A protein has 5 Tryptophan (Trp) and 10 Tyrosine (Tyr) residues. If the absorbance of protein at 280 nm is 0.5 and molecular weight is 7 kDa, what is the approximate concentration... A protein has 5 Tryptophan (Trp) and 10 Tyrosine (Tyr) residues. If the absorbance of protein at 280 nm is 0.5 and molecular weight is 7 kDa, what is the approximate concentration of the protein? Read more

Steps to Solve

1. Calculate Total Extinction Coefficient ($\varepsilon_{280}$)

Combine the extinction coefficients for tryptophan and tyrosine based on their respective contributions: [ \varepsilon_{280} = (5 \times 5500) + (10 \times 1500) ] This gives: [ \varepsilon_{280} = 27500 + 15000 = 42500 , \text{M}^{-1} \text{cm}^{-1} ]

2. Use Beer-Lambert Law to Find Concentration

Apply Beer-Lambert Law, which states: [ A = \varepsilon \cdot c \cdot l ] Where (A) is absorbance, (\varepsilon) is extinction coefficient, (c) is concentration, and (l) is path length.

Rearranging for concentration gives: [ c = \frac{A}{\varepsilon \cdot l} ] Substituting values into the equation: [ c = \frac{0.5}{42500 \cdot 1} ]

3. Calculate Concentration

Now calculating concentration: [ c = \frac{0.5}{42500} \approx 1.176 \times 10^{-5} , \text{M} ]

4. Convert Concentration to Appropriate Units

Since the molecular weight is given in kDa, convert the concentration to mg/mL:

• The molecular weight is 7 kDa = 7000 g/mol.

Concentration in mg/mL is: [ c_{mg/mL} = c_{M} \times \text{molecular weight} \times 1000 ] [ c_{mg/mL} = 1.176 \times 10^{-5} \times 7000 \times 1000 ] [ c_{mg/mL} \approx 82 , \mu g/mL ]