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Topic: Calculating Dosage Given Volume and Molarity  (Read 5727 times)

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Offline ChrisPowers

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Calculating Dosage Given Volume and Molarity
« on: January 13, 2014, 05:52:40 PM »
Hey guys,

I was hoping one of you chemistry whizzes might be able to help me.

I, for all intents and purposes, flunked high school chemistry.  Now, I'm trying to analyze a medical journal article, but to do so, I need to figure out how much of a specific substance is present in a given dosage, and every time I do the math, the result I obtain doesn't seem feasible.  Here's what I'm trying figure out:

A subject is given a dose of 100 μL of 10 μM via subcutaneous injection of a substance with molar mass 302.236.  I'd like to know how much of the substance is present in each dose.

The article mentions that the substance is (presumably dissolved) in 10% dimethyl sulfoxide in phosphate-buffered saline.  I didn't feel that part was relevant with regard to calculating the amount in each dose, but maybe this is the source of my error?

Anyway, when I do the math, it yields 0.000000302236 g, or 3.02236 x 10-7 g.

Now, granted, the subjects in this experiment were lab mice (which I estimate at 24 g to 25 g at the time of administration), but 0.000302236 mg per dose (one daily) seems absurdly low to me when you consider that the suggested oral dose in humans is on the order of a gram and up.

Assuming my 24 g estimate of mouse weight, 0.000302236 mg / 0.024 kg bodyweight yields 0.012593167 mg/kg, while the suggested dose for humans is in the range of 12.5 to 25 mg/kg.  No conversion from mouse to human or vice-versa that I can find would justify these results.

I realize that subcutaneous injection probably results in much greater bioavailability, but I still can't imagine that explaining such an enormous disparity in relative dose.

I'm assuming that my calculation must be woefully incorrect.  Can anyone confirm this result?  I'm happy to take you through my steps so you can all have a laugh.

Thanks,
Chris

Offline Borek

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Re: Calculating Dosage Given Volume and Molarity
« Reply #1 on: January 14, 2014, 03:14:09 AM »
μ means 10-6.

So you have 100×10-6 (100 μL) times 10×10-6 (10 μM) times 302 g/mol.

100×10-6 L × 10×10-6 mol/L × 302 g/mol = 3.02×10-7 g

Nothing wrong about your calculations. And you are right about solvent being irrelevant to the dose calculations.

I wonder if they really used μM concentration. Why not something unusual, mM would put the result in the expected mg/kg range.
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Offline ChrisPowers

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Re: Calculating Dosage Given Volume and Molarity
« Reply #2 on: January 17, 2014, 12:26:02 AM »
Hey Borek,

Thanks a lot for replying and double-checking my math.

I wasn't clear on your last comment regarding the mM solution: "Why not something unusual, mM would put the result in the expected mg/kg range."  Are you saying that that concentration is or is not unusual?

Taking that theory to its conclusion, I get roughly 12 mg/kg for a mouse dosage.  Now, 12.5 to 25 mg/kg is the range I mentioned, but that's a human dosage.  Do you happen to know if it's common practice to not do any conversion between species?

The problem I run into is that all the conversion recommendations I see result in the human dose in mg/kg being much smaller than the mouse dose.  For instance, an FDA Guidance document recommends multiplying the mouse dose (in mg/kg) by 0.08 to obtain the human equivalent dose (in mg/kg).  Even going to the source for these recommendations and correcting for my estimated mouse and human weights still yields a very small number.

Even assuming the mM concentration, once converted, the human dosage still seems way too low to be meaningful.  I wonder if, at that point, such a difference in efficacy can be attributed to the route of administration alone (subcutaneous in the mouse experiment vs. oral in typical human usage)?

Thanks again,
Chris

Offline Borek

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Re: Calculating Dosage Given Volume and Molarity
« Reply #3 on: January 17, 2014, 02:59:38 AM »
Sorry, I should be more careful about what I wrote. It was intended to mean "While use of μM is not unusual in this context, mM would put the result in the expected range".
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