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I would like to know why do we not need to take account of Bradford agent solution in dilution calculation?

For example, $\pu{2.5 \mu g mL-1}$ BSA will calculate from $\pu{0.4 \mu L}$ of $\pu{1 mg mL-1} \ \text{BSA} + \pu{159.6 \mu L}$ of distill water. The additional Bradford reagent ($\pu{40 \mu L}$ in my experiment) does not count in the dilution calculation. This seems mysterious to me since the total volume is $\pu{200 \mu L}$. If the volume of Bradford reagent counted, the concentration should have been $\pu{2 \mu g mL-1}$.

I try to search for publications regarding to the protocol and most of them said just ignore the Bradford volume with no explanation. Could someone kindly explain the mechanism behind this?

Edited: I aim to determine protein concentration via Bradford assay. The standard curve is prepared by serial dilution of BSA + distill water + Bradford serial diluion was as below:

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I understand completely how to calculate the unknown concentration. However, I just want to know that why 'Bradford solution' volume can be ignored during the calculation.

Regards

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  • $\begingroup$ How did you prepare your calibration curve? Is BSA your standard? If so, describe how you prepare standards to make the calibration curve. So, we know what you are doing. $\endgroup$ – Mathew Mahindaratne May 27 at 6:10
  • $\begingroup$ I've edited the question, thanks. $\endgroup$ – Pete Kittinun May 27 at 6:48
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    $\begingroup$ It doesn't matter you included Bradford reagent or not in your calculation as long as you treated the same for all reading including unknowns as shown in the data. That is because, when you find the concentration of unknown sample from calibration curve, you back calculate to get the initial concentration of undiluted solution. This means you are going to substract the volume of Bradford reagent at the end. $\endgroup$ – Mathew Mahindaratne May 27 at 7:45
  • $\begingroup$ Now I get it. Thank you so much. :) $\endgroup$ – Pete Kittinun May 27 at 8:06

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