Working through some sample data

This exercise will take you through the initial steps needed to use the clearance approach to  investigate the renal response to a diuretic. In this chapter we will calculate the control values (in the absence of the diuretic).  In the next chapter you will be able to have a go at the calculations in the presence of the diuretic.

In the experiment, urine and plasma concentrations and urine flow rate were evaluated in an individual under control conditions and on ingestion of acetazolamide (a carbonic anhydrase inhibitor). In the control experiment subjects drank 1 litre of water at the start of the experiment and then each hour drank a  volume of water equivalent to the urine lost. In the acetazolamide experiment subjects repeated the water loading but also ingested 400 mg of acetazolamide at 140 minutes.

The following table shows the urinary and plasma parameters measured, and the urine flow rate in the control experiment. Make sure you look carefully at the units in the tables as you go through the calculations. Be particularly careful with creatinine!

Control data

Let’s use these data to examine the renal handling of water, Na⁺ and HCO^-₃.

Step 1 - calculate the GFR in ml/min

Calculate the clearance of creatinine to work out the GFR.

[latex]GFR =   \frac{UV}{P}[/latex]

U = 0.8 μmoles/ml, V = 12.5 ml/min, P = 83 μmole/l

We cannot simply put these into the equation as there is a mismatch. U and P are different (mls  versus litres). So we need to convert to standard units.

U = 0.8 μmoles/ml = 800 μmoles/l (multiply by 1000 to go from ml to litres). V = 12.5 ml/min = 12.5 x10^-3 litres per minute (or 0.0125)

P = 83 μmole/l = 83 μmoles/litre

Now we could convert the μmole to moles for complete standard units, but as both U and P are in μmole we don’t need to.

[latex]GFR =   \frac{800 \cdot 0.0125}{83}[/latex]

GFR = 0.12 l /min

GFR = 120.5 ml/min

Step 2 - calculate the total water reabsorbed in ml/min and %

This is the difference between GFR and V. So……

120.5 – 12.5 = 108.0 ml/min

To get the percentage reabsorption of water, divide by the GFR and multiply by 100 = 89.6%.

Step 3 - calculate the amount of Na⁺ and HCO^-₃ filtered

Amount filtered [latex]= P \cdot GFR[/latex]

Na⁺  = 139 μmole/ml x  120.5 ml/min

No need to convert here as both volumes are the same. So

[latex]139 \cdot 120.5 = 16749.5[/latex] μmoles/min

HCO^-₃ = 25 μmole/ml x 120.5 ml/min

No need to convert here as both volumes are the same. So

[latex]25 \cdot 120.5 = 3012.5[/latex] μmoles/min

Step 4 - calculate the amount of Na⁺ and HCO^-₃ excreted

Amount excreted [latex]= U \cdot V[/latex]

Na⁺ = 13.8 umole/ml x 12.5 ml/min

[latex]13.8 \cdot 12.5 = 172.5[/latex] umoles/min

HCO^-₃ = 2.4 umole/ml x 12.5 ml/min

[latex]2.4 \cdot 12.5 = 30.0[/latex] umoles/min

Step 5 - calculate the amount of Na⁺ and HCO^-₃ reabsorbed

This is the difference between the filtered and excreted.

For Na⁺ = 16749.5 – 172.5 = 16577.0  μmoles/min.

For HCO^-₃ 3012.5 – 30.0 = 2982.5 μmoles/min

We can simply subtract one from the other as the units are the same.

Step 6 - calculate the fractional excretion of Na⁺ and HCO^-₃ (as a %)

This is the amount excreted divided by the amount filtered, then multiplied by 100 to get the value as a percent.

FE [latex]=  \frac{(U \cdot V)}{(P \cdot GFR)}[/latex]

FENa⁺ =  [latex]\frac{172.5}{16749.5} \cdot 100[/latex] = 1.03%

FEHCO^-₃  =  [latex]\frac{30.0}{3012.5} \cdot 100[/latex] = 1.00%

Step 7 - calculate the fractional reabsorption of Na⁺ and HCO^-₃ (as a %)

This is the amount reabsorbed divided by the amount filtered, then multiplied by 100 to get the value as a percent.

FR [latex]=  \frac{ (P \cdot GFR)-(U \cdot V)}{(P \cdot GFR)}[/latex]

FRNa⁺ = [latex]\frac{(16577.0)}{(16749.5)}\cdot 100[/latex] = 99.0%

FRHCO^-₃= [latex]\frac{(2982.5)}{(3012.5)}\cdot 100[/latex] = 99.0%