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Help me understand sample dilution in ELISA

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#1 Orpheus

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Posted 04 February 2019 - 10:03 PM

1. In the ELISA that I'm using, I create a 4000 pg/mL high standard and 1:2 dilute down to a 62.5 pg/mL low standard. However, when I add the standards (as well as the samples) to my plate, they are added in 50 ul to another 50 ul of assay diluent (100 ul total volume per well). This makes me think that each sample/standard is AGAIN diluted 1:2. Is this dilution valid or do you ignore the 50 ul of assay diluent since each sample is treated that way (i.e. you do not use a dilution factor)? The kit seems to imply that this "dilution" is ignored when it goes on to explain creation of the standard curve and subsequent calculations.
2. As stated above, this ELISA assay begins with 50 ul of assay diluent in each well, to which I add 50 ul of sample/standards. My question is, what if you do not have 50 ul of sample? Is it valid to add amounts not equal to 50 ul? For example, if you only added 25 ul of sample, would you then just multiply by 3 at the end (75/25 = dil. factor of 3). Or do you not multiply by anything since you added your sample "neat"? And lastly, what if I add amounts less than 50 ul AND also use different amounts for different samples (e.g. 25 ul added for one sample, and 10 ul added for another sample). Is this okay so long as I take into account a relevant dilution factor? And what would that factor be given the issue described in #1 above?
Thanks a lot!

#2 Sotirios

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Posted 05 February 2019 - 11:58 AM

Hello Orpheus,

First of all, when a method treats the sample and standards in the same way, there is no dilution factor. So, there is no reason to multiply by a factor the final result.

When a method uses a diluent, it is because there is a need to make the samples/standards more compatible with the method. Maybe it is due to the matrix effect or to form a better standard curve in terms of the range. This means that you do not use different volumes and then calculate the result with factors respectively. For instance, the only way to do this, is only if you add 10μl of the sample, 40μl of the standard 1 (in order to keep the appropriate matrix) and then 50μl of the diluent. In this case, indeed you have a dilution factor 5. However, the final volume must be 100μl because it is crucial for the reaction inside the well.

Sotirios

Edited by Sotirios, 05 February 2019 - 11:59 AM.

Athanasiou Sotirios

#3 Orpheus

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Posted 05 February 2019 - 01:04 PM

"First of all, when a method treats the sample and standards in the same way, there is no dilution factor. So, there is no reason to multiply by a factor the final result.

When a method uses a diluent, it is because there is a need to make the samples/standards more compatible with the method. Maybe it is due to the matrix effect or to form a better standard curve in terms of the range."

Okay, this part I can understand. The rest I'm not so sure of (I can understand the idea that the reaction must have 100 ul (all of them) as this intuitively makes sense). But I'm not sure I understand your example and also why you say 40 ul of STANDARD 1 (why would I mix sample and standard??).

Maybe you could explain it again another way?

Thanks a lot!

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