, when you hydrolyze the phosphate off, why they're more exergonic. So I think there's perhaps an example with phosphocreatine. So look at this. So when you think about it, hydrolysis of phosphate out of ATP gives a rise in resonance stabilization of just the inorganic phosphate and not the ADP. So you still have this sort of phosphoester, a phosphoenhydride bond in the phosphate. It's not really optimal resonance there, but the phosphate, we refer to it as inorganic phosphate. That has beautiful resonance. So with ATP hydrolysis, it's just one molecule, one of the two products increases in resonance. Whereas with hydrolysis of these other molecules, in general, both products of the hydrolysis reaction increase in resonance. So, for example, creatine can have a resonance form in addition to the resonance with inorganic phosphate. And so this gives rise to a larger negative change in standard free energy. And because this is so exergonic, you can literally have enough energy to do the inorganic process of putting a phosphate on an ADP molecule. And we'll see a few examples of this substrate level phosphorylation today, later on. And so we have this idea of coupling reactions. So here we have one reaction where we can attack the carboxylacryl with this ammonia to form glutamine. So this is one way you could do this reaction. An alternative pathway would be to first attach a phosphate. And so in terms of the organic chemistry, what is attaching the phosphate accomplish for you? Yes. Absolutely. So increasing the nucleo-fugacity of O-minus is not a very good leaving group. Phosphate is a great leaving group. So this is driven for this process is exergonic because you have highly exergonic hydrolysis of ATP. Of course, putting a phosphate on a carboxylate is endergonic. These are coupled together. And then that process is exergonic. And of course, nucleophilic attack of ammonia on the carboxylate with release of phosphate, that's also exergonic, right? Because you have great resonance of the product. And so these are two different pathways. And we'll see throughout today's lecture, for example, where we're coupling reactions together to put difficult chemical transformations forward by coupling them to exergonic chemical transformations. Okay, so today's lecture has lots of topics. It's important that you remember the oxidation state of carbon atoms. So I actually had a question for you. So after the last lecture, I was informed that they actually dropped sugar organic chemistry from Orgo. So what, did you guys get that anywhere else? Good God. Well, so remember that I gave you a lot of hints. And I'm sorry I didn't know that because in the past they did have sugar chemistry. Look for the carbon with two oxygens bound. I mean, that's trick and the trick for specifying alpha and beta. Those are really important things. But yeah, definitely review that lecture. I'm in office hours. If you're still confused, it's totally understandable. You realize it's the first time I was throwing this at you. So today also will probably be a lot of similar things. Hi, how are you? Should we know some of the arbitrary names? Well, so technically, legally you're responsible for knowing everything on every slide. But I surely know probably maybe 10% of what's on every slide. And I don't have to take the exam, fortunately. But I would say that, like, am I going to ask you which sugars are in sucrose? Probably not. Am I going to ask you to draw a structure like draw galactose? Probably not. Things that I would not tend to remember. If you see me, if you go through the video and you see me talking really fast and getting excited, I'm more likely to think of those topics when I'm writing an exam. But I haven't finished writing the exam, so I don't know what to do. So these are good questions. Let's move on. We have a lot to cover. So we've looked at different types of enzymes so far. I'm going to introduce a new class today. And these are this class of oxido reductases. These enzymes are going to help to catalyze the transfer of electrons between two reactions. And so these also are commonly called dehydrogenases. So if you see the word dehydrogenase in the name of an enzyme, it's called oxido reductases. And so we need to think about this transfer. So the chemical process here is moving electrons from one molecule to another. And so you have, obviously, if you're moving one to another, one of the molecules is donating the electrons, the other is receiving the electrons. And so we can call these reducing agents and oxidizing agents. But you never have, like, removal of electrons from a molecule and then you just sort of toss them in solution because that wouldn't work. You always need to transfer those electrons to some other molecule. Because electrons are not stable. Okay. And so the way we can think of this, we need to eventually think about, okay, what's the standard change in free energy of these types of reactions? Because that's the type of energy we're going to harness in burning fuels. And so to begin to think about that, we need to think about the affinity of certain molecules for electrons. And so this is a table. This table is derived from an experiment I'll show you in a moment. In this map, how high of an affinity each of these molecules has for an electron. So for example, oxygen has the highest affinity for electrons on this table. And ferrodoxin has the lowest affinity for electrons. And so as this number increases, there's a higher affinity. We call them stronger oxidants, stronger reductants. But in my mind, it's a lot easier to think about electron affinity. It wants an electron, or it doesn't want an electron as much. Okay. And so each of these are half reactions. Inzons are not just catalyzing, you know, electrons coming from nowhere and being deposited on a molecule. They're transferring electrons between molecules. You always have one molecule being reduced and the other molecule being oxidized. Right. And so, moving along or not. There we go. Did I skip one? Maybe. Right. And so these are standard changes in free energy. And so these are from our same standard state, right? One molar concentration of everything except for water and protons, which are 10 to the minus 7 molar. Okay. But these are relative to a standard state and that you might think, okay, well, that's nice, but the cellular state is not likely to be that. And so these, electron affinity can be modulated by the concentration of the acceptor and the donator of electrons. And so here's the experimental setup to calculate these voltages. So what was on that table was simply putting certain substances in this setup and looking at the voltmeter to see the voltage red. So this voltmeter doesn't know how many electrons are being released per molecule. It just measures the flow of electrons. And the way we do this is that we have a reference standard. And that is a hydrogen gas bubble that this partial pressure gives one molar concentration of protons. Right. And so this is our reference state. So for example, if we put the same hydrogen gas on this side, the same partial pressure, the voltage measured would be zero. Right. So that's set arbitrarily at zero. We can calibrate our voltmeter and that reaction is set at zero. But now we can put other substances in their standard state, one molar concentration over here and we can measure the flow of electrons. So if what's over here has a higher affinity for electrons than hydrogen, then you'll get a positive voltage. And if they have a lower affinity, electrons will go in the opposite direction. You'll have a negative voltage. Now if you just had volt, the electrons moving in a one-way process, that would tend to accumulate electrons. That would not be bad. So you have this bridge in