Pearson Edexcel International A Level Biology: Homeostasis, Kidney, Osmoregulation and Gene Switching (YBI11) – 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 7.16 to 7.22 (feedback, homeostasis, the kidney, osmoregulation and gene switching) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2. Topic 7 is International A2 (IA2) and there are no tiers. Learn the content in the study guide and revision notes first. The course hub, checklist and free diagnostics show the wider course.

Questions

1. (a) Explain what is meant by negative feedback. [2] (b) Explain why positive feedback is not used to maintain a steady internal environment. [1]

2. (a) Define homeostasis. [1] (b) Explain how exercise disturbs the internal environment and why this is described as a dynamic equilibrium. [2] (c) State why a small rise in core temperature during exercise is acceptable but a large rise is not. [1]

3. A person walks from a warm room into a cold outdoor space. Describe how the hypothalamus and effectors bring about a response to keep core temperature near its set point. [5]

4. (a) State the region of the kidney that contains the Bowman’s capsule and the region through which most loops of Henle and collecting ducts run. [2] (b) Name the tube that carries urine from the kidney to the bladder. [1] (c) Name the blood vessel that carries blood into the glomerulus. [1]

5. A micrograph shows a section through a kidney tubule. The image of one tubule is 52 mm wide and its real width is 130 µm. A second structure measures 18 mm on the image. (a) Calculate the magnification. [2] (b) Calculate the real width of the second structure in µm. [1]

6. (a) Describe how urea is produced in the liver from excess amino acids. [3] (b) A person has plasma urea of 4.0 mmol dm⁻³ and urine urea of 220 mmol dm⁻³. Calculate the factor by which urine urea concentration exceeds plasma urea concentration. [1]

7. The values below are supplied as data. Hydrostatic pressure in a glomerular capillary is 9.0 kPa. Hydrostatic pressure in the Bowman’s capsule is 2.5 kPa. The plasma proteins give an opposing pressure of 3.8 kPa. (a) Calculate the net filtration pressure. [2] (b) Explain how a high hydrostatic pressure is produced in the glomerulus. [1] (c) Describe the three layers that separate blood from the space in the capsule. [3]

8. The glomerular filtration rate of a person is 110 cm³ min⁻¹ and plasma glucose is 6.0 mmol dm⁻³. (a) Calculate the glucose filtered each minute. [2] (b) The urine contains no glucose. Calculate the amount reabsorbed in one hour. [1] (c) Explain how glucose is reabsorbed in the proximal convoluted tubule. [3]

9. The table gives data supplied for three mammals.

Mammal	Loop of Henle length (mm)	Maximum urine concentration (mOsm kg⁻¹)
P	6	1200
Q	10	2100
R	16	3600

(a) Calculate the maximum urine concentration per millimetre of loop for mammal R. [1] (b) Describe the relationship shown by the data. [2] (c) Explain how the loop of Henle acts as a countercurrent multiplier. [4]

10. A volunteer drinks no water for 18 hours. Data for the volunteer are shown.

Measurement	Normal hydration	After 18 hours
Urine flow (cm³ min⁻¹)	1.5	0.5
Urine concentration (mOsm kg⁻¹)	400	1100
Plasma ADH (pmol dm⁻³)	2.0	6.5

(a) Calculate the percentage change in urine flow. [2] (b) Explain how the change in ADH produces the change in urine flow. [4] (c) Explain how this is an example of negative feedback. [2]

11. (a) Explain what is meant by a transcription factor. [2] (b) Describe how a steroid hormone such as oestrogen switches on a gene. [3] (c) Explain why a peptide hormone must act differently. [2]

12. A runner loses sweat at a mean rate of 0.8 dm³ per hour for 2.5 hours. The runner’s total body water is 42 dm³. (a) Calculate the total volume of sweat lost and express it as a percentage of body water, to one decimal place. [3] (b) Explain how sweating lowers body temperature. [2] (c) Explain how the runner’s kidneys respond to this water loss. [4]

Answers

1. (a) A change away from the set point is detected by receptors [1]. The response reverses the change and returns the variable to the set point [1]. [2] (b) Positive feedback increases the change, so the variable would move further from the set point [1]. [1] Examiner insight: In (a), say what the response does to the original change. Writing only “it keeps things the same” does not show that the change is reversed.

2. (a) Maintaining a stable internal environment within narrow limits [1]. [1] (b) Respiring muscle cells release extra heat and carbon dioxide and use glucose and oxygen, which moves the values away from the set point [1]. Corrective responses are constantly switched on and off, so values fluctuate either side of the set point [1]. [2] (c) A large rise would reduce enzyme activity as proteins denature, while a small rise is within the range that enzymes tolerate [1]. [1] Examiner insight: In (c), link the temperature rise to enzymes instead of saying only that the body “overheats”.

3. Central thermoreceptors in the hypothalamus detect lower blood temperature, and skin receptors send impulses to it [1]. The hypothalamus coordinates the response and sends impulses to effectors [1]. Skin arterioles vasoconstrict, so the skin receives less blood and loses less heat to the air [1]. Skeletal muscle shivers, releasing heat from respiration [1]. As temperature rises, the stimulus is reduced and the responses stop, which is negative feedback [1]. [5] Examiner insight: Name the hypothalamus as the coordinator and the muscles or arterioles as effectors. Answers that say the hypothalamus itself shivers or narrows vessels are describing the wrong part.

4. (a) Cortex [1]. Medulla [1]. [2] (b) Ureter [1]. [1] (c) Afferent arteriole [1]. [1] Examiner insight: Give the specific vessel name. “Renal artery” is the vessel into the kidney, whereas the afferent arteriole leads to a single glomerulus.

