Important parameters in the Fish Tank 2: Half-life, Serum Concentration, and AUC

Let’s take a look at half-life as a concept. And also percent remaining as functions of the elimination rate constant. Now, half-life is a common parameter that everyone who deals with drugs is familiar with it’s simply the time it takes for the serum concentration to drop in half or the time that it would take for C2 the second serum concentration to be 50% of C1 the first concentration. If we represent the time equation being natural log of C1 over C2 divided by K equals time.

If C1 is twice C2 or C2 is 50 percent of C1 than the half-life equation would simply become the natural log of 2 divided by K and the natural log of 2 is always 0.693 So the half-life of a drug is always 0.693 divided by K. It’s also interesting to keep in mind however is that this relationship using time equals natural log of C1 over C2 divided by K that always holds true. It isn’t just for when C1 is twice C2. So we can identify any percent comparison between C1 and C2 mathematically. The equation for that is e to the minus KT and then e to the minus KT represents the percent that’s remaining after time T.

Now if time is the half-life, that percent is 50% but time could be different from the half-life, and in that situation e to the minus KT, no matter what that time is will tell us what percent of drug would be remaining after that amount of time. So we can break that down into a very straightforward equation C2 equals C1 times e to the minus K T where T is the time between C1 and C2. We can also take that same relationship and put it in reverse to estimate a time that occurred previously. If we take C2 and divide it by e to the minus KT that will tell us what C1 was, T hours into the past.

Now, let me review how we can identify the best equation when we’re pursuing one of those four variables that we’ve just been talking about C1, C2, K or the elimination rate constant and time. If we know C1 and C2 and we need to calculate what the K is for this drug or we need to calculate what the time interval is between C1 and C2, we would use the equation that I’ve shown on this slide. Natural log of C1 over C2, and if we’re trying to find K then we would divide by time. If we’re trying to find time we would divide by K. So this is when we know both C1 and C2.

Let’s say we are trying to find C1 or C2, but we know the K and the time interval between C1 and C2. Then it would be easier to use a variation of the equations above shown in red box and instead use C2 equals C1 times e to the minus KT or C1 equals C2 divided by e to the minus KT. These are all essentially the same equation it’s just a matter of using the form, that’s best suited to the variable that you’re trying to calculate. Now, let’s shift gears and talk about area under the curve from a single dose. The area under the curve is a representation of the total exposure of drug during the elimination phase.

So we give a dose of drug and based on how rapidly that drug is eliminated from the body. The exposure of the patient to that drug will vary. The units of area under the curve are milligram hours per liter, it also represent those units as milligrams per liter, per hour. The equation for area under the curve, this can be done in a couple of different ways, dose divided by clearance. Now since clearance equals volume times K, we can also modify that to be dose over volume times K or the initial serum concentration or the serum concentration at time 0 divided by k.

Now these are actually the same thing, the dose over volume is equal to the initial serum concentration C time at time 0 is equal to dose over V. So these equations are essentially identical. Now if we take a look at what happens, in two different cases, in both cases we gave the patient a dose of 150 milligrams. We can see that the initial serum concentration at time zero seven point five milligrams per liter but the red curve represents a much higher clearance. 15 liters per hour as opposed to only 5 liters per hour, you can see how much more rapidly the drug concentration declines, and the area under the curve is much smaller.

In fact, it’s one third as much because the clearance is three times larger so we have an area under the curve under the red curve of only 10 milligram hours per liter. Under the blue curve it’s 30 milligram hours per liter. When we evaluate the area under the curve between multiple doses now, after one dose, the area under the curve of in this case, 80 milligram hour as per liter represents all the area under that red curve of one dose of 200 milligrams with a concentration at time zero of ten milligrams per liter.

If we give that same 200 milligram dose as part of a multiple dosing regimen with a clearance of 2.5 liters per hour, the concentration at time zero is still ten milligrams per liter. But now we’re giving that 200 milligrams every six hours, we’ll obtain a C max of 18.9 and a C min of 8.9 and what’s important is that the area under the curve its steady state is also going to be 80 milligram hours per liter.

In other words the area under the curve, the entire red curve, the curve at the top for a single dose is the same as the area under the curve shown in the yellow portion between two of those same 200 milligram doses and this same relationship would hold true if the dosing interval was eight hours or 12 hours. The shape of that yellow area would change but the actual area itself would not change as long as the dose is 200 milligrams and the clearance is 2.5 liters per hour.