23. An important step in glycolysis is the formation of ATP and pyruvate from phosphoenol-pyruvate (PEP) and ADP. PEP ADP pyruvate+ATP The equilibrium constant (Keq) for this reaction is approximately 2.5x10°. Calculate standard free energy change (AG°') for this reaction. Show your work. onege b. Is the reaction exergonic or endergonic at standard conditions? If, at equilibrium, the concentrations of ADP and ATP are 0.2 mM and 2.0 mM, respectively, what is the equilibrium concentration ratio of [pyruvate] to [PEP]? Show your work. с.

23. An important step in glycolysis is the formation of ATP and pyruvate from phosphoenol-pyruvate (PEP) and ADP. PEP ADP pyruvate+ATP The equilibrium constant (Keq) for this reaction is approximately 2.5x10°. Calculate standard free energy change (AG°') for this reaction. Show your work. onege b. Is the reaction exergonic or endergonic at standard conditions? If, at equilibrium, the concentrations of ADP and ATP are 0.2 mM and 2.0 mM, respectively, what is the equilibrium concentration ratio of [pyruvate] to [PEP]? Show your work. с.

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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1. a. Calculate the physiological DG of the reaction shown below at 37°C, as it occurs in the cytosol ofneurons, with phosphocreatine at 4.7 mM, creatine at 1.0 mM, ADP at 0.73 mM, and ATP at 2.6mM. The standard free energy change for the overall reaction is –12.5 kJ/mol. Phosphocreatine + ADP ® creatine + ATP b. The enzyme phosphoglucomutase catalyzes the conversion of glucose 1-phosphate to glucose6-phosphate. Calculate the standard free energy change of this reaction if incubation of 20 mMglucose 1-phosphate (no glucose-6 phosphate initially present) yields a final equilibrium mixtureof 1.0 mM glucose 1-phosphate and 19 mM glucose 6-phosphate at 25°C and pH 7.0. c. If the rate of a nonenzymatic reaction is 1.2 x 10–2 μM s–1, what is the rate of the reaction at 37℃ inthe presence of an enzyme that reduces the activation energy by 30.5 kJ/mol?
The enzyme triosephosphate isomerase (TIM) catalyzes the following reaction in glycolysis, where it converts dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (GAP). CH₂OH C=O CH₂OPO²- DHAP triose phosphate isomerase [DHAP] (MM 1.00 2.00 3.00 6.00 O V. (mM/s) 3.700 6.727 9.250 14.800 = H HCOH CH₂OPO²- Kinetics experiments were performed on TIM. Enzyme activity (initial velocity, V.) was measured at varying concentrations of DHAP. The enzyme kinetics were also measured in the presence of two inhibitors, A and B. The enzyme concentration used in all experiments was 25.00 μM. The data are shown below. GAP V. (mM/s) + A 1.452 2.794 4.038 7.281 V₁ (mM/s) + B 0.755 1.379 1.905 3.077
1. Provide the best coenzyme(s) for each step shown below. Give a 1–2 sentence rationale for your choice. 2. Reaction (C) can require multiple coenzymes, or the enzyme may only require one. Briefly describe how the reaction that uses only one coenzyme works. Be specific about the mechanism.
An enzyme catalyzes a reaction with a Km of 7.50 mM and a Vmax of 2.90 mM - s-1. Calculate the reaction velocity, un, for each substrate concentration. [S] = 2.75 mM mM · s- * TOOLS x10 [S] = 7.50 mM mM · s-! [S] = 11.0 mM mM - s-
Explain the difference between ΔG°’ and ΔG and calculate the free energy (delta G) of hydrolysis of ATP in a rat liver cell in which the ATP, ADP, and Pi concentrations are 3.4, 1.3, and 4.8 mM.
Assume that in a certain cell, the ratio of products/reactants or Keg = 809.5 (Keq is dimensionless) for the reaction Glucose + 2ATP > Glucose-1,6-diP + 2ADP, at a particular instant, the concentrations of each compound were Glucose =2.4M, ATP =11.1M, ADP -12.8M and G-6-P -28.4M. Calculate the difference (dimensionless) between Keq and the ratio of products/ractants at this instance, in this cell, to five decimal places
6. Examine the plot of the enzyme-catalyzed reaction below: Wheat-germ acid phosphatase shows Michaelis-Menten kinetics when acting on para-nitrophenol phosphate 16 14 12 10 8 4 2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 (S] / mM a) Is this reaction better characterized by Michaelis-Menten kinetics or simple mass action? b) What is the approximate value of Vmax? Km? c) Assuming that this plot was made at 100 nM enzyme, draw the curve, including axis labels, for 300 nM enzyme. d) Imagine that this enzyme is inhibited by a competitive inhibitor with Ki (for inhibitor) = Km (for substrate). Note that v = (kcat [E] tot [S]/Km) / (1 + [S]/Km + [I]/Ki). Draw the curve of reaction velocity (v) versus [S] for [I] = 1 mM. e) Now consider an experiment where you simultaneously introduce substrate and competitive inhibitor at equal concentrations. Make a plot of v versus concentration of [S] = [I], i.e. the X-axis should be the concentration of both I and S, which are equal. v/ uM min-
8
The free energy change of each step of glycolysis is given in the table below. ∆G°’ is the free energy under standard conditions (25°C, 1M each reactant, pH 7), while ∆G is the free energy change at presumed physiological conditions. Why must no step have a positive ∆G under physiological conditions?
Two enzyme catalyzed reactions, RA (red) and Rp (blue), have kinetic profiles shown below. Report which reaction has the higher maximal rate and which reaction has the smaller Km. в A 1/ [S] Higher Vm [ Select ] Lower Km [ Select ] A/T
Select the incorrect statement. With regards to free energy ΔG of the reaction below E+S ⇌ ES Negative ΔG mean the reaction toward is facourable More negative value of ΔG indicates stronger binding to S to E It is possible to compute disassociation constant from the ΔG value alone It is possible to calculate the term ( ΔH – T ΔS) from the value of ΔGalone ΔG = 0 indicates (ES)/(E)(S) =1 None of the above
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