A review of the multi-level adaptations for maximizing aerobic dive duration in marine mammals: from biochemistry to behavior

Abstract

Marine mammals exhibit multi-level adaptations, from cellular biochemistry to behavior, that maximize aerobic dive duration. A dive response during aerobic dives enables the efficient use of blood and muscle oxygen stores, but it is exercise modulated to maximize the aerobic dive limit at different levels of exertion. Blood volume and concentrations of blood hemoglobin and muscle myoglobin are elevated and serve as a significant oxygen store that increases aerobic dive duration. However, myoglobin is not homogeneously distributed in the locomotory muscles and is highest in areas that produce greater force and consume more oxygen during aerobic swimming. Muscle fibers are primarily fast and slow twitch oxidative with elevated mitochondrial volume densities and enhanced oxidative enzyme activities that are highest in areas that produce more force generation. Most of the muscle mitochondria are interfibriller and homogeneously distributed. This reduces the diffusion distance between mitochondria and helps maintain aerobic metabolism under hypoxic conditions. Mitochondrial volume densities and oxidative enzyme activities are also elevated in certain organs such as liver, kidneys, and stomach. Hepatic and renal function along with digestion and assimilation continue during aerobic dives to maintain physiological homeostasis. Most ATP production comes from aerobic fat metabolism in carnivorous marine mammals. Glucose is derived mostly from gluconeogenesis and is conserved for tissues such as red blood cells and the central nervous system. Marine mammals minimize the energetic cost of swimming and diving through body streamlining, efficient, lift-based propulsive appendages, and cost-efficient modes of locomotion that reduce drag and take advantage of changes in buoyancy with depth. Most dives are within the animal’s aerobic dive limit, which maximizes time underwater and minimizes recovery time at the surface. The result of these adaptations is increased breath-hold duration and enhanced foraging ability that maximizes energy intake and minimizes energy output while making aerobic dives to depth. These adaptations are the long, evolutionary legacy of an aquatic lifestyle that directly affects the fitness of marine mammal species for different diving abilities and environments.

Appendix

Estimated energy efficiency of a mid-water foraging dive for a Weddell seal in McMurdo Sound during the austral spring

Assumptions.

• 1. Body mass is 430 kg (Davis et al. 2013).

• 2. Resting post-prandial metabolic rate is 5.53 ml O₂ min⁻¹ kg⁻¹ (Williams et al. 2004).

• 3. The metabolic cost of a flipper stroke is 0.036 ml O₂ kg⁻¹ stroke⁻¹ (Williams et al. 2004).

• 4. Energy equivalence of oxygen is 18.6 kJ L O₂ ⁻¹ assuming a protein diet with a respiratory quotient of 0.81 (Schmidt-Nielsen 1997).

• 5. Mean energy content of an Antarctic silverfish is 325 kJ (Davis, unpublished results).

• 6. Mean duration of a mid-water foraging dive is 16.8 min (Davis et al. 2013).

• 7. Mean number of silverfish consumed per dive is 15.7 fish (Davis et al. 2013).

• 8. Mean total number of strokes during a mid-water foraging dive is 821 strokes (Davis, unpublished results).

• 9. Foraging efficiency (% successful) is 96 % (Davis et al. 2013).

Calculated energy efficiency of a mid-water foraging dive

Energy expenditure during a dive:

• 1. [(16.8 min × 5.53 ml O₂ min⁻¹ kg⁻¹ × 430 kg) + (821 strokes × 0.036 ml O₂ kg⁻¹ stroke⁻¹ × 430 kg)]/1,000 = 52.7 L O_2.

• 2. $52.7\;\text{L} \;\text{O}_{2} \times 18.6\;\text{kJ} \;\text{L} \;\text{O}_{2}^{ - 1} = 979\;\text{kJ}$.

Energy ingested as silverfish during a dive:

$$ 15.7\;\text{fish} \times 325\;\text{kJ} \,\text{fish}^{ - 1} = 5,103\;\text{kJ }$$

Mean net energy gain per successful dive (energy ingested - energy - expenditure):

$$ 5,103\;\text{kJ} - 978\;\text{kJ} = 4,123\;\text{kJ} $$

Overall energy efficiency (percent energy consumed to energy expended):

$$ [(5,103\;\text{kJ} \times 0.96)/ 979\;\text{kJ} ] \times 100 = 500\,\% $$