OCR GCSE Biology A: Osmoregulation: water balance, the kidney and ADH (J247)

This guide teaches statements B3.3f to B3.3j of module B3.3 (Maintaining internal environments) in OCR GCSE (9-1) Gateway Science Biology A (J247), specification Version 4.0 (August 2026), for first assessment in 2018. Statements B3.3i and B3.3j are in bold type in the specification, so they are tested only in the Higher Tier papers; B3.3f, B3.3g and B3.3h are assessed in both tiers. Temperature control by the skin and blood glucose control are separate statements in the same module and are not taught here. Course pages: OCR GCSE Biology hub | checklist | all free 10-minute diagnostics.

What this unit covers

Statement	What you must be able to do	Tier
B3.3f	Explain the effect on cells of osmotic changes in body fluids: higher, lower or equal water potentials leading to lysis or shrinking	Both tiers
B3.3g	Describe how the kidneys maintain water balance by varying the amount and concentration of urine, and so the water excreted	Both tiers
B3.3h	Describe the gross structure of the kidney and the structure of the kidney tubule: Bowman’s capsule, proximal convoluted tubule, loop of Henlé, collecting duct	Both tiers
B3.3i	Describe the effect of ADH on the permeability of the kidney tubules: the amount of water reabsorbed, and negative feedback	Higher Tier only
B3.3j	Explain the response of the body to temperature and osmotic challenges: sweating and dehydration, excess water intake, high salt intake; responses include the mechanism of kidney function and thirst	Higher Tier only

The specification says no mathematical use of water potentials is required. You describe water potential as higher, lower or equal. The mathematical outcome for this module is to extract and interpret data from graphs, charts and tables, so the worked examples below use data.

B3.3f: Osmotic changes and animal cells

In osmosis, water has a net movement through a partially permeable membrane, passing from a dilute solution (higher water potential) into a more concentrated one (lower water potential). The cell surface membrane is partially permeable, so water crosses it freely but many dissolved substances do not. For the basic idea of osmosis, see the transport across membranes guide.

Cells of the body are surrounded by tissue fluid, which comes from blood plasma. If the water potential of that fluid changes, water moves into or out of the cells.

Fluid around the cell compared with the cytoplasm	Net water movement	Result for an animal cell
Higher water potential (more dilute)	Into the cell	Cell swells and may burst: lysis
Equal water potential	None	Cell keeps its normal shape and volume
Lower water potential (more concentrated)	Out of the cell	Cell shrinks

An animal cell has no cell wall, so nothing stops it swelling until the membrane tears. This is why the body must keep the water potential of blood and tissue fluid close to constant. A red blood cell is a good example: in very dilute fluid it lyses, and in concentrated fluid it shrinks and can no longer carry oxygen properly.

Worked example 1

Red blood cells were placed in sodium chloride solutions. After 30 minutes the mean cell volume was recorded as a percentage of the starting volume. These data are supplied for the example.

Sodium chloride (%)	0.0	0.6	0.9	1.5	2.5
Cell volume (% of start)	all burst	122	100	78	58

(a) Which solution has the same water potential as the cytoplasm? The volume does not change at 0.9%, so there is no net movement of water: 0.9%.

(b) Calculate the percentage change in volume in 2.5% solution. Change = 58 - 100 = -42, so the volume fell by 42%.

(c) Explain the result in 0.6% solution. The outside solution is more dilute than the cytoplasm, so water moves in by osmosis and the cells swell to 122% of their starting volume (a rise of 22%).

(d) Explain why no cells survived in 0.0% solution. Pure water has the highest possible water potential, so water kept entering by osmosis. With no cell wall, the membrane could not resist the swelling and the cells burst (lysis).

B3.3h: Gross structure of the kidney and the kidney tubule

The renal artery brings blood to each kidney and the renal vein takes it away. Urine leaves each kidney by a ureter, is stored in the bladder and is released through the urethra.

Inside a kidney you can see three regions:

• Cortex: the outer layer. It contains the Bowman’s capsules and the convoluted tubules.
• Medulla: the inner region, made of the loops of Henlé and collecting ducts, which gives it a striped look.
• Pelvis: the hollow centre where urine collects before it drains into the ureter.

Each kidney contains many thousands of tiny tubules, called nephrons. The path of the fluid is:

Blood in a ball of capillaries (glomerulus)
   -> Bowman's capsule (filtration)
   -> proximal convoluted tubule (reabsorption)
   -> loop of Henlé (water reabsorbed)
   -> collecting duct (water reabsorbed, controlled by ADH)
   -> pelvis -> ureter -> bladder

Part	Where	What happens
Bowman’s capsule	Cortex	Cup around a knot of capillaries. High pressure forces water, glucose, urea and ions out of the blood into the tubule. Blood cells and large proteins stay in the blood. The fluid is called filtrate.
Proximal convoluted tubule	Cortex	All the glucose, and much of the water and useful ions, is reabsorbed into the blood (selective reabsorption).
Loop of Henlé	Medulla	Dips into the medulla and back. More water is reabsorbed here, so the fluid becomes more concentrated.
Collecting duct	Medulla	Passes through the medulla to the pelvis. The amount of water reabsorbed here is varied.

Urea, made in the liver from excess amino acids, is filtered out and most of it is not reabsorbed, so it leaves in the urine.

B3.3g: How the kidneys maintain water balance

Water balance means that water gained equals water lost. We gain water from drinks, food and respiration. We lose it in urine, sweat, breath and faeces. Sweating and breathing are not adjusted to fix the balance, so the kidneys do it. They vary both the volume of urine and its concentration, and so the amount of water excreted.

