Comparative Anatomy and Physiology I: Respiration, Circulation, Digestion and Excretion
2. Respiratory surfaces and ventilation
Animals breathe through the skin (earthworms, and frogs in part), gills (fishes, crustaceans, molluscs), book lungs (spiders, scorpions), tracheae or lungs. Insects pipe air through spiracles, tracheae and fine tracheoles directly to the cells, so their haemolymph carries little oxygen; the reliance on diffusion along the tubes is one reason insects stay small. Water holds far less oxygen than air and is denser and more viscous, so water-breathers drive a one-way stream across the gills, while air-breathers can afford tidal, in-and-out ventilation.
| Animal | Organ and pump | Exchange arrangement |
|---|---|---|
| Teleost fish | Gill lamellae; buccal and opercular pumps drive water one way | Countercurrent: blood flows opposite to water, so a gradient persists along the whole lamella and blood can leave with a PO2 close to that of the incoming water |
| Frog | Lungs filled by a positive-pressure buccal pump; moist skin | Lungs are a mixed pool; the skin is a major route for CO2 loss |
| Mammal | Alveolar lungs; negative pressure from the diaphragm and intercostal muscles | Tidal: fresh air mixes with residual air, so alveolar PO2 (about 100 mmHg) stays well below atmospheric |
| Bird | Rigid lungs of parabronchi; air sacs act as bellows; a breath takes two cycles to pass through | One-way flow with cross-current capillaries: blood leaving the lung can have a higher PO2 than the exhaled air |
In mammals only part of each breath reaches the alveoli; the rest fills the anatomical dead space of the conducting airways. Alveolar ventilation V̇A = (VT − VD) × f. With tidal volume VT = 500 mL, dead space VD = 150 mL and f = 12 min⁻¹, V̇A = 350 × 12 = 4,200 mL min⁻¹ out of a minute ventilation of 6,000. Breathing 250 mL at 24 min⁻¹ keeps minute ventilation at 6,000 but cuts V̇A to 100 × 24 = 2,400 mL min⁻¹: rapid shallow breathing is inefficient. The rhythm comes from the medulla. In mammals the main stimulus is a rise in PCO2, sensed as H⁺ by central chemoreceptors, while the carotid and aortic bodies respond chiefly to a fall in PO2. Fishes, for which oxygen is the scarce gas and CO2 escapes easily into water, ventilate mainly in response to oxygen.
3. Transport of oxygen and carbon dioxide
Haemoglobin (two α and two β chains, each with a haem) binds four O2 cooperatively: binding at one haem raises the affinity of the others, which gives the sigmoid dissociation curve, steep over the 20 to 40 mmHg of working tissues and flat near the 100 mmHg of the lungs. P50, the PO2 at 50% saturation, measures affinity. A rise in PCO2 or H⁺ (the Bohr effect), temperature or 2,3-bisphosphoglycerate (2,3-BPG) shifts the curve right and unloads more O2 in active tissue. Myoglobin, with one haem, has a hyperbolic curve and higher affinity and stores O2 in muscle; diving mammals are rich in it. Fetal haemoglobin (α2γ2) binds 2,3-BPG weakly, and its higher affinity draws O2 across the placenta. CO binds haem more than 200 times as strongly as O2.
Oxygen content = 1.34 × [Hb] × saturation + 0.003 × PO2, in mL O2 per dL. Textbooks use 1.34 to 1.39 mL per g of Hb, so a question must state its value. With Hb 15 g dL⁻¹ at 98% saturation and PO2 100 mmHg: 1.34 × 15 × 0.98 + 0.3 ≈ 20 mL dL⁻¹, almost all of it bound. CO2 travels about 70% as bicarbonate, made by carbonic anhydrase in red cells while Cl⁻ enters in exchange (the chloride shift), about 20% bound to haemoglobin as carbamino compounds, and the rest dissolved. Deoxygenated haemoglobin carries more CO2 and H⁺ (the Haldane effect), so loading CO2 in tissues and unloading O2 help each other.
