Hill Equation - Video Tutorials & Practice Problems
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Hill Equation
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In this video, we're going to begin our lesson on the Hill Equation. So before we talk directly about the Hill Equation, there's first some background information and a little bit of history that we need to tell you guys about Cooperative Ligand Binding in allosteric proteins. And so way back in 1913 before any knowledge of hemoglobin structure even existed, this scientist named Archibald Hill tried to study hemoglobin's cooperative oxygen binding. And so Archibald Hill knew what we already know from our previous lesson videos, so recall that coefficients in a reaction, which are the numbers in front of a molecule, are included into the dissociation equilibrium constant kd as exponents. And so from this, along with not knowing hemoglobin structure at the time, it seemed to Archibald Hill that for proteins with an unknown number of ligand binding sites, an n number of ligand binding sites, the protein ligand reaction and equations for k d and theta would be as follows down below. And so if we say that n is equal to the number of ligand binding sites on a protein, then, of course, the n would be included as a coefficient in front of the ligand. And of course the n in the protein ligand complex wouldn't be included as a subscript. And then so for the kd and the fractional saturation, all we would need to do is make sure to include the coefficients as exponents. And so, of course, that means that whenever we see the free ligand, which has a coefficient, we would need to include the, coefficient here as an exponent. So we would need to include n here, here, and here. And then of course, for whenever we see the protein lagging complex, we need to include n as a subscript. So we would include n here. And so these equations here, it seemed to Archibald Hill, would be appropriate for a protein with an n number of ligand binding sites. And so really this is the background information that we need, as we move forward and talk about the Hill Equation. So I'll see you guys in our next video.
2
concept
Hill Equation
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3m
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Video transcript
In this video, we're going to introduce the Hill equation, which is named after the scientist Archibald Hill. Now, we're not gonna talk all the details about how to derive the Hill Equation in this video. However, I can tell you that the equation up above from our last lesson video for the fractional saturation theta, essentially this equation right here, is the equation that can actually be algebraically rearranged and reformatted to get the hill equation. And so the benefit of rearranging this equation right here to get the Hill equation is that this algebraically rearranged Hill equation actually resembles the equation of a line which we all know is y equals mx+b. And so lines are just very straightforward and any data that we can get to form a line is just gonna be overall easier to understand. Equation resembles the equation of a line, we can say that the Hill equation allows us to graph protein ligand binding data on a linear plot called the Hill plot, also named after Archibald Hill. And we'll talk more details about the Hill plot later in our course. But for now, let's take a look down below at the Hill equation. And so notice over here on the left what we have is the equation of a line, y equals mx+b, where, recall that the m here is just the slope of the line and the b here is just going to be the y intercept. And so, what's important to know here is that all we need to do to get the Hill equation is literally just to substitute in variables, substitute in for these variables. And so for the y value, all we need to substitute in is the log of this ratio of theta over 1 minus theta. For the slope m, all we need to substitute in is the variable n, which recall is just the number of ligand binding sites on a protein. For the variable x, all we need to substitute is the log of the concentration of ligand. And recall for myoglobin and hemoglobin, the ligand is actually oxygen gas. And so the ligand concentration can be replaced with the partial pressure of oxygen when it comes to myoglobin and hemoglobin. And then, of course, b, the y intercept, is just gonna be replaced with the value of n, the same value of n, times the log of the dissociation equilibrium constant k d. And so it's this equation right here that is the Hill equation that, again, resembles the equation of a line and allows us to plot protein ligand binding data on a linear plot called the Hill plot. So now that we've introduced the Hill equation, we'll be able to apply it more later in our course when we talk about the Hill plot. But in our next lesson video, we're going to introduce the Hill constant and cooperativity. So I'll see you guys in that video.
3
concept
Hill Equation
Video duration:
4m
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Video transcript
In this video we're going to introduce the hill constant n h and how it relates to cooperativity. And so contrary to what it may have seemed to Archibald Hill, when he actually plotted hemoglobin's experimental data onto a hill plot, he quickly realized that all of the equations above that we mentioned in our previous lesson videos must actually replace the variable n with the hill constant n h. And so as we'll see very shortly, this hill constant, n h, is different than the variable n. And so the hill constant is sometimes referred to as the hill coefficient and again it's abbreviated with n h. And so this hill constant or hill coefficient n h is really just the degree of cooperativity of a ligand binding reaction. And so again we have to replace the variable 'n' and all of the equations up above with the Hill constant 'nh'. And so we can go ahead and do that now. And so here with the hill equation, notice that we have the variable n showing up twice here. And again, we have to replace the variable n with the hill constant n h. So we can go ahead and do that now. So, we can put an NH right here and we can also put an NH right here and we also have to do that with the equations up above as well. Again, replacing all of the ends that we see with NHs. And so we can do that over here as well. And we can also do that over here. And so what you might notice is that now we have these blanks here for, the k d. And the reason we have that is because it turns out that a protein's affinity for its ligand, represented by the k d, is actually affected by cooperativity. And so k d is affected by cooperativity. And because the hill constant or hill coefficient n h is the degree of cooperativity, therefore, the k d is also affected by the Hill constant, n h, in the fashion that we have up above. And so we can say that the 'kd', is going to be raised to the power of the nh', in both of these equations here, as well as here. And so this will be important later on when we're trying to determine hemoglobin's fractional saturation. Now down below, what I want you guys to notice is that the hill constant n h is always going to have a value between 0 at its minimum and the maximum number of ligand binding sites on the protein n. And so essentially what we're trying to say right here is that the hill constant, n h, is always going to be greater than or equal to a value of 0 and less than or equal to the value of n, which recall from our previous lesson videos, n is just the number of ligand binding sites on a protein, which again we can see, n is defined here, number of ligand binding sites on a protein. And so down below we have this table where on the left hand side we have the hill constant n h, which we know again is going to be greater than or equal to 0 and less than or equal to n. And on the right hand side what we have is the degree of cooperativity. And so it's important to know that when the hill constant NH is exactly equal to 1, there's absolutely no cooperativity that takes place. When the hill constant NH is greater than 1, then that means that positive cooperativity is going to take place. And then, of course, if the hill constant NH is less than a value of 1, that means that negative cooperativity is place. And so, really, just by, determining the hill constant NH, biochemists are able to quickly determine the degree of cooperativity of a protein. Whether there's no cooperativity, positive cooperativity, or negative cooperativity is going to depend on the value of the hill constant. And so we'll be able to learn more about the hill constant as we move forward in our course and talk about the Hill plot and in our next lesson video, we're going to break down the Hill Constant just a bit further. So, I'll see you guys in that video.
