HomeLearning HubA Level BiologyA2 14: Homeostasis
A2 14

Homeostasis

A Level · Topic 14 · Papers 4 and 5

🎯What you need to be able to do

  • Explain homeostasis and its importance, and describe it in terms of stimuli, receptors, coordination systems, effectors and negative feedback.
  • State that urea is produced in the liver by deamination of excess amino acids.
  • Describe the structure of the kidney and identify the parts of a nephron and its blood vessels.
  • Describe ultrafiltration in the Bowman’s capsule and selective reabsorption in the proximal convoluted tubule, and relate structure to function.
  • Describe the roles of the hypothalamus, posterior pituitary, ADH, aquaporins and collecting ducts in osmoregulation.
  • Describe cell signalling using the control of blood glucose by glucagon, including G proteins, adenylyl cyclase, cAMP, protein kinase A and amplification.
  • Explain negative feedback control of blood glucose by insulin and glucagon, and the principles of test strips and biosensors.
  • Explain stomatal opening and closing, guard cell function, daily rhythms, and the role of abscisic acid and calcium ions.

📚The biology

The principle

Homeostasis is the maintenance of a constant internal environment within narrow limits, despite changes outside. It matters because cells work only within a limited range of temperature, pH and solute concentration — enzymes denature or work slowly outside their optimum, and water potential changes damage cells by osmosis.

The general scheme, and the vocabulary questions expect:

\[ \text{stimulus} \rightarrow \text{receptor} \rightarrow \text{coordination system} \rightarrow \text{effector} \rightarrow \text{response} \]

The coordination system is the nervous system or the endocrine system; effectors are muscles or glands. Control is by negative feedback: a deviation from the set point triggers a response that reverses it, returning the value towards normal. The value therefore fluctuates around the set point rather than staying perfectly fixed, and questions that ask you to describe a graph want that fluctuation described, not denied.

Excretion of urea

Excess amino acids cannot be stored. In the liver they are deaminated — the amino group is removed — forming ammonia, which is highly toxic and is converted to the much less toxic urea. Urea is carried in the blood to the kidneys and excreted. The remaining carbon-containing part is respired or converted to carbohydrate or fat.

Kidney structure

Outer fibrous capsule, then the cortex, the medulla, and the renal pelvis leading into the ureter. Branches of the renal artery and renal vein run through it.

A nephron consists of: glomerulus, Bowman’s capsule, proximal convoluted tubule, loop of Henle, distal convoluted tubule and collecting duct. Bowman’s capsule and both convoluted tubules lie in the cortex; the loop of Henle and collecting duct run down into the medulla.

Ultrafiltration

Blood enters the glomerulus through an afferent arteriole that is wider than the efferent arteriole leaving it. The narrower exit creates a high hydrostatic pressure in the glomerular capillaries, which forces fluid out through the filtration barrier.

The barrier has three layers: the capillary endothelium, which is perforated by pores; the basement membrane, a mesh of glycoproteins which is the actual molecular sieve; and the podocytes of the capsule wall, whose foot processes leave filtration slits between them.

Small molecules — water, glucose, amino acids, urea, mineral ions — pass through into the capsule, forming the glomerular filtrate. Blood cells and plasma proteins are too large to cross the basement membrane and remain in the blood. Finding protein in urine therefore indicates damage to the basement membrane, and finding blood cells indicates worse damage still.

Selective reabsorption

The proximal convoluted tubule reabsorbs all the glucose and amino acids and most of the water and ions. Its cells are adapted for it:

  • Microvilli on the luminal surface, giving a very large surface area for absorption;
  • many mitochondria, supplying the ATP for active transport;
  • many carrier and cotransporter proteins in the membranes;
  • infoldings of the basal membrane, increasing the area for transport into the blood.

Sodium ions are actively pumped out of the cell into the blood, lowering the sodium concentration inside; sodium then enters from the lumen through cotransporter proteins, each bringing a glucose or amino acid molecule with it against its own gradient — the same cotransport principle as phloem loading in Topic 7. The solutes then leave the cell by facilitated diffusion, and water follows by osmosis.