between to allow you to have a current through this system so you can flow the electrons. So this is the type of experimental setup that gave the values in that table. And so you always have an oxidized form of the molecule, plus some electrons giving the reduced form. And these reduction potentials as I just described are relative to this particular reaction of hydrogen. So that's set at a voltage of zero. And so we always need to transfer electrons between substances. And so it's not enough to just say, to characterize the reaction, to just say what is the voltage measured in this setup when I release electrons? Because in a real biochemical reaction those electrons will be received. And the affinity of the receiver will affect the process. And so what's important here is a change in free energy, or a change in this reduction potential. And this is defined as the reduction potential of the molecule being reduced minus the reduction potential of the molecule being oxidized. And so if you look back at that table, it's always something plus electrons give something. So if you're being reduced, you're gaining electrons. But if you're transferring molecules, one molecule is going to gain electrons, one's going to lose. You have to subtract the one that's being oxidized. And I'll give you an example. It's a bit abstract, hard to conceive of. Let's look at a particular example. Let's look at the transfer of electrons from ethanol to iron. So iron is being reduced, ethanol is being oxidized. And so we have in this table a reduction potential for each of these reactions. So how can we describe the overall process? The standard change in reduction potential is the molecule being reduced, iron, right? 0.771 minus the reduction potential of the molecule being oxidized here. So you see how this is actually over here, reversed, ethanol is on this side, acid albohide is here. Now you have to subtract that reduction potential. So for the overall process, the change in reduction potential is nearly a volt. And so this is the type of chemistry that's going to occur when we do glycolysis. So we're going to be removing electrons from carbon-containing molecules. And so this is a lot of reduction potential change, but you obviously want to think about, okay, that's nice, but spontaneity of a reaction is determined by delta G. This just says the flow of electrons. This gives you a general sense for the relative affinity for electrons of these two half reactions. But we need to understand how we can think about the standard change in free energy. We have to be able to be really solid on this slide. So this is a very important slide. So you need to be able to look at a molecule. And this is sort of like the importance of finding the Hemiacetal Hemiaketal. This is really core to this lecture. So you need to be able to look at each of these carbon atoms and determine their oxidation state. And so the way this works is that the most electronegative atom owns the bonding electrons. And as it turns out, although there isn't much difference between the electronegativity of carbon and hydrogen, carbon is slightly more electronegative. So it owns those bonding electrons. So for example, methane, hydrogen doesn't get any of these electrons. So that carbon's oxidation state is 8. Here you have 2, 4, 6. Now here we have the same atom, so they equally share. They have the same electronegativity. And so that's an oxidation state of 7. But we can also think of, so here we have the numbers of the oxidation state of the carbon that's highlighted in pinkish-reddish sort of color. But you can also think of, you know, throughout the sugars, each carbon, and what each carbon's oxidation state is. So let's just focus on this one on the end here. So here we have oxidation state of 7. Here do you see it going down by 2 electrons? Do you see that? So you get 1, 2, 3, 4, 5. So you've gone from 7 to 5 because now you, instead of having a bond to hydrogen, you have a bond to oxygen. Oxygen is more electronegative. It hogs those electrons. And so this is a 2-electron oxidation to the alcohol. And so you might think of fatty acids. Those are very reduced forms of carbon. And these in sugars, you know, you begin to have a lot of hydroxyl groups. And so those are more oxidized forms of carbon. But you can keep going with this. You can now form a double bond. And now you just get 1, 2, 3 electrons for that carbon. And here you just have one electron. So going from the aldehyde to the carboxylic acid, you only have one electron. So these are two electron hops for each of these processes. Does that make sense so far? And I find it useful. As you look at molecules in the lecture slides, look at all the carbons and try to think about the oxidation state of each one. Get some practice at this. Because it's important to map where the electrons are going. The energy we're harnessing is coming from the oxidation of these fuels. And that oxidation is the movement of electrons. That's literally providing the chemical energy here. And so the standard change in free energy, the standard reduction potential change, is insufficient to fully explain a particular reaction, because that's the standard state. And you might guess biochemistry rarely occurs in a standard state. It's useful because you can have a table and it can be predetermined and you can look up these things. But what's really important is the cellular state. So remember we had a similar conversation where we were talking about the change in Gibbs free energy. You had the standard state, delta G, and you had delta G. Here you have E and E prime naught. And so if you, for example, increase the concentration of an electron acceptor, you're increasing the affinity for electrons. So by modulating these concentrations, you're affecting the reduction potential. And so this happens in a cell. For example, NAD concentrations are not one-to-one with NADH. They're manipulated. And so NAD tends to be higher concentration than NADH. And this makes this electron potential higher, more positive. And as we'll see in the next slide, it translates into a more exergonic process. So we can play the same tricks we did with Gibbs free energy with these reduction potentials by manipulating the concentrations of the acceptor and the donor. So here's another critical slide. This is a very important equation. So here we're going from an experimental piece of data, a reduction potential from this experimental setup, to a prediction of spontaneity of the reaction. Whether at equilibrium, a reaction will tend to favor the products or the reactants. And so delta G is sort of inversely related to the change in this reduction potential. So as the reduction potential becomes more positive, delta G becomes more negative. And therefore, the process is more exergonic. At equilibrium, you will have the accumulation of products. And this, if you actually plug the numbers in here, look at this. This is freaking amazing. One volt. Plug that in here times 96. Here we have the number of electrons transferred. So before we didn't have that, because our experiment couldn't tell where did that electron come from. It was just a volt meter. But here we need to factor that in. So we have nearly 200 kilojoules per mole. Well, it's just a simple two-electron oxidation of one carbon atom. So if you think about the glucose molecule, each on average, each of those carbon atoms are an oxidation state of four. The ends sort of average out if you map that out. So we have four electrons times six carbon atoms. Every two electrons coming out generates a flash of energy. There's a tremendous amount of energy released here. So positive standard changes in reduction potential equate to negative changes in free energy. So this is the process that's going