Tannin

________________________________________________________ Tannin assay for use with Spectrophotometer James F. Harbertson and Douglas O. Adams ________________________________________________________ Tannin assay Adapted from A.E. Hagerman and L.G. Butler. Protein precipitation method for the quantitative determination of tannins. J. Agric. Food Chem. 26:809-812 (1978) This method utilizes protein precipitation to allow measurement of tannins using a spectrophotometer. Materials required All materials required for colormetric assay are listed below. All reagents and materials are available separately from Sigma or Fisher. Name in Procedure

Description

Storage

Buffer A (Washing Buffer)

200 mM acetic acid 170 mM NaCl pH adjusted to 4.9 with NaOH

Stable at room temperature

Buffer B (Model Wine)

5 g/L potassium bitartrate (KHTa) 12% EtOH pH adjusted to 3.3 with HCl

Stable at room temperature

Buffer C (Resuspension Buffer)

5% triethanolamine (v/v) 5% SDS (w/v) pH adjusted to 9.4 with HCl

Stable at room temperature

Protein Solution

1 mg/mL bovine serum albumin dissolved into Buffer A

Stored at 4 ºC

Ferric Chloride Reagent

0.01 N HCl 10 mM FeCl3

Stable at room temperature

Catechin Standard

1 mg/mL (+)-catechin solution dissolved in a 10% EtOH

Make fresh each time

Hardware necessary for assay: Spectrophotometer, 1.5 mL cuvets, a set of micropipets capable of measuring in the range of 1-1000 µL, 1.5 mL microfuge tubes, a microfuge test-tube rack, benchtop vortex and a Microcentrifuge capable of at least 14,000 RPM.

Procedure Standard Curve Perform all incubations at room temperature. 1. Prepare standard curve (+)-Catechin samples by taking from 50 µL to 300 µL of standard Catechin solution and adjusting the volume to 875 µL with Buffer C. (e.g. 100 µL Catechin solution plus 775 µL of Buffer C) 2. Add 125µL of the Ferric Chloride Reagent and mix. Make a zero tannin sample with 875 µL Buffer C and 125 µL of Ferric Chloride Reagent. Read the absorbance at 510 nm and record the value. This absorbance must be subtracted from the absorbance obtained from the standard tannin samples and the wine samples. 3. Incubate the standard samples and the zero tannin for 10 minutes. 4. Read the absorbance at 510nm. Wine Samples This procedure describes how to measure tannins in wine samples. 1. Prepare Protein Solution by dissolving BSA into Buffer A. 2. Dilute wine samples in Buffer B. Wines typically need to be diluted from 1:1 to 1:9 to give tannin values in the range of the standard curve. 3. For each wine sample, pipette 1mL of the Protein Solution into a microfuge tube and then add 500 mL diluted wine sample. Incubate for 15 to 30 minutes with slow agitation. 4. Centrifuge samples for 5 minutes in the microfuge (14,000 RPM). 5. Carefully pour off the supernatant, retaining the pellet in the microfuge tube. 6. Slowly add 250 µL of the Buffer A to the pelleted sample. 7. Centrifuge for 1 minute (14,000 RPM). 8. Repeat steps 5, 6 and 7 to wash the pellet a second time. 9. Carefully pour off the supernatant and then add 875µl of Buffer C and incubate for 10 minutes. 10. Mix the tube until the pellet is completely dissolved. 11. After disolving the pellet let the solution stand for 10 minutes and then read the absorbance at 510 nm before adding the Ferric Chloride Solution. Record the value. Then add 125 µL of the Ferric Chloride Reagent, mix the sample, incubate 10 minutes and re-read the absorbance at 510 nm. 12. The amount of tannin in wine samples is calculated using the standard curve, after the background absorbance (the one obtained before the Ferric Chloride Reagent addition) is subtracted from the final absorbance (after adding the Ferric Chloride Reagent). The absorbance due to tannin is: [(Final A510) - ( A510 from zero tannin sample)]- (background A510 *0.875)

Sample Standard Curve for Tannin Determination

The General Equation to determine the concentration of tannin in a wine is:

2x

(

Abs - Intercept Slope

)(

Dilution

)

For example if you use this standard curve and your final absorbance is 0.500 and dillution is 5:

2 x(

0.500 - 0.0075 0.0053

)(

5

)

= 929 mg/L CE (Catechin Equivalents)