Tuesday, August 6, 2019
Thermo Answers Essay Example for Free
Thermo Answers Essay Answer 8: I am not getting answer correct as the electron donor used by Alcohol                  Dehydrogenase is NADH and FAD is not evolved in it. Answer 9: FAD is used in neutralization of free radicals as it has higher oxidation                   potential than other and can easily reduced by free radicals. Answer 10: The answer will be when ∆H=0 as based on equations                     ∆G °=∆H- T∆S and ∆G °= -RT In Keq the value of Keq remains greater                     than 1 when value of ∆G ° remains negative and that is only possible in two                    Condition 1) when ∆H remains 0 or negative. And 2) ∆H remains negative                    along with ∆S becomes 0. Answer 11: Here rest of the answers is according to biologic standard for example H+  Concentration as all biological processes occurred at pH 7, the standard was taken as [H+] = 10-7 similarly biological processes conceder to be worked at atmospheric pressure 1atm. As per the standard convention concentration of each reactant was taken as unity or 1M. Since all three answer were part of the biological standard condition only answer d remains choice as the biological processes occurs at temperature 37 C (our body temperature) and not at 0 C. Answer 12:         Here among all four compound only glucose -6- phosphate will not give rise to any energy but converted to fructose 6 phosphate in glycolysis, While 1,3-bisphosphoglycerate converted to 3-phospho glycerate and give rise to 2ATP, similarly phosphoenol pyruvate converted to pyruvate and gives 2ATP (both in process of glycolysis). Phosphocreatine act as energy storage in skeletal muscles where there is fluctuation in energy requirement, here in high energy requirement phosphocreatine converted to creatine by converting ADP to ATP. Answer 13: here answer is e, because if you see the reactions for the formation of Glucose-6 –phosphate 1 ATP has to be hydrolyzed in two step 1) first ATP gets converted to AMP and PPi by releasing energy equivalent to 45.6 kCal and as this reaction give rise to energy (liberation of energy) the value is indicated as negative (-45.6). In second step PPi again get hydrolyzed to Pi with energy release equivalent to 19.6 (i.e -19.6).      Now glucose converted to Glucose-6- phosphate by utilizing energy released from above mention reaction and it requires 13.8 Kcal, here the reaction requires energy and that’s why value is positive (i.e 13.8). In conclusion the energy balance sheet for formation of glucose-6-phosphate from Glucose will be -45.6-19.6+13.8 = -51.6 (favorable forward reaction as ∆G ° is negative) Answer 14: here the answer is e, The free energy of ATP hydrolysis is depend on 4 parameters 1) ratio of ATP/ADP (higher the ration lesser the hydrolysis) 2) pi concentration as being end product accumulation of it leads to more hydrolysis 3) H+ being slightly Acidic hydrolysis of ATP is depend o n pH (H+), and 4) concentration of Mg which act as cofactor for enzyme adenylate kinase which plays important role in ATP synthesis as well as ATP hydrolysis. Answer 15: As the phosphate group transfer from ATP to other Nucleotide is catalyzed by enzyme Nucleoside diphosphate kinase, and this is a reversible reaction the dynamics of reaction almost remains in steady state. For example in case of higher concentration of ATP the reaction goes in forward direction by generating NTP and ADP, But once ATP crises arises ADP converted back to ATP by the action of adenylate kinase. In conclusion the ratio of two nucleotide ATP and ADP+ NTP remains constant and hence Keq remains around 1. Answer 22: here the total out put or free energy is calculated based redox potential of electron except – electron donor and hence for answer d it will be highest  ie     ∆E °Ã¢â‚¬â„¢= -0.219- (.320) = 0.529 V Answer 23: In all these above mention reactions conversion of PEP to pyruvate have highest free energy changes of -61.9 compared to other reactions for example ATP hydrolysis gives rise to -30.5, and 45.6 similarly glucose 6 phosphate gives -13.8 Kj/mol. This high energy is due to direct transfer of Phosphate group from PEP to ADP.
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