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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover statements B3.3f to B3.3j of module B3.3 in OCR GCSE (9-1) Gateway Science Biology A (J247), specification Version 4.0 (August 2026), for first assessment in 2018. Questions labelled Higher Tier only draw on B3.3i (ADH) or B3.3j (responses to challenges), which are in bold in the specification. The others suit both tiers. Learn the content from the study guide or the revision notes. Course pages: OCR GCSE Biology hub | checklist | all free 10-minute diagnostics.

Questions

1. A red blood cell is placed in a salt solution that has a lower water potential than its cytoplasm. Describe and explain what happens to the cell. [3]

2. Cultured animal cells were placed in three salt solutions. The table shows their volume after 30 minutes as a percentage of the starting volume.

Salt solution (%)	0.3	0.9	1.8
Cell volume (% of start)	131	100	84

(a) Calculate the percentage change in volume in the 1.8% solution. [2] (b) State which solution has the same water potential as the cells. [1] (c) Explain the result in the 0.3% solution. [2]

3. The kidney tubule has four named parts. (a) Put these in the order in which filtrate passes through them: collecting duct, Bowman’s capsule, loop of Henlé, proximal convoluted tubule. [2] (b) Describe what happens in the Bowman’s capsule. [2]

4. Name the structure described in each case. (a) It carries blood into the kidney. [1] (b) The outer region of the kidney, which contains the Bowman’s capsules. [1] (c) The tube linking the kidney to the bladder. [1] (d) The hollow centre of the kidney where urine collects. [1]

5. On a mild day a person gains 1800 cm³ of water from drinks, 600 cm³ from food and 200 cm³ from respiration. They lose 300 cm³ in breath, 400 cm³ in sweat and 100 cm³ in faeces. The rest is lost in urine and the person stays in water balance. (a) Calculate the volume of urine. [2] (b) On a hot day, sweat loss is 1900 cm³ and everything else is unchanged. Calculate the volume of urine. [1] (c) Calculate the percentage decrease in urine volume from the mild day to the hot day. [1] (d) Describe how the urine differs in volume and concentration on the hot day. [1]

6. (a) Name two components of blood that cannot enter the Bowman’s capsule. [2] (b) State what happens to the glucose that enters the filtrate. [1] (c) Urea is found in urine. Explain why. [1]

7. (Higher Tier only) Describe the effect of ADH on the kidney tubules and explain why this is an example of negative feedback. [4]

8. (Higher Tier only) The table shows the ADH level in the blood and the volume of urine produced in one hour in four situations.

ADH level (arbitrary units)	2	5	9	14
Urine volume (cm³ per hour)	420	240	110	50

(a) Describe the relationship shown by the data. [2] (b) Calculate the percentage fall in urine volume between ADH levels 2 and 14. [2] (c) Explain the relationship. [2]

9. (Higher Tier only) A hiker has a mass of 72.0 kg at the start of a hot day and 70.4 kg at the end, having drunk very little. (a) Calculate the percentage of body mass lost. [2] (b) Explain how the hiker’s body responds to the loss of water, including thirst and the kidneys. [4]

10. (Higher Tier only) A volunteer drinks 2.0 dm³ of water in 30 minutes. Over the next 4 hours the volunteer produces 1.25 dm³ of urine. (a) Calculate the percentage of the water drunk that was lost as urine. [2] (b) Explain why the urine was a large volume and dilute. [3] (c) Explain why drinking much more water than this, very quickly, could harm the body. [2]

11. (Higher Tier only) After a very salty meal a person feels thirsty and passes only a small volume of urine. Explain these responses. [4]

12. (Higher Tier only) After exercise, a dehydrated volunteer drank 1 dm³ of one of three drinks. The urine collected over the next 3 hours is shown.

Drink	Water	Low-salt drink	High-salt drink
Urine volume (cm³)	480	330	190

(a) Calculate how much less urine was produced after the high-salt drink than after water, as a percentage of the volume after water. [2] (b) Explain why the high-salt drink produced the least urine. [2] (c) A student concludes that the high-salt drink is best for athletes. Give two reasons why this conclusion is not safe. [2]

Answers

1. Water moves out of the cell by osmosis [1] from the higher water potential inside the cell to the lower water potential outside, through the partially permeable membrane [1]. The cell shrinks [1]. [3] Examiner insight: Say which way the water moves and name the process. “The cell shrinks” alone describes the result but does not explain it.

2. (a) 84 - 100 = -16 [1]. The volume fell by 16% [1]. [2] (b) 0.9% [1] because the volume did not change. [1] (c) The outside solution is more dilute than the cells, so water moves in by osmosis [1] and the cells swell to 131% of their starting volume [1]. [2] Examiner insight: Give a sign or a word such as fall or rise with a percentage. Dividing by the original volume, not the final one, is the usual method.

