In: Chemistry
You need to validate the concentration of a 50 μL stock solution reported it be 25 mg/mL based single measurement. To conserve sample you decide to use the UV-vis plate reader to make an absorbance measurement using a 100 μL well-volume.
Given:
The protein has a molecular weight of 6,343 g/mol. and a molar extinction coefficient or 113,268 M-1cm-1 (Effective pathlength = 0.3 cm for a 100 μL well volume). The instrument you have available provides accurate absorbance readings for sample absorbance in the range 0.1 - 1. 0. You have P20 and P200 pipettes and 20 mM phosphate buffer available.
a) What fold dilution of the sample would be needed get and absorbance of 0.5?
b) Describe how you dilute 2 μL of the 25 mg/mL sample so that this absorbance reading can be made.
c) After completing a single measurement of the sample your calculations indicate the sample is 26.7 mg/mL. What percent agreement exists between your determination and the previously reported value? What factors are most likely?
d) You need to assay the purity of the sample from parts A-C by SDS-PAGE. Explain how to prepare a 20 μL sample containing 30 μg for this experiment.
After combining with 2x sample loading buffer and heating, you transfer 25 μL of the denatured protein into a well of the resolving gel.
Compute the total amount of protein that was added to the gel in micrograms and nanomoles.
Ans. #A. Given, [Protein] = 25 mg/ mL
= 25 g/ L ; [1 mg = 10‑3 g ; 1 mL = 10-3 L]
Moles of protein in 1.0 L solution = Mass/ Molar mass
= 25.0 g / (6343 g/ mol)
= 0.0039414 mol
Now,
Molarity of protein solution = Moles of protein / Volume of solution in liters
= 0.0039414 mol / 1.0 L
= 0.0039414 M
# Beer-Lambert’s Law, A = e C L - equation 1,
where,
A = Absorbance
e = molar absorptivity at specified wavelength (M-1cm-1)
L = path length (in cm)
C = Molar concentration of the solute
Put the value A = 0.5 to get the concentration of protein sample-
Or, 0.5 = (113268 M-1cm-1) x C x 0.3 cm
Or, C = 0.0000147 M
Therefore, [Protein] = 0.0000147 to give an absorbance of 0.5 units.
# So far, we have-
Original [Protein] = 0.0039414 M
Desired, diluted [protein] = 0.0000147 M - that gives desired OD of 0.5 units
Now,
Required dilution factor = Original [protein] / diluted [protein]
= 0.0039414 M / 0.0000147 M
= 267.86
So, the original solution must be 267.86 times diluted to yield a solution that gives abs of 0.5 units.
# B. Given, C1 = 25 mg/mL = 0.0039414 M ; V1 = 2 uL
C2 = 0.0000147 M ; V2 = ?
Using C1V1 = C2V2 - equation 1
C1= Concentration, and V1= volume of initial solution 1 ; i.e. original solution
C2= Concentration, and V2 = Volume of final solution 2 ; diluted, desired
Putting the values in above equation-
V2 = (0.0039414 M x 2 uL) / 0.0000147 M = 535.715 uL
Preparation: Take 2 uL of original protein solution in 1 mL Eppendorf tube. Make the volume upto 535.72 uL by adding sufficient water. It is the desired solution.
#C. % agreement = (experimental value / theoretical value) x 100
= (26.7 mg mL-1 / 25 mg mL-1) x 100
= 106.80 %
Note: The value above 100 % indicates that the experimental values is greater than the theoretical value.
The error, increase in concertation, is most likely due to loss of solvent from solution. For example, the solution was left open for some time. Other factors like inaccurate weighing of protein or adding less than required water or making up the volume at elevated temperatures may also result similar errors.
#D. Given, concentration of desired solution (say, solution 2) = 30 ug/ 20 uL
= 1.5 ug/ uL
Let the quantity required for the specified solution = 20 uL
Given, original protein concentration (solution1) = 25 mg/ mL = 25 ug/ uL
Putting the values in equation 1-
(25 ug/ uL) x V1 = (1.5 ug/ uL) x 20 uL
Or, V1 = [(1.5 ug/ uL) x 20 uL] x (25 ug/ uL) = 1.2 uL
Preparation: To 1.2 uL original protein solution (i.e. 25 mg/ mL = 25 ug/ uL) add sufficient water to make the final volume to 20 uL. The resultant solution consists of 30 ug protein in a volume of 20 uL.