4. Circulatory systems and the vertebrate heart
In an open system (arthropods, most molluscs) the heart pumps haemolymph into sinuses that bathe the organs, at low pressure. A closed system (annelids, cephalopods, vertebrates) keeps blood in vessels at higher pressure, giving faster and adjustable delivery. Vertebrate and molluscan hearts are myogenic, the beat arising in the muscle itself; the hearts of the horseshoe crab Limulus and of decapod crustaceans are neurogenic, driven by a cardiac ganglion.
| Group | Chambers | Circulation and key feature |
|---|---|---|
| Fishes | 2: atrium and ventricle, with sinus venosus and conus or bulbus | Single: heart → gills → body → heart; only deoxygenated blood passes through the heart |
| Amphibians | 3: two atria, one ventricle | Double but incomplete; a spiral valve in the conus arteriosus partly separates the streams |
| Lizards, snakes, turtles | 3, ventricle partly divided by an incomplete septum | Double but incomplete; blood can be shunted past the lungs during diving or breath-holding |
| Crocodilians | 4 | Double; the foramen of Panizza links the left and right aortas |
| Birds and mammals | 4 | Complete double circulation; birds keep the right systemic arch, mammals the left |
The aortic arches explain the last row. Of the six embryonic pairs, the third forms the carotid arteries, the fourth the systemic arch (both sides in reptiles, the right in birds, the left in mammals) and the sixth the pulmonary arteries, whose dorsal part is the fetal ductus arteriosus, closing after birth to the ligamentum arteriosum.
In the mammalian heart the SA node sets the pace; the impulse spreads over the atria, is delayed at the AV node so the ventricles can fill, and runs down the bundle of His and the Purkinje fibres. On the ECG the P wave is atrial depolarisation, QRS ventricular depolarisation and T ventricular repolarisation. Cardiac output Q = heart rate × stroke volume, and stroke volume = end-diastolic − end-systolic volume; greater diastolic filling gives a stronger beat (Frank–Starling law). The Fick principle measures Q from oxygen uptake: Q = V̇O2 ÷ (CaO2 − CvO2), with arterial and mixed venous (pulmonary artery) blood. With V̇O2 = 280 mL min⁻¹, CaO2 = 200 mL L⁻¹ and CvO2 = 144 mL L⁻¹, Q = 280 ÷ 56 = 5 L min⁻¹. Flow = pressure difference ÷ resistance, and by Poiseuille's law resistance varies as 1/r⁴, so halving an arteriole's radius raises its resistance 16-fold; arterioles set peripheral resistance.
5. Digestive systems: gut design follows diet
Sponges and protists digest inside food vacuoles; cnidarians and flatworms have a sac-like gut with one opening; from nematodes onwards a complete gut with mouth and anus allows one-way processing and regional specialisation. Birds store food in the crop, secrete acid and pepsin in the glandular proventriculus and grind food with grit in the muscular gizzard. Herbivores have longer guts than carnivores of similar size: the long coiled intestine of the plant-eating tadpole shortens at metamorphosis into that of the carnivorous frog.
No vertebrate makes cellulase, so plant-eaters house microbes that do. Foregut fermenters such as cattle have a four-part stomach: rumen, reticulum, omasum and the abomasum, the true glandular stomach. Rumen microbes ferment cellulose to short-chain fatty acids (acetate, propionate, butyrate), the ruminant's main energy source, and the microbes themselves are digested further down as protein; the cud is regurgitated and re-chewed (rumination). Hindgut fermenters (horse, rabbit) ferment in an enlarged caecum and colon beyond the small intestine, so their microbial protein would be lost in the faeces; rabbits recover it by eating their soft caecal pellets (caecotrophy).
