Intro Bio Exam 2 Oct. 22, 2015 ID code_____KEY________________ NAME ___________ANSWER KEY___________ Name on cover sheet only, to keep grading anonymous CLASS REGISTERED FOR (check one): One point deducted if you do not check a class! _____ C2005-sect. 001 [morning] _____ C2005-sect. 002 [afternoon] _____ F2401 SEAT NO. ___________ PLEASE READ THIS PAGE!! There are 4 pages with questions. All information is the same in all exam copies, but the question order may differ. Each page is worth approximately 25%. To receive proper credit for your answer, you must write your name on the first page only, and on all other pages write only an ID code consisting of your birth date plus a 5-letter word of your choice (e.g., 08/16/90, gauss). Pages will be separated for grading and later reassembled. It is essential that your ID code be on each question page and the cove page. Please do it now. Please do it legibly. One point will be deducted for each page without an ID code. The use of ID codes makes the grading anonymous. Please check the class for which you are registered; one point will be deducted for failure to do so. Information you will need is provided on the last page. That information page can be torn off and used as scrap paper, and you may use the back of this cover page as scrap paper. But be sure to turn in this cover page. No electronic devices other than simple (scientific) calculators (non-graphing) are allowed, either visible or on your person-- no cell phones, Blackberries, IPods, etc. All electronic devices must be stored in briefcases or backpacks, not on your person. Please keep these items under your seat. Students found to have a cell phone on their person (e.g., in a pocket) will receive a zero for the exam. Use pen and no white-out. If you must change your answer, cross it out and use the back of the same page. If you use pencil a regrade is not possible. However, regrades are rare anyway. When choices are presented, unless instructed otherwise, circle what you consider to be the best answer or answers and provide an explanation of your reasoning for choosing your answers. Your explanation should show how you arrived at your answer, not just repeat it or the question. Think of trying to explain your answer to a classmate. Try to list or explain all the critical steps in your reasoning that led you to your answer. If you present some correct reasoning, you may earn partial credit. But no partial credit can be given if you provide no explanation at all, so try to write something for every question requiring an explanation (which is almost all of them). Don’t leave blank white spaces, write something. On the other hand: You should limit your answer to the space provided for each question. If you must cross out some writing, use additional space for your answer on the back of the SAME page on which the question appears, but limit your total writing to occupy the same amount of space as was originally provided. Use of more space for irrelevant information may result in the deduction of credit. Pages are separated for grading and the last page without questions will be discarded, so if you do need more space for your answer because of crossing out, be sure to use the back of the SAME page. Do not use the last information page for any answers, as these will be discarded. Note that academic dishonesty is dealt with severely at Columbia. The instructors give an F for the course and refer the matter to the Dean. Expulsion is a common outcome, with no second chances. Please complete the following pledge: “I affirm that I will not plagiarize, use unauthorized materials, or give or receive illegitimate help on assignments, papers, or examinations. I will also uphold equity and honesty in the evaluation of my work and the work of others. I do so to sustain a community built around this Code of Honor.” Student signature_____________________________________________ Intro Bio Exam 2 Oct. 22, 2015 ID code_____KEY________________ 1) E. coli contains the enzyme L-alanine-dehydrogenase that catalyzes the reaction: + L-ala + NAD à pyruvate + NH3 + NADH + H and can grow in a minimal medium containing alanine as the sole source of carbon and energy in the presence of oxygen. 1A. How many net moles of ATP could be produced from ADP per mole of L-ala? _18___. 1 from substrate level phosphorylation in the KC (GTP) 3 from the 1 NADH from the ala dehydrogenase reaction 3 from the 1 NADH produced in the pre-entry reaction to the KC 9 from the 3 NADH’s produced in the KC 2 from the 1 FADH2 produced in the KC 18 total. 