in this video, we're going to begin our discussion on the Hill plot. So we already know from our previous lesson videos that the Hill plot is actually a linear graph that plots the Y value and the X value of the hill equation, respectively on the Y axis and the X axis of the hill plot. And so notice down below we have a little refresher of the Hill equation, which we know resembles the equation of a line allowing us to obtain the hill plot, which is a linear graph. And so all we need to do is take the Y value of the Hill equation, which is the log of theta over one minus data and plot that on the y axis of the hill plot. And then we can take the X value of the Hill equation, which is the log of the concentration of login and plot that onto the x axis of the hill plot. And so notice that the slope of the line on the hill plot is actually going to denote the hill constant and H So we could say that the slope is equal to N H. And so we can see that down below because we know that the variable M is going to represent the slope in the equation of a line. And the M corresponds with the NH in the hill equation showing again that the slope is going to equal the hill constant NH, which we know denotes the degree of ligand binding site interactions. And so ultimately, it's the slope of the line and H that's going to determine the degree of lining binding site interactions. And so notice down below in our hill plot, we're not actually showing you guys any line here. We're just showing you guys a blink canvas, if you will. However, moving forward, we will be able to show you guys some lines on these help. Lots. Now what I want you guys to recall from our previous lesson videos is that both myoglobin and hemoglobin, uh, lie again is going to be oxygen gas. And of course, the concentration of oxygen gas can be expressed with the partial pressure of oxygen or the P 02 which means that we could take the log of the concentration of ligand and replace it with log of the partial pressure of oxygen so we can take this log of the concentration of Lagan and replace it with log of the partial pressure of oxygen. And that's why we have the log of the partial pressure of oxygen here on this X axis. Now what I want you guys to note is that as we go up on this y axis, uh, the ligand binding is increasing in this direction. And so the further we are up on this y axes, the mawr lie again is bound to the protein and the further we are at the bottom, the less lie again is bound to the protein. And then with this X X is over here, the partial pressure of oxygen is increasing from left to right. So on the left of the X axis, we have low partial pressure of oxygen. And on the right of the X axis, we have high partial pressures of oxygen. And so I also want you guys to know is that on a hill plot, the X intercept is always going to reveal where the value of data is equal to 0.5 and so recall from our previous lesson videos that the X intercept is always gonna be the X value when the Y value is equal to zero. And so notice that the why value again here is gonna be the log of this ratio of data over one minus data. And so the why value is gonna equal zero right here at this point, which corresponds with the middle of our hill plot instead of corresponding with the bottom like some of our previous graphs did. And so what this means is that when we do have a line here when it crosses this point this pink dotted line that is going to represent, um, the X intercept and so note that this why value of the hill equation, the log of theta over one minus data. It's actually going to equal zero when the value of theta itself is equal to 0.5. And so notice that when the y value the log of this ratio is equal to zero. This dotted line that we have here eyes going to represent where theta is equal to 0.5 and so we can see that right here, and we can actually figure this out if we were to take this log of data over one minus data and replace The state is here with 0.5. We can calculate and see what happens. So what we'll get is the log of 5/1 minus 05 Again. Since we're replacing the status here with we could go ahead and continue to calculate. So what we get is log of zero point 5/1 minus 0.5 is, of course, 0.5. And then, of course, 0.5 divided by 0.5 is going to be one. So what we get is the log of one. And then if you take your calculator and type in the log of one, which you'll get is an answer of zero, and so you can see that again. The log of fate over one minus data is gonna equal zero when theta itself is equal to 0.5. And this is gonna be important because we know already from our previous lesson videos that when theta is equal to 0.5 that is associated with the K D. And we'll be able to talk more about that in our next lesson video. So I'll see you guys there

2

concept

Hill Plot

Video duration:

