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In part C i found that the intracellular concentration f glow is 5.35, PLEASE help with the rest of the questions very confused D. Your diagnostic tools allow you to see glow when its intracellular concentration is at or above 1 uM. When resting, these fish are not visibly luminescent. From your calculation in part C is this reaction at equilibrium? Is the resting intracellular Q less than, equal to, or greater than Keq : Why? E. Studies of the fish show that there is a specific enzyme that is activated during specific stimuli to rapidly switch from a dark to a glowing fish. You discover this enzyme facilitates the spontaneous reaction by coupling the reaction of dark to glow with another reaction,the hydrolysis of ATP in the following simplified reaction: Dark + ATP -> Glow + ADP +P Knowing the Gof the hydrolysis of ATP to be -32.2 kJ/mol calculate the total Gof the coupled reaction. F. Calculate the Kea of this net coupled reaction. G. Using the new Kea of the coupled reaction calculated in part F,repeat part C to calculate how much glow would now be created with 50.0 M and 0 M as the initial concentrations of dark and glow respectively in the cell. (Assume that the intracellular amounts of ATP,ADP,and Piremain unchanged from this reaction and their relative ratios in the equation for Keg are close to one and do not affect the calculation.Would this be enough to detect using your diagnostic tools?

          In part C i found that the intracellular concentration f glow is 5.35, PLEASE help with the rest of the questions very confused
D. Your diagnostic tools allow you to see glow when its intracellular concentration is at or above 1 uM. When resting, these fish are not visibly luminescent. From your calculation in part C is this reaction at equilibrium? Is the resting intracellular Q less than, equal to, or greater than Keq : Why?
E. Studies of the fish show that there is a specific enzyme that is activated during specific stimuli to rapidly switch from a dark to a glowing fish. You discover this enzyme facilitates the spontaneous reaction by coupling the reaction of dark to glow with another reaction,the hydrolysis of ATP in the following simplified reaction:
Dark + ATP -> Glow + ADP +P
Knowing the Gof the hydrolysis of ATP to be -32.2 kJ/mol calculate the total Gof the coupled reaction.
F. Calculate the Kea of this net coupled reaction.
G. Using the new Kea of the coupled reaction calculated in part F,repeat part C to calculate how much glow would now be created with 50.0 M and 0 M as the initial concentrations of dark and glow respectively in the cell. (Assume that the intracellular amounts of ATP,ADP,and Piremain unchanged from this reaction and their relative ratios in the equation for Keg are close to one and do not affect the calculation.Would this be enough to detect using your diagnostic tools?

in part c i found that the intracellular concentration f glow is 535 please help with the rest of the questions very confused d your diagnostic tools allow you to see glow when its intracell 65687

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