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George Washington University<ECE 3820<unit_5_ps1_blood_answer_

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PROBLEM SET 5.1: BLOOD ANSW ER KEY 1. The value of L p for the red blood cell is about 1.8 x 10-11 cm3 dyne-1 s-1. Its surface area is about 1.35 x 10-6 cm2 (Solomon, Methods in Enzymology, 192-222 , 1989). A. W hat is the initial osmotic flow if the osmolarity inside is initially 300 mOsM and the osmolarity outside is 275 mOsM? (Assume F = 1.0 for all solutes) The appropriate equation here is QV = A Lp ()P - F)B) In the particular example no information is given about the hydrostatic pressure difference across the mem brane, so we will assume there is none. We are given values for A and for Lp and F and for the osm olarity. So we calculate Q v = 1.35 x 10-6 cm2 x 1.8 x 10-11 cm3 dyne-1 s-1 (RT (0.300 - 0.275)) W e can calculate )B = RT)C as .082 L atm m ol-1 °K-1 x 310°K x 0.025 = .6355 atm W e need to convert this to units of dyne cm-2 in order to use the units of Lp that are given. Fortunately, we already did this last semester: 1 mm Hg = 1334 dyne cm-2. Since 1 atm = 760 m Hg, we have 1 atm = 1.01 x 106 dyne cm-2 so we can multiply this in our equation for Qv: Q v = 1.35 x 10-6 cm2 x 1.8 x 10-11 cm3 dyne-1 s-1 x .6355 atm x 1.01 x 106 dyne cm-2 atm-1 = 1.56 x 10-11 cm3 s-1 = 15.6 x 10-12 cm3 s-1 B. If the volume of the cell is 100 x 10-12 cm3, how long would it take to double its volume provided that the osmotic pressure and area of the membrane and Lp did not change? The cell would double its volume if another 100 x 10-12 cm3 of water cross the mem brane. If the initial flow is maintained, this would occur in 100 x 10-12 cm3 / 15.6 x 10-12 cm3s-1 = 6.41 s The purpose of this calculation is to convince you that these fluid shifts are very fast. C. In the case described, how much water would be required to enter the cell to equilibrate the osmotic pressure between inside and outside? Assume that the outside bath is essential infinite so that its osmotic pressure is kept constant. In this case there has to be enough water entering the cell to dilute its contents from 300 mOsM to 275 mOsM, at which point the flow of water would cease. To calculate this volume of water, we assume that the intracellular contents cannot leave the cell. The total amount of osmolytes is just V0 x C, where C is the concentration of osmolytes and V0 is the initial volume of the cell. After some volume has entered the cell, the new concentration is V0 x C/V, where V = V0 + )V, and )V is the volume of water that enters the cell. This new concentration would be equal to the outside osm olarity. Here we write V 0 x C/ V = Cout5.PS1.2 Plugging in the values, we get 100 x 10-12 cm3 x 300 mOsM / V = 275 mO sM Solving for V we get V = 109.1 x 10-12 cm3 According to this calculation, only 9.1 x 10-12 cm3 of water has to enter the cell to equilibrate the osmotic pressures. If the initial flows are maintained, this would occur in 9.1 x 10-12 cm3/15.6 x 10-12 cm3 s-1 = 0.58 s. 2. A. Assume that water movem ent is fast compared to solute movement when red blood cells are placed in contact with solutions of different tonicity. Derive an eq

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