| Site | Secretion and action | Control |
|---|---|---|
| Mouth | Saliva: α-amylase begins starch digestion | Nervous reflex |
| Stomach | Parietal cells: HCl and intrinsic factor; chief cells: pepsinogen, activated to pepsin by acid and by pepsin itself | Gastrin (G cells), vagus |
| Duodenum: pancreatic juice | Bicarbonate; trypsinogen activated by brush-border enteropeptidase, then trypsin activates chymotrypsinogen and procarboxypeptidase; amylase; lipase with colipase; nucleases | Secretin (bicarbonate), CCK (enzymes) |
| Duodenum: bile | Bile salts emulsify fat and form micelles; bile has no digestive enzyme | CCK contracts the gall bladder |
| Small intestine | Brush-border disaccharidases (maltase, sucrase, lactase), aminopeptidases, dipeptidases | Local |
Glucose and galactose enter enterocytes with Na⁺ through SGLT1 (secondary active transport, powered by the basolateral Na⁺/K⁺-ATPase), fructose by facilitated diffusion through GLUT5, and all three leave through GLUT2. Oral rehydration solution works because glucose drags Na⁺ in, and water follows. Long-chain fatty acids and monoglycerides from micelles are re-esterified, packed into chylomicrons and enter the lacteals (lymph), not the portal blood. Vitamin B12 is absorbed with intrinsic factor in the terminal ileum.
6. Excretion, osmoregulation and the nephron
| Waste | Properties | Typical excretors |
|---|---|---|
| Ammonia | Very toxic, very soluble; needs much water; costs nothing to make | Most aquatic invertebrates, bony fishes, tadpoles |
| Urea | Less toxic, soluble; made in the liver by the urea (ornithine) cycle at an ATP cost | Mammals, adult amphibians, cartilaginous fishes (which retain it as an osmolyte) |
| Uric acid | Least toxic, nearly insoluble; leaves as a paste with little water | Birds, most reptiles, insects, land snails; suits embryos sealed in a shelled (cleidoic) egg |
Excretory organs range from the contractile vacuole of freshwater protists, which bails out water, through protonephridia with flame cells (flatworms), metanephridia opening into the coelom (annelids), antennal glands (crustaceans) and Malpighian tubules (insects: they secrete K⁺ into the lumen, water and wastes follow, and the hindgut reclaims water so that uric acid precipitates), to the vertebrate kidney. A freshwater teleost, saltier than its medium, gains water: it does not drink, makes copious dilute urine and takes up ions actively at the gills. A marine teleost loses water: it drinks seawater, excretes salt through gill chloride cells and makes little urine. Sharks keep urea and TMAO in the blood so as to be slightly hyperosmotic to seawater, and secrete NaCl from a rectal gland. Only birds and mammals make urine hyperosmotic to plasma, because only they have loops of Henle.
The nephron filters, reabsorbs and secretes. Net filtration pressure = glomerular capillary pressure − Bowman's capsule pressure − plasma oncotic pressure. The proximal tubule reabsorbs about two-thirds of the filtered Na⁺ and water, isosmotically, and normally all the filtered glucose and amino acids. The descending limb of the loop is permeable to water; the thick ascending limb is impermeable to water and pumps out NaCl through the Na⁺–K⁺–2Cl⁻ cotransporter, which loop diuretics such as furosemide block. This countercurrent multiplier builds the medullary osmotic gradient, and the vasa recta, acting as countercurrent exchangers, preserve it. ADH inserts aquaporin-2 into the collecting duct so that water leaves into the hypertonic medulla and the urine is concentrated; aldosterone increases Na⁺ reabsorption and K⁺ secretion in the distal nephron. Desert rodents, with very long loops, make extremely concentrated urine.
Clearance C = (U × V̇) ÷ P is the volume of plasma cleared of a substance per minute. Inulin is freely filtered and neither reabsorbed nor secreted, so its clearance equals the GFR, about 125 mL min⁻¹ (180 L day⁻¹) in an adult human. PAH is filtered and secreted almost completely in one pass, so its clearance estimates renal plasma flow (about 650 mL min⁻¹), and the filtration fraction GFR ÷ RPF is about 0.2. Worked example: inulin U = 30 mg mL⁻¹, P = 0.25 mg mL⁻¹ and V̇ = 1 mL min⁻¹ give C = 30 × 1 ÷ 0.25 = 120 mL min⁻¹. For any freely filtered solute, filtered load = GFR × P; a clearance below the GFR means net reabsorption (glucose has zero clearance until its filtered load exceeds the transport maximum), one above it net secretion.
