One of the four reactions of the ß-oxidation cycle for a fatty acid is shown below. What are the missing co-substrate(s) and co-product(s) in this reaction? OH || CH3-(CH₂)6 CH -CH₂-C-S-CoA В Co-substrate(s)? 3-Hydroxyacyl CoA dehydrogenase C Co-product(s)? || CH3 (CH₂)6-C-CH₂-C-S-CoA B α • Co-substrate(s): HS-COA • Co-product(s): H₂O • Co-substrate(s): FAD •Co-product(s): FADH2 • Co-substrate(s): FADH2 Co-product(s): FAD ● • Co-substrate(s): NADH + H¹ Co-product(s): NAD+ • Co-substrate(s): NAD+ • Co-product(s): NADH + H*

One of the four reactions of the ß-oxidation cycle for a fatty acid is shown below. What are the missing co-substrate(s) and co-product(s) in this reaction? OH || CH3-(CH₂)6 CH -CH₂-C-S-CoA В Co-substrate(s)? 3-Hydroxyacyl CoA dehydrogenase C Co-product(s)? || CH3 (CH₂)6-C-CH₂-C-S-CoA B α • Co-substrate(s): HS-COA • Co-product(s): H₂O • Co-substrate(s): FAD •Co-product(s): FADH2 • Co-substrate(s): FADH2 Co-product(s): FAD ● • Co-substrate(s): NADH + H¹ Co-product(s): NAD+ • Co-substrate(s): NAD+ • Co-product(s): NADH + H*

Introduction to General, Organic and Biochemistry

11th Edition

ISBN:9781285869759

Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Chapter23: Enzymes

Section: Chapter Questions

Problem 23.35P

See similar textbooks

Similar questions

In one of the steps in this pathway fructose 1,6-biphosphate (F1,6BP) is converted to gylceraldehyde-3-phosphate (G-3-p) and dihydroxyacetone phosphate (DHAP). This reaction is catalyzed by the enzyme aldolase. For this reaction at 25 celcuius and Ph7 we have: Keq= 10-4M and Delta G = +5456 cal/mol Calculate the following: The concentration of F1, 6BP, DHAP and G-3-P at equilibrium when the initial F1,6BP is (A) 1M, (b) 10-2 M, (c) 2 X 10*4 M and (d) 10-5 M. I been trying to solve this problem but I dont even know where to begin with. Atleast help me with problem A, I would be able to guide myself through.
Which of the reactions are spontaneous (favorable)? 1)DHAP↽−−⇀glyceraldehyde-3-phosphateΔ?=3.8 kJ/mol 2)L-malate+NAD+⟶oxaloacetate+NADH+H+Δ?=29.7 kJ/mol- 3) glutamate+NAD++H2O⟶NH+4+α-ketoglutarate+NADH+H+Δ?=3.7 kcal/ 4) C6H13O9P+ATP⟶C6H14O12P2+ADPΔ?=−14.2KJ/mol 5) −Rh(I)C2H6Δ?=−150.97 kJ/molC2H4+H2→Rh(I)C2H6ΔG=−150.97 kJ/mol 6) C4H4O5⟶C4H2O4+H2OΔ?=3.1 kJ/mol
TPP is a coenzyme for transketolase, the enzyme that catalyzes the conversion of a ketopentose (xylulose5-phosphate) and an aldopentose (ribose-5-phosphate) to an aldotriose (glyceraldehyde-3-phosphate) and a ketoheptose (sedoheptulose-7-phosphate). Notice that the total number of carbons in the reactants and products is the same (5 + 5 = 3 + 7). Propose a mechanism for this reaction.
One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of xylulose 5-phosphate with ribose 5-phosphate in the presence of a transketolase to give glyceraldehyde 3-phosphate and sedoheptulose 7-phosphate. (a) The first part of the reaction is nucleophilic addition of thiamin diphosphate (TPP) ylide to xylulose 5-phosphate, followed by a retro-aldol cleavage to give glyceraldehyde 3-phosphate and a TPPcontaining enamine. Show the structure of the enamine and the mechanism by which it is formed. (b) The second part of the reaction is addition of the enamine to ribose 5-phosphate followed by loss of TPP ylide to give sedoheptulose 7-phosphate. Show the mechanism.
The first step of the metabolic process known as glycolysis is the conversion of glucose to glucose- 6-phosphate. This process has a positive value for rG' Glucose + Pi Glucose-6-phosphate + H2O rG' = +13.8 kJ/mol-rxn This reaction is coupled to the hydrolysis of ATP ATP + H2O ADP + Pi rG' = -30.5 kJ/mol-rxn What is the sum of these two equations and the value of rG' for the coupled reaction? Is the coupled reaction product-favored at equilibrium?
Write down the reactions: D-Glucopyranose + (CH3CO)2 O →
From Table 1.1 indicates, among other things, the melting points of various fatty acids. Analyze the trend that the melting points of saturated fatty acids follow, from n-dodecanoic (CH3-(CH2) 10-COOH) to n-tetracosanoic (CH3-(CH2)22-COOH), and answer both next questions. Briefly justify your answers. i) What value would you expect the melting point of triacontanoic acid (CH3-(CH2) 28-COOH) to have? Choose one of the following options:a) More than 200 ° Cb) About 94 ° Cc) Less than 86 ° Cd) Exactly 86.1 ° C ii) What value would you expect the melting point of octanoic acid (CH3-(CH2) 6-COOH) to have? Choose one of the following options:a) 43.5 ° Cb) -20 ° Cc) About 20 ° C
Iis D-glucose capable of forming mutarotation in solid state?
Determine the branch points and reducing ends of amylopectin. A 0.5g sample of amylopectin was analyzed to determine the fraction of the total glucose residues that are branch points in the structure. The sample was methylated and then digested, yielding 50 μmol of 2,3-methylglucose and 0.4 μmol of 1,2,3,6-tetramethyl glucose. (MWglucose = 162 g/mol, minus H2O) What fraction of the total residues are branch points? How many reducing ends does the amylopectin have?
The standard free energy change (ΔG°') for hydrolysis of fructose-l,6-bisphosphate (FBP) to fructose-6-phosphate (F-6-P) and Pi is -16.7 kJ/mol: FBP + H2O → Fructose-6-P + Pi The standard free energy change (ΔG°') for ATP hydrolysis is -30.5 kJ/mol: ATP + H2O → ADP + Pi a What is the standard free energy change for the phosphofructokinase reaction: ATP + Fructose-6-phosphate → ADP + FBP ΔG°' = -13.8 kJ/mol b What is the equilibrium constant for this reaction at 25.0 °C? K'eq =
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?
Suppose that you made some wine whose alcohol content was 10% w/v (i.e., 10 g of ethanol per 100 mL of wine). The initial fermentation mixture would have had to contain what molar concentration of glucose or its equivalent to generate this much ethanol? Is it likely that an initial fermentation mixture would contain that much glucose? In what other forms might the fermentable carbon appear?