I use pH-differential method to measure Anthocyanin content in red cabbage.
Im just not sure if i calculated it right
This is how i do it:
20 gram fresh red cabbage in 200ml solvent (ethanol:water 1:1 Hcl) => filted => 200ml extract
Put 5ml of extract in 20ml buffet pH 1.0 and pH 4.5 (total volumn 25)
The absorbance of the extract at two different wavelengths (520 nm and 700 nm) and two different pH values (1 and 4.5) are:
I use this formula:
a= (A*MW*DF*1000)/(e*l)
a: Monomeric anthocyanin pigment (mg/liter)
A = (Aλ520 – A700)pH 1.0 – (Aλ520 – A700)pH 4.5
MW = 449.2
DF is the dilution factor (is the DF is 5? (25/5))
ε = 26,900
Can anyone help me calculate %anthocyanin in this fresh weight of redcabbge?
If i got The moisture content is 87.47% what % anthocyanin?