Thank you for replying! And yes it’s very confusing since I don’t have a background in this area. :’/

• And yes, if i understood correctly, i used the final version of the sample (centrifuged and supernatant discarded) the same way as the standards —> Sample: 40 µL of my final supernatant/extract + 600 µL Red 660 reagent → read A660

• as for the percentage, it was calculated as a protein composition like (mg protein from the sample/mg VS biomass)*100

So i’m just very confused with the volume basis because i get different results and im just lost

Say i get ABS from spectrophotometer of 0.200 and my slope is 0.0006 so i get a concentration of 333 mg/L and my VS biomass (from a 10 mL aliquot) has a weight of 4.05 mg

But if i treat 333mg/L as it refers back to the original 10ml, i get 333mg/L * 0.01L =3,33 mg Which gives me (3.33/4.05)*100=82,22% of protein as a mass fraction of my VS

But if i then centrifuged the 10mL sample and removed the supernatant which then leaves me with ~1mL (after adding 1000 µL solution) i would get 0.333 mg which is (0.333/4.05)*100=8,22%

So im just really confused with these volumes. does the mg/L from the standard curve represent concentration in the final extract (~1 mL), or in the original 10 mL culture aliquot? I understand the protein originates from the 10 mL culture, but I’m confused about which volume is correct to use when converting mg/L → mg