Intro Bio Exam 2 Oct. 22, 2015 ID code_____KEY______NAME ______ANSWER KEY______Name on cover sheet only, to keep grading anonymous

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_____ 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%.

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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 L--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 . 1A. How many net moles of ATP could be produced from ADP per mole of L-ala? _18___.

1 from 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. They produce a single band in a native gel because they remain associated as a single multimeric molecular species, e.g., a dimer. Intro Bio Exam 2 Oct. 22, 2015 ID code_____KEY______4. Consider the mutants of E. coli described below. In each case there is an amino acid change that disrupts the protein’s normal function as indicated. Predict the energy consequences in the long run (i.e., do not consider very short term effects). And assume there is an enzyme that simply hydrolyzes AcetylCoA to acetate and CoA in an exergonic reaction. 4A. Consider a mutation of succinic dehydrogenase, such that succinate can no longer be converted to fumarate. How many net moles of ATP can be formed per mole of glucose metabolized for this E. coli mutant growing in air? __14__

ATP from substrate level phosphorylation in glycolysis: 2 ATP from 2 NADH formed in glycolysis: 6 ATP from 2 NADH formed in the pre-enry reaction to the Krebs Cycle: 6 ATP from the running of Krebs cycle: 0, as oxaloacetate cannot be regenerated. Total: 14

4B. Consider a mutant in which the protein that is disrupted is the enzyme within Complex II of the electron transport chain that transfers electrons from the FADH2 produced by succinate dehydrogenase to form reduced Coenzyme Q. The net moles of ATP that can be formed per mole of glucose metabolized for this E. coli mutant growing in air is (less than) (greater than) (the same as) the answer to 4A?

Same as 2A, since the Krebs Cycle will stop at succinate because the cell will run out of FAD, which has all accumulated as FADH2. Flow of electrons through Complexes I, III, and IV should not be affected.

4C. Consider a mutation in the ATP synthetase that destroys the for phosphate in the αβ subunit complexes. How many net moles of ATP can be formed per mole of glucose metabolized for this E. coli mutant growing in air? _4_

No oxphos but the KC and the ETC should run normally and the protons returned to the matrix normally. So all ATPs would be generated by substrate level phosphorylation, 2 ATPs from glycolysis and 2 from the GTP step in the KC.

4D. Consider a mutation that disrupts the function of the that acts in step 5 of glycolysis. Consider this mutant growing in a minimal medium containing glucose-6-phosphate rather than glucose, and that glucose-6-phosphate can freely diffuse into the cell. Can this mutant E. coli grow in glucose-6-phosphate in the absence of air? (yes) (no) (can’t predict)

Only a single ATP need be invested in the conversion of G6P to G3P. In the subsequent conversion of one G3P to pyruvate, 2 ATPs are generated. Thus the net ATP production is one. The NAD used in this conversion will be regenerated by reducing pyruvate to lactate as in regular E. coli fermentation. The DHAP could be excreted as an unused by-. Another possible but less straightforward answer is No, if you propose that the DHAP is not excreted but rather accumulates in the cell and then prevents the conversion of G6P to G3p and DHAP by mass action. UN2005/2401 ‘16 Exam #2 Oct. 20, 2016 ID code______KEY 26___ ID = birthdate plus 5-letter word

6. There is a pathway in Drosophila melanogaster that leads from cholesterol (A) to 20-hydroxyecdysone (H), an important developmental hormone. Mutations (inherited changes) that cause changes in amino acids that inactivate any pathway enzyme lead to a “Halloween” phenotype: Drosophila larvae with a featureless cuticle that looks like a ghost. The mutants have names like phantom, disembodied, spooky, and spookier. J3(for36B3only)

Reaction #: 1 2 3 ! 4 5 6 7 A ! B ! C ! D ! E ! F ! G ! H

6A. Assuming only A and H are important biological molecules (i.e., B-G are intermediates with no other purpose), 4 which enzyme is likely to be allosterically inhibited and by which molecule? The enzyme that catalyzes reaction: (1) (2) (3) (4) (5) (6) (7) is likely to be inhibited by molecule (A) (B) (C) (D) (E) (F) (G) (H). The0end0product020 !hydroxyecdysone0(H)0likely0feedback0inhibits0the0first0committed0step0in0its0synthesis,0the0 1 reaction0A0 ! B0catalyzed0by0enzyme01,0once0H0is0present0in0high0enough0levels0that0no0more0needs0to0be0made.

