BIS102 - Final

BIS102 - Final - Name Blue Key Last, First BIS102-02,...

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Name Blue Key BIS102-02, Spring '09, Page 1 of 11 Last, First C. S. Gasser BioSci 102-02 June 6, 2009, Final Exam Instructions: There are eleven pages in this exam including the cover sheet, please count them before you start to make sure all are present. Write your name on each page of the exam. Write your answers in the space provided below each question. If you need more space use the back of the page and indicate clearly that you have continued your answer on the back. Do not use additional paper. For calculation problems a clear line of calculations leading to your final answer must be shown to obtain full credit. Use the following pK a values for the exam. Amino Acid α -NH 2 α -COOH R-group log 1.0 = 0.00 Ala, Gly, Ile, Leu, Val 9.7 2.3 log 1.5 = 0.18 Asn, Gln 9.0 2.1 log 2.0 = 0.30 Met, Ser, Trp 9.3 2.3 log 2.5 = 0.40 Phe 9.1 1.8 log 3.0 = 0.48 Pro 10.6 2.0 log 3.5 = 0.54 Thr 10.4 2.6 log 4.0 = 0.60 Asp 9.8 2.1 3.9 log 4.5 = 0.65 Glu 9.7 2.2 4.2 log 5.0 = 0.70 His 9.2 1.8 6.0 log 5.5 = 0.74 Cys 10.8 1.8 8.3 log 6.0 = 0.78 Tyr, Lys 9.1 2.2 10.0 log 6.5 = 0.81 Arg 9.0 2.2 12.5 log 7.0 = 0.85 log 7.5 = 0.88 Question Val 1 ue Score log 8.0 = 0.90 1 30 log 8.5 = 0.93 2 12 log 9.0 = 0.95 3 20 log 9.5 = 0.98 4 18 log 10.0 = 1.00 5 12 6 14 R = 8.3 x 10 -3 kJ/°K•mol 7 20 25°C = 298K 8 21 9 16 10 8 11 6 12 8 13 15 Total 200
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Name Blue Key BIS102-02, Spring '09, Page 2 of 11 Last, First C. S. Gasser 1. You have been given two samples of blood. One sample comes from a child that is believed to have consumed plant material containing malonate, an inhibitor of succinate dehydrogenase (SDH) an essential metabolic enzyme. The second sample comes from a healthy child. Unfortunately, the samples were only labeled only with the letters “A” and “B” and you don’t know which is which. You go ahead and perform enzyme assays to test the effects of addition of portions of the two samples on SDH activity to try to detect the effects of inhibitor. The results for sample A are plotted below, and the results for sample B are given in the table. Use this information to answer the questions below. 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 1/[S] 1/V A (min • ml/nmole) (1/μM) B Data for Sample B [S] (μM) V (nmole/ml•min) 1/[S] 1/V 2.5 2.0 0.4 0.5 5 3.3 0.2 0.3 10 5.0 0.1 0/2 20 6.7 0.05 0.15 a) (6) Plot the data for sample B on the axes above using the units as labeled. Plot 1/V vs. 1/[S] (calculations at right in table above) b) (4) Which sample appears to be the control sample not containing malonate (circle one)? B has an apparent increase in the Km. So A has no inhibitor. B
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Name Blue Key BIS102-02, Spring '09, Page 3 of 11 Last, First C. S. Gasser c) (8) What are the K m and V max for SDH in the assay that does not contain malonate? -1/K m = -0.5, K m = 2 μM 1/V max = 0.1 so V max = 10 nmol/min•ml K m = 2 μM V max = 10 nmol/min•ml units units d) (8) You determine that the concentration of malonate in the inhibited samples is 2 μM. What is the K I of malonate for SDH? -1/K mapp = -0.1, K mapp = 10 μM K mapp = Km(1 + [I]/K I ) 10 = 2(1 + 2/K I ); 5 = 1 + 2/K I ; 4 = 2/K I ; K I = 2/4 = 0.5 μM K I = 0.5 μM units e) (4) The diagram below shows the mechanism of an enzyme catalyzed reaction. Complete the diagram by showing how malonate (I) would interact with SDH (E) and S?
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This note was uploaded on 01/29/2010 for the course BIS 102 taught by Professor Hilt during the Spring '08 term at UC Davis.

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BIS102 - Final - Name Blue Key Last, First BIS102-02,...

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