1. Metabolic Rate, Taxonomic Group and Body Size

Early in this century comparative physiologists measured the metabolic rates for a vast variety of vertebrate species (Huxley 1972, Kleiber 1947, Brody 1945). As much as possible, these measurements were made on thermoneutral, post-absorptive, non-stressed animals. When the results of these studies were plotted it was found that the energy required for minimal, resting body functions (breathing, heart beat, peristalsis, muscular support, etc.) did not increase linearly in proportion to increases in mass of individual animals. When plotted on a log-log scale it was apparent that for the vast majority of terrestrial vertebrates basal metabolism was proportional to their body mass taken to the 3/4 (or 0.75) power. This indicated that smaller species had higher metabolic rates per unit mass than larger ones. (There are workers who claim that the exponent is closer to 0.67 and relate things to body surface area, but the basic result is the same.)

Those of us who work with birds realize that these organisms are both smaller and have faster metabolic processes than domestic mammals. Observation reveals that birds also eat more food and excrete wastes more frequently than their equivalently sized mammalian counterparts.

When the lines relating metabolic rate to mass for various taxa are compared, they form an almost parallel set of lines having a slope of about 0.75, and differing only in their Y intercepts.

For virtually all groups of terrestrial vertebrates these lines can be described by an equation of the form:

Y = the resting animal's energy output in kilocalories (kcals) per 24 hours

K = a taxonomically dependent constant, and

M = animal's body mass in kilograms.

*(at a species' preferred optimum temperature)

In practice, this equation allows us to calculate the basal energy requirement for almost any species we might encounter. We refer to this as MEC or minimum energy cost.

2. Specific Metabolic Rate

When allometric rates are expressed per unit of body mass they are termed "mass specific" or "specific". Thus, MEC per kilogram is termed specific MEC, or SMEC. Mass specific scaling exponents are formed by dividing the rate equation by M, eg:

Thus, the mass-specific metabolic rate of an animal is a measure of the energy required per unit mass to carry out its essential metabolic functions. In general we can say that smaller animals require more energy per unit mass than large animals.

This can also be used to make comparisons between taxa. Comparing the SMEC (MEC divided by mass, expressed in kcal/kg), of an 850 g passerine bird with that of a placental mammal of the same mass:

The bird has nearly twice (1.8 x) the metabolic rate as the mammal of the same mass. This can also be seen by dividing the equations for MEC for the two species: 129 M^0.75 /70 M^0.75= 1.8

3. Allometric Calculation of Dietary Energy Requirements

Minimum energy cost (MEC) is the amount of energy required in order to just sustain vital functions when a resting animal is in a thermally neutral environment. Obviously, in real-life situations a diet must provide more energy than that, especially if the animal is actively growing or attempting to heal, as after an injury or surgery, or if it has fever. The following correction factors are based on clinical observation and on data from a variety of studies.

4. Size, Scaling and Nutrition

We have already shown that one can easily calculate the daily energy requirements of raptorial birds. References such as Fowler (1986), Exler (1987) or Watt and Merrill (1963) can then be consulted to find out how much caloric energy a given food item contains, and a simple calculation will tell you how much a given bird should be eating. For example:

1. bald eagle: weight = 5 kg >>> MEC = 78 (5)^0.75 = 260 kcal/day
2. assume a low level of activity in captivity >>> 260 x 1.5 = 390 kcal/day
3. how many adult rats should it be eating daily?
4. rats contain +/- 2 kcal/ gm (from table in Fowler, 1986).
5. therefore the bird should eat about 195 gm of rat per day just for maintenance energy in captivity.

Sedgwick and Kirkwood have introduced to veterinary medicine the concept that specific nutrient requirements can also be scaled according to metabolism. For example, work by several authors on water turnover rates has shown that, for a variety of avian species, water turnover (Y=liters per day) scales to body mass with an exponent of 0.75 (Y = 0.119 M^0.75). Other nutrients, both macro and micro, must be utilized in proportion to the metabolic needs of an animal. Therefore, once the MEC for an animal has been calculated, the daily amount of every essential nutrient, including water, can be determined allometrically per 100 kcal of daily energy cost.

Estimating water and energy deprivation in injured wild birds is the most important preliminary consideration in treating such animals because birds must maintain high blood sugars compared to mammals, and because of their relatively high metabolic rates. Since injured, smaller birds may rapidly succumb to hypoglycemia and dehydration, the most critical nutrients to be supplied are sugar and fluid with electrolytes. For injured wildlife the clinician must use judgement and experience in selecting the appropriate combination of intravenous fluids and nutrients, as there is seldom a good history available.

5. Metabolic Approach to Drug Dosage

Rates of metabolism can critically influence the uptake, distribution and excretion of drugs as well as nutrients. Some researchers have therefore begun extrapolating drug dose and treatment interval or frequency for non-domestic animals based on calculated metabolic rates rather than the traditional extrapolation based on body mass. Note that the dose per kilogram body weight decreases rapidly with an increase in animal size. We include a sample calculation for deriving a drug dose based on metabolic rate.

6. Physiologic (Metabolic) Time

Biologists have long made the association that large animals tend to live longer than small ones. In addition, larger animals tend to have slower heart rates, and reach sexual maturity at a later age. In 1945 Brody originated the term physiological time to indicate that smaller animals seem to live according to a faster time scale than large animals. Linstedt and Calder (1981) have shown that for homeotherms, a large number of developmental, physiological and ecological cycles clearly scale in proportion to the body mass. Calder (1984) states,"...there are two kinds of time scales, absolute time and size-dependent relative time (physiological time).".

