Pearson Edexcel International A Level Biology: Homeostasis, Kidney, Osmoregulation and Gene Switching (YBI11) – Revision Notes

These are condensed revision notes for statements 7.16 to 7.22 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2: homeostasis, the kidney, osmoregulation and gene switching. Topic 7 is International A2 (IA2), and there are no tiers. Full explanations are in the study guide, and exam-style questions are in the practice set. The course hub, checklist and free diagnostics cover the wider course.

Key definitions (7.16 and 7.17)

Term	Meaning
Homeostasis	Maintaining a stable internal environment, as a dynamic equilibrium around a set point
Negative feedback	A change is detected and the response reverses it, so the variable stays within narrow limits
Positive feedback	A change is detected and the response increases it, so the system moves further from the start
Dynamic equilibrium	Values fluctuate a little either side of the set point as corrections switch on and off

Control sequence: stimulus, receptor, coordinator, effector, response, then the response reduces the stimulus.

Positive feedback examples: oxytocin and contractions in labour; sodium ion channels opening during depolarisation of a neurone.

Thermoregulation in steps

1. Central thermoreceptors in the hypothalamus detect blood temperature, and skin thermoreceptors send impulses to it.
2. The hypothalamus coordinates the response.
3. Effectors act: skin arterioles, sweat glands, skeletal muscle.
4. Temperature returns to the set point and the stimulus fades.

Too hot	Too cold
Vasodilation of skin arterioles	Vasoconstriction of skin arterioles
Sweating, with heat taken for evaporation	Shivering releases heat
More heat lost by radiation	Less sweat and a raised metabolic rate

During exercise, extra heat from muscle respiration is removed by these responses, so core temperature rises only slightly.

Kidney structure (7.18)

Level	Structures
Gross	Fibrous capsule, cortex, medulla with pyramids, renal pelvis, ureter, renal artery, renal vein
Microscopic	Bowman’s capsule and glomerulus, then proximal tubule, loop of Henle, distal tubule, collecting duct

Where things are: Bowman’s capsule, glomerulus and both convoluted tubules are in the cortex. Loops of Henle and collecting ducts run through the medulla.

Blood flow: afferent arteriole (wide) to glomerulus to efferent arteriole (narrow) to vasa recta.

Magnification = image size ÷ actual size. Convert both to the same unit before dividing.

Urea and ultrafiltration (7.19)

• Excess amino acids are deaminated in the liver. The amino group becomes ammonia, which is combined with carbon dioxide to form urea. Ornithine cycle details are not required.
• Urea is carried in plasma to the kidneys.
• High hydrostatic pressure in the glomerulus is due to the afferent arteriole being wider than the efferent arteriole.
• Filtration barrier: capillary endothelium with fenestrations, basement membrane, podocytes with slits.
• Passes through: water, urea, glucose, amino acids, ions. Stays in blood: blood cells and large plasma proteins.
• Net filtration pressure = capillary hydrostatic pressure − capsule hydrostatic pressure − plasma opposing pressure.

Reabsorption and the loop of Henle (7.20)

Proximal tubule features: microvilli, many mitochondria, sodium–potassium pumps in the basal membrane.

1. Sodium ions are pumped out of the cell into the blood.
2. Sodium ions enter from the filtrate with glucose or amino acids on co-transporters.
3. Glucose and amino acids leave to the blood by facilitated diffusion.
4. Water follows by osmosis.

Loop of Henle summary:

Part	Permeability	What happens
Descending limb	Permeable to water	Water leaves by osmosis, filtrate becomes more concentrated
Ascending limb	Impermeable to water	Ions leave (active transport in thick part), filtrate becomes more dilute, medulla becomes more concentrated
Collecting duct	Controlled by ADH	Water leaves into the concentrated medulla if permeable

Countercurrent: the two limbs carry fluid in opposite directions, so a small difference at each level adds up to a large gradient. Vasa recta remove water without destroying the gradient. Longer loops give a steeper gradient and more concentrated urine.

ADH feedback loop (7.21)

Plasma too concentrated	Plasma too dilute
Osmoreceptors in hypothalamus shrink and stimulate neurosecretory cells	Osmoreceptors are not stimulated
ADH released from posterior pituitary	Less ADH released
Aquaporins inserted in collecting duct membrane	Aquaporins removed
More water reabsorbed, small volume concentrated urine	Less water reabsorbed, large volume dilute urine
Plasma and blood volume return towards normal	Plasma and blood volume return towards normal

Gene switching (7.22)

	Steroid hormone	Peptide hormone
Solubility	Lipid-soluble	Water-soluble
Crosses the membrane	Yes, by diffusion	No
Receptor	Inside the cell	On the cell surface membrane
Route to the gene	Hormone–receptor complex is itself a transcription factor	Second messenger cascade activates transcription factors

A transcription factor binds a specific DNA sequence. An activator helps RNA polymerase begin transcription, and a repressor prevents it.