Key takeaways
- Homologous organs share origin (the pentadactyl forelimb), analogous organs share function (insect and bird wings). Surface area falls relative to volume as size rises, and Fick's law of diffusion explains why large animals need thin, folded exchange surfaces, pumps and vessels.
- Fish gills are countercurrent, bird lungs one-way and cross-current, mammalian lungs tidal. Alveolar ventilation = (VT − VD) × f, so rapid shallow breathing wastes ventilation on dead space.
- Haemoglobin binds O2 cooperatively; CO2, H⁺, temperature and 2,3-BPG shift the curve right. O2 content = 1.34 × Hb × saturation + 0.003 × PO2 with the factor stated; about 70% of CO2 travels as bicarbonate.
- Fishes have a two-chambered heart and single circulation; amphibians and most reptiles three chambers with incomplete separation; crocodilians, birds and mammals four. The fourth arch is the systemic arch, right in birds and left in mammals. Q = HR × SV = V̇O2 ÷ (CaO2 − CvO2).
- Gut design follows diet: ruminants ferment before the abomasum, horses and rabbits after the small intestine. Enteropeptidase activates trypsinogen and trypsin activates the rest; glucose enters by SGLT1, fructose by GLUT5, and fat leaves in chylomicrons to the lymph.
- Ammonia, urea and uric acid track water availability. The thick ascending limb's NaCl pump builds the medullary gradient that ADH exploits. C = UV/P; inulin clearance = GFR, PAH clearance ≈ renal plasma flow, and filtered load − excreted load = amount reabsorbed.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The wing of a bat and the wing of a butterfly both generate lift in flapping flight. How are these two wings correctly classified?
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Answer: C — Analogous organs, because they share a function but arise from different structures
The bat wing is a modified pentadactyl forelimb, while the butterfly wing is an outgrowth of the body wall with no skeletal elements, so the two share a function without a common origin: analogy, the product of convergent evolution. Calling them homologous is the tempting slip because both are flight appendages, but homology requires shared ancestral and embryonic origin. Serial homology refers to repeated parts within one body, and neither wing is a reduced remnant.Six pairs of aortic arches form in the vertebrate embryo. From which arch does the systemic aorta of an adult bird develop?
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Answer: B — The right fourth arch
The fourth arch forms the systemic arch: birds retain the right one, mammals the left one and reptiles both. The left fourth arch is the tempting answer because it is the mammalian condition. The sixth arch forms the pulmonary arteries and the fetal ductus arteriosus, and the third forms the carotids.Blood leaving the gill lamellae of a teleost can have a PO2 close to that of the water entering the gills, which a tidal lung cannot achieve. What arrangement makes this possible?
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Answer: A — Blood in the lamellae flows opposite to the water, so a PO2 gradient favouring uptake persists along the whole exchange surface
In countercurrent exchange, blood about to leave a lamella meets the freshest, most oxygen-rich water, so at every point the water is richer than the blood and uptake continues along the full length; the blood's PO2 can approach that of the incoming water. Water carries far less oxygen than air, not more, and fish ventilation is one-way, not tidal. High haemoglobin affinity alone cannot raise blood PO2 above that of the water it meets; with flow in the same direction the two streams would only equilibrate at an intermediate value.In birds the PO2 of arterial blood can exceed the PO2 of the air leaving the lung, which does not happen in mammals. Which feature of the avian respiratory system accounts for this?
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Answer: B — Air flows one way through rigid parabronchi, and capillaries cross that airstream at many points along its length
The parabronchial lung is a cross-current exchanger: air moves one way along each parabronchus and many capillaries draw from it in turn, so blood pooled from the capillaries near the air inlet carries a PO2 higher than the depleted air finally exhaled. The air sacs are poorly vascularised bellows that do not exchange gas, which rules out the tempting air-sac option; birds also have no alveoli and no muscular diaphragm.A subject breathes with a tidal volume of 600 mL at 10 breaths per minute and then switches to 300 mL at 20 breaths per minute. The anatomical dead space stays 150 mL. By how much does alveolar ventilation fall, in mL per minute?
