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 log and binding in Alice Derek proteins. And so way back in 1913 before any knowledge of hemoglobin structure even existed, this scientist name Archi Bald Hill, tried to study. Hemoglobin is cooperative oxygen binding and so aren't you. Bald Hill knew what we already know from our previous lesson videos. So recall that coefficient 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 Thio Archibald Hill, that for proteins with an unknown number of ligand binding sites and end number of ligand binding sites, the protein Ligon reaction and equations for Katie 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 end would be included as a coefficient in front of the likened. And, of course, the end and the protein legging 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 lie again, which has a coefficient, we would need to include the coefficient here as an exponents. So we would need to include And here, here and here. And then, of course, for whenever we see the protein leg in complex, we need to include n as a sub script. So we would include. And here and so these equations here, it seemed to Archibald Hill would be appropriate for a protein with an end number of ligand binding sites. And so, really, this is the background information that we need, Um, as we move forward and talk about the hill equation. So I'll see you guys in our next video

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Hill Equation

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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 data. Essentially, this equation right here is the equation that can actually be algebraic lee 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 algebraic Lee Rearranged Hill equation actually resembles the equation of a line which we all know is why equals M X plus B and so lines air just very straightforward. And any data that we can get to form a line is just gonna be overall easier toe, understand? So that is definitely the benefit of the Hill equation. Now again, because the Hill equation resembles the equation of a line, we could say that the Hill equation allows us to graph protein, leg and binding data on a linear plot called the Hill Plot, also named after Archi Bald 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. Why equals M X plus B where recall that the M here is just the slope of the line and the be 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 two substitute and variables, um, 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 one minus data for the slope em. 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 lie gand is actually oxygen, gas, and so the like and concentration can be replaced with the partial pressure of oxygen. When it comes to myoglobin and hemoglobin. And then, of course, be the Y intercept is just going to be replaced with the value of n this 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 like in binding data on a linear plot called the Hill Plot. So now that we've introduced the Hill equation will be able to apply it mawr later in our course when we talk about the hill plot. But in our next lesson video, we're going toe introduced the Hill constant and cooperative ity, So I'll see you guys in that video.

3

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Hill Equation

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in this video, we're going to introduce the Hill Constant and H and how it relates to cooperative ity. And so, contrary to what it may have seemed toe Archibald Hill when he actually plotted hemoglobin 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 end with the hill Constant and H. And so, as we'll see very shortly, this hill, Constant and H is different than the variable end. And so the hill Constant is sometimes referred to as the hill coefficient. And again it's abbreviated with an H. And so this hill constant or hill coefficient and H is really just the degree of cooperative ity of a ligand binding reaction. And so again, we have to replace the variable end and all of the equations up above with the hill constant and H, and so we could go ahead and do that now and so here with the hill equation, notice that we have the variable and showing up twice here and again, we have to replace the variable end with the Hill constant N. H. So we could go ahead and do that now so we can put an N H right here. And we can also put an age 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 an H is. And so we could 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, uh, the K D. And the reason we have that is because it turns out that a proteins affinity for its like in represented by the K D, is actually affected by cooperative ity. And so K d is affected by cooperative ity and because the hill constant or hill coefficient and H is the degree of cooperative ity. Therefore, the K D is also affected by the hill constant and H in the fashion that we have up above. And so we can say that the K D eyes going to be raised to the power of the N. H 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 is fractional saturation now down below. What I want you guys to notice is that the Hill Constant and H is always going to have a value between zero at its minimum and the maximum number of ligand binding sites on the protein. And and so essentially, what we're trying to say right here is that the Hill constant and H is always going to be greater than or equal to a value of zero and less than or equal to the value of n, which recall from our previous lesson videos and is just the number of ligand binding sites on a protein which again we can see, uh and 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 and H, which we know again is going to be greater than or equal to zero and less than or equal to end. And on the right hand side. What we have is the degree of cooperative ity, and so it's important to know that when the hill constant NH is exactly equal toe one. There's absolutely no cooperative ity that takes place when the Hill constant NH is greater than one, then that means that positive cooperative it is going to take place. And then, of course, if the Hill Constant and H is less than a value of one, that means that negative cooperative ITI is taking place. And so, really, just by determining the Hill constant, NH biochemists are able to quickly determine the degree of cooperative ity off a protein. Whether there is no cooperative ity, positive cooperative ity or negative cooperative, ITI is going to depend on the value of the Hill constant. And so we'll be able to learn Maura 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

