1. The diagram below plots the free energy of glycolytic intermediates in mature red blood cells.+10AG°' =- 130.9 kJAGU-10AGO130 9 kJ(-31.3 kcal)or-20- 31.3 kcal-30-40AG =AG103.8 KJ(-24.6kcal)-50AG103.8 kJ-60or- 24.6 kcal-70GLUG6PF6PFBPGAPGBPPG3PG2PEPPYRLACNote that Gor free energy in kcal is indicated along the vertical axis. The abbreviations in upper caseletters along the horizontal axis at the bottom represent the various intermediates in the glycolyticconversion of glucose to pyruvate and to lactate. Therefore, the difference in any two points on thegraph represents the difference in free energy between the two points. For instance, the AG'and theAG indicated with arrows on the left-hand end of the graph represent the two types of free energychanges associated with the conversion of glucose to glucose-6-phosphate catalyzed by hexokinase.bB SdD, write the equation(a)showing the relationship of AG and AG°' with respect to the mass-action ratio. Define AG, AG°, andAG°'.For a typical chemical reaction aA+cC+(b)equilibrium. Explain which data in the graph reflect reactions far from equilibrium and which reflectreac- tions at or near equilibrium. How do these two types of metabolic reactions differ with respectto regulation?Some glycolytic reactions lie far from equilibrium while others are near or essentially atG (free energy) (kcal)

Metabolic processes involve chemical reactions and these reactions usually involve changes in energy…

bB 5 cC dD, write the equation (a) showing the relationship of AG and AG°' with respect to the mass-action ratio. Define AG, AG°, and AG°'. For a typical chemical reaction aA + (b) equilibrium. Explain which data in the graph reflect reactions far from equilibrium and which reflect reac- tions at or near equilibrium. How do these two types of metabolic reactions differ with respect to regulation? Some glycolytic reactions lie far from equilibrium while others are near or essentially at

bB 5 cC dD, write the equation (a) showing the relationship of AG and AG°' with respect to the mass-action ratio. Define AG, AG°, and AG°'. For a typical chemical reaction aA + (b) equilibrium. Explain which data in the graph reflect reactions far from equilibrium and which reflect reac- tions at or near equilibrium. How do these two types of metabolic reactions differ with respect to regulation? Some glycolytic reactions lie far from equilibrium while others are near or essentially at

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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3. (a) The beakers below represent different conditions for measuring the initial velocity (vo) of an enzyme-catalyzed reaction under the steady-state approximation. The gray "donut-shaped" struc- tures represent the enzyme. The blue, filled circles represent free substrate molecules. The red cir- cles represent substrate molecules bound in the active site of the enzyme forming the Michaelis (ES) complex. Indicate in the diagram of the double reciprocal plot which kinetic parameters or variables each of the three "beaker conditions" represent either alone or in combination with another. A B 455 C V₁™¹ [So]-¹
10) The graph shown below is the pre-steady state data for the serine protease (chymotrypsin) enzyme. Explain how this result suggests the formation of a covalent enzyme intermediate in the active site during the first step in the pathway. 3.0 2.0 1.0 1 3 Time (min) p-Nitrophenol (mol/mol of enzyme) 2.
How will a graph of reaction rate (V) vs. [substrate] for an allosteric enzyme differ from the hyperbolic plot characteristic of enzymes having Michaelis-Menten kinetics?
Given the following reaction and equation for the initial velocity of the reaction: k₁ k3 E+S ES E + P V=Keat [ES] = k3 [ES] k₂ where keat is the rate constant for the reaction which forms the product from the ES complex. Explain in words why the velocity is directly proportional to theamount of enzyme added in the presence of saturating substrate levels.
You are working on an enzyme that obeys standard Michaelis-Menten kinetics. What variable is the V, dependant on if the concentration of the substrate is substantially higher than the concentration of the enzyme? [S] [E] [ES] O [P] O not enough information provided
An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-s
At what substrate concentration would an enzyme with a kcat of 33.0 s-1 and a Km of 0.0046 M operate at one-tenth of its maximum rate? Assume the enzyme obeys Michaeli s-Menten kinetics.
Given that the AG" values for the hydrolysis of glucose 1-phosphate and glucose 6-phosphate are approximately -21 kJ/mol and –14 kJ/mol, respectively, which statement is true regarding the isomerization shown? glucose 1-phosphate → glucose 6-phosphate O The reaction can likely proceed in either direction and will therefore depend on the concentrations of glucose 1-phosphate and glucose 6-phosphate. O Only the anabolic direction will occur because the free energy change is positive. O The reaction will proceed in both directions simultaneously. This reaction can only proceed in one direction and must be the rate limiting reaction for this point in metabolism.
Derive an Equation that explains the realtionship between kE and kN with respect to the equilibrium constants provided in the reaction scheme provided below. Assume that the enzyme must bind with A before it binds with B. The two reactions are related by the following reaction scheme:provided below.
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 equilibrium constant for the hydrolysis of the peptide alanylglycine (Gly-Ala in the reaction from Part B) by a peptidase is K = 9.0 × 10² at 310 K. Calculate AG for this reaction. Express the Gibbs free energy to three significant figures. AG = Submit ΠΑΠΙ ΑΣΦ Request Answer ? kJ/mol Keq [Gly] [Ala] [Gly-Ala]
For an enzyme catalyzed reaction of the form: S + E → P + E, the rate of product formation, [P], is given by: d[P]/dt = k2[E}total [S] /(Km + [S]) = For the enzymatically catalyzed hydrolysis of ATP at 25 °C and pH 7.0, the Michaelis Menten constant, Km was found to be equal to 16.8 μmol L 1 and the value of k2[E]total was found to be 0.220 μmol L-¹s¹. Find the initial rate at an initial ATP concentration of 30.0 μmol L-1