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Physiology General physiology ffa1c0b0

If the plasma concentration of a freely filterable substance is 2 mg/mL, GFR is 100 mL/min, urine concentration of the substance is 10 mg/mL, and urine flow rate is 5 mL/min, we can conclude that the kidney tubules

A
Reabsorbed 150 mg/min
B
Reabsorbed 200 mg/min
C
Secreted 50 mg/min
D
Secreted 150 mg/min
High-Yield Explanation
Reabsorbed 150 mg/minThe filtered load of the substance is P x GFR = 2 mg/mL x 100 mL/min = 200 mg/min. The rate of excretion is UxV=10 mg/mL x5 mL/min = 50 mg/min. Hence, more substance X was filtered than was excreted, and the difference, 200 mg/min - 50 mg/min = 150 mg/min, gives the rate of tubular reabsorption of substance X.(Note : amount or quantity = volume x concentration)Ref: Ganong - Review of Medical Physiology 23rd Ed Page 888

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