Absorbance Spectroscopy
[1] Using Protein Extinction Coefficients
Thanks to a few sizable blood donations from your roommate (and knowledge of protein purification gained from completing biochemistry), you have been able to isolate a solution of pure human insulin in dilute aqueous buffer. The volume of your insulin solution is 35 mL. Using 1.0 mL of that solution, you measure an absorbance of 0.67 at 280 nm in a 1.0 cm cuvette in the spectrophotometer. Looking up standard information about human insulin, you discover that the molecular weight of insulin is 5808 g/mol and the molar extinction coefficient at 280 nm is 6020 M-1cm-1. What is the mass amount of human insulin you have remaining?
Show Answer
Since we have an absorbance, a pathlength and a molar extinction coefficient, we can apply the Beer-Lambert Law directly to calculate the concentration of the solution.
$$A=\varepsilon \times l \times c$$
Substitutioning numerical values,
$$0.67 = 6020~M^{-1}cm^{-1} \times 1~cm \times c$$
and solving for c, we discover that the concentration of human insulin in solution is 1.1 x 10-4 M (to two significant figures). To calculate the mass amount of human insulin, this reduces to a dimensional analysis problem:
$${1.1 \times 10^{-4}~moles\over1~L} \times {1~L \over 1000~mL} \times {34~mL \over \space} \times {5808~g \over 1~mole} \times {1000~mg \over 1~g} = 21.72192~mg$$
With two significant figures, that's 22 mg of human insulin. Thank you, roommate!
[2] Measuring DNA Concentration
Having just completed an alkaline lysis mini-prep, you measure the absorbance of a ten fold dilution of your product. The absorbance is 0.84 at 260 nm. What is the concentration of your product?
Show Answer
There are two steps to answer this question. The first is to calculate the concentration of DNA in your dilution (the sample for which you have an absorbance reading) and then use the dilution factor to calculate the concentration of your original solution. Remembering that a DNA solution of 50 μg/mL has an absorbance of 1.0 at 260 nm, we can calculate the concentration of the diluted sample:
$${50~\mu g/mL \over 1.0} = {c \over 0.84}$$
where c = 42 μg/mL. Since we measured the absorbance of a 10 fold dilution, the concentration of the original solution is ten times concentration we measured, 420 μg/mL.
[3] Interpreting Circular Dichroism Spectra
Below is a CD sprectrum belonging to either myoglobin (
1mbn) or azurin (
1azu). Using the Protein Data Bank for structures, is the spectrum of myoglobin or azurin? Why?
Show Answer
Looking at the structures of myoglobin and azurin, they have very different compliments of secondary structure (what circular dichroism detects in the far UV where the peptide bond is the chromaphore). The secondary structure of myoglobin is all α helix while azurin is largely β strand. The CD spectrum shows a strong double minimum around 222 nm, the characteristic of α helical secondary structure. Therefore, myoglobin is the most likely protein.