Osmoregulation

  1. Osmoreceptors in the hypothalamus detect a fall in the water potential of the blood.
  2. The hypothalamus signals the posterior pituitary gland to release antidiuretic hormone (ADH) into the blood.
  3. ADH binds to receptors on the cells of the collecting duct (and distal convoluted tubule), causing vesicles containing aquaporins to fuse with the cell surface membrane.
  4. The membrane becomes more permeable to water, so more water is reabsorbed from the filtrate into the blood as it passes through the concentrated medulla.
  5. A small volume of concentrated urine is produced, and the water potential of the blood rises — negative feedback.

When blood water potential is high, less ADH is released, fewer aquaporins are inserted, less water is reabsorbed, and a large volume of dilute urine is produced.

Cell signalling: glucagon and the cascade

This is the syllabus’s worked example of cell signalling, and it must be known in order:

  1. Glucagon binds to a cell surface receptor on a liver cell, causing a conformational change in the receptor.
  2. A G protein is activated, which stimulates adenylyl cyclase.
  3. Adenylyl cyclase converts ATP into the second messenger cyclic AMP (cAMP).
  4. cAMP activates protein kinase A, which initiates an enzyme cascade.
  5. The signal is amplified at each step, because each activated enzyme phosphorylates and so activates many molecules of the next enzyme.
  6. The final enzyme in the pathway is activated and catalyses the breakdown of glycogen to glucose, which leaves the cell.

Amplification is the reason a very small number of hormone molecules produces a large cellular response, and it is the point most often asked about.

Blood glucose control

  • Glucose too high → β cells of the islets of Langerhans secrete insulin → more glucose transporters in muscle and other cells, so more glucose is taken up and respired; liver and muscle convert glucose to glycogen → blood glucose falls.
  • Glucose too low → α cells secrete glucagonliver cells break glycogen down to glucose (and make glucose from other substrates) → blood glucose rises.

Note the asymmetry the syllabus is careful about: insulin acts on both muscle and liver cells; glucagon acts on liver cells only. Muscle glycogen is for the muscle’s own use and is not released into the blood.

Test strips and biosensors

Both use the same chemistry. A strip carries glucose oxidase and peroxidase immobilised on a pad. Glucose oxidase catalyses the oxidation of glucose, producing hydrogen peroxide; peroxidase then uses the hydrogen peroxide to oxidise a colourless chemical into a coloured one. The intensity of colour is proportional to the glucose concentration and is compared against a chart.

Because glucose oxidase is specific to glucose, other sugars do not interfere. A biosensor replaces the colour comparison with an electrode that detects the electrons released, and converts the current into a numerical display — faster, more precise, and not dependent on the user’s judgement of colour.

Homeostasis in plants: stomata

Stomata balance two conflicting needs: carbon dioxide uptake for photosynthesis and minimising water loss by transpiration. They show daily rhythms, opening in the day and closing at night, and continue to do so for a time even in constant conditions, which shows the rhythm is endogenous.

Opening. Guard cells actively pump H+ out using ATP; potassium ions enter through channels, and other solutes accumulate. The solute concentration rises, so the water potential falls, and water enters by osmosis. The guard cells become turgid. Because the inner wall is thicker and less elastic than the outer, and the cellulose microfibrils are arranged around the cell, the cells bend apart and the stoma opens. Closing is the reverse: solutes leave, water follows, the cells become flaccid and the pore closes.

Abscisic acid (ABA) is produced under water stress. It binds to receptors on the guard cell membrane; calcium ions act as a second messenger, causing channels to open so that ions leave the guard cells. Water potential rises, water leaves by osmosis, the cells become flaccid and the stomata close, conserving water even at the cost of photosynthesis.

