Question The following test results were obtained over a 24-hourperiod:urine volume = 1.4 Lurine [inulin] = 100 mg%urine [urea] = 220 mmol L-1urine [PAH] = 70 mg mL-1plasma [inulin] = 1 mg%plasma [urea] = 5 mmol L-1plasma [PAH] = 0.2 mg mL-1hematocrit = 0.40 We know that inulin is freely filtered but NOTreabsorbed/secreted.PAH is freely filtered, and it is secreted, but NOTreabsorbed.Urea is freely filtered, and it is reabsorbed, but NOTsecreted.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 themass balance: Input = Output) 1pt

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} ...