3m

Play a video:

in this video we're going to talk about Myoglobin is hill plot. And so since myoglobin on Lee has one single sub unit, it's not actually an Alice Terek protein and therefore has no cooperative ity now recall from our previous lesson videos that when a protein has no cooperative ity, this means that the hill constant NH, which is also equal to the slope of the line on the hill plot, is going to equal a value of exactly one. And so notice down below. We have an image of myoglobin to remind us that myoglobin structure has Onley one single sub unit which again means that it's not an Alice Derek protein and therefore can display no cooperative ity. And of course, no cooperative ity corresponds with the slope of the line on the hill plot and the hill constant NH equaling a value of one. And so notice over here on the right, what we have is a hill plot where on the Y axis we have the log of the ratio of data over one minus data and on the X axis, what we have is the log of the partial pressure of oxygen and so notice that myoglobin is hill plot here is actually relatively easy because its data forms a single straight line with a slope of one. And so, of course, the slope of one that we see here tells biochemist visually that there is no cooperative iti in myoglobin in the myoglobin protein. And so, of course, because the slope of this black line is equal toe one, we can also say that the hill constant and age is also going to equal a value of one for myoglobin. Now, notice that with this straight line that we see here as we increase the partial pressure of oxygen from left to right on this hill plot, myoglobin lie again. Binding or myoglobin is binding. Toe oxygen also increases. And so also recall from our last lesson video that since the X intercept is going to indicate the exact lie again concentration where the value of theta is equal to 0.5 Uh, this, of course we know I's gonna mean that the X intercept is going to indicate the k d for the protein and so recall from our previous lesson video that when the why value here is equal to zero right here in the middle. Where are line intersects this dotted line here, we'll see, is that it's going to correspond with an X value in this X value here represents the X intercept. And because we know that the value of Fada is equal to 05 here at this dotted line, we can say that this X value that we see down here is gonna be the K D. And recall that the K D is going to determine the proteins affinity for its lie again. And so, by using a hill plot, Ah, biochemist can also determine the proteins affinity for the lie again. And so we'll be able to compare the K D here for myoglobin with the K D of hemoglobin as we move forward in our course. And so really, that's it for my Global's Hill plot. Here on the main takeaways are that it forms a single straight line with a slope of one because it displays no cooperative ity, and we'll be able to compare my Global's Hill plot to hemoglobin. Still plot a zoo. We move forward in our course and we'll cover hemoglobin still plot in our next video. So I'll see you guys there

3

concept

Hill Plot

Video duration:

5m

Play a video:

So now that we've introduced Myoglobin Hill plot in our last lesson video in this video, we're going to introduce hemoglobin is hill plot. And so, unlike myoglobin is protein like and data, which was relatively easy considering that it formed just a single straight line with a slope of one on a hill plot. Hemoglobin protein like and data, on the other hand, is much more complicated, since it seems to form three identifiable lines rather than just a single line when plotted onto a hill plot. And so if we take a look at the hill plot down below in our image over here on the right hand side, notice that it's showing you myoglobin is protein, leg and data here with this black line, which is again relatively easy, considering that it forms just a single straight line with a slope of one. But notice that we're also showing you hemoglobin, protein, leg and data here in red, which is clearly much more complex since it does not form a single straight line. And so the complexity of hemoglobin help lot has to do with the complexity of hemoglobin structure and so recall that hemoglobin is an Alice Derek Protein with multiple sub units that display positive cooperative ity. And so over here, what we're saying is that in hemoglobin help lot, there are these three identifiable lines. And so really, what we're saying is that hemoglobin help lot has three identifiable regions. The first region is this region down below here. The second region is this region here in the middle and the last region is this region up here at the top. And so as we move forward in our course, we're going to break down. Each of these three different regions of hemoglobin is hill plot in separate videos. But for now, what I want you guys to know is that these three different regions form three different identifiable lines. And so, of these three identifiable lines on hemoglobin is hell plot. Two of the hemoglobin lines are actually parallel to myoglobin is line. And so what you'll notice is that this line here highlighted in green if we extend it out and this line highlighted here in light blue if we extend it out, are both parallel to myoglobin is line here in the middle. And so what? We need to realize here is that lines that are parallel are always going to have the same exact slope. And so because myoglobin has a slope of one, that means that the slope of lines and these other regions of hemoglobin hill plot are also going to have a slope of one as well. And so, of course, the slope of the line on the hill plots correspond with the hill constant and H. And so what we're saying is that this green region and this light blue region of hemoglobin still plot. Both have a slope in a hill constant equal toe one. And so what we need to recall is that a slope or a hill constant of one means that there's absolutely no cooperative iti in those regions. And so this green region displays no cooperative ity, and this light blue region displays no cooperative ity. Now that might seem kind of strange, since we know from our previous lesson videos that hemoglobin displays positive cooperative ity. And so to say that two of the three identifiable lines display no cooperative ity seems kind of weird, but it turns out that the reason this is is because hemoglobin will actually bind to its first and to its last oxygen non co operatively. And so, as we'll find out moving forward in our course, this light blue region down below corresponds with hemoglobin binding its first oxygen molecule, non co operatively and this green region up above corresponds with hemoglobin binding its last oxygen non cooperatively. And so really, what allows hemoglobin to display the positive cooperative ity that we've been talking about in our previous lesson videos is this third hemoglobin line. And so although two of the three hemoglobin lines display no cooperative ity, this third hemoglobin line has a different and a greater slope than the slope of myoglobin. And so the slope of this yellow region down below in hemoglobin still plot eyes going to be a slope of three. And of of course, a slope in a hill constant greater than one, which would include three is going to suggest positive cooperative ity. And so really, it's this yellow region highlighted in, uh, Hemoglobin Hill plot that shows the positive cooperative ity that we talked about in our previous lesson videos. And so really, there's a lot of information to take in here, and so that's why moving forward In our course, we're going to break down each of these Three different regions of hemoglobin still plot one by one, so we'll start off with the light blue region. Uh, then we'll move into the yellow region and then we'll talk about the green region all by itself. So I'll see you guys in our next video when we talk about this light blue region, so I'll see you guys there.