✏️Worked example

The table shows the concentration (g dm−3) of four substances in the blood plasma, the glomerular filtrate and the urine of a healthy person. Protein: plasma 80, filtrate 0, urine 0. Glucose: plasma 1.0, filtrate 1.0, urine 0. Urea: plasma 0.3, filtrate 0.3, urine 20.0. Sodium ions: plasma 3.2, filtrate 3.2, urine 3.5. (a) Explain the protein values. (b) Explain the glucose values. (c) Urea is not reabsorbed at all in this person, yet its concentration rises 67-fold. Explain. (d) A patient’s urine contains 0.9 g dm−3 glucose. Suggest two causes.

(a) Protein is present in the plasma at 80 but absent from the filtrate. Protein molecules are too large to pass through the basement membrane of the filtration barrier, which acts as a molecular sieve, so they are not filtered and remain in the blood. Since none enters the filtrate, none can appear in the urine.

(b) Glucose is small enough to be filtered, so its concentration in the filtrate equals that in the plasma at 1.0. It is then completely reabsorbed in the proximal convoluted tubule, by cotransport with sodium ions through carrier proteins, followed by facilitated diffusion into the blood. In a healthy person all of it is recovered, so the urine contains none.

(c) The rise is caused by the reabsorption of water, not by any addition of urea. About 99% of the water filtered is reabsorbed — in the proximal convoluted tubule, the loop of Henle and, under the influence of ADH, the collecting duct. Removing the water while leaving the urea behind concentrates the urea that remains. A concentration is an amount per unit volume, so shrinking the volume raises the concentration without changing the quantity of solute.

Sodium makes the same point in the opposite direction: its concentration barely changes, from 3.2 to 3.5, because most of the sodium is reabsorbed along with the water, in roughly matching proportions.

(d) Either: (i) the plasma glucose concentration is abnormally high, as in untreated diabetes mellitus, so the quantity filtered exceeds the capacity of the carrier proteins in the proximal convoluted tubule — the transport system is saturated and the excess passes on into the urine; or (ii) the carrier or cotransporter proteins are faulty or reduced in number, or the tubule cells are damaged, so glucose cannot be reabsorbed even at a normal plasma concentration.

