• $U_x$$U_x$ = concentration of substance X in urine

• $P_x$$P_x$ = concentration of substance X in plasma

• $K_{\text{fil}}$$K_{\text{fil}}$ = filtration coefficient

• $U_{\text{PAH}}$$U_{\text{PAH}}$ = concentration of para-aminohippuric acid ( PAH ) in urine

• UFR = urine flow rate

• $P_{\text{PAH}}$$P_{\text{PAH}}$ = concentration of PAH in plasma

• $C_{\text{PAH}}$$C_{\text{PAH}}$ = total clearance of PAH

• $U_x$$U_x$ = concentration of substance X , in the urine

• this is equal to 1 for creatine , because creatinine is freely filtered and NOT reabsorbed or secreted by the kidneys

• for water though :

• $P_{\text{cap}}$$P_{\text{cap}}$ = hydrostatic pressure in capillaries

• $P_{\text{isf}}$$P_{\text{isf}}$ = hydrostatic pressure in interstitial fluid

Table 1 - Renal Function Chart

1. Indicate how glomerular filtration rate ( GFR ) changed during this day ( Table 1 )

   $${\text{GFR}}_{\text{morning}}=\frac{14.3mM\ast \frac{0.7mL}{min}}{0.08mM}=\frac{125mL}{1minute}$$
   $${\text{GFR}}_{\text{afternoon}}=\frac{0.65mM\ast \frac{15.1mL}{min}}{0.076mM}=\frac{129.145mL}{1minute}$$

2. Indicate how fractional excretion of water ( Table 1 ) changed during this day.

   $${\text{FE}}_{\text{morning}}^{\text{creatinine}}=\frac{14.3mM\ast \frac{0.7mL}{min}}{0.08mM\ast \frac{125mL}{min}}=1$$
   $${\text{FE}}_{\text{morning}}^{\text{H2O}}=\frac{\frac{0.7mL}{min}}{\frac{125mL}{min}}\ast 100\mathrm{\%}=0.559\mathrm{\%}$$
   $${\text{FE}}_{\text{afternoon}}^{\text{creatinine}}=\frac{0.65mM\ast \frac{15.1mL}{min}}{0.076mM\ast \frac{129mL}{min}}=1$$
   $${\text{FE}}_{\text{afternoon}}^{\text{H2O}}=\frac{\frac{15.1mL}{min}}{\frac{129mL}{min}}\ast 100\mathrm{\%}=11.705\mathrm{\%}$$

3. Indicate how fractional reabsorption of water ( Table 1 ) changed during this day.

   $$F{R}_{\text{morning}}^{\text{H2O}}=(1-\frac{\frac{0.7mL}{min}}{\frac{125mL}{min}})\ast 100\mathrm{\%}=99.44\mathrm{\%}$$
   $$F{R}_{\text{afternoon}}^{\text{H2O}}=(1-\frac{\frac{15.1mL}{min}}{\frac{129mL}{min}})\ast 100\mathrm{\%}=87.92\mathrm{\%}$$

4. Calculate filtered load for Na+ ( Table 1 ) during this day.

   $$P_{\text{morning}}^{\text{Na}}=\frac{147millimoles}{1Liter}\ast \frac{1mol}{1\ast {10}^{3}millimoles}\ast \frac{1Liter}{1\ast {10}^{3}milliliters}=\frac{147\ast {10}^{-6}mol}{milliliter}$$
   $${\text{Filtered Load}}_{\text{morning}}^{\text{Na}}=\frac{125mL}{min}\ast \frac{147\ast {10}^{-6}mol}{milliliter}=\frac{0.018375mol}{min}$$
   $$P_{\text{afternoon}}^{\text{Na}}=\frac{144millimoles}{1Liter}\ast \frac{1mol}{1\ast {10}^{3}millimoles}\ast \frac{1Liter}{1\ast {10}^{3}milliliters}=\frac{144\ast {10}^{-6}mol}{milliliter}$$
   $${\text{Filtered Load}}_{\text{afternoon}}^{\text{Na}}=\frac{129mL}{min}\ast \frac{144\ast {10}^{-6}mol}{milliliter}=\frac{0.018576mol}{min}$$

5. Calculate Na+ excretion rate ( Table 1 ) during this day.

   $$U_{\text{morning}}^{\text{Na}}=\frac{196millimoles}{1Liter}\ast \frac{1mol}{1\ast {10}^{3}millimoles}\ast \frac{1Liter}{1\ast {10}^{3}milliliters}=\frac{196\ast {10}^{-6}mol}{milliliter}$$
   $${\text{ER}}_{\text{morning}}^{\text{Na}}=\frac{196\ast {10}^{-6}mol}{milliliter}\ast \frac{0.7mL}{1min}=\frac{0.000137199mol}{min}$$
   $$U_{\text{afternoon}}^{\text{Na}}=\frac{9millimoles}{1Liter}\ast \frac{1mol}{1\ast {10}^{3}millimoles}\ast \frac{1Liter}{1\ast {10}^{3}milliliters}=\frac{9\ast {10}^{-6}mol}{milliliter}$$
   $${\text{ER}}_{\text{afternoon}}^{\text{Na}}=\frac{9\ast {10}^{-6}mol}{milliliter}\ast \frac{15.1mL}{1min}=\frac{0.0001359mol}{min}$$

