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

plasma [inulin] = 1 mg%

plasma [urea] = 5 mmol L-1

plasma [PAH] = 0.2 mg mL-1

hematocrit = 0.40

We know that inulin is freely filtered but NOT reabsorbed/secreted.

PAH is freely filtered, and it is secreted, but NOT reabsorbed.

Urea is freely filtered, and it is reabsorbed, but NOT secreted.

**1 mg% = 0.01 mg/mL**

**calculate:**

**A)** Glomerular filtration rate (GFR)
(**Hint.** GFR equals to the clearance of
?)

**B) **The rate of urea filtration

**C) **The rate of urea excretion

**g)** The rate of urea reabsorption
(**Hint.** Urea is NOT secreted. Please use the
mass balance: Input = Output) **1
pt**

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a) \begin{aligned} \text { RATE OF URINE EXCRETION } & =\frac{1.4 \mathrm{~L}}{24 \mathrm{H}} \\ & =\frac{1400 \mathrm{ml}}{1440 \mathrm{~min}} \\ & =0.972 \mathrm{ml} / \mathrm{min}\end{aligned}b)\begin{array}{l}C_{\text {inulin }}=\frac{1.4 \mathrm{~L}}{1440 \mathrm{~min}} \times \frac{1 \mathrm{mg} \mathrm{ml}}{0.1 \mathrm{mgml}-1} \\C_{\text {inulin }}=97.2 \mathrm{~mL} \mathrm{min-1} \\C_{\text {urea }}=\frac{1.4 \mathrm{~L}}{1440 \mathrm{~min}} \times \frac{220 \mathrm{mmolL}}{5 \mathrm{mmolL}-1} \\C_{\text {urea }}=42.8 \mathrm{~mL} \mathrm{min-1} \\C_{\text {PAH }}=\frac{1.4 \mathrm{~L}}{1440 \mathrm{~min}} \times \frac{70 \mathrm{mg} \mathrm{mL}-1}{0.2 \mathrm{mg} \mathrm{L}-1} \\C_{\text {PAH }}=340.3 \mathrm{~mL} \mathrm{min-1.}\end{array}c) EFFECTIVE RENAL PLASMA FLOW = CpAH.\text { ERPF }=C_{P A H}=340.3 \mathrm{~mL} \mathrm{~min}-1d) GLOMERULAR FILTRATION RATE = C_{\text {insulin }}G F R=C_{\text {insulin }}=97.2 \mathrm{~mL} \mathrm{min-1}e)F L=G F R \times P_{x}\begin{array}{l}F_{L}=\text { Filtered load } \\G F R=\text { Glomerular filtration rate }=97.2 \\P_{x}=\text { plasma concentration }=5 \mathrm{mmol} / \mathrm{L}\end{array}\begin{aligned}5 \mathrm{mmol} / \mathrm{L} & =5 \times 18 \mathrm{mg} / \mathrm{dl} \\P_{x} & =90 \mathrm{mg} / \mathrm{l}\end{aligned}\begin{array}{l}F L=97.2 \times 90 \\F L=8748 \mathrm{mg} / \mathrm{min}\end{array}\text { F) } \begin{array}{l} E_{x}=U_{x} \times V \quad \begin{aligned}E_{x} & =\text { Excretion Rate } \\U_{x} & =\text { Urine concentration (urea) }=220 \mathrm{mmol} / \mathrm{L} \\V & =\text { urine Flow rate }\end{aligned} \\\quad 220 \mathrm{~m} \mathrm{~mol} / \mathrm{L}=220 \times 18 \\E_{x}=3960 \times 0.972 \\E_{x}=3849 \mathrm{mg} / \mathrm{min}\end{array}G.R_{x}=F L-E_{x}R_{x}= Reabsorption Rate (mg/min)E_{x}= Excretion Rate (mg/min)F_{L}= Filtered load ( \mathrm{mg} / \mathrm{min} )\begin{array}{l}R_{x}=8748-3849 \mathrm{mg} / \mathrm{ml} \\R_{x}=4899 \mathrm{mg} / \mathrm{ml}\end{array} ...