 So to understand beta sheets, we're gonna need to focus on the anti-panel case. Let me draw a couple of alternatives for you. Starting out from a long coil, right? The first thing we can do is form what I call a single hairpin. We have a well-defined turn up there and then a number of hydrogen bonds there. Leave a little bit of room there. I can also imagine having three chains or individuals beta strands. And then I have a well-defined turn there and a well-defined turn. And eventually, I will get to something that has three turns and so on. And I'm mostly gonna be interested in these long structures. Now, between these two, as an example, I'm gonna need to have something that goes up, has formed one, and then it has just taken a turn. But I haven't formed any new hydrogen bonds in the second strand yet. So that's gonna be some sort of transition state that I will talk about. This is an example where you need to sit down and define things with paper and pen. And if we do things in smart ways, it's gonna be slightly easier. Let's put some energy numbers here. I already talked about turns. I say that the turn has to be bad so that we'll say that U is the turn energy, free energy, but I'm gonna need to save some writing here. We know that that has to be larger than zero. Otherwise, it would be advantageous to just keep making turns and that would be absurd. Then I'm gonna need to put some numbers here. In particular, if I look at very large beta sheets, just to simplify things in my head, I'll say that there is some sort of free energy beta that is the free energy of a residue to be in a beta sheet compared to being in a coil. In principle, I should have deltas here everywhere, but sorry, it becomes too cluttered in writing, so I need to simplify it. Can we say something about F beta? Is it smaller or greater than zero? Well, if this was greater than zero, it would never be advantageous to have large sheets. It would prefer to be in coil. So we know that F beta must be smaller than zero, at least on average. But then we have this first and the last strand, and here there is some sort of extra effect, right? They can't form as many hydrogen bonds, well, they form that with water, but not inside the sheet. So there will be some sort of extra surface effect. I could call that something completely different, but since these residues, they have the same entropy that I had to put them in a beta sheet. So I'm gonna call that F beta plus delta F beta. For some sort of change that it's on the surface. For now we don't know whether that whole number is positive or negative. So let's reason a little bit about that. One possibility is that that's negative, and that means that it's gonna be good. But in that case, if I just take a single hairpin and make it long enough, that would be stable. And you know what? That happens now in them. It's a very simple and quick forming secondary structure. You've seen it in some of the structures I showed you and you will see it in the future. You can easily make that derivation and it's gonna look roughly like an alpha helix. So in the interest of time, I won't show you that. But if that was the only case we ever had, there would be no point never going to more than a hairpin. And we do know that nature occasionally does form larger structures. And that's because we need a large structure here to make it more stable, even though we're losing entropy. So maybe the more interesting case is when this is positive. Meaning that the single hairpin is still not stable. Something around here is gonna be stable. And I would argue that by the time we get, I'm gonna look at the case when it's stable by the time we're adding a third strand. You could of course imagine that even the three one is still stable and it's not much more complicated. But it turns out that it's not gonna change things by an order of magnitude. And then I argue, and here's why you might have to think a little bit about it that in that case, the transition state is gonna be that one. Why do I talk about transition states? Well, remember your kinetics. So what the kinetics told us is that first, there might be more than one way to form a beta sheet. And in that case, what I'm interested in, then we have multiple parallel opinions, and I'm only interested in the lowest free energy pathway because that's gonna dominate the motion. The book, after having shown what the free energy will go through some detail to prove that there is no reasonable lower free energy path, I'm gonna skip that in the interest of time. So then we're assuming that we're gonna follow this particular path. Along this particular path, what's gonna determine the kinetics is gonna be the least favorable free energy, the barrier. So I need to find what is the worst possible state along the path. This one was still not stable, but because the turn itself is also bad, this is gonna be even worse because here I'm also paying the energy of a turn. But by the time I start forming hydrogen bonds in the third strand here, then I'm going downhill. And then I'm past the barrier. There's one more complication though that this will depend on the length of the strand because if the length here would just be one residue, I would have one residue turn, one residue turn, one residue turn, and that's not gonna be stable either. So the turns are bad and I need to gain enough on the inside here to balance that. So the first thing we're gonna need to assess is how long does the hairpin have to be at least for this to happen? Well, how do we determine that? Well, that would be the case if I have this two strands paired up. And if I keep adding a third strand, for each strand I add, I'm adding n residues here. If that ends up being a plus minus zero effect, then I'm exactly balanced. If the hairpin is shorter than that, it's gonna be bad because I don't have enough energy to compensate the turns. And if it's longer than that, it's gonna be good because I gain more than I pay for in the turn. So for a single strand, I would have delta G or hairpin. That would be roughly U plus, that's the turn, plus the number of residues in each strand multiplied by F beta plus delta F beta because all of them are at the surface, right? But if I add one strand to this, delta G, let's say, call this add one, then I'm adding one turn and I'm adding n residues on the surface. But when I'm adding those on the surface, I'm taking the previous strand and turning that into the inside. So that's just gonna be F beta, right? Do the math, if you don't trust me. So the free energy change for adding one strand is U plus n F beta. And I'm interested in the smallest number of residues this can happen for is when this is zero. So then I say zero equals U plus n min multiplied by F beta. Or if I just solve for n min, that means that n min equals minus U divided by F beta. So that the smallest number of residues I can have on the height here is gonna depend on the energy for the turn divided by the stability of being on the inside of a beta sheet. But let's then look at the specific transition state. So delta F hash equals what? Well, that's almost the same here, but now I have two turns plus two number, the number residues, but that was n min. This is why we needed it. And then multiply that by F beta plus delta F beta. Now I can take this expression for n min. Enter that here, n min is U divided by F beta. But you see the first term there, it says F beta. So that U is gonna cancel that U. Pretty cool. So delta F beta is just gonna be equals to minus two U delta F beta divided by F beta. That term remains, right? And there was a minus sign. So minus two U F beta, delta F beta divided by F beta. This might not make you a whole lot wiser now, but this is pretty cool because we know that the free energy barrier here, remember for the alpha helix, there I said there was roughly always two or four K cal. And here we have a free energy barrier that depends on the turn. I'm not gonna worry about the turn energy, but it's something that also depends on the surface energy. We're not gonna worry about that either. But in the denominator here, we have something that will depend on the specific stability of a residue to be on the inside of a beta sheet. And this will eventually turn up in an exponent. So let's look at what that's gonna do. I'm gonna need to erase this. So give me two seconds.