What is the GFR of each patient? Patient 1: V = rate of urine production (ml/min) = 1.2 U_inulin = amount of inulin excreted in urine each minute (mg/mL) = 15.8 P_inulin = plasma inulin (mg/mL) = 0.151 Patient 2: V = rate of urine production (ml/min) = 0.75 U_inulin = amount of inulin excreted in urine each minute (mg/mL) = 25.2 P_inulin = plasma inulin (mg/mL) = 0.155 Patient 3: V = rate of urine production (ml/min) = 15 U_inulin = amount of inulin excreted in urine each minute (mg/mL) = 1.23 P_inulin = plasma inulin (mg/mL) = 0.154
What is the GFR of each patient? Patient 1: V = rate of urine production (ml/min) = 1.2 U_inulin = amount of inulin excreted in urine each minute (mg/mL) = 15.8 P_inulin = plasma inulin (mg/mL) = 0.151 Patient 2: V = rate of urine production (ml/min) = 0.75 U_inulin = amount of inulin excreted in urine each minute (mg/mL) = 25.2 P_inulin = plasma inulin (mg/mL) = 0.155 Patient 3: V = rate of urine production (ml/min) = 15 U_inulin = amount of inulin excreted in urine each minute (mg/mL) = 1.23 P_inulin = plasma inulin (mg/mL) = 0.154
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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What is the GFR of each patient?
- Patient 1:
- V = rate of urine production (ml/min) = 1.2
- U_inulin = amount of inulin excreted in urine each minute (mg/mL) = 15.8
- P_inulin = plasma inulin (mg/mL) = 0.151
- Patient 2:
- V = rate of urine production (ml/min) = 0.75
- U_inulin = amount of inulin excreted in urine each minute (mg/mL) = 25.2
- P_inulin = plasma inulin (mg/mL) = 0.155
- Patient 3:
-
- V = rate of urine production (ml/min) = 15
- U_inulin = amount of inulin excreted in urine each minute (mg/mL) = 1.23
- P_inulin = plasma inulin (mg/mL) = 0.154
-

Transcribed Image Text:Substance, S, has renal clearance CS is the same as the amount of S excreted in
urine each minute, US.
When a substance is only filtered its clearance value equals the GFR (inulin, not
insulin, is the standard).
Substances both filtered and secreted will have clearances >GFR.
Reabsorbed substances have a clearance <GFR.
GFR=inulin clearance=(Uinulin/Pinulin) x V (ml/min)
% filtered water reabsorbed=(GFR-V)/GFR
Ls=GFR x Ps
Quanity of S secreted/min =Us x V
%filtered S reabsorbed=(Ls-quantity excreted/min)/Ls
CH2O=V x (1-Uosm/Posm) (ml/min)
Curea=V x (Uurea/Purea)
Test your calculations against the pre-calculated red answers
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