Check it. Read down each column asking one question: was it filtered, and was it reabsorbed? Plasma-to-filtrate tells you about filtration — equal means filtered, zero in the filtrate means too large. Filtrate-to-urine tells you about reabsorption — zero in urine means completely reabsorbed, a large rise means water was removed. Applying those two questions in order answers every table of this kind, and it also catches the classic error of concluding that urea is ‘secreted’.
Saying urea is added to the filtrate. A 67-fold rise in concentration looks like something was put in, but nothing was: the volume fell. Confusing concentration with quantity is the single most common error on kidney data, and it is worth writing the phrase ‘because water was reabsorbed’ explicitly. Second trap in part (d): do not write simply ‘the person has diabetes’. The mark is for the mechanism — the carriers are saturated because the filtered load exceeds their capacity.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Explain how ultrafiltration occurs in the Bowman’s capsule, naming the structures involved.
Blood enters the glomerulus through the afferent arteriole, which is wider than the efferent arteriole that leaves it. The narrower exit restricts outflow and so generates a high hydrostatic pressure in the glomerular capillaries. This pressure forces fluid out through a three-layered barrier: the capillary endothelium, which is perforated by pores; the basement membrane, a mesh of glycoproteins that acts as the molecular sieve; and the podocytes of the capsule wall, whose foot processes leave filtration slits between them. Small molecules — water, glucose, amino acids, urea and mineral ions — pass through into the capsule as glomerular filtrate, while blood cells and plasma proteins are too large to cross the basement membrane and remain in the blood.
2. Describe how the cells of the proximal convoluted tubule are adapted for selective reabsorption.
(i) Microvilli on the surface facing the lumen, greatly increasing the surface area for absorption. (ii) Many mitochondria, supplying the ATP needed for the active transport of sodium ions, which drives the whole process. (iii) Many carrier and cotransporter proteins in the luminal membrane, allowing glucose and amino acids to be brought in with sodium against their own concentration gradients. (iv) Infoldings of the basal membrane next to the blood capillary, increasing the area over which substances pass into the blood. (v) The cells are closely associated with capillaries, giving a short diffusion distance and maintaining the gradients. The mechanism: sodium is pumped out of the cell into the blood, lowering internal sodium; sodium then enters from the lumen through cotransporters, bringing glucose or an amino acid with it; these leave by facilitated diffusion and water follows by osmosis.
3. Describe the sequence of events by which a fall in the water potential of the blood leads to the production of more concentrated urine.
Osmoreceptors in the hypothalamus detect the fall in water potential of the blood. The hypothalamus stimulates the posterior pituitary gland to release antidiuretic hormone (ADH) into the blood. ADH is carried to the kidney and binds to receptors on the cell surface membranes of the collecting duct cells. This causes vesicles containing aquaporins to fuse with the membrane, inserting more water channels, so the membrane becomes more permeable to water. As filtrate passes down the collecting duct through the concentrated medulla, more water is reabsorbed by osmosis into the blood. The result is a smaller volume of more concentrated urine, and the water potential of the blood rises back towards normal — negative feedback, which then reduces ADH secretion.
4. Explain what is meant by amplification in cell signalling, using the glucagon pathway.
Amplification means that one signal molecule produces a very large cellular response, because each step of the pathway activates many molecules of the next. When glucagon binds to its cell surface receptor on a liver cell, the receptor changes shape and activates a G protein, which activates adenylyl cyclase. One adenylyl cyclase molecule converts many ATP molecules into cyclic AMP. Each cAMP molecule activates protein kinase A, and each activated protein kinase A phosphorylates and so activates many molecules of the next enzyme — an enzyme cascade. At each stage the number of activated molecules multiplies, so a handful of glucagon molecules at the cell surface leads to the activation of very large numbers of the final enzyme, which catalyses the breakdown of glycogen to glucose. This is why hormones are effective at extremely low blood concentrations.
5. Explain the mechanism by which guard cells open a stoma, and how abscisic acid closes it.
Opening: guard cells use ATP to actively pump hydrogen ions out of the cell. This causes potassium ions to enter through channel proteins, and other solutes accumulate. The rise in solute concentration lowers the water potential of the guard cells, so water enters by osmosis from neighbouring cells and the guard cells become turgid. Because the inner wall (next to the pore) is thicker and less elastic than the outer wall, and the cellulose microfibrils are arranged in hoops around the cell, the cells cannot expand uniformly: they curve apart, opening the pore between them. Closing by ABA: under water stress, abscisic acid binds to receptors on the guard cell membrane. Calcium ions act as a second messenger, entering the cytoplasm and causing channels to open so that ions leave the guard cell. Water potential rises, water leaves by osmosis, the cells become flaccid and the stoma closes, conserving water.
6. Explain how a glucose test strip works, and give two advantages of a biosensor over a colour chart.
The strip carries two immobilised enzymes. Glucose oxidase catalyses the oxidation of glucose in the sample, producing hydrogen peroxide. Peroxidase then catalyses a reaction in which the hydrogen peroxide oxidises a colourless chemical into a coloured product. The intensity of the colour is proportional to the glucose concentration, and is compared against a printed chart. Because glucose oxidase is specific to glucose, other sugars present do not react and do not interfere. Advantages of a biosensor: (i) it gives a numerical reading from an electrode detecting the electrons released, rather than requiring the user to judge a colour, which is subjective and impossible for someone who is colour-blind or in poor light; (ii) it is more precise and more sensitive, distinguishing values a colour chart would group together, and it is faster and can store readings over time so that trends can be monitored.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Any labelled diagram set of the nephron showing which region lies in the cortex and which in the medulla — the geography is examined directly
  • Diabetes UK — clear material on blood glucose control, test strips and continuous monitoring, which grounds the theory in practice
  • Animations of the G protein / cAMP cascade — the amplification step is much clearer when you can see the numbers multiply at each stage