In his famous experiment, Anfinsen denatured bovine ribonuclease using urea and
ID: 65593 • Letter: I
Question
In his famous experiment, Anfinsen denatured bovine ribonuclease using urea and mercaptoethanol (a reducing agent), and then slowly removed the denaturant and reducing agent. He observed that bovine ribonuclease lost all its activity upon denaturation but regain almost all of its activity as the denaturant was removed. Anfinsen interpreted this information to imply that the amino acid sequence alone contained the information necessary to fold ribonuclease into its native active form. Bovine ribonuclease has 8 cysteine residues which form four disulfide bonds in the final native structure. As the denaturant was removed how many different disulfide bond pairs could be formed during the refolding of ribonuclease? Provide a brief explanation of how you arrived at your answer. A polypeptide has the sequence LLWYSG. Using our discussions of the characteristics of aromatic residues: Estimate the extinction coefficient of this peptide at 280 nm. Show your calculations. A solution of this peptide is placed in a 1 cm thick cuvette and its absorbance is found to be 1.3 at 280 nm. What is the concentration of the polypeptide in solution? You perform a gel filtration experiment and measure the activity using your nifty colorimetric assay. The following results are obtained (the elution position of molecular weight standards is superimposed on the profile). What is the estimated molecular mass of the molecules in the sample with the highest activity?Explanation / Answer
1.
1-------2--------3-------4-------5------6--------7--------8
1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 2-3, 2-4, 2-5, 2-6, 2-7, 2-8, 3-4, 3-5, 3-6, 3-7, 3-8, 4-5, 4-6, 4-7, 4-8, 5-6, 5-7, 5-8, 6-7, 6-8, 7-8 (total 28)
2.
a) Extinction coefficient (E) = aE (tyrosine) + bE (tryptophan) where a and b are number of amino acid residues
= 1 (5690) + 1 (1280) [some write 5500 for tryptophan and 1490 for tyrosine]
= 6970 CM-1M-1
b) A = Ecl
1.3 = 6970 x C x1.0
or C = 1.3 / 6970 = 1.8 x 10-4 mol / litre
3.
a. Fraction 50 has the highest activity and contains molecules of approximately 50 kDa.