Life Science 1A Final Exam. January 19, 2006

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1 ame: TF: Section Time Life Science 1A Final Exam January 19, 2006 Please write legibly in the space provided below each question. You may not use calculators on this exam. We prefer that you use non-erasable pen when writing your answers. A significant number of completed exams will be photocopied by the teaching staff. Please write your name on each page of the exam. There are TE multi-part questions in this exam. We recommend that you first read through all questions and begin with the questions that are easiest for you. Be sure to take a look at all questions before the end of the exam! Please write your TF s name and section time on the front page only. Question 1 / 20 Question 6 / 38 Question 2 / 20 Question 7 / 44 Question 3 / 36 Question 8 / 28 Question 4 / 26 Question 9 / 36 Question 5 / 36 Question 10 / 16 Total /300

2 1. You decide to use FRAP (Fluorescent Recovery After Photobleaching) to study microtubule dynamics. Your photobleaching set-up is as follows: Microtubule growth w/ fluorescent tubulin Photobleach one section Further growth w/? fluorescent tubulin Centrosome Centrosome Centrosome a. (6 points) f the two choices below, which does the microtubule resemble after continued growth (assume catastrophe does not occur)? Explain briefly. Choice A: Choice B: Centrosome Centrosome 1

3 Shown below is the drug colchicine, which affects microtubule dynamics: 3 C C 3 3 C Colchicine 3 C C 3 b. (4 points) Circle any chiral centers present on colchicine c. (4 points)draw a box around any amide bonds in colchicine. d. (6 points) Colchicine inhibits the formation of microtubules by binding to tubulin dimers. With which phase of the cell cycle would you expect colchicine to interfere the most? 2

4 2. Scientists can use analogs of ATP and GTP to analyze protein function. Shown below are GTP, and two closely related analogs of GTP. The first analog, GMppCp, has a terminal (gamma) phosphorous-carbon bond that cannot be hydrolyzed. The second analog, 3 deoxy GTP, is missing the 3 hydroxyl group. Assume that GTP, GMppCp, and 3 deoxygtp bind to the proteins below with the same affinity (same K d ). GTP P P P GMppCp P P C 2 P 3' deoxygtp P P P a. (4 points) What effect should GMppCp have upon the activity of Ras? Explain. b. (4 points) What effect should 3 deoxygtp have upon the activity of Ras? Explain. 3

5 c. (4 points) Could RA polymerase use GMppCp in the transcription of mra? Explain. d. (4 points) Could RA polymerase use 3 deoxygtp in the transcription of mra? Explain. e. (4 points) GTP modulates the propensity of microtubules to grow (by adding subunits) or shrink (by losing them). Explain how GMppCp would affect microtubule elongation and contraction. 4

6 3. Adenosine triphosphate (ATP) is used by the cell to fuel many biochemical reactions. a. (4 points) Describe the chemical features of phosphate esters that allow ATP hydrolysis to be regulated in time and space in the cell. b. (6 points) Describe three methods that protein kinases use to catalyze the transfer of a phosphate group from ATP to an amino acid sidechain. c. (8 points) Draw an energy diagram for the uncatalyzed and enzymecatalyzed hydrolysis of ATP. Explain how the enzyme affects the kinetics and/or thermodynamics of this reaction. Free Energy Reaction Coordinate 5

7 d. (8 points) XP is a high-energy, phosphate-containing compound. Phosphate hydrolysis of XP to X + P i releases 30 kcal/mol of energy (compared to 7 kcal/mol released in the hydrolysis of ATP to ADP +P i ). The reaction diagrams for the uncatalyzed hydrolysis of ATP and XP are shown below. Based on your understanding of kinetics and thermodynamics together with your answer to part (a), would XP be a better or worse energy currency for the cell? Explain your answer. 6

8 e. (4 points) In the amino acid sequence shown below, circle and name any amino acid residues that could serve as substrates for phosphorylation by protein kinases. S 3 C C C C C 2 C 3 C 2 C C C C C C 3 C 2 C C C C C 2 C 2 C 2 C f. (6 points) Draw the phosphorylated versions of the amino acids that you indicated in part (e) above. You only need to draw those residues that are modified by phosphorylation, but be sure to include the entire side chain. 7

