Human Physiology

1. Digestion

Overview

Digestion is the mechanical and chemical breakdown of food into small, soluble molecules that can be absorbed across the intestinal wall into the blood or lymph.

Mechanical Digestion

• Mouth: chewing (mastication) by teeth increases surface area; tongue mixes food with saliva.
• Stomach: churning by muscular contractions mixes food with gastric juice.
• Small intestine: peristalsis (rhythmic muscular contractions) propels and mixes chyme.

Chemical Digestion

Mouth

Salivary amylase (secreted by salivary glands, optimal pH $\approx 6.8$) hydrolyses starch into maltose.

Stomach

Gastric juice contains:

• Hydrochloric acid ($\mathrm{HCl}$, pH $\approx 1.5$--$2.0$): denatures proteins, kills ingested microorganisms, provides optimal pH for pepsin.
• Pepsin (secreted as inactive pepsinogen by chief cells; activated by $\mathrm{HCl}$): endopeptidase that hydrolyses peptide bonds, breaking proteins into polypeptides.
• Mucus (secreted by goblet cells): forms a protective barrier against $\mathrm{HCl}$ and pepsin.

Small Intestine

The majority of digestion occurs here. Pancreatic juice and bile are released into the duodenum.

Enzyme	Source	Substrate	Products	Optimal pH
Pancreatic amylase	Pancreas	Starch	Maltose	$\approx 7$--$8$
Trypsin	Pancreas (as trypsinogen)	Polypeptides	Smaller polypeptides	$\approx 7$--$8$
Lipase	Pancreas	Triglycerides	Fatty acids + glycerol	$\approx 7$--$8$
Maltase	Intestinal wall	Maltose	Glucose	$\approx 7$--$8$
Lactase	Intestinal wall	Lactose	Glucose + galactose	$\approx 7$--$8$
Sucrase	Intestinal wall	Sucrose	Glucose + fructose	$\approx 7$--$8$

Bile (produced in the liver, stored in the gallbladder) emulsifies fats: bile salts break large fat globules into smaller droplets, increasing the surface area for lipase action. Bile is not an enzyme.

Large Intestine

Absorption of water and mineral ions. Undigested material forms faeces, eliminated via the rectum and anus.

Absorption

The small intestine is the primary site of absorption. Adaptations for absorption:

• Villi: finger-like projections of the intestinal wall, increasing surface area.
• Microvilli: projections on the epithelial cells of villi (the "brush border"), further increasing surface area.
• Single layer of epithelial cells: short diffusion distance.
• Dense capillary network: maintains a steep concentration gradient.
• Lacteals (lymphatic capillaries): absorb fatty acids and glycerol (re-esterified into triglycerides and packaged as chylomicrons).

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2. The Cardiovascular System

Blood Vessels

Feature	Arteries	Veins	Capillaries
Wall structure	Thick, muscular, elastic	Thinner, less muscular, with valves	Single layer of endothelial cells
Lumen	Narrow (relatively)	Wide (relatively)	Very narrow ($\approx 8\;\mathrm{\mu m}$)
Pressure	High	Low	Low (drops from arteriolar end)
Direction	Away from heart	Toward heart	Between arteries and veins
Valves	Absent	Present (prevent backflow)	Absent

The Heart

The heart is a double pump: the right side pumps deoxygenated blood to the lungs (pulmonary circulation); the left side pumps oxygenated blood to the body (systemic circulation).

Cardiac cycle:

1. Atrial systole: the atria contract, forcing blood through the AV valves (tricuspid on the right, bicuspid/mitral on the left) into the ventricles.
2. Ventricular systole: the ventricles contract forcefully; AV valves close (producing the first heart sound, "lub"); semilunar valves open; blood is ejected into the aorta and pulmonary artery.
3. Diastole: the heart muscle relaxes; semilunar valves close (producing the second heart sound, "dub"); blood flows passively from the atria into the ventricles.

Cardiac output:

$$\mathrm{Cardiac output} = \mathrm{stroke volume} \times \mathrm{heart rate}$$

Typical resting cardiac output: $\approx 5\;\mathrm{L/min}$.

Blood Components

Component	Function
Red blood cells (erythrocytes)	Transport $\mathrm{O}_2$ (bound to haemoglobin) and $\mathrm{CO}_2$ (as bicarbonate, $\mathrm{HCO}_3^-$).
White blood cells (leucocytes)	Defence: phagocytes (neutrophils, macrophages) engulf pathogens; lymphocytes produce antibodies.
Platelets (thrombocytes)	Cell fragments involved in blood clotting (initiate the cascade leading to fibrin mesh formation).
Plasma	Liquid matrix: transports dissolved substances (glucose, amino acids, urea, hormones, antibodies).

Haemoglobin and Oxygen Transport

Haemoglobin ($\mathrm{Hb}$) is a quaternary protein with four subunits, each containing a haem group with an iron ion ($\mathrm{Fe}^{2+}$) that binds one $\mathrm{O}_2$ molecule.

Oxygen dissociation curve: an S-shaped (sigmoidal) curve plotting percentage saturation of $\mathrm{Hb}$ against partial pressure of $\mathrm{O}_2$ ($\mathrm{pO}_2$).

• High $\mathrm{pO}_2$ (lungs): $\mathrm{Hb}$ has high affinity for $\mathrm{O}_2$; loading occurs (steep part of curve).
• Low $\mathrm{pO}_2$ (respiring tissues): $\mathrm{Hb}$ affinity decreases; unloading occurs.
• The curve shifts right (lower affinity, more unloading) with increased $\mathrm{CO}_2$, decreased pH (Bohr effect), increased temperature, and increased $23$-BPG (produced in active tissues).

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3. The Immune System

Non-Specific (Innate) Immunity

First-line and second-line defences that act regardless of the pathogen.

Barrier / Response	Mechanism
Skin	Physical barrier; sebum inhibits microbial growth.
Mucous membranes	Trap pathogens; mucus contains lysozyme (enzyme that destroys bacterial cell walls).
Phagocytes	Neutrophils and macrophages engulf and digest pathogens by phagocytosis.
Inflammation	Histamine release increases capillary permeability, causing swelling, redness, heat, and pain; increased blood flow brings phagocytes to the site.
Fever	Elevated body temperature inhibits pathogen replication and enhances immune cell activity.

Phagocytosis steps:

1. Chemotaxis: phagocyte moves toward the pathogen following chemical signals.
2. Recognition: phagocyte binds to the pathogen surface.
3. Engulfment: phagocyte extends pseudopodia around the pathogen, enclosing it in a phagosome.
4. Digestion: phagosome fuses with a lysosome; hydrolytic enzymes degrade the pathogen.
5. Exocytosis: indigestible material is expelled.

Specific (Adaptive) Immunity

Cell-Mediated Immunity (T Cells)

• Helper T cells ($\mathrm{CD4}^+$): recognise antigens presented by antigen-presenting cells (APCs: macrophages, dendritic cells) via MHC class II molecules. Once activated, they release cytokines that stimulate B cells and cytotoxic T cells.
• Cytotoxic T cells ($\mathrm{CD8}^+$): recognise antigens presented by all nucleated cells via MHC class I molecules. They kill infected or cancerous cells by releasing perforin (forms pores in the target cell membrane) and granzymes (induce apoptosis).
• Memory T cells: provide long-term immunity; rapidly respond upon re-exposure.

Humoral Immunity (B Cells and Antibodies)

1. B cells recognise specific antigens via surface antibodies.
2. Helper T cells activate the B cell.
3. The B cell differentiates into:
   • Plasma cells: secrete large quantities of antibodies specific to the antigen.
   • Memory B cells: provide long-term immunity; rapid antibody production upon re-exposure.

Antibodies (Immunoglobulins)

Antibodies are Y-shaped proteins produced by plasma cells. Each antibody has:

• Two identical heavy chains and two identical light chains.
• A variable region at the tips of the Y: antigen-binding site, specific to one antigen (complementary shape).
• A constant region: determines the antibody class ($\mathrm{IgG}$, $\mathrm{IgM}$, $\mathrm{IgA}$, $\mathrm{IgE}$, $\mathrm{IgD}$).

Mechanisms of antibody action:

• Neutralisation: antibodies bind to pathogens or toxins, blocking their ability to enter cells or cause damage.
• Opsonisation: antibodies coat pathogens, making them more easily recognised by phagocytes.
• Agglutination: antibodies bind to multiple pathogens, clumping them together for easier phagocytosis.
• Complement activation: antibodies trigger the complement cascade, leading to pathogen lysis.

Vaccination

Vaccines introduce a weakened, inactivated, or subunit form of a pathogen (or its antigen), stimulating the adaptive immune system to produce memory cells without causing disease. Upon subsequent exposure, memory cells mount a faster and stronger secondary immune response.

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4. Gas Exchange

Alveolar Structure and Adaptations

The lungs contain millions of alveoli (air sacs), the sites of gas exchange between air and blood.

Adaptation	Significance
Large total surface area	Maximises rate of diffusion ($\approx 70\;\mathrm{m}^2$).
Thin alveolar wall (one cell thick)	Short diffusion distance.
Dense capillary network	Maintains concentration gradient; large blood supply.
Moist inner surface	Gases dissolve before diffusing across the membrane.
Ventilation (breathing)	Constantly refreshes air, maintaining $\mathrm{O}_2$ and \mathrm{CO}_2} gradients.

Diffusion of Gases

Oxygen: diffuses from the alveolar air ($\mathrm{pO}_2 \approx 13.3\;\mathrm{kPa}$) into the blood ($\mathrm{pO}_2 \approx 5.3\;\mathrm{kPa}$ in deoxygenated blood arriving at the lungs).

Carbon dioxide: diffuses from the blood ($\mathrm{pCO}_2 \approx 6.0\;\mathrm{kPa}$) into the alveolar air ($\mathrm{pCO}_2 \approx 5.3\;\mathrm{kPa}$).

Ventilation

Phase	Muscles involved	Volume change	Pressure change	Air movement
Inspiration	External intercostal muscles contract (ribs up and out); diaphragm contracts and flattens	Thoracic volume increases	Intrapulmonary pressure decreases below atmospheric	Air flows in
Expiration	External intercostal muscles relax; diaphragm relaxes and domes upward	Thoracic volume decreases	Intrapulmonary pressure increases above atmospheric	Air flows out

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5. Neurones and Synapses

Neurone Structure

• Cell body (soma): contains the nucleus and metabolic machinery.
• Dendrites: branched extensions that receive signals from other neurones.
• Axon: long, thin extension that transmits action potentials away from the cell body. Surrounded by a myelin sheath (formed by Schwann cells) with gaps called nodes of Ranvier.
• Axon terminals (synaptic knobs): release neurotransmitters at synapses.

