Good afternoon
Question is related to phenolic compounds. The method was performed as follows:
- First, 20 ml of ethanol were added to 0.1 g of dry sample. Second, the extract (0.1 ml) was used to react with Folin Ciocalteu and Na2CO3. Finally, the absorbance (0.96) was measured at 725 nm.
- Equation from a standard curve of gallic acid was used: y = 3.6554x + 0.0893. Where y is Absorbance and x is gallic acid concentration (mg/ml).
- The result was obtained as follows: 0.96 = 3.6554x + 0.0893 and finally
x = 0.238 mg/ml.
How do I express this in mg/g of dry weight?
Thank you very much in advance
Dear Dr. Omar
To express concentration from mg/ml to mg/g dry weight it is needed to multiply 0.238 by the dilution factor, which is 20 / 0.1.
Thank you.
To express concentration from mg/ml to mg/g dry weight:
the total phenolic content was calculated as gallic acid equivalent (GAE) by the following equation:
T =Cx V/M .
T is the total phenolic content in mg·g–1 of the extracts as
GAE, C is the concentration of gallic acid established
from the calibration curve in mg·ml–1, V is the volume of
the extract solution in ml and M is the weight of the extract
in g
Dear Dr. Amrani
It is similar to multiply the result by dilution factor using volume (20 ml) and sample weight (0.1).
Thank you.
Regards.
Dear sir
The total phenolic content was calculated as ug gallic acid equivalent / mg exttact (ug GAE/ mg E) " the same of mg EGA/ gE" by the following equation:
TPC=(0.96-0.0893)/(3.6554*5)=47 mg EGA/g E
When the concentration ofsample is 0.1/20=5 mg/ml (not 20/0.1)
Thanks
Dear Dr. Benslama
If initial concentration of sample is known (0.1 g/20 ml = 0.005 g/ml) it will be replaced in 0.238 mg/ml. So, 0.238 mg will be divided by 0.005 g = 47.6 mg/g dry weight. You're right!
But, it is similar to know ratio between 0.1 g and 20 ml. So it is achieved by dividing 20 ml/0.1g = 200. So, 0.238 mg/ml * 200 = 47.6 mg/g. It's like Dr. Amrani's equation:
T = C * V / M
T is the total phenolic content in mg·g–1 of the extracts as GAE, C is the concentration of gallic acid established from the calibration curve in mg·ml–1, V is the volume of the extract solution in ml and M is the weight of the extract in g.
Thank you!
Dear Andrés
It is important that you also take into account the humidity of the material so that reports on dry basis
Hello Cristian
How could I use humidity to express mg GAE/g dry weight? Are you saying that without using humidity, conversion is not complete?
Regards
Dear Dr. Amos
Final result is 47.6 mg/g (T = C x V / M) but it is obtained using 0.1 g (mass, M) of plant sample. I understand that there are 47.6 mg in 1 g of plant sample according to formula. But, if 47.6 mg/g was obtained using 0.1 g, then for 1 g of plant sample should be 476 mg/g. Let me know if I'm wrong. Thank you in advance.
Your answer is already per gram of sample. Olaniyi is talking nonsense. His answer will only contribute towards confusion.
Dear Omar
You're right. My answer is already per g of sample. I think it's not necessary to work starting from 1 gram because phenolics in food are different and require certain amounts of mass to be extracted and quantified.
Thank you.
The total phenolic content was calculated as ug gallic acid equivalent / mg exttact (ug GAE/ mg E) " the same of mg EGA/ gE" by the following equation:
TPC=(0.96-0.0893)/(3.6554*5)=47 mg EGA/g E
When the concentration ofsample is 0.1/20=5 mg/ml (not 20/0.1)
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