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Hill Equation

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5m

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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 and H is equal to end, there's no cooperative ity that takes place in the protein. And so it turns out that the hill constant NH can Onley equal the number of light and binding sites and under two circumstances. The first is, of course, if the protein displays no cooperative ity such as a protein like myoglobin, and the second circumstance, where the Hill Constant and H will equal end, is when the protein Onley follows. The concerted model of cooperative ity, however, recall from our previous lesson videos that hemoglobin oxygen binding behavior is explained via a combination of both the concerted and sequential model, So hemoglobin does not Onley follow the concerted model of cooperative ity. It follows both concerted and sequential models, and that means that Hemoglobin Hill Constant and H is not going to equal. It's, uh and so we can say hemoglobin. NH is not going to equal hemoglobin and and so in hemoglobin is cooperative state hemoglobin is Hill Constant is going to range from 2.8 to about three. So we're gonna go ahead and say that Hemoglobin Hill Constant and H is gonna be about three as we move forward in our course. And this is true even though hemoglobin has four like and binding sites. So hemoglobin n H equals three, even though it's n is equal to four. And so if we take a look down below at our oxygen bun and curve notice on the y axis, we have the fractional saturation data. Or why and on the X axis, we have the partial pressure of oxygen and units of Taurus and notice that we have these three curves. We have this black curve here that resembles a rectangular hyperbole and noticed that the number one here is indicating that it has no cooperative iti just like up above. We could say that the protein has no cooperative ity such as myoglobin and notice that this blue curve right here represents, um, the curve. If hemoglobin displayed the concerted model on Lee and so if hemoglobin displayed on Lee the concerted model, then it's an H is going to equal. It's n. And of course we know that Hemoglobin N is equal to four. But again, this is if hemoglobin Onley followed the concerted model, it would have this Blucher. But we know again that hemoglobin does not only follow the conservative model, it follows a combination of the conservative and sequential models. And so when we actually plot hemoglobin is data onto this plot. What we'll get is this red curve that we see here and notice that hemoglobin is going to have ah Hill, Constant of three and NH equal to three. And so notice that this example problem here is saying to assume hemoglobin is Katie is equal to 26 tours and calculate the fractional saturation of hemoglobin at a partial pressure of oxygen equal to 100 tours and hill coefficient equal to three. And so, in order to solve for 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, um, the fractional saturation data equation and raise the Lagan concentration to NH and the K D. We raised to the N H as well, and so notice that were given the partial pressure of oxygen as 100 tours and that is going to represent the concentration of lie again so we could go ahead and start to substitute this end. So what we have is the concentration of lie again is gonna be on and we can raise this to the N H, which again is given to us as three. So 100 over three. And this is gonna be over again. 100 over three. I'm sorry. 100 race to the third power. Uh, plus the K D, which is also given to us as 26 tours. So we can put that in here 26 tours and this is also going to be raised Thio 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 cube and divided by the sum of 100 cubed plus 26 cube, what we'll get is our answer, which is theta is equal to 0.983 which is of course, equal to about 0.98 And so we can say that data is gonna be equal to 0.98 which matches with answer option C here. And so we can indicate that sea is the correct answer for this example problem. And so, in our next couple of videos will be able to get some practice utilizing thes equations, and we'll also be able to talk. Maura, about hemoglobin is Hill constant as we talk about the hill plot. So I'll see you guys in our next video.

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Problem

Problem

At a CO_{2} partial pressure of 5 torr, the p_{50} value for hemoglobin is 26 torr. What is the fractional saturation when n_{H} = 3 and pO_{2} = 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 p_{50} = 26 torr, n_{H} = 3, and pO_{2} = 100 torr, a typical pO_{2} in the lungs?