 To study a particular process it's instructive to pick some sort of state that we can characterize experimentally. And an obvious such state is that when we have the equilibrium midpoint here that the helix does not want to elongate anymore. It doesn't want to shrink and it doesn't want to elongate. At that moment we have delta G for elongations equals zero because if it was positive or negative it would like to go in one of the directions. We already know that this was F8 bond minus Ts alpha. So that should be zero. So that the energy for that hydrogen bond at that point corresponds to temperature times the entropy of putting a residue in that state. We're also going to need to consider the entropy here. How many places can I put that helix? In principle that means I could put the start somewhere n and I could put the end anywhere. That's also n positions so that would be uppercase n square if I have total n uppercase n residue here. But the end has to come before it starts. In principle it's n square divided by 2. Well we're only going to be working with order of magnitude estimates here so to make life slightly easier I'm going to skip that factor of 2. It's not as bad as you might think because it will only show up inside the logarithm not outside it. So that the number of states would be n2 divided by 2 formation places. So that would mean that the entropy for the helix if I skip that factor 2 to make my life easier that would then be roughly 2 for the 2 in the exponent 2 ln uppercase n sorry Boltzmann's constant 2 k ln n. And then if I then want to look at the midpoint of this transition when I see no net change in helix I'm not initiating any more helices either and I don't see any helices disappearing. Then I also know that the total delta g should be zero. So delta g total is zero that would mean that f in it minus the entropy 2 k t ln n should be zero. And based on this I can start to draw some conclusions here right that the full mixing here I would have that maybe say half the residues or so uppercase let's say if I say n lowercase n residues are helical. If I say in the ballpark of 50 percent of them but you know 50 percent that's just another factor of 2. So I would say that that is in the same ballpark as uppercase n. And in that case I can actually solve for it there and that would means that the typical length of these residues would be roughly e raised to plus f in it divided by 2 k t. Very rough ballpark estimate and I don't care about the factors of 2. So I just solved the equation here but replaced uppercase n with lowercase n. But this number I can get from CD spectroscopy. If I have a chain and that I know that roughly 50 percent of them are helical at a particular midpoint I can solve for that. The book actually mentioned that it is possible to show that it's not 50 percent but it should be 12 percent but again ballparks. But if I know n I can calculate a finite. So from that simple CD spectroscopy experiment and a very simple say pure polyalanine helix I can estimate what the initiation free energy for an alpha helix is. And it turns out that it's going to be roughly in the ballpark of say 2 hydrogen bonds or so maybe 4k cal. So it's a relatively low energy. 4k cal remember that we should compare it to 0.6k cal which is k t. So that's ballpark maybe of e raised to the power of minus 6 is the probability of forming that. And that's the barrier that we can then use to describe how quickly things will happen in terms of kinetics. It turns out that helix is formed quickly not only for the form quickly therefore very very quickly. I'll take you through that on the next screen. The other thing we can say about elongation. The elongation if we calculate that the same way the elongation might end up being roughly half of this for a typical residues and maybe 1 or 2k cal but then negative. So helices form quickly because the free energy barriers and for that matter the free energy advantage of being helix are relatively close to delta g equals zero. And that means that they will be stable but they will also form in reasonably short times.