Question

Absorption by a protein ( at 100 μM concentration) in a 1 cm path-length cell eliminates 92.06 % of the UV light passing through the solution at 280 nm. The composition of the protein is unusual - it contains no aromatic residues besides tryptophan, and no disulfide bonds. The protein is well behaved and does not scatter UV radiation.

1. What is the absorbance of the protein solution?

2. What is the molar absorption coefficient (M-1cm-1) of the protein?

3. Estimate how many Tryptophan residues this protein contains.

No more information was provided.

Answer 1

Background information:

Concentration (c) of protein given is 100 micromolar

Path length (d) of the cuvette given is 1cm

92.06% of the light is eliminated, in other words, is absorbed by the protein, so % transmittance is 100-92.06%= 7.24%

There are no disulphide bonds and no aromatic amino acids other than tryptophan, also, there is no scattering of UV.

(1) Absorbance of protein solution: Absorbance of solution can be easily calculated with the help of % transmittance by the formula A (Absorbance) = 2 - log(%T) (log of percentage transmittance)

So, A = 2 - log (7.24)

= 1.14 Absorbance unit

(2) Molar absorption coefficient of the protein (): This can calculated using the formula = A/ C.d; molar absorption coefficient = Absorbance/ Concentration of protein path length

So, = 1.14 / 100 1 = 1.14 10-2  (10 raise to -2) M-1 cm-1 = 1.14 10-8 M-1 cm-1

(3) Tryptophan residues this protein contains: The molar absorption coefficient is also calculated using the formula :

Molar absorption coefficient = (Number of trytophan residues 5500) + (Number of Tyrosine residues 1490)

But, the question mentions that there are no aromatic amino acids except tryptophan, so, Molar absorption coefficient = (Number of trytophan residues 5500)

Rearranging the equation : Number of trytophan residues = molar absorption coefficient / 5500

So, Number of trytophan residues = 1.14 10-8 / 5500

= 2.07   10-12 residues