• Too much water in the body: a large volume of dilute urine.
• Too little water in the body: a small volume of concentrated urine.

Most of the filtered water is reabsorbed. The fine adjustment is made in the collecting duct, by changing how much water is taken back.

Worked example 2

On a mild day a person gains 1500 cm³ from drinks, 700 cm³ from food and 300 cm³ from respiration. They lose 400 cm³ in sweat, 400 cm³ in breath, 100 cm³ in faeces and the rest in urine.

(a) Total gain = 1500 + 700 + 300 = 2500 cm³. If water balance is kept, total loss is 2500 cm³. Non-urine losses = 400 + 400 + 100 = 900 cm³. Urine = 2500 - 900 = 1600 cm³.

(b) On a hot day the sweat loss rises to 1500 cm³ and the intake stays the same. Urine = 2500 - (1500 + 400 + 100) = 500 cm³.

(c) Percentage fall in urine volume = (1600 - 500) / 1600 x 100 = 68.75%, so about 69%.

B3.3i: ADH and the kidney tubules (Higher Tier only)

ADH (antidiuretic hormone) is the hormone that adjusts the collecting ducts. It is made in the hypothalamus and released into the blood by the pituitary gland. Its target cells are in the walls of the kidney tubules, mainly the collecting duct. For how hormones work in general, see the endocrine system guide.

ADH raises the permeability of the collecting duct walls to water. More water is reabsorbed from the filtrate into the blood, so less is lost.

	High ADH	Low ADH
Permeability of collecting duct to water	Increased	Decreased
Water reabsorbed	More	Less
Urine	Small volume, concentrated	Large volume, dilute

This is a negative feedback loop:

1. Water potential of the blood falls (blood is too concentrated).
2. Receptors in the hypothalamus detect the change.
3. More ADH is released into the blood.
4. Collecting ducts become more permeable, so more water is reabsorbed.
5. Water potential of the blood rises back towards normal.
6. The change is detected and less ADH is released.

If the blood becomes too dilute, the opposite sequence happens: less ADH, less permeable ducts, less water reabsorbed and more dilute urine. The loop works in both directions, which keeps the level close to a set value.

Worked example 3

A volunteer drinks 1000 cm³ of water quickly. Urine was collected each hour. The data are supplied for the example.

Hour after drinking	0 to 1	1 to 2	2 to 3	3 to 4
Urine volume (cm³)	60	290	410	120

(a) Total urine = 60 + 290 + 410 + 120 = 880 cm³. Percentage of the drink excreted = 880 / 1000 x 100 = 88%.

(b) Percentage rise in urine volume from the first to the third hour = (410 - 60) / 60 x 100 = 583% (about 580%).

(c) Explain the pattern. The water made the blood more dilute, so less ADH was released. The collecting ducts became less permeable, so less water was reabsorbed and a large volume of dilute urine was made. As the blood water potential returned to normal, ADH rose again and the volume fell to 120 cm³.

B3.3j: Responses to temperature and osmotic challenges (Higher Tier only)

Each challenge changes the water potential of the blood. The response combines the kidney mechanism (B3.3g to B3.3i) with thirst, a behavioural response triggered by the hypothalamus that makes you drink.

Challenge	Effect on blood	Response
High temperature, sweating, dehydration	Sweat is lost, so the blood becomes more concentrated (lower water potential)	Thirst; more ADH; ducts more permeable; small volume of concentrated urine
Excess water intake	Blood becomes more dilute (higher water potential)	Less ADH; ducts less permeable; large volume of dilute urine
High salt intake	Blood becomes more concentrated	Thirst; more ADH; more water reabsorbed; salt excreted in a small volume of concentrated urine

If the response fails, cells are at risk. Severe dehydration shrinks cells. Drinking far more water than the kidneys can excrete dilutes the body fluids and cells take in water and swell.

Worked example 4

A runner has a mass of 68.0 kg before a 2.5 hour run in hot weather and 66.3 kg after it, having drunk 0.6 dm³ of water. Assume that every kilogram of mass lost is one dm³ of water and ignore other mass changes.

(a) Mass lost = 68.0 - 66.3 = 1.7 kg, which is 1.7 / 68.0 x 100 = 2.5% of body mass.

(b) Water lost = 1.7 + 0.6 = 2.3 dm³ (the drink replaced part of the loss). Sweat rate = 2.3 / 2.5 = 0.92 dm³ per hour.

(c) Explain the runner’s response. Water lost in sweat makes the blood more concentrated. The hypothalamus detects this, causing thirst and the release of more ADH. The collecting ducts become more permeable, so more water is reabsorbed and little, concentrated urine is made.

Common errors

• Writing that cells “explode” in dilute fluid without saying water enters by osmosis. Animal cells burst (lysis) because they have no cell wall.
• Saying ADH is made by the kidney. It is released by the pituitary gland and acts on the kidney.
• Saying ADH makes the kidney “produce more water”. It increases permeability, so more water is reabsorbed.
• Mixing up permeability and the amount of water: high ADH means high permeability and low water loss.
• Describing urine as “concentrated” without saying the volume is small.
• Omitting the feedback step: the rise in blood water potential reduces the release of ADH.

Next, use the revision notes for fast recall, then test yourself with the practice questions. Check your readiness with the free diagnostics.

Official syllabus

OCR GCSE (9-1) Gateway Science Biology A (J247), specification Version 4.0 (August 2026), for first assessment in 2018, published by OCR.