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ps10_solutions

ps10_solutions - Harvard-MIT Division of Health Sciences...

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Harvard-MIT Division of Health Sciences and Technology HST.542J: Quantitative Physiology: Organ Transport Systems Instructors: Roger Mark and Jose Venegas MASSACHUSETTS INSTITUTE OF TECHNOLOGY Departments of Electrical Engineering, Mechanical Engineering, and the Harvard-MIT Division of Health Sciences and Technology 6.022J/2.792J/BEH.371J/HST542J: Quantitative Physiology: Organ Transport Systems PROBLEM SET 10 SOLUTIONS May 6, 2004
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Problem 1 Inulin and para-amino hippuric acid are infused into the blood stream of a 40 kg ape. The fol- lowing measurements were made during the course of the infusion, and were noted to be constant throughout the experiment: arterial plasma concentration of inulin 0.15 mg/ml, of para-amino hippuric acid 0.2 mg/ml; renal venous plasma concentration of para-amino hippuric acid 0.0015 mg/ml; urine flow 2 ml/min; urinary concentration of inulin 9.0 mg/ml, of para-amino hippuric acid 50 mg/ml. [ IN ] p = [ PAH ] p = [ PAH ] v = U ˙ = [ IN ] u = [ PAH ] u = A. What is the glomerular filtration rate? 0 . 15 mg/ml . 2 mg/ml . 0015 mg/ml 2 ml/min 9 . 0 mg/ml 50 mg/ml GFR inulin clearance = C L ( IN ) = U ˙ [ IN ] u = [ IN ] p ( 2 )( 9 ) 120 ml/min = . 15 = B. What is T m (transport maximum) for PAH secretion? In this problem, some PAH is still found in venous blood from the kidney. This implies two important facts: The tubular secretion mechanism is saturated; that is, the secretion rate is at T m . Since [ PAH ] v is non-zero, the simple equation for renal plasma flow (RPF = C L [ PAH ] ) is not completely valid, and the more complete formula based on conservation of mass must be used. secretion rate T m = = excretion rate filtration rate U [ PAH ] u GFR [ PAH ] p = ˙ = ( 2 )( 50 ) ( 120 )( 0 . 2 ) = 100 24 = 76 mg/min 6.022j—2004: Solutions to Problem Set 10 2
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(undefined) [PAH] p mg/L Secreted mg/min 100 76 24 0.2 d e t e r c x E d e r e t l i F C. What is the renal plasma flow?
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