QUESTION 63 The reaction catalyzed by UDP-glucose pyrophosphorylase, UTP + glucose-1-phosphate <-> UDP-glucose + PP; proceeds with approximately AGO = 0. Meanwhile, the reaction PP; + H20 --> 2 Pj, catalyzed by pyrophosphatase, proceeds with AGO = -19.3 kJ mol1. If these two reactions are coupled in the cell, wh- AGO' for the net reaction: UTP + glucose-1-phosphate + H20 <--> UDP-glucose + 2 P; O 19.3 k) mol-1 O Approximately 0 kj mol1 O +19.3 k) mol1 O Cannot tell from the information given.

Question 1: In some microorganisms, carbon fixation occurs by reversal of the citric acid cycle. This reversal is accomplished in part by the use of a strong reductant (ferredoxin) to drive the alpha-ketoglutarate dehydrogenase reaction in the reductive direction. Part a: ΔG°‘ for reaction as it occurs in the ‘normal’ (oxidative) citric acid cycle is -30.1 kJ/mol. The standard reduction potential for NADH is -0.32 V. In order to drive the reaction in the reverse direction, the reductant (a ferredoxin) must have a lower reduction potential than NADH/NAD+. Remembering that this is a two-electron reduction, and using the numbers given just above, compute the value of the ferredoxin reduction potential that would be needed to make the standard free energy zero (so that the reductive reaction is enegetically just as favorable as the oxidative reaction). Assume that all of the other reactants are the same in the reductive as in the oxidative reaction. Write out the steps in your calculation;…

When rotenone is added to actively respiring bovine mitochondria, the ratio of NADH to NAD+ increases but the FADH2 to FAD ratio remains unchanged.            a. What step in the system is being inhibited by rotenone?                  b. What would be the effect of rotenone on pyruvate oxidation to CO2?                              ____________________ Why?

Remaining V Ha Question 17 Triose phosphate isomerase catalyzes the conversion of Glyceraldehyde-3- phosphate to Dihydroxyacetone phosphate. Glyceraldehyde-3-P Dihydroxyacetone phosphate The Km of this enzyme for its substrate Glyceraldehyde-3-phosphate is 1.8 x 10 M. When [Glyceraldehyde-3-phosphate] = 30 uM the rate of the reaction (v) was 82.5 umol/mL/sec and the Vmax was 132 umol/mL/sec. Assuming 3 nanomoles/mL of enzyme was used ([Eotal = 3 nanomol/mL) the kcat for this enzyme is: Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a 44000/sec b 1100/sec 88/min d. 120/hour e 11/min Im f 450000/sec g 52800/sec 175/sec

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Question #2: Consider the following data for Cell G run in 1 L chemostat, with a feed containing 30 g/L substrate: Feedrate (ml/hr) (c) Calculate 27.5 24.2 22.1 18.8 Steady State Cell Mass Concentration (g/L) 12.0 14.7 15.8 16.7 Steady State Substrate Concentration (g/L) 10.0 5.56 (d) Using the calculated kinetic parameters, 3.70 (a) Determine the kinetic parameters for growth (b) At each dilution rate, calculate Yx/s. Is it constant with dilution rate and substrate concentration? 2.23 (i) the critical dilution rate, D. (ii) the dilution rate corresponding to maximum productivity, Dm (iii) the maximum cell productivity (cells/L/hr) (i) Generate a plot of X and S vs D for a feed concentration of 25 g/L. (ii) What is the critical dilution rate for this condition? (iii) What is the dilution rate corresponding to maximum productivity at this condition? (iv) What is the maximum cell productivity at this condition? (v) For a reactor operating at this maximum productivity, what would be the…

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The answer to part a is -22.4 I need help with part b thank you

Question 1: The amount of energy effectively stored in each ATP (the actual energy of ATP hydolysis) will depend on the concentrations of ATP, ADP, and Pi, according to equations discussed in class and in the book. If the ATP synthase is operating at close to equilibrium (as might be the case if ATP is being used slowly), we can compute the maximum possible energy of ATP hydrolysis by assuming balance between the energy provided by the proton gradient and the energy used to make the ATP. (Since we care about the magnitude of the energy it’s convenient to talk in terms of absolute values; the maximum possible ATP hydrolysis energy is then the most negative value possible.) Using what you know about the mechanism of the ATP synthase, and assuming that the membrane gradient consists of a Δψ of -.15V and a ΔpH of 1 unit (alkaline inside), compute the maximum possible ATP-hydrolysis energies for synthases with: a.) 8 c-subunits   b.) 13 c-subunits