to drive the creation of, or the harnessing of energy. And so we have a variety of things that are important in this process. And so remember, we learned ATP hydrolysis. Gibbs changes free energy from standard C around 30 from the cellular state generally around 50 or 60. That's not enough. These electrons are coming off two at a time. That's 200 kilojoules per mole. So we can't catch that energy just in forming, for example, a phospho-anhydride bond to ADP with a phosphate. We need something else to catch this energy. If we don't catch it, it's just going to be given off as heat. That'll be wasted. We won't be able to use it for biochemical processes. So we have a variety of electron catchers. And isn't it surprising? Look at the placement, where nature has selected these catchers from. Remember, this is the scale of electron affinity. So the things that are, you might guess, well, if you want to catch something, you've got to have a good grip. You've got to grab it really tightly. But these have very low electron affinity. But when you release the electron from these, you can release a tremendous amount of energy. But it might be a little counterintuitive. So let's look at the structures of some of these pyridine nucleotide coenzymes. So here, there's NAD, nicotinamide adenine dinucleotide. We can have a variant of NAD with a phosphate here. But the business end of this molecule is this pyridine ring. This nicotinamide ring. And so this molecule, when you put two electrons into this molecule, you go to this form of that nicotinamide ring. Do you see how you're destroying aromaticity? Do you see why it makes sense you probably would have a low electron affinity? Because this thing's like, I really sort of like being aromatic. You're making me be not aromatic. But this is stable enough. There's an activation barrier to this energy that it's still sufficient to catch those electrons. But you can imagine, if I remove these electrons from NAD, that's going to release a lot of energy because you're going to go from non-aromatic back to aromatic. So do you see the ADP molecule here? So you have ADP attached to another ribose here. So here we have a five-position linkage of these two phosphate. But then we have this ring. And so NAD cannot accept one electron at a time. It can only accept two at a time. What's with me so far? And it's useful to have two different forms of this because they can be used separately. So if you have two pools, they can be at different concentrations. So in general, NAD is used in the production of energy, especially in aerobic respiration. It generally transports its electrons into something called an electron transport train, and that generates a large amount of ATP. NADP, on the other hand, is used for a variety of processes that build molecules. So we call these anabolic processes. And so if we have separate pools, we can have separate concentrations and we can have a bit of independence in those. So the other electron catcher is FAD or FAD, and so this is a one-electron catcher. And so again, you have ADP. So if you know the structure of ADP, by the end of the class, you're going to have to know ADP because we're going to go all into the pathways that make it. And so here again, you have these two phosphates, but you're attached to an isoloxazene ring. And this is aromatic, but then you can, again, disrupt the aromaticity by gently one electron at a time, putting those electrons in the molecule. And so this has generally a somewhat low electron affinity because you're going from aromatic to not aromatic. And some biochemical processes, it's useful to be able to transfer one electron at a time. For glycolysis, we're not going to see it, but for other processes, you'll see it throughout the semester. Any questions on this so far? I'm sorry, I'm going so fast, yes. Only two. NAD is two or nothing. FAD can go to two. It can accept either one or two. That's right. It's more flexible. And one other question from the audience. What is a positive delta E imply negative delta G? It's just because the equation delta G equals negative NFE delta E. That's a good question. Okay, so looking at NAD and NADP, they have a useful biochemical property. So if you want to measure the rate of a oxidation reduction chemistry occurring with an enzyme, you can measure the appearance or disappearance of the reduced form of NADH. It has a unique absorbance at 340 that you don't see when you're lacking those two electrons. And so this is useful if you want to do some kinetics, for example, on enzymes that are pushing electrons away. So this is sort of a side. Okay, so there's lots of different ways that this redox chemistry is indicated on the slides throughout the semester. So you can directly transfer electrons between atoms, especially metal ions. So here we have one being reduced, the other being oxidized. You can transfer electrons that can hit a ride on a hydrogen atom or a hydride ion. So for example, for NAD, you can think of it as a transfer of a hydride ion to NAD. And you can also change the oxidation state of a molecule by forming a bond to oxygen. But in the case of these carbon oxidations we're going to be seeing today, those are two electron oxidations so these other ones don't necessarily have to be two electron oxidations reductions. And so this is the general process that we're looking at, some nomenclature. Here's some more nomenclature that you might see throughout the semester. You can think of this as transfer of protons and electrons, hydrogen atoms or hydride ions. These are just different notations. When you see these types of notations you typically see that these are dehydrogenase enzymes. So this is just to make you familiar with notations. So in general, there's two separate processes. You can either destroy things or do construction. So generally, today we're going to be looking at destruction. The disassembly of a sugar molecule into a simpler form. And these catabolic, so your metabolism is changed throwing the Latin origin. So catabolism, this is the destructive processes. You take fuels and you disassemble them. And anabolism are the construction, the constructive processes. You're building from simple precursors more complex molecules. And so the next three lectures, it's all going to be about destruction, destructive lectures. But then for the rest of the semester we'll be looking a lot at anabolism, the building up of molecules. And so that if you look at, coming back, what we're doing is we're transferring molecules. As we're destroying molecules, we're gaining energy. And that energy is being pumped into constructive processes of building new molecules. So this energy is transferred by hitching a ride on ATP or some of these reduced cofactors. And then anabolic processes will use this energy. For example, harness the energy by hydrolyzing ATP or oxidizing an ADH. And then these molecules will be ready again for catabolism. So these smaller molecules are shuttling energy around these two processes. And so catabolic processes, you can consume all kinds of fuels. You can destroy lots of stuff. And so here you can disassemble fat, sugars, or polymers of sugars. Even amino acids can be disassembled. But all of these catabolic pathways converge onto a very limited number of intermediates. So acetyl-CoA is one great example. Now we can burn these things and make ATP molecules. Or we can build up them into more complex molecules. So do you see how the arrows are all pointing to one molecule here and then the anabolic processes? The diversity is increasing. So that's a general theme you'll see of anabolic and catabolic pathways. Okay, so today we're doing glycolysis. And this is sort of getting ready for the rampant burning of fuel. So today we're just going