# Problem Set 5 Renal

## 5.1 A back of the envelope calculation of GFR

1. 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

$$$C_s = \frac{\dot{M}_s}{P_s}$$$

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 $$s$$.

1. 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)?

2. 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 $$\dot{M}_s$$.

3. 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

$$$GFR = \frac{U_{in} \dot{V}}{P_{in}}$$$

(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.