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S5 Biology | Kidney and excretion prep guide (for tutors) | nephron, filtration, ADH

S5 Biology | Kidney and excretion prep guide (for tutors) | nephron, filtration, ADH

  • 2026-11-01

Students think the kidney manufactures urine, so they cannot explain how 180 litres of filtrate a day becomes 1.8 litres of urine. This lesson separates filtration from reabsorption.

Objectives — Students can: (1) label the glomerulus, Bowman capsule, proximal tubule, loop of Henle, distal tubule and collecting duct; (2) explain filtration as size-based sieving and reabsorption as selective recovery, with two examples each; (3) use ADH to explain why urine becomes smaller and more concentrated when water intake falls.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill Quick check Pass mark
Blood composition plasma versus cells protein stays in blood
Osmosis water moves to higher solute get direction right
Active transport reabsorption needs ATP against the gradient

02 Lesson flow (55 minutes)

Time Stage Tutor does Student does
0-10 min Nephron structure cortex versus medulla label six parts
10-24 min Filtration GFR 125 mL/min 180 L filtrate per day
24-38 min Reabsorption 99% recovered only 1.8 L out
38-48 min ADH and collecting duct low versus high water intake 3.6x urine volume
48-52 min Integrated question urea concentration, dialysis 80x concentrated
52-55 min Exit check three questions collected independent work

03 Teaching script (read it out)

Scene 1: A nephron is a sieve followed by a salvage line

The glomerulus is a capillary knot; blood pressure pushes water, salts, glucose and urea into Bowman capsule while cells and large proteins stay behind. This step is filtration, passive and pressure-driven.
At a GFR of 125 mL/min the daily filtrate is 180 litres. Voiding all of it would dehydrate a person, so reabsorption must follow. The proximal tubule recovers glucose and most water and salts, the loop of Henle builds the medullary gradient, and the distal tubule and collecting duct fine-tune under hormone control.

Student task: on a blank nephron, mark filtration, reabsorption and secretion, with one substance for each.

Scene 2: Reabsorb 99 per cent and 1.8 litres remain

From 180 litres of filtrate, recovering 99 per cent leaves 1.8 litres. For glucose, 5 mmol/L across 180 L filters 900 mmol a day, all recovered in the proximal tubule, so normal urine contains none. Glucose in urine means blood glucose has passed the transport maximum or the tubule is damaged.
Urea goes from 5 mmol/L in plasma to 400 mmol/L in urine, a 80-fold concentration, because water leaves while urea stays in the tubule. Reabsorption is either active (glucose, amino acids, against the gradient) or passive (water by osmosis).

Student task: compute the 900 mmol glucose load and explain why a diabetic produces more urine.

Scene 3: ADH: less water in, more concentrated urine out

Rising plasma osmotic pressure is detected by the hypothalamus, the posterior pituitary releases ADH, the collecting duct becomes more permeable to water, and urine falls from 1.8 L to about 0.50 L while total urea output stays similar. After a large drink the opposite happens: ADH falls, little water is recovered, urine is copious and dilute.
Stress that ADH controls water recovery, not urea excretion; long questions award marks for that distinction. Dialysis replaces filtration and ultrafiltration but not reabsorption, which is why dialysis patients lose nutrients and swing between electrolyte levels.

Student task: draw two flow lines (low intake, high intake) marking ADH level, duct permeability, urine volume and concentration.

04 Question types and worked steps

Type Example Steps Answer
Structure labelling nephron diagram split cortex and medulla first six parts each
Calculation GFR 125 mL/min multiply by 1440 minutes, divide by 1000 180 L per day
Explanation why no glucose in urine filtered load against transport maximum 900 mmol all recovered
Hormonal control urine after low water intake ADH up, water recovery up down to 0.50 L
Sanity check (teach this): Filtrate at 180 L must exceed final urine at 1.8 L, otherwise the reabsorption figures are reversed. A 80-fold concentration looks dramatic, but 5 and 400 mmol/L are standard textbook values; a student reporting 8-fold has usually mixed litres and millilitres.

05 Independent practice (3 questions)

Question Answer What to watch
GFR 100 mL/min, daily filtrate in litres? 144 L did they multiply by 1440
150 L filtrate, 98.5 per cent recovered, urine volume? 2.25 L convert the percentage first
Insufficient ADH, what happens to urine? large volume, dilute separate volume from concentration

06 Common mistakes and what to say

Mistake Student writes Say this
Kidney as a urea factory writes that the kidney makes urea “The liver makes urea; the kidney only sieves and recovers.”
Filtration and reabsorption swapped says glucose is recovered in the glomerulus “Recovery happens in the tubule; the glomerulus only sieves.”
Units not unified treats 180000 mL as 180000 L “Write everything in litres first, then divide by 1000.”
ADH described as urea control writes that ADH raises urea output “ADH moves water; urea simply follows the water.”

07 Homework (paper, section and question numbers)

08 Exit check (last 5 minutes)

Question Answer Decision
180 L filtrate, 99 per cent recovered: urine volume? 1.8 L wrong: redo act two
Glucose in urine: most likely cause? high blood glucose or tubule damage wrong: redo transport maximum
Effect of rising ADH on the collecting duct? higher water permeability, concentrated urine two wrong: back to act three

09 Tutor FAQ

Which year is this for?

S5 to S6; it covers the HKDSE excretion and homeostasis topics.

How many structures must be recalled?

Six in flow order: glomerulus, Bowman capsule, proximal tubule, loop of Henle, distal tubule, collecting duct.

Is there practice attached?

The homework list links to the biology chapter index, pairing with respiration and homeostasis.

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