Question

I mixed 160 ul of serum with 40ul of water. I then pipetted 1000ul of glucose oxidase reagent into the tube. I ran the test and got a glucose result of 107 mg/dL.

Using that result what is the expected glucose concentration of the original sample.

Answer 1

Concentration of glucose in final solution (M₂) = 107 mg/dL.

Volume of the serum solution (V₂) = (160+40) ul = 200 ul.

Initial volume of the serum (V₁) = 160 ul.

Initial concentration of glucose in serum = M₁ .

For dilution,

Moles of glucose = M₁V₁= M₂V₂

Or , M₁ × 160 (ul) = 107 (mg/dL) × 200(ul)

Or, M₁ = {(107×200)/160)}

Or, M₁ = 133.75 mg/dL.

Therefore,

expected glucose concentration of the orginal sample is 133.75 mg/dL.