# Problem Set 5 Renal

## 5.1 A back of the envelope calculation of GFR

- Estimate GFR using a back-of-the-envelope calculation. The calculation is

GFR = cardiac output X renal fraction X plasma fraction X filtration fraction

Look up reasonable values for the four variables to parameterize this equation. Do the computation in your Google Sheet. Insert the units of GFR in the adjacent cell.

## 5.2 Using renal clearance to measure GFR in an indivudal

\[\begin{equation} C_s = \frac{\dot{M}_s}{P_s} \end{equation}\]where \(C_s\) is the clearance of solute \(s\), \(M_s\) (“m dot”) is the mass of \(s\) excreted in the urine per unit time, and \(P_s\) is the plasma concentration of \(s\).

What are the units of \(C_s\)? These are the units of what kind of measure (for example Force per Area are the units of a pressure)?

Remember that a dot over a variable is a first derivative; here we assume that this is constant and so \(\dot{M}_s = \frac{\Delta Mass}{\Delta Time}\). What are the units of \(\dot{M}_s\)? This kind of measure is “kinda like” the kind of measure in #2. Google around to see what we call \(\dot{M}_s\).

The clearance of a solute is useful in pharmacology but we can also use the concept to measure the GFR in a person. This is done using a solute \(s\) that is filtered but no amount is either 1) secreted into the nephron, or 2) is not reabsorbed from the nephron). Inulin is an example. We could give a person some inulin and then measure the urine concentration of inulin (\(U_{in}\)), the volume of urine generated per time (\(\dot{V}\)), and the plasma concentration of inulin (\(P_{in}\)) to compute the GFR

(Note that I use \(\dot{V}\) and not \(V\) to make it crystal clear that this is a measure of the volume of urine produced per time not simply a volume).

Using this information, compute the GFR for a person in which 1) inulin was given continuously to generate a constant plasma concentration of 1.0 mg/dL. 1.6 L of urine was collected over a 10 hour period. The urinary concentration of inulin was 462 mg/L.