In the 1950s, Christian Anfinsen demonstrated the renaturation of the protein ribonuclease (RNase) in vitro. After reduction and the addition of urea, the protein was in an unfolded state. After removing the urea and then the reducing agent, the protein oxidized and refolded, with greater than 90 % activity. If reducing agent removal occurs before removing the urea, the protein showed less than 5 % activity. Why would the protein not refold correctly if the urea were removed after the reducing agent was removed? (In other words, what would happen if the urea were removed after oxidation?) Choose the best answer.

a. Urea would participate in weak bonding interactions with RNase, preventing oxidation of Cys.

b. The protein would not fully unfold (denature).

c. Disulfide bonds are not positioned correctly unless weak bonding interactions are present.

d. Contaminants in the RNase preparation would form covalent bonds with the protein, preventing reactivation.

Question

Aubrie Berg

Biology

28 April, 02:08

In the 1950s, Christian Anfinsen demonstrated the renaturation of the protein ribonuclease (RNase) in vitro. After reduction and the addition of urea, the protein was in an unfolded state. After removing the urea and then the reducing agent, the protein oxidized and refolded, with greater than 90 % activity. If reducing agent removal occurs before removing the urea, the protein showed less than 5 % activity. Why would the protein not refold correctly if the urea were removed after the reducing agent was removed? (In other words, what would happen if the urea were removed after oxidation?) Choose the best answer.

a. Urea would participate in weak bonding interactions with RNase, preventing oxidation of Cys.

b. The protein would not fully unfold (denature).

c. Disulfide bonds are not positioned correctly unless weak bonding interactions are present.

d. Contaminants in the RNase preparation would form covalent bonds with the protein, preventing reactivation.

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Answers (1)

Know the Answer?

Charisma · Answer 1

The Correct Answer is "C".

Explanation:

Because poor holding communication appropriately place the - SH deposits to frame disulfide spans while expelling the diminishing operator (permitting oxidation) while evacuating the urea. prior to the evacuation of urea, oxidation happens so the cysteine buildups may not be found accurately and the disulfide spans that structure in an inappropriate spots bringing about inert protein