OCR A Level Biology A: The kidney, osmoregulation and renal failure (H420) – 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 learning outcomes 5.1.2(c) to 5.1.2(f) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. Learn the content first in the study guide or the revision notes. Course pages: the OCR A Level Biology hub and the printable checklist. Where a question gives data, use those values. Not sure where to start? Try the free diagnostics.

Questions

1. Describe how the structure of the renal corpuscle allows ultrafiltration of blood plasma. [4]

2. The question supplies these pressures for a glomerulus. Hydrostatic pressure in the capillary is 6.8 kPa, hydrostatic pressure in the capsule is 2.1 kPa, and plasma proteins pull water in with an effect equal to 3.2 kPa. (a) Work out the net pressure driving fluid out of the capillary. [2] (b) Explain what would happen to filtration if the afferent arteriole constricted, and why. [3]

3. Explain how glucose in the filtrate is reabsorbed in the proximal convoluted tubule. [5]

4. Filtrate forms at 0.090 dm³ min⁻¹. The glucose carriers in the proximal tubules can reabsorb at most 270 mg of glucose per minute. A patient has a plasma glucose concentration of 4.0 g dm⁻³. (a) Find how much glucose is filtered each minute, in mg. [2] (b) Find how much glucose is left in the filtrate. [1] (c) Explain why this patient also produces an unusually large volume of urine. [2]

5. The table shows supplied data for three mammals. The data are not measurements of real species.

Mammal	Length of loop of Henle (mm)	Maximum urine concentration (mOsm kg⁻¹)
P	6	1 100
Q	12	2 900
R	22	5 400

(a) Calculate how many times more concentrated the urine of R is than that of P. [1] (b) Describe the relationship shown. [1] (c) Explain how a longer loop of Henle allows more concentrated urine, and suggest which mammal lives in the driest habitat. [3]

6. A person eats a very salty meal and drinks no water for several hours. Explain how the body responds to restore the water potential of the blood. [6]

7. A patient with kidney disease has a plasma creatinine concentration of 0.060 mg cm⁻³. Urine creatinine is 0.90 mg cm⁻³ and urine flow is 1.0 cm³ min⁻¹. Creatinine is filtered and not reabsorbed. (a) Calculate the patient’s GFR. [2] (b) The healthy value for this person was 120 cm³ min⁻¹. Calculate the present GFR as a percentage of this. [1] (c) Explain why the blood potassium concentration of this patient may rise, and why this is dangerous. [3]

8. During haemodialysis, blood enters a dialyser at 250 cm³ min⁻¹. Its potassium concentration is 7.2 mmol dm⁻³ on entering and 4.4 mmol dm⁻³ on leaving. (a) Calculate the mass of potassium removed during a 4 hour session, in mmol. [3] (b) Explain why dialysis fluid and blood flow in opposite directions. [2] (c) Explain why the dialysis fluid contains glucose at the normal plasma concentration. [2]

9. A 34-year-old with kidney failure is offered haemodialysis now or a place on a transplant waiting list. Evaluate the two treatments. [8]

10. A home pregnancy test strip uses monoclonal antibodies. (a) Explain how the strip shows a positive result. [4] (b) Suggest why monoclonal antibodies are used rather than a mixture of antibodies. [1] (c) State how the same technology can be used to screen an athlete for a banned drug. [1]

11. A student dissected a pig kidney and examined stained sections. (a) The cortex is 9 mm thick in the kidney and 27 mm thick on the student’s drawing. Calculate the magnification. [2] (b) State two safety precautions for the dissection. [2] (c) A glomerulus measures 56 mm on a micrograph with magnification ×400. Calculate its actual diameter in µm. [2] (d) Describe how the student could tell a proximal tubule from a distal tubule in a stained section. [2]

Answers

1. The efferent arteriole is narrower than the afferent one, so pressure builds up inside the glomerulus [1]. The capillary endothelium has gaps (fenestrations) so plasma can leave but blood cells cannot [1]. The basement membrane acts as a filter that holds back large proteins [1]. Podocytes of Bowman’s capsule have slits between their pedicels so filtrate passes into the capsule [1]. [4] Examiner insight: Answers that name the three layers and give the function of each are more convincing than “it is leaky”.

2. (a) Net outward pressure = 6.8 − 2.1 − 3.2 [1] = 1.5 kPa [1]. (b) Less blood enters the glomerulus, so capillary pressure falls and filtration decreases [1]. Net pressure would fall towards zero [1] because the capsule pressure and plasma protein effect stay the same [1]. [5] Examiner insight: Show the subtraction and include the unit; “filtration falls” needs the cause (lower hydrostatic pressure) to be convincing.

3. Sodium ions are actively transported out of the cell into the blood by the sodium-potassium pump [1], which lowers the sodium ion concentration inside the cell [1]. Sodium ions diffuse in from the filtrate through co-transporter proteins [1], carrying glucose in against its concentration gradient [1]. Glucose then leaves the cell by facilitated diffusion into the blood [1]. [5] Examiner insight: Say which step is active and which is passive; “glucose is actively transported” with no mention of sodium is a common half-answer.