9 4. The er2 receptor is a receptor tyrosine kinase whose activation leads to cell growth and proliferation. In many breast cancers, er2 receptor mutations have been isolated. a. (4 points) The ligand for the er2 receptor causes receptor dimerization. What is the effect of receptor dimerization for many receptor tyrosine kinases? b. (4 points) A point mutation in the transmembrane domain of the er2 Receptor converting a Val to a Gln causes the receptor to dimerize in the absence of ligand. What effect would this mutation have upon the activation state of the receptor? c. (4 points) In order to study receptor tyrosine kinases, scientists can genetically modify the mutant er2 receptor described in part b and eliminate the intracellular domain. If this modified receptor is the only one expressed in a cell, will it respond to the er2 ligand? Explain. d. (8 points) If a heterodimer were to form containing one subunit of the receptor described in part b and the other subunit of the receptor in part c, would this complex be able to activate downstream targets? Explain. e. (6 points) Activation of er2 leads to the phosphorylation of MAP kinase (MAPK). Why do normal cells require either a phosphatase that removes the phosphate group(s) from MAP kinase or a MAP kinase that is rapidly degraded by proteolysis? 8

10 5. Chronic Myelogenous Leukemia (CML) is a cancer caused by a specific chromosomal translocation between chromosomes 9 and 22. The result of this translocation is the rapid proliferation of immature lymphoblasts (white blood cell precursors). a. (6 points) In healthy individuals, immature lymphoblasts represent on average less than 5% of the cells made in the bone marrow. The percentage of immature lymphoblasts is relatively constant in the bone, and individual lymphoblasts have a lifetime of about a week. In untreated CML patients, the percentage of immature lymphoblasts increases to greater than 30% of the bone marrow in approximately 4-5 years. If the lymphoblast population is maintained at 30% for one year, have the lymphoblast populations of these untreated CML patients now reached a new steady state or a new equilibrium? Briefly explain. b. (6 points) Upon treatment with Gleevec, how do you predict the lymphoblast populations will change? In your answer mention if after successful treatment with Gleevec for one year the populations will have reached a new steady state, a new equilibrium, or neither. c. (6 points) Based on its mechanism of action why does Gleevec cause fewer side effects than traditional chemotherapeutic agents like methotrexate? Explain in three sentences or less. d. (6 points) Is Gleevec a cure for CML? Explain your answer. 9

11 In order to gain a deeper insight into CML, scientists studied, c-src a kinase related to the Abl kinase affected in CML. Below is the scheme showing the normal regulation of c-src. Scientists have found a mutant of c-src. When the DA from the mutant Src gene was sequenced, the reading frame was determined and it read as follows (in both wild type and mutant Src, TAC is the codon at position 499, and CAG and TAG are the codons at position 500, respectively): WILD TYPE Src: 5 TAC CAG 3 MUTAT Src: 5 TAC TAG 3 e. (4 points) ow does this particular mutation affect the translation of src? f. (8 points) This mutation affects the amino acid at position 500 (refer to schematic for location of Tyr 527 in protein). Why did this mutation cause the mutant Src to be constitutively active? 10

12 6. Growing and dividing human cells take 24 hours to proceed through the cell cycle and complete division into two identical daughter cells. a. (8 points) Label the four phases of the cell cycle on the diagram below and indicate which of the phases are part of interphase. Daughter cell b. (4 points) Cells can enter and exit the cell cycle. Indicate with arrows from which phase this entry and exit occurs, and name this resting state. c. (8 points) Growing human cells take 24 hours to double their mass. What will happen to the average mass of each cell in a population of proliferating cells if the duration of the cell division cycle is reduced to 18 hours, without affecting the time it takes for them to double each cell s mass? 11

13 By measuring the amount of DA inside each cell in a population of cells, it is possible to construct a graph that show how many cells in the population contain various amounts of DA. Such a graph is shown below for a population of cells that are growing and proliferating. The population is asynchronous, meaning that different individual cells are in different phases of the cell cycle. The graph has been divided into three sections: A, B, and C. umber of Cells A B C Relative Amount of DA d. (6 points) Which phase/phases of the cell cycle does section C of the graph correspond to? Explain in two or fewer concise sentences. e. (6 points) Which phase/phases of the cell cycle does section A of the graph correspond to? Explain in two or fewer concise sentences. f. (6 points) Which phase/phases of the cell cycle does section B of the graph most likely correspond to? Explain in two or fewer concise sentences. 12

14 7. Ras is a member of a large family of small GTPases and plays a key role in many intracellular signaling pathways. When Ras is bound to GTP, it is capable of activating many protein kinases, while when Ras is bound to GDP, Ras is inactive. For this reason, the binding of GTP is a molecular switch for many biochemical reactions inside the cell. The figure below shows GTP in the active site of Ras, where the numbers represent the amino acid residues of Ras P - P - Mg 2+ P

