g&g_ch_5_p_1-5

# g&g_ch_5_p_1-5 - G&G Ch 5 Problem 1(pg 145 This...

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Unformatted text preview: G&G Ch. 5 Problem 1 (pg. 145) This problem asks you to speculate as to quaternary structure after performing a surprisingly simple calculation. IMO, this is a nicely designed question that gets your juices flowing and thinking about higher order protein structure. MW of Mo : 96g/mol MW of nitrate reductase : 240,000g/mol Remember that we are dealing with a real biological system, which is an aqueous environment, meaning Mo exists in its cationic form (Mo 2+ ). If Mo 2+ makes up 0.08% of the weight of the protein, then (0.0008)(240,000) = 192g/mol of the MW of nitrate reductase is composed of Mo 2+ . That means there are 192g/mol 96g/mol = 2 Mo 2+ ions per nitrate reductase protein. Now here is where making some conjecture is necessary. Because we are dealing with a divalent cation, the protein most likely uses only one to occupy the active site. But because we have determined that there are 2 Mo 2+ cations per protein, we can speculate that the protein exists as a dimer . G&G Ch. 5 (cont’d) Problem 2 (pg. 145) This is a great problem. Moving through steps a-d leads you nicely to the final sequence. Considering what each observation tells you is the key. In a step-wise manner, you build the final 1° structure. a. Trypsin treatment had no effect. What does trypsin do? It cleaves at the C-terminal ends of lysine (Lys, K) and arginine (Arg, R) residues. The fact that it does not cleave our peptide tells us that the Lys residue must lie at the C-terminus of the protein, since this is the only location this amino acid can occupy without being affected by trypsin treatment. b. The PTH derivative released in the Edman degradation is given. What does the Edman degradation do? It converts the N-terminal amino acid into a PTH derivative. Since our PTH derivative contains a phenylalanine (Phe, F) side chain, we can safely say that Phe lies at the N-terminus. H 3 N + –X–X–X–X–X–X– K –COO- H 3 N + – F –X–X–X–X–X– K –COO- G&G Ch. 5 (cont’d) Problem 2 (cont’d) c. Brief chymotrypsin treatment yielded a dipeptide and a tetrapeptide. The tetrapeptide contains Leu, Lys, and Met. What does chymotrypsin do? It cleaves at the C-termini of F, Y, and W. The fact that treatment yields both a dipeptide and a tetrapeptide is a clue: the tetrapeptide is composed of only three different amino acids, meaning one must be represented twice. Because we know from a. that there is only one Lys residue (trypsin had no effect), we know the tetrapeptide must lie at the C-terminus of the protein, with the Lys residue being the most C-terminal. Furthermore, since a Tyr residue is not represented in this C-terminal tetrapeptide, it must be the third residue, since we know the first is F, and the production of the dipeptide is a result of cleavage at the C-termini of the F and Y residues....
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## This note was uploaded on 07/16/2008 for the course BIS 102 taught by Professor Hilt during the Summer '08 term at UC Davis.

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g&g_ch_5_p_1-5 - G&G Ch 5 Problem 1(pg 145 This...

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