1B. Would you expect E. coli to be able to grow in a minimal medium containing alanine as the sole source of carbon and energy in the absence of oxygen? (yes) (no) (can’t predict). Whether you answered yes or no, approximately how many net moles of ATP could be produced from ADP per mole of L-ala? (Consider only the long run (i.e., do not consider very short term effects). _0_ One mole of NAD is consumed in the conversion of 1 mole of alanine to pyruvate and a mole of NADH is consumed in the conversion of pyruvate to lactate, which is the only way E. coli could shuttle the NADH back to NAD. Thus the NAD could be used and regenerated but there is no energy that is recoverable from these two reactions in the absence of air. 1C. Suppose the ΔG° of the alanine dehydrogenase reaction (measured in vitro) is +8 kcal/mole. If this reaction starts with 0.001M of each component and is allowed to reach equilibrium the final ratio of pyruvate to L-alanine in the absence of alanine dehydrogenase, would be (<1) (1) (>1) and in the presence of the enzyme it would be (<1) (1) (>1). A positive ΔG° indicates that the reaction is endergonic, requiring free energy to proceed, so little pyruvate would accumulate. An enzyme does not affect this energetic relationship, which is governed solely by the relative stabilities of the reactants and products. 1D. Suppose E. coli cells are growing on alanine as the only carbon source. Within these cells, you would expect the ΔG° for the alanine dehydrogenase reaction to be (<8) (8) (>8) (can’t predict how many) kcal/mole and the ΔG for this reaction to be (<8) (8) (>8) (how should I know??) kcal/mole. ΔG° is the standard change in free energy for the reaction above under set standard conditions and is a constant regardless of the environment. Since the E. coli cells are growing using alanine, alanine is being metabolized and the alanine DH reaction is the first step. Therefore the alanine DH reaction must be flowing to the right despite the unfavorable ΔG°, undoubtedly being pulled forward by more favorable downstream reactions. Intro Bio Exam 2 Oct. 22, 2015 ID code_____KEY________________ 2. 2. Consider an enzyme for which k3<<< k2 and its Km is 0.0001M. It is studied at a concentration [Eo]. Suppose a non- competitive inhibitor at a concentration of [N] reduces the measured turnover number of this enzyme from 1000 sec-1 to 500 sec-1. Further suppose that a competitive inhibitor of this enzyme at a concentration of [C] increases the apparent (measured) Km of the enzyme to 0.0004M. 2A. In the presence of the 2 inhibitors at [N] and [C], the apparent (measured) Km should be: (0.0001M) (between 0.0001M and 0.0004M) (0.0004M) (>0.0004M) and the calculated apparent turnover number should be: (0) (<500) (500) (between 500 and 1000) (1000) (>1000) sec-1. In the presence of both inhibitors, the maximum Vo will be ½ the Vmax in the absence of the inhibitor due to the effect of the non-competitive inhibitor on k3. The approach to this Vmax will be determined by the apparent Km in the presence of the competitive inhibitor, reaching ½ of the new Vmax at a [S} of 0.0004M. 2B. Draw a diagram showing the shape of two curves for Vo vs. [S] for this enzyme: 1) in the absence of any inhibitors and 2) in the simultaneous presence of both of these inhibitors at concentrations [N] and [C] as was used above. Plot only these 2 conditions. Label the Km’s and Vmax’s. The majority of the credit for this question 2 lies in the diagram. 3.A protein contains no cysteines (sulfhydryls). When subjected to native polyacrylamide gel electrophoresis or SDS- polyacrylamide gel electrophoresis it yields a single band, but when electrophoresed in a similar polyacrylamide gel containing 6M urea it yields 2 bands. Migration is toward the anode (+ pole) in all cases. This protein probably has subunits with: (the same molecular weight) (different molecular weights) (can’t predict) AND (the same net charge) (different net charges) (can’t predict). Circle an answer from each group of 3 and explain the results of all 3 electrophereses. Since 2 bands are evident in one of the separations there must be at least 2 subunits in this protein. Since the 2 subunits migrated identically (single band) in SDS, they must have the same MW. Since they migrated differently (2 bands) in urea where there was no SDS to impart a uniform charge, they would migrate according to their different inherent net charge despite their same MW.