The isoelectric point, pI, of the protein horse liver alcohol dehydrogenase is 6.8 , while that of aldolase is 6.1 . What is the net charge of horse liver alcohol dehydrogenase at pH 5.1 ?| What is the net charge of aldolase at pH 6.7 ? The isoelectric point of serine is 5.68 ; valine , 5.96 . During paper electrophoresis at pH 7.3 , toward which electrode does serine migrate? During paper electrophoresis at pH 6.5 , toward which electrode does valine migrate?

The isoelectric point, pI, of the protein horse liver alcohol dehydrogenase is 6.8 , while that of aldolase is 6.1 . What is the net charge of horse liver alcohol dehydrogenase at pH 5.1 ?| What is the net charge of aldolase at pH 6.7 ? The isoelectric point of serine is 5.68 ; valine , 5.96 . During paper electrophoresis at pH 7.3 , toward which electrode does serine migrate? During paper electrophoresis at pH 6.5 , toward which electrode does valine migrate?

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter29: The Organic Chemistry Of Metabolic Pathways

Section29.SE: Something Extra

Problem 23MP: One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of...

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3) The enzyme aldolase catalyzes the conversion of fructose-1,6-diphosphate (FDP)to dihydroxyacetone phosphate (DHAP) and glyceraldehide-3-phosphate (G3P). Thereaction isFDP ⇄ DHAP + G3P , ∆rG0(298.15 K) = 23.8 kJ mol−1In red blood cells the concentration of these species are [FDP] = 35 µM, [DHAP] =130 µM, and [G3P] = 15 µM. (Remember that 1.0 µM = 1.0 × 10−6 mol L−1. Thestandard state for reactions in solution can be taken as c0 = 1.0 mol L−1).Calculate ∆rG in a red blood cell at 25 0C. Will the reaction occur spontaneouslyin the cell at this temperature?
Assume you are studying a hydrolase that catalyzes the following reaction which has a ΔGO of + 27 kJ/mol. A-B + H2O ↔ A-H + B-O− + H+ a. What is Keq for this reaction? b. What are Keq’ and ΔGO’ for this reaction? c. Is this reaction favorable under standard biochemical conditions at pH 7? Justify your answer.
One step in glycolysis is the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate. The direct reaction of fructose-6-phosphate with inorganic phosphate fructose-6-phosphate + Pi ⇌ fructose-1,6-bisphosphate + H₂O is very unfavorable, with ∆G'°= 16.3 kJ/mol and K'eq = 1.39×10⁻³. However, in glycolysis the reaction is coupled to ATP hydrolysis: fructose-6-phosphate + ATP ⇌ fructose-1,6-phosphate + ADP and the reaction if favorable. What is the equilibrium [fructose-1,6-bisphosphate]/[fructose-6-phosphate] ratio in the coupled reaction when [ATP] = [ADP]? For ATP hydrolysis ∆G'°= -30.5 kJ/mol at 25°C.
Draw the product that valine forms when it reacts with CH3COCl, pyridine; then H2O. You do not have to consider stereochemistry. Do not draw organic or inorganic by-products. Draw the product in neutral form unless conditions are clearly designed to give an ionic product. Include cationic counter-ions, e.g., Na+ in your answer, but draw them in their own sketcher. Do not include anionic counter-ions, e.g., I-, in your answer.
How would you find the initial [SCN-] and the Equilibrium [FeSCN2+]?
Polyunsaturated fatty acids (PUFA's) in oils 0.g. sunflower oil are prone to rancidity because they contain conjugated double bond systems which are easily oxidized in the presence of light or peroxides. BHT Is used as an antioxidant in oils to prevent oxidative rancidity. How do BHT prevent oils from reacting with peroxides so that they do not become rancid? Show the mechanism.
The overall adenylate kinase reaction (2 ADP = AMP + ATP) is a combination of 3 reactions: ATP hydrolysis, pyrophosphate hydrolysis, and ATP formation. Show each step and provide the standard free energy (ΔG°′) for each and the overall reaction. If the adenylate kinase reaction is at equilibrium and intracellular [ATP] = 5 mM and [ADP] = 0.5 mM, calculate the concentration of AMP at pH 7 and 25 °C. Show all work. ATP (ADP+P) = -30.5 kj/mol; ATP (AMP+PP) = -45.6 kj/mol; PP = -19.2 kj/mol
Aldolase catalyzes the glycolytic reaction The standard free-energy change for this reaction in the direction written is +23.8 kJ/mol. The concentrations of the three intermediates in the hepatocyte of a mammal are: fructose 1,6-bisphosphate, 1.4 × 10−5 M; glyceraldehyde 3-phosphate, 3 × 10−6 M; and dihydroxyacetone phosphate, 1.6 ×10−5 M. At body temperature (37 °C), what is the actual free-energy change for the reaction?
How can the Michaelis constant (Km) be identified and used to represent the affinity between the enzyme and the substrate?
7please help explain in detail the spec
An enzyme reaction takes place in a 10mL solution that has a total citrate concentration of 120mM and an initial pH of 7.00. During the reaction, 0.2mmoles of acid are produced. (pKa values for citric acid: pKa1=3.15, pKa2=4.77, pKa3=6.40) a) At pH 7.00 (before the enzyme reaction occurs), determine the concentrations of the predominant forms of citrate in solution. b) calculate the pH of the solution after the enzymatic reaction has occurred.
Consider the hydrolysis reactions of creatine phosphate and adenosine-5-monophosphate. Creatine Phosphate + H2O Creatine + HPi rG = 43.3 KJ/mol-rxn Adenosine-5-Honophosphate + H2O Adenosine + HPi rG = 9.2 KJ/mol-rxn Which of the following combinations produces a reaction that is product-favored at equilibrium: for creatine phosphate to transfer phosphate to adenosine or for adenosine-5-monophosphate to transfer phosphate to creatine?