Highland and lowland rodents: what a common-garden experiment can and cannot show
Two populations of a single rodent species live at 4,200 m in the Andes and at 50 m on the coast. Pups from both populations were raised for six months in one laboratory at sea level on an identical diet, then measured as adults (body mass 210 g vs 205 g). Results, highland versus lowland: alveolar surface area 0.087 m² versus 0.062 m²; blood–gas barrier thickness 0.31 µm versus 0.42 µm; pulmonary capillary density 1.9 versus 1.3 (relative units); haematocrit 58% versus 44%; haemoglobin P50 (the partial pressure at which haemoglobin is 50% saturated) 26 mmHg versus 34 mmHg. Lung mass was 1.62 g versus 1.58 g, a non-significant difference. Resting cardiac output was identical. Both groups breathe the same sea-level air at 760 mmHg. A visiting student writes: 'The highland animals have bigger, thinner lungs and more haemoglobin because they need more oxygen at altitude.'
1. Using Fick's law for diffusion, explain the contribution of three of the measured differences to oxygen uptake, and identify the point at which this explanation is incomplete.
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For gas exchange across the alveolar membrane, Fick's law gives: rate of diffusion ∝ (surface area × difference in partial pressure) / thickness.
- Alveolar surface area, 0.087 vs 0.062 m² (+40%): for the same gradient and barrier, oxygen flux rises roughly in proportion to area. This is the largest single structural difference.
- Barrier thickness, 0.31 vs 0.42 µm (−26%): flux is inversely proportional to thickness, so conductance across the membrane rises by about 1.3–1.35×.
- Capillary density (1.9 vs 1.3) and haematocrit (58% vs 44%): denser capillaries shorten the diffusion path in tissue and keep capillary PO₂ low during transit, preserving the alveolar-to-capillary gradient; a higher haematocrit raises the oxygen-carrying capacity per millilitre of blood, so more O₂ can be taken up per unit blood flow.
Where the explanation is incomplete
- The membrane is only one resistance in series. Plasma, the red-cell interior and the reaction with haemoglobin also limit uptake, so a 1.35× thinner barrier does not translate into a 1.35× whole-animal gain.
- The P50 shift to 26 mmHg is not an increase in uptake capacity; it is a change in affinity. A left-shifted curve loads O₂ well at low alveolar PO₂ but unloads it less readily in tissue - a genuine trade-off, and arguably a disadvantage at sea level.
- At 760 mmHg in the laboratory, all animals are at sea-level PO₂, so the differences cannot be explained by the gradient term; they are intrinsic properties of the animals.
2. Explain the trade-off created by a haematocrit of 58%, and predict how these highland animals would perform if kept at sea level for a year.
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The trade-off
- Oxygen-carrying capacity per unit blood volume rises roughly linearly with haematocrit, but blood viscosity rises non-linearly and steeply above about 45–50%.
- Higher viscosity increases cardiac work: the heart must generate more pressure for the same flow, and pulmonary vascular resistance rises, loading the right ventricle. Chronic severe polycythaemia in mammals is associated with pulmonary hypertension and right heart failure.
- There is therefore an optimum haematocrit; pushing past it buys extra O₂ content at the cost of increased cardiac workload and reduced flow, particularly in the smallest capillaries.
Prediction at sea level
- The animals keep a haematocrit and haemoglobin affinity adapted to hypoxia. At sea-level PO₂ they are over-equipped on capacity but handicapped on two counts: higher viscosity (raised cardiac workload, possible hypertension) and a left-shifted dissociation curve that releases O₂ to tissues less readily, reducing the arterio-venous difference.
- Expect no advantage in maximal oxygen consumption, and possibly a reduced exercise capacity and higher cardiac mass relative to lowland controls; the traits are only favourable in the environment that selected for them. Over a year, some reversal would be expected through acclimation (reduced erythropoiesis), but the structural lung differences would persist because they are developmental.
3. Evaluate the student's sentence as a piece of biological reasoning. Distinguish what the common-garden design allows you to conclude from what it does not.
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The sentence has three faults.
- 'Bigger lungs' is contradicted by the data. Lung mass was 1.62 g vs 1.58 g, not significantly different. The highland lung is not larger; the same mass of tissue is subdivided into more, smaller alveoli, which is where the extra surface area comes from. That is a difference in internal partitioning, not in organ size.
- 'Because they need more oxygen' is teleological. A need does not produce a structure. The correct statement is that within the ancestral population, variants with finer alveolar partitioning and higher haemoglobin affinity left more surviving offspring at altitude, so those alleles increased in frequency. Need is a consequence of selection, not its cause.
- 'Thinner lungs' is imprecise - the blood–gas barrier is thinner, not the lung.
What the design allows
- Because pups of both populations were raised in one laboratory, on one diet, at one PO₂, the differences cannot be simple acclimatisation to the ambient conditions of the two field sites. They must reflect heritable differences (or maternal effects carried over from wild-caught mothers).
- What it does not allow: it cannot separate genetic from maternal/epigenetic effects - cross-fostering or a second-generation common-garden breeding design is needed. It also cannot quantify the strength of selection in the wild, which requires fitness data at both altitudes.