to extract a very small amount of energy out of a glucose molecule. But then in the next lecture, we're going to be harnessing a tremendous amount of energy from the remaining carbon, the reduced carbons in the glucose molecule. And so, and then in the third lecture, we're going to be looking at how do we make ATP molecules through electron transport? How do we get the energy out of those reduced cofactors? So today we're doing glycolysis. So Latin is sugar splitting. So it's important that we be able to see the point here is to break this glucose molecule into two, three carbon sugars, two trioses. And because the next step, after you make the trios, you can decarboxylate and then you have a two-carbon molecule and that's what you're going to really work on and through the Krebs cycle, which is coming up later. Okay, so let's look at glycolysis. We're going to go through this. This is the most important pathway in terms of evolution out there. Every organism has glycolysis. It's in all of yourselves. It's central core skeleton of all the metabolism, glycolysis and then the Krebs cycle. So we're going to go really into the details. Later in the semester when we're talking about other pathways, we're not going to have this type of depth. So let's look at this. Let's think about the oxidation state of all the carbons and glucose. Do you see I've written them here? Do you see how it makes sense? So here you have two hydrogens. So you have four or five oxidation states. Here you have a double bond to an oxygen. You have three. But do you see how the average of these numbers are four? So you have four electrons on average from each carbon atom that can be removed. Every two electrons you remove releases about 200 kilojoules per mole. But if you break this down to the simplest part, you have for every carbon atom, you're getting four electrons out of it. And today glycolysis, so what is that? Six times four, 12, 24. So in all, you can remove 24 electrons from glucose, but today we're only going to be removing two. So we're going to be splitting the sugar and only removing two electrons from the whole molecule. And so that never works. That's fuel, fire. So what we don't want to do here is if you take a log and you set it on fire, it's the exact same chemistry. And it's good for roasting marshmallows making s'mores. Not good for driving biochemical processes. It's hard to catch that heat. You'll burn yourself. So today instead of rampant uncontrolled oxidation of glucose, we're going to disassemble this molecule very, very slowly and carefully, catching that energy on these electron carriers. Right, so yay, burn. Or not. Okay, so oxygen, but the end point here is if you fully oxidize the glucose molecule, well, you're going to transfer those electrons ultimately onto the molecule with the highest electron affinity, the oxygen molecule. Okay, and so we're transferring these electrons in the end game from glucose to oxygen to form carbon dioxide. And so on average, you're releasing, you know, approaching nearly 500 kilojoules per mole per carbon atom. Right, and so we have six carbon atoms. But we can't catch those on a 50 kilojoule per mole carrier. You know, you just, there's no enzymes that can phosphorylate like 5 ADPs simultaneously. That doesn't exist. So we're going to use these reduced cofactors to catch a large amount of the energy. And we're going to be very, it's amazing how evolution has figured this out, very strategic in the way that we manipulate the molecules. So we're going to leave, you know, we might take an ATP off, but we're going to leave a lot of energy in the molecule and that will create some very unstable intermediates. And that way, in two different steps, we can slowly take energy out. You'll see that in a moment. It's amazing how strategic this is. So this is the preparatory phase of glycolysis. So here you have the, it seems like you're kind of productive. So the purpose here is to burn fuels, make energy, catch that energy. But here, you're hydrolyzing two ATP molecules. Does that make any bit of sense? You're actually losing in the game. But you have to invest to make a dividend, right? So if you don't invest any energy, you're not going to get these molecules to the stage where you can extract energy. They're not ready to extract energy as they come in as glucose molecules. So we're going to phosphorylate various positions on the sugar and do some rearrangements. And then we have an enzyme called aldolase, which is the sugar lysing step. And it's very important that you see which bonds are breaking. The whole point of this pathway is to break this hexose into two trioses. And so we're going to break this molecule into two, shown here in this schematic, into two trioses. So that's the preparatory phase. In terms of catching energy, not productive so far. We've actually invested two ATPs. But then in the payoff phase, the show me the money phase, we're going to not only regain those two ATPs we initially invested, but we're going to yield a dividend of two ATPs. So every glucose molecule that goes into glycolysis, you have the net production of two ATPs. And you have the reduction of the NAD oxidized cofactor to its reduced form. So some of our energy is ending up in ATPs, some is ending up in NADH. So this is overall schematic. Okay, one at a time, nice and simple. Ease into it. So can you imagine that it makes sense to put a phosphate, to make a new chemical bond of phosphate on a glucose, that's going to be endergonic, right? How happy? I mean, so you're putting this negative charge on the glucose molecule. That's endergonic. But the enzyme is coupling two reactions, ATP hydrolysis and phosphorylation of glucose. And this ATP hydrolysis is more exergonic than glucose phosphorylation is endergonic. Do you see that? So when you couple these together, you have a pretty highly exergonic overall process starting for the standard state. Okay, and so ATP hydrolysis is driving this reaction. The enzyme is called hexokinase. Hexokinase is a C6 non-specific kinase. So you'll see examples of other sugars that can be phosphorylated at the C6 position. Remember, you got to go back to that sugar lecture. You got to get the numbering of the sugar molecules down to be able to understand glycolysis. So this is C6 phosphorylated, and hexokinase is the enzyme that does that. So we've now phosphorylated glucose. And this has some strategic benefits through evolution to say, you know, those blue transporters, they sort of, they're nice, they get the glucose in, but they're in equilibrium. So if you build up the concentration of glucose in the cell, if it gets too high, you'll start exporting the glucose back out of the cell. It'll see, hey, glucose transporter, cool, let's go back out. But if you phosphorylate it, when the phosphorylated sugar molecule comes to the glucose transporter and it's like, I don't know what that would go away. I'm not giving you a ride. You're phosphorylate. So by phosphorylating the glucose, you're trapping it, helping to accumulate glucose within the cell because phosphorylate glucose doesn't go on those transporters. You're also putting some energy into this molecule, and you're going to use that energy to break a carbon-carbon bond. How strong is the carbon-carbon bond? That's pretty strong. So you're going to have to configure this molecule to have a lot of energy invested in it. It's going to have to be a pretty unstable molecule to break a carbon-carbon bond. And so also by adding the phosphate, you can think about it. All of the substrates here are going to have this negative charge throughout the pathway. And that's handy because you can have pretty tight binding of your enzyme to these metabolites because you could strategically place a positively