Question 3 of 6

Question 3

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Functional anatomy of the kidney

Each kidney is organised into an outer cortex and inner medulla, the medulla arranged as 8–18 renal pyramids whose papillae drain into minor, then major, calyces and the renal pelvis, which narrows into the ureter. The functional unit is the nephron (≈1–1.3 million per kidney): a glomerulus (a tuft of fenestrated capillaries) invaginated into Bowman’s capsule, filtering plasma into the tubule — proximal convoluted tubule → loop of Henle (descending thin limb, ascending thin/thick limb) → distal convoluted tubule → collecting duct. Blood enters each glomerulus via an afferent arteriole and leaves via an efferent arteriole (the only capillary bed in the body fed and drained by arterioles, letting the kidney control glomerular pressure independently of downstream pressure); the efferent arteriole then supplies a second capillary bed, the peritubular capillaries (and, for juxtamedullary nephrons, the vasa recta running alongside the loop of Henle) that reabsorb tubular fluid. Where the thick ascending limb passes back between its own afferent and efferent arterioles, specialised macula densa cells and juxtaglomerular (granular) cells of the afferent arteriole form the juxtaglomerular apparatus, the sensor/effector unit for tubuloglomerular feedback and renin release.

(b) Filtered load, Tm-limited reabsorption, excretion and urine concentration of glucose

Given.

Find. Filtered load of glucose, rate of tubular reabsorption, rate of excretion, and glucose concentration in the urine.

Approach. Glucose is freely filtered at the glomerulus (filtered load $= \text{GFR}\times P_{glu}$) and then actively reabsorbed by a carrier-mediated transporter with a finite capacity $T_m$; whatever is filtered but exceeds $T_m$ cannot be reabsorbed and is excreted, and the excreted mass rate divided by urine flow gives the urine concentration.

1. Filtered load of glucose. $$\text{Filtered load} = \text{GFR}\times P_{glu} = 125\ \text{mL/min}\times 10\ \text{mg/mL} = \boxed{1250\ \text{mg/min}}$$
2. Rate of glucose reabsorption. Since the filtered load (1250 mg/min) exceeds the stated transport maximum $T_m=400$ mg/min, the glucose transporters are fully saturated and can reabsorb no more than $T_m$, regardless of how much more is filtered: $$\text{Reabsorption rate} = T_m = \boxed{400\ \text{mg/min}}$$
3. Rate of glucose excretion. By mass balance across the nephron, whatever is filtered but not reabsorbed is excreted: $$\text{Excretion rate} = \text{Filtered load} - \text{Reabsorption rate} = 1250 - 400 = \boxed{850\ \text{mg/min}}$$
4. Concentration of glucose in the urine. The excreted mass rate divided by the urine flow rate gives the urine concentration: $$U_{glu} = \frac{\text{Excretion rate}}{\dot V_{urine}} = \frac{850\ \text{mg/min}}{10\ \text{mL/min}} = \boxed{85\ \text{mg/mL}}$$