Question

Human myoglobin is a protein with a molar extinction coefficient at 280nm of 1.40 x 104 M-1cm-1 . 100.0 uL of the stock solution of myoglobin (the tube was marked as 0.050 mg/mL) was mixed with 0.900 mL of buffer, giving an absorbance at 280nm of 0.810 (cuvette is 1.00 cm pathlength). From this information, what is the molecular weight of human myoglobin? Do you think this is a reasonable value? Explain WHY or WHY NOT?

Answer 1

Knowing the absorbance of the final solution, we can calculate its concentration using the Lambert-Beer law:

$A=\varepsilon \cdot l\cdot c$

Where A is the absorbance, epsilon the molar absorptivity, l the optical path length and c the concentration. We can rearrange it to calculate the concentration:

$c=\frac{A}{\varepsilon \cdot l}=\frac{0.810}{1.40x10^{4}\frac{1}{M\cdot cm}\cdot 1.00cm}=5.79x10^{-5}M$

But this solution is a dilution of the stock solution, obtained after mixing 100.0 microliters of stock with 900 microliters of buffer. The initial concentration can be calculate using the dilution formula:

$C_{stock}=\frac{C_{final}\cdot V_{final}}{V_{stock}}=\frac{5.79x10^{-5}M\cdot 1000\mu L}{100.0\mu L}=5.79x10^{-4}M$

This means that in the stock solution there were 5.79x10^-4 moles per liter. The mark on the tube says the concentration is 0.050 mg/mL, which is equivalent to 0.050 g/L. This would mean that 5.79x10^-4 moles of myoglobin weight 0.050 g. This means that 1 moles would weight:

$m_{molar}=\frac{0.050g}{5.79x10^{-4}mol}=86.4g/mol$

Which doesn't make sense, since this is an extremely low molar weight for a protein.