AQA A-level Biology: Osmoregulation and the Kidney (7402) – 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 follow AQA A-level Biology (7402), section 3.6.4.3, Control of blood water potential, in the AQA AS and A-level Biology specification for first teaching 2015. The section is A-level only. Explanations are in the study guide and the revision notes. Course pages: AQA A-level Biology hub and the printable checklist. The marks shown are for practice and are not an official tariff.

Questions

1. (a) Define osmoregulation. [1] (b) Name the part of the brain that contains osmoreceptors. [1] (c) Name the structure from which ADH is released into the blood. [1]

2. A hiker sweats heavily and does not drink for several hours. Explain how ADH production and action lead to a small volume of concentrated urine. [5]

3. Data for one glomerulus are supplied. Hydrostatic pressure in the capillary is 8.4 kPa, pressure in the capsule space is 1.9 kPa, and the plasma proteins give an inward pressure of 3.9 kPa. (a) Calculate the net outward pressure. [2] (b) Explain why glucose enters the capsule but albumin does not. [2] (c) State the effect on the net outward pressure if the efferent arteriole were widened. [1]

4. A micrograph of a section through a Bowman’s capsule is 38 mm wide on the image. The real width is 190 µm. (a) Calculate the magnification. [2] (b) The gap between a capillary and the capsule wall measures 4 mm on the image. Calculate its real width in µm. [1]

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

6. The following values are supplied as data. Filtrate forms at 0.120 dm³ min⁻¹. The glucose carriers in the proximal convoluted tubules can reabsorb at most 300 mg of glucose per minute. A patient has plasma glucose of 2.6 g dm⁻³. (a) Calculate the mass of glucose filtered per minute, in mg. [2] (b) Calculate how much glucose is left in the filtrate each minute. [1] (c) Explain why this patient may pass a larger volume of urine than usual. [1]

7. Explain how the loop of Henle maintains a gradient of sodium ions in the medulla and why this is needed to produce concentrated urine. [6]

8. The table gives data supplied for three mammals. Urine concentration is shown as a multiple of plasma concentration.

Mammal	Loop of Henle length / mm	Maximum urine concentration (× plasma)
X	3	1.8
Y	9	3.5
Z	21	7.2

(a) Calculate how many times longer the loop of Henle of Z is than that of X. [1] (b) Calculate the percentage increase in maximum urine concentration from Y to Z, to the nearest whole number. [2] (c) Z lives in a desert. Explain how its loop of Henle allows it to save water. [2]

9. An investigation measured urine flow rate at different concentrations of ADH in the blood, in arbitrary units.

ADH concentration	0	1	2	4	8
Urine flow rate / cm³ min⁻¹	11.0	6.5	3.4	1.5	1.2

(a) Calculate the percentage decrease in urine flow from ADH concentration 0 to 4, to the nearest whole number. [2] (b) Describe how urine flow changes as ADH concentration rises. [2] (c) Suggest why there is little change between ADH concentrations 4 and 8. [1]

10. A volunteer drank 1.2 dm³ of pure water. Urine flow rose from 0.9 cm³ min⁻¹ before drinking to a peak of 8.0 cm³ min⁻¹. (a) Calculate the percentage increase in urine flow, to 1 decimal place. [2] (b) Explain how the change in blood water potential after drinking led to this increase. [5]

11. A drug stops the active transport of sodium ions out of the ascending limb of the loop of Henle. Predict and explain the effects on the medulla, on water reabsorption and on the volume and concentration of urine, and explain what would happen to blood water potential if the person did not drink more. [8]

Answers

1. (a) Control of the water potential of the blood [1]. (b) Hypothalamus [1]. (c) Posterior pituitary gland [1]. [3] Examiner insight: Naming where ADH is made (hypothalamus) is a different point from where it is released (posterior pituitary), and answers that blur the two usually lose the mark.

2. Blood water potential falls and osmoreceptors in the hypothalamus shrink as water leaves them by osmosis [1]. Neurosecretory cells send impulses to the posterior pituitary [1], so more ADH is released into the blood [1]. ADH binds to receptors on the cells of the DCT and collecting duct, and more aquaporins are inserted into their membranes, which become more permeable to water [1]. More water leaves the filtrate by osmosis into the medulla tissue fluid and the blood, so urine is small in volume and concentrated [1]. [5] Examiner insight: A strong answer follows the pathway in order, from stimulus to effector to outcome.

3. (a) 8.4 − 1.9 − 3.9 [1] = 2.6 kPa [1]. (b) Glucose molecules are small enough to pass through the gaps in the capillary wall, the basement membrane and between podocytes [1]. Albumin is a large protein and the basement membrane acts as a filter that retains it in the blood [1]. (c) The net outward pressure would fall, because hydrostatic pressure in the glomerulus would be lower [1]. [5] Examiner insight: Show the subtraction with the supplied figures and include the unit; a bare answer with no unit or working puts the marks at risk.

4. (a) Magnification = image size ÷ real size; 38 mm ÷ 0.190 mm [1] = ×200 [1]. (b) 4 ÷ 200 = 0.020 mm = 20 µm [1]. [3] Examiner insight: Put both lengths into one unit before dividing; mixing mm with µm gives an answer wrong by a factor of 1000.

