 Okay, so today we need to get ready for the Uber lecture coming up, which is glycolysis. And so a lot of what we're going to be talking about today is sugars. We have a little bit to finish up on fat, but we need to learn another whole language in a class. Let's finish the last one. So a bit to do. Let me go this way. Am I going the wrong way? One moment. Oh, that's nice. All right. So there's some announcements. So these haven't really changed. The only thing that has changed is that the conference sessions has posted in the first lecture and last time, Monday and Tuesday is a holiday, so TAs need time off as well. But actually one of the TAs still wanted to have conference sections. So he was afraid my previous announcement would have scared everybody off. So the conference session for Monday noon to one is still on. The rest are not for Monday and Tuesday since it's a holiday. And again, we're split up A to M in this place for exam one through Z and Ernie and Burtville. That's not cool. All right. So we need to finish up some of our thinking of the thermodynamics of transport across a bilayer. Remember delta G prime knot is zero because you're not chemically transforming the molecule. It's the same molecule chemically on either side of the barrier. But the concentration on each side affects the equilibrium. So this equation, so the change in free energy is C2, which is the concentration of where the molecule is moving to. And C1, concentration of where the molecule is coming from. So if you move from a zone of low concentration to high concentration, delta G is positive. And if you go the opposite way, delta G is negative if you look at this. So these are arbitrarily defined. It's sort of similar to products and reactors. In this case, the product is where you're going to because that's the outcome, the hopeful outcome of this reaction. So primary active transport is where you couple the highly hexagonic hydrolysis of ATP to the endragonic movement of molecules against this concentration gradient. So those two react, or that hexagonic hydrolysis allows you to do the endragonic process. And so primary active transport is the power plant. So every cell has a power plant and it's not very diverse. Like in our world, at one point in time there is AC and DC power. But it makes sense and it's intuitive that why would you want to make things more complicated than they need to be? Why should every transporter have its own power plant? Why not just plug into the existing power provided by the power plant? And so the electricity is this equilibrium between the concentration of a molecule. And so we'll see how that works in a moment. But the things that we're plugging into the power plant are the secondary active transport. And this is where one of the molecules that's moving down the electrochemical gradient, that was the molecule that was put in that orientation by the power plant, primary active transport, the other molecule is moving up the gradient. So it's very much how you would plug an iPhone into a USB plug. Nature has a similar mechanism. Okay, so this is just a schematic to help you to understand this. So here's our power plant. We're coupling the highly hexagonic hydrolysis of ATP to the endogonic movement of a molecule from a zone of low concentration to a zone of high concentration. We then plug into this power source in terms of difference in concentrations. So as this molecule returns back from the zone of high concentration to low concentration, that provides enough energy to move a second molecule across the bilayer against its gradient because here the second molecule is in low concentration outside of the cell and high concentration inside of the cell. Alex, could you try to get this advancing somehow? And I'm just going to hang out here for a little while. Thank you. Okay, so the types of power plants that you could plug into are basically protons, sodium ions, and potassium ions. And so we'll see that one fourth of your cellular energy, so that breakfast burrito you had this morning, one fourth of that went to a single primary active transporter that sets up this gradient, this power source on all your membranes driving the majority of transport across those membranes. So it's a pretty important guy. And that guy moves sodium and potassium ions, sets them up to provide the energy for secondary active transport. But the things that are moving here in the secondary active transport are very diverse. So here we have simplicity, a limited number of power sources, and here we have basically every molecule that's got to go across the bilayer for metabolism. So different parts of metabolism occur in different locations. And so it's important to pay attention to whether the molecule that we're wanting to move across the bilayer is charged or not. So if the molecule is charged, then that factors into these equilibrium because the membranes that we're going across are impervious both to molecules, but also to electrons, they're resistors. And so we need to think about the movement of charge and any non-equilibrium between charges on either side of a bilayer. So here is the power plant. One fourth of your breakfast burrito is right in this bad guy. And so this literally sets a voltage on your membranes. Do you know that every one of your cellular membranes has a voltage of about 50 to 70 millivolts? I think that's pretty amazing. And that happens because this particular transporter couples the highly exergonic hydrolysis of ATP to the movement of three sodium ions out and two potassium ions in. And both of these molecules are moving against their gradients because the concentration of potassium inside the cell is higher than outside and the concentration of sodium outside the cell is higher than inside. And so both of these things set up a disequilibrium between molecules, but also between charges. Every cycle of this transporter moves three positive charges out, two positive charges in, so there's a net efflux of positive charge. And because of that, the outside of this membrane is positively charged. The inside is negatively charged. If this molecule were able to move more ions and you could set up 200 millivolts of voltage, you would actually get arcing across the membrane. And so, I mean, this membrane is very, very thin. I think it's amazing that it can hold, resist, you know, 50 millivolts. It seems like a lot, or even 100 millivolts. And so let's look at how this guy... So really, these are the battery. I'm not really here, right? So the matrix is true to some degree. Each one of your cells has a potential across its membranes provided by lunch. Okay, so here is this actual molecule. And now watch what's happening with the ions. So first, is it going? Well, you can watch it online. So you can see... You'll have to trust me that this thing is moving things against the gradient. So from a zone of low concentration to high concentration. But basically, the binding of an ion induces a conformational change combined with the hydrolysis of ATP. So you can see the timing of this if you got the movie. Is it working? Okay, that'll work or not? Nope. Yep, sweet. Alex is the man. All right, so we've described the power plant. It's using one quarter of your energy. Let's see the power plants in action. So here is the transport of glucose. That's an important molecule. We'll learn all about that today and on Tuesday or on Thursday. So here we're setting up this disequilibrium between sodium potassium ions. So because of that primary active transport, the concentration of sodium inside of this cell is very high. But what we need to do is move the glucose from your stomach into this cell. And the concentration of glucose in the cell is high. And so we couldn't just do that if we didn't have something to help us. So we have this co-transport of sodium ions down the electrochemical gradient and the movement of glucose ions up the electrochemical gradient. But