charged amino acid, which would bind in a charge-charge interaction with the phosphate. If you didn't have a charge on here, you wouldn't have the possibility of strong charge-charge interactions. So remember, it's a balance. You don't want it to bind too tightly, but it does allow you to specifically help to improve the specificity of the interaction of the various enzymes with the metabolites. So the next step is a summarization. So you see the aldose, C1. Remember, this can equilibrate with a linear form. And then you've isomerized to a ketone. So now you have carbonyl group here if you open this up. These are all equilibrium. They're still mutorotation. The hexapionates in a lot of these enzymes are somewhat indiscriminating in the anomer that they take as substrates. But it doesn't really matter. Even if they had a preference for one, these are all equilibrating. They're still hemiacetel and hemiketel. So even if each of these enzymes could take one form, well, that's okay because the other form would interchange to that. And so we've isomerized. We're now at ketose. We still have six carbons. Summarization, you're just rearranging bonds so that it's pretty close to equilibrium starting from the standard state. And now we're going to put a little bit more energy in here. You see the strategy? You're amping this molecule. You're now going to phosphorylate the C1 position with a second kinase called PFK1, phosphofructokinase. Transfer a phosphate onto this molecule, makes this fructose 1,6 bisphosphate. Okay, and so this is exergonic because ATP hydrolysis is more exergonic than phosphorylation of the sugar is endergonic. These reactions, again, are coupled. And starting from the standard state, you tend to favor formation of products. See what's been so far? All right, I know, I know. I went through it too as a student. So let's think about throughout the semester, we're going to think about the regulation of pathways. And as it turns out, the most exergonic steps are the rate limiting steps. Not because of the thermodynamics, but because evolution has decided let's decrease the abundance of the enzymes in these steps. That would be cool because not only, so if they're very exergonic, that means in effect they're not in equilibrium, right? So they're grossly favoring the formation of products. Once you've made a product, it's not going to go back. So that's a one-way bell, right? So that helps us to commit into this pathway. And so if we limit the enzymes here, we can really regulate the process. So the two kinases, the two steps where we added a phosphate to the sugar molecule are enzyme limited. It's just a very small amount of enzyme. And these steps occur very far from equilibrium, starting from the standard state. So you tend to accumulate a bit of molecules here in this process. So we want to regulate the flux of sugar molecules through this process because the energy demand of the cells is not a constant. There's an environment. Sometimes if I'm running around, I need to make more energy. If I stop and veg out on the couch, I need less energy. If I eat a meal, then I've got tons of fuel. What am I going to do with that? I need to store some of that. So you need to balance the amount of energy available to whatever is necessary for the organism. And so these are the steps that are regulated. Let's look at some of this. And so this is what I just mentioned in words. We also have a variety of fuels, and different fuels have different efficiencies for extraction of energy. Lipids have tons of reduced carbons, and so those you can get even more energy out. And those are useful for storage of energy. So there's different characteristics of these allosterically regulated enzymes. Remember, the regulation is going to occur other than the active site changing the shape of the enzyme. So these regulated steps are going to control the rate of the entire pathway. Because if you stop one step, the rest of the steps are just going to have to wait for that slow step to proceed. So these are one-way valves. Effectively, irreversibly, nature said, hey, this is a one-way valve. Once it's gone past this step, so much energy has been released, it's very unlikely to go in the reverse direction. And in general, these regulated steps are going to have different enzymes in different directions. That makes sense. So we'll see later on in this semester we can take simple molecules and build up sugars. That's called gluconeogenesis. But you don't want this to be a circle. You don't want to break down sugars, build them up, break them down, build them up. Nothing productive is happening. So the enzymes, you have to independently control whether the pathways are going up or down. And this will become more clear throughout the semester. Okay. So these two steps, step one and three, as I've already mentioned, highly hexagonal, see there's just one arrow, not really much of a back arrow. And those are the steps that are actually regulated. Let's look at the logic of the regulation. It's pretty cool. So here's hexokinase. This is the active site of the enzyme where you're going to be putting a phosphate at the C6 position of a variety of different sugars, including glucose. And this is the allosteric site. This is where allosteric regulators are going to bind. It's at a site other than the active site. Okay. And so in this sort of interesting, hexokinase is actually inhibited by its own product, glucose 6-phosphate. So let's look at this. So you have ATP and citrate. So ATP is a general indicator of the amount of energy already available to the cell. Keep that at a certain level so that your processes in your cells can occur. And, oh, sorry. I thought we had a slide showing. Yeah. So I guess the glucose 6-phosphate is the only negative regular of hexokinase. I thought there was a slide that described that. This is the second step, the allosteric regulation of this other kinase. Remember, it's one-way highly hexagonic because of ATP hydrolysis. This step, the substrate inhibits the enzyme. Really? Doesn't that seem counterproductive? It would be if that regulation were occurring at the active site, but it's not. It's occurring at the allosteric site. And the amount of ATP necessary to turn the enzyme off allosterically is higher than the amount that you actually need for the reaction. So the ATP levels get really high. We're not going to actually turn the enzyme off, but at a lower level, there'll still be enough for us to push this reaction forward. So it's a bit surprising to think about. Okay. And these are the curves. And you can see at low ATP, you're grabbing your substrates tighter. The apparent affinity is increased because K.5 is decreased. Okay. So this is the critical step. This is where we've built up to this moment. This is the time, man. This is when we're going to break this sugar molecule into two pieces. So if we're going to disassemble molecules, we're going to eventually break some carbon-carbon bonds. And so we're taking fructose 1,6-bisphosphate, a very unstable molecule, and in one step, breaking it into two triophosphates. It's a ketose and an aldose, the two products. And you have to look... This is a good slide for you later on to practice oxidation states. Ask yourself the question, is there oxidation or reduction here? We could sort of cheat, and say I didn't see any NAD, didn't see any electron carriers, yes. But if you actually look at it, we've redistributed electrons amongst these two parts of the molecule, but no electrons have been removed. So this is not an oxidation reduction reaction. This is just a cleavage of this sugar. And this, even though we've put all these phosphate here, phosphates are going to tend to repel, you put them on a small molecule, it's a good thing