Shalini Gupta

I don't fully understand your experiment Muhammad. I presume that the 1,5-diphenylcarbazide reacts with Cr(III). It may be that other constituents of the microbial system could produce coloured species that would absorb nearby the Cr (III) chromophore, and/or there might be a competing complexing agent. That is why it is good practice to spike Cr (III) into the meduim and compare the colorimetric response with Cr (III) without the medium.

Respected Sir Paul Milham, Thank you for your kind consideration. I'm using bacteria for the reduction of Cr (VI) into Cr (III) as literature read by me i.e., Young Hak et al., 2003, support that 1,5-diphenylcarbazide forms colored complex with Cr (VI). I'm using US EPA 7196A method which is for Cr (VI) detection and has been used by many authors for similar work of my interest. So dear sir, can you please recommend me some literature or protocols to detect Cr (III) with 1,5-diphenylcarbazide so I can further enhance my knowledge with respect to your directions and apply in my experimental work.

Thank you in anticipation.

The blank could be the nutrient broth with the bacteria inoculated but without any Cr(VI) spiked in it. If it is a specific bacterial species and not a mixed culture then this blank will definitely be appropriate. This is because any colored compound or complex formation by the specific species will remain uniform in each condition. Moreover, the DPC will only be complex with Cr(VI) to produce purple diphenylcarbazone. So there is no need to spike the blank.

However, if you are studying a particular concentration (let say 200 ppm of Cr(VI)), then just use the blank as the uninoculated nutrient broth spiked with 200 ppm of Cr(VI). You will get the readings in comparison to blank as negative values. In this case, ignore the negative sign and the values give you the reduction percentage.