5. (a) Magnification = image size ÷ actual size = 52 mm ÷ 0.130 mm [1] = ×400 [1]. [2] (b) 18 mm ÷ 400 = 0.045 mm = 45 µm [1]. [1] Examiner insight: Show the unit conversion, since dividing 52 by 130 directly mixes mm and µm.

6. (a) Excess amino acids are deaminated in the liver, removing the amino group [1]. The amino group becomes ammonia, which is combined with carbon dioxide to form urea [1]. The remaining carbon skeleton is respired or converted to carbohydrate or fat [1]. [3] (b) 220 ÷ 4.0 = 55 times [1]. [1] Examiner insight: The ornithine cycle is not required. Cover deamination, ammonia to urea, and the fate of the carbon skeleton.

7. (a) 9.0 − 2.5 − 3.8 [1] = 2.7 kPa [1]. [2] (b) The afferent arteriole is wider than the efferent arteriole, so blood enters faster than it leaves [1]. [1] (c) Capillary endothelium has gaps (fenestrations) [1]. A basement membrane of glycoproteins acts as a filter [1]. Podocytes with slits form the inner layer of the capsule [1]. [3] Examiner insight: For (b), compare the two arterioles. Saying only “the pressure is high” restates the question.

8. (a) Filtered = 6.0 × 0.110 [1] = 0.66 mmol min⁻¹ [1]. [2] (b) 0.66 × 60 = 39.6 mmol [1]. [1] (c) Sodium–potassium pumps in the basal membrane move sodium ions out of the cell, lowering sodium concentration inside it [1]. Sodium ions enter from the filtrate with glucose on a co-transporter, so glucose enters against its concentration gradient [1]. Glucose leaves to the blood by facilitated diffusion, and the cells have microvilli and many mitochondria [1]. [3] Examiner insight: In (a), convert 110 cm³ to 0.110 dm³ before multiplying by a concentration in mmol per dm³.

9. (a) 3600 ÷ 16 = 225 mOsm kg⁻¹ per mm [1]. [1] (b) As the loop gets longer, maximum urine concentration increases [1]. R’s loop is about 2.7 times that of P and its urine is 3.0 times as concentrated [1]. [2] (c) Ions are transported out of the ascending limb, which is impermeable to water, so the medulla has a low water potential [1]. Water leaves the descending limb by osmosis, so the filtrate becomes more concentrated towards the bend [1]. The flow in the two limbs is in opposite directions, so a small difference at each level is multiplied into a steep gradient [1]. The collecting duct passes through this gradient, so water can leave by osmosis [1]. [4] Examiner insight: Use figures from the table when describing data. A statement such as “R is longest” does not show a relationship, so quote a comparison.

10. (a) (0.5 − 1.5) ÷ 1.5 × 100 [1] = −66.7% [1]. [2] (b) Osmoreceptors in the hypothalamus detect the low water potential of the plasma [1]. ADH made in hypothalamic cells passes to the posterior pituitary, which secretes it into the blood [1]. ADH binds to collecting duct cells, which insert aquaporins in the membrane [1]. More water leaves by osmosis into the concentrated medulla, so a small volume of concentrated urine forms [1]. [4] (c) Reabsorbing more water makes plasma less concentrated and raises blood volume [1]. Osmoreceptors are less stimulated, ADH release falls, and the system returns towards its normal state [1]. [2] Examiner insight: Give the sign in (a). In (b), keep the sequence in order: detector, hormone release, target cells, effect.

11. (a) A protein that binds to a specific DNA sequence near a gene [1]. It affects whether RNA polymerase can begin transcription, switching the gene on or off [1]. [2] (b) Oestrogen is lipid-soluble and diffuses through the cell surface membrane [1]. It binds a receptor inside the cell and the hormone–receptor complex enters the nucleus [1]. The complex binds DNA as a transcription factor, so transcription of the gene starts [1]. [3] (c) A peptide hormone is water-soluble and cannot cross the phospholipid bilayer, so it binds a receptor on the cell surface [1]. This triggers a second messenger, such as cyclic AMP, that activates transcription factors inside the cell [1]. [2] Examiner insight: Contrast the two hormone types by where the receptor is. A peptide hormone does not enter the nucleus, so do not write that it does.

12. (a) 0.8 × 2.5 = 2.0 dm³ [1]. 2.0 ÷ 42 × 100 [1] = 4.8% [1]. [3] (b) Sweat evaporates from the skin [1] and takes latent heat from the body, lowering temperature [1]. [2] (c) The plasma becomes more concentrated and osmoreceptors in the hypothalamus detect it [1]. More ADH is released from the posterior pituitary [1]. The collecting duct becomes more permeable because of aquaporins, so more water is reabsorbed [1]. A small volume of concentrated urine is produced, conserving water [1]. [4] Examiner insight: Make the link explicit: water loss raises plasma concentration, which is the stimulus.

Where marks are usually lost

• Not stating that negative feedback reverses the change.
• Naming the hypothalamus when the effector is asked for.
• Not converting units before calculating magnification or glucose.
• Writing that ultrafiltration is selective for useful substances.
• Mixing up which limb of the loop is permeable to water.
• Saying ADH is made in the pituitary gland.
• Omitting aquaporins when explaining the permeability change.
• Saying a peptide hormone enters the nucleus.
• Quoting no figures when describing a table.

Next steps

• Revision notes
• Study guide
• Edexcel A Level Biology hub
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Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology (YBI11) specification, Issue 2, published by Pearson Education Limited.