Alright. So here we have an example problem that says the following rate constants were measured for a simple enzyme-catalyzed reaction. Determine the Michaelis constant or the kilometers for the enzyme, and notice that we're given the values for the three relevant rate constants for a simple enzyme-catalyzed reaction. And then we've got 4 potential answer options for the Kilometers of the enzyme. So recall from our previous lesson videos, we talked about 2 predominant ways to calculate the Kilometers of an enzyme.

And so notice the first way, which is the algebraic rearrangement of the Michaelis-Menten equation. Notice that we're not actually given the values for the initial reaction velocity, the v, or the substrate concentrations, which means that we cannot use the first method here of algebraically rearranging the Michaelis-Menten equation to determine the Km. And so, for that reason, we can actually just go ahead and just remove this completely and, essentially what that means is we're going to need to use our second method of calculating the kilometers. And, recall that the second method of calculating the Kilometers is related to the rate constants and expressing Kilometers with rate constants.

And so, we know that a simple enzyme-catalyzed reaction, shown here is going to have 3 relevant rate constants, and we know that the kilometers is going to be the rate constants for the enzyme substrate complex dissociation over the association of the enzyme substrate complex. So, essentially, what we need to recall is that the enzyme substrate complex can dissociate in 2 different ways. It can dissociate backwards via k-1 to form the free substrate and free enzyme, and it can dissociate forwards via k2 to form the free enzyme and the free product. And so it's going to be the dissociation over the association, so these two dissociations here, it's gonna be the sum of them, so it's gonna be k-1 + k2, and then it's gonna be over the rate constant for the association of the enzyme substrate complex. So that would be via k1 to form the enzyme substrate complex.

And so, what we can do is we can put the k1 on the bottom. And so we know that these rate constants here, the dissociation of the enzyme substrate complex is associated with the free enzyme and the free substrate, and the association of the enzyme substrate complex is, of course, associated with the concentration of the enzyme substrate complex, and so this is another way to express this equation. Now, notice that, we're not actually given the values for the concentrations of the enzyme or the substrate or the enzyme substrate complex. So we're not gonna be able to use that portion of the equation and we're gonna be focusing specifically on this portion right here.

And so, notice that we're given k1, as 2×109 inverse molarity, inverse seconds, and then we're given, the value of k1 and k2 as well. So all we need to do is plug in these values into our expression here to calculate for the Kilometers. So, essentially what we see here is that the Kilometers is going to be equal to, k-1, which is given to us as 1×103 inverse seconds, plus the value of the k2, which is 5×103 inverse seconds. And then, of course, this is all going to be divided by k1, the association rate constant, which is 2×108, and units of inverse molarity, inverse seconds. And so, essentially, if you take your calculators and add 1×103 plus 5×103 and divide that answer by 2×108, what you'll get is the answer for our Km, which is equal to 3×10-5 M. And so we can see that 3×10-5M matches with answer option a. So we can go ahead and indicate that a here is the correct answer for this example problem. And that concludes this example problem, and we'll be able to apply these concepts moving forward in our practice problem. So I'll see you guys there.