An elegant statement from Mordenti is :

If one looks at groups of similar animals that vary only in size (eg: felines, raptors, etc.), certain generalizations can be made:

1. smaller species have higher heart rates and respiratory rates
2. smaller species have higher basal metabolic rates and consume a higher percentage of their body weight in food daily
3. smaller species mature faster, have shorter gestation periods, reproduce more frequently, and have shorter lifespans
4. smaller species have shorter induction and recovery times from anesthesia, and
5. smaller species become hypothermic more rapidly during anesthesia.
6. large species are more prone to malignant hyperthermia and capture myopathy.

One way to unify all these concepts is to consider that smaller species are living at a faster rate and have a greater relative surface area than larger ones. The challenge for those clinicians treating a broad taxonomic diversity of animals is to appreciate and quantify this variation.

One of the most useful indicators for measuring depth of anesthesia is heart rate (HR). The resting HR of placental mammals can be estimated from the equation 241 M ^-0.25 (M=mass in kg), and that of birds scales according to 156 M ^-0.23. These formulae allow us to calculate the resting HR for most species. A functional definition of bradycardia may be considered as any HR more than 20% below the calculated rate.

A knowledge of metabolic time is critical for successful anesthetic technique. It is said that humans can live for about 5 minutes without oxygen before sustaining brain damage and dogs can survive for about 4 minutes under the same conditions. According to metabolic time, this indicates that any vertebrate should be able to live for about 500 heartbeats before serious damage occurs. We know that birds have slower HR than mammals of the same mass. Calculating the HR for a 6 kg raptor reveals that it should have a resting HR of around 100 beats per minute. A 50 gm raptor, on the other hand would have a calculated HR of 310 beats per minute. The large raptor should survive about 5 minutes without oxygen, but our smaller patient would suffer irreversible damage in 2 minutes or less. Physiologically then, every minute on the surgical table for our 6 kg raptor is metabolically equivalent to 3 minutes of surgical time for the 50 gm bird. Therefore in planning the length of stressful procedures with small patients we must remember to pay attention to the metabolic clock - not the one on the wall. A one hour surgery on the tiny bird could be expected to cause the same amount of physiologic stress as subjecting the larger raptor to a 3 hour procedure.

Our smaller patients have proportionately more surface area from which to lose heat than our larger patients. A surgical plane of anesthesia eliminates our patients' ability to generate heat by movement, shivering, or other adaptive actions. In addition, we often ventilate these small patients with relatively cool, dry gases. Thus before we make our initial incision these small animals may be hypothermic and rapidly losing water from respiratory surfaces. Surgical incision eliminates the insulative protection of an animal's fur or feathers, and drastically increases the surface area available for cooling and evaporative water loss. It is apparent that the provision of external sources of heat and moisture are especially critical to surgical success with small patients. Most manufacturers of anesthesia equipment sell modules for the provision of warmed, humidified inspired gases.

Size Range of Organisms

Linear Dimensions, Surface Area & Volume

1. surface area proportional to L²
2. volume proportional to L³
3. surface area proportional to V^2/3

The Math >>> straight lines can be described by the equation: y = ax^b

b = slope

a = Y intercept

7. Scaling Drug Dosages

See Allometric Scaling Worksheet in Supplementary Material .

Assume that you would like to administer drug x to an unusual species: a 450 gm turtle.

1. Find a model species in the literature, one for which laboratory pharmacokinetic studies of this drug have been performed.
   Example: in 10 kg dogs it was determined that this drug needed to be administered BID at 15 mg/kg to maintain effective blood levels.
2. Convert the typical (mg/kg) dose to a metabolically based dose (mg/Kcal). Remember that the constants (K) are 129 for passerine birds, 78 for nonpasserines, 70 for placental mammals, 49 for marsupials, and 10 for reptiles at their optimal body temperature and that: MEC = K times (mass in kilograms) to the 3/4 power
   1. calculate MEC for 10 kg dogs - 70(10)^0.75 = 393.6 Kcal/day (where 70 is the constant for placental mammals and 10 is the animals' weight in kilograms)
   2. dose for 10 kg dog = 15 mg x 10 kg = 150 mg/treatment.
   3. MEC dose = 150 mg / 393.6 kcal = 0.38 mg/kcal
3. Determine required periodicity of dosing.
   1. Calculate MEC for patient (turtle) - 10(0.45)^0.75 = 5.4 Kcal/day
   2. Calculate the mass specific metabolic rate (SMEC)for both control species (dog) and subject species (turtle):
      1. dog - SMECc = 393.6 Kcal/day / 10 kg = 39.4 Kcal/kg
      2. turtle - SMECs = 5.4 Kcal/day / 0.45 kg = 12 Kcal/kg
   3. The treatment interval for the dog (Tc) was 12 hours. Therefore, the treatment interval for the turtle Ts is:
      1. Ts = Tc x (SMECc / SMECs) = 12 hr x (39.4 / 12)= 39.4 hours
4. We would then dose this 450 gm turtle a total of: 5.4 Kcal x 0.38 mg/Kcal = 2 mg once every +/- 36 hours

{Note: This method for calculating drug dosages can in no way predict metabolic peculiarities of particular species.} Sedgwick & Pokras, 1994

8. Formulas

This material was adapted from a worksheet developed by Dr. Charles C. Sedgwick.

9. References and Resources