For an experiment I am trying to calculate the reducing sugar content as a percentage of the original weight of a sample of ripe bananas and am having some difficulty doing so.
Some background info:
First, I prepared a calibration curve of the reduction of DNS reagent by the reducing sugar glucose (absorbance vs. concentration of glucose in µg/mL) and obtained the equation y = 0.0025x - 0.0322. This equation will later be used to solve for the concentration of reducing sugar in the supernatant.
The mass of the sample of ripe banana, 1 g, was immersed in distilled water and following centrifugation, the sample was decanted. The total supernatant volume was 10 mL. Next, I prepared a test tube containing 0.2 mL of supernatant, 0.8 mL water, and 2.0 mL DNS reagent. The absorbance of the banana sample was measured to be 0.706 via spectrophotometry.
Here is my attempt at the calculation:
First, plugging the absorbance into the equation y = 0.0025x - 0.0322 to solve for the concentration of reducing sugar in the supernatant, the x value. The result is 295.28 µg/mL.
To account for the dilution and solving for the concentration of reducing sugar in the test tube used to measure absorbance:
(295.28 µg/mL)(0.2 mL) = (C2)(0.2 + 0.8 + 2.0 = 3.0 mL)
C2 = 19.68 µg/mL
Next, multiplying this concentration by the total supernatant volume:
19.68 µg/mL x 10 mL = 196.8 µg = 1.968 x 10-4 g
Finally, calculating the % of reducing sugar content in ripe banana sample as follows:
(grams of reducing sugar in banana / grams of banana)(100) = (1.968 x 10-4 g / 1 g)(100) = 0.01968 %.
According to literature values, the reducing sugar content in ripe bananas is usually around 10-15% on average. Thus, my answer does not seem accurate which is why I am questioning if my method for solving the problem is correct. Is there anything in my attempt at the solution that seems incorrect?
All guidance is very much appreciated. Thank you.