Question

1) Tell me how you would prepare 200 mL of a 200 mM MOPS (3-[N-morpholino]propanesulfonic acid, FW 209.3) buffer at pH 7.5 using the Henderson Hasselbalch equation. You have only the solid acid form of MOPS available to you to make the buffer. How else could you prepare this buffer in a more speedy and pratical way? Be specific as possible.

2) I isolated a pure protein (W) and determined the protein content using the UV, Lowry, BCA, and Coomassie protein assay methods using BSA as a standard. I obtained the following protein concentrations for the isolated sample W: UV -0.5 mg/mL ; Lowry - 0.8 mg/mL ; BCA -1.0 mg/mL ; Coomassie -2.0 mg/mL. Provide a reasonable explanation for these results.

Answer 1

Question 1.

The no. of millimoles of buffer = (200/1000) L * 200 mmol/L = 40 mmol

i.e. [MOPS-O^-] + [MOPS-OH] = 40 mmol ........... Equation 1

According to Henderson-Hasselbulch equation:

pH = pKa + Log([base]/[acid])

i.e. 7.5 = 7.2 + Log([MOPS-O^-]/[MOPS-OH])

i.e. Log([MOPS-O^-]/[MOPS-OH]) = 7.5-7.2 = 0.3

i.e. [MOPS-O^-]/[MOPS-OH] = 10^0.3 ~ 2

i.e. [MOPS-O^-] = 2*[MOPS-OH] ........... Equation 2

From equations 1 and 2: 2*[MOPS-OH] + [MOPS-OH] = 40 mmol

i.e. [MOPS-OH] = 40/3 = 13.333 mmol = reamining amount (unreacted amount)

i.e. The actual mass of MOPS required to make the buffer = 40 mmol * 209.3 g/mol = 8.372 g

i.e. [MOPS-O^-] = 40 - 13.333 = 26.667 mmol = [NaOH] (say)

The mass of NaOH required to make the buffer = 26.667 mmol * 40 g/mol = 1.067 g