prob_set9_solns

prob_set9_solns - prob_set9 Page 1 of 8 SOLUTIONS TO...

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SOLUTIONS TO PROBLEM SET #9 1. E.g., [inulin]urine X volumeurine = [inulin]plasma X volumeplasma 127 mM x 2 ml urine/min = 20 mM x X ml plasma filtered/min Plasma filtered = GFR = (127 x 2)/20 = 12.7 ml/min Notice that you must be careful of the time units. The urine samples were taken every 10 minutes, so the rate of urine production per minute needs to be calculated. a. The calculated values ranged from 10.6 ml/min to 12.7 ml/min with an average value of 11.7 ml/min. Because there was no obvious systematic change in the values with time, the average value is probably the best number to represent the glomerular filtration rate. b. The first measurement was taken only 10 minutes after infusion began, a time when the inulin would still be equilibrating in the plasma and interstitial fluid compartments, so unusually high or unusually low numbers probably represent inadequate mixing at this time. Furthermore, the tubular fluid within the nephrons at the time when the infusion began would not contain any inulin, nor would much inulin be filtered into the nephrons for several minutes as mixing occurred. c. This GFR is about 1/10 of the normal expected GFR of about 125 ml/minute filtered. This person appears to have some kind of problem. d. The tubular maximum for glucose is about 375 mg/min of glucose. In other words, the proximal tubules can reabsorb only 375 mg of glucose/min. If more glucose enters the tubules per minutes, the glucose will be excreted in the urine. How much glucose is entering the proximal tubules in this person? The molecular weight of glucose is 180, so 425mMolar glucose would have 0.425 moles/liter X 180 grams/mole = 76.5 grams of glucose/liter of blood (or 76.5 mg/ml). This person is filtering 12.7 ml of plasma per minute, or 25.4 ml of whole blood per minute (assuming a hematocrit of 50%). 25.4 ml of whole blood would contain 76.5 mg glucose/ml X 25.4 ml = 1943 mg of glucose entering the proximal tubules per minute. This value is very much above the tubular maximum and you would, indeed, expect to find glucose in the urine. e. Some glucose is reabsorbed in the proximal tubules, so the clearance of glucose would be lower than the clearance of inulin. (Please be certain that this statement makes sense to you.) f. PAH is very nearly entirely removed from the blood entering the renal artery in one pass through the kidney, because it is both filtered into Bowman's capsule and efficiently secreted into the proximal tubules. Therefore, the clearance of PAH very closely reflects renal plasma flow. Page 1 of 8 prob_set9 5/28/2009 http://biology.ucsd.edu/classes/bipn100.SP09/prob_set9_solns.htm
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Renal plasma flow=clearance of PAH = {[PAH]urine X Vurine}/ [PAH]plasma = {(375 mMoles/liter) X 2 ml/min} / (15 mMoles/liter) = 50 ml plasma/min, Correcting for hematocrit, this is approximately 100 ml blood/min. Normally the kidneys receive 20-25% of the cardiac output, and the heart pumps about 5 liters per minute. So under normal conditions you would expect the kidneys to receive about 1 to 1.25 liters of blood per minute. The value that you just calculated is about one tenth of that. It looks as though there is some blockage in the flow of blood to the kidneys. (You probably would want to check the real cardiac
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