Resting Potential

At rest, the inside of the neurone is negatively charged relative to the outside ($\approx -70\;\mathrm{mV}$).

• The sodium-potassium pump ($\mathrm{Na}^+/\mathrm{K}^+$ ATPase) actively transports $3\;\mathrm{Na}^+$ out and $2\;\mathrm{K}^+$ in, using ATP.
• The membrane is more permeable to $\mathrm{K}^+$ than to $\mathrm{Na}^+$ (more $\mathrm{K}^+$ leak channels), so $\mathrm{K}^+$ diffuses out, carrying positive charge.
• The membrane also contains impermeable anions (proteins, organic phosphates) inside the cell.

Action Potential

A rapid reversal and restoration of the membrane potential.

1. Depolarisation: a stimulus raises the membrane potential to the threshold ($\approx -55\;\mathrm{mV}$). Voltage-gated $\mathrm{Na}^+$ channels open; $\mathrm{Na}^+$ rushes in, making the inside positive ($\approx +40\;\mathrm{mV}$).
2. Repolarisation: voltage-gated $\mathrm{Na}^+$ channels close; voltage-gated $\mathrm{K}^+$ channels open; $\mathrm{K}^+$ rushes out, restoring the negative internal charge.
3. Hyperpolarisation: $\mathrm{K}^+$ channels remain open briefly, making the inside temporarily more negative than resting potential.
4. Restoration of resting potential: the $\mathrm{Na}^+/\mathrm{K}^+$ pump restores the original ion concentrations.

Propagation: the action potential travels along the axon. In myelinated axons, the action potential "jumps" between nodes of Ranvier (saltatory conduction), which is much faster than continuous propagation in unmyelinated axons.

Synapses

Definition. A synapse is the junction between two neurones (or between a neurone and an effector cell) where signals are transmitted chemically.

Steps of synaptic transmission:

1. An action potential arrives at the presynaptic terminal.
2. Voltage-gated $\mathrm{Ca}^{2+}$ channels open; $\mathrm{Ca}^{2+}$ ions flow in.
3. $\mathrm{Ca}^{2+}$ triggers exocytosis of synaptic vesicles, releasing neurotransmitter into the synaptic cleft.
4. Neurotransmitter diffuses across the synaptic cleft and binds to specific receptors on the postsynaptic membrane.
5. This opens ligand-gated ion channels, causing an excitatory (depolarising) or inhibitory (hyperpolarising) postsynaptic potential (EPSP or IPSP).
6. The neurotransmitter is broken down by enzymes (e.g., acetylcholinesterase breaks down acetylcholine) or reabsorbed (reuptake), terminating the signal.

Acetylcholine (ACh) is a common excitatory neurotransmitter at neuromuscular junctions. Noradrenaline and dopamine are neurotransmitters in the central nervous system.

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6. Hormones and Homeostasis

Endocrine System

Hormones are chemical messengers secreted by endocrine glands into the bloodstream, acting on target cells with specific receptors.

Gland	Hormone(s)	Function
Hypothalamus	Releasing hormones (e.g., GnRH, TRH)	Regulate anterior pituitary; connects nervous and endocrine systems
Anterior pituitary	FSH, LH, ACTH, TSH, GH, prolactin	Stimulate other glands; growth, reproduction
Posterior pituitary	ADH (vasopressin), oxytocin	Water reabsorption in kidneys; uterine contraction
Thyroid	Thyroxine ($\mathrm{T}_4$), triiodothyronine ($\mathrm{T}_3$)	Basal metabolic rate, growth, development
Adrenal cortex	Cortisol, aldosterone	Stress response; $\mathrm{Na}^+$ and water reabsorption
Adrenal medulla	Adrenaline (epinephrine)	"Fight or flight": increased heart rate, bronchodilation, glycogenolysis
Pancreas	Insulin, glucagon	Blood glucose regulation
Ovaries / Testes	Oestrogen, progesterone / testosterone	Secondary sex characteristics, reproductive cycle

Blood Glucose Regulation

Normal fasting blood glucose: $\approx 5\;\mathrm{mmol/L}$.

Condition	Sensor	Hormone	Response
Blood glucose high (after a meal)	Beta cells of pancreatic islets	Insulin	Stimulates uptake of glucose by liver, muscle, and adipose cells; promotes glycogenesis (glycogen synthesis) in the liver; inhibits glycogenolysis and gluconeogenesis.
Blood glucose low (fasting / exercise)	Alpha cells of pancreatic islets	Glucagon	Stimulates glycogenolysis (glycogen breakdown) in the liver; stimulates gluconeogenesis (synthesis of glucose from non-carbohydrate sources).

Thermoregulation

The hypothalamus acts as the body's thermostat, receiving input from thermoreceptors in the skin and core.

Condition	Mechanisms of Heat Loss	Mechanisms of Heat Conservation / Production
Body temperature above set point ($37^\circ\mathrm{C}$)	Vasodilation of skin blood vessels; sweating (evaporative cooling); decreased metabolic rate	---
Body temperature below set point	Vasoconstriction of skin blood vessels	Shivering (involuntary muscle contraction); increased metabolic rate (hormonal: thyroxine, adrenaline); piloerection (goosebumps --- vestigial in humans)

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7. Reproduction

Male Reproductive System

• Testes: produce sperm (spermatogenesis in seminiferous tubules) and testosterone (Leydig cells).
• Epididymis: sperm maturation and storage.
• Vas deferens: transports sperm from epididymis to urethra.
• Seminal vesicles and prostate gland: produce seminal fluid (fructose for energy, alkaline pH to neutralise vaginal acidity, prostaglandins for sperm motility).

Spermatogenesis:

1. Diploid spermatogonia ($2n$) divide by mitosis to maintain the germ cell population.
2. Primary spermatocytes undergo meiosis I to form secondary spermatocytes ($n$).
3. Secondary spermatocytes undergo meiosis II to form spermatids.
4. Spermatids differentiate into spermatozoa (sperm cells), acquiring a head (containing the haploid nucleus and acrosome with digestive enzymes), midpiece (packed with mitochondria for ATP), and tail (flagellum for motility).

Female Reproductive System

• Ovaries: produce ova (oogenesis) and hormones (oestrogen, progesterone).
• Fallopian tubes (oviducts): site of fertilisation; transport the ovum to the uterus via ciliary action and peristalsis.
• Uterus: site of implantation and foetal development; lined by the endometrium.
• Cervix: entrance to the uterus; produces mucus (changes consistency during the menstrual cycle).
• Vagina: receives the penis during copulation; birth canal.

Oogenesis:

1. Diploid oogonia ($2n$) divide by mitosis in the foetal ovary.
2. Primary oocytes begin meiosis I but arrest at Prophase I until puberty.
3. Each month, one primary oocyte completes meiosis I, producing a secondary oocyte (large, haploid) and a first polar body (small, degenerates).
4. The secondary oocyte begins meiosis II but arrests at Metaphase II; it is released during ovulation.
5. Meiosis II is completed only if fertilisation occurs, producing the ovum and a second polar body.

The Menstrual Cycle

Phase	Days (approx.)	Ovarian events	Hormonal changes	Uterine events
Menstruation	1--5	---	Low oestrogen and progesterone	Endometrium is shed (menstrual flow).
Follicular phase	1--13	Several follicles develop; one becomes dominant	FSH stimulates follicle growth; oestrogen rises (positive feedback triggers LH surge)	Endometrium thickens and becomes vascularised (oestrogen).
Ovulation	Day 14	Mature follicle ruptures; secondary oocyte released	LH surge triggers ovulation; oestrogen peak	---
Luteal phase	15--28	Ruptured follicle becomes corpus luteum	Progesterone rises (maintains endometrium); LH and FSH remain low (negative feedback)	Endometrium is maintained; secretes glycogen.
(If no implantation)	Day 28	Corpus luteum degenerates	Oestrogen and progesterone drop	Endometrium breaks down; cycle restarts.

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Common Pitfalls

• Confusing insulin and glucagon: insulin lowers blood glucose; glucagon raises it.
• Stating that veins carry "deoxygenated blood": the pulmonary vein carries oxygenated blood from the lungs to the heart.
• Confusing antigens and antibodies: antigens are molecules that trigger an immune response (found on pathogen surfaces); antibodies are produced by the immune system in response to antigens.
• Misunderstanding the refractory period: after an action potential, the neurone cannot be immediately restimulated (absolute refractory period), ensuring unidirectional propagation.
• Confusing oogenesis and spermatogenesis: oogenesis produces one ovum and polar bodies (asymmetric division); spermatogenesis produces four functional sperm cells (symmetric division).

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Practice Problems

Question 1: Oxygen Dissociation Curve

Explain why the oxygen dissociation curve shifts to the right in actively respiring muscle tissue. Describe the consequence for oxygen unloading.

Answer

In actively respiring muscle tissue, $\mathrm{CO}_2$ concentration increases, pH decreases (Bohr effect), and temperature rises. These factors reduce haemoglobin's affinity for $\mathrm{O}_2$, shifting the dissociation curve to the right. At any given $\mathrm{pO}_2$, a lower percentage of $\mathrm{Hb}$ is saturated with $\mathrm{O}_2$, meaning more $\mathrm{O}_2$ is unloaded from haemoglobin into the tissues. This ensures that active muscles receive a greater supply of oxygen to sustain aerobic respiration.

Question 2: Action Potential Calculation

The resting membrane potential of a neurone is $-70\;\mathrm{mV}$. During an action potential, the membrane potential reaches $+40\;\mathrm{mV}$. Calculate the magnitude of the change in membrane potential during depolarisation.

Answer

Change in membrane potential: $+40\;\mathrm{mV} - (-70\;\mathrm{mV}) = +110\;\mathrm{mV}$.

The membrane potential changes by $110\;\mathrm{mV}$ during depolarisation.

Question 3: Immune Response Pathway

A person is bitten by a mosquito carrying the malaria parasite (Plasmodium), which enters their liver cells. Describe how cell-mediated immunity would respond to this infection, naming the specific cells and molecules involved.