Numerical answer — type the value.
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Answer: 1500
V̇A = (VT − VD) × f. Before: (600 − 150) × 10 = 4,500 mL min⁻¹. After: (300 − 150) × 20 = 3,000 mL min⁻¹. The fall is 1,500 mL min⁻¹, although minute ventilation is 6,000 mL min⁻¹ in both cases: the dead space takes a larger share of each shallow breath.A ventricle has an end-diastolic volume of 130 mL and an end-systolic volume of 58 mL, and the heart rate is 70 beats per minute. What is the cardiac output in L per minute? Give the answer to two decimal places.
Numerical answer — type the value.
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Answer: 5.04
Stroke volume = 130 − 58 = 72 mL. Cardiac output = 72 mL × 70 min⁻¹ = 5,040 mL min⁻¹ = 5.04 L min⁻¹. The ejection fraction, 72 ÷ 130, is about 55%.During cardiac catheterisation a patient's oxygen consumption is 260 mL per minute. Haemoglobin is 15 g/dL, arterial saturation 98% and mixed venous (pulmonary artery) saturation 73%. Take 1.34 mL O2 per g of haemoglobin and ignore dissolved oxygen. Using the Fick principle, what is the cardiac output in L per minute? Give the answer to two decimal places.
Numerical answer — type the value.
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Answer: 5.17
Arteriovenous O2 difference = 1.34 × 15 × (0.98 − 0.73) = 20.1 × 0.25 = 5.025 mL per dL = 50.25 mL per L. Cardiac output = V̇O2 ÷ (CaO2 − CvO2) = 260 ÷ 50.25 = 5.174 L min⁻¹, which is 5.17 to two decimal places. Forgetting to convert per dL to per L gives an answer ten times too large.Which change shifts the oxygen–haemoglobin dissociation curve to the left, that is, raises the affinity of haemoglobin for O2?
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Answer: C — A fall in PCO2 and a rise in pH as blood passes through the lungs
Loss of CO2 and the resulting rise in pH in the lung capillaries shift the curve left (the Bohr effect in reverse), helping haemoglobin load O2. Raised 2,3-BPG, temperature and H⁺ all shift it right and favour unloading. The 2,3-BPG option is tempting because acclimatisation to altitude sounds like a gain in O2 uptake, but 2,3-BPG lowers affinity.Which of the following statements about circulatory systems are correct?
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Answer: A — In a teleost fish only deoxygenated blood passes through the heart; C — Crocodilians have a completely divided ventricle, yet blood can pass between the left and right aortas through the foramen of Panizza
A fish heart receives venous blood and pumps it to the gills, so it handles only deoxygenated blood. Crocodilians have four chambers, and the foramen of Panizza links the two aortas outside the ventricles. Insects have an open system in which the tracheae, not the haemolymph, deliver oxygen. Mammals keep the left fourth arch; the right one is retained in birds.In a cow, which compartment of the stomach secretes hydrochloric acid and pepsinogen, working like the stomach of a mammal with a simple stomach?
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Answer: D — Abomasum
The abomasum is the true glandular stomach. The rumen, reticulum and omasum are non-glandular chambers used for microbial fermentation, sorting of particles and absorption of water and fatty acids. The rumen is the tempting answer because it is the largest and busiest chamber, but its work is fermentation, not acid-pepsin digestion.A child is born without functional enteropeptidase (enterokinase) in the duodenal brush border. Which digestive defect is expected to be the primary one?
