[daveC2012]

Hi,
Ive to develop an extraction procedure based on the following example but I can’t figure out how the original author came up with the concentrations:

Method 1:
20 ul of compound x (1000ng/20ul or 50,000ng/ml) is added to milk (1g).  4 mls of methanol is added.  After shaken and centrifugation the methanol layer is removed and evaporated under nitrogen.  The extract is resuspended in 1 ml of ethyl acetate.   100 ul of this is diluted in 900 ul of buffer before analysis.
This results in a final concentration 50 ng/ml in the extract.

Method 2:
20 ul of compound x (1000ng/20ul or 50,000ng/ml) is added to milk (1g).  4 mls of methanol is added.  After shaken an centrifugation the methanol layer is removed and evaporated under nitrogen.  The extract is resuspended in 1 ml of hexane and 250 ul of water/solvent mix.  After shaken and centrifugation 200 ul of water/solvent mix is removed and evaporated to dryness.  The extract is resuspended in 1 ml of ethyl acetate.   100 ul of this is diluted in 900 ul of buffer before analysis.
This resulted in a final concentration of 40 ng/ml in the extract.

I can’t understand the dilution factor.  It looks like compound x was diluted by 1000 in method 1 and 1250 in method 2 but how?

[JGK]

The dilution factor for extraction process 1 is 10

The dilution factor for extraction process 2 is 50


Oh, and hexane and water are immiscible so I don't know how extraction 2 could be made to work

[MarkHolland]

Dilution factor is the final volume / aliquot volume.aliquot volume is the measure of sub volume of original sample . final volume is the total volume.

dilution factor =final volume /aliquot vol.

for example ; what is the df when you add 2ml sample to 8m???

total vol is 2+8=10

DF=total vol/aliquot.

10/2=5

so 5 is dilution factor

[Hunter2]

The dilution factor for extraction process 2 is 50

I got Dilutionfactor 62,5 for example 2.

20 µl is dissolved in 1 ml hexane and 250 µl of the aquaeous mixture. Even it is not mixable, but I think it should show Nernst distribution the total volume is 1250 µl; from this solution 200 µl taken means 3.2 µl . This diluted by ten gives 0.32 µl in the final. And this is a dilutionsfactor of 62.5

[DrCMS]

I think the first one is 1 in 10 dilution and the second is 1 in 12.5 dilution, assuming compound x has zero solubility in hexane; which is what the analysis data suggests.  Both methods seem to only have a 50% detection efficiency.

[Hunter2]

But why to do the step with hexan. In your calculation it looks like you take only 250 µl of the mixture. And get later 200 µl of it. What gives 12.5 dilution.

[DrCMS]

As the only difference between the two methods is the hexane extraction and that is method 2 I guess the hexane step was added to remove one or more hexane soluble compounds from the milk that interferes with the analysis.  I'm not sure if compound x is an internal standard or it reacts with some component in the milk but the analysis only detects about half of the amount added.