6. Calculate fractional Na+ excretion ( Table 1 )

   $${\text{FE}}_{\text{morning}}^{\text{Na}}=\frac{\frac{0.000137199mol}{min}}{\frac{0.018375mol}{min}}\ast 100\mathrm{\%}=0.7466666666666666\mathrm{\%}$$
   $${\text{FE}}_{\text{afternoon}}^{\text{Na}}=\frac{\frac{0.0001359mol}{min}}{\frac{0.018576mol}{min}}\ast 100\mathrm{\%}=0.7315891472868218\mathrm{\%}$$

7. Calculate fractional Na+ reabsorption ( Table 1 )

   $${\text{FR}}_{\text{Na}}=100\mathrm{\%}-{\text{FE}}_{\text{Na}}$$
   $${\text{FR}}_{\text{morning}}^{\text{Na}}=100\mathrm{\%}-0.7466666666666666\mathrm{\%}=99.25333333333333\mathrm{\%}$$
   $${\text{FR}}_{\text{afternoon}}^{\text{Na}}=100\mathrm{\%}-0.7315891472868218\mathrm{\%}=99.26841085271317\mathrm{\%}$$

8. Calculate osmolar clearance ( $C_{\text{osm}}$$C_{\text{osm}}$ , Table 1 )

   $$\text{Osmolar Clearance}\left(C_{\text{osm}}\right)=\frac{U_{\text{osm}}\ast \text{UFR}}{P_{\text{osm}}}$$
   $$\text{Plasma Osmolarity}\left(P_{\text{osm}}\right)\approx 2\ast [N{a}^{+}]$$

   • because each sodium ion is accompanied by anions ?

     • under normal conditions

   $$P_{\text{morning}}^{\text{osm}}\approx 2\ast 147mM=\frac{294mOsm}{L}$$
   $$C_{\text{morning}}^{\text{osm}}=\frac{\frac{800mOsm}{L}\ast \frac{0.7mL}{min}}{\frac{294mOsm}{L}}=\frac{1.9047619047619047mL}{min}$$
   $$C_{\text{afternoon}}^{\text{osm}}=\frac{\frac{50mOsm}{L}\ast \frac{15.1mL}{min}}{\frac{288mOsm}{L}}=\frac{2.6215277777777777mL}{min}$$

9. Using the renal function data ( Table 1 ) , speculate on what likely occurred for this individual during this day.

   • Observation:

     • Significant Increase in UFR: From 0.7 mL / min to 15.1 mL / min.

     • Decrease in Urine Osmolarity: From 800 mOsm / L to 50 mOsm / L.

     • Decrease in Urine Na⁺ Concentration: From 196 mM to 9 mM.

     • GFR Remained Relatively Constant: Slight increase from 125 to 129 mL / min.

   • Speculation:

     • The individual likely consumed a large volume of water during the day , leading to:

       • Water Diuresis: Increased urine output to eliminate excess water.

       • Dilute Urine: Low urine osmolarity due to excretion of free water.

       • Conservation of Na⁺: Low fractional excretion of Na⁺ indicates the kidneys reabsorbed most Na⁺ to maintain electrolyte balance.

   • Answer:

     • The individual probably ingested a large amount of water, causing water diuresis characterized by high urine flow rate, dilute urine, and low urinary Na⁺ concentration, while maintaining stable GFR and efficient Na⁺ reabsorption.

10. Plasma protein concentration is ~40 g/L and glomerular filtrate contains 0.01 to 0.1 g/L , calculate the daily filtered load of plasma protein.

    $$\text{Daily GFR}=\text{GFR}\ast \frac{60min}{1hour}\ast \frac{24hours}{1day}$$
    $$\text{Daily GFR}=\frac{125mL}{1minute}\ast \frac{1Liter}{1000milliliters}\ast \frac{60min}{1hour}\ast \frac{24hours}{1day}=\frac{180Liters}{Day}$$
    $${\text{Filtered Load}}_{\text{minimum}}=\frac{180Liters}{Day}\ast \frac{0.01grams}{L}=\frac{1.8grams}{Day}$$
    $${\text{Filtered Load}}_{\text{maximum}}=\frac{180Liters}{Day}\ast \frac{0.1grams}{L}=\frac{18grams}{Day}$$

11. A healthy individual reports a plasma creatinine concentration of 100 µM and GFR of 120 mL / min. Assume that creatinine is freely filtered , not secreted , and not reabsorbed. Calculate the daily excretion rate for creatinine by the kidney. Determine the daily production rate for creatinine.