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Answer: A — Pancreatic proteases stay as inactive zymogens, so protein digestion in the small intestine largely fails
Enteropeptidase cleaves trypsinogen to trypsin, and trypsin then activates chymotrypsinogen, procarboxypeptidase and the other pancreatic zymogens; without it intestinal protein digestion collapses. Pepsinogen is activated by gastric acid and by pepsin itself, not by enteropeptidase, so the pepsinogen distractor confuses two zymogen steps. Emulsification by bile salts and Na⁺–glucose cotransport by SGLT1 do not depend on enteropeptidase.Which of the following statements about absorption in the mammalian small intestine are correct?
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Answer: B — Glucose enters enterocytes with Na⁺ through SGLT1, driven by the Na⁺ gradient that the basolateral Na⁺/K⁺-ATPase maintains; D — Fructose crosses the apical membrane by facilitated diffusion through GLUT5
SGLT1 is a secondary active Na⁺–glucose cotransporter and GLUT5 carries fructose by facilitated diffusion. Long-chain fatty acids are re-esterified and leave in chylomicrons through the lacteals, not the portal vein, and bile contains bile salts but no lipase; the lipase is pancreatic.Birds and most reptiles excrete their nitrogen mainly as uric acid. Which explanation best accounts for this?
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Answer: D — Their embryos develop sealed in shelled eggs, where an insoluble, low-toxicity waste can be stored without water
Uric acid is nearly insoluble and of low toxicity, so it precipitates harmlessly inside a cleidoic egg and leaves the adult as a paste, saving water. It is the least soluble nitrogenous waste, not the most, and making it costs more energy than making ammonia, not none. Diet does not explain uricotely, which is found in plant-eating birds and reptiles as well.A marine teleost fish lives in seawater that is more concentrated than its body fluids. Which of the following does it do to stay in water and salt balance?
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Answer: A — It drinks seawater to replace the water it loses by osmosis; C — It excretes monovalent ions actively through chloride cells in the gills
Being hypo-osmotic to seawater, a marine teleost loses water across the gills; it drinks seawater, excretes the excess NaCl through gill chloride cells and makes small volumes of urine. Copious dilute urine is the freshwater strategy, and retention of urea and TMAO is the strategy of sharks and rays.During a constant inulin infusion, plasma inulin is 0.30 mg per mL, urine inulin is 36 mg per mL and urine flow is 1.2 mL per minute. What is the glomerular filtration rate in mL per minute?
Numerical answer — type the value.
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Answer: 144
Inulin is freely filtered and neither reabsorbed nor secreted, so its clearance equals the GFR: C = U × V̇ ÷ P = 36 × 1.2 ÷ 0.30 = 43.2 ÷ 0.30 = 144 mL min⁻¹.A solute is freely filtered at the glomerulus and is not secreted by the tubules. The GFR is 125 mL per minute, the solute's plasma concentration 2 mg per mL, its urine concentration 20 mg per mL and the urine flow 2 mL per minute. What percentage of the filtered solute is reabsorbed?
Numerical answer — type the value.
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Answer: 84
Filtered load = GFR × P = 125 × 2 = 250 mg min⁻¹. Excreted load = U × V̇ = 20 × 2 = 40 mg min⁻¹. Reabsorbed = 250 − 40 = 210 mg min⁻¹, which is 210 ÷ 250 = 84% of the filtered load. Equivalently, the solute's clearance is 20 × 2 ÷ 2 = 20 mL min⁻¹, 16% of the GFR, so 84% is reabsorbed.A drug blocks the Na⁺–K⁺–2Cl⁻ cotransporter of the thick ascending limb of the loop of Henle. Even with high circulating ADH, the kidney can no longer produce strongly concentrated urine. Why?
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Answer: C — Without NaCl transport out of the ascending limb the medullary gradient collapses, leaving no hypertonic interstitium
NaCl pumping by the thick ascending limb is the single effect that the countercurrent multiplier amplifies into the medullary gradient. Block it and the medulla becomes nearly iso-osmotic, so even collecting ducts made fully water-permeable by ADH cannot concentrate the urine; this is how loop diuretics such as furosemide act. The aquaporin option is tempting, but the cotransporter blocker leaves the ADH-dependent water channels intact: the missing gradient, not missing channels, is the problem.