Question

-1,5-anhydro-D-glucitol (AG) is one of the main sugar alcohols in human blood. In plasma, the normal AG level is 24.6 ± 7.2 mg/L, while patients with diabetes show reduced AG levels of 7.3 ± 7.1 mg/L. Based on this fact, a diagnostic enzyme assay for diabetes has been developed.

-AG + O2  AG-Lactone + H2O2 (PROD = pyranose oxidase) EX ABTSred + H2O2  ABTSox + 2 H2O (EX = enzyme X)

-The oxidized form of ABTS absorbs at 420 nm (ε = 3.48 x 104 M-1 cm-1 ). The AG assay involves the measurement of this product following a one-hour incubation period at 37°C. A 1 mL patient sample was diluted to 10 mL and measured at 420 nm in a 1 cm cuvette. An absorbance value of 0.171 was observed.

QUESTION:  Is the patient diabetic? Use calculations to support your answer.

Answer 1

From the law of Lambert-Beer:

A = E * L * C

we clear the concentration:

C = A / E * L = 0.171 / 3.48x10 ^ 4 M-1 cm-1 * 1 cm = 4.91x10 ^ -6 M

We know that this concentration is for the 10 mL of diluted sample, then by means of a ratio we calculate the concentration for 1 original:

c: concentrated

d: diluted

Cc * Vc = Cd * Vd

Cc = Cd * Vd / Vc = 4.91x10 ^ -6 M * 10 mL / 1 mL = 4.91x10 ^ -5 M

We apply the conversions and next to the molar weight of AG we have:

4.91x10 ^ -5 mol / L * (164.16 gr / mol) * (1000 mg / 1 g) = 8.06 mg / L

Yes, we are dealing with a diabetic patient.