Answer

1. Infected liver cells present Plasmodium antigens on their surface via MHC class I molecules.
2. Cytotoxic T cells ($\mathrm{CD8}^+$) recognise the antigen-MHC I complex.
3. The cytotoxic T cell binds to the infected cell and releases perforin, which forms pores in the infected cell's membrane.
4. Granzymes enter through the pores and activate caspases, inducing apoptosis (programmed cell death) in the infected liver cell.
5. Helper T cells ($\mathrm{CD4}^+$) are activated by antigen-presenting cells (e.g., macrophages presenting Plasmodium antigens via MHC class II). Helper T cells release cytokines (e.g., interleukin-2) that stimulate the proliferation and activation of cytotoxic T cells and B cells.
6. Memory T cells are generated for a faster secondary response upon re-infection.

Question 4: Hormonal Feedback

A tumour on the anterior pituitary causes excessive secretion of TSH. Predict the effect on the thyroid gland and on blood levels of thyroxine ($\mathrm{T}_4$). Explain whether this is positive or negative feedback.

Answer

Excessive TSH (thyroid-stimulating hormone) would overstimulate the thyroid gland, causing it to enlarge (goitre) and secrete excess thyroxine ($\mathrm{T}_4$). Blood thyroxine levels would be elevated (hyperthyroidism), leading to increased metabolic rate, weight loss, rapid heart rate, and other symptoms.

Normally, thyroxine exerts negative feedback on both the hypothalamus (reducing TRH secretion) and the anterior pituitary (reducing TSH secretion). In this case, the tumour disrupts this feedback loop: the pituitary is secreting TSH autonomously, independent of thyroxine levels. The negative feedback mechanism fails because the pituitary is no longer responsive to the elevated thyroxine concentration.

Question 5: Digestion and Absorption

A patient has a condition that prevents the pancreas from secreting lipase. Explain the consequence for fat digestion and absorption, and predict what would be found in the patient's faeces.

Answer

Without pancreatic lipase, triglycerides in the small intestine cannot be hydrolysed into monoglycerides and free fatty acids. Although bile salts still emulsify the fat into smaller droplets, the large triglyceride molecules cannot be absorbed across the intestinal epithelium (they are too large and hydrophobic). The undigested fat passes into the large intestine and is excreted in the faeces. The faeces would be pale, bulky, greasy, and difficult to flush (steatorrhoea). The patient would also experience fat-soluble vitamin (A, D, E, K) deficiencies due to malabsorption, since these vitamins require dietary fat for absorption.

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Worked Examples

Worked Example: Cardiac Output and Oxygen Delivery

A $70\;\mathrm{kg}$ athlete at rest has a heart rate of $60\;\mathrm{bpm}$ and a stroke volume of $70\;\mathrm{mL}$. During maximal exercise, her heart rate increases to $190\;\mathrm{bpm}$ and her stroke volume increases to $120\;\mathrm{mL}$. Her arterial $\mathrm{O}_2$ content is $200\;\mathrm{mL\;O_2/L}$ blood and her venous $\mathrm{O}_2$ content is $150\;\mathrm{mL\;O_2/L}$ at rest, decreasing to $50\;\mathrm{mL\;O_2/L}$ during exercise. Calculate the cardiac output and oxygen consumption at rest and during maximal exercise.

Solution

At rest: $\mathrm{Cardiac\;output} = \mathrm{stroke\;volume} \times \mathrm{heart\;rate} = 70\;\mathrm{mL} \times 60\;\mathrm{bpm} = 4200\;\mathrm{mL/min} = 4.2\;\mathrm{L/min}$

$\mathrm{O_2\;consumption} = \mathrm{cardiac\;output} \times (\mathrm{arterial\;O_2} - \mathrm{venous\;O_2})$ $= 4.2\;\mathrm{L/min} \times (200 - 150)\;\mathrm{mL\;O_2/L} = 4.2 \times 50 = 210\;\mathrm{mL\;O_2/min}$

During maximal exercise: $\mathrm{Cardiac\;output} = 120\;\mathrm{mL} \times 190\;\mathrm{bpm} = 22800\;\mathrm{mL/min} = 22.8\;\mathrm{L/min}$

$\mathrm{O_2\;consumption} = 22.8\;\mathrm{L/min} \times (200 - 50)\;\mathrm{mL\;O_2/L} = 22.8 \times 150 = 3420\;\mathrm{mL\;O_2/min}$

The athlete's cardiac output increases $5.4$-fold and oxygen consumption increases $16.3$-fold from rest to maximal exercise. The larger increase in $\mathrm{O}_2$ consumption relative to cardiac output is explained by the increased oxygen extraction (a-v difference doubles from $50$ to $150\;\mathrm{mL\;O_2/L}$), demonstrating that both increased blood flow and increased tissue oxygen extraction contribute to meeting exercise demands.

Worked Example: Blood Glucose Homeostasis Calculation

A patient's fasting blood glucose is measured at $12\;\mathrm{mmol/L}$ (normal: approximately $5\;\mathrm{mmol/L}$). The total blood volume is $5\;\mathrm{L}$. The molar mass of glucose is $180\;\mathrm{g/mol}$. Calculate the mass of excess glucose in the patient's blood. Assuming the renal threshold for glucose is approximately $10\;\mathrm{mmol/L}$, explain why glucose would appear in the urine (glycosuria).

Solution

Excess glucose concentration: $12 - 5 = 7\;\mathrm{mmol/L}$ above normal.

Excess glucose in blood: $7\;\mathrm{mmol/L} \times 5\;\mathrm{L} = 35\;\mathrm{mmol}$

Mass of excess glucose: $35 \times 10^{-3}\;\mathrm{mol} \times 180\;\mathrm{g/mol} = 6.3\;\mathrm{g}$

Total glucose in blood: $12\;\mathrm{mmol/L} \times 5\;\mathrm{L} = 60\;\mathrm{mmol} = 10.8\;\mathrm{g}$

The blood glucose exceeds the renal threshold of $10\;\mathrm{mmol/L}$, meaning the proximal convoluted tubule's $\mathrm{Na}^+$-glucose co-transporters (SGLT2) are saturated. The excess glucose ($12 - 10 = 2\;\mathrm{mmol/L}$, or $10\;\mathrm{mmol}$ total, equivalent to $1.8\;\mathrm{g}$) cannot be reabsorbed and remains in the filtrate, appearing in the urine. This is a hallmark of diabetes mellitus. The osmotic effect of unreabsorbed glucose in the renal tubule also draws water into the urine, causing polyuria (excessive urination).

Worked Example: Neurone Conduction Velocity

A myelinated motor neurone has nodes of Ranvier spaced $1.5\;\mathrm{mm}$ apart. The time for an action potential to travel between two adjacent nodes is $0.02\;\mathrm{ms}$. Calculate the conduction velocity between nodes. If the axon length is $0.6\;\mathrm{m}$, calculate the total transmission time and compare this with an unmyelinated axon of the same diameter with a conduction velocity of $2\;\mathrm{m/s}$.

Solution

Conduction velocity between nodes (saltatory conduction): $v = \frac{d}{t} = \frac{1.5\;\mathrm{mm}}{0.02\;\mathrm{ms}} = \frac{1.5 \times 10^{-3}\;\mathrm{m}}{0.02 \times 10^{-3}\;\mathrm{s}} = 75\;\mathrm{m/s}$

Total transmission time for myelinated axon ($0.6\;\mathrm{m}$): $t = \frac{0.6}{75} = 0.008\;\mathrm{s} = 8\;\mathrm{ms}$

Total transmission time for unmyelinated axon ($v = 2\;\mathrm{m/s}$): $t = \frac{0.6}{2} = 0.3\;\mathrm{s} = 300\;\mathrm{ms}$

The myelinated axon transmits the signal $37.5$ times faster than the unmyelinated axon ($300/8$). This demonstrates why myelination is critical for rapid reflexes and coordination in vertebrate nervous systems. Multiple sclerosis, which involves autoimmune destruction of the myelin sheath, dramatically slows conduction velocity, producing the characteristic motor and sensory deficits.

Worked Example: Immune Response Kinetics

A person receives their first dose of a vaccine on Day 0 and a booster dose on Day 28. The primary response produces a peak antibody titre of $100\;\mathrm{AU/mL}$ on Day $14$. The secondary response produces a peak titre of $2000\;\mathrm{AU/mL}$ on Day $7$ after the booster. Calculate the fold increase in peak titre and the time to peak, and explain the cellular basis for these differences.

Solution

Fold increase in peak titre: $\frac{2000}{100} = 20$-fold increase.

Time to peak: decreased from $14$ days (primary) to $7$ days (secondary).

Cellular basis:

• Primary response: naive B cells must be activated by helper T cells, clonally expand (taking $3$--$5$ days), differentiate into plasma cells (taking additional days), and then secrete IgM followed by IgG. Memory cells are generated during this process.
• Secondary response: memory B cells are already present in much higher numbers than the original naive B cell clone. Upon re-exposure, memory B cells rapidly differentiate into plasma cells (within $1$--$3$ days), producing predominantly high-affinity IgG. This is the basis for vaccination: the booster dose "reminds" the immune system, generating a faster, stronger, and more sustained antibody response that provides long-lasting protection.

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Common Pitfalls (Expanded)

• Confusing insulin and glucagon: insulin lowers blood glucose (stimulates uptake, glycogenesis); glucagon raises blood glucose (stimulates glycogenolysis, gluconeogenesis). Both are produced by the pancreas but by different cell types (beta vs alpha cells).
• Stating that veins carry "deoxygenated blood": the pulmonary vein carries oxygenated blood from the lungs to the heart. The rule is: arteries carry blood away from the heart; veins carry blood toward the heart.
• Confusing antigens and antibodies: antigens are molecules that trigger an immune response (found on pathogen surfaces); antibodies are Y-shaped proteins produced by plasma cells in response to antigens.
• Misunderstanding the refractory period: after an action potential, the neurone cannot be immediately restimulated during the absolute refractory period (voltage-gated $\mathrm{Na}^+$ channels are inactivated). This ensures unidirectional propagation and limits firing frequency.
• Confusing oogenesis and spermatogenesis: oogenesis produces one ovum and polar bodies (asymmetric division, arrested at prophase I until puberty); spermatogenesis produces four functional sperm cells (symmetric division, continuous from puberty).
• Describing the Bohr effect incorrectly: increased $\mathrm{CO}_2$ decreases pH, which shifts the oxygen dissociation curve to the right, causing haemoglobin to release $\mathrm{O}_2$ more readily. The Bohr effect does not reduce the total oxygen-carrying capacity of blood.
• Confusing ventilation and gas exchange: ventilation is the mechanical movement of air in and out of the lungs; gas exchange is the diffusion of $\mathrm{O}_2$ and $\mathrm{CO}_2$ across the alveolar-capillary membrane.

