Bovine pancreatic trypsin inhibitor (BPT I) contains six cysteine residues that form three disulfide bonds in the native structure of BPT I. Suppose BPT I is reduced and unfolded in urea. If the reduced unfolded protein were oxidized prior to the removal of the urea, what fraction of the resulting mixture would you expect to possess native disulfide bonds?

Express your answers using three significant figures.

Question

Finnegan Waller

Biology

10 November, 05:06

Bovine pancreatic trypsin inhibitor (BPT I) contains six cysteine residues that form three disulfide bonds in the native structure of BPT I. Suppose BPT I is reduced and unfolded in urea. If the reduced unfolded protein were oxidized prior to the removal of the urea, what fraction of the resulting mixture would you expect to possess native disulfide bonds?

Express your answers using three significant figures.

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Answers (1)

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Nicole Blankenship · Answer 1

Answer and Explanation:

Bovine pancreatic trypsin inhibitor (BPT I) is a small globular protein and it inhibits the proteolytic enzymes like trypsin. BPTI is composed of ∝ helices, β sheets and 3 disulfide bonds. Due to these BPTI is a stable protein in its tertiary structure. It is almost inert to denaturation by urea and exhibits denaturation below 100 degree, only in highly acidic solutions. When all the disulfide bonds in BPTI are reduced, the protein is unfolded at room temperature and can reform three correct S-S pairings in native confirmation. if the 6 cysteine residues are reduced and unfolded in urea, the re-oxidation would yield 3 pairs with probability of first pair with 5, second pair with 3 and the third pair with 1 cysteine residues. Therefore, 5 x 3 x 1 = 15 combinations are possible accounting for 7% of protein refolding.