b) What is the gradient at room temperature? Du = J/molK * 298 K * ln (1/1000) = kj/mol

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1 Chem350 Practice Problems Membranes 1. a) What is the chemical potential generated by the movement of glucose by passive diffusion established by a 1000 fold concentration gradient at physiological temperature? Du = RT ln C2/C1 Du = J/molK * 310 K * ln (1/1000) = kj/mol b) What is the gradient at room temperature? Du = J/molK * 298 K * ln (1/1000) = kj/mol c) Is the gradient at either temperature sufficient to phosphorylate ADP? Phosphorylation of ADP requires approximately kj/mol at 298 K. So this is not sufficient alone d) How large a gradient would be required to produce ATP from ADP? J/mol = J/molK * 298 K * ln (x) x = 1.78 E- 6 = C2/C1 C2 = 1 C1 = 561,725 or more than ½ million (a gradient not particularly practical!) 2. The concentration of chloride ion in serum is about 100 mm. In the kidneys the concentration is about 160 mm. a) Ignoring the electrostatic effects, how much work must be done to pump the chloride ion out of the blood into the kidneys? In this case the work can be estimated by the chemical potential as the heat change is negligible (essentially 0). Du = J/molK * 298 K * ln (160/100) = kj/mol b) How many chloride ions are transported per ATP molecule hydrolyzed? ATP hydrolysis provides J/mol Chloride ion requires 1240 J/mol Since molecule: molecule is the same ratio as mole:mole J/1164 J = 28.2 or 29 molecules c) Including the electrostatic energy recalculate a and b. Assume a voltage gradient of approximately 120 mv for the ion. We must add in the voltage gradient term Du = RT ln C2/C1 + nfdv Du = J/molK * 298 K * ln (160/100) + (- 1)(96500J/V)(0.120V)

2 =- 10,416 J (highly favorable!) - the voltage gradient drives the reaction! ATP is not required to drive this reaction! 3. a) ATP synthase (ATPase) is a critical transport protein in the mitochondrian. ATPase pumps protons. Express the proton motive force of chemiosmotic theory (formula for the chemical potential of proton transport across a membrane) in terms of ph ) Du = RT ln C2/C1 + nfdv ln C2/C1 = ln[h+]i/[h+]0 = ln [H+]i - ln[h+]0 ph = - log H+ relationship between ln and log (2.3) ln [H+]I - ln/[h+]0 = 2.3log [H+]i 2.3 log[h+]0 = 2.3DpH so Du = RT ln C2/C1 + nfdv = 2.3 RT log DpH + nfdv b)in the liver the mitochondrial voltage gradient (inside to out) is about 170 mm. The ph differential is about 0.75 (higher inside than outside). What is the chemical potential for the transport out of the mitochondrial matrix? Du= 2.3 RT log DpH + nfdv = 2.3 * 8.314J/molK * 298 L (- 0.75) + (1*96500 J/Vmol *0.17 V) = 12kJ/mol 4. At 298 K a ph differential of one unit gives a free energy change of kj/mol. a) Show that this is true. Du= 2.3 RT log DpH = J/molK *298K T log DpH log DpH = DpH = ~ 1 (given round off error) b) In a membrane the situation is actually more favorable. In addition to a ph gradient, the actual physiological situation is even more favorable as membrane potential exists whereby the membrane is more negative on the inside relative to the outside. For a membrane potential of mv how much energy would be added to the free energy of transport by the electrical gradient? Du= 2.3 RT log DpH + nfdv For a single charge difference nfdv = 1 (96500 J/Vmol) (- 0.1V) = kj/mol

3 Total = kj/mol (in reality there are three protons transferred so the electrical term is and the total is kj/mol!!!! c) The standard free energy for the synthesis of ATP from ADP and Pi is 32.5 kj/mol. The ratio of ATP to ADP is about 100 physiologically and the concentration of phosphate can be estimated as 10 mm. What is the free energy at physiological temperature for ATP synthesis? ADP + Pi < == > ATP DG = DG0 + RTlnKeq DG = 32500J/mol J/molK (310K)ln (100/10) DG = 32500J/mol J/mol = J/mol = 38.4 kj/mol d) Is the transport energy provided by a and b sufficient to form a mole of ATP from ADP and Pi? Explain It is sufficient when it is realized that 3 protons are moved across the membrane that provides an energy of kj/mol (38.4 kj/mol is needed) 5. Na+/K+ ATPase simultaneously pumps ions and hydrolyzes ATP. The process can be written as the sum of two reactions: 2K + out + 3Na + in < == > 2K + in + 3 Na + out ATP + H2O < == > ADP + Pi 2K + out + 3Na + in + ATP + H2O < == > 2K + in + 3 Na + out + ADP + Pi a) Write the free energy expression for the first step (mathematical equation) DG = RTln{ [Na + out] 3 *[K + in ] 2 /[Na + in ] 3 * [K + out ] 2 } + 3FDV - 2FDV b) If the voltage gradient is 70 mv (unfavorable for Na+ transport, favorable for K+ transport); sodium gradient 10 mm inside to 140 mm outside; potassium gradient 100 mm inside to 5 mm outside, calculate the free energy associated with the ion transport. DG = RTln{ [Na + out] 3 *[K + in ] 2 /[Na + in ] 3 * [K + out ] 2 } + 3FDV - 2FDV DG = J/molK (298)* ln{0.140m 3 *0.100M 2 /0.01M 3 * 0.005M 2 } + 1(96500J/Vmol)(- 0.07V) (math check: ln term = 13.9) DG = 34463J/mol J/mol = 27708J/mol = 27.7 kj/mol (positive as expected for an ATP coupled reaction) c) If the overall free energy of the coupled process is kj/mol calculate the energy associated with ATP hydrolysis. DG = DG ATP + DG Na/K kj/mol = DG ATP