now once we have a high concentration of glucose, all we need to do is provide a specific way for them to exit the cell. So they bind this uniporter, that glued transporter we looked at last time, and that binding event is specific so only glucose can exit here. And this is driven by the accumulation of the concentration of glucose in the cell. The charge on the membrane doesn't affect this process because glucose is polar but not charged. If glucose were charged, then you would have to consider that voltage. So this is an example of transport of glucose being plugged into the power plant. Glucose cannot simply diffuse between these cells. It's really tight with each other. That'll work. So there's all kinds of different things that set up the primary active transporters. This one works in a way that I've shown you, moving ions against the gradient and typically involving the hydrolysis of ATP. But these other ones tend to work in reverse. So they feed on a disequilibrium of protons across the membrane to drive the synthesis of ATP. These are reversible processes. And so this is, for example, the F0F1HP synthase. And this thing allows protons to move down to the electrochemical gradient while chemically transforming ADP to ATP, which is an endergonic process. So there's all kinds of different primary active transporters. But when the solute ion is moving, it's not accompanied by the movement of conorion. It generates this electric potential. So we saw that for the case of sodium-potassium transporter. And so this is an electrogenic. In other words, it generates, literally, generates a charge across the bilayer. If you're moving a molecule that's not charged, such as sodium, or if you're moving an equal number of charges, maybe one positive charge goes out, one positive charge comes in. That's electroneutral. And so we saw an example of that in the last lecture. Okay. But we need to factor in these change in charge. So let's think about this. So remember, the big point from the last lecture is that the membrane is impervious for the free diffusion of the majority of molecules. CO2 can get through, but really, most other molecules cannot. So at equilibrium, there would be an equal concentration of molecules if you're just considering the concentration and not the charge on the membrane. There would be an equal concentration at equilibrium. But this is the cellular state. The cellular state prevents this equilibrium from happening and regulates the process. So we build up concentration gradients, and those can be used to drive transport. But the similar thing can be thought of in terms of the movement of charges across the impervious membrane. So this membrane is impervious to electrons. So if we move, if there's an equal concentration of charges on either side, well, there's no voltage on the membrane. But in the cellular state, there is a disequilibrium of charges set up predominantly by the sodium-potassium ATPase. And so that means that, well, if I, for example, wanted to move an additional positive charge this way, that would be endergonic, or if I wanted to move an additional positive charge this way, that would be exergonic. So that could, that change in voltage, you could plug right into that if you're moving a charged molecule down this electrochemical gradient. Okay, so it's a bit surprising that these membranes are impervious to electrons. And so we need to consider, so remember we talked before that delta G prime knot is zero because there's no chemical transformation going on here, but delta G has to do with the concentration of molecules on either side of the membrane. So delta G is zero at equilibrium, right? We all know that. And so if you set this equal to zero and this equal to zero, then you discover that at equilibrium, there's an equal concentration of molecules on either side of the bilayer. And that, we saw that previously, just from a logical perspective, not looking at this equation when we considered that membrane that I showed you a few slides back. So if we had a charge, how would that alter things? Well, let's put a voltage. So we can literally describe that by a volt meter. And so the equation gets changed. So remember delta G prime knot still is not changing because there's no chemical transformation. But now we have a component that is regulated by disequilibrium of concentration, but also disequilibrium of charge. And so this is a little bit confusing. Let's define the terms. Z is the charge of the solute. So if the charge is zero, so for glucose, there is no charge. It's not ionic. It's not ionized, right? And so Z would be zero for glucose. So when you're considering the movement of glucose, it doesn't matter what kind of voltage is on the membrane because this is zero and this term goes away. But if there is a charge, you need to consider both the voltage set up across the membrane and the number of charges that you want to move. And the surprising outcome of this is that, well, what happens for the movement of charged molecules at the equilibrium? Would there be an equal concentration on either side of the membrane as we saw in the case of the movement of an uncharged molecule? Well, if you run the mass substituting delta G equals zero, which is equilibrium, so uncharged doesn't matter and the equation reduces, but if we move a charge, so we substitute zero into here, set these two, move this term to the other side, becomes negative, and then solve for C2 over C1. We see that if you're moving a charged molecule, the concentration ratio on either side of the membrane does depend on the number of charges moved, the types of charges, and the voltage across the membrane. Okay, so this is surprising, but it is the reality. We are batteries. All right, question. That's not just the number, it's the type. What if you move a negative charge? Then Z would be negative one. The number and the sign. Yep, yep. And Psi is the voltage, right? And R is the gas, law constant. T is the temperature in Kelvin, okay? Which goes along with R, okay? So I'm not gonna say, you know, generally there's no calculators on exams, so mindless sort of plugging numbers in. We might do that in class because you do have calculators, but anytime you have something on an exam and you feel like you need to solve it for a number, it's gonna be okay for you just to set it up. Okay, so if you set it up correctly, we don't care if you know how to use a calculator. That's not the topic of this class. Any other questions? This is difficult for people sometimes. This idea of moving charged molecules. So maybe review that a little bit after lecture. So let's move on to sugars. So we need to consider, yes, hi. Diffusion, there's no diffusion. There's no diffusion. Diffusion is blocked. So these gradients are providing energy for transporter molecules. But free diffusion, so yeah, the word diffusion is sort of a sacred term that has to do with things in an unregulated fashion passing through. But yeah, it's an important point, right? You have to consider both the difference in concentration and the difference in charge. And both factors matter. Because moving something, a positive charge out to where there's already a lot of positive charges is difficult, right? The more charge you accumulate, the harder it gets. Not sure I answered your question. Yes, okay. So let's move on. So let's think about sugar. So we're going to go through a very short phase of descriptiveness. And then we're going to get into all kinds of organic chemistry. There's a lot of organic chemistry. This is the fundamental slide on the sideboard. If you don't get this, it's hopeless. You're not going to get sugar chemistry. And you're not going to be able to, you're going to get really lost throughout the semester. And so I'm going to give you all kinds of clues along the way to help you to remember things. So let's consider sugars. So we can also call sugars residues, sugar residues, like we have