7 - BCH 451 Fall 2000 Exam #3 Some constants: F = 96.48...

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Unformatted text preview: BCH 451 Fall 2000 Exam #3 Some constants: F = 96.48 kJ/mol-V; k = 1.381 x 10-23 J/K; N o = 6.022 x 10 23 /mol; R = 8.315 J/mol-K (2) Cells are described as being in a "steady state condition". What does that mean? 1. Draw the predominate structures of the following molecules at pH = 7. Be sure to include all carbon and hydrogen atoms! 1. (4) Structure #1 in anti conformation: A = pTpdG; B = dUpAp 1. (4) Structure #2 C = 1-myristoyl-2-arachidonoyl-phosphatidyl ethanolamine; 2. D = 1-stearoyl-2-oleoyl-phosphatidyl inositol (2) Puzzler A pure DNA sample was treated with a newly isolated DNase. The enzymatic digest was analyzed by determining the nature of the two ends of these fragments, and the results are given below: From these data, determine the type of cleavage for this DNase. Full credit requires a statement of your reasoning. 1. Base % composition 3' end as OH 3' end as PO 4 5' end as OH 5' end as PO 4 A 31 31 T 26 25 96 G 22 21 2 2 C 21 23 Total 100 100 100 (3) In the following structure, circle each bond/linkage. Beside each bond/linkage, give the general name/class and(3) In the following structure, circle each bond/linkage....
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This note was uploaded on 12/01/2010 for the course BCH 451 taught by Professor Knopp during the Spring '08 term at N.C. State.

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7 - BCH 451 Fall 2000 Exam #3 Some constants: F = 96.48...

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