4
concept
Hill Equation
Video duration:
5m
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Video transcript
In this video, we're going to continue to talk about the Hill constant n h. And so we already know from our previous lesson video that when the Hill constant n h is equal to n, there's no cooperativity that takes place in the protein. And so it turns out that the hill constant n h can only equal the number of ligand binding sites n under 2 circumstances. The first is, of course, if the protein displays no cooperativity, such as a protein like myoglobin. And the second circumstance where the hill constant NH will equal n is when the protein only follows the concerted model of cooperativity. However, recall from our previous lesson videos that hemoglobin's oxygen binding behavior is explained via a combination of both the concerted and sequential models. So hemoglobin does not only follow the concerted model of cooperativity. NH is not going to equal its, n. And so we can say hemoglobin's NH is not going to equal hemoglobin's n. And so in hemoglobin's cooperative state, hemoglobin's hill constant is going to range from 2.8 to about 3. So we're we're gonna go ahead and say that hemoglobin's hill constant n h is gonna be about 3 as we move forward in our course. And this is true even though hemoglobin has 4 ligand binding sites. So hemoglobin's n h equals 3 even though it's n is equal to 4. And so if we take a look down below at our oxygen binding curve, notice on the y axis we have the fractional saturation theta or y and on the x axis, we have the partial pressure of oxygen in units of torus. And notice that we have these three curves. We have this black curve here that resembles a rectangular hyperbola. And notice that the number 1 here is indicating that it has no cooperativity. Just like up above, we could say that the protein has no cooperativity, cooperativity, such as myoglobin. And notice that this blue curve right here represents, the curve if hemoglobin, displayed the concerted model only. And so if hemoglobin displayed only the concerted model, then its NH is going to equal its n. And, of course, we know that hemoglobin's n is equal to 4. But, again, this is if hemoglobin only followed the concerted model, it would have this blue curve. But we know again that hemoglobin does not only follow the concerted model, it follows a combination of the concerted and sequential models. And so when we actually plot hemoglobins data onto this plot, what we'll get is this red curve that we see here. And notice that hemoglobin is going to have a hill constant, of 3 and NH equal to 3. And so notice that this example problem here is saying to assume hemoglobin's KD is equal to 26 tors and calculate fractional saturation the fractional saturation, we need to recall from our previous lesson video, the fractional saturation of hemoglobin. And so, what we need to recall is this equation right here from our previous lesson video where we took the fractional saturation theta equation and raise the ligand concentration to NH and the kd, we raise to the NH as well. And so notice that we're given the partial pressure of oxygen as 100 tors and that is gonna represent the concentration of ligand. So we can go ahead and start to substitute this in. So what we have is the concentration of ligand is gonna be 100, and we can raise this to the NH which again is given to us as 3. So, 100 over 3 and this is gonna be over. Again, 100 over 3. I'm sorry, 100 race to the 3rd power. Plus, the k d, which is also given to us as 26 tours. So we can put that in here, 26 6 tors, and this is also going to be raised to, the third power. And so all we need to do now is just type all of this into our calculator. So if we take 100 cubed and divide it by the sum of 100 cube plus 26 cubed, what we'll get is our answer, which is theta is equal to 0.9 83, which is, of course, equal to about 0.98. And so we can say that theta is gonna be equal to 0.98, which matches with answer option c here. And so we can indicate that c is the correct answer for this example problem. And so in our next couple of videos, we'll be able to get some practice utilizing, these equations. And, we'll also be able to talk more about Hemoglobin's Hill Constant as, we talk about the Hill plot. So I'll see you guys in our next video.
5
Problem
Problem
At a CO2 partial pressure of 5 torr, the p50 value for hemoglobin is 26 torr. What is the fractional saturation when nH = 3 and pO2 = 25 torr, a typical venous oxygen partial pressure?
A
1.1
B
0.68
C
0.12
D
0.88
E
0.47
6
Problem
Problem
What is Hb's fractional saturation when p50 = 26 torr, nH = 3, and pO2 = 100 torr, a typical pO2 in the lungs?