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%**

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