4

concept

Hill Plot

Video duration:

4m

Play a video:

in this video we're going to talk about The first region of hemoglobin is hill plot, which is highlighted down below in light blue and corresponds with hemoglobin, lowest oxygen, affinity state. And so recall. We said in our last lesson video that hemoglobin is not always displaying positive cooperative ity. And that's because hemoglobin will actually bind to its first and to its last oxygen gas molecules, non co operatively. And of course, no cooperative ity we know corresponds with the slope of the line and the hill constant NH Equalling to a value of one. And so this first region here of hemoglobin is hill. Plot actually corresponds with hemoglobin binding to its very first oxygen gas molecule, non co operatively. And so you can see that at the very, very beginning of this region over here notice we have a hemoglobin molecule in the full T state. And of course, the T state we know is going to be the 10th state which is going to bind toe oxygen inefficiently. And so the full T state here is going to have the lowest possible oxygen affinity. And so notice that in this region hemoglobin goes from the full t state a two very beginning of the region with zero oxygen's bound on. Then by the end of this region, it ends up with one oxygen gas molecule bound to the hemoglobin sub unit. And again, this first oxygen molecule that binds is gonna buy non co operatively, meaning that the slope of the line in this region is going to equal one in the hill constants going to equal one in this region. And so again, because this first region corresponds with the full T state this is going to represent hemoglobin is lowest affinity state where the slope of the line is going to be one until it binds its first oxygen. And so the reason for this is because when hemoglobin is in the full T state here at the beginning, all four of the hemoglobin sub units are going to independently, uh, compete for oxygen binding without any cooperative ity whatsoever. And it will do that until the very first oxygen gas molecule binds. And so this first region right here represents hemoglobin binding to its first oxygen gas molecule. And once it does, buying this first oxygen gas molecule right here in this region. At that point, hemoglobin transitions into its cooperative state, which is the next region off hemoglobin is hill plot, which we'll talk about in our next video. But for now, as we've said, this first region represents this line for the first region represents hemoglobin, lowest oxygen affinity state again because it's starting with the full T state. And of course, the lowest oxygen affinity is going to correspond with the highest value of the K. D. And again, this is going to be the K D for just the very first oxygen gas molecule that binds to the hemoglobin molecule. And so, if we were to take this line right here of hemoglobin, help plot and we were to project it forward an imaginary line that projects this line forward so that it can actually intersect this dotted line that we have here in the middle. Then the line going down here, we know is going to correspond with the K D. And of course, the K D represents, uh, hemoglobin affinity for oxygen. And so this is specifically gonna be hemoglobin affinity for its very first oxygen gas molecule represented by this region right here and so What you'll notice is that this K D right here corresponds with log of the partial pressure of oxygen being to a value of two. And this is going to be the highest k d. Um, uh, when we compare it to the K. D s of the other regions of human global's help lot later, in our course on dso, Really, this concludes our lesson on this very first region here of hemoglobin help plot which corresponds with its lowest oxygen affinity state. And the main take away is that the first region represents hemoglobin binding to its very first oxygen non co operatively. And so, in our next video, we're going to talk about the second region here of hemoglobin still plot, which represents hemoglobin cooperative state. So I'll see you guys in that video.

