It seems that you have a problem on the expression of the standard solutions concentrations when preparing the standard curve. The concentration using the equation should have been expressed as μg GAE / ml and not μl/ml, since, when you prepared the GA stock solution you diluted a given quantity (mg) of solid gallic acid to water. If you performed further dilutions of this stock solution to prepare the standard ones, you should not express the concentration as the amount of stock solution (in μl) to the final amount of solution. For example, if you diluted ten times a stock solution of GA with a concentration of 2.5g GA/l, you have a standard solution of 0.25 g GA/l. If you used standard solutions and reagents (e.g. FCR and sod. carbonate) at the same amount and the same procedure of measuring when you prepared the standard curve, as you did with the samples, the concentration from the equation (expressed at μg GAE/ ml, lets say 20.03 μg/ml) is the concentration of GAE within the extract of the sample you measured. That means that at 50ml of distilled water you have 20.03 X 50 = 1001.5 μg or 1,001mg GAE per 5 g of sample, and finally 20.02 mg GAE / 100 g sample.

It seems that you have a problem on the expression of the standard solutions concentrations when preparing the standard curve. The concentration using the equation should have been expressed as μg GAE / ml and not μl/ml, since, when you prepared the GA stock solution you diluted a given quantity (mg) of solid gallic acid to water. If you performed further dilutions of this stock solution to prepare the standard ones, you should not express the concentration as the amount of stock solution (in μl) to the final amount of solution. For example, if you diluted ten times a stock solution of GA with a concentration of 2.5g GA/l, you have a standard solution of 0.25 g GA/l. If you used standard solutions and reagents (e.g. FCR and sod. carbonate) at the same amount and the same procedure of measuring when you prepared the standard curve, as you did with the samples, the concentration from the equation (expressed at μg GAE/ ml, lets say 20.03 μg/ml) is the concentration of GAE within the extract of the sample you measured. That means that at 50ml of distilled water you have 20.03 X 50 = 1001.5 μg or 1,001mg GAE per 5 g of sample, and finally 20.02 mg GAE / 100 g sample.

I got 20.03µg/ml.

Yes the caluation of Karapanos is correct

According to your calculation 0.5 ml (sample) contains 20.03 µg/ml as GAE

And you have stock conc. 5gm/50ml so

So 20.03 * 50 = 1001.5 µg/ml or 1.0015 mg in 50ml (5gm)

1.0015 * 20 = 20.03 mg/100 gm

So total phenolic content in 100 gm of your sample is 20.03 mg as GAE

@ajay rana

I took 5g sample and dilute it to 50ml distilled water. From this I took 0.5ml and then mix it with 2.5ml 0f 0.2N FCR for 5 minutes and 2ml of sod. Carbonate was then added. After incubation for 2 hrs absorbance was measured at 760nm against methanol blank. Using the equation y= 0.016x+0.021. I got 20.03µg/ml. How to convert it to now mgGAE/100g sample?

5 g sample in 50 ml distilled water means 0.1 g sample per ml distilled water.Thus

20.03 µg/ml = 20.03 µg/0.1 g sample = 20.03 x 10 µg/g sample

= 20.03 x 10 x 100 µg/100 g sample

= 20.03 x 10 x 100 mg/(100 x 1000) g sample

= 20.03 mg/100 g sample

20.03 µg/ml ( 1 ml of extract ) remember you take 0.5ml

and which units for equation

The answer that you are getting is on '1 ml of extract' basis and not for 0.5 ml. As 1 ml contains 0.1 g of sample you replace 1 ml in the denominator by 0.1 g and the result follows.

Calculating back all the dilutions along an extraction series to get the correct units for a calorimetric assay is one of the trickiest things to get right. I draw all the steps along the extraction/dilution pathway as little pictures of tubes, with the g sample added to the first tube, the total volume in each tube, and the sub-sample volumes taken out from one and placed into the next. It sounds childish, but it works for me (30 years in biochem).

