Hill Equation - Online Tutor, Practice Problems & Exam Prep

The Hill Equation, developed by Archibald Hill, describes cooperative ligand binding in allosteric proteins, particularly hemoglobin. It allows for the analysis of protein-ligand interactions by transforming the fractional saturation equation into a linear form, facilitating data interpretation. The Hill constant (n_H) indicates the degree of cooperativity, with values ranging from 0 to the number of binding sites (n). A value of n_H greater than 1 signifies positive cooperativity, while less than 1 indicates negative cooperativity. Understanding these concepts is crucial for analyzing protein behavior in biochemical contexts.

Hill Equation

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 \( k_d \) 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 \( k_d \) 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 ligand 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.

Hill Equation

In this video, we're going to introduce the Hill equation, which is named after the scientist Archibald Hill. Now, we're not going to talk about all the details about how to derive the Hill equation in this video. However, I can tell you that the equation 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=mx+b. And so, lines are just very straightforward, and any data that we can get to form a line is just going to be overall easier to understand. When the 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=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 θ/1-θ. 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. 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 going to 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.

Hill Equation

In this video, we're going to introduce the Hill constant \( n_h \) and how it relates to cooperativity. 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 mentioned in our previous lesson videos must actually replace the variable \( n \) with the Hill constant \( n_h \). As we'll see very shortly, this Hill constant, \( n_h \), is different from the variable \( n \). The Hill constant is sometimes referred to as the Hill coefficient, and again, it's abbreviated with \( n_h \). The Hill constant or Hill coefficient \( n_h \) is really just the degree of cooperativity of a ligand binding reaction. We have to replace the variable \( n \) in all of the equations above with the Hill constant \( n_h \). We can do that now.

Here, with the Hill equation, notice that we have the variable \( n \) showing up twice. We have to replace the variable \( n \) with the Hill constant \( n_h \). We can go ahead and do that now. So, we can put an \( n_h \) right here and we can also put an \( n_h \) right here, and we also have to do that with the equations above as well, replacing all of the \( n \)'s that we see with \( n_h \)'s. What you might notice now is that we have these blanks here for \( K_d \). The reason we have that is because it turns out that a protein's affinity for its ligand, represented by \( K_d \), is actually affected by cooperativity. Since the Hill constant or Hill coefficient \( n_h \) is the degree of cooperativity, \( K_d \) is also affected by the Hill constant \( n_h \) in the fashion we have discussed. We can say that \( K_d \) is going to be raised to the power of \( n_h \) in both of these equations.

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 \( n \). Essentially, 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. 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 \( n \). On the right-hand side, we have the degree of cooperativity. It's important to know that when the Hill constant \( n_h \) is exactly equal to 1, there's absolutely no cooperativity that takes place.

When the Hill constant \( n_h \) is greater than 1, then that means that positive cooperativity is going to take place. Of course, if the Hill constant \( n_h \) is less than a value of 1, that means that negative cooperativity takes place. By determining the Hill constant \( n_h \), 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. We'll be able to learn more about the Hill constant as we move forward in our course and talk about the Hill plot. 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.

Hill Equation

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 n_h 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. n_h is not going to equal its n. And so we can say hemoglobin's n_h 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 going to go ahead and say that hemoglobin's Hill constant n_h is going to 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 its 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 θ or y and on the x-axis, we have the partial pressure of oxygen in units of torr. 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, 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 n_h 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 hemoglobin's 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 n_h equal to 3.

And so notice that this example problem here is saying to assume hemoglobin's K_D is equal to 26 torr 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 θ equation and raise the ligand concentration to n_h and the K_D, we raise to n_h as well. And so notice that we're given the partial pressure of oxygen as 100 torr and that is going to 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 going to be 100, and we can raise this to the n_h which again is given to us as 3. So, 100 raised to the third power. Plus, the K_D, which is also given to us as 26 torr. So we can put that in here, 26 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 cubed plus 26 cubed, what we'll get is our answer, which is θ = 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.

At a CO₂ partial pressure of 5 torr, the  p₅₀ value for hemoglobin is 26 torr. What is the fractional saturation when n_H = 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

What is Hb's fractional saturation when p₅₀ = 26 torr, n_H = 3, and pO₂ = 100 torr, a typical pO₂ in the lungs?

• A. 0.65
• B. 1.6
• C. 0.89
• D. 0.98
• E. 0.33