amino acid residues. And so the most simple sugars are the monosaccharides. And so here's the most common monosaccharides that occur in nature. We have glucose, fructose, D ribose, and D2 deoxy D ribose, right? And so each of these, you have a carbonyl group either at the C1 position or the C2 position. All naturally occurring sugars have, and you number the carbons starting from the most oxidized carbons. It's just a semantic definition. That's how everybody has decided to do this. So this is carbon 1, 2, 3. Here you could actually remember it each way, right? But if there's more, we'll see, like this example, this would not be 1, 2, 3, 4, 5, reset, you know, 6. It's 1, 2, 3, 4, 5, 6. And these sugars, the stereochemistry, the DNL stereochemistry only talks about the carbon, the highest numbered chiral carbon. If you don't know how to number it, you're not going to be able to figure out what DNL is. And so that would be this carbon. We'll have lots of slides describing this. This is just a general introduction. All right, so here's the stereochemistry. Remember, for amino acids, we said glyceraldehyde was the founding member that helped us to arbitrarily assign DNL. Remember, L is substituent on the left, D is substituent on the not left, or duot if you're French. And so here, glyceraldehyde naturally occurring, all naturally occurring sugars are D sugars. And maybe at the end of the lecture, you can begin to think about why that might be. And so, for example, we have D glyceraldehyde, our most simple aldose. And so this one, you could consider L glyceraldehyde. It just doesn't occur in nature. These can be represented as fissure projections where the vertical bonds recede away from you and the horizontal bonds come out at you. And so we can also do these in perspective representation. If all chiral centers are inverted, those molecules are enantiomers. How many enantiomers are there of naturally occurring sugars? Integrating two things I've mentioned so far. How many enantiomers are there of naturally occurring? I'm not saying possible sugars, zero. There are no naturally occurring enantiomers because nature never made any L sugars. Just decided not to do that. It wasn't necessary for the functions in the cell. So there is no, all sugars are in the D configuration. And for something to be an enantiomer, all chiral centers need to be inverted. And if it's a D sugar, there is no such thing as a naturally occurring L sugar. So it's a little bit of a trick question. We have two classes of sugars depending on where the oxidized carbon is. If it's an aldehyde, carbon one, these are called aldoses. And easy to remember, aldehyde, aldose, right? The most simple member is D glyceraldehyde. Now as we introduce another carbon, there's two possibilities for that carbon. And so actually in nature, the only thing that's restricted is the stereochemistry of these highest numbered carbon. Those are always D. But these other carbons as we add them, they can be in either orientation. And we could have used more systematic and perhaps logical R&S designations for the stereochemistry, but instead we decided to come up with just a whole flu of different names to describe these diastereomers. So these are diastereomers because just one of two chiral centers are inverted. And so there's two four carbon aldoses, right? Because one of these is fixed in nature at least, conceptually not, but in nature it is. And then this other carbon can have either the hydroxy group on the right or the left. And so we give different names. The red boxes here are to give you a hint of what is most important in nature. So I tend to value what nature thinks is important. And so for the five carbon sugars, we added another chiral center. And so that chiral center can have the hydroxy group on either side. So you can see this is growing in an exponential fashion to the one, to the two, to the three possibilities. Each chiral center multiplies by two. And so we have a total of four five carbon D aldoses, right? And the most common are D ribose and D xylose. And for the six carbon sugars, we have eight possibilities. So two to the three possibilities of the D six carbon aldoses. Half of the carbon atoms on our planet are in this one molecule. Do you think that's important? I do. Every tree, it's just full of that particular sugar. Your body is full of this sugar. And so D glucose is a very stable molecule. And we can think about that as we look at the stereochemistry of that. But there's other common or perhaps less common, but six carbon aldoses and mannose and black dose, for example. So those are the aldoses. The other group have ketones, right? And so those are called ketoses. Easy to remember. Hard to advance. There we go. And so the founding member of that is a chiral. There is no chiral synthesis. So dihydroxyacetone has the carbonyl group here. But this is not chiral. This is not chiral. This is not chiral. So that's a bit unusual. It is a ketose. You cannot describe it as D or L because D or L specifies stereochemistry. There is none here. It's a chiral. But then there's one D four carbon sugar, right? So carbon two is occupied by a carbonyl group. That doesn't have two possibilities. You can't switch which side the doubly bonded oxygen is on. That would be ridiculous. So there's only one naturally occurring D four carbon ketose. There's two five carbon ketoses and four six carbon ketoses. So fructose is the most common of all of these. But the ones in the red box we'll see throughout the semester as well. You with me so far? So so far we're just becoming introduced to a dizzying array of somewhat random names for these sugars that have been generated through history. So you guys have probably thought about stereochemistry and organic chemistry. But you might not have heard of the term epimer. Does that sound familiar? So epimer is something generally that people talk about when they're thinking about sugars. And epimer is when one and only one chiral center is inverted. So these two molecules are epimers. Carbon two is inverted. These two molecules are epimers. Carbon four is inverted. But these two are not epimers because both carbon two and carbon four are inverted. So this is a definition. Enantimer is when all chiral centers are inverted. And epimer is when just one chiral center is inverted. So that's a definition. So let's get to the fundamental chemistry. We'll ease into this of sugars. And so the reaction of an alcohol with an aldehyde gives a hemisatel. And this equilibrium is absolutely relevant. So as soon as you make the hemisatel, that can with relative ease revert back to these two molecules. Now a ketone can also react with an alcohol to make a hemiketel. And the most difficult part of today's lecture is finding the carbons that are at the center of a hemisatel and hemiketel. Let me give you a really easy hint. It's the only carbon bound to two oxygens and sugars. The carbon bound to two oxygens is at the center of a hemisatel and a hemiketel. All the other carbons and sugar molecules won't be bound to two oxygens. And it's very important that within seconds you be able to pinpoint in the molecule where is that hemisatel and hemiketel. Because it affects the naming and it affects considerations of the stereochemistry. Because we're going here from an achiral state to a potential of adding chirality at this center. The equilibrium is real and important. Hemisatels have a little bit of instability. They can revert back to their original reactants. And so you might imagine, okay, where can I find an alcohol and a carbonyl group? They're both in the same molecule. You have a carbonyl group either at the aldehyde or the ketone for the ketose sugars. We have in aldose glucose, our favorite molecule. And this alcohol can psych-wise with this carbonyl group making a hemisatel. And so this is achiral. Here we've created a new chiral center. So the carbon at the center of a hemisatel