H H O H H H но C C C C C C он H. O H Molecule A Molecule B In the reaction shown above, is Molecule A gaining or losing electrons? Is Molecule B gaining or losing energy?
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- For the reaction B A at standard conditions with [B] =1 M and [A] =1 M, AG is initially a large negative number. As the reaction proceeds, [B] decreases and [A] increases until the system reaches equilibrium. How do the values of AG and AG" change as the reaction moves toward equilibrium? both AG and AG stay the same O AG reaches zero and AG" becomes more negative AG becomes less negative and AG" stays the same AG stays the same and AG becomes less negative AG becomes positive and AG" becomes positiveFor the following reaction[FAD] + 2 cyt c (Fe 2+) + 2 H+ —> [FADH2] + 2 cyt c (Fe+3)determine which of the redox couples is the electron acceptor and which is the electron donor under standard-state conditions, calculatethe value of ΔE and determine the free energy change for the reaction.NH2 fofota HO-P- ÓH ОН ОН Using the graphic above, answer the following questions. Is removing phosphates from this molecule considered oxidation or reduction? Is removing phosphates from this molecule considered an exogenous or endogenous reaction?
- Calculate ΔG°′ for the reaction A + B ⇌ C + D at 25°C when the equilibrium concentrations are [A] = 10 μM, [B] = 15 μM, [C] = 3 μM, and [D] = 5 μM. Is the reaction exergonic or endergonic under standard conditions?Inside a body, this reaction has a value of positive ∆G C6H12O6 + H3PO4 → C6H14O12P2 + H2O ∆G= +13.4 kJ/mol Can you show me how to keep this reaction going in living things? Prove it with calculationsНн O HO OH HO но, A OH O reducing nonreducing Other: CH2OH CH2OH H H он ОН H НО CH2OH OH H OH OH H O reducing O nonreducing Other:
- Consider the following equilibrium at 25ºC :Glucose-1-Phosphate Glucose-6-PhophateUsing the equilibrium concentrations of [Glucose-1-Phosphate] = 0.35 M and [Glucose-6-Phosphate] = 1.65 M, calculate BOTH K′eqand Gº′ for this reaction. Is this reaction exergonicor endergonic? R = 8.314 J/K·molA biochemical reaction transfers 60 kJ mol-1(15 kcal mol-1) of energy. What general process most likely wouldbe involved in this transfer? What cofactor (or cosubstrate) likelywould be used? Which cofactor probably would not be used?Carbonic anhydrase catalyzes the hydration of CO. CO2 + H2O ¬ H½CO3 The Km of this enzyme for CO, is 1.20×104 µ.M. When [CO,] = 3.60×104 µM, the rate of reaction was 4.50 umol·mL! sec-1 a What is Vmax for this enzyme? umol·mL-!sec-!
- Carboxylic acids, such as the one shown below, can be difficult to reduce Provide the name of a reductant A. Explain why this reductant is able to react with the carboxylic acid whereas others cannot (Hint: Draw and discuss the potential reactive intermediates/transition states).His388 Glu357 His388 Glu357 Ring opening Proton HO HO abstraction HO но- G6P He NH- NH His388 His388 His388 Glu357 Glu357 Glu357 HO но HO но- но OH cis-enediol F6P Ring closure intermediate OH Describe the mechanism shown above for phosphoglucose isomerase. Describe the chemistry of each step • How the enzyme appears or might facilitate the chemistry How the enzyme increases the reaction rate.ver is given. A +B C+ D 1) Calculate AG for the above react un and indicáte whether the reaction is favorable or unfavorable [A] = 0.9 M 20°C AG° = 4 KJ/mol %3D [B] = 15mM [C] = 4mM [D] = 3 M R= 8.314 J/moleK %3D %3D 4.7 and explain how you know.
