The following test results were obtained over a 24-hour period: urine volume = 1

Question

The following test results were obtained over a 24-hour period: urine volume = 1.4 L urine [inulin] = 100 mg% urine [urea] = 220 mmol L^-1 urine [PAH] = 70 mg mL^-1 (PAH: Para-aminohippurate) plasma [inulin] = 1 mg% plasma [urea] = 5 mmol L^-1 plasma [PAH] = 0.2 mg mL^-1 hematocrit = 0.40 Calculate: (A) C_inulin, C_urea, C_PAH; (B) The rate of urea filtration; (C) The rate of urea excretion; (D) The rate of urea reabsorption; (E) The rate of PAH filtration; (F) The rate of PAH excretion; (G) The rate of PAH filtration secretion. (H) C_PAH has been widely used to estimate the effective renal plasma flow, ERPF. Why can we use C_PAH to estimate ERPF?

Explanation / Answer

answers:

A) C inulin = 1.4L /1440 min X 1mg mL / .01 mg mL -1 = 97.2 mL min-1

     C urea = 1.4 L / 1440 min X 220 mmol L / 5 mmol L -1 = 42.8 mL min-1

     C PAH = 1.4L/1440min X 70 mg mL-1 / 0.2 mg L -1 = 340.3 mL min-1

B) The rate of urea filtration : C inulin X Plasma urea

                                                     97.2 mL min X 5 mmol L-1 = 0.486 mmol min -1

C) The rate of Urea excretion : Urine volume / 24hr X Urine urea

                                                : 1.4L/ 1440min X 220 mmol L = 0.214 mmol min-1

D) The rate of Urea reasborption : It is difference between the rate of filtered and rate of excreted

                                                     :0.486 mmol min -1 -- 0.214 mmol min-1 = 0.272 mmol min-1

E) The rate of PAH filtration :   C inulin X Plasma (PAH)

                                              :       97.2 mL min-1 X 0.2 mg mL -1 = 19.44 mg min-1

F) The rate of PAH excretion : Q U X Urine (PAH)

                                                 0.972 mL min-1 X 70 mg mL-1 = 68.06 mg min-1

G) The rate of PAH filtration secretion : (assuming no PAH reasborbed) is

     The rate of PAH excreted -- Rate of PAH filtered = 68.06 mg min-1 - 19.44 mg min-1 = 48.06 mg min-1

H) The measurement of CPAH can thus be used to estimate RPF, since PAH is filtered by the glomerulus, secreted by the tubule (PT), considered to be almost completely excreted in the urine by the kidney. The RPF is estimated to approximately 625ml/min.   However, plasma PAH is actually is not 100% clered by kindeys. The RPF measured by using CPAH is generally referred to as effective RPF ( ERPF).

ERPF = 340.3 ml min

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