and hemiketel is called the anomeric carbon. And that's because there are two different stereo anomers, two different stereo configurations about this that are created in the process of cyclization. Is this a condensation reaction? I'm full of trick questions today. No, it's a trick question. The next step will be a condensation. There's no water molecules on either side of those reactions. So this is not a condensation, it's just a cyclization. It's the reaction of an alcohol with an aldehyde. And so you see that there's a free sort of equilibrium between all of these forms of the sugar, the linear form, which is about 1% of the molecules, and then the two different stereo isomers of the cyclized form. And we can describe the stereochemistry at the anomeric carbon arbitrarily by this letter alpha and beta. We could have called it Mickey and Mouse. These are semantic definitions so we can communicate with each other about which stereo isoform it is. And in a few slides I'll give you a trick to be able to just look at a molecule and figure out whether it's alpha or beta. But for now, look at this. So you seem to have this preference for this stereo configuration compared to this stereo configuration. Do you have any intuition why that might be? I can say that this is more thermodynamically stable. Little hint. That's right. So it puts having this hydroxyl in that position, if you draw it in the proper chair configuration, you can see that that maximizes the number of equatorial substituents. You'll see that in a moment, but that's great intuition. The main thing that's driving this equilibrium to slightly favor this, but there's always one percent of the linear form. So you can say that a cyclized, this Hemiacetel is reactive. It's reactive because it's in equilibrium with an aldehyde and a ketone. Aldehyde and ketone are reducing agents, as we'll see in a moment. So these are reactive molecules. So these are not planar. This is the only way to represent in the last slide. Here's a more truthful way. We know that it's not a boat. We don't have bulky substituents making it do weird things. It's in a chair configuration. And so look at this. Here's alpha-d glucopurinose. We're like, good, good, good, not good. That is, and is that at the anomeric carbon? One second. You should be able to answer that. It's bound to two oxygens. It stands for oxygen. So the alpha is not, I'm trying to make it as simple as possible. I could say it's the one that's in ether linkage, but that would make you think too much. So here we have one axial. One axial. And so that's, you know, really, that makes that much of a difference. And so in organic chemistry, you might have received knowledge. It gets you confused at this point. I love you. You're just glad to be done with that. But some of you might think, wait a minute. I remember this sugar lecture. And we learned about orbital overlap here. Right? So the lone pairs in this oxygen overlap with the orbitals when the hydroxyl group is in the axial position. And that helps to stabilize that. And I say to that, true. But the equatorial, the sterics, is driving the equilibrium. It's trumping this thing you learned about in Orgo. If you don't remember that, that's fine. But it is confusing for the few that might remember this anomeric effect. Okay, so let's get it the way that we arbitrarily name the stereochemistries at the anomeric carbon. So we have here, so again, every one of these molecules, one second each, where's the anomeric carbon? Okay, so here is it here? No, that's not two oxygens. Here it is, the anomeric carbon. And you can say, okay, would you imagine as a ketose or aldose? So the anomeric carbon for a ketose is carbon two. And for an aldose is carbon one. Just to sort of look back over to some of the slides. So here we have a hydroxyl group pointing down. And here we have the highest numbered carbon substituent on the sugar pointing up. And the way we're going to remember this is if the hydroxyl group is down and the highest numbered carbon substituent is up, you could think of that in terms of trans and that it's the most distant. And so trans has letter A. Whereas if the hydroxyl group is on the same face of this ring and the H was prospective, and then that sits. It's not trans, right? It's sort of like D is not L. And so this is beta. Arbitrally, it could have been Mickey and Mouse, but we decided alpha and beta. And when it's trans, it's alpha. When it's beta, here, it's cis. And we're only considering, so here we're just, the only thing that matters is to wear the hydroxyl groups of the hydrogen atoms we don't really care about. Okay, so in the Hayworth perspective, remember these are in equilibrium between the linear form, the fissure projection, and the cyclic form, typically represented in this Hayworth's perspective. And there's another, I'm just full of bags of tricks today. Another trick is that the substituent's pointing down to figure out where those would be in the fissure projection, those would be on the right side, I believe, if I remembered it. Yes, it's right side. Okay, so if you want to switch between those, there's another trick to do that. So this is semantics. Here we have alpha one, right? Now, in the exam, I'm mocking the number of the carbons for you. That's your job. Look for that one with two oxygens. Alpha, right? So this is trans to that. This is trans to that. Okay, this is cis to this. That's beta one. Cis, beta one. Right? How are we doing so far? Here we go. Where is it? Where is it? I'm failing. Big time. Right here. This one is cis as well. So that's beta. Even I am slow sometimes. Okay, so this is, we're just defining what alpha and beta are. Yes? The fissure projection, which side is it on the left? It's on the right. Sorry to be uncertain. Right. Okay. So because there's a viable equilibrium between the linear form and the cycloid form of these sugars, we need to consider the possibility of interchange between the stereoisomers. Every time you go to the linear form, you could either go to either of the two anomeric forms, the alpha or the beta configuration. So you can consider this muto rotation. The alpha and beta forms are inter-converting because of the relevant equilibrium. Okay, but you know, we saw that that's only about the linear form. Even though it's just about one percent, you still can get there. And every time you get there, you can go to alpha or beta. So another semantics, when we're describing these sugars, we need to... So you can imagine, the sugars have alcohol groups all up and down the carbon skeleton. And any of those alcohol groups, in theory, could cyclize with either the aldehyde or the ketone. So you might think, okay, thermodynamically, a six-membered ring is going to be the most favored. But we could also get a five-membered ring. And so we need to name the cyclized sugars. So we call these, according to a molecule, that is not a sugar, but looks somewhat similar. So a pyran molecule does have five carbons in an oxygen, for sure, just like a sugar. And so we call these pyrinoses for a six-membered ring and pyrinoses for a five-membered ring. Okay, so for example, fructose historically is typically represented in a five-membered ring, even though thermodynamics favors an equilibrium, where in nature, it's more common to exist in a six-membered ring configuration. So for each possibility here, you have two different ring sizes and two different stereochemistries. Two times two is four, plus one, the linear forms. There's five forms of these molecules. So you can have alpha, beta pyrinose, alpha, beta pyrinose. So when you look at this, you see these pictures, you have to be aware underneath, okay, there's some more possibilities. Now for glucose, almost exclusively in the six-membered form in the equilibrium. But the fructose pyrinose is mostly in the six-membered form but also in the five-membered form. Okay, so cyclization can be described with