A certain metabolic pathway can be diagrammed as:X Y ZA B C DC²Dwhere A, B, C, and D are the metabolic intermediates and X, Y, and Z are the enzymes responsible for each conversion. The physiological freeenergies for each enzyme catalyzed reaction are:XAG = -0.2 kJ/molYAG¹ = -1.3 kJ/molZAG' = -12.3 kJ/mola. Which reaction likely represents a major regulatory point for this pathway?b. If your answer for part a. was, in fact, the case, if an inhibitor of Enzyme Z was present, would the concentrations of metabolites A, B, C,and D be increased, decreased, or not affected?c. Identify which of the enzymes X, Y, and Z, if any, would likely need to be bypassed for the metabolite D to be converted to A.

Metabolism is the total of all chemical transformation that takes place in a living cell. One…

Answered: A certain metabolic pathway can be…

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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A particular reaction has a ΔG‡ of 30.0 kJ mol-1 at 25.0 °C. In the presence of an enzyme, the same reaction has a ΔG‡ of 1.50 kJ mol-1 at the same temperature. Calculate the rate enhancement of this enzyme. (R = 8.3145 J mol-1 K-1)
The enzyme phosphoglucomutase catalyzes the conversion of glucose 1-phosphate to glucose 6-phosphate. After the reactants and products were mixed and allowed to reach equilibrium at 25°C, the concentration of glucose 1-phosphate was 4.5 mM and that of glucose 6-phosphate was 86 mM. Calculate Keq' and AG for this reaction. The reaction coordinate diagram for an enzyme-catalyzed reaction is shown below. How many transition states and intermediates are in the reaction? Is the reaction thermodynamically favorable? Which step is the rate-determining step of the reaction? G Reaction coordinate
You will perform the protocol below for the calf intestinal alkaline phosphatase (CIP) provided. For each reaction, your final enzyme concentration should be 10 nM CIP. Note: Enzymes purchased are typically labelled with their “units of activity” (U), as this relates to how much enzyme is needed to catalyze a reaction. The 100 nM CIP provided has approximately 3 U/mL and was diluted 1 in 1,000 from a 500 U/mL purchased enzyme. 1) Create a table (similar to the one below) to help you determine and keep track of what to add to each of the cuvettes in which your reactions will be measured. The five different concentrations of PNPP should be: 25, 50, 100, 200, 300 μM. Each reaction will be in a final volume of 1 mL and contain 10 nM alkaline phosphatase. Concentrations of stock solutions: 1.0 mM PNPP, 100 nM calf intestinal phosphatase
What terms would best describe the above coupled reaction? (If the DGo for ATP hydrolysis into ADP + inorganic phosphate is -7.3 kcal/mole, and the DGo for maltose synthesis from glucose + glucose is +3.7 kcal/mole, calculate the standard free energy change for the combined reaction of ATP + glucose + glucose g ADP + maltose + inorganic phosphate.) it is non-spontaneous and endothermic (because the overall DGo is negative) it is spontaneous and exothermic (because the overall DGo is negative) it is non-spontaneous and endothermic (because the overall DGo is positive) it is spontaneous and exothermic (because the overall DGo is positive) it is non-spontaneous and exothermic (because the overall DGo is negative)
The Michaelis-Menten equation for the enzyme chymotrypsin is 0.14[S] v : 0.015 + [S] where v is the rate of an enzymatic reaction and [S] is the concentration of a substrate S. Calculate dv/d[S] and interpret it.
The protein catalase is an enzyme that catalyzes the decomposition of hydrogen peroxide:2 H2O2 (aq) → 2 H2O (l) + O2 (g)and has a Michaelis-Menten constant of 25 × 10-3 mol·dm-3 and a turnover number of 4.0×107s-1.The total enzyme concentration is 0.016×10-6 mol·dm-3 and the initial substrate concentration is4.32×10-6 mol·dm-3 Calculate the maximum reaction rate (????) for this enzyme, and the initial rateof this reaction. Note that catalase has a single active site.
In biological systems, enzymes are used to accelerate the rates of certain bio- logical reactions. Glucoamylase is an enzyme that aids in the conversion of starch to glucose (a sugar that cells use for energy). Experiments show that 1 pg mol of glucoamylase in a 4% starch solution results in a production rate of glucose of 0.6 ug mol/(mL)(min). Determine the production rate of glucose for this system in the units of Ib mol/(ft³)(day). Ibmol Answer: 0.0639 ft day True False
Enzymes Calculate the AG between glucose and ATP catalyzed by hexokinase: glucose + ATP glucose-6-phosphate + ADP Is the reaction spontaneous? Given data: Glucose-6-phosphate + H,O glucose + P, AG' = -3138 cal/mole ATP+H,O=ADP + P: AG'= -7700 cal/mole
Most glucose in the human diet comes from plants. Unlike humans, plants transport sucrose instead of glucose. The conversion of sucrose to glucose and fructose is a reversible reaction as shown in Reaction 25.1: Reaction 25.1 Sucrose (aq) + H2O (l) ⇄ Glucose (aq) + Fructose (aq) Under conditions typical of human metabolism, the Keq for Reaction 25.1 is 140,000. At equilibrium, would you expect more of the reactant sucrose or more of the products glucose and fructose to be present? Explain reasoning.
The enzyme aldolase catalyzes the reaction shown in the glycolytic pathway: Fructose 1,6-bisphosphate dihydroxyacetone phosphate + glyceraldehyde 3-phosphate The AG" for the reaction is +23.8 kJ mol¯¹ (+5.7 kcal mol−¹), whereas the AG in the cell is −1.3 kJ mol¯¹ (−0.3 kcal mol¯¹). Calculate the ratio of products to reactants under standard (equilibrium) conditions at 37°C. [products] [reactants] 7 x10-5 [products] [reactants] Incorrect Aldolase ===== Calculate the ratio of products to reactants under intracellular conditions at 37°C. 4 ×10-5 Incorrect Complete the statement using your results. under standard conditions under intracellular conditions A reaction that is endergonic under standard conditions can be converted into an exergonic reaction by maintaining the ratio of products to reactants below the equilibrium value.
The first reaction in Glycolysis is the phosphorylation of Glucose:Pi + Glucose → Glucose 6-Phosphate + waterThis is a thermodynamically unfavorable process with a ∆G0’ = +13.8 kJ/mol. In a liver cell at370C the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be the equilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state [H2O] = 1.0 a. 0.00000012 b. 0.0000025 c. 0.00000025 d. 0.00474 e. 0.005 f. 1.0 g. 0.1 h. 0.25 i. 2.4
The first reaction in Glycolysis is the phosphorylation of Glucose:Pi + Glucose → Glucose 6-Phosphate + waterThis is a thermodynamically unfavorable process with a ∆G0’ = +13.8 kJ/mol. In a liver cell at370C the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be the equilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state [H2O] = 1.0 (this will help with the question below) When the reaction described in question number 3 is coupled to the hydrolysis of ATP the equilibrium concentration of Glucose 6-Phosphate goes to: a. 10.5 b. 350 c. 3000 d. 77 e. 4.5 f. 55 g. 550 h. 652 i. 1120

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