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# ELISA dilution factor calculation and result interpretation-URGENT

ELISA kit dilution factor

Best Answer DRT, 09 April 2014 - 11:48 AM

I understand the df=5, but how do I get df=3? By approximating the volume of the weighed sample(1g~ 1mL)?

Yep

Doesn't help the problem you have with the 10x higher concentrations though.

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### #1 Antonija

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Posted 08 April 2014 - 09:51 AM

Dear all,

I' m in a need of an urgent assistance I'm a newbie, working at a reference laboratory for food analysis and I recently got an assignment to validate a method for detection of an antibiotic in muscle tissue in 5 days period, using an Euro-Proxima kit with ready to use standards for the calibration curve. I have set up the method for my spectrometer (Tecan), using the Magellan programm.

For the purpose of validation, I had to spike my negative samples with a standard received from Sigma Aldrich to a certain concentration. The real concentration is of course calculated using the method dilution factor (the antibiotic equivalents obtained from the graph are multiplied by it)The dilution factor for muscle tissue in the kit manual is f=15. If I use that factor, my concentrations end up cca 10 x higher than they should be (instead of 140 ppb, I get about 1500 ppb). Thus, I tried to calculate the factor myself and obtained f=4, used it for the calculation and the results around the value of spiking.

So I don't know if the dilution factor written in the manual is wrong, or my calculation or maybe something else.

The procedure:  1g of muscle sample, 2ml of a solution added, centrifuged and 40 microliters of supernatant soluted in 160 microliters of a solvent. In the end, 50 microliters of the end solution used for the test. ( my calculation: 1g/2 mL * 40 microL/ 160 microL= 0.125 g/mL ---- c1= 0.5 g/mL, c2= 0.125 g/mL----- c1/c2= 4)

the test procedure: after the washing process, 100 microliters of substrate added and 100 microliters of stop solution-- I didn't take the test into consideration for my dil.factor)

What should I do? What is correct?  Help!

Edited by Antonija, 08 April 2014 - 09:57 AM.

### #2 tkf

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Posted 08 April 2014 - 10:50 AM

I don't completely understand your question, but...

40uL added to 160uL solvent is a df =5, (160+40)/40 = 5

### #3 Antonija

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Posted 08 April 2014 - 11:41 AM

Thnx for you're answer, but I'm asking about the method dilution factor needed for the calculation of real concentrations from absorbance values (how many times the sample is diluted during the purification step) because I spiked the samples (added a certain concentration of an antibiotic standard) which is needed for the validation of the method. So, the factor is needed to calculate the antibiotic conc. because both the sample and the antibiotic standard added to the sample got diluted during purification.

Edited by Antonija, 08 April 2014 - 11:46 AM.

### #4 DRT

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Posted 08 April 2014 - 12:44 PM

You have not been dividing by the total volume.

Thus there is a 3 fold dilution during the extraction from muscle  multiplied by the 5 fold dilution that tkf pointed out = 15 fold dilution.

### #5 Antonija

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Posted 08 April 2014 - 10:14 PM

I understand the df=5, but how do I get df=3? By approximating the volume of the weighed sample(1g~ 1mL)?

### #6 DRT

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Posted 09 April 2014 - 11:48 AM   Best Answer

I understand the df=5, but how do I get df=3? By approximating the volume of the weighed sample(1g~ 1mL)?

Yep

Doesn't help the problem you have with the 10x higher concentrations though.

### #7 Antonija

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Posted 11 April 2014 - 04:16 AM

OK, thank you! At least I know that is not he problem