5

concept

Hill Plot

Video duration:

4m

Play a video:

in this video, we're going to talk about the second region of hemoglobin is hill plot, which is this region right here, highlighted in yellow that corresponds with hemoglobin is cooperative state. And so what we need to realize is that after the very first oxygen gas molecule binds to hemoglobin at that point, hemoglobin subunits are going to begin to display positive cooperative ity where the slope of the line in the hill constant NH are gonna equal value of three. Now, recall from our previous lesson videos that hemoglobin, oxygen binding behavior, uh, in terms of its positive cooperative, it is really best explained via a combination of both the concerted and sequential models. And so, of course, what this means is that the hill constant NH, which is equal to a value of three, is not going thio equal The variable end, which recall is the number of ligand binding sites on the protein which we know hemoglobin has unequal to four. So these are not equal to each other, and that's totally okay. So if we take a look down below at our image of the hill plot over here, uh, recall that in this first region of hemoglobin still plot that corresponds where hemoglobin goes from having zero oxygen's bound up until having its very first oxygen bound. And so, as soon as the first oxygen molecule binds to hemoglobin again at that point is when the hemoglobin subunits begin to display positive cooperative ity. And so you can see that right here at this region which again corresponds with hemoglobin binding its first oxygen is when hemoglobin transitions into its cooperative state here in this region. And so notice that the slope of the line in the hill constant that corresponds with this line here eyes going to be a value of three. Now we can visually see that positive cooperative ITI is taking place in this cooperative region here because notice that, uh, the sub units begin to take on this confirmation that we see here, which represents a sub unit with an increased oxygen affinity, which of course, is going to correlate with positive cooperative ity. And so you can see that the confirmation is shown right here in this, uh, sub in these two sub units when the first oxygen molecule binds and it's also shown right here when the second oxygen molecule bonds to. So there's positive cooperative ity occurring in this region up until the third oxygen gas molecule binds. And then at that point, noticed that the, um this confirmation is not present anywhere, uh, in this molecule. And so what we're saying here is that hemoglobin is going to continue to display positive cooperative ity from the moment that it binds its very first oxygen up until the third oxygen gas molecule binds. And then as soon as the third oxygen gas molecule binds here, um, then it transitions again into a non cooperative state, which is this final region, which we'll talk about in our next lesson video. But what we can also see is that, uh, in this region right here in this thes two hemoglobin subunits, the hemoglobin subunits are not equally competing for oxygen binding. And so you can clearly see that here with this, uh, hemoglobin molecule, because you have one sub unit in the full. Our state you have to sub units in this altered confirmation will state and you have one sub unit in the full t state, and so these sub units are going to have different affinities for oxygen and therefore are not going. Thio equally compete for oxygen. And so because there is unequal competition for oxygen binding, that is more evidence to show that positive cooperative ITI is taking place here. And so really, this concludes our lesson on hemoglobin second region, which is again, it's, uh, Cooperative State. And so we'll be able to talk about the final region of hemoglobin. Is hill plot up here in Green and our next lesson video, So I'll see you guys there.

