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Solution for problem 10 .(1)Given that,\begin{array}{l}\text { * Urine volume }=1.4 \mathrm{~L}=1400 \mathrm{~mL} . \\* \text { Urine }[\text { inulin }]=100 \mathrm{mg} \% \\* \text { urine }[\text { urea }]=220 \mathrm{mmol} / \mathrm{l} . \\* \text { Urine }[\mathrm{PAH}]=70 \mathrm{mg} \mathrm{m} / \mathrm{L} . \\* \text { Plasma[inulin] }=1 \mathrm{mg} \% \\* \text { plasma[urea }]=5 \mathrm{mmoll} \\* \text { Plasma[PAH] }=0.2 \mathrm{mg} \mathrm{m} / \mathrm{L} \\* \text { Hematocrit }=0.40 .\end{array}T0 find out,Cirulin, Curea and CPAHNow,clearance of inulin \left[c_{\text {inulin }}\right]urine flow rate (volume) =1400 \mathrm{~mL} / 1440=0.972 \mathrm{ml} / \mathrm{min}.\text { * urine }[\text { inulin }=100 \mathrm{mg} / 100 \mathrm{ml}=1 \mathrm{mg} / \mathrm{mk} \text {. }plasma [ inulin] =1 \mathrm{mg} / 100 \mathrm{ml}=0.01 \mathrm{mg} / \mathrm{ml}\therefore \quad ilearance of inulin =\frac{0.972 \times 1}{0.01}=97.2 \mathrm{ml} / \mathrm{min}.(2)clearance of urea\begin{array}{l}\text { * urine (urea) }=\frac{220 \mathrm{mmol} / \mathrm{L}}{1000 \mathrm{ml}}=13.21 \mathrm{mg} / \mathrm{ml} \\* \text { plasma (urea) }=\frac{5 \mathrm{mmol} / \mathrm{L}}{1000 \mathrm{ml}}=0.3 \mathrm{mg} \mathrm{ml}\end{array}clearance of urea=\frac{0.972 \times 13.21}{0.30}=42.8 \mathrm{mg} / \mathrm{ml}clearance of \mathrm{pAH} :-\begin{aligned}\text { * Urine }(P A H)=70 \mathrm{mg} / \mathrm{ml} . \\* \quad \text { plasma }(P A H)=0.2 \mathrm{mg} / \mathrm{ml} . \\* \text { clearance of } P A H=\frac{0.972 \times 70}{0.2} \\=340.2 \mathrm{mg} / \mathrm{ml}\end{aligned} ...