Its also simplest if you base your calculation on the milligram or millimole amount of standard in the cuvette that corresponds to the absorbance of your sample, not the concentration of standard (mol/L).

i think you may need this equation

mg/100 gm Dry Wight = O.D X Solvent volume up X 100 / slope X sample weight X 100

In you case, if you back calculate for solid content, it is 0.5 ml belongs 50 mg of solids.

then,

50 mg should give 10.01 ug of GAE (w/w)

100 g belongs to 100000 mg.

If 50 mg gives 10.01 ug GAE

then

100 g should give 20020 ug of GAE

that means

20.02 mg GAE per 100 g sample.

I think, Ioannis Karapanos , Shanker Lal Shrivastava are got the same answer and it is correct.

Can i ask you my analyse.

i extracted 2,5 g dry weight in 250 ml solution. then i used 0,5 ml extract and diluted it with dH2O, FCR and sodium carbonate to 12,5 ml total value. i read absorbance at 750 nm. i prepared gallic acid standard curve as mg/l and my exuation is y=0,0059x. so when i use the curve i calculated the value as 34 mg/l. can anyone help me to find the final concantration as mg/100 g dry weight? thanks!

No answer for this question?

0.5 ml extract equivalent 50 mg of sample 

so 20.03µg/ml = 20,03 ug / 100mg

0.203 mg / g

= 20.3mg /100g

 i need to help in determining my TPC. i used a Gallic acid in preparing my calibration curve with conc ranging from 0.02-0.1mg/ml and obtained a GAE of 0.1548mg/ml from the curve, but am unsure of how to convert it to GAE mg/g. i used 0.5ml of my standard/extract, 2.5ml FCR, 2ml of Na2SO3 making a final assay volume of 5ml. my extract concentration was 1mg/ml. thanks

 Dear Ohikhena,

For you to convert 0.1548mg/mL to GAE mg/g. All you need to do is to multiply the value with dilution factor. In this case, the dilution factor is 1000

Therefore TPC = 0.1548* 1000 = 154.8 mg GAE/g.

Regards.

Dear Sir

I have same problem.

I used stock solution of 1mg/ml of both extract and standard...then i used I ml of different concentrations (5ug-100 ug) of Standard and extract and then added 1.25 ml of FC reagnt and 1 ml of Na2CO3 and final volume of reaction mixture was 3.25.

I got equation:  y=0.0226x+0.0304. and i get O.D.- 0.085,0.173,0.267,0.351 for 5 ug,10ug,20 ug and 40 ug respectively.

How to solve it..please help me.

 Dear Kumar,

The equation y = 0.0226X + 0.0304 is what you are going to use to find gallic acid equivalent and then multiply with dilution factor to give you the TPC

Regards

Thanks Alkasim

How can I convert mg 476 GAE/100 g in mg/100g?

mmol/100g GA eq. to percent converter????

Hi guys, I have a question.

I have extracted 0.5 g biomass using 5 mL of solvent. From this, 40 uL were used to calculate the total phenolics content (TPC).

40 uL extract + 1560 uL distilled water + 100 uL Folin reagent + 300 uL Na2CO3. Using absorbance of 725 nm, and based on gallic acid calibration curve, we have only got 25 mg/L GAE. How can I convert it to mg/g biomass?

Does the dilution using 1560 uL distilled water affect the calculations?

Thank you.

Hi..I need to calculate TPC in mg/GAE (g) by Folin Ciocalteu colorimetric assay.

Standard gallic acid of 0mM, 0.25mM, 0.50mM, 0.75mM, and 1mM was used for calibration curve. I got 0.057 absorbance of my extract ( 0.05mg/mL) whereas the straight line equation in MS excel is Y= 0.0224x+0.0398 while R^2 =0.9987.

i don't know how to use this data for determining TPC value. Kindly help me regarding this issue.

Looking forward for response.

Thankyou.