6

concept

Hill Plot

Video duration:

5m

Play a video:

in this video, we're going to talk about the third and final region of hemoglobin is hill plot, which is this green highlighted region up above here that corresponds with hemoglobin is highest oxygen affinity state. And so what's important to know is that after the third oxygen molecule binds to hemoglobin, the fourth and the final oxygen molecule is going to bind non co operatively again, just like the first oxygen molecule did. It's a recall from our previous lesson videos that, uh, hemoglobin is binding to its first oxygen molecule is represented by this first region down below here of Hemoglobin Hill plot. And so we also know that this second region here corresponds with positive cooperative ity. But what we're saying is that after the third oxygen molecule binds to hemoglobin again, the fourth oxygen molecule is going to bind non co operatively just like this first one did down here in this blue region. And so that's saying that the slope of the line in this region and the hill constant are both going to have a value of one, just like the slope of the line and the hill. Constant had a value of one in this light blue region from our previous lesson videos. And so what we can say is that once again, just like in this light blue region, the hemoglobin subunits up here in this green region are going to equally and independently compete for binding to the last oxygen molecule without any cooperative ity whatsoever. And so it's important to note here is that here, in this beginning of this green region, you can see that hemoglobin is bound to just three oxygen molecules. And so this last unoccupied hemoglobin sub unit, it's important to notice that it's actually in the full our state. And so because this last unoccupied hemoglobin subunits is in the full, our state, it's actually experiencing features of the concerted model which recall from our previous lesson videos, means that it's displaying the cemetery role, which of course, says that all of the sub units in the hemoglobin molecule must be in the same exact confirmation. And so this is important because when we zoom in on this little area up here, like we are over here on the left hand side, we can see that when there are three oxygen molecules bound to the hemoglobin sub unit. This last unbound sub unit right here is going to be in the full our state, which means that it's going to equally compete for oxygen binding and independently of other sub units. And so there is no sub unit here that's in the altered confirmation. I'll state like, uh, like we have over here in this second region. Uh, we have these, uh, altered confirmation, All states, That's not what's happening here in this final region. So that means there's no positive cooperative ity up here in this final region. And so again, I noticed that all of the sub units are in the full our state, and because they're all in the full, our state and the our state has the highest affinity for oxygen. What we're saying is that hemoglobin is in its highest affinity state in this green region up above. And of course, we already know that the slope of the hill constantly gonna have a value of one. And so, if we were to take this line here and projected backwards here just so that we can see where it intersects this dotted horizontal line where y is zero here at this point here, it's going to correspond with the ex access. So instead of drawing down towards the X axis here in order to avoid intersecting lines, I didn't wanna, uh, Intersect lines. I'm just drawing this dotted line up. But that's basically the same thing is drawing it down here. And so where this value corresponds on the X axis, of course, we know is going to reveal the K D. And because this is representing the binding of the last oxygen, this represents the K D that hemoglobin has for its last oxygen. And so when we compare where this value is on the X axis, it's relatively low, close, close to the left hand side, over here of the X axis. And so what we're saying is that this lineup here, highlighted in green, represents hemoglobin, highest oxygen affinity state. But of course, since oxygen affinity and K d have an inverse relationship, this means the highest oxygen affinity is going to correspond with the lowest value of the K D. For this fourth oxygen. And so really, this concludes our introduction to this final region here of hemoglobin till plot that corresponds with its highest oxygen affinity state. And in our next lesson, video will be able to put together all of the pieces that we've talked about briefly. So I'll see you guys in that video.

7

concept

Hill Plot

Video duration:

4m

Play a video:

