Treatment of the nonapeptide question 23 with cnbr

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25. Treatment of the nonapeptide (question 23) with CNBr produced a tetrapeptide containing the N- terminal amino acid and a pentapeptide. The pentapeptide after one round of the Edman procedure produced a product that had a aliphatic R group. This means that the pentapeptide has: a. N-terminal I. b. N-terminal S. c. C-terminal E. d. N-terminal H. e. C-terminal M. 26. The tetrapeptide (question 25) at pH 6 had a charge of -0.5. This indicates that the tetrapeptide could be:
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27. The second and third rounds of the Edman procedure (question 25) on the pentapeptide produced products with aliphatic alcohol groups. This means that the pentapeptide had: 28. Hydrazinolysis of the pentapetide (question 25) produced modified amino acids and the only free amino acid had an aromatic-alcohol group. This means that the pentapeptide is most likely: 29. As shown in lecture, cytochrome c primary structures have conserved and non-conserved (amino acid) sequences. This means that changes (mutations) in primary structure has: a. caused DNP derivatives of internal amino acids. b. been lethal in non-conserved sequences. c. lethal in conserved sequences. d. changed only the N-terminal end of the protein. 30. Acid (such as 6M HCl) hydrolysis of proteins at 105 o C results in: PLEASE TURN OVER TO THE NEXT SIDE FOR THE WRITTEN ANSWER QUESTIONS.
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Written Answer Questions: Answer only in the space provided.. 1. You have 100 mL of 0.08 M phosphate buffer at pH 6.0. By accident, 3.2 mL of 6 M HCl was added. What is the resulting pH? The pka’s of phosphate are 2.14, 6.86, 12.4. (10 points) 2. Diagram, showing all atoms, the artificial sweetener aspartame (L-aspartyl-L-phenylalanine-methyl ester) at pH 6 (in your soft drink). 10 points.
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