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Exam-Style Problems

Problem 1: Data Analysis -- Oxygen Dissociation Curves

The following data show the percentage saturation of haemoglobin at different partial pressures of oxygen:

$\mathrm{pO}_2$ (kPa)	Normal Hb (%)	Fetal Hb (%)
1.0	10	30
2.0	25	55
4.0	55	80
6.0	75	90
8.0	85	95
10.0	92	97
13.3	98	98

(a) Plot both curves on the same axes. (b) Explain the physiological advantage of fetal haemoglobin having a higher affinity for oxygen at any given $\mathrm{pO}_2$. (c) Calculate the difference in $\mathrm{O}_2$ saturation between normal and fetal Hb at $\mathrm{pO}_2 = 4\;\mathrm{kPa}$, and explain the significance of this difference at the placental interface.

Problem 2: Extended Response -- Digestive Enzyme Regulation

Pepsin is secreted as inactive pepsinogen by chief cells in the stomach lining and is activated by $\mathrm{HCl}$. (a) Explain why pepsin is secreted in an inactive form, relating this to the structure of the stomach wall. (b) Describe the role of enterokinase (enteropeptidase) in activating pancreatic enzymes in the small intestine. (c) Compare and contrast the activation mechanisms of pepsinogen and trypsinogen, explaining how these prevent autodigestion of the digestive tract.

Problem 3: Quantitative -- Respiratory Volumes and Alveolar Ventilation

A patient has the following lung volumes: tidal volume $= 500\;\mathrm{mL}$, vital capacity $= 4500\;\mathrm{mL}$, residual volume $= 1500\;\mathrm{mL}$, respiratory rate $= 15\;\mathrm{breaths/min}$. The dead space volume is $150\;\mathrm{mL}$. Calculate: (a) the minute ventilation, (b) the alveolar ventilation rate, and (c) explain why alveolar ventilation is more physiologically relevant than minute ventilation.

Problem 4: Extended Response -- Synaptic Transmission and Neuropharmacology

Organophosphate pesticides inhibit acetylcholinesterase. (a) Describe the normal role of acetylcholinesterase at a cholinergic synapse. (b) Explain the effect of organophosphate poisoning on synaptic transmission, including the effect on postsynaptic membrane potential. (c) Describe two symptoms that would be expected and explain the physiological basis for each. (d) Explain how atropine (a muscarinic acetylcholine receptor antagonist) can be used as an antidote.

Problem 5: Data Analysis -- Menstrual Cycle Hormone Profiles

The following data show hormone concentrations across a $28$-day menstrual cycle:

Day	Oestrogen (pg/mL)	Progesterone (ng/mL)	LH (mIU/mL)	FSH (mIU/mL)
1	40	0.5	5	8
7	80	0.5	6	10
12	250	1.0	15	12
14	350	1.5	45	15
16	200	5.0	10	5
21	150	15.0	5	3
25	80	8.0	4	5
28	40	1.0	5	8

(a) Identify the day of ovulation and justify with reference to the data. (b) Describe the feedback mechanism responsible for the LH surge. (c) Explain why progesterone peaks after oestrogen, and what would happen to the cycle if progesterone secretion were artificially maintained at high levels.

Problem 6: Extended Response -- Type 1 and Type 2 Diabetes

Compare and contrast the aetiology (cause), pathophysiology, and treatment of Type 1 and Type 2 diabetes mellitus. In your response, include the role of beta cells, insulin receptors, and the concept of insulin resistance. Explain why a patient with Type 2 diabetes may not require insulin therapy initially, while a Type 1 patient always requires exogenous insulin.

Problem 7: Extended Response -- Ventilation Mechanics

A patient with emphysema has lost alveolar elasticity, reducing the elastic recoil of the lungs. (a) Explain how this affects the pressure gradients driving expiration. (b) Describe the consequence for residual volume and vital capacity. (c) Explain why this patient would have an elevated respiratory rate at rest compared with a healthy individual, relating your answer to alveolar ventilation and gas exchange efficiency.

Problem 8: Extended Response -- Kidney Function and Osmoregulation

The loop of Henle creates a concentration gradient in the medulla of the kidney. Describe how the countercurrent multiplier mechanism in the descending and ascending limbs of the loop of Henle establishes this gradient. Explain the role of antidiuretic hormone (ADH) in regulating water reabsorption in the collecting duct, and predict the appearance of urine (volume and concentration) in a person who has been drinking large volumes of water versus a person who is dehydrated.

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If You Get These Wrong, Revise:

• Cell membrane and transport mechanisms --> Review ./cell-biology
• DNA and gene expression --> Review ./molecular-biology
• Enzymes and metabolism --> Review ./molecular-biology
• Genetics and inheritance --> Review ./genetics
• Plant hormones and transport --> Review ./plant-biology

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8. The Excretory System

Structure and Function of the Kidney

The kidneys filter blood, remove metabolic waste (urea, uric acid, creatinine), regulate water and ion balance, and maintain blood pH.

Gross anatomy:

• Renal cortex: outer region containing glomeruli (Bowman's capsules) and convoluted tubules.
• Renal medulla: inner region containing the loop of Henle, collecting ducts, and renal pelvis.
• Renal pelvis: funnel-shaped chamber that collects urine and drains into the ureter.

The Nephron

The nephron is the functional unit of the kidney. Each kidney contains approximately $1$ million nephrons.

Structure:

1. Glomerulus: a knot of capillaries inside Bowman's capsule. Blood pressure forces filtrate (water, glucose, amino acids, ions, urea) into the capsule (ultrafiltration).
2. Bowman's capsule: cup-shaped structure surrounding the glomerulus; receives the filtrate.
3. Proximal convoluted tubule (PCT): reabsorbs $\approx 65\%$ of filtered $\mathrm{Na}^+$, water, $\mathrm{K}^+$, $\mathrm{Ca}^{2+}$, $\mathrm{HCO}_3^-$, glucose, and amino acids. Secretes $\mathrm{H}^+$ and $\mathrm{NH}_4^+$.
4. Loop of Henle: a U-shaped tubule that creates a concentration gradient in the medulla.
   • Descending limb: permeable to water (aquaporins); impermeable to solutes. Water exits by osmosis as the filtrate descends into the increasingly concentrated medulla.
   • Ascending limb (thin then thick segment): impermeable to water; actively transports $\mathrm{Na}^+$, $\mathrm{K}^+$, and $\mathrm{Cl}^-$ out (diluting the filtrate). The countercurrent multiplier mechanism concentrates the medullary interstitium to $\approx 1200\;\mathrm{mOsm/kg}$ (four times plasma osmolarity).
5. Distal convoluted tubule (DCT): fine-tunes ion balance. Reabsorbs $\mathrm{Na}^+$, $\mathrm{Ca}^{2+}$, and $\mathrm{HCO}_3^-$ under the control of aldosterone. Secretes $\mathrm{K}^+$ and $\mathrm{H}^+$.
6. Collecting duct: permeable to water under the control of ADH. Water exits by osmosis into the concentrated medulla, producing concentrated urine.

Ultrafiltration

Driven by hydrostatic pressure in the glomerular capillaries ($\approx 55\;\mathrm{mmHg}$) opposed by:

• Bowman's capsule hydrostatic pressure ($\approx 15\;\mathrm{mmHg}$)
• Glomerular oncotic pressure ($\approx 30\;\mathrm{mmHg}$, from plasma proteins)

Net filtration pressure: $\approx 55 - 15 - 30 = 10\;\mathrm{mmHg}$

The glomerular filtration rate (GFR) is approximately $125\;\mathrm{mL/min}$ per kidney ($250\;\mathrm{mL/min}$ total), producing $\approx 180\;\mathrm{L}$ of filtrate per day.

Selective Reabsorption

Substance	Filtered/day	Reabsorbed (%)	Excreted/day	Mechanism
Water	$180\;\mathrm{L}$	$\approx 99\%$	$\approx 1.5\;\mathrm{L}$	Osmosis (PCT, descending limb, collecting duct with ADH).
$\mathrm{Na}^+$	$\approx 600\;\mathrm{g}$	$\approx 99.5\%$	$\approx 3\;\mathrm{g}$	Active transport (PCT, ascending limb, DCT).
Glucose	$180\;\mathrm{g}$	$100\%$ (normally)	$\approx 0\;\mathrm{g}$	Secondary active transport ($\mathrm{Na}^+$-glucose co-transport, SGLT2, in PCT).
Urea	$\approx 50\;\mathrm{g}$	$\approx 50\%$	$\approx 25\;\mathrm{g}$	Passive diffusion and some secretion.
Amino acids	$\approx 65\;\mathrm{g}$	$\approx 100\%$	$\approx 1\;\mathrm{g}$	Secondary active transport with $\mathrm{Na}^+$.

Osmoregulation and ADH

Antidiuretic hormone (ADH / vasopressin) is produced by the hypothalamus and released from the posterior pituitary.

• When blood osmolarity is high (dehydration): osmoreceptors in the hypothalamus detect the increase. ADH is released, causing insertion of aquaporin-2 channels into the collecting duct apical membrane. Water is reabsorbed, producing concentrated urine (dark yellow, low volume).
• When blood osmolarity is low (overhydration): ADH secretion is suppressed. The collecting duct remains impermeable to water, producing dilute urine (pale, high volume).

Kidney Failure

• Acute kidney injury (AKI): sudden loss of kidney function; may be reversible.
• Chronic kidney disease (CKD): progressive loss of nephron function over months to years.
• Treatment: dialysis (haemodialysis or peritoneal dialysis) or kidney transplant.