to have a split up of this molecule. But still, starting from the standard state, highly endergonic. And so you might imagine, wow, that's going to be tough. And the way that evolution is figured out to deal with this, is to strategically place one of the fastest enzymes known to biochemistry right after this step. And so that has the effect of rapidly whisking the products away, in effect decreasing the concentration of one of the products, and making delta G close to zero. So this process actually occurs, delta G is around zero. So it's very close to equilibrium because of the strategic placement of Superman enzyme in the next step. And so here is Mr. Superman. This thing, the instant, it's just a matter of diffusion. As soon as this substrate touches this enzyme, bam, it turns it into this other molecule. And so now, from the one hexose, we have two glyceraldehyde 3-phosphates. And this is just an assomeration. So the standard change in free energy is just slightly undergone. It can actually, delta G is very close to zero in this step because of some of the subsequent steps. So we've accomplished our goal. And now we're going to, but we still haven't harvested any ATP. We've put two ATPs in, we've split the molecule in two. We haven't made any, captured any energy yet. We need to do that. So this is trial phosphate. It's diffusion limited. Two times 10 to the 8 per molar second is the Kcat over Km. So it's an amazing, amazing enzyme. Not coincidentally placed after a really tough chemical transformation. So it's important for the clicker that you be able to track which bond is broken. The whole point of this pathway is to break a molecule into two. So let's just remember where we're breaking the molecule. If you're making two trioses, you could guess where it's going to break, right? Between C3 and C4. And so if you map this, so you're making from fructose 1, 6, 5 phosphate, these two molecules and eventually two molecules of glycerol by 3 phosphate. So C1 and C6 have now become C3 because of our conventions of naming. Remember the most oxidized carbon is where we start the numbering. So this is 1, 2, 3 in this glyceraldehyde 3 phosphate, right? So carbon 1 and carbon 6 end up over here. We broke, obviously, carbon 3 and 4 because we make, just remember, we make two trioses. The only way you're going to make two trioses is you break carbon 3 and 4. So we invested two ATPs and now we've got to show me the money. Come on, man, the payoff phase. So here we go. So this is the big payout, man. This is the Jerry McGuire moment right here. So we're taking glyceraldehyde 3 phosphate plus inorganic phosphate and we're oxidizing. Do you see this? The aldehyde to the carboxylic acid. Do you see this? So this is a two-electron oxidation releasing a flaming amount of energy. Some of that energy is captured in NADH but even NADH alone is not enough to catch it all. You've also caught some more energy by driving a very unfavorable process. You've taken a 3-carbon sugar that already has one phosphate and you stuck another one on. So this now is a trios bisphosphorylated molecule. So some of that oxidation energy is captured in NADH. Some is still in this molecule. And in the next step, we're going to synthesize a molecule of ATP because the hydrolysis of this insanely energetic molecule is more exergonic than ATP hydrolysis. So we can do this substrate-level phosphorylation that we've been talking about. This molecule is so must-up, so unstable that these two enzymes have to huddle together. If this molecule were to be released into the cytosol, it would probably spontaneously decompose, spontaneously hydrolyze. We have to very carefully, as soon as this is made, whisk it away to the next step. Give it to the next enzyme where we're going to then catch the rest of that energy on an ADP molecule making the ATP. So these two reactions, the two enzymes are physically associated and there's a channel between them. Let's look at the energy. We've already thought about this a lot. So NADH, remember there's a large amount of energy captured in NADH. But then we're driving the highly endergonic addition of a phosphate to this glyceraldehyde-3 phosphate. So that is endergonic to the tune of 50 kilojoules per mole. But when we couple all of these things together, starting from the standard state, we're reasonably close to equilibrium. The energy here is this oxidation and reduction chemistry. That's what's driving this. So this is the moment of harnessing the energy. But then in this next step, we're tramping and we can look at the free energy as well. So hydrolysis of a second phosphate off of this trios is more exergonic than hydrolysis of ATP. So adding a phosphate onto ADP is endergonic. And this is more exergonic. The reason it's more exergonic is because there's more, in orgo it's either steric hindrance or resonance. So 50-50 chance, man. There's more resonance here. Why? Look, it's a carboxylate. Both products have increased resonance, whereas with ATP hydrolysis, just the phosphate has increased resonance. So it's more exergonic. It can drive substrate level phosphorylation. So here it is up here. We can literally transfer the phosphate directly onto an ADP because both the inorganic phosphate and this molecule have a high degree of resonance. It's doubly stabilized. So these things are literally coupled together. And when you add everything up, it's not exergonic the whole process. So it tends to be actually pretty reasonably close to equilibrium. And here's a picture of this idea of not letting the hot potato get away. So here we have one scenario where we could potentially release our this phosphorylated triodes into solution and see what happens. It's not going to be good. What nature has come up with is, let's get this right to the next enzyme and then immediately make an ATP molecule. So we have substrate channeling between these two enzymes through a physical interaction of them. And so here we can continue on. So we now have three phosphoglycerates. And this is sort of finishing up and getting to the final state, which is pyruvate. So we have three phosphoglycerates. You can rearrange this through a mutate. Intermediate in this step is 2-3-bisphosphoglycerate. Remember where we saw this? This hemoglobin sits in the donut. It's a donut hole, man. It's a donut hole right there. That's an intermediate in this step. And so three phosphoglycerates mutated to two phosphoglycerates. We're setting things up again here for another highly exergonic process. So now we have an enolase, right, which is catalyzing the elimination of water, forming a double bond. Phospholeno-pyruvate. So both of these steps are close to equilibrium, starting from the standard state. And now we're going to take that PEP, our Phospholeno-pyruvate, and do another substrate level phosphorylation, transfer the phosphate from PEP to ATP. And you can then think there, obviously the hydrolysis, or the removal of phosphate from PEP is more exergonic than the removal of phosphate from ATP. And so you can think about, okay, why might that be? So here is, that's actually the most exergonic process, this transfer of phosphate. And the reason is because of this enolketal potomerization. So you have the enol form of pyruvate, which in a blink of an eye, transforms into the keto form, which is the most stable form of pyruvate. And this helps to make this energy that's released is tremendous here because of this isomerization. So it sort of strategically places a flash of release of energy at the very end of the pathway, because up to this point, equilibrium, equilibrium, equilibrium, pull, equilibrium, equilibrium, you see it, at the end of the pathway, it's very exergonic. So that pulls flux through the pathway. Okay, everything here is