 So my goal in the next step is to understand if protein folding is indeed a phase transition if the entire small domain has to fold at once. To understand this we're going to take you back to the structural transitions we looked at. That was not an excursion into physics that we completely forgot about later, but if you might remember that I actually never proved that anything beyond the beta sheet was a proper phase transition. Now we're going to try to do that. You remember that I showed you a couple of plots. We for instance looked at energy as a function of temperature where we had a temperature one and a temperature two and there was an increase in energy over a relatively short span here and that related to at least two different processes as we can draw this as a function of temperature or energy or state is really states we're looking If you started out in state one and I ended up in state two for a continuous process if I stopped that halfway through I would literally be in the middle, but for an all or non-transition I would start in one and up in two but halfway between them I would have half the population in each. We don't really know that protein folding is this type of process and if it is that it has to involve the entire domain. How do we prove that? Well it's not entirely easy because I don't really have a whole lot to work with, but I might be able to use this plot. So there's going to be a fairly narrow span here that if I'm increasing the temperature and if I'm really denaturing unfolding the protein with temperature here I should go from roughly 0% to 100% and I to the first approximation I could estimate this with the linear process then I know how much energy I had to add to get the protein to unfold. I can also compare that a bit with the energy required in the lab. It's another one of these simple devices called the calorimeter which is really just a large well insulated box where I can measure for instance how much heat it requires to heat the sample one degree. Then I can change how the specific heat changes when we undergo a phase transition and again because it's isolated I know exactly how much energy I had to put into the box and I know the concentration that is how many molecules I had in the box and by putting these roughly equal to each other and also using the Boltzmann distribution I will be able to get there. Why is the specific heat interesting? Well the specific heat literally measures how much energy it requires to increase the temperature of the substance. So normally this will go up if I have a protein it will go up weekly. The reason why this goes up weekly is that I'm increasing the temperature the hydrophobic effect becomes a little bit worse all the time as we've seen before right so that if I move from 30 to 40 to 50 degrees in general it will be harder and harder to heat the sample but this is not a significant change cp but then something happens. At some point I will start to go up fairly abruptly and there will be a peak over a fairly narrow span and then I will go down but I will not go down all the way to the original curve but I will start out with some sort of curve that's parallel but slightly higher. Depending on the conditions such as pH or salt concentration or something this curve can appear earlier and in that case I might start to deviate already here there will be a curve and then I'm going to be down on that slope again. The transition we're seeing here that is the actual phase transition when I'm unfolding the protein. The difference between these two slopes for proteins that would correspond to that we've exposed many more hydrophobic groups to the solvent and with all these groups exposed it's going to be more difficult to heat the system compared to the well-ordered one but this also means that I can calculate how much heat I'm putting into the system by using this type of machine. So let's try to do that and I'm going to use both that curve and cp as a concept but before we do that we're going to need to have something to check what actually happens in the transition and that's going to be the Boltzmann distribution looking at before and after. Let me magically erase this. So what I'm going to try to derive is something called the Fanthoff criterion. Fanthoff which says that if the melting unit or folding unit it's the same in opposite directions if that is roughly the same as the size of the entire domain that means that it's a cooperative all or none process a phase transition. If the melting unit is smaller then it's a gradual transition and if the melting unit is larger then it would have to be some amyloid like aggregate or so we're not going to go there. Right now I don't have much to work with so I'm going to start to see what I can do just by looking at say an unfolded state and a folded state. So normally I'm going to start by unmelting things here so we start from the native one and before I draw the Boltzmann distribution here we're going to need a couple of definitions so in the native state I will just say that I have energy E and entropy S and then in an unfolded or molten state let's call them E prime and S prime. You can use any notion you want and that of course means you can also define differences between these and then I just write down the Boltzmann distribution. I'm also going to normalize this with the partition function that's trivial there are just two states so the probability of being molten large down here that's going to use the terms from the second expression right E raised to minus E prime minus TS prime divided by kT and to normalize this I will now need both the native and the molten state E raised to minus E prime minus TS prime divided by kT same as above plus the term without primes E minus TS divided by kT. I know this looks