Question 470a1

Apr 12, 2015

I made some assumptions, but here's how to do it.

Step 1. Calculate the concentration of the original protein

This comes from Beer's Law:

A = εcl,

where $A$ is the absorbance, ε is the extinction coefficient, $c$ is the concentration, and $l$ is the path length.

For proteins, $c$ is typically measured in units of "mg/mL" and $l$ is typically 1 cm.

I will assume that your protein had ε = "0.66 mL·mg"^-1"cm"^-1. Then

c = A/(εl) = 0.231/("0.66 mL·mg"^-1 cancel("cm⁻¹") × 1 cancel("cm")) = "0.35 mg/mL"

Step 2. Calculate the concentration of the diluted protein

You took 15.0 µL of the sample and diluted it with 15.0 µL of buffer to get 30.0 µL of solution.

You get the new concentration from a standard dilution calculation.

${c}_{1} {V}_{1} = {c}_{2} {V}_{2}$

c_2 = c_1 × V_1/V_2 = "0.35 mg/mL" × (15.0 cancel("µL"))/(30.0 cancel("µL")) = "0.175 mg/mL"#

Step 3. Calculate the mass of protein loaded on the gel

You loaded 20.0 µL of the dilute solution onto the gel.

$\text{Mass of protein loaded" = 20.0 cancel("µL soln") × "0.175 mg protein"/(1 cancel("µL soln")) = "3.5 mg protein}$