3. (a) Bowman’s capsule, proximal convoluted tubule [1], loop of Henlé, collecting duct [1]. [2] (b) Blood is under high pressure and small molecules such as water, glucose, urea and ions are forced out into the capsule [1]. Blood cells and large proteins stay in the blood [1]. [2] Examiner insight: Name what passes through and what does not. “Waste is filtered” is too vague for two marks.

4. (a) Renal artery [1]. (b) Cortex [1]. (c) Ureter [1]. (d) Pelvis [1]. [4] Examiner insight: Do not confuse the ureter, which leaves the kidney, with the urethra, which leaves the bladder. Spell each term clearly.

5. (a) Total gain = 1800 + 600 + 200 = 2600 cm³ [1]. Urine = 2600 - (300 + 400 + 100) = 1800 cm³ [1]. [2] (b) 2600 - (300 + 1900 + 100) = 300 cm³ [1]. [1] (c) (1800 - 300) / 1800 x 100 = 83.3% [1]. [1] (d) The urine is a smaller volume and more concentrated [1]. [1] Examiner insight: Show the total gain and the total loss in your working, so that each step of the method is visible and checkable.

6. (a) Any two of: red blood cells, white blood cells, platelets, large proteins [1] [1]. [2] (b) All of it is reabsorbed into the blood (in the proximal convoluted tubule) [1]. [1] (c) Most of the urea in the filtrate is not reabsorbed, so it passes out in the urine [1]. [1] Examiner insight: Plasma proteins are not a cell, so list the two items carefully. Glucose “removed” is wrong because it is taken back into the blood.

7. ADH raises the permeability of the collecting duct walls to water [1], so more water passes back into the blood [1]. This raises the water potential of the blood back towards normal [1]. The change then reduces the release of ADH, so the response reverses the original change [1]. [4] Examiner insight: Two marks are for the effect, two for the feedback. State that the correction switches the response off.

8. (a) As ADH level increases, urine volume decreases [1]. Quote data, for example 420 cm³ per hour at ADH 2 and 50 cm³ per hour at ADH 14 [1]. [2] (b) 420 - 50 = 370 [1]. 370 / 420 x 100 = 88% [1]. [2] (c) ADH increases how easily water crosses the collecting duct [1], so more is taken back and less leaves as urine [1]. [2] Examiner insight: A description of a trend is stronger with two quoted values and units. Do not explain in part (a) when the command word is describe.

9. (a) 72.0 - 70.4 = 1.6 kg [1]. 1.6 / 72.0 x 100 = 2.2% [1]. [2] (b) Sweating lowered the water potential of the blood [1]. The hypothalamus detects this and causes thirst, so the hiker drinks [1]. More ADH is released [1]. The collecting ducts become more permeable, so more water is reabsorbed and a small volume of concentrated urine is produced [1]. [4] Examiner insight: Thirst is a response too, so a complete answer links the sensation of thirst to the kidney response.

10. (a) 1.25 / 2.0 x 100 [1] = 62.5% [1]. [2] (b) The blood became more dilute (higher water potential) [1]. Less ADH was released [1]. The collecting ducts were less permeable, so less water was reabsorbed [1]. [3] (c) The body fluids would become too dilute [1], so water would enter cells by osmosis and they would swell [1]. [2] Examiner insight: Explain the chain in order. Saying only “less ADH” without the change in permeability leaves the mechanism incomplete.

11. The salt makes the blood more concentrated (lower water potential) [1]. The hypothalamus triggers thirst, so the person drinks and dilutes the blood [1]. More ADH is released and the collecting ducts become more permeable [1]. More water is reabsorbed, so a small volume of concentrated urine, containing the excess salt, is made [1]. [4] Examiner insight: Link each response to its cause. A list of “thirst, ADH, small urine” without the chain gains less credit.

12. (a) 480 - 190 = 290 [1]. 290 / 480 x 100 = 60% [1]. [2] (b) The salt made the blood more concentrated, so more ADH was released [1] and more water was reabsorbed, leaving less urine [1]. [2] (c) Two of: only one volunteer, with no repeats [1]; urine volume does not show how hydrated the body is, because blood water level or body mass was not measured [1]. [2] Examiner insight: Evaluation answers need a limitation of the evidence, not just a restatement of the result.

Where marks are usually lost

• Writing that cells “burst” or “shrink” without naming osmosis and the direction of water movement.
• Calculating a percentage change using the new value as the divisor.
• Giving a percentage with no sign, direction or the word fall or rise.
• Describing the urine as “concentrated” without mentioning the small volume, or the reverse.
• Saying ADH makes the kidney “hold on to” water without stating that the permeability of the collecting duct changes.
• Leaving out the final feedback step, where the blood returning to normal reduces ADH release.
• Forgetting thirst in answers about dehydration or high salt intake.
• Mixing up the ureter and urethra, or the cortex and medulla.
• Using vague terms such as “waste” where the question needs a named substance such as urea.

Next steps

• Reread the revision notes and the study guide.
• Course pages: OCR GCSE Biology hub and checklist.
• Find weak spots with the all free 10-minute diagnostics.
• Want a teacher to go through your answers? Book a free trial class.

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.