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9. Liver Function

Roles of the Liver

1. Detoxification: converts ammonia (toxic) to urea (less toxic) via the ornithine cycle; breaks down alcohol and other toxins.
2. Bile production: synthesises bile salts (from cholesterol) and bile pigments (bilirubin from haem breakdown), stored in the gallbladder and released into the duodenum for fat emulsification.
3. Protein synthesis: synthesises plasma proteins (albumin, fibrinogen, prothrombin, most clotting factors).
4. Carbohydrate metabolism: stores glycogen (glycogenesis) and releases glucose (glycogenolysis); performs gluconeogenesis.
5. Lipid metabolism: synthesises lipoproteins (VLDL, HDL); breaks down fatty acids ($\beta$-oxidation); produces ketone bodies during fasting.
6. Iron storage: stores iron in ferritin; transports it as transferrin.
7. Vitamin storage: stores vitamins A, D, B$_{12}$.
8. Immune function: Kupffer cells (resident macrophages) phagocytose pathogens and old red blood cells.

Jaundice

Jaundice (yellowing of the skin and sclera) results from elevated bilirubin in the blood:

• Pre-hepatic (haemolytic): excessive RBC breakdown (e.g., sickle-cell disease, malaria).
• Hepatic: liver damage impairs bilirubin conjugation (e.g., hepatitis, cirrhosis).
• Post-hepatic (obstructive): blockage of bile flow prevents bilirubin excretion (e.g., gallstones, pancreatic cancer).

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10. Skin as an Organ

Structure

• Epidermis: outer layer; keratinised stratified squamous epithelium. Contains melanocytes (produce melanin for UV protection), Langerhans cells (antigen-presenting cells), and Merkel cells (touch receptors).
• Dermis: inner layer; contains connective tissue (collagen, elastin), blood vessels, nerve endings, hair follicles, sebaceous glands, and sweat glands.
• Hypodermis (subcutaneous layer): adipose tissue for insulation and energy storage.

Functions

• Thermoregulation: vasodilation/vasoconstriction of dermal blood vessels; sweating (evaporative cooling).
• Protection: keratinised epidermis provides a physical barrier; melanin absorbs UV radiation; sebum has antimicrobial properties; Langerhans cells provide immune surveillance.
• Sensory reception: Meissner's corpuscles (light touch), Pacinian corpuscles (pressure and vibration), nociceptors (pain), thermoreceptors (temperature).
• Vitamin D synthesis: UV radiation converts 7-dehydrocholesterol in the skin to cholecalciferol (vitamin D3), which is converted in the liver and kidneys to the active form ($1,25$-dihydroxyvitamin D).
• Excretion: sweat contains water, $\mathrm{Na}^+$, $\mathrm{Cl}^-$, urea, and lactic acid.

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11. Extended: Lung Volumes and Ventilation Mechanics

Lung Volumes and Capacities

Volume	Definition	Typical value (adult male)
Tidal volume (TV)	Volume of air inhaled/exhaled during normal breathing	$\approx 500\;\mathrm{mL}$
Inspiratory reserve volume (IRV)	Maximum additional air inhaled after normal inspiration	$\approx 3100\;\mathrm{mL}$
Expiratory reserve volume (ERV)	Maximum additional air exhaled after normal expiration	$\approx 1200\;\mathrm{mL}$
Residual volume (RV)	Air remaining in lungs after maximal expiration	$\approx 1200\;\mathrm{mL}$
Vital capacity (VC)	TV + IRV + ERV (maximum air that can be moved)	$\approx 4800\;\mathrm{mL}$
Total lung capacity (TLC)	VC + RV	$\approx 6000\;\mathrm{mL}$
Dead space	Air in conducting airways not involved in gas exchange	$\approx 150\;\mathrm{mL}$
Minute ventilation	TV $\times$ respiratory rate	$\approx 7500\;\mathrm{mL/min}$ (at 15 breaths/min)
Alveolar ventilation	(TV - dead space) $\times$ respiratory rate	$\approx 5250\;\mathrm{mL/min}$

Alveolar Gas Exchange

Fick's Law applied to alveolar exchange:

$\text{Rate of diffusion} \propto \frac{A \times \Delta P}{d}$

• $A$ (surface area) $\approx 70\;\mathrm{m}^2$ in adult lungs.
• $\Delta P$ (partial pressure gradient) maintained by ventilation.
• $d$ (diffusion distance) $\approx 1\;\mathrm{\mu m}$ (alveolar epithelium + capillary endothelium).

Alveolar gas equation:

$P_A\mathrm{O}_2 = P_I\mathrm{O}_2 - \frac{P_A\mathrm{CO}_2}{R}$

where $R$ is the respiratory exchange ratio ($\approx 0.85$ at rest).

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12. ECG and the Cardiac Cycle (Extended)

The Electrocardiogram (ECG)

The ECG records the electrical activity of the heart using surface electrodes.

Wave / Segment	Duration (s)	Amplitude (mV)	Cardiac event
P wave	$0.08$--$0.12$	$< 0.25$	Atrial depolarisation (contraction).
PR interval	$0.12$--$0.20$	---	Atrioventricular conduction delay (AV node).
QRS complex	$0.06$--$0.10$	$1.0$--$1.5$	Ventricular depolarisation (contraction). Bundle of His + bundle branches + Purkinje fibres.
ST segment	$0.08$--$0.12$	Isoelectric (flat)	Ventricular repolarisation beginning. Elevation = ischaemia (STEMI).
T wave	$0.16$	$0.1$--$0.5$	Ventricular repolarisation.

Clinical significance:

• Tachycardia: heart rate $> 100\;\mathrm{bpm}$.
• Bradycardia: heart rate $< 60\;\mathrm{bpm}$.
• Atrial fibrillation: irregular P waves; loss of coordinated atrial contraction.
• Ventricular fibrillation: chaotic, disorganised ECG; no cardiac output; fatal without defibrillation.
• Myocardial infarction (heart attack): ST segment elevation (STEMI); inverted T waves; elevated cardiac enzymes (troponin, CK-MB).

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Exam-Style Problems (Extended)

Problem 9: Extended Response -- Nephron Function and Osmoregulation

Describe the process of ultrafiltration in the glomerulus, identifying the forces that favour and oppose filtration and calculating the net filtration pressure. Explain how the loop of Henle establishes a concentration gradient in the medulla using the countercurrent multiplier mechanism. Explain the role of ADH in regulating water reabsorption in the collecting duct, and predict the volume and concentration of urine produced by a person who has been exercising vigorously in a hot environment (sweating heavily) versus a person at rest who has been drinking water excessively.

Problem 10: Quantitative -- Kidney Filtration

A patient's GFR is measured at $80\;\mathrm{mL/min}$ (normal: $125\;\mathrm{mL/min}$). Plasma glucose concentration is $5\;\mathrm{mmol/L}$ (MW $= 180\;\mathrm{g/mol}$). (a) Calculate the mass of glucose filtered per day. (b) If the patient has diabetes mellitus and the renal threshold for glucose is $10\;\mathrm{mmol/L}$, calculate the mass of glucose excreted per day. (c) Explain why glucosuria (glucose in urine) causes increased urine volume (polyuria).

Problem 11: Extended Response -- Liver Function and Jaundice

Describe the biochemical pathway by which bilirubin is produced from haemoglobin, transported in the blood (conjugated vs unconjugated), excreted in bile, and ultimately eliminated in faeces. Explain how each of the following conditions causes jaundice: (a) haemolytic anaemia, (b) hepatitis B infection, (c) gallstones blocking the common bile duct. For each, explain whether the jaundice is pre-hepatic, hepatic, or post-hepatic.

Problem 12: Data Analysis -- Spirometry

A spirometry test produces the following results for a patient: FVC = $3.0\;\mathrm{L}$ (predicted: $4.5\;\mathrm{L}$), $\mathrm{FEV}_1 = 1.8\;\mathrm{L}$ (predicted: $3.6\;\mathrm{L}$), $\mathrm{FEV}_1/\mathrm{FVC}$ ratio $= 0.60$ (predicted: $> 0.75$). (a) Calculate the percentage predicted values for FVC and $\mathrm{FEV}_1$. (b) Interpret the $\mathrm{FEV}_1/\mathrm{FVC}$ ratio and diagnose the type of respiratory disorder (obstructive vs restrictive). (c) Explain the physiological basis for the reduced $\mathrm{FEV}_1$ in obstructive lung disease (e.g., asthma, COPD). (d) Explain why $\mathrm{FEV}_1/\mathrm{FVC}$ is normal or increased in restrictive lung disease (e.g., pulmonary fibrosis).

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Additional Worked Examples

Worked Example: Cardiac Output and Blood Pressure

A patient has a heart rate of $72\;\mathrm{bpm}$ and a stroke volume of $70\;\mathrm{mL}$. (a) Calculate the cardiac output. (b) If the mean arterial pressure is $93\;\mathrm{mmHg}$, calculate the total peripheral resistance (TPR). (c) During exercise, heart rate increases to $150\;\mathrm{bpm}$ and stroke volume increases to $120\;\mathrm{mL}$. Calculate the new cardiac output and the percentage increase. (d) Explain how the baroreceptor reflex responds to a sudden drop in blood pressure.

Solution

(a) $\text{CO} = \text{HR} \times \text{SV} = 72 \times 70 = 5040\;\mathrm{mL/min} = 5.04\;\mathrm{L/min}$.

(b) $\text{MAP} = \text{CO} \times \text{TPR}$, so $\text{TPR} = \text{MAP} / \text{CO}$. Need consistent units. $\text{CO} = 5040\;\mathrm{mL/min} = 0.084\;\mathrm{L/s}$. $\text{TPR} = 93 / 0.084 = 1107\;\mathrm{mmHg \cdot s/L}$ (or $\text{PRU}$, peripheral resistance units). Normal TPR $= 1000$--$1200\;\mathrm{PRU}$.

(c) $\text{CO}_{exercise} = 150 \times 120 = 18000\;\mathrm{mL/min} = 18.0\;\mathrm{L/min}$. Percentage increase $= \frac{18.0 - 5.04}{5.04} \times 100 = 257\%$.

(d) Baroreceptor reflex (drop in blood pressure):

1. Baroreceptors in the aortic arch and carotid sinus detect decreased stretch (lower pressure).
2. They send fewer action potentials via the vagus (CN X) and glossopharyngeal (CN IX) nerves to the vasomotor centre and cardiovascular centre in the medulla oblongata.
3. The medulla increases sympathetic outflow and decreases parasympathetic outflow.
4. Effects: increased heart rate (noradrenaline on SA node), increased stroke volume (noradrenaline on ventricular muscle), and vasoconstriction (noradrenaline on arterioles).
5. These changes increase cardiac output and TPR, restoring blood pressure toward normal.
6. This is a negative feedback mechanism.

