Can someone help with this question? Hexokinase catalyzes the following reaction in glycolysis:Glucose + ATP -→• glucose-6-phosphate + ADPThe standard free-energy change for this reaction in the direction written is -16.7 kJ/mol.The concentrations of the related reaction components in the hepatocyte of a mammalian cell are:Glucose: 5 mMGlucose-6-phosphate: 0.083 mMATP: 1.85 mMADP: 0.14 mMWhat is the actual free-energy change for the reaction at body temperature (37 C)?-14.3 kJ/mol-33.3 kJ/mol+33.3 kJ/mol+14.3 kJ/mol Use the following table for all questions in the problem set related to glycolysis:TABLE 18.1 Reactions and Thermodynamics of GlycolysisAG"AGReactionEnzyme(k/mol)at 25°c(kJ/mol)HexokinaseGlucokinasea-D-Glucose + ATPt-=-16.7850-33.9*glucose-6-phosphate?- + ADP- + H*Glucose-6 phosphate²-=fructose-6-phosphate-Fructose-6-phosphate- + ATP-=fructose-1,6-bisphosphate- + ADP- + H*Fructose-1,6-bisphosphatedihydroxyacetone-P- + glyceraldehyde-3-P:-Dihydroxyacetone-P=glyceraldehyde-3-P-Glyceraldehyde-3-P- + P?- + NAD*=1,3-bisphosphoglycerate- + NADH + H*1,3-Bisphosphoglycerate + ADP-PhosphoglucoisomerasePhosphofructokinase+1.670.51-2.92-14.2310-18.8Fructose bisphosphatealdolase+23.96.43 X 10-3-0.23Triose phosphate isomeraseGlyceraldehyde-3-PdehydrogenasePhosphoglycerate kinase+7.560.0472+2.41+6.300.0786-1.29-18.92060+0.13-Pglycerate- + ATP-3-Phosphoglycerate =2phosphoglycerate2-Phosphoglycerate-=Phosphoglycerate mutase+4.40.169+0.83Enolase+1.80.483+1.1phosphoenolpyruvate-+ H,0Phosphoenolpyruvate- + ADP- + H* =Pyruvate + ATpt-Pyruvate + NADH + H*= lactate" + NAD*Pyruvate kinase-31.73.63 x 10-23.0Lactate dehydrogenase-25.22.63 x 10-14.8

The above reaction is the first step in glycolysis where one molecule of glucose is phosphorylated…

Hexokinase catalyzes the following reaction in glycolysis: Glucose + ATP-- glucose-6-phosphate + ADP The standard free-energy change for this reaction in the direction written is -16.7 kJ/mol. The concentrations of the related reaction components in the hepatocyte of a mammalian cell are: Glucose: 5 mM Glucose-6-phosphate: 0.083 mM ATP: 1.85 mM ADP: 0.14 mM What is the actual free-energy change for the reaction at body temperature (37 C)? -14.3 kJ/mol -33.3 kJ/mol +33.3 kJ/mol +14.3 kJ/mol

Hexokinase catalyzes the following reaction in glycolysis: Glucose + ATP-- glucose-6-phosphate + ADP The standard free-energy change for this reaction in the direction written is -16.7 kJ/mol. The concentrations of the related reaction components in the hepatocyte of a mammalian cell are: Glucose: 5 mM Glucose-6-phosphate: 0.083 mM ATP: 1.85 mM ADP: 0.14 mM What is the actual free-energy change for the reaction at body temperature (37 C)? -14.3 kJ/mol -33.3 kJ/mol +33.3 kJ/mol +14.3 kJ/mol