4. (a) Amount filtered = 4.0 × 0.090 [1] = 0.36 g min⁻¹ = 360 mg min⁻¹ [1]. (b) 360 − 270 = 90 mg min⁻¹ [1]. (c) The remaining glucose lowers the water potential of the filtrate in the tubule [1], so less water is reabsorbed by osmosis and more is excreted [1]. [5] Examiner insight: Convert g to mg before comparing with the carrier limit; many slips come from mixing units.

5. (a) 5 400 ÷ 1 100 = 4.9 times [1]. (b) The longer the loop, the more concentrated the urine (positive correlation) [1]. (c) The ascending limb pumps ions into the medulla but water cannot follow it out; a longer loop does this over a greater length, so the medulla has a lower water potential [1]. The collecting duct passes through it, so more water leaves by osmosis down a steeper gradient [1]. Mammal R lives in the driest habitat because it conserves most water [1]. [5] Examiner insight: Linking length to gradient and gradient to water leaving the collecting duct gains more than “longer loops reabsorb more”.

6. The salt lowers the water potential of the blood, which osmoreceptors in the hypothalamus detect [1]. Nerve impulses cause the posterior pituitary gland to release more ADH [1]. The blood carries ADH to the collecting duct, where it attaches to receptors in the wall cells [1]. More aquaporins are placed in the cell-surface membrane, so the walls become more permeable to water [1]. More water is reabsorbed by osmosis, producing a small volume of concentrated urine [1]. The blood water potential rises towards normal, so less ADH is released (negative feedback) [1]. [6] Examiner insight: Answers that follow the sequence stimulus, detector, hormone, target, effect are easiest to credit; put the steps in order and name the organs.

7. (a) GFR = (0.90 × 1.0) ÷ 0.060 [1] = 15 cm³ min⁻¹ [1]. (b) 15 ÷ 120 × 100 = 12.5% [1]. (c) Fewer nephrons filter the blood, so less potassium is excreted and it accumulates [1]. A high potassium concentration disrupts the electrical activity of muscle and nerve cells [1], so the heartbeat can become irregular [1]. [6] Examiner insight: Give a GFR with its units; name the organ affected (the heart) rather than saying “it is harmful”.

8. (a) Potassium removed per minute = 0.250 × (7.2 − 4.4) [1] = 0.70 mmol; for 240 min, 0.70 × 240 [1] = 168 mmol [1]. (b) Counter-current flow keeps blood always next to dialysis fluid that has a lower concentration of waste [1], so a diffusion gradient is maintained along the whole membrane [1]. (c) Glucose in the blood is at the same concentration as in the fluid, so there is no net diffusion [1], so the patient does not lose glucose [1]. [7] Examiner insight: Convert cm³ to dm³ before using mmol dm⁻³; state the time conversion in the working.

9. Haemodialysis: available immediately without a donor [1]; removes urea and excess ions but only during sessions, so levels fluctuate [1]; needs several hours several times a week, limiting work and travel, and the blood access site can become infected [1]. Transplant: the new kidney filters continuously, allowing a more normal diet and fluid intake [1]; needs tissue-type matching and a donor, so there is a wait [1]; risk of rejection, so immunosuppressant drugs are needed [1], which raise the risk of infection [1]. A balanced conclusion is that dialysis is a bridge and a transplant is preferred long-term for a young patient [1]. [8] Examiner insight: An evaluation needs points for both treatments and a stated judgement; a list of facts for one treatment only does not answer the question.

10. (a) Urine moves along the strip and meets mobile monoclonal antibodies to hCG on coloured beads [1]. If hCG is present, it binds to these antibodies [1]. The complex is captured by immobilised antibodies at the test line, which turns coloured [1]. Unbound beads are captured at the control line, so two lines mean positive [1]. (b) A monoclonal antibody is specific to one antigen, so it binds only hCG [1]. (c) Use monoclonal antibodies specific to the drug in an immunoassay on a urine sample [1]. [6] Examiner insight: Explain what each line means; a control line shows the test has worked, not that the person is pregnant.

11. (a) Magnification = image ÷ actual [1] = 27 ÷ 9 = ×3 [1]. (b) Any two: wear gloves and eye protection; cut away from the body with a sharp scalpel on a board; disinfect the bench and wash hands afterwards [1] [1]. (c) Actual = 56 ÷ 400 [1] = 0.14 mm = 140 µm [1]. (d) The proximal tubule has a brush border of microvilli on its cells [1] and the cells appear denser because they contain many mitochondria; the distal tubule has fewer microvilli and a wider, clearer lumen [1]. [8] Examiner insight: Convert mm to µm explicitly in a magnification answer, and give a feature that can be seen under the microscope rather than a function.

Where marks are usually lost

• Naming the kidney or “the nephron” without the glomerulus, Bowman’s capsule or the collecting duct.
• Describing ultrafiltration as selective by carriers.
• Missing the sodium co-transport step in glucose reabsorption.
• Muddling the hypothalamus (detection) and the posterior pituitary (release).
• Leaving out aquaporins in an ADH answer.
• Missing unit conversions between g and mg, cm³ and dm³, or mm and µm.
• Evaluating only one treatment, or not reaching a judgement.
• Describing the control line of a pregnancy test as the result.

Next steps

• Revisit the revision notes and the study guide.
• Tick off outcomes on the printable checklist and see the course hub.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Biology A (H420), specification Version 4.1 (April 2026), first assessment 2025, published by OCR. Section 5.1.2, outcomes (c) to (f).