5. Sodium ions are actively transported out of the PCT cells into the tissue fluid, which keeps the sodium ion concentration in the cell low [1]. Sodium ions move into the cell from the filtrate down a concentration gradient [1] through co-transport proteins that carry glucose in with them [1]. Glucose concentration builds up inside the cell and it moves into the tissue fluid by facilitated diffusion [1], then into the capillary. Microvilli and many mitochondria help by giving a large surface area and ATP [1]. [5] Examiner insight: The idea to secure is that the active step moves sodium ions, and the glucose uptake is indirect through co-transport; saying glucose is “actively transported” without this link loses detail marks.

6. (a) 2.6 × 0.120 [1] = 0.312 g min⁻¹ = 312 mg min⁻¹ [1]. (b) 312 − 300 = 12 mg min⁻¹ [1]. (c) The glucose left in the filtrate lowers its water potential, so less water is reabsorbed by osmosis and more urine is produced [1]. [4] Examiner insight: Take care with the unit change from g to mg; the explanation in (c) needs the link between solute, water potential and osmosis.

7. In the ascending limb, sodium ions (with chloride ions) are moved out into the medulla tissue fluid by active transport [1]. The ascending limb wall is impermeable to water, so water cannot follow [1]. The tissue fluid of the medulla therefore has a low water potential [1]. Fluid flows in opposite directions in the two limbs, which multiplies the effect so that the gradient is steepest at the base of the loop [1]. Water leaves the descending limb and then the collecting duct by osmosis, down the water potential gradient [1]. This means the filtrate can be made more concentrated than the blood, so a small volume of concentrated urine can be produced [1]. [6] Examiner insight: Mention both the impermeable wall and the active transport; one without the other leaves the explanation incomplete.

8. (a) 21 ÷ 3 = 7.0 times [1]. (b) (7.2 − 3.5) ÷ 3.5 × 100 [1] = 106% [1]. (c) A long loop sets up a steep sodium ion gradient over a great depth in the medulla [1], so much more water is reabsorbed from the collecting duct by osmosis and the urine is very concentrated [1]. [5] Examiner insight: Back a comparison with figures and units taken from the table; a statement without numbers is weaker evidence.

9. (a) (1.5 − 11.0) ÷ 11.0 × 100 [1] = −86.4%, so a decrease of 86% [1]. (b) As ADH concentration increases, urine flow decreases [1]. The fall is steep at low concentrations and levels off at higher ones, so the relationship is not linear [1]. (c) Almost all the water that can be reabsorbed has been, or the aquaporins in the membranes are almost all in use [1]. [5] Examiner insight: Describe the shape of a curve as well as its direction; a negative correlation alone is only a partial description.

10. (a) (8.0 − 0.9) ÷ 0.9 × 100 [1] = 788.9% [1]. (b) Drinking raises the water potential of the blood [1]. Osmoreceptors in the hypothalamus swell and are stimulated less [1], so less ADH is released from the posterior pituitary [1]. Fewer aquaporins are in the membranes of the DCT and collecting duct, so these are less permeable to water [1]. Less water is reabsorbed by osmosis, so a large volume of dilute urine is produced and blood water potential returns to normal [1]. [7] Examiner insight: Give the percentage to the number of decimal places asked, and finish the explanation by showing the negative feedback outcome.

11. Sodium ions are no longer pumped out of the ascending limb, so fewer remain in the medulla tissue fluid [1]. The water potential of the medulla is therefore higher than normal and the gradient is smaller [1]. Less water leaves the descending limb [1] and the collecting duct, because the water potential gradient between filtrate and tissue fluid is smaller [1]. More water stays in the filtrate, so a larger volume of dilute urine is produced [1]. Water loss without replacement lowers the blood water potential [1]. Osmoreceptors shrink and more ADH is released [1], but ADH has little effect because the gradient needed to draw water out of the collecting duct is not present [1]. [8] Examiner insight: Structure the answer as a chain of cause and effect from the site of action to the urine to the blood; each link is a separate point.

Where marks are usually lost

• Saying ADH is made in or released by the “pituitary” without specifying the posterior pituitary, and giving the hypothalamus no role.
• Writing that water is “actively transported” or “pumped” rather than moving by osmosis.
• Forgetting that the ascending limb is impermeable to water, so the sodium ion gradient is left unexplained.
• Writing “the kidney filters out glucose” instead of explaining that glucose is filtered and then reabsorbed.
• Mixing up units: mm and µm in magnification, g and mg in glucose load.
• Giving a percentage change with no sign or with the wrong base value, instead of dividing by the original.
• Describing a trend in a table as simply “goes down” without quoting data values.

Next steps

• Go back over the revision notes and the study guide.
• Tick off the topic on the printable checklist and visit the course hub.
• Take any of the free 10-minute diagnostics to find your next weak spot.
• Book a free trial class if you would like a teacher to go through your answers.

Official syllabus

AQA, AS and A-level Biology specification (7401 and 7402), for first teaching 2015, section 3.6.4.3 Control of blood water potential (A-level only). Published by AQA.