in this video, we're going to break down the hill plot. And so notice over here on the left, what we have is the hill equation, which we know resembles the equation of a line. And so this right here, the log of the ratio of data over one minus data represents the why value and why value of the hill equation is plotted onto the y axis of the hill plot. And then, of course, the X value of the hill equation the log of the concentration of lie again is gonna be plotted onto the X axis of the hill plot and notice that on this help plot, we actually have the line for myoglobin here in black. And then we have the lines for hemoglobin here in red and noticed that myoglobin only forms one single straight line, which is pretty straightforward because it has a slope of one, meaning that it displays no cooperative ity whatsoever. However, hemoglobin seems to form three distinct lines. The ones that we talked about in our previous lesson videos this blue region, this yellow region and this green region up here, and notice that the green region and the blue region here are parallel to myoglobin is line and of course, parallel lines have the same slope and the slope on a hill plot represents the hill constant. And so, um, that means that these regions all have a hill constant of one displaying no cooperative ity. And so that's because the very first oxygen binds to hemoglobin, non co operatively and the very last oxygen binds to hemoglobin non cooperatively as well. And so the blue region represents hemoglobin, lowest affinity state and the green region up above at the top here represents hemoglobin. Highest affinity state on they both have a slope of one and a hill. Cops that of one. Now notice that the k D or the X intercept for the lowest affinity state region here actually has the highest KD value on the X axis and the highest KD value corresponds with lowest affinity and then notice that this green region right here has the lowest KD value on the X axis. And of course, the lowest KD value corresponds with the highest affinity. And then, of course, we have this yellow region in between. And this yellow region represents positive cooperative ity where hemoglobin is in its cooperative state, and the hill constant is equal to three. So the region, this region right here, has a slope of three. And so basically what a hill plot allows biochemist to do is visually display the protein Ligon affinities or the K DS, as well as visually display the degree of cooperative ity or the hill constant and age in a protein leg in interaction. And so, by this visual display, a biochemist could look at my globe and see a straight line and notice it has a slope of one and know that myoglobin has no cooperative ity and were able to determine my global's k d by looking at its X intercept here. So my global's K d would be somewhere over here and then for hemoglobin. We can see that it has these three different states. It has a non cooperative state down here, where it binds its first oxygen. It has a positively cooperative state here, where it binds its 2nd and 3rd oxygen's, and then it has another non cooperative state where it binds its fourth and final, the last oxygen. And so now that we can better understand the hill plot will be able to get some practice with all of these concepts. But before we get there, we have a little analogy for you guys in case you're looking for an easier way to remember this stuff. So I'll see you guys in our next video.

8

concept

Hill Plot

Video duration:

2m

Play a video:

all right, so in this video, we have a very interesting analogy on a car's velocity and acceleration to help you guys better understand myoglobin and hemoglobin hill plots. And so both myoglobin and hemoglobin, oxygen binding can be thought of as a car's velocity to represent the oxygen affinity and a car's acceleration to represent cooperative ity. Now notice down below. We have this black line here for myoglobin and notice that myoglobin is always going to maintain the same relatively high slash medium velocity or the same oxygen affinity, regardless of what the partial pressure of oxygen is. However, uh, my global is not going to be accelerating, so there is absolutely no acceleration, which means it has no cooperative ity. So notice that myoglobin here is just gonna be going on this straight line here with a medium steady medium velocity but no acceleration Now hemoglobin, on the other hand, is going to start off with a very low velocity. So it's going very steady, low velocity, and it's gonna maintain this steady low velocity until it begins accelerating or displaying positive cooperative ity once it binds its first oxygen so again, once it binds its first oxygen at this point right here. It turns on its thrusters and accelerates up to a very, very high speed here. A very high velocity, uh, where it is bound to its third oxygen. But once hemoglobin binds its third oxygen, it has reached and maintained its maximum velocity. So it's already going. It's full velocity and has its highest oxygen affinity. Um, and it's going to stop accelerating at this point once it reaches this high velocity. And so this is going to be representing when the fourth Oxygen is going to bind without any cooperative ity. And so this is just a little analogy to help you guys better understand Hemoglobin and myoglobin hill plot. So hopefully this was useful for you guys, but you can let us know in the comments below. So I'll see you guys in our next video.

9

Problem

Problem

Label the axis of the Hill Plot below & fill-in the graph with Mb's & Hb's approximate O_{2}-binding data.

Video duration:

2m

Play a video:

Was this helpful?

10

Problem

Problem

The slope of a Hill plot for hemoglobin _______________; whereas that for myoglobin ________________.

A

is about 3 in its cooperative state; is 1.0.

B

decreases at low pO_{2} ; is constant at all pO_{2}.

C

increases at high pO_{2} ; curves upward for all pO_{2}.