Substrate cycling does not violate the laws of thermodynamics in making both directions of a reaction favorable (i,e Step 3 glycolysis). How is this so? Explain
Q: In the microbial community of the bovine rumen, the actual AG value has been calculated for glucose…
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- 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 GlycolysisSubstrate 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 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.The metabolic reactions and enzymes that require NAD/NADH are shown in Figure 1. However, it is not specified whether oxidized or reduced NAD is used in each reaction, nor what form of NAD is produced as a product. Add this specificity to the attached figure. You are also welcome to draw your own figure.1. (a) The reaction catalyzed by citrate synthase 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 is similarly coupled to an unfavorable reaction in the TCA cycle. Write the unfavorable TCA reaction using structural formulas and write the key step that drives the two coupled reactions forward. What is the overall AG" of the coupled reactions? (b). K(yM) 25.7 Inhibitor Bromoacetyl-CoA ATP NADH 6800 8300 The inhibitor constants for three inhibitors of por- cine citrate synthase are summarized in the table on the right. The compounds were all determined to bind in the ac- tive site as competitive inhibitors of acetyl-CoA. Because they bind as competitive inhibitors, all three inhibitors must exhibit structural similarity to some part of acetyl-CoA. Look up in the textbook the structural formulas for…
- 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/molIn 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?A direct measurement of the standard free-energy change associated with the hydrolysis of ATP is technically demanding because the minute amount of ATP remaining at equilibrium is difficult to measure accurately. The value of ΔG′° can be calculated indirectly, however, from theequilibrium constants of two other enzymatic reactions having less favorable equilibrium constants:Using this information for equilibrium constants determined at 25 °C, calculate the standard free energy of hydrolysis of ATP.
- Some reaction components are shown on the left. Match them to the reactions catalyzed by glutamine synthetase and glutaminase in the cell. (Multiple components on the left may match to the same option on the right. Some righthand options may have no corresponding component.) Glutamine ATP NAD+ H₂O Clear All Glutamine synthetase only Glutaminase only Both glutamine synthetase and glutaminase Neither glutamine synthetase nor glutaminaseWithin biological systems, there are always reactions that seem to occur when thermodynamically, they should not. An example is in the process of glycolysis (the conversion of glucose to pyruvate) which has ΔG°' = 2183.6 kJ/mol. How is glycolysis possible with such a large, positive ΔG°', when cells are governed by the laws of thermodynamics?Example: Oxidation of ethanol by NAD+ in the presence catalyzed by alcohol dehydrogenase Calculate the standard free energy change for the reaction below: Ethanol + NAD* 2 acetaldehydye + NADH + H*