new names. So sugars are reactive because they're in viable equilibrium in the linear form. Aldehydes and ketones are pretty reactive. So these aldehydes and ketones could potentially be oxidized. So we could transfer electrons from this sugar molecule to some other molecule. They can act as reducing agents. We'll come back to oxidation reduction a little bit in the beginning of the next lecture. So if you don't see the electrons moving and the naming there, they're going to be a sin. But for now, we can see this is obviously becoming oxidized. It's becoming a carboxylate. Those electrons can be transported to a metal ion. And that oxidation actually can occur, predominantly occurs at C1 and C6. But once you start oxidizing things, you can also get a complex decomposition. So you can start breaking bonds. So in terms of glucose, I said that. So you can indicate that sugars are reactive by adding in a colorless, oxidized form of molecule that becomes reduced and the reduced form is colored. So silver precipitates out and you see a plume of silver in your sugar water. You say, yikes. That's why we need to watch out for sugar in the blood because it's a reactive little molecule. So here are all the different oxidation forms. Glucose is a mess. It's very reactive. And once it starts getting oxidized, it tends to fall apart. And we have all kinds of different forms of this molecule. Gluconic acid is oxidation at carbon 1. Glucurinonic acid is oxidation at carbon 6. And so here are some of the structures of the molecules that are formed. So it's a very reactive molecule. You can, diabetics have problems regulating the amount of free glucose in their blood. Because that thing is a great reducing agent. It's going to start reducing your body, basically. And we can measure the amount of glucose by taking that glucose, oxidizing it to a cyclized form. So once we oxidize at carbon 1, we can cyclize. That's not entirely surprising. And this oxidation could be catalyzed by an enzyme. But this reaction produces H2O2. And at H2O2, we can add a second oxidase, which converts another color's compound to a colored product. So in a sense, as the amount of H2O2 goes up in these glucose tests that you see diabetics using, the color is produced. And so that color, the amount of colors coincident with the concentration of glucose. And the reason the test works is because the molecules react if it tends to get oxidized. Okay, so these, here's some more nomenclature. Gluconate, or gluconic acid, another way to say this, can cyclize here and make a glucono, D-glucono-delta lactone. So a delta nomenclature for the cyclization of these oxidized molecules specifies this is a six-membered ring. And so it's another nomenclature. And so there's all kinds of different cyclization, predominantly occurring at C1 and C6. So oxidation at C6, we can still cyclize forming our hemiasatel, but oxidation at C1 can lead to this new cyclization. And here, this is beta. Why isn't this beta? There's no stereochemistry. It's a carbonyl group. Whereas here, we have stereochemistry, beta-cyst. This is on the same side. This is the highest-numbered carbon atom. So far? Yeah, so sort of like at this point in the course, we take the car from first gear and we're kicking it up into second. But I'll have pauses periodically so we can catch our breath. Nature has created a lot for us to discuss. Any questions? We do have one question from the audience. What's the difference between an epimer and a diastereomer? They can be overlapping venn diagrams. So diastereomer is a more generic term because if you have a more than one chiral center, if you don't have them all inverting, that's a diastereomer. Whereas an epimer, you have to have them all inverting. For an epimer, you can have only one. So epimers are also diastereomers. And we have one more that just came in. Are betas always cis and alphas always trans? Yes, in the way I've defined it. Now, it's not a double bond. Usually you say cis and trans about a double bond, but it has the same. I'm just trying to help you to remember the arbitrary Mickey Mouse definition of these things. Very good. Okay, here we go. So now we're going to make a polymer. We've got some words. Let's make a language. Let's make a polymer. We made polymers of lipids, polymers of amino acids, polymers of sugars. So we take our aldehyde. We've already seen this. We react, not condense this molecule with an alcohol. This is a cyclization in the case of sugars to make a hemiacetal for the albosis and a hemiaketal for the ketosis. But now we can react with another alcohol. Sugar molecules are chock full of alcohols, but here this alcohol is not going to be from the same molecule. It comes from another molecule. And so if we react a hemiacetal with another alcohol, we get an acetal. And so the difference here is you've gone from an alcohol to an ether. And so this is going to be important in glycosidic linkages we'll see in a moment. And so the ketal, what have we done? Converted the hydroxyl group to an ether as well by reacting. Now the important thing here, the reason that we really can exist is this is more stable than this. This is in a relevant equilibrium with the un-cyclized linear form. This is not, okay, so this is not going to just spontaneously equilibrate. Okay, so this is a safer place to put your glucose molecules because you've damped down the reactivity. And so if you want to understand the chemical basis behind why there's differences in reactivity, I've posted a 10-page thing on the internet. It has to do with the types of ketalysis that's involved in the breakdown. Okay, but we're not going to get into that. You just do need to know that acetyls are more stable than hemiacetals, ketels are more stable than hemiacetals. Okay? All right. So the way that we're going to get this reaction to occur is take our hemiacetal and react it with the alcohol on a second sugar residue, forming a disaccharide. So we've looked at monosaccharides, one sugar residue. Now we're going to start looking at disaccharides. So this is a condensation reaction. There's water coming off. Think frosty beverage container. And so this condensation reaction forms an acetal. O-R-O-R-O-H-O-R. See it? And again, it's really important that you find those things. The rule hasn't changed. The animal hair carbon is a carbon with two oxygens bound. That hasn't changed. We just changed from alcohol to an ether. And so here we now have an acetal linkage. We can still specify stereochemistry. You've got to turn your head a bit, right? So this is down, this is up. That would be alpha. We can also specify the carbon atom number involved in this glycosidic bond. That's alpha 1. And then we also would want to specify what the other side is to this O-R group. And that would be carbon 4. So we describe this as alpha 1, 4. This is not alpha 1, beta 4, or alpha 4. The alpha and beta designation are exclusive privilege of the elite club of carbon atoms at the anomeric carbon. Other carbons are not elite enough to have an alpha or a beta. They get a whole different name for their sugar. And so here we've condensed these two molecules together. Is it still reactive? It's less reactive, but still reactive in a weird sort of way. This can open up and be linear even though it's got something attached to its middle bits. So this is still a Hemiacetel. The fact that we've done some chemistry over here does not affect the chemistry on this side. This is still in free equilibrium with the linear form. So we can have muta rotation. This is still a reducing agent. But instead of having two reducing molecules, we now have one. So as we begin to polymerize, every time we add a glycosidic linkage, at least one anomeric carbon is going to be involved. And as each anomeric carbon is involved in a glycosidic linkage, the reactivity of that anomeric carbon is eliminated. So we can attach these alpha 1, 4, and we can also flip this around