Method boxes for calculations

Magnification. Write the formula, convert the actual size to the unit of the image (1 mm = 1000 µm), divide, and give the answer as ×number. To find a real size, rearrange to actual size = image size ÷ magnification.

Percentage reabsorbed. (filtrate volume − urine volume) ÷ filtrate volume × 100. Both volumes must be per the same time and in the same unit.

Percentage change. (new − old) ÷ old × 100. A fall gives a negative answer, and you should state whether the value rose or fell.

Rate. Change in quantity ÷ time. For a rate of temperature rise, divide the temperature change by the minutes taken.

Concentration factor. Urine concentration ÷ plasma concentration of the same substance. It shows how much water has been removed relative to the solute.

Must-know distinctions

• Afferent and efferent: afferent arteriole is wider and carries blood in. Efferent arteriole is narrower and carries blood away, which raises pressure in the glomerulus.
• Filtration and reabsorption: filtration is by size in the glomerulus, while reabsorption is selective and uses carrier proteins in the tubule.
• Descending and ascending limbs: the descending limb loses water, and the ascending limb loses ions.
• Hypothalamus and pituitary: the hypothalamus contains the osmoreceptors and makes ADH. The posterior pituitary releases it.
• Coordinator and effector: the hypothalamus coordinates. Sweat glands, skin arterioles and muscles are effectors.
• Negative and positive feedback: negative reverses a change, and positive amplifies it.
• Vasoconstriction and vasodilation: narrowing of arterioles reduces blood flow to the skin, and widening increases it.

Command words in this topic

• Describe means give the structure or the order of events, without needing reasons.
• Explain needs a cause linked to a consequence, such as “the efferent arteriole is narrow, so the pressure in the glomerulus is high”.
• Suggest is used when you apply the same ideas to an unfamiliar situation, for example a person who cannot make ADH.
• Calculate needs working. Show the numbers substituted into the formula.

Quick self-test

1. State what is meant by dynamic equilibrium.
2. Give one example of positive feedback and explain why it is not used to keep a system steady.
3. Name the three structures that form the filtration barrier in the nephron.
4. A micrograph of a glomerulus is 36 mm wide and the real width is 240 µm. Calculate the magnification.
5. Glomerular capillary hydrostatic pressure is 8.0 kPa, capsule pressure is 2.0 kPa and the opposing plasma protein pressure is 3.0 kPa. Calculate the net filtration pressure.
6. In a day, 180 dm³ of filtrate forms and 1.5 dm³ of urine is produced. Calculate the percentage of the filtrate reabsorbed to one decimal place.
7. Explain why urine contains no glucose in a healthy person.
8. State the role of the ascending limb of the loop of Henle.
9. Explain what happens to urine volume when plasma is too dilute.
10. Plasma urea is 6.0 mmol dm⁻³ and urine urea is 270 mmol dm⁻³. State the factor by which urea is more concentrated in urine than in plasma.
11. Explain why a steroid hormone can switch on a gene without a cell surface receptor.

Answers

1. Values fluctuate slightly either side of a set point as corrections are constantly switched on and off.
2. Oxytocin and contractions in labour, or depolarisation. The change increases, so the system moves further from the set point.
3. Capillary endothelium (with fenestrations), basement membrane, podocytes.
4. 36 ÷ 0.240 = ×150.
5. 8.0 − 2.0 − 3.0 = 3.0 kPa.
6. (180 − 1.5) ÷ 180 × 100 = 99.2%.
7. All the glucose is reabsorbed in the proximal tubule by co-transport and facilitated diffusion.
8. Ions are transported out into the medulla while water stays in, which lowers the water potential of the medulla.
9. Less ADH is released, collecting duct walls are less permeable to water, so a large volume of dilute urine is produced.
10. 270 ÷ 6.0 = 45 times.
11. It is lipid-soluble, so it diffuses through the membrane and binds a receptor inside the cell. The complex acts as a transcription factor.

Where marks are usually lost

• Describing the hypothalamus as the effector instead of the coordinator.
• Writing that blood vessels “move” closer to the skin.
• Saying ultrafiltration is selective for useful substances.
• Giving the wrong limb as water-permeable.
• Not mentioning aquaporins, or saying ADH enters the cell.
• Forgetting that ADH is released from the posterior pituitary, not made there.
• Using different units in magnification.
• Saying peptide hormones “enter the nucleus”.

Official syllabus

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