Question

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 = 0.0024x - 0.0529. 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, 0.94 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 0.767 via spectrophotometry.

How do I find the % of reducing sugar given this information??

Answer 1

According to Beer-Lambert's law: A = cl

Where is the molar absorption coefficient, c is the concentration of solution and l is the path length.

Now, 0.767 = 0.0024 mL/g * c - 0.0529

i.e. c = (0.767 + 0.0529)/0.0024 = 341.625 g/mL

The total volume of the solution = 0.2 mL + 0.8 mL + 2.0 mL = 3.0 mL

Therefore, c = 341.625 g/mL * 3 mL = 1024.875 g = 1.024875*10^-3 g

Now, the percentage of glucose in the sample of ripe bananas = (mass of glucose/mass of sample of

ripe banana) * 100 = (1.024875*10^-3 g/0.94 g)*100 = 0.11%