TABLE 3-LACTATE PRODUCTION IN FORTIFIED HEMOLYSATES OF HUMAN ERYTHROCYTES* Substrate Glucose Glucose Lactate production† No. of experiments pH 6 7.1 2.03 ± 0.91 6 7.8 4.76 ± 1.09 7-1 10-73 +1-88 5 7.8 12.34 ±2.92 5 7.0 7-15±0.73 5 7-7 (b)( ) In mature erythrocytes (red blood cells) the end product of glycolysis is lactate because of the absence of mitochondria. On the right is a table comparing the rate of lac- tate production in hemolysates (lysed cells) of human RBCs as a function of pH with dif- ferent substrates introduced into the glyco- lytic pathway. The hemolysate was fortified with 30 μmoles substrate, 7.5 μmoles MgCl2, 10 μmoles disodium phosphate, 1.5 μmoles NAD and 5 μmoles ATP in a volume of 5 mL. The rate of lactate production is given as μmoles of lactate/g Hb/hr at 37° C, buffered to either pH 7.1 or 7.8, as indicated. According to the results in the table which glycolytic enzyme is rate-limiting? Explain. Glucose-6-phosphate Glucose-6-phosphate Fructose-1,6-diphosphate Fructose-1,6-diphosphate 7-15±0.80 (c) (10 pts) Human red blood cell hexokinase I is known as a bisubstrate enzyme meaning that there are two substrates that must be in the active site together for the reaction to proceed. The reaction illustrated below is balanced according to elemental stoichiometry and charge. CH₂OH CH₂OPO hexokinase I OH + ATP4 → OH +ADP3 + H+ HO OH HO OH OH AG' = 16.7 kJ/mol.K OH x-D-glucose α-D-glucose 6-phosphate The diagram on the right illustrates a Lineweaver-Burk plot of the ac- tivity of the human red blood cell hexokinase. The initial velocity measurements were made under the conditions that the substrate glucose was held at a constant concentration while the concentration of ATP was varied. Measurements were made in the presence of three concentrations of glucose-6-phosphate, one of the products of the reaction. The three concentrations are TO (O), 0.06 (●), and 0.12 mM (X). From the diagram it is apparent that glucose-6-phosphate not only is a product of the reaction but is also an inhibitor of the enzyme. On the basis of the diagram, what kind of kinetic inhibitor pattern can be most likely best ascribed to glucose-6-phosphate? Name the pattern and write its reaction scheme using the letters E, enzyme; S, substrate; P, product; and I, inhibitor. VELOCITY,' (CPM × 10-4)-1 B 9 8 L L OJ 02 03 04 05 06 07 I/ATP, (mM)-' | Question #1, continued: (d) ( ) It is generally stated in textbooks that the hexokinase reaction does not proceed to equilib- rium in the cell and that the reaction under the reactant and product concentrations in the cell is not reversible. For any reaction the total free energy change for the reaction is given by the expression AG = AG'° + RT In T where R is the universal Gas constant (8.315 J/mol·K) and T is the temperature in Kelvin (310) and г (gamma) is the mass action ratio, in this case [ADP] x [glucose-6-phosphate] [ATP] x [glucose] in which expression the brackets represent the actual concentrations of the reactants and products. At equilibrium, AG = 0. Calculate AG, assuming г = 0.08 (heart muscle; see Table 13-10, p. 502, 8th edition) and compare I in the heart to that, were the reaction to proceed to equilibrium. What would happen to the ATP in the cell if the reaction proceeded to equilibrium and the initial concentration of ATP were 5 mM? Assume in the latter case that the ratio of [glucose-6-phosphate]/[glucose] = 1.

TABLE 3-LACTATE PRODUCTION IN FORTIFIED HEMOLYSATES OF HUMAN ERYTHROCYTES* Substrate Glucose Glucose Lactate production† No. of experiments pH 6 7.1 2.03 ± 0.91 6 7.8 4.76 ± 1.09 7-1 10-73 +1-88 5 7.8 12.34 ±2.92 5 7.0 7-15±0.73 5 7-7 (b)( ) In mature erythrocytes (red blood cells) the end product of glycolysis is lactate because of the absence of mitochondria. On the right is a table comparing the rate of lac- tate production in hemolysates (lysed cells) of human RBCs as a function of pH with dif- ferent substrates introduced into the glyco- lytic pathway. The hemolysate was fortified with 30 μmoles substrate, 7.5 μmoles MgCl2, 10 μmoles disodium phosphate, 1.5 μmoles NAD and 5 μmoles ATP in a volume of 5 mL. The rate of lactate production is given as μmoles of lactate/g Hb/hr at 37° C, buffered to either pH 7.1 or 7.8, as indicated. According to the results in the table which glycolytic enzyme is rate-limiting? Explain. Glucose-6-phosphate Glucose-6-phosphate Fructose-1,6-diphosphate Fructose-1,6-diphosphate 7-15±0.80 (c) (10 pts) Human red blood cell hexokinase I is known as a bisubstrate enzyme meaning that there are two substrates that must be in the active site together for the reaction to proceed. The reaction illustrated below is balanced according to elemental stoichiometry and charge. CH₂OH CH₂OPO hexokinase I OH + ATP4 → OH +ADP3 + H+ HO OH HO OH OH AG' = 16.7 kJ/mol.K OH x-D-glucose α-D-glucose 6-phosphate The diagram on the right illustrates a Lineweaver-Burk plot of the ac- tivity of the human red blood cell hexokinase. The initial velocity measurements were made under the conditions that the substrate glucose was held at a constant concentration while the concentration of ATP was varied. Measurements were made in the presence of three concentrations of glucose-6-phosphate, one of the products of the reaction. The three concentrations are TO (O), 0.06 (●), and 0.12 mM (X). From the diagram it is apparent that glucose-6-phosphate not only is a product of the reaction but is also an inhibitor of the enzyme. On the basis of the diagram, what kind of kinetic inhibitor pattern can be most likely best ascribed to glucose-6-phosphate? Name the pattern and write its reaction scheme using the letters E, enzyme; S, substrate; P, product; and I, inhibitor. VELOCITY,' (CPM × 10-4)-1 B 9 8 L L OJ 02 03 04 05 06 07 I/ATP, (mM)-' | Question #1, continued: (d) ( ) It is generally stated in textbooks that the hexokinase reaction does not proceed to equilib- rium in the cell and that the reaction under the reactant and product concentrations in the cell is not reversible. For any reaction the total free energy change for the reaction is given by the expression AG = AG'° + RT In T where R is the universal Gas constant (8.315 J/mol·K) and T is the temperature in Kelvin (310) and г (gamma) is the mass action ratio, in this case [ADP] x [glucose-6-phosphate] [ATP] x [glucose] in which expression the brackets represent the actual concentrations of the reactants and products. At equilibrium, AG = 0. Calculate AG, assuming г = 0.08 (heart muscle; see Table 13-10, p. 502, 8th edition) and compare I in the heart to that, were the reaction to proceed to equilibrium. What would happen to the ATP in the cell if the reaction proceeded to equilibrium and the initial concentration of ATP were 5 mM? Assume in the latter case that the ratio of [glucose-6-phosphate]/[glucose] = 1.

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter18: Glycolysis

Section: Chapter Questions

Problem 8P

See similar textbooks