A patient has a GFR of 100 ml/min, her urine flow rate is 2.0 ml/min, and her plasma glucose concentration is 200 mg/100 ml. If the kidney transpo maximum for glucose is 250 mg/min, what would be her approximate rate of glucose excretion?
Correct Answer: 0 mg/min
Description: The filtration rate of glucose in this example is: = GFR x plasma glucose concentration =(100 ml/min) x (200 mg/100 ml, or 2 mg/ml) =200 mg/min. Because the transpo maximum for glucose in this example is 250 mg/min, All of the filtered glucose would be reabsorbed and the renal excretion rate for glucose would be zero(0 mg/min).
Category:
Physiology
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