    • since creatinine is freely filtered and neither reabsorbed nor secreted , the excretion rate equals the filtered load

    $$\text{Filtered Load}\left(F{L}_{\text{Cr}}\right)=\text{GFR}\ast P_{\text{Cr}}$$
    $$F{L}_{\text{Cr}}=\frac{120mL}{min}\ast \frac{1Liter}{1000mL}\ast \frac{1\ast {10}^{-4}mol}{1Liter}\ast \frac{1.2\ast {10}^{-5}mol}{min}$$
    $$\text{Daily Excretion Rate}=\frac{1.2\ast {10}^{-5}mol}{min}\ast \frac{1440min}{1Day}=\frac{0.01728mol}{Day}$$

    • in steady state , the daily production equals daily excretion ,

      • therefore , daily production rate also equals 17.28 milli moles / day

12. In a patient with acute renal failure, GFR is reduced to almost zero. Determine how long it will take for the plasma concentration of an end-product of metabolism ( creatinine for example ) to be doubled.

    • with GFR near zero , creatinine is not being excreted. its being produced continuously

    • Volume of Distribution $\left(V_d\right)$$\left(\ V_d \ \right)$ = creatine distributes in total body water ( ~60% of body weight ). For a 70 kg person :

      • $V_d=70kg\ast 0.6=42Liters$$V_d = 70\ kg * 0.6 = 42\ Liters$

    • Initial amount of creatine in the body $([Cr{]}_{initial})$$\left(\ [\ Cr\ ]_{initial} \ \right)$ :

    • $[Cr{]}_{initial}=P_{\text{Cr}}\ast V_d=\frac{100\ast {10}^{-6}mol}{Liter}\ast 42Liters=4.2\ast {10}^{-3}mol$$[\ Cr\ ]_{initial} = P_{\text{Cr}} * V_d = \frac{100 * 10^{-6}\ mol}{Liter} * 42\ Liters = 4.2 * 10^{-3}\ mol$

    • From Question 11 , the creatine production rate $\left(R_p\right)=\frac{0.01728mol}{Day}\ast \frac{1Day}{24Hours}=\frac{0.00072mol}{Hour}$$\left(\ R_p \ \right) = \frac{0.01728\ mol}{Day} * \frac{1\ Day}{24\ Hours} = \frac{0.00072\ mol}{Hour}$

    • Time $\left(t\right)$$\left(\ t \ \right)$ it takes to double $[Cr{]}_{initial}$$[\ Cr\ ]_{initial}$ :

      • $2\ast [Cr{]}_{initial}=[Cr{]}_{initial}+R_p\ast t$$2 * [\ Cr\ ]_{initial} = [\ Cr\ ]_{initial} + R_p * t$

      • $\therefore ,t=\frac{[Cr{]}_{initial}}{R_p}$$\therefore\ ,\ t = \frac{[\ Cr\ ]_{initial}}{R_p}$

    $$t=\frac{4.2\ast {10}^{-3}mol}{\frac{0.00072mol}{Hour}}=5.833333333333333hours$$

13. The healthy individual has one kidney removed for transplant to a close blood relative. Estimate their rate of creatinine excretion immediately after surgery. Estimate their excretion rate of creatinine after 24 hours. Comment on changes in their plasma creatinine concentration after surgery?

14. The plasma concentration of para-aminohippurate ( PAH ) was measured as 68 µM and urine PAH concentration was 31 mM , together with a urine flow rate of 1.2 mL/min. Calculate PAH clearance. Estimate renal plasma flow. Calculate the filtration fraction.

15. Using the filtration fraction, calculate the mean glomerular oncotic pressure if systemic oncotic pressure was 25 mmHg.

16. The proximal tubular osmotic gradient consists of luminal hypo-osmolarity ( 287 mOsM ) and interstitial hyper-osmolarity ( 293 mOsM ) , compared with intracellular osmolarity ( 290 mOsM ). Determine the osmotic gradient in mmHg between the lumen and cell, as well as between cell and peritubular interstitium.

17. The transport maximum for PAH is 0.4 mmole/min , and when plasma PAH concentration is increased to 1.2 mM , calculate the rate of PAH return to the systemic circulation.

18. After two hours of copious vomiting, a previously healthy individual is found to have a plasma pH of 7.55 , and a PaCO2 of 40mmHg.

    Indicate the cause of the alkalosis.

    Estimate the plasma bicarbonate concentration in the circumstance described.

    Subsequent respiratory responses to this alkalosis bring plasma pH back down to 7.45 :

    Indicate the respiratory response and its effect on PaCO2 and bicarbonate concentration.

    Indicate the name for this response.

    Indicate how acid-base balance and a normal PaCO2 and bicarbonate concentration are achieved over a longer period of time.