Basic Clinical Laboratory Techniques 6E

6th Edition

ISBN:9781133893943

Author:ESTRIDGE

Publisher:ESTRIDGE

Chapter6: Basic Clinical Chemistry

Section6.5: Blood Glucose And Hemoglobin A1c

Problem 8RQ

See similar textbooks

Similar questions

The standard free energy change for this reaction in the direction written is +23.8 kuimol. The tabie shows the concentrations of the three intermediates in the hepatocyte of a mammal. Intermediate Concentration (M) Fructose 1.0-bisphosphate 0.000028 Gyoeraldehyde 3phosphate 0.0000068 Ditydroxyacetone phosphate 0.000032 At body temperature (37 "C). what is the actual free energy change for the reaction (in kimol) ?
In the third step of glycolysis, the given reactions are coupled. reaction 1: fructose-6-phosphate + Pi ⟶ fructose-1,6-bisphosphate + H2O (Δ? = −28 kJ/mol) reaction 2: ATP + H2O ⟶ ADP + Pi (Δ? = +13.8 kJ/mol) Calculate the overall ΔG (kJ/mol) for the coupled reaction.
Explain thermodynamics of glycolysis in relation to Step 3 as shown in the Table and how it does not violate thermodynamic laws?
The first reaction in glycolysis is the phosphorylation of glucose: Pi + glucose → glucose-6-phosphate + H2O This is a thermodynamically unfavorable process, with ∆G°′ = +13.8 kJ/mol. (a) In a liver cell at 37 °C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. What would be the equilibrium concentration of glucose-6-phosphate, according to the above data? (b) This very low concentration of the desired product would be unfavorable for glycolysis. In fact, the reaction is coupled to ATP hydrolysis to give the overall reaction ATP + glucose → glucose-6-phosphate + ADP + H+ What is ∆G°′ for the coupled reaction? (c) If, in addition to the constraints on glucose concentration listed previously, we have in the liver cell ATP concentration = 3 mM and ADP concentration = 1 mM, what is the theoretical concentration of glucose6-phosphate at equilibrium at pH = 7.4 and 37 °C? The answer…
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?
The first reaction in Glycolysis is the phosphorylation of Glucose: Pi + Glucose - Glucose 6-Phosphate + water This is a thermodynamically unfavorable process with a AG,' = +13.8 kJ/mol. In a liver cell at 37°c 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 Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a 0.00000012 0.0000025 0.00000025 0.00474 e 0.005 1.0 0.1 h 0.25 i 2.4
The glucose/glucose-6-phosphate substrate cycle involves distinct reactions of glycolysis and gluconcogenesis that interconvert these two metabolites. Assume that under physiological conditions, [ATP] = [ADP] and [Pi] =1 mM. Consider the following glycolytic reaction catalyzed by hexokinase: ATP + glucose = AG' = -16.7 kJ/mol ADP + glucose-6-phosphate (a) Calculate the equilibrium constant (K) for this reaction at 298 K, and from that, calculate the maximum [glucose-6-phosphate]/[glucose] ratio that would exist under conditions where the reaction is still thermody- namically favorable. (b) The reverse of this interconversion in gluconeogenesis is catalyzed by glucose-6-phosphatase: glucose-6-phosphate + H,0 = glucose + P, AGr = -13.8 kJ/mol K= 262 for this reaction. Calculate the maximum ratio of [glucose]/ [glucose-6-phosphate] that would exist under conditions where the reaction is still thermodynamically favorable. (c) Under what cellular conditions would both directions in the…
Coupled reactions occur where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. This approach is common in biological settings. Determine if ATP could be generated by this biochemical reaction. You have calculated that cell potential is +0.637V. An example of a coupled reaction is the first step of glycolysis, the phosphorylation of glucose to form glucose-6-phosphate shown below. kJ/mol The net AG° for this reaction is 1 2 3 н он H. H- H- H H H H 4 6. H. glucose phosphate anion glucose-6-phosphate AG = +14.0 kJ/mol 7 8 9. АТР ADP phosphate anion AG = -30.5 kJ/mol +/- LO
The first reaction in glycolysis is the phosphorylation of glucose to form glucose 6-phosphate: P₁+ glucose glucose 6-phosphate + H₂O This is a thermodynamically unfavorable reaction, with AG° = +13.8 kJ/mol. a) In a liver cell at 37°C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. What is the equilibrium concentration of glucose 6- phosphate, according to the above reaction? b) This very low concentration of glucose 6-phosphate is unfavorable for glycolysis. In fact, the reaction is coupled to the hydrolysis of ATP: ATP + H₂O2 ADP + P₁ +H* AG°¹ = -32.2 kJ/mol Write the expression for the overall reaction and calculate AG°¹. c) In addition to the glucose and phosphate concentrations listed in part (a) above, assume that the liver cell concentrations of ATP and ADP are 3 mM and 1 mM, respectively. Calculate the theoretical equilibrium concentration of glucose 6-phosphate at pH = 7.4 and 37 °C. C ZOOM +
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
Which of the reactions are spontaneous (favorable)? C6H130,P + ATP → › C6H14º₁₂P2 + ADP AG = -14.2 kJ/mol L-malate + NAD+ → oxaloacetate + NADH + H+ AG = 29.7 kJ/mol glutamate + NAD+ + H₂O → NH‡ + α-ketoglutarate + NADH + H+ AG = 3.7 kcal/mol → CH2O4 + H2O AG = 3.1 kJ/mol * CąHẠO, — CH,O4 + H,O DHAP C₂H + H₂ glyceraldehyde-3-phosphate AG = 3.8 kJ/mol Rh(I) C2H6 AG-150.97 kJ/mol