In a major metabolic pathway involving the monosaccharide glucose, one of the reactions involve the conversion of glucose to glucose-6-phosphate summarizedbelow together with accompanying free energy change:glucose + phosphate =glucose-6-phosphate + H,0 AG = 13.8 kJ · mol-1 (Reaction 1)In cells, the production of G6P (Reaction 1 above) is coupled to a reaction that involves the hydrolysis of ATP to ADP (shown below, Reaction 2) which makes theoverall reaction much more favorable to the production of glucose-6-phosphate.ATP + H,O = ADP + phosphate(Reaction 2)Why do you think coupling the production of glucose-6-phosphate to the hydrolysis of ATP makes the overall reaction spontaneous? What can you say aboutthe free energy change accompanying the hydrolysis of ATP (Reaction 2)?

The given reaction represents the first reaction of the glycolysis pathway. In this reaction, the…

Answered: In a major metabolic pathway involving…

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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Write the coupled reaction that would occur spontaneously for the following pairs of molecules under standard condition: Oxaloaxcetate and NAD+
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.
The primary source of phosphate for chemical reactions in cells comes as ATP, ADP, or AMP. The ΔG˚’ for the hydrolysis of ATP into ADP and Pi is -30.5 kJ/mol. Calculate the equilibrium constant for the hydrolysis of ATP.
The reaction below occurs in nearly all human cells. It is catalyzed by the enzyme enolase. CO- CO- enolase -OPO3 OPO3 + H20 ČH,OH AGO'=+1.7 kJ/mol CH2 2PGA PEP Under cellular conditions, the AG is -3.3 kJ/mol. Which of the following could account for these changes in cellular AG compared to the standard biological free energy? The concentration of 2PGA is 7 times higher than PEP. The concentration of PEP is 7 times higher than 2PGA Enolase lowers the activation energy of the transition state. The release of water from 2PGA is a form of energy coupling.
Under standard conditions, is the oxidation of free FADH2 by ubiquinone sufficiently exergonic to drive the synthesis of ATP? (F = 96.5 kJ/V • mol) ubiquinone + 2H++ 2e¯ → ubiquinol E' = 0.045 V FAD + 2H++ 2e → FADH2 = E' -0.219 V ○ no, as oxidation of FADH2 by ubiquinone generates 33.6 kJ/mol ○ no, as oxidation of FADH2 by ubiquinone generates 50.9 kJ/mol ○ no, as oxidation of FADH2 by ubiquinone generates -25.5 kJ/mol ○ yes, as oxidation of FADH2 by ubiquinone generates -33.6 kJ/mol ○ yes, as oxidation of FADH2 by ubiquinone generates -50.9 kJ/mol
Glucose 1-phosphate is converted into fructose 6-phosphate in two successive reactions. Reaction 1: Glucose 1-phosphate → glucose 6-phosphate AG'° = -7.3 kJ/mol Reaction 2: Glucose 6-phosphate → fructose 6-phosphate AGʻ° = +1.7 kJ/mol Using the values given, calculate the standard free-energy change, AGʻ°, ,for the sum of the two reactions: sum > Sum: Glucose 1-phosphate → fructose 6-phosphate AG'°, kJ/mol sum Calculate the equilibrium constant, K', eq > for the sum of the two reactions. K'e eq II
The conversion of dihyroxyacetone phosphate to glyceraldehyde 3-phosphate, catalyzed by triose phosphate isomerase, has an equilibrium constant of Keq’ = 0.0475. Calculate the standard free-energy change for this reaction at 25ºC and pH 7.0.
The hydrolysis of ATP has an enthalpy and entropy of -24.3 kJ/mol and +21.6 J.K-1.mol-1, respectively, at 37 o C. What is the free energy change for the hydrolysis of 5 mols of ATP? Explain what contributes to the negative enthalpy change and positive entropy change in this reaction. What physical characteristics of the reaction would change if an ATP hydrolase enzyme is added to the solution?
(b) You are investigating the effects of several agents on the activity of alcohol dehydrogenase. The enzyme activity data are shown in the table below. Construct a [substrate] vs. activity plot and a double-reciprocal plot for this enzyme. Be sure to label all axes. Determine the Vmax and KM for AD from the graphs in each type of plot. AD activity (nM/min) AD activity + agent A (nM/min) AD activity + agent B (nM/min) [Alcohol] (nM) 0.1 14 2 0.5 50 7 8. 1.0 65 10 30 2.0 72 12 45 4.0 80 14 62 8.0 85 15 75 32.0 90 16 90
Calculate AG for this reaction under the following conditions: 37°C, pH 7, [Pyruvate] = [CO₂] = 4.0 mM, [OAA] = 2.0 mM, [ATP] = 3.5 mM, [P;] = 5.0 mM, and [ADP] = 1.8 mM. Use 2 signficant figures. AG= i kJ.mol-1 Under these conditions, the reaction is (Tolerance is +/- 2%)
Complete the balanced equation for the overall reaction by selecting an answer choice in the brackets. Sucrose + [2 Pi, 4Pi]+[4 ADP, 2 ADP, 4 ATP, 2ATP]+[2 NAD+, 4 NAD+, 6 NAD+]+[H2O, 5 H2O, 3 H2O] --> [2 cirate, 2 oxaloacetate, 2 pyruvate, 2 acetyl-coA]+[4 ADP, 2 ADP, 4 ATP, 2ATP] + [2 NAD+, 4 NAD+, 6 NAD+] + [2H+, 8H+, 6 H+, 4 H+, 10 H+] Does the commercial process require aerated culture medium—that is, is this a fermentation or an aerobic process? A. a fermentation process, because A. niger cells must use O2O2 to continuously regenerate NAD+ B. an aerobic process, because A. niger cells must use O2O2 to continuously regenerate NAD+ C. a fermentation process, because A. niger cells cannot use O2O2 to continuously regenerate NAD+ D. an aerobic process, because A. niger cells cannot use O2O2 to continuously regenerate NAD+
Consider the following chemical reaction: Glucose + ATP → Glucose-6-Phosphate + ADP + Pi Given the following information, calculate the actual free energy change (G’) of this reaction. G’ of Glucose + ATP → Glucose-6-Phosphate + ADP + Pi = -16.7 kJ/mol [Glucose] = 5.0 mM [ATP] = 1.85 mM [Glucose-6-Phosphate] = 0.083 mM [ADP] = 0.14 mM [Pi] = 1.0 mM Temperature = 37 C R (Gas Constant) = 8.314 J/mol•K

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