Briefly describe two strategies used in metabolism so that a thermodynamically unfavorable (a reaction with a +AG") reaction can be made thermodynamically possible.
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- Thermodynamically unfavourable reactions can be “made favourable” if they are "coupled" to strongly exergonic reactions. Some phosphate compounds with a strong tendency to transfer their phosphate group are frequently used in this context. The free energies of hydrolysis of these compounds are relatively very negative (see table below); they are sometimes called "high energy compounds ". The table below gives the standard Gibbs (free) energy of hydrolysis of some phosphorylated compounds. Compounds Phosphoenolpyruvate Carbamyl-phosphate Acetyl-phosphate Creatine-phosphate Pyrophosphate ATP Glucose-1-phosphate Glucose-6-phosphate Glycerol-3-phosphate ATP + créatine A,Gº' of hydrolysis (kJ.mol´¹) - 61.7 - 51.4 -43.0 - 43.0 - 33.4 - 30.4 - 20.8 - 13.8 The group transfer potential is the Gibbs (free) energy of reaction when a group is transferred from a donor compound to a water molecule (i.e. when the group is detached from the donor compound by hydrolysis). 1. Which phosphorylated…Within 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?Although graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction, sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained:
- The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min-1 when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme in µmol⋅min−1?Which of the following statements are true for BOTH the "transition state" and an "intermediate" of reaction? (This is a multi-select question, select all that apply.). Both are only observed in enzyme-catalyzed reactions. Both can be converted to product(s) or might decompose back to the reactant(s). Neither are part of the "net equation" for the reaction. Both contain covalent bonds are in the process of breaking and/or forming. Both are part of every chemical reaction. (i.e. the mechanisms of all chemical reactions, whether enzyme catalyzed or not, will have involve both a transition state and an intermediate).
- The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V for an enzyme-catalyzed, single-substrate reaction E + SES →E + P. The model can be more readily understood when comparing three conditions: [S] > Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity V, where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] > Km Reaction rate is independent of [S]. Not true for any of these conditions The rate is half of the maximum rate.True or False In the presence of enzymes, the value of free energy of activiation (delta G°‡) for the reaction is more likely to become negative.Given the assumption used to derive the Michaelis-Menten equation for enzyme kinetics that the back reaction from P + E back to ES does not occur, why is it important in measuring experimental data for Michaelis-Menten analysis to determine the initial rate of the reaction just after adding substrate to the sample being measured? O Because faster rates can be measured more accurately than slower rates later in the process. Because the Michaelis-Menten equation also contains the position of the equilibrium of the reaction. ● By using this approach, the amount of P at the start of the experiment is extremely low, causing the back reaction to be essentially absent. Because in this way the [ES] concentration has not yet reached steady state.
- The following questions deal with a fundamental understanding of enzyme catalysis.a. Why is the rate of an enzyme-catalyzed reaction proportional to the amount of (ES) complex?b. What do you think is meant by saturation of the enzyme?c. What do you think is meant by the term “saturation kinetics”?d. How does the Michaelis-Menten equation explain why the rate of an enzyme-catalyzed reaction reachesa maximum value at high [S]?Why the endergonic reactions are thermodynamically unfavorable?This is a general question regarding kinetics and inhibition. In the space below, provide a clearly labeled plot, on one graph (no graph paper needed here), what you would expect for an uninhibited, competitively inhibited and non-competitively inhibited reactions. Points will depend on how clearly you label everything and indicate that you know what you are talking about.