Spiking solutions

Discussions about sample preparation: extraction, cleanup, derivatization, etc.

8 posts Page 1 of 1
I am conducting an experiment that requires me to spike a biological matrix, let's say 2 mL urine, with a drug. I have read some papers where the author says that the drug standard (in 2 mL of urine) was prepared at conc. 250, 500, 1000 ng/mL by adding 50, 100, and 200 uL of drug standard mixture (conc. 20 ug/mL). I fail to understand how they got the final concentration of 250, 500 and 1000 ng/mL.

I also don't understand why we can't simply spike a matrix (mine will be 0.5 mL) with a drug std that is 100 ug/mL. If I spike it with 100 uL or 1 mL, the conc. will be still 100 ug/mL.

I would greatly appreciate your help!
How much drug (micrograms) is in 1 ml of standard ?. If you add the 1 ml of standard to 0.5 ml of matrix, what is the final volume that the drug is now dissolved in ?

Peter
Peter Apps
Thank you for your reply.

Initially my drug was 1 mg/mL but I have decided to dilute it to 100 ug/mL. Therefore there is 100 ug of drug in 1 mL of methanol. My matrix volume is 0.5 mL therefore spiking it with 1 mL of drug is pointless. I want to spike it with 100 uL of 100 ug/mL drug (or perhaps 10 uL?). Does it make sense? This is not the way authors do or write in their papers which means I am missing something.
One more thing:

I have found a procedure of extraction of a drug from a biological matrix where it says:
Cal 2: 50 (ng/mL)-> 100 uL of 1 ug/mL standard
Cal 3: 100 -> 200 uL of 1 ug/mL standard
Volume of sample: 2 mL

That means that when I spike my solutions with 1 ug/mL STD, the concentration of drug is not 1 ug/mL (in the matrix). It is (1*100)/2000 = 50 ng/mL

Is this a correct approach to prepare drug standards in biological matrices?
This is really very simple:

Calculate the mass of drug you have in the volume of standard that you use.

Calculate or measure the final volume of sample plus standard.

Divide the mass by the volume to get the concentration in the spiked sample (assuming the sample had no drug in it to begin with).

Peter
Peter Apps
Something else to consider, when you add a relatively large volume of methanol to your urine you no longer have urine but a mixture.
This is one reason to keep the spike volume low compared to the sample volume.
amateur23 wrote:
One more thing:

I have found a procedure of extraction of a drug from a biological matrix where it says:
Cal 2: 50 (ng/mL)-> 100 uL of 1 ug/mL standard
Cal 3: 100 -> 200 uL of 1 ug/mL standard
Volume of sample: 2 mL

That means that when I spike my solutions with 1 ug/mL STD, the concentration of drug is not 1 ug/mL (in the matrix). It is (1*100)/2000 = 50 ng/mL

Is this a correct approach to prepare drug standards in biological matrices?


1ug/ml is equal to 1ng/ul. If you spike 100ul into your 500ul matrix you now have 100ng in 600ul total volume which would be equal to 166.6ng/1ml.

If you make your spiking solution 10x more concentrated, then you have 10ng/ul and you can spike 10ul into the 500ul of matrix which gives you 100ng/510ul which would be 196ng/1ml.

If you make your spiking solution 100x more concentrated, then you have 100ng/ul and you can spike 1ul into the 500ul of matrix which gives you 100ng/501ul which would be 199.6ng/1ml.

When you inject onto the instrument, it does not read total mass per total amount of matrix, it reads mass per volume injected. The only way to make it work without having to calculate exact concentrations for your standards in you matrix would be to inject the full amount of matrix plus standard in to the instrument in a single injection. You also have to take into account what effects on instrument response are caused by injecting matrix alone, or matrix plus methanol from the standard. By adding less of a more concentrated spiking solution you introduce less effect of the mixed matrix.
The past is there to guide us into the future, not to dwell in.
Thank you all for your answers. It seems very straightforward now, I appreciate your help!
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