and attach this alpha 1, beta 1. In this case, for maltose, we're very confusing. There are different kinds of different names for different combinations of sugars. So this is maltose. That's alpha D glucopyrinosil 1, 4D glucopyrinose. So we have two glucose molecules, different anomers of glucose molecules, condensing to form this disaccharide. You have to have a naming convention. So in proteins, naming conditions, you start the naming at the end terminus. For sugars, the naming convention at the sugar residue most distant from the hemiacetal or hemiketal. So if you can't find them, you can't name them. So here, for example, this is alpha D glucopyrinosil 1, 4D glucopyrinose, not vice versa. Right? It's a convention. Let's give some more examples. Let's practice some of our nomenclature. So here we have beta. This is up at 5th. Here we have down, up. Oh, bad, bad, bad. Here we have up, up. So these are both beta. We've connected an anomeric carbon to a not anomeric carbon. Still have one reducing in. This is lactose. And so we call this beta D galactopyrinosil, beta D glucopyrinose, not vice versa. So here we've attached two anomeric carbons together. So we can't name the molecule from the sugar residue most distant from the anomeric carbon or from the hemiacetal or hemiketal because there isn't one. So in the case of the condensation of two anomeric carbons, we can name the molecule in either direction, right? Just to be entertaining. And so here we have, for example, trellos. The condensation of alpha one with a glucose with alpha one glucose. And so this is not reactive at all. It's all acetals, man. Here we have, here we have not reactive at all as well, right? And so these sugars also have a special name that they're not reducing. In other words, if they don't have a hemiacetal or hemiketal, if they're not in free equilibrium with the linear formant and aldose or ketos, then we call them tyrannocides or furanocides. And that name is telling you something about the orientation of the molecules. But so is this designation. So we're specifying the stereochemistry, the anomeric stereochemistry, only for anomeric centers. So this is alpha one, one alpha. Okay, but these, you can go either direction. And this one is reducing because we call this beta one four, not beta one, beta four, just beta one four, because your anomeric carbon is still a hemiacetal. So if you're like drowning in this chemistry at this point, I think it's just helpful to go over this. But it's important for what's about to happen. Okay, so we've gone through, here's some more examples. You can go through these on your own time. I'll let you do that. Any questions so far on naming? We've done disaccharide. What's next? Tri-saccharides and so forth. So we're getting to the polysaccharides, polymers. Proteins are polymers of amino acids. Polysaccharides are polymers of sugar residues. And we can have what's called a homo-polysaccharide. And that's when you have just one type of sugar residue. So half the carbon atoms on Earth are in a glucose, in a polysaccharide of glucose, a homo-polysaccharide of glucose called starch. And so hetero-polysaccharide is when you have an odd assortment of different types of sugar residues. You might have some glucose, some other sugar residues. And these linkages between the residues can be alpha-1,4 or alpha-1,6. And later after today's lecture, you can go home and think, why might that be the preference? Why might there not be other attachments? Think about spacing things out. So any place we have an alpha-1,6 linkage, we've condensed a reducing in with an alcohol group on the sugar residue. And we'll see that it's helpful to actually look at the molecule. So this is just nomenclature homo-polysaccharides and hetero-polysaccharides. So homo-polysaccharides of glucose are called starches, glycogen, and cellulose. So starches occur very commonly in plants. You have alpha-1 link homo-polysaccharides to amylose. So they're just linear. But then amylopectin is alpha-1,4 and alpha-1,6, with the alpha-1,6 occurring at each of the branch points. We'll see that in a moment. Glycogen is similar to amylopectin, but it happens in a different type of organism where you have branching. And this glycogen tends to be a little bit more branched, and we'll think about why in a moment. Cellulose has a different stereochemistry in this linkage. And the difference in the stereochemistry gives a different fold to the sugar. So we thought about the secondary structure of proteins. We have some structure here as well in sugars. We can also attach these polysaccharides to proteins. And we'll see some examples of that in a moment. Okay, so here is starch. Do you see it? Alpha-1,4, alpha-1,4, alpha-1,4. They're all glucose molecules. In other words, look at the position of all the substituents to invariant amongst the repeating units here. And as it turns out, these things are helical. And it's not because of hydrogen bonding. Helicies and proteins are driven by hydrogen bonding. Here we have rigid structures. We have cyclized sugars. And because it's alpha, that confers a certain angle between each sugar residue. And so we have helical segments. But then when we have a branch point, we have alpha-1,4 length here. But on this same sugar residue, we have alpha-1,6. So reducing in here, yeah, reducing in is attached to carbon-6. These are nice and spaced out. And so the branch points mean that we have these parallel helices. So if it's all just alpha-1,4, that just makes a nice, pretty helix. Every time you have alpha-1,6, you branch and you branch. And there's a lot of engineering that has gone through and through evolution to create this. And so here is the angle I was talking about between the two planes of neighboring sugar residues. Do you see how each one has a turn? So these things start to orbit each other, but it's much more opened up than the alpha-helices we're used to with proteins. It's not really hydrogen bonds holding this together. But the other structure, okay, so you can think about the engineering, the miraculous engineering that nature has done here. What's the advantage here, can you guys imagine, of storing these glucose as a polymer? So say, why is it better to have one polymer molecule compared to a thousand monomer molecule? So it's less reactive and? Oh, who said that? You rock, osmolarity, right? So if you have a thousand molecules, what is a mole per liter, right, molarity? So if you decrease the number of parts, you decrease the concentration. And solubility is, if we're going to store a bunch of glucose, solubility is a legitimate concern. By making a polymer, you decrease the osmolarity, the number of parts of the molecule. Decrease the concentration as well of glucose. So that's an important thing to think about, okay? Four questions. What's so good about having one reducing end from what we've learned so far? So we have a thousand monosaccharides compared to one reducing end containing polysaccharide. Why might that be good? What's happening at that one sugar residue at the reducing end? I guess another legitimate question is, what is the reducing end? Reducing end is the end that has the ability to be a reducing agent. It has a propensity to be oxidized. That would be the only place in this polymer where you have a Hemiacetel or Hemiaketel. So this huge molecule, you went to, from thousands of Hemiacetels and Hemiaketels to just one. The rest are these very stable acetels and ketels. That's good stuff. In the case of glucose, it's all acetels and Hemiacetels. Okay, with me so far? Okay, cool. This one is a little bit, you have to think, sort of in terms of extraction of energy. This is a storage form of energy. So why might it be good to have all these questions very commonly end up on exams? So it's good to discuss these together. I'm not guaranteeing. So why is it good to have so many ends in terms of rapid extraction of sugar residues from