Question

Given a glucose standard concentration of 200 mg/dL, with an absorbance of .640. Your patient's serum has an absorbance of .252. Calculate the concentration of glucose in this patient's serum.

Answer 1

concentration of glucose = 78.75 mg/dL

Step-by-step explanation:

The question essentially wants to test the ability to calculate the concentration of a patient's test result done on a spectrophotometer using the absorbance from Beer-Lambert's law, which states that when incident light passes through a medium, the absorbance is directly proportional to the concentration of the medium and inversely to the length of the light path.

Mathematically it is represented as

Absorbance (A) ∝ Concentration (C) (or length of path)

A ∝ C

A = kC

where "k" represents the factors that are kept constant.

As a result we can rewrite the formula as:

A₁ = C₁ - - - - - (1)

A₂ = C₂ - - - - -(2)

And dividing both equations:

`(A_1)/(A_2) = (C_1)/(C_2)`

Next, let us define what is "standard" is; in analytical chemistry, a standard solution is one containing a precisely known concentration of the analyte in question, and it can be applied into the Beer-Lamberts law as follows:

`(A_T)/(A_S) = (C_T)/(C_S) \\Where: \\A_S = Absorbance\ of\ test\\A_T = Absorbance\ of\ standard\\C_T = Concentration\ of\ test\\C_S = Concentration\ of\ standard`

`Making\ C_T the\ subject\ of\ the\ equation\ by\ cross-multilication\\C_T * A_S = A_T * C_S\\C_T = (A_T)/(A_S) * C_S\\where:\\ A_T = 0.252\\A_S = 0.640\\C_S = 200mg/dL\\\therefore C_T = (0.252)/(0.640) * 200\\C_T = 78.75\ mg/dL`