Use the following table for all problems: Intracellular and Extracellular lon Co

Question

Use the following table for all problems: Intracellular and Extracellular lon Concentrations for a Typical Mammalian Cell Ion Coutside, mM Cinside, mM Na 145 10 140 Mg? 0.5 Ca I1 10H ICI 110 10 Transmembrane potential Apa -60 mV,T 310 K 1. Using the table above, calculate the Gibbs free energy for transfer of 2 moles of Mg2 from the cell to extracellular environment. Consider the transmembrane potential of -60 mV. (5) 2. Calculate the Gibbs free energy for transporting 1moleof Cl-into the cell. (5) 3. Using the Nernst equation, calculate the equilibrium transmembrane protential for a) Meg ,and b) Cl ions. Based on your data and the actual value of Ap, make a conclusion regarding the state of the ion channels for each of these ions (open or closed). (15) 4. Calculate the Gibbs free energy for transfer of 1 mole of K out of the cell. Describe possible mechanism (via molar ratio of ions) of coupling the transport of K'and Mg into the cell, assuming that the maximum coupling efficiency of 100% (you may use your calculations from Problem #1 for Mg (10) 5. In the "box in the sea" model, assume 3 cal of at was transferred from the box to the sea at 310 K. What is the entropy change of the sea (in e.u.)? (5) 6. During generation of action potential a) Chloride channels are responsible for depolarization of the membrane b) Potassium channels open first, then sodium channels c) Sodium channels open first to let Na out; later potassium channels open to let K in d) Sodium channels open first to let Na nto the cell later potassium channels open to let K* out; e) The amplitude of the action potential depends on the length of the axon. (5) 7. Non-competitive inhibitor of an enzyme will affect a) both the slope of the Lineweaver-Burk plot, and the y-intercept; b) both the slope of the Lineweaver-Burk plot, and the x-intercept;

Explanation / Answer

1. G = (z)(F)(Vm)

where, z = the charge on the ion (+2 Mg+ in this case)

F = 23,062 = the calories released as one mole of charge moves down a voltage gradient of 1 volt (1000 mV)

Vm = the membrane potential, 60 mV

G =2X23062X(-.06)

=-2767.44 cal/mole= -2.7 kcal/mole

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