Worked Example: Digestion and Absorption Calculations

A meal contains $50\;\mathrm{g}$ of protein, $80\;\mathrm{g}$ of carbohydrate, and $20\;\mathrm{g}$ of fat. (a) Calculate the total energy content of the meal in kcal. (b) Calculate the volume of gastric juice needed to digest the protein, assuming pepsin requires a pH of $2$ and the stomach secretes $\mathrm{HCl}$ at $0.15\;\mathrm{mol/L}$. (c) Calculate the minimum volume of bile needed to emulsify the fat, assuming each gram of fat requires $10\;\mathrm{mL}$ of bile for adequate emulsification. (d) Calculate the number of ATP molecules theoretically produced from the absorbed glucose (from the carbohydrate).

Solution

(a) Energy content (using Atwater factors):

• Protein: $50 \times 4 = 200\;\mathrm{kcal}$
• Carbohydrate: $80 \times 4 = 320\;\mathrm{kcal}$
• Fat: $20 \times 9 = 180\;\mathrm{kcal}$
• Total: $200 + 320 + 180 = 700\;\mathrm{kcal}$

(b) This requires knowledge of the buffering capacity of the meal. In practice, the stomach secretes approximately $2$--$3\;\mathrm{L}$ of gastric juice per meal. The question as stated cannot be precisely answered without additional information (the buffering capacity of the meal, the initial pH of the stomach contents). The key point is that parietal cells secrete $\mathrm{HCl}$ at approximately $0.15\;\mathrm{mol/L}$, and approximately $2$--$3\;\mathrm{L}$ is produced per meal.

(c) Volume of bile $= 20 \times 10 = 200\;\mathrm{mL}$. Note: the liver produces approximately $500$--$1000\;\mathrm{mL}$ of bile per day, stored and concentrated in the gallbladder. The gallbladder releases bile in response to CCK (cholecystokinin) when fat enters the duodenum.

(d) Glucose from $80\;\mathrm{g}$ carbohydrate (assuming all is digestible starch/sugar): $80\;\mathrm{g} / 180\;\mathrm{g/mol} = 0.444\;\mathrm{mol}$ glucose. Complete oxidation: $1$ glucose $\to$ $30$--$32\;\mathrm{ATP}$ (using modern P/O ratios). Total ATP $= 0.444 \times 32 \times 6.022 \times 10^{23} = 8.56 \times 10^{24}$ ATP molecules.

Worked Example: Renal Function and Clearance

A patient produces $1440\;\mathrm{mL}$ of urine per day. The urine creatinine concentration is $1.5\;\mathrm{g/L}$ and the plasma creatinine concentration is $0.010\;\mathrm{g/L}$. (a) Calculate the creatinine clearance. (b) The patient's inulin clearance is $110\;\mathrm{mL/min}$. What does this value represent? (c) The urine glucose concentration is $2.0\;\mathrm{g/L}$ and the plasma glucose is $1.0\;\mathrm{g/L}$. Calculate the glucose clearance and explain the significance of this value. (d) If the GFR is $125\;\mathrm{mL/min}$ and $99\%$ of the filtered water is reabsorbed, calculate the expected daily urine volume.

Solution

(a) Creatinine clearance $= \frac{U \times V}{P}$ where $U = 1.5\;\mathrm{g/L}$, $P = 0.010\;\mathrm{g/L}$, $V = 1440\;\mathrm{mL/day} = 1.0\;\mathrm{mL/min}$.

$\text{Clearance} = \frac{1.5 \times 1.0}{0.010} = 150\;\mathrm{mL/min}$.

Normal creatinine clearance (men): $105$--$125\;\mathrm{mL/min}$. This patient has an elevated clearance, which may indicate high muscle mass (creatine is converted to creatinine, which is filtered by the glomerulus and not significantly reabsorbed or secreted -- actually, creatinine IS slightly secreted, so clearance slightly overestimates GFR).

(b) Inulin clearance of $110\;\mathrm{mL/min}$ represents the glomerular filtration rate (GFR). Inulin is an ideal marker for GFR because it is freely filtered at the glomerulus and neither reabsorbed nor secreted by the renal tubules. Normal GFR is approximately $120$--$130\;\mathrm{mL/min}$; $110$ is slightly reduced but within or near normal range.

(c) Glucose clearance $= \frac{2.0 \times 1.0}{1.0} = 2.0\;\mathrm{mL/min}$.

This is much lower than GFR ($110\;\mathrm{mL/min}$), indicating that glucose is almost completely reabsorbed in the proximal convoluted tubule. Normally, glucose clearance is approximately $0$ (all filtered glucose is reabsorbed). A clearance of $2.0\;\mathrm{mL/min}$ suggests some glucose is spilling into the urine, which occurs when plasma glucose exceeds the renal threshold (approximately $180\;\mathrm{mg/dL} = 1.8\;\mathrm{g/L}$). This patient's plasma glucose ($1.0\;\mathrm{g/L}$) is below the threshold, so the small clearance may reflect tubular dysfunction or incomplete reabsorption.

(d) Expected daily urine volume: Filtration rate $= 125\;\mathrm{mL/min} \times 60 \times 24 = 180\,000\;\mathrm{mL/day} = 180\;\mathrm{L/day}$. If $99\%$ is reabsorbed: urine volume $= 180 \times 0.01 = 1.8\;\mathrm{L/day}$.

This matches normal urine output ($1$--$2\;\mathrm{L/day}$).

Worked Example: Hormonal Control of Blood Glucose

After a carbohydrate-rich meal, blood glucose rises from $5.0$ to $8.5\;\mathrm{mmol/L}$. (a) Describe the hormonal response. (b) Calculate the total mass of glucose in the blood if blood volume is $5\;\mathrm{L}$ (molar mass of glucose $= 180\;\mathrm{g/mol}$). (c) Explain how insulin promotes glucose uptake into muscle and adipose cells. (d) A type 1 diabetic patient has a fasting blood glucose of $15\;\mathrm{mmol/L}$. Explain why glucose appears in the urine.

Solution

(a) Hormonal response to rising blood glucose:

1. Beta cells of the pancreatic islets detect increased blood glucose (via GLUT2 transporters and glucokinase, which phosphorylates glucose, triggering ATP production and closure of $\mathrm{K}^+_{ATP}$ channels, depolarising the cell and opening voltage-gated $\mathrm{Ca}^{2+}$ channels).
2. $\mathrm{Ca}^{2+}$ influx triggers exocytosis of insulin-containing secretory vesicles.
3. Insulin is released into the blood and acts on target cells (liver, muscle, adipose tissue).
4. Glucagon secretion by alpha cells is suppressed (paracrine inhibition by insulin and direct effect of high glucose on alpha cells).
5. Blood glucose returns to normal ($4$--$6\;\mathrm{mmol/L}$) within approximately $2$ hours.

(b) At $8.5\;\mathrm{mmol/L}$: mass of glucose $= 8.5 \times 10^{-3}\;\mathrm{mol/L} \times 5\;\mathrm{L} \times 180\;\mathrm{g/mol} = 7.65\;\mathrm{g}$. At $5.0\;\mathrm{mmol/L}$: $5.0 \times 10^{-3} \times 5 \times 180 = 4.5\;\mathrm{g}$. Increase: $7.65 - 4.5 = 3.15\;\mathrm{g}$ of glucose in the blood.

(c) Insulin promotes glucose uptake by:

1. Binding to insulin receptors (tyrosine kinase receptors) on the cell membrane.
2. Triggering an intracellular signalling cascade involving IRS-1, PI3K, and Akt.
3. Akt promotes the translocation of GLUT4 vesicles from intracellular stores to the cell membrane.
4. GLUT4 transporters increase the membrane's permeability to glucose, allowing facilitated diffusion of glucose into the cell down its concentration gradient.
5. In muscle, glucose is used for glycogen synthesis or glycolysis. In adipose tissue, glucose is used for triglyceride synthesis.

(d) In type 1 diabetes, the beta cells are destroyed (autoimmune), so no insulin is produced. Without insulin:

• GLUT4 translocation does not occur, so glucose cannot enter muscle and adipose cells efficiently.
• Blood glucose remains elevated ($15\;\mathrm{mmol/L}$, well above the normal renal threshold of approximately $10\;\mathrm{mmol/L}$).
• The filtered load of glucose exceeds the maximum tubular reabsorption capacity ($T_m$).
• The excess glucose is excreted in the urine (glycosuria), creating an osmotic diuresis (the glucose in the tubule draws water with it, producing large volumes of urine -- polyuria).

Worked Example: Oxygen-Haemoglobin Dissociation and Bohr Effect

At sea level, alveolar $\mathrm{pO}_2 = 13.3\;\mathrm{kPa}$, tissue $\mathrm{pO}_2 = 4.0\;\mathrm{kPa}$. Tissue $\mathrm{pCO}_2 = 6.0\;\mathrm{kPa}$ and tissue pH $= 7.2$. (a) Using the oxygen dissociation curve, estimate the percentage saturation of haemoglobin in the lungs and in the tissues (under normal conditions and Bohr-shifted conditions). (b) Explain the molecular mechanism of the Bohr effect. (c) Calculate the oxygen delivery per litre of blood if haemoglobin concentration is $150\;\mathrm{g/L}$ and each gram carries $1.34\;\mathrm{mL\;O}_2$, comparing normal and Bohr-shifted conditions.

Solution

(a) Normal conditions (pH 7.4):

• Lungs ($\mathrm{pO}_2 = 13.3\;\mathrm{kPa}$): $\approx 97\%$ saturated.
• Tissues ($\mathrm{pO}_2 = 4.0\;\mathrm{kPa}$): $\approx 55\%$ saturated.
• $\mathrm{O}_2$ unloaded: $97 - 55 = 42\%$.

Bohr-shifted conditions (pH 7.2, high $\mathrm{pCO}_2$):

• Lungs: $\approx 97\%$ (the curve shift has minimal effect at high $\mathrm{pO}_2$).
• Tissues: $\approx 35\%$ (the curve is shifted to the right, so saturation is lower at the same $\mathrm{pO}_2$).
• $\mathrm{O}_2$ unloaded: $97 - 35 = 62\%$.