the polymer storage form? So you might imagine the assembly and the disassembly. Yes, you can come visit me soon. Oh, that's weird. Someone came up in the iPad. So what's good about this? So if we're going to store molecules, we need enzymes to make these glycosidic linkages. The enzyme, this is not spontaneous. You don't just put things together and have starch fall out of the test tube. You have enzymes put together, enzymes tearing them apart. If you have lots of these non-reducing ends, you can simultaneously extract sugar residues from each of the ends at the same time. You just put one enzyme molecule at each of the ends and start chipping away. And so because there's so many reducing ends, it's good to have this branching because you can more rapidly provide bursts of glucose. Glucose is what we need to make energy. We're going to spend next lecture burning some glucose and capturing the energy. We need to get it out of this polymer. That one's less intuitive, I think. But think about the other possible stereochemistry, the beta 1,4 linkage. And the beta 1,4 linkage, the angle between these two planes is completely different. Instead of having a bend and entering a helix, here we have a planar structure, rife with hydrogen bonds. You might predict one of the properties of this beta 1,4-linked homopolysaccharide of glucose is that it would be very strong and rigid. So if you go outside today, look at the trees. Those are sugar sticks. They're cellulose. They have beta 1,4. All those trees, the majority of what you're seeing there is just a sugar stick of beta 1,4-linked glucose. And they don't flop over because it's beta 1,4. If they had alpha configuration, they would flop over. So we need strength when we're building a tree and other things that need to be rigid. So stereochemistry matters. That's why you've got to be able to specify it and describe it. Here's another beta 1,4-linked polysaccharide. And in this one, we have some elaborate decoration of different positions within the sugar residues. And so did you know that losters are sugar-coated? When you look at them, you think, that must be a rock. It seems like it's hard, right? If you put your finger in that claw, bad things would happen. That's hard stuff. But it's just sugar. It's beta 1,4-linked. If it was alpha 1,4-linked, you could put your finger in there and nothing bad would happen. Beta 1,4 turns it into a rock. I mean, you hit the thing on the table. It's like making a lot of noises there. So beta 1,4-linked stereochemistry affects the physical properties of the polymer. And it's sort of funny too. And even in your gut right now, your gut's full of beta 1,4-linked cell wall around E. coli cells, right? Sort of gross. So these cell walls are rife. This needs to be an impermeable barrier. These E. coli bacteria, they're undergoing warfare here. So they need to have a defensive armor. So they have beta 1,4-linked heteropolysaccharides. And they need this even to be even stronger. They can't just be a tree trunk. They're going to have a real cell wall here. So they've attached these beta 1,4-linked heteropolysaccharides by polypeptide cross-links. And so some amino acids have amino groups or hydroxyl groups on their side chains that could react with sugar residues and make these cross-links. Look how crafty these guys are. So these peptides that are attaching these heteropolysaccharides have L amino acids. But they also have D amino acids. Does that violate the law of physics? Why would they want to evolve such a bizarre orientation? How good of a substrate would these peptides be for the assembly of proteases in our cells? Wouldn't be a substrate at all. Stereo... Remember that binding pocket and kinetripsin? If that thing were on the other side, the whole thing wouldn't work. So these have evolved bizarre, weird, unnatural stereo configurations around the alpha carbon to evade destruction by us. I mean, they're battling us. Going to war. Okay. All right. So we can think of the combination of proteins and polysaccharides. And so proteoglycans are where you have a large amount of polysaccharides. Potentially tens of thousands of residues of polysaccharides are attached onto a relatively small protein. And you can imagine, wow, if it had so many sugar residues, it'd be sort of wet. Sugar residues are wet. They have more hydroxyl groups than amino acid side chains. So it would be a mucous-like consistency. So it's various parts of your body. It's important to lubricate. You know, you don't want to be doing this and have intense amount of pain. You need to have lubricants. And those are proteoglycans with a high amount of polysaccharide. It's highly hydrated because of all the hydroxyl groups there. And so that's one class of proteins attached to polysaccharides. The other are these glycoproteins. And you might say, okay, when does it cross over, man? When does it become, go from the last thing to a glycoprotein? It's not been defined. It's just sort of, people say, that's one thing, that's the other. In general, glycoproteins have a little bit less sugar residues. They don't have tens of thousands of sugar residues per protein molecule. And sometimes it can be very simple. So here's an immunoglobulin. You probably recognize the Y-shaped structure. It just has a simple carbohydrate, covalently bonded to one of the amino acids. But here we have another protein that has a little bit more complex polysaccharide decoration. And these types of glycosylations are very important to any protein ever exposed to the outside world. When you send Johnny protein out into the world, you want it to be able to interact with other molecules. And so if you put a bunch of sugar residues all over that protein, you create a very unique type of molecule. Amino acids only provide a little bit of diversity. But now you're adding a whole world, a very elaborate types of polysaccharides that can also be there. And so these are important in the binding of molecules outside of the cell to other proteins, for example. So let's look at how these links happen. So we can have alcohol containing groups, or we have an amino group here, and we can make a new linkage, right? That's an anomeric carbon, right, at this case, right? So here we have serine amino acid. It can be serine or serine. A spherigine. This is called olinked. This is called inlinked. The oxygen or nitrogen is involved in the actual bond of the side chain, so the sugar. Here you can have, when we have these colored representations, each color and shape is a different type of residue. There's no standardized assignment. If blue is one sugar and yellow is a different. But you can have different degrees of branching and so forth. So this is a very, very complex set of modifications. And as I mentioned, these are important. And the proteins that tend to spend time away from home, either out on the cell surface or actually released from the cell. And so that's a very, you know, polar environment. And also, it's an environment where things need to meet each other with specificity. And so for example, a T cell, you know, needs to have a binding event between some other cell that's in trouble. So the cell is saying, I'm in trouble, puts an assortment of polysaccharide decorated molecules on its surface and the T cell says, okay, I received the trouble signal. So this is just a very general description of the ways in which we can make covalent attachments of amino acids and polysaccharides. So I want to get to this and this is setting us up for the next lecture. So we need to think of thermodynamics to complete a change of gear and that's why I have a firefly. Okay. So here we're going to think about this. Standard change in free energy. Remember the definition is a one molar concentration of everything except for protons, 10 to the minus seven molar, and water molecules. So in the standard state, we're not talking about one molar water. We're talking about water