Triose phosphate isomeraseIn a key reaction of glycolysis, dihydroxyacetone phosphate (DHAP) is isomerized intoglyceraldehyde 3-phosphate (G3P) by the action of the enzyme triose phosphate isomerase:CH₂OHC=OCH₂OPO3²-DHAPCHOHC-OHCH₂OPO3²-G3PAG°¹ +7.5 kJ/molBecause AG is positive, the equilibrium lies to the left.(a) Calculate the equilibrium constant for this reaction, assuming a temperature of 37 °C.(b) In the cell, depletion of G3P makes the reaction proceed. What is the value of AG if theconcentration of G3P is kept at 1/100 of the value of the concentration of DHAP?CZOOM+

Given Values: ∆G°=7.5 kJ/mol or 7500 J/molT = 37°C or 310 KR = 8.314

Answered: Triose phosphate isomerase In a key…

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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In the hexokinase reaction (in glycolysis) Keq caclualtion, what is the OBVIOUS problem with getting a Keq of 678? what should keq be normally? how can you EXPLAIN a Keq of 678?
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Please provide an explanation for how to work through this practice problem: Using table 12.1, calculate the free energy change for the synthesis of ATP from cAMP and inorganic phosphate. (Note: cAMP is hydolyzed to AMP, and the free engery of hydrolysis for ATP and ADP is approximately equal.) Table 12.1: (Compound: Change in Free Energy in kJ/mol) cAMP: -50.4 Creatine phosphate: -43.3 ATP: -30.5 Glucose 6-phosphate: -13.9 AMP: -9.2
Triosephosphate isomerase (TIM) catalyzes the conversion of dihydroxyacetone phosphate to glyceraldehyde-3-phosphate. The enzyme’s catalytic groups are Glu 165 and His 95. In the first step of the reaction, these catalytic groups function as a general-base and a general-acid catalyst, respectively. Propose a mechanism for the reaction.
24. Hexokinase catalyzes the phosphorylation of glucose from ATP, yielding glucose-6-P and ADP. The standard-state free energy change for hydrolysis of glucose-6-P is -13.9 kJ/mol. Calculate the standard-state free energy change and equilibrium constant for the hexokinase reaction. AG" for ATP hydrolysis -31.5 kJ/mol
[AktivGrid] Draw the product of the reaction of isocitrate catalyzed by isocitrate dehydrogenase in the TCA (citric acid) cycle. Provide the structure in the protonation state found in physiological conditions. 9 H-C-OH 800- -H CH₂ ° Coo of NAD+ dehydrogen isocitrate ase NADH, H*, Drawing CO2
C Phosphoglycerate enolase coo O H-C-O-P=O CH₂OH 2-Phosphoglycerate This is a (an) d Tyrosine hydroxylase HO coo NH3 L-Tyrosine This is a (an) 0₂ coo C-O-P-C || CH₂ Phosphoenolpyruvate H₂O + HO HO L-DOPA H₂O coo NH3
Given the tri-acylglyceride (TAG) below as source of energy when administered from an organism's diet, H,C-O HC-0 H,C-o [A] Show the calculated maximum amount of ATP that could be generated by the full oxidation of the compound. Illustrate where the ATPS are derived from (show the reactions and enzymes involved). You can use general structure of the FFA in the reactions involved in the process. (B) Discuss the fate of the glycerol. (C] One of the FFAS derived from the TAG has the same carbon number as glucose. Compare the maximum yield of ATP derivable from this FFA as compared to the ATP yield from glucose.
Chymotrypsin has the highest affinity for which of the following substrates: Table. The values of KM and kcat for some Enzymes and Substrates Enzyme Chymotrypsin Ки (М) 4.4 x 10-1 8.8 x 10-2 6.6 x 104 Kcat (S-1) 5.1 x 10-2 1.7 x 10-1 1.9 x 102 Substrate N-acetylglycine ethyl ester N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester Catalase H2O2 2.5 x 10-2 1.0 x 107 Urease Urea 2.5 x 10-2 4.0 x 105 OA. N-acetylglycine ethyl ester OB. N-acetylvaline ethyl ester OC. N-acetyltyrosine ethyl ester D. Urea

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