Substrate cycling does not violate the laws of thermodynamics in making both directions of a reaction favorable. How is this so? Explain in regards to Step 3 of Glycolysis
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- The condensation reaction catalyzed by ß-ketoacyl-ACP synthase synthesizes a four-carbon unit by combining a two-carbon unit and a three-carbon unit, with the release of CO₂. What is the thermodynamic advantage of this process over one that simply combines two two-carbon units? The reaction is reversible and does not require the input of ATP. The exergonic release of CO₂ drives the irreversible reaction in the direction of fatty acid synthesis. The release of CO₂ is endergonic and results in the production of ATP. The reaction is endergonic and requires the input of ATP.Substrate cycling does not violate the laws of thermodynamics in making both directions of a reaction favorable. Explain in regards to the Table with Step 3 of GlycolysisExplain thermodynamics of glycolysis in relation to Step 3 as shown in the Table and how it does not violate thermodynamic laws?
- The reaction catalyzed by citrate syn- thase, shown on the right, is the first step of the TCA cycle. In glycolysis, two key reactions to produce ATP occur because an unfavorable reaction is coupled to another reaction that is thermodynamically favorable. The reaction catalyzed by citrate synthase, shown on the right, is similarly coupled to an unfavorable reac- tion in the TCA cycle. Write the unfavorable reaction using structural formulas and write the key step that drives the two coupled reactions forward. What is the overall AG'o of the coupled reactions? CH3-C >=0 + S-COA Acetyl-CoA 0-C-COO- CH₂-COO Oxaloacetate H₂O COA-SH J citrate synthase CH₂-C HỌ—C—COO SO CH₂-COO Citrate AG'= -32.2 kJ/molOrganisms growing anaerobically cannot perform glycolysis for long without reducing the pyruvate from glycolysis into another compound, most commonly to lactate or to ethanol plus CO2. Both of these reactions are given below in their unbalanced forms. Explain in one sentence why one of these reducing steps is needed to sustain anaerobic glycolysis.Describe the energy transformation steps that occur within an electron transport chain during the process of oxidative phosphorylation. Please include a drawing of the explanation as well.
- Calculate the ATP yield when glucose is catabolized completely to six CO2 by a eukaryotic microbe. How does this value compare to the ATP yield observed for a bacterium? Suppose a bacterium used the Entner-Doudoroff pathway to degrade glucose to pyruvate and then completed the catabolism of glucose via the TCA cycle. How would this affect the total maximum ATP yield? Explain your reasoning.In the microbial community of the bovine rumen, the actual AG value has been calculated for glucose fermentation to acetate: CaH12O6 + 2H20 → 2CH:03 + 2H+ + 4H2 + 2CO; AG =-318 kJ/mol If the actual AG for ATP formation is 44 kJ/mol and each glucose fermentation yields four molecules of ATP, what is the thermodynamic efficiency of energy gain? Where does the lost energy go? Why is the AG not 31 kJ/mol?Indicate at what step in the glycolysis pathway each of the following events occur 1. First phosphorylation of ADP occurs 2. First “energy rich” compound is produced 3. Second “energy rich” compound undergoes reaction 4. First isomerization reaction occurs 5. Second formation of ATP occurs 6. Second “energy rich” compound is produced 7. ATP is converted to ADP for the second time 8. A dehydration reaction occurs
- 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?Draw OUT THE arrow pushing reaction mechanism for the following steps of glycolysis: SHOW/IDENTIFY the arrow pushing mechanism in at least 2 steps. CH,OPO CH,OH hexokinase 2+ H H H H H H ОН + ATP OH H ADP + H+ H ОН + НО HO OH H ОН H ОН Glucose-6-phosphate (G6P) GlucoseThe 9th step of glycolysis involves the conversion of 2PG to PEP. The reason why this step is important is because it helps to make the eventual end product of glycolysis (pyruvate) a very stable molecule. Please answer the following two questions regarding this: 1. WHY is it useful thermodynamically to have the final product of glycolysis be very stable? 2. What specific feature of pyruvate, that we discussed in class, makes it so stable and what did step 9 of glycolysis have to do with this?