The Bohr effect increases $\mathrm{O}_2$ unloading by $20$ percentage points ($62\%$ vs $42\%$).

(b) The Bohr effect is caused by:

1. $\mathrm{H}^+$ (from carbonic acid: $\mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^-$) binds to haemoglobin, stabilising the T (tense, deoxygenated) state.
2. $\mathrm{H}^+$ binds to specific amino acid residues (His residues on the beta chains), forming salt bridges that stabilise the deoxygenated conformation.
3. Increased $\mathrm{CO}_2$ also promotes carbamino formation ($\mathrm{CO}_2$ binding to the N-terminal amino groups of haemoglobin chains), further stabilising the T state.
4. The combined effect shifts the oxygen dissociation curve to the right, reducing haemoglobin's affinity for $\mathrm{O}_2$ and facilitating $\mathrm{O}_2$ release in metabolically active tissues (which produce $\mathrm{CO}_2$ and $\mathrm{H}^+$).

(c) Total $\mathrm{O}_2$-carrying capacity per litre: $150 \times 1.34 = 201\;\mathrm{mL\;O}_2/L$.

Normal conditions: $\mathrm{O}_2$ delivered = 201 \times 0.42 = 84.4\;\mathrm{mL\;O}_2/L\;blood}. Bohr-shifted conditions: $\mathrm{O}_2$ delivered = 201 \times 0.62 = 124.6\;\mathrm{mL\;O}_2/L\;blood}.

The Bohr effect increases oxygen delivery by approximately $47\%$ during exercise.

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Additional Common Pitfalls

• Confusing tidal volume and vital capacity: tidal volume is the volume of air in a normal breath ($\approx 500\;\mathrm{mL}$); vital capacity is the maximum volume that can be exhaled after maximum inhalation ($\approx 4.5\;\mathrm{L}$).
• Confusing the hepatic portal vein and hepatic vein: the portal vein carries nutrient-rich blood from the gut to the liver (for processing); the hepatic vein carries processed blood from the liver to the vena cava.
• Forgetting that the loop of Henle creates a concentration gradient, not directly concentrated urine: the loop of Henle establishes the medullary osmotic gradient; water reabsorption in the collecting duct (under ADH control) actually produces concentrated urine.
• Confusing spermatogenesis and oogenesis: spermatogenesis produces 4 sperm from each primary spermatocyte continuously from puberty; oogenesis produces 1 ovum (plus 2--3 polar bodies) from each primary oocyte, with meiosis I arrested at prophase I until puberty and meiosis II at fertilisation.
• Stating that the SA node initiates contraction of the ventricles: the SA node initiates atrial contraction; the impulse reaches the ventricles via the AV node, bundle of His, and Purkinje fibres, after a delay that allows the atria to contract first.
• Confusing antibodies and antibiotics: antibodies are produced by the immune system (B cells) against specific antigens; antibiotics are drugs that kill or inhibit bacteria.

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Additional Exam-Style Problems with Full Solutions

Problem 13: Extended Response -- Liver and Metabolic Integration

The liver plays a central role in metabolic integration. Describe the liver's role in: (a) carbohydrate metabolism (glycogenesis, glycogenolysis, gluconeogenesis), (b) lipid metabolism (lipogenesis, beta-oxidation, lipoprotein synthesis), (c) protein metabolism (deamination, transamination, urea synthesis), and (d) detoxification (alcohol metabolism, drug metabolism). Explain how these processes are hormonally regulated.

Answer 13

(a) Carbohydrate metabolism:

• Glycogenesis: conversion of excess glucose to glycogen for storage. Stimulated by insulin (after a meal). The enzyme glycogen synthase is activated by insulin (via dephosphorylation by protein phosphatase 1).
• Glycogenolysis: breakdown of glycogen to glucose-6-phosphate. Stimulated by glucagon and adrenaline. The enzyme glycogen phosphorylase is activated by phosphorylation (protein kinase A, activated by cAMP).
• Gluconeogenesis: synthesis of glucose from non-carbohydrate precursors (lactate, glycerol, alanine). Stimulated by glucagon and cortisol. Key enzymes (PEPCK, fructose-1,6-bisphosphatase, glucose-6-phosphatase) are induced by glucagon and repressed by insulin.

(b) Lipid metabolism:

• Lipogenesis: conversion of excess glucose to fatty acids and triglycerides. Stimulated by insulin. Acetyl-CoA carboxylase (rate-limiting enzyme) is activated by insulin.
• Beta-oxidation: breakdown of fatty acids to acetyl-CoA for energy (in liver mitochondria, the acetyl-CoA cannot enter the Krebs cycle at high rates; instead, it is converted to ketone bodies). Stimulated by glucagon (when blood glucose is low).
• Lipoprotein synthesis: the liver synthesises VLDL (very-low-density lipoprotein) to transport triglycerides to peripheral tissues, and HDL (high-density lipoprotein) for reverse cholesterol transport. VLDL synthesis is stimulated by insulin.

(c) Protein metabolism:

• Deamination: removal of the amino group from amino acids, producing ammonia ($\mathrm{NH}_3$) and a keto acid. The keto acid can enter the Krebs cycle or be used for gluconeogenesis.
• Transamination: transfer of an amino group from one amino acid to a keto acid, producing a different amino acid. Catalysed by transaminases (aminotransferases), which require pyridoxal phosphate (vitamin $\mathrm{B}_6$) as a cofactor.
• Urea synthesis (ornithine cycle): ammonia (highly toxic) is converted to urea (less toxic) in the liver. The urea cycle occurs partly in mitochondria and partly in the cytoplasm. Urea is excreted by the kidneys. The cost is $4\;\mathrm{ATP}$ per urea synthesised (2 ATP are used to make carbamoyl phosphate; 1 ATP is used to make argininosuccinate; 1 ATP is hydrolysed to AMP).

(d) Detoxification:

• Alcohol metabolism: ethanol is oxidised by alcohol dehydrogenase (ADH) to acetaldehyde (toxic), then by aldehyde dehydrogenase (ALDH) to acetate (harmless), which enters general metabolism. Acetaldehyde accumulation causes the unpleasant symptoms of hangover and is responsible for alcohol-related liver damage (cirrhosis).
• Drug metabolism (phase I and II): phase I reactions (cytochrome P450 enzymes in the smooth ER) modify the drug by oxidation, reduction, or hydrolysis; phase II reactions conjugate the drug or its metabolites with polar groups (glucuronic acid, glutathione, sulphate) to increase water solubility for renal excretion.

Hormonal regulation: insulin promotes storage (glycogenesis, lipogenesis, protein synthesis); glucagon and adrenaline promote mobilisation (glycogenolysis, gluconeogenesis, lipolysis). Cortisol promotes gluconeogenesis and protein catabolism during prolonged stress.

Problem 14: Quantitative -- ECG Interpretation

An ECG recording shows the following intervals:

• P wave duration: $120\;\mathrm{ms}$ (normal: $< 120\;\mathrm{ms}$)
• PR interval: $200\;\mathrm{ms}$ (normal: $120$--$200\;\mathrm{ms}$)
• QRS complex duration: $80\;\mathrm{ms}$ (normal: $80$--$120\;\mathrm{ms}$)
• QT interval: $400\;\mathrm{ms}$ (normal: $350$--$440\;\mathrm{ms}$)
• Heart rate: $75\;\mathrm{bpm}$

(a) Correlate each ECG component with the corresponding cardiac event. (b) Are any of the intervals abnormal? (c) Calculate the corrected QT interval ($\mathrm{QTc} = \mathrm{QT} / \sqrt{RR}$). (d) Explain what first-degree heart block looks like on an ECG.

Answer 14

(a) ECG correlations:

• P wave: atrial depolarisation (SA node impulse spreads through both atria).
• PR interval: time from the start of atrial depolarisation to the start of ventricular depolarisation; includes AV node delay.
• QRS complex: ventricular depolarisation (impulse spreads through the bundle of His, bundle branches, and Purkinje fibres). The atria also repolarise during this interval, but the signal is masked by the large ventricular depolarisation.
• QT interval: ventricular depolarisation plus repolarisation (represents the duration of ventricular action potential).
• T wave: ventricular repolarisation.

(b) All intervals are within or at the upper limit of normal:

• P wave: $120\;\mathrm{ms}$ (at the upper limit; $< 120$ is strictly normal).
• PR interval: $200\;\mathrm{ms}$ (at the upper limit of normal; $> 200\;\mathrm{ms}$ indicates first-degree heart block).
• QRS: $80\;\mathrm{ms}$ (normal).
• QT: $400\;\mathrm{ms}$ (normal).

The PR interval of $200\;\mathrm{ms}$ is borderline and may indicate first-degree heart block.

(c) RR interval $= 60 / 75 = 0.80\;\mathrm{s}$. $\mathrm{QTc} = 400 / \sqrt{0.80} = 400 / 0.894 = 447\;\mathrm{ms}$. Normal $\mathrm{QTc}$ is $< 440\;\mathrm{ms}$ (men) or $< 450\;\mathrm{ms}$ (women). This is borderline prolonged. A prolonged QTc increases the risk of torsades de pointes (a type of ventricular tachycardia).

(d) First-degree heart block: every P wave is followed by a QRS complex (no dropped beats), but the PR interval is prolonged ($> 200\;\mathrm{ms}$). This indicates delayed conduction through the AV node. It is often benign but may progress to higher-degree blocks. Causes include ischaemic heart disease, drug effects (beta-blockers, calcium channel blockers, digoxin), and age-related fibrosis of the conduction system.

Problem 15: Extended Response -- Kidney and Osmoregulation

Describe the mechanism by which the kidney produces concentrated urine under the influence of antidiuretic hormone (ADH, vasopressin). Include: (a) the role of the loop of Henle in creating the medullary osmotic gradient, (b) the role of the vasa recta in maintaining the gradient, (c) the mechanism of ADH action on the collecting duct, and (d) the difference between water reabsorption in the presence and absence of ADH.