horrible actually it's not horrible at all. To simplify this I'm not interested in the absolute numbers here I'm interested in the difference so I'm going to so I'm going to take and divide everything by this number that means that the nominator becomes one this term becomes one this term does not become one that would admit a bit too easy so here I'm going to hand up with expressions that E and then divide everything that's going to be differences in the exponent so here I'm going to have E minus E prime well let's make our life simple that just means we have delta E minus T delta S divided by kT right much simpler a nicer expression this is something that's a function of the temperature and everything and also the difference in energy between these two states the difference in energy between the two states is not going to vary so I will typically be looking at this as a function of temperature I'm going to continue on that expression in a second but I'm going to need to compare this to what we had on the previous slide first that I don't have a simple way to calculate this rather I have a way to calculate it but it depends on the temperature and I would need a way where I could somehow compare this to what I measure in the experiment but I kind of had that on the last slide remember this curve when I had energy as a function of temperature right and in a relatively short region things went from roughly 0% to 100% folded that means that over this range 0% folded and 100% folded that's kind of the same thing as the probability right so this means that going from 0 to 100% in that range that's kind of the derivative so I know here again I'm going to come back to that later here I know that dp dt is roughly equals to delta probability divided by delta temperature but the probability in this area goes from between zero and one and that really means it's roughly equals to one over the temperature gap there surprisingly simple right so that is what I could hope to get from an experiment that means I should try to calculate that same concept in this equation and see what I can I can work with this so I'm going to now I will use the equation there and calculate dp molten dt well that's that expression it's not that horrible because those two t's cancel and that means I'm going to have a delta e divided by t there it gets a bit ugly so I'm not going to write down every single step but you could do this if you had 10 minutes this is going to be one over one plus e raised to minus delta e minus t delta s divided by kt that entire term squared multiplied by e raised to minus delta e minus t delta s by kt these are just the chain terms coming out and then eventually the inner derivative here is going to be a minus delta e divided by kt squared that's it but if we look at this this is p molten squared right and it turns out that you can actually write the other term we're multiplying here as try to take one minus that expression so what this will actually simplify actually it's not simplified I can just write it in a slightly different way this corresponds to p molten multiplied by one minus p molten multiplied by delta e divided by kt squared if you don't believe me take that expression or that expression in particular enter in that equation and you will see that it matches that one but it's a bit of mathematics but the point is that this is now a closed expression that depends on p molten the delta energy and I will be able to get some of these from the experiments what is this well we're going to need to simplify a bit of course this term will vary as I'm going through the change here right here p molten is going to be zero and here p molten is 100 percent but let's pick something simple let's pick the midpoint here in the midpoint p molten is 0.5 one minus p molten is also 0.5 so at the midpoint the derivative at midpoint dp dt is equal to delta e divided by 4k t squared 0.5 multiplied by 0.5 is 0.25 but I also know that dp dt is equal one over delta t so those two terms are the same in fact that is how we derived that very narrow temperature gap for water melting but a much wider gap for proteins melting but I didn't tell you about that in lecture two was a bit a bit too much math then but if we now put these equal and use just use measurements here and compare to the energy we will be able to solve this let me erase it so what I carried over from the previous slide was just that from the equation we got that dp dt at the midpoint temperature t0 added a zero there was delta e divided by 4k t0 where delta e was a change in energy from the transition and we also had the estimate that that should correspond to one over delta t if I put these two as equal I get that delta e equals 4k t0 squared divided by delta t for some sort of melting unit meaning that if I know the time span sorry the temperature gap over which it melts and the temperature at it melts I know the amount of energy the temperature gap I could get from that simple experiment where I checked the width of the curve for this specific heat right in the calorimeter but now I need to know delta e well delta e I also get from the calorimeter so in the calorimeter delta e is equal delta h the amount of energy I'm putting in divided by n that is the amount per actual domain for instance and now I just need to compare it to that number and it turns out that they match not perfectly not always but in general they do and that means that protein folding it is a phase transition for the entire domain involved again these small units that we've talked about it will be complicated for even larger aggregates but it means that all the physics we learned in particular all the properties of phase transitions I took you through in the first few lectures they apply and that's of course why we did it