6B. For this part only assume that instead D is also a precursor for J, another molecule. Which enzyme in the 20- hydroxyecdysone pathway is most likely to be allosterically inhibited and by which molecule? In this case, the 4 enzyme that catalyzes reaction: (7) (6) (5) (4) (3) (2) (1) is likely to be inhibited by molecule (H) (G) (F) (E) (E) (C) (B) (A). The0end0product020!hydroxyecdysone0(H)0likely0feedback0inhibits0the0first0committed0step0in0its0synthesis,0this00time0 1 reaction0D0! E0catalyzed0by0enzyme04.0Inhibition0of0enzyme010would0prevent0J0from0being0made.

6C-D. There are multiple different Drosophila mutants on the ecdysone pathway, each with a change in its genetic material that inactivates one of the 1-7. One way to identify which step of the pathway is affected by mutation of an enzyme gene is to feed the Drosophila mutant various intermediate compounds and determine whether the mutant phenotype is “rescued” (i.e., the mutant organisms now develop normally).

6C. A mutant for the enzyme 5 gene will develop normally when fed (A) (B) (C) (D) (E) (F) (G) (H) but not when 4 fed (A) (B) (C) (D) (E) (F) (G) (H). Circle all that apply. Mutants0with0inactive0enzyme0for0 rxn 50cannot0make0F,0and0thus0normally0would0not0make0G0or0H0either.0Because0 1 the0enzyme0for0 rxns 60and070are0working0properly,0providing0F,0G,0or0H0will0allow0development0to0proceed0normally.0 Mutant0for0 enz.050are0not0“rescued”0by0any0intermediate0before0the0block0because0it0cannot0make0F0or0ultimately0H. 6D. Several uncharacterized mutants exist in this pathway. One such mutant, phantom, is rescued by being fed intermediate E but not intermediate D. The phantom mutant has inactivating alterations in the enzyme that catalyzes 2 reaction (1) (2) (3) (4) (5) (6) (7).

The0fact0that0the0phantom mutant0is0rescued0by0intermediate0E0means0that0the0block0is0in0enzyme01,02,030or04.0The0

fact0that0it0is0not0rescued0by0intermediate0D0narrows0the0choice0to04,0since01!30would0be0rescued0by0feeding0D.0 1

7. The degradation of the carbon skeleton of uracil to CO2 proceeds through malonate semialdehyde, at the expense 8 of an NADH. Malonate semialdehyde then forms acetyl-CoA, CO2, and NADH (see attached info sheet 2). How many net moles of ATP could be produced from ADP per mole of uracil? Show your calculations.

The0cell0has0everything0needed0to0complete0the0reactions0on0info0sheet02.0This0means0Krebs0and0ETC0must0be0

working0or0you0wouldn’t0get0CO 2.0 In0the0process0of0going0from0uracil0to0acetyl !CoA,0one0NADH0is0used0and0one0is0produced,0cancelling0each0other0out.0 Thus,0there0is0no0net0production0of0ATP0for0that0part0of0uracil0degradation.0Krebs0is0as0normal,0producing: 10ATP0(from0substrate0level0phosphorylation,0a0GTP0that0gets0converted0to0ATP) 90ATP0from030NADH 20ATP0from010FADH2 120net0ATP0(120moles0of0ATP0per0mole0of0uracil) Intro Bio Exam 2 Oct. 22, 2015 ID code_____KEY______