# Can you find the mass of drug X remaining in the body after an oral dose with only bioavailability and half-life?

I'm reviewing this question and think you can't find the answer with the information given.

A patient receives 1g of drug X as a single oral dose. Drug X has a half-life of 12 hours and an oral bioavailability of 40%. What mass (in milligrams) of drug X will still be present in the body 24 hours after a 1g dose?

I think the fact that it is an oral dose means the question is more complicated, involving absorption and distribution of the drug, meaning more factors are needed which are provided in the question.

Is it as easy as: $$1000mg \times 0.4 \times 0.5 \times 0.5 = 100mg$$

Or there is is more it than that as I suspect? Many thanks!

• I'm no pharmacist at all, but I think you have given yourself a red-herring. The bioavailability surely means that this is the proportion absorbed?
– bob1
Commented May 22, 2021 at 21:36
• @bob1 It does but it doesn't tell you over what time-scale the drug is absorbed. You could answer what mass of drug in total will make it into and out of the body (400mg) but not how much at a specific time point (24 hours).
– Tom
Commented May 23, 2021 at 11:00

Unfortunately, it is not so easy, the easy calculation you suggested might be a good approximation if the drug was administered intravenously via bolus, but for oral administration, as you say in the comment, we need to know something about the dynamics of absorption from the gastrointestinal tract, such as Time to Peak Concentration (how long it takes for Drug X to be fully absorbed and reach peak concentration in the bloodstream.), and Absorption Rate (the rate at which drug is absorbed, as this influences the concentration profile in the initial hours after administration.

Well $$40\% \times 1 g = 400 \space mg$$ That's how much drug is in your body's fluid compartments.

If the drug's half-life is $$12$$ hrs, the amount of drug in your body (after $$24$$ hrs) = $$\frac{1}{2} \times \frac{1}{2} \times 400 = 100 \space mg$$.

If dose = $$d$$ and bioavailability = $$b\%$$ and half-life = $$h$$ hours ...

The amount of drug left in your body after $$t$$ hrs = $$\left(\frac{1}{2}\right)^{\frac{t}{h}} \times b\% \times d$$

EDIT 1 START

This fancy geometric/mixed (?) series might be of interest ...

$$1 - \left(1 \times \left(\frac{1}{2}\right)^1 + \frac{3}{2} \times \left(\frac{1}{2}\right)^2 + ... \right) = 0$$ where $$1$$ = bioavailability, the exponent is $$\frac{t}{h}$$ and the rate of drug absorption = $$1$$/hr.

EDIT 1 END

• No, that's not how it works. Bioavailability is how much of the drug ends up in the body ever; your answer is making an unstated assumption that it all enters immediately at time 0. Robert's answer explains already why this is a problem. You're just restating the attempt OP mentioned ("is it as easy as") without acknowledging this is what you did. Do you understand the question? Commented Nov 13, 2023 at 5:10
• Ah, a simple error. You mean to say that at no point in time (after drug ingestion) is there around $40\%$ of the drug in the body? We must then discuss concentration of the drug in the body as a time function $f(t)$ so that by integration we get the average concentration which we can then use to find bioavailability which is exactly what cogito is $40\%$. What's bioavailability then if it isn't that simple. Why calculate it if it serves no purpose? Also, the $40\%$ may be the maximum amount of the drug in the body which will be important for toxicity assessment I suppose. Commented Nov 13, 2023 at 5:24
• No, that is not what bioavailability is. Bioavailability is the amount that ever gets into the body. If it takes time to be absorbed and is eliminated at any rate, it will never all be in the body at the same time. That doesn't mean it's not a useful concept, it just means it's not sufficient information to answer the question. Commented Nov 13, 2023 at 5:27
• Indeed, but $40\%$ is the grand total amount of the drug that can exert a therapeutic effect. What do you make of, $1 \times \left(\frac{1}{2}\right)^1 + \frac{3}{2} \cdot \left(\frac{1}{2}\right)^2..$ Commented Nov 13, 2023 at 5:33
• @BryanKrause, much obliged for the feedback. I edited my answer. Commented Nov 13, 2023 at 5:41