BIS102 Hilt Sp03 FINAL

BIS102 Hilt Sp03 - 1 of 7 Biological Sciences 102 Spring 2003 K Hilt Name_Key(2 pts Last First Final Exam pH = pKa log cfw[base[acid pH =(pKa1

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1 of 7 Biological Sciences 102 Name _______________ Key _________________________ Spring, 2003 (2 pts.) Last, First K. Hilt Final Exam Score (200): ______________ pH = pK a + log {[base]/[acid]} G = H - T S F = (q 1 q 2 )/( ε r 2 ) v o = {V max [S]} / {K m + [S]} pH = (pK a1 + pK a2 )/2 (198 pts.) Answer each of the following questions as “True” or “False” (one point) and then explain your answer (5 points). If your answer is “False”, then give the correct answer. If your answer is “True”, then elaborate further about why it is true. For all questions involving calculations, show your calculations. 1) __ F __ An oligopeptide is suspended in water. Its sequence of amino acids allows it to form an α -helix. The only thing that allows this helix to form is van der Waals forces. H-bonds (+2) Steric hindrance from R-groups (+1) . Keeps H 2 O out (+1) and restricts phi, psi angles (+1) . 2) __ F __ If I use 1 nmole of enzyme X in each test tube, in a series of test tubes to determine K m and V max , using a Lineweaver-Burk plot, then repeating the experiments using 10 nmole of enzyme X in every tube, will give a new K m value that is ten times greater and a new V max that is ten times greater. K m will be the same (+3) , but V max will be 10X higher (+2) . 3) __ T __ I need to make up a 10 mM buffer at pH 7.7 to control the pH of an enzyme catalyzed reaction that produces H + . In terms of just controlling pH, it is better for me to choose phosphate (pK a 7.2) than to use Tris (pK a 8.2). We need a buffer with a greater reservoir of base (+2) . The pH is higher than phosphate’s pK a , hence the [b] will be greater. (+3) 4) __ F __ The pH of a 3.45 x 10 -12 M HCl solution is approximately 11.5. pH 7 (+2) since H 2 O contributes approximately 1 x 10 -7 M H + , far greater than the concentration coming from HCl. (+3) 5) __ F __ The pH of a glycine solution must be 7.8 for glycine to have a net charge of “–1.8”. The pK a ’s of glycine are 2.3 and 9.6, for the carboxyl and amino groups, respectively. 2.3 9.6 gly +1 ± gly 0 ± gly -1 (+1) pH = pKa + log {[b]/[a]} = 9.6 + log (0.8/0.2) (+2)
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2 of 7 = 10.2 (+2)
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3 of 7 BIS 102 Name _______________ Key _________________________ Remember: if your answer is “False”, then give the correct answer. If your answer is “True”, then elaborate further about why it is true. For all questions involving calculations, show your calculations. 6) _ F ___ The strategy that we use today to sequence proteins is identical to that used by Dr. Sanger in the 1950’s. There are many similarities. However, today we generate large protein fragments (+3) , rather than small ones, in order to take advantage of the abilities of the protein sequenator (+2) (to yield ~ 40 a.a. residues). 7)
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This note was uploaded on 10/07/2009 for the course BIS 52192 taught by Professor Enochbaldwin during the Spring '09 term at UC Davis.

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BIS102 Hilt Sp03 - 1 of 7 Biological Sciences 102 Spring 2003 K Hilt Name_Key(2 pts Last First Final Exam pH = pKa log cfw[base[acid pH =(pKa1

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