15 You make mutations in the GTP-binding pocket of Ras and examine the effects on the binding of GTP. For parts a-c, first draw the new amino acid. Then considering the size and the nature of the substitutions give the most likely reason why each mutation has the stated effect. a. (4 points) Residue 16 is mutated to an Arginine, resulting in Ras that still binds GTP. b. (4 points) Residue 119 is mutated to a Leucine, resulting in Ras that cannot bind GTP. c. (4 points) Residue 146 is mutated to a Tyrosine. o effect on GTP binding is seen. 14

16 ATP and GTP are structurally very similar however Ras is extremely selective in only being activated by GTP and not ATP. Biochemical analyses have determined that the K d for GTP binding to Ras is 5 x M and the k on is 2 x 10 6 M -1 s -1. The binding of ATP to Ras is much weaker, however the k on for ATP is identical at 2 x 10 6 M-1 s-1. d. (8 points) Write the chemical equilibrium for the dissociation of Ras-GTP to Ras and GTP. Label k on next to the appropriate arrow in your reaction. e. (8 points) Write the rate equations below for both the forward and reverse reactions of Ras-GTP dissociating to Ras and GTP. Forward rate = Reverse rate = f. (8 points) At equilibrium, how do these rate equations relate to the equilibrium constant for the dissociation of Ras-GTP to Ras and GTP (K d ). g. (8 points) Given your answers above, explain why Ras preferentially binds GTP over ATP in terms of specific kinetic or thermodynamic parameters. 15

17 8. Recall the Dupont-Merck IV protease inhibitors and examine the analogs below: S Ile 50 Ile 50' S Dupont-Merck Inhibitor + IV protease Ile 50 and Ile 50' partially shown S S Analog #1 S S Analog #2 a. (8 points) ow would you expect the K d value for the complex between Analog #1 and IV protease to compare with that of the Dupont-Merck inhibitor? Explain your answer in terms of specific enthalpic and/or entropic differences. 16

18 b. (8 points) ow would you expect the K d value for the complex between Analog #2 and IV protease to compare with that of the Dupont-Merck inhibitor? Explain your answer in terms of specific enthalpic and/or entropic differences. c. (8 points) Inhibition of the IV protease prevents the formation of mature viral proteins. ne of the proteins affected is integrase. ow would an inability to make integrase would this affect future viral lifecycles? d. (4 points) IV protease itself must be cleaved from a gag/pol precursor polyprotein substrate by another IV protease molecule. What type of feedback loop does this represent for untreated wildtype virus? 17

19 9. The enzyme triose phosphate isomerase, or TIM, catalyzes the interconversion between two metabolic intermediates, as shown below: P 3 2- K eq =.042 Compound A TIM (enzyme) Compound B P 3 2- Energy via metabolism a. (5 points) If compound A and B come to equilibrium in a flask, which compound is present in higher concentrations? For each circumstance below, choose the set of arrows (A, B, or C) that best describes the relative rates of the forward and reverse reactions. A B A B A B Choice A Choice B Choice C b. (5 points) Enzyme has JUST been added to a solution containing equal amounts of compound A and compound B. Choice: c. (5 points) A solution containing compound A, compound B, and the enzyme has reached equilibrium. Choice: d. (5 points) The enzyme is present within a cell, where compound B is constantly being consumed to generate metabolic energy. Choice: 18

20 The first step of the reaction catalyzed by TIM is shown below: Enyzme glutamate 2- P 3 Enyzme Enyzme histidine glutamate 2- P 3 Enyzme histidine Enyzme glutamate 2- P 3 Enyzme histidine e. (8 points) ow does the enzyme glutamate assist in catalyzing the reaction? ame the specific type of catalysis in your answer and explain how it accelerates the reaction. f. (8 points) ow does the enzyme s histidine assist in catalyzing the reaction? ame the specific type of catalysis in your answer and explain how it accelerates the reaction. 19

21 10. Arachidonic acid is one of the essential fatty acids required by most mammals. Arachidonic acid a. (4 points) Label each double bond as cis or trans b. (4 points) ow many total geometric isomers of arachidonic acid are possible? c. (4 points) Give one physical property you could use to distinguish the geometric isomers of arachidonic acid from one another. d. (4 points) Using the following graph determine the pk a for the carboxylic acid group of arachidonic acid. 20

22 The Genetic Code 21

Life Science 1A Final Exam. January 19, 2006

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