strategically designed. Pull, it's like a tug of war. And so this is the overall process. We invested two ATPs in the preparatory phase and the payoff phase. We've harvested the two and invested, one of two, and we've captured some of those electrons on NAD. Now, in the presence of oxygen, we could do something with those electrons. We could make ATP. But in the absence of oxygen, if we keep doing glycolysis, every time you do one round of glycolysis, you have one less NAD molecule. So say you had a million molecules of glucose and 10,000 molecules of NAD+. You would get to a point where you couldn't process any more glucose because there would be no NAD+, left. There would be nothing to do the catching. And so if you're only going to do glycolysis and you're not going to bring things further along, you need to regenerate the NAD. So you need to put the electrons back onto the metabolic intermediate so that you can have some more NAD molecules. Here, if you want to start thinking about the cellular state, these are the cellular states for the variety of metabolites we've seen throughout this pathway. You can see they're not one-to-one. They're strategically manipulated so that this pathway is at equilibrium except for those two, step one and step three in the beginning and at the end of the pathway pulling things through. And so if you look at delta G, this is the standard change in free energy. What step is this? What's the strongest bond we're breaking here? Carbon-carbon bond, aldolase. So this is where from the standard state, this pathway would have troubles. But remember we put the super enzyme right after that. We strategically placed it. So that starting from the cellular state, you have the two valve functions, the two regulatory steps, and then you have the end of the line. You've got this tautomerization process pulling metabolites through. So this is highly exergonic starting from the cellular state. This is also exergonic. This is the next step that we're going to see. We haven't seen this step yet, step 11. So overall, we've oxidized, we removed two electrons from glucose. You can run through these numbers. This sort of summarizes what we've done so far. Some of that energy has been deposited in ATP. A large amount has been deposited in ADH. And overall, the pathway of glycolysis, glucose going to pyruvate is exergonic even starting from the cellular state. So equilibrium tends to favor the forward direction of this pathway. But individual steps are manipulated so they're either at equilibrium or exergonic. So starting from the cellular state, delta G would be more negative. All right. So now we've done something good. We got two electrons out. We made some ATP. But now, we've got this problem with NADH and we're running out of NAD. So we need some way to reoxidize this molecule. Reoxidization of NADH to NAD is important because that's one of the substrates in the steps of this pathway. And the way that we're going to deal with this problem is a process called fermentation. And so fermentation is literally the deposition of electrons from NADH onto either pyruvate or another molecule. So there's two different types of fermentation. One makes lactate. So that happens when we're running around. We make lactate if we're not getting enough oxygen in our muscles when we're really going for it. And in yeast, you can put those electrons onto ethanol. And that makes you happy. There we go. So here's the two different processes, two different ways in an anaerobic state where we don't have oxygen. We need to put those electrons either onto pyruvate or onto acid aldehydes. We'll see in a moment. But if we have oxygen, we'll see in the next lecture that we can literally throw that into a furnace and just start making a tremendous number of ATP molecules. And now we don't have oxygen. So let's look at this fermentation. So in us, in our muscles, pyruvate, you can literally put these electrons directly back onto pyruvate to make lactate. See the oxidation state change? Do you see it? So here you have what? Oxidation state of 2. And here you have oxidation state 1, 2, 3, 4. So you put those two electrons from NADH onto lactate. And the critical thing is that we made, we create, we solved the problem. We made some NAD. We can do some more glycolysis. But we created another problem. We begin to accumulate lactate. So when you're running around and you get a cramp, that's this molecule right here, building up in your muscles. And so slowly it's cleared away. It actually gets transported to your liver and gets metabolized there. So this process is a transfer of electrons. And so it's actually very exergonic. It's not quite as exergonic as the oxidation reduction reaction we saw previously. Now in yeast, you can transform pyruvate to acetaldehyde. So you can decarboxylate. So when you drink beer, there's bubbles, right? And there's ethanol. And so this is what these organisms are doing. They're doing anaerobic fermentation. So this is not oxidation reduction. So if you can look at each of the carbons and see electrons have not come to either of these or been removed. This step is. So here this is oxidized. And you reduce it down to ethanol. So this is the step where you're actually putting the electrons back on to acetaldehyde. So different organisms, different way to deal with this problem. So for each turn of glycolysis, you're just making two ATP molecules. And the absence of oxygen, you're sort of being kind of productive, right? So you put all this energy in the NADH and then you just sort of put it back on the molecule. So all you've done is made two ATPs. And the cell needs a certain amount of energy. And so if we had oxygen, then we could make tremendous amounts of ATP from each glucose molecule. But here we're only making two ATPs per molecule. So glycolysis obviously has to go very, very quickly. And so this is the past turn effect. The rate of glycolysis is extremely fast so that we can keep up with the cellular demand for ATP. We're very inefficiently only making two ATPs per round of this cycle. So aerobic restoration or catabolism of sugars in the presence of oxygen is much more efficient. And so for each glucose molecule there, you're going to be making large amounts of ATP. So you need less flux through the pathway. And so this is something that you can measure in microorganisms and yeast as well. You need a rapid flux for this process to regenerate that NAD. Okay, so we've come to the end. We've made pyruvate. This is glycolysis. Here in this lesson on that. Let's move to another pathway. So yes, we need to make energy from glucose, but we also need to feed other pathways from glucose. And so the pentose phosphate pathway is one of the scariest pathways I saw as a student. Don't panic. I don't know the steps. You don't need to know them. I'll tell you what you need to know as we go through this. The output of pentose phosphate ribose 5-phosphate and NADPH. Remember, that's the reduced cofactor that's used in an holy Christmas. And that's that. And the next thing. Okay, so here's pentose phosphate. Mr. iPad will... Okay, so we're taking our glucose 6-phosphate. How did we make glucose 6-phosphate? Hexokinase. You remember? Woohoo! Mr. Hexokinase made glucose 6-phosphate. And now we're going to immediately start removing electrons. So glucose 6-phosphate, we have the release of NADPH here. Two electrons transferred. The second two electrons transferred. Bubbles coming out. Carbon dioxide released. I'm so excited about this pathway I keep going for. So the output is two NADPH molecules and a ribose 5-phosphate. You're like, hey, I know what I can do with that. I can make some DNA and RNA. So now we're sort of transitioning. This is catabolic pathway, but it's leading into an anabolic