4 DG ATP = kj/mol (a number we have seen before!) d) In which direction does the summary equation above proceed? The net DG is negative so it proceeds to the right. Process: 6. Explain (don t describe, explain) the following figure with respect to the dielectric constant of the medium

5 This figure shows the dependence of the electrostatic energy on the distance between two charges. The energy is inversely proportional to the distance so repulsion is greatest at shortest distances. The energy also relies on the medium (the stuff between the charges). As the dielectric constant of the medium increases, the stuff can cushion the charge as the charged species are not forced to interact solely with each other they can interact with the medium. So to interact with each other the charges must get very close in water. The largest effect of ion- ion distance is seen in a vacuum and the least in water (high dielectric constant). In the interior of the lipid bilayer a hydrophobic neutral area the effect is intermediate between the two but with a shape more parallel to vacuum as the potential dipolar interactions from the bilayer interior are miniscule. Thus charged particles will interact over a longer distance. In a vacuum particles do not need to be very close to interact (there is nothing between them) 7. Mg 2+ ions interact with ATP under physiological conditions. What is the likely effect of this on the free energy of hydrolysis of ATP? Why? That magnesium ions bind to ATP tells us that the bound state has a lower free energy than the unbound state. Assuming that magnesium ions bind neither to ADP nor to Pi, the free energy following cleavage is the same in the presence and absence of ion. Thus, the free energy difference between the ion- bound state and the cleaved state must be lower than that between unbound ATP and the cleaved state. This raises the question of the biological advantage conferred by the binding of metal ions to ATP. 8. The release of insulin from pancreatic beta cells on uptake of glucose is multistep. The membrane potential of a beta cell is determined by open ATP sensitive K + channels in the plasma membrane. After nutrients are ingested, glucose is taken into the cell and phosphorylated. The result is an increase in the ATP to ADP ratio in the cell. This increase causes closure of the K + channels. The membrane then depolarizes stimulating the opening of Ca +2 channels. Calcium enters the cell causing the release of insulin through exocytosis of secretory grannules. Describe each step of this process in thermodynamic terms. An electrical potential is a form of stored energy. In the case of a biological membrane, the electrical potential depends on the asymmetric distribution (unequal concentrations) of ions and small molecules that, at least at certain times, cannot penetrate the membrane. It is essential, though, that the solvent, water, can readily pass through the membrane. For a specific species of solute, an asymmetric distribution across a membrane is a higher free energy state than the concentration being the same on both sides. Because of the natural tendency of differences in concentration to dissipate as a particle moves spontaneously down its concentration gradient, energy is required to maintain concentration asymmetry. A membrane potential depends completely on asymmetry, so energy is required to maintain a membrane potential.

6 In the case of the ATP- sensitive potassium channels, the passage of potassium through this type of membrane channel is regulated by the association of ATP with the channel. By the way, other channels, for example the sodium- potassium pump, are similar: ATP is hydrolyzed in the coordinated transport of sodium in one direction through the membrane and potassium in the other. After a meal, there is plenty of glucose in the blood. It enters cells by moving down its concentration gradient. Phosphorylation within the cell makes the sugar a charged particle, and this strongly inhibits its movement back out of the cell because of the high energetic unfavorability of something charged passing through a region of low dielectric, viz. the membrane. Glycolysis and oxidative phosophorylation convert glucose into ATP, increasing [A TP]/[ADP] in the cell. The increased level of [A TP] means that more A TP molecules are available to interact with ATP- sensitive molecules, particularly the potassium channels. Binding of ATP to the channel, a thermodynamic phenomenon, closes it and prevents passage of potassium ions. Closure of the potassium channels leads to a less sharp difference in concentration of potassium on opposite sides of the membrane and therefore a depolarization of the membrane potential. This, however, stimulates the opening of calcium channels, and calcium enters the cells as it moves down its concentration gradient. A variety of cellular processes are sensitive to changes in cytosolic calcium concentration. One of these is the fusion of insulin- filled secretory granules to the plasma membrane. A possible mechanism for this is that the calcium ions neutralize the charge effect of negatively charged phospholipid head groups, reducing the energy barrier to fusion. Once fusion has occurred, the contents of the granule are released from the cell. Insulin binds in a highly specific manner to the extracellular portion of a transmembrane receptor molecule. Such binding is of course a thermodynamic phenomenon.

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