HPLC post-calculations...

[esbena]

HI all.

I could really use some help here. I´m currently working on a method for measuring creatinine in serum from mice, and have decided to use HPLC for this.

I´m using the Zorbax SCX300 2.1 x 50 mm, 5 micron, with in-front guard column, and a 5 mM sodium acetate pH 4.1 (adjusted with glacial acetic acid) as mobile phase.

My standard curve is composed of 9 concentrations of creatinine powder from Sigma (128 - 64 - 32 - 16 - 8 - 4 - 2 - 1 - 0.5 uM) dissolved in mobile phase.

This is my prep of std. and samples:

25 ul std./sample
100 ul cold Acetonitrile (AcN) + 0.5% glacial acetic acid to precipitate proteins
vortex 15 s
15 min at -20 degrees
spin 10000 rpm at 4 degrees for 10 min
transfer sup. to new epp. (100 ul of the initial 125 ul ~ 80%)
speedvav to (~ 40 -45 min)
resuspend in 25 ul mobile phase
spin 10000 rpm at 4 degrees for 10 min
transfer all 25 ul to 96-well Abgene PCR plate for Aqurity autosampler
load 3 ul/run (all samples in duplicates)

run time: 10 min
flow: 1 ml/min
column temp: 50 degrees
autosampler temp: 18 degrees
backpressure: ~ 2300 psi
UV at 225 nm

My standard curves are pretty similar with an R square of 0.96-0.99

But, and this is my real question, when I interpolate my samples with an unknown concentration of creatinine how do I calculate the final concentration of creatinine in the initial serum sample??? In either umol/L or mg/dL??

[tom jupille]

If you run your standards through the same workup as the samples, then there is no adjustment necessary: uM is uM (is micromoles/Liter).

[esbena]

Is the value for the samples directly comparable to the initial serum samples even though I only use a fraction of the initial serum?

I don´t get it!!

[shaun78]

yes, it is directly comparable. However, you need to account for your dilution factor.

It looks like you take 25 uL and dilute to a final volume of 125 uL, which is a 5x dilution of your sample. So, what ever result you come up with from the testing of your sample, you then multiply by 5.

If you want to get technical here, you actually need to divide by your dilution factor, which is actually 0.2 (25 uL/125 uL). Simple math tells you that division by 0.2 = multiplication by 5.

[tom jupille]

It looks like you take 25 uL and dilute to a final volume of 125 uL, which is a 5x dilution of your sample. So, what ever result you come up with from the testing of your sample, you then multiply by 5.

But if she runs the standards through the same procedure, that will cancel out.

Is the value for the samples directly comparable to the initial serum samples even though I only use a fraction of the initial serum?

You're measuring concentration, not mass. Again, so long as the standards and samples are run through the identical procedure, any dilution factors will cancel out.