Answer 15

(a) Loop of Henle and the medullary gradient:

• The descending limb is permeable to water but not to solutes. As filtrate descends into the increasingly hypertonic medulla, water leaves by osmosis, concentrating the filtrate.
• The thin ascending limb is impermeable to water but permeable to $\mathrm{Na}^+$ and $\mathrm{Cl}^-$, which diffuse passively into the medullary interstitium (countercurrent multiplication).
• The thick ascending limb actively transports $\mathrm{Na}^+$, $\mathrm{K}^+$, and $\mathrm{Cl}^-$ out of the filtrate via the $\mathrm{Na}^+/2\mathrm{Cl}^-/\mathrm{K}^+$ co-transporter (NKCC2), further diluting the filtrate and concentrating the medulla. This segment is impermeable to water.
• The result is a gradient from approximately $300\;\mathrm{mOsm}$ (cortex) to $1200\;\mathrm{mOsm}$ (inner medulla).

(b) Vasa recta (countercurrent exchange):

• The vasa recta are hairpin-shaped capillaries that run alongside the loop of Henle.
• As blood descends into the medulla, water leaves the blood (drawn out by the high osmolarity) and solutes enter (from the concentrated interstitium).
• As blood ascends, the reverse occurs: solutes leave the blood and water enters.
• This countercurrent exchange prevents the vasa recta from "washing out" the medullary gradient. The vasa recta carry away the excess water and solutes while maintaining the gradient.

(c) ADH action on the collecting duct:

1. ADH is released from the posterior pituitary in response to increased plasma osmolarity (detected by osmoreceptors in the hypothalamus) or decreased blood volume (detected by baroreceptors).
2. ADH binds to V2 receptors on the basolateral membrane of principal cells in the collecting duct.
3. This activates a G-protein-coupled signalling cascade, increasing intracellular cAMP.
4. cAMP activates protein kinase A (PKA), which triggers the insertion of aquaporin-2 (AQP2) water channels into the apical membrane via vesicle trafficking.
5. Water exits the collecting duct through AQP2 channels and enters the blood through constitutively present AQP3/AQP4 channels on the basolateral membrane.

(d) With vs without ADH:

• With ADH: the collecting duct is highly permeable to water. As filtrate passes through the hypertonic medulla, water is reabsorbed, producing concentrated urine (up to $1200\;\mathrm{mOsm}$, as low as $500\;\mathrm{mL/day}$).
• Without ADH (e.g., diabetes insipidus, or after large fluid intake): AQP2 channels are removed from the apical membrane. The collecting duct is impermeable to water. Dilute urine is produced (approximately $50\;\mathrm{mOsm}$, up to $20\;\mathrm{L/day}$), regardless of the medullary gradient.

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Cross-References to Related Topics

• Cell membrane and transport: Review ./cell-biology for membrane proteins, osmosis, and active transport.
• Gas exchange and haemoglobin: Review ./human-physiology for the oxygen dissociation curve (within this document).
• Nervous system and heart regulation: Review ./nervous-system for autonomic control of heart rate and reflex arcs.
• Immunology and defence: Review ./immunology for fever, inflammation, and antibody production.
• Metabolism and ATP: Review ./metabolism-cell-biology for cellular respiration and energy production in tissues.

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Supplementary: Endocrine System in Detail (HL Extension)

Hormone Classification

Hormones are classified by their chemical structure, which determines their mechanism of action:

Class	Examples	Solubility	Receptor location	Mechanism
Peptide/protein	Insulin, glucagon, ADH, FSH, LH, TSH, GH	Hydrophilic	Cell surface (GPCR or tyrosine kinase)	Second messenger cascade (cAMP, IP3/DAG, Ca2+)
Steroid	Cortisol, testosterone, oestradiol, progesterone, aldosterone	Lipophilic	Intracellular (cytoplasmic or nuclear)	Directly modulate gene transcription
Amino acid derivative	Thyroxine (T4), triiodothyronine (T3), adrenaline, noradrenaline	Mixed	Adrenaline: cell surface; T3/T4: intracellular	Second messenger (adrenaline) or gene transcription (T3/T4)

Peptide Hormones: Insulin Signalling Pathway (Detailed)

Insulin is a peptide hormone (51 amino acids, two chains linked by disulphide bonds) produced by beta cells of the pancreatic islets of Langerhans.

Synthesis and secretion:

1. Preproinsulin is synthesised on ribosomes bound to the rough ER.
2. The signal peptide is cleaved in the ER lumen, producing proinsulin.
3. Proinsulin is transported to the Golgi, where it is packaged into secretory vesicles.
4. In the vesicles, proinsulin is cleaved by prohormone convertases (PC1/PC3) and carboxypeptidase E, producing insulin (A chain + B chain) and C-peptide.
5. Glucose enters the beta cell via GLUT2 transporters and is phosphorylated by glucokinase.
6. Increased ATP closes $\mathrm{K}^+_{\mathrm{ATP}}$ channels, depolarising the cell.
7. Voltage-gated $\mathrm{Ca}^{2+}$ channels open; $\mathrm{Ca}^{2+}$ influx triggers exocytosis of insulin vesicles.

Insulin receptor signalling:

1. Insulin binds to the insulin receptor (a receptor tyrosine kinase) on the target cell surface.
2. The receptor autophosphorylates on tyrosine residues.
3. Insulin receptor substrates (IRS-1, IRS-2) bind to the receptor and are phosphorylated.
4. IRS activates PI3K (phosphoinositide 3-kinase), which generates PIP3 from PIP2.
5. PIP3 activates PDK1, which phosphorylates Akt (protein kinase B).
6. Akt phosphorylates multiple targets:
   • GLUT4 translocation: Akt promotes the movement of GLUT4 vesicles to the cell membrane (in muscle and adipose tissue), increasing glucose uptake.
   • Glycogen synthesis: Akt inactivates GSK-3 (glycogen synthase kinase 3), which normally inhibits glycogen synthase. Net effect: increased glycogen synthesis.
   • Protein synthesis: Akt activates mTOR, stimulating protein synthesis and cell growth.
   • Lipogenesis: Akt activates SREBP (sterol regulatory element-binding protein), promoting fatty acid and triglyceride synthesis.

Steroid Hormones: Cortisol Action

Cortisol is a glucocorticoid produced by the zona fasciculata of the adrenal cortex. It is lipophilic and diffuses directly through the plasma membrane.

Mechanism of action:

1. Cortisol diffuses through the cell membrane and binds to the glucocorticoid receptor (GR) in the cytoplasm. The GR is bound to heat shock proteins (HSPs) that keep it inactive.
2. Cortisol binding causes a conformational change, releasing HSPs and exposing the nuclear localisation signal.
3. The cortisol-GR complex translocates to the nucleus.
4. The complex binds to glucocorticoid response elements (GREs) in the promoter regions of target genes, recruiting co-activators or co-repressors and modulating transcription.
5. Alternatively, the cortisol-GR complex can interact with other transcription factors (e.g., NF-$\kappa$B) in the cytoplasm, inhibiting their activity (transrepression). This is how cortisol suppresses inflammation (by blocking NF-$\kappa$B, which activates inflammatory gene expression).

Effects of cortisol:

• Metabolic: increases blood glucose by promoting gluconeogenesis (in the liver), reducing glucose uptake by peripheral tissues, and promoting protein catabolism (muscle wasting) and lipolysis.
• Anti-inflammatory: suppresses the immune response by inhibiting cytokine production, reducing leukocyte migration, and promoting apoptosis of lymphocytes.
• Stress response: cortisol is the end product of the HPA (hypothalamic-pituitary-adrenal) axis, released in response to physical or psychological stress.

The Hypothalamic-Pituitary Axis

The hypothalamus controls the anterior pituitary via releasing and inhibiting hormones, which are transported through the hypophyseal portal system:

Hypothalamic hormone	Effect on anterior pituitary	Target gland	Main hormone
GnRH	Stimulates	Gonads	FSH, LH
TRH	Stimulates	Thyroid	TSH
CRH	Stimulates	Adrenal cortex	ACTH
GHRH	Stimulates	Liver, bones	GH
Somatostatin	Inhibits	--	GH, TSH
Dopamine (PIH)	Inhibits	--	Prolactin

Negative feedback: hormones produced by the target glands (thyroid hormones, cortisol, sex steroids) feed back to the hypothalamus and pituitary to inhibit further release of releasing hormones and tropic hormones. This maintains hormone levels within a narrow physiological range.

Example (thyroid axis):

• Low $\mathrm{T}_3/\mathrm{T}_4$ $\to$ hypothalamus releases TRH $\to$ pituitary releases TSH $\to$ thyroid releases $\mathrm{T}_3/\mathrm{T}_4$ $\to$ blood levels rise $\to$ $\mathrm{T}_3/\mathrm{T}_4$ inhibit TRH and TSH release $\to$ thyroid hormone production decreases.

Worked Example: Hormone Dosage and Half-Life

A drug containing cortisol is administered intravenously at a dose of $200\;\mathrm{mg}$. The elimination half-life of cortisol is $90\;\mathrm{minutes}$. (a) Calculate the plasma concentration after $6$ hours if the volume of distribution is $10\;\mathrm{L}$. (b) How long until the plasma concentration falls below $5\;\mathrm{mg/L}$? (c) Explain why cortisol must be administered more frequently than aldosterone (half-life $\approx 20\;\mathrm{minutes}$).

Solution

(a) Initial concentration: $C_0 = 200 / 10 = 20\;\mathrm{mg/L}$.

After $6$ hours ($360\;\mathrm{minutes}$): number of half-lives $= 360 / 90 = 4$. $C = C_0 \times (0.5)^4 = 20 \times 0.0625 = 1.25\;\mathrm{mg/L}$.

(b) We need $C < 5\;\mathrm{mg/L}$: $20 \times (0.5)^n < 5$. $(0.5)^n < 0.25$, so $n > 2$ half-lives. Time $= 2 \times 90 = 180\;\mathrm{minutes} = 3\;\mathrm{hours}$.

(c) Aldosterone has a shorter half-life ($20\;\mathrm{min}$) than cortisol ($90\;\mathrm{min}$), so aldosterone must be administered more frequently or by continuous infusion to maintain therapeutic levels. The shorter half-life means aldosterone is cleared from the blood more rapidly. However, aldosterone is also produced locally in some tissues (paracrine action), and its effects are mediated by mineralocorticoid receptors that have high affinity (requiring lower concentrations). In practice, aldosterone is replaced by fludrocortisone (a synthetic mineralocorticoid with a longer half-life) in patients with adrenal insufficiency.