The question is: "The molar extinction coefficient of pure antibody is 1.4 absorbance units per milligram antibody @280nm. You have a concentrated antibody solution, so you make a 5-fold dilution and measure @280, getting an absorbance of 0.845. What is the concentration of that solution?"
Using Beer's Law this is my work:
0.845 = (1.4)x(1)x(c)
solve for c:
c = 0.604 mol/L
Then I think you need to factor in the dilution?
(5)x(0.604)= 3.02 mol/L concentration
Is this correct?
If molar extinction coefficient is 1.4 A units/mg then wouldn't c come out in mg?. But that confuses me too. The usual unit is M which is mols/L and this unit you quote is just mg. Yes you should make a correction for dilution. So whatever c is the original solution will be 5c. That part is ok except for the starting value of 0.604 mol/L
To determine the concentration of the antibody solution after the dilution, you need to take into account both the molar extinction coefficient and the dilution factor. Let's break down the steps to calculate the concentration correctly:
1. Calculate the concentration before dilution:
Using Beer's Law, the equation can be written as: A = εlc, where A is the absorbance, ε is the molar extinction coefficient, l is the path length (usually 1 cm), and c is the concentration.
In this case, the molar extinction coefficient (ε) of the pure antibody is given as 1.4 absorbance units per milligram antibody @280nm.
To convert the units to Molar (M) units, you need to convert milligrams to grams and take into account the molecular weight of the antibody.
2. Convert molar extinction coefficient to Molar units:
Given that the molecular weight of the antibody (MW) is not provided, I will assume a standard value of 150,000 g/mol.
1.4 absorbance units per milligram antibody @280nm can be converted to Molar units as follows:
(1.4 absorbance units/mg) x (1 g/1000 mg) x (1 mol/150,000 g) = 9.33 x 10^(-9) M^(-1) cm^(-1)
Now we have the molar extinction coefficient (ε) in appropriate units.
3. Calculate the concentration before dilution:
Using the Beer's Law equation: A = εlc
Given that the absorbance after dilution (A) is 0.845, the path length (l) is typically 1 cm, and the concentration (c) is unknown, the equation becomes:
0.845 = (9.33 x 10^(-9) M^(-1) cm^(-1)) x (1 cm) x (c)
Solve for c:
c = (0.845) / (9.33 x 10^(-9) M^(-1) cm^(-1)) = 9.074 x 10^7 M (before dilution)
4. Factor in the dilution:
Since you made a 5-fold dilution, you need to multiply the concentration before dilution by the dilution factor.
Concentration after dilution = (5) x (9.074 x 10^7 M) = 4.537 x 10^8 M
Therefore, the concentration of the antibody solution after the 5-fold dilution is approximately 4.537 x 10^8 M.