process. We need some ribose 5-phosphates to make DNA and RNA. We also need NADPH for an anabolic processes because those reduced cofactors will be used later on in the course to build up molecules. But do you see how there's two options here? One, circle, the other, down. So we can either make ribose 5-phosphate or NADPH, but as it turns out, the cell needs more NADPH than ribose 5-phosphate. So every time we do this non-oxidative phase of the pathway, we make two additional NADPHs and release a carbon dioxide. So we can regulate the amounts by regulating the enzymes at these decision points. We can either have an enzyme going this way or an enzyme going this way. And these are allosterically regulated. If I tell you, you feel like you have to know it, just realize that we can have different rates through this cyclic pathway or down into making DNA and RNA. So let's look at the steps real quick. This is digestible. So here we have glucose 6-phosphate. You remove... So this is a aldose. Glucose is an aldose. So it's an... It's in equilibrium with linear form that's an aldehyde. We've oxidized by two electrons to a carboxylic acid. Do you see that? You see the carboxylic acid. And so that's a 6-phosphoglucos delta-lactome. We have an enzyme that can open up that ring, making the 6-phosphogluconate. And then in the second step, we can transfer two more electrons by B-carboxylating here and forming a new ketone. So you see the two electrons went into that ketone. That gets isomerized into an aldehyde. So we're moving electrons around. So you might say, okay, what do I got to know, man? I got to know these structures. You're sick, man. You're torturing us. No, you need to know what's going in, what's going out, and that you can regulate both independently. Okay, so what's coming out? Let's see. What's coming out? Ribos, 5-phosphate, and we have NADP. Anything else coming out? Bubbles, yep. Three things coming out. Okay. Now, what if we want to regenerate this and make more NADPA? So we can actually take our 5-carbon sugar and transform it back into a 6-carbon sugar. And this is sick. I mean, look at this. I don't even feel like you have to know this. Just know that you have... I put a little clue. This is what you got to know. This is for me. I'm not too sharp. So you have 6-5-carbon sugars coming in. You got a food fight in between. Carbons are being flicked between molecules back and forth. And then you have, out the other side, you have 6-5-carbon sugars coming out. 6-5-carbon sugars going in. And you've got 5-6-carbons coming out. The number of carbons hasn't changed. There's not... I mean, many of these steps are at equilibrium. You're just sort of isomerizing, transferring pieces apart. So I just numbered the number of carbons at each step. I don't know any of these names. You don't either. You can look it up. You can even reduce it to this. It's just the number of carbons. You have these 2-5-carbon sugars coming together. Make a 6, a 7, and a 3-carbon yada-yada-yada at the end. You've got 6 times 5 is 30, and 5 times 6 is 30. That's about as deep as you need to know this. And it's important because each time this turns around, you've taken a 5-carbon ribose and transformed it back into what you're going to feed into this pathway. The only way you're going to make more NADPH is to make that a 6-carbon sugar again. But it's a bit confusing because you're like, well, there's 6 of them. Each time you go to the top, you've fully oxidized 6 carbons. That's confusing. So don't worry about it. Just what goes in, what goes out, and it's logically regulated because you need different levels. Don't panic. It's all good. Okay, one last thing. It's painful. So this is a catabolic pathway. You can burn one fuel, glucose, but there's all kinds of other fuels in your beverages, the ones that didn't have the dipeptide sweetener. If you have sugar in your beverage, that's sucrose. That's not a double or a disaccharide of glucose. It's a disaccharide of glucose and fructose. So what are we going to do with the fructose? We're just going to excrete it, take it, cut off the glucose throughout the rest. No. We can take those other molecules and feed them at different places into the pathway. So we can metabolize glucose, fructose, lactose, and mannus, and lactose. So fructose, here are the pathways. You can either burn it as a fuel or convert it into glycerol and make some lipids. So you've got some choices. Nice to have choices in life. So fructose, remember, this is that cheating kinase. It can phosphoryte the C6 position of glucose, but also of fructose. So that same enzyme can make fructose 6-phosphate. We can bisphosphorylate it, and this is heading to glycolysis. Alternatively, we can monophosphorylate with a different enzyme at the C1 position. Hexokinase cheese, but only at C6. Fructokinase is needed to phosphoryte C1. Then we're going to split a sugar that just has one phosphorylation. So it just has one phosphorylation. One of the triose products is going to have a phosphate. The other isn't. So we make dihydroxyacetone phosphate, same as in glycolysis, and we're going to make glyceraldehyde. Now, we have an option. Do we want to make some lipids? Almost all lipids have a glycerol molecule, so we can take our glyceraldehyde, convert it into glycerol, or we can phosphorylate glyceraldehyde and feed it into glycolysis. We have options, and obviously there's going to be allosteric regulation. If we need more lipids, then this step is going to be inhibited and this enzyme is going to be activated. Galactose is an ephemer of glucose at the C4 position. What is an ephemer? Just one and only one center has... Do you see how galactose might be less stable than glucose? Then why might that be thermodynamically? Because necessarily one of those hydroxyl groups needs to be in a not equatorial position. Whereas glucose, one of the animers, has all groups equatorial. So glucose is most stable. But galactose can occur. If we want to feed this into glycolysis, we're going to have to ephemerize. We're going to have to switch that stereochemistry at that C4 position. So we phosphorylate position one, and then we have this UDP, uredine diphosphate glucose. We're going to transubstitute the UDP group with the phosphate. It's still a phosphoester bond. Now we have a UDP. UDP is a big honking molecule. The reason you're like, where did this come from? UDP is something you can grab onto. It's got some heft to it. Phosphate is a little thing. UDP is big. And so you can really grab your substrate with high specificity once you put the UDP on. And now while you're holding it tight, you take that hydroxyl group and you oxidize it to carbonyl, eliminating the stereochemistry. And then you reduce in a stereoselective manner. Do you see what it's doing here? So you're going through this achyrointermediate metabolite, carbonyl group, and then you're reducing on the other side. And NAD is participating. It's like, I got your electron. And then it gives it back to the molecule in a stereoselective manner. So this is an epimerase. It's inverted the stereochemistry at that position. And now we're good to go. We basically got UDP glucose. We just got to get rid of this thing that we're holding onto, this UDP group. And so we transfer it back there. Make glucose one phosphate. Not quite what we need. We need glucose six phosphate for glycolysis. So we have a esomerizing enzyme, phosphoglucomutase. Okay, we made it. So this is a very hard question. And I'm going to give you a hint because I'm not trying to make you all fail this class. So can you guys remember the question? Because I'm going to show you a slide that is helpful. And I'm going to keep the voting on because this is too many people missing it. So the answer is there.