concentration, like 55 molar water. I didn't mention that last time. Yeah, so if you're chemically transforming something, you can describe how far the standard state is from the equilibrium state. This term describes this arbitrary just one molar of everything, whatever, and this is the equilibrium state. How far is that? So if delta G prime knot is zero, the standard state is the equilibrium state. If you start with the one molar concentration of everything, the other assumptions, you're at equilibrium. And so this equilibrium constant is defined as products of a reactant. So as the concentration of products goes up, this goes down, or goes up. Delta G becomes more positive, and if the concentration of reactants is higher, it's more exergonic. It's more negative, I should say. And so we've already covered that, but let's take it one step further. Would you imagine that every single enzyme in your body is experiencing a one molar concentration of products and reactants? Possible. Highly unlikely. It depends on how many significant digits you might use. Delta G describes the true cellular state, the physiological state. Physiological state, contrived standard state, equilibrium state. So at equilibrium in the cellular state, Delta G is zero. So that concentration... Delta G is zero. And as this term becomes more negative, Delta G is decreased. So say you have something that is endergonic starting from the standard state. You can manipulate the concentration of products. You can withdraw products to push the reaction forward. Delta G would be more negative than Delta G prime naught if you're removing products, decreasing the concentration of C and D. This might be a review. A very important concept we're going to see throughout this semester is this idea of coupling reactions. Take a complex transformation, break it into pieces parts. Perhaps one step is tough, the other step is easier, and when you combine the two, it's not spontaneous. So what does that mean? So you have A becomes B, a standard change of free energy. Pretty easy to tabulate. You can look up in a book. B goes to C. That has another change in free energy. If you want to consider the transformation of A to C, you just sum these two free energies together. So if this one has a positive Delta G prime naught, this has a negative. If the magnitude of the negative Delta G is larger than the positive Delta G, it's going to be not spontaneous. It's going to tend to proceed towards reaction starting from the standard state. So usually when we think about the pathways, all the numbers are going to see a Delta G prime naught. We're looking at the way the sequence of enzymes is engineered to manipulate the concentrations, to change Delta G. So in a couple of reactions together, here's a perfect example, phosphorylation of glucose. So you can add a phosphate to glucose and you can hydrolyze ATP. So these two reactions, one is exergonic, pretty highly exergonic. One is endergonic. I don't want to phosphate. So this is endergonic. You add these two together. It's not spontaneous starting from the standard state. So by coupling them together, you can cause something that's endergonic to occur spontaneously. In some cases, these two reactions can share a common intermediate, but this transfer of energy can also occur through a conformational change in the protein. So sometimes the ATP hydrolysis winds the protein up, causing it to have the energy necessary to push an endergonic process forward. In many cases, these things are coupled by sharing a molecule. So here's an endergonic process, a highly exergonic process. This is good news. It's spontaneous from the standard state. Also, this highly exergonic process, we need to regulate it. And so this activation barrier prevents the spontaneous release of energy. If we put all our energy in ATP, and it just spontaneously fell apart, that would be bad. This little activation barrier and an enzyme lowers that to help the rate of equilibrium achievement to be accelerated. ATP is the most important molecule of the semester. It has an adenine, a ribose, phosphoester bond to a phosphate. That's an adenosine residue. So this is called AMP. Add another one ADP, triphosphate, or diphosphate. Add another one, triphosphate. So this is phosphoester bond, and then these are phosphoanhydride bonds. Remember, carbon-containing anhydride is sort of similar structures to what we have here. We have a lot of energy in this molecule. It has a whole lot of charge. That's really stressful. You have four negative charges. This molecule has a lot of energy pent up inside. We can tame some of that energy by chelating it to magnesium ions. But why is the hydrolysis of a phosphate from ATP so exergonic? The reason is, and this is a very important site, come back to this when you're studying, hydrolysis is exergonic for three different reasons. One is that there's more resonance in the products than the reactants. Here we've got a little bit of resonance out here, but when we clip the phosphate off through hydrolysis, we have more full resonance. That is thermodynamically favorable. One of the reactants is a water molecule. At the standard state, it's 55 molar. That is going to help to push the reaction forward. Then, obviously, as you clip off a phosphate, what have you done? You've taken four charges on a tiny little molecule and reduced it by one. Three factors come together to make delta G prime naught pretty exergonic, 30 kilojoules per mole. Starting from here, we can see the general hydrolysis reactions are very exergonic. Some hydrolysis reactions are more exergonic than the hydrolysis of phosphate from ATP. For example, this one, you have resonance in both of the products of this reaction. You have increased resonance, whereas with hydrolysis of phosphate from ATP, the majority of gain in resonance is just one of the two products. In general, high-concentration water is making these pretty exergonic, but you can tabulate at delta G prime naught of any hydrolysis reaction. We'll see this throughout the semester, these types of numbers. Starting from the cellular state, the hydrolysis of ATP is even more exergonic than starting from the standard state. The cellular state does not have a one molar concentration of each of the products in the reactants. ATP reactant products is ADP. There's a higher amount of the reactants in the product. Delta G is actually more negative. Delta G is actually from negative 50 to negative 65 kilojoules per mole, depending on the type of cell. If these were one-to-one ratio, then delta G would be a negative 30.5. We're manipulating these concentrations. We have pretty high concentrations of ATP relative to ADP to make this an even better provider of energy in the cell. Just one more. We can also describe the rearrangement of molecules. Every reaction you see for the rest of the semester will typically have a standard change in free energy. This makes sense. Here we're just rearranging. It might not take that much more energy compared to adding a nasty phosphate. Look at this. This is the whole topic of the next lecture. The burning of trees and glucose to greenhouse gases releases a flash of energy. If you set a tree on fire, you can roast marshmallows. It's hot. Do you realize in your body now that you're actually burning glucose? The same exact reaction. But instead of a run-of-way fire, it's happening in a very stepwise process. You're receiving quantum energy out of the glucose molecule, in a stepwise oxidation. In capturing that energy, instead of making heat, you're making ATP. This is it. We have a clicker. Everybody done? Everybody done? No. I'm going to call it. Everybody voted? Did you turn the clicker off? Okay. This one is plug-in chug, right? What's the answer? You guys are awesome. I have office hours now, and I tend to get lonely. Feel free to... I went for it a little bit faster than I had today. Please stop. I'd love to talk to you.