The study was performed in Moscow Region at Zvenigorod Biological Station (55°44′N, 36°51′E). Two common species—one of passerine, European Greenfinch (Chloris chloris) and one of non-passerine, Budgerigar (Melopsittacus undulatus) birds with similar body masses were chosen for analysis.
All birds were maintained in large indoor aviaries at natural daylengths and temperatures for no less than 1 year. The aviaries were heated in the winter to maintain temperatures of 5–10 °C. These conditions facilitate accurate assessments of seasonal acclimatization [23]. The energy values for non-molting birds were measured in the winter (November–January, February) and summer (late May–June, late August–September). Studies of seasonal variation in energy expenditure, both at rest and during performance, were conducted at experimentally controlled temperatures, where T
A was varied from −20 °C to +45 °C to construct a thermal energy profile of each of the studied species.
Measurements of body mass variations in birds
The following experiments were performed for more precise ascertainment of the speed at which food passes through the ali-mentary canal, and the character of mass variation at night. Post-absorptive birds were permitted to equilibrate in the dark for at least 3–4 h at the desired T
A and humidity before measurements were made. All determinations were performed in darkened chambers to minimize the activities of the birds. Groups of five European greenfinch, or just greenfinch (Chloris chloris) and budgerigars (Melopsittacus undulatus) were placed immediately after the evening feeding in a light low cage with a net floor to facilitate droppings of excrements into a cuvette with liquid mineral oil, in which the excrement sank (to prevent evaporation of water). The cage and cuvette were connected to scale-levers, for the registration of mass variation at night.
Metabolic rate measurements
To improve the efficiency of determining the level of metabolism, I used three different methods of measurement of oxygen consumption and carbon dioxide exhaled. Oxygen consumption is determined using Kalabukhov’s closed-loop respirometer system [24] with my modifications [25, 26] in all the birds and at all studied ambient temperatures. The apparatus operates on the following principle: oxygen is consumed by the bird and the expired carbon dioxide is immediately absorbed. The decrease in gas pressure causes oxygen to be drawn from the container into the bird chamber (Fig. 1). Then an equal amount of water flows from the burette into the container through the water-type pressure valve to replace the oxygen. Oxygen consumption can be read from the water level in the burette. During the experimental period the pressure inside the respirometer is slightly lower than the atmospheric pressure.
The birds were placed in a small cage with a wire mesh floor, which was subsequently placed in a sealed Plexiglas chambers in the dark. The chambers were connected to a ventilation pump [Fig. 1(2)], and the volume of a chamber was 4 L (250 × 145 × 110 mm). After 2–3 h, when the birds were asleep and the temperature in the chamber had stabilized, the chamber was connected to a device to measure the oxygen consumption [Fig. 1(5)]. The alkali CO2 absorbent KOH was placed in the chamber under the cage floor. The chambers were equipped with calibrated thermistors [Fig. 1(6)], and the temperature in the chamber was remotely monitored using an Electronik-16 potentiometer recorder (Electronschik, Russia). The chamber was placed in a thermostat or in the refrigerator. The measurements were obtained at no more than two temperatures for one night during the winter, and each measurement was obtained after acclimatization for 2–3 h. During the summer, the measurements were obtained at only one temperature because of short nights. The measurements were continuously obtained for 2–4 h, and the data were recorded every hour. The accuracy of measurements was ±0.1 °C for temperature, ±0.5 mm Hg for atmospheric pressure and ±0.2 mL for consumed oxygen volume. Measurements obtained using the respirometer were based on air pressure; therefore, they were sensitive to temperature changes. The data were not used when the temperature fluctuations with an hour in the sealed chamber exceeded 0.3 °C for at least. The average volume of consumed oxygen, calculated while obtaining the metabolic rate measurements, was transformed into volume at standard temperature and pressure and converted to kJ/day according to the following equation: 1 L of O2 = 15.97 + 5.16RQ (kJ) [27]. Thus, the oxygen consumption was measured at rest at different ambient temperatures (−20 °C to +45 °C).
After measuring the metabolic rate (MR), the birds were weighed to the nearest 0.1 g, assessed for molt stage (the molt score was recorded) and released from the aviary at dawn. The body temperatures were registered after measuring the MR using a thermistor taped inside the cloaca and connected to a remote readout device.
In a series of experiments, the respiratory quotient RQ was determined using a Haldene gas analyzer, designed on the successive absorption of the components of the gas mixture (carbon dioxide is absorbed by alkali and oxygen through pyrogallol), to measure the volume of residual gas. Air samples from the sealed chamber, in which the birds breathed for 15 min, were collected into a special bag (Syringe A-7, 70 mL, using a soft tip). The volume of breathed air was measured in a gas meter, and the samples were analyzed for the O2 and CO2 concentrations. A known volume of the gas sample was first treated with KOH solution. The CO2 concentration in the expired air was determined as the amount of CO2 uptake through KOH corresponding to a decrease in the original volume of the gas sample analyzed. The remaining gas was exposed to alkaline pyrogallate (pyrogallic acid in KOH), which absorbs O2, to determine the concentration of oxygen in the expired air. Because the bird inhaled atmospheric air of a known composition (practically constant), determining the amount of O2 utilized and CO2 exhaled was a simple task. Based on these values, the RQ was calculated using the following formula: RQ = Volume of CO2 exhaled/Volume of O2 utilized. Benadé et al. [28] conducted experiments, in which the oxygen and carbon dioxide content and the RQ values obtained from the expired air samples using the Haldane technique, were compared with those obtained using paramagnetic and infrared analysis. No significant bias was observed between the Haldane and paramagnetic analyses of oxygen content. Infrared analysis showed more consistent results for CO2 than those obtained using the Haldane apparatus. According to the calculated RQ values, the chemical and physical methods were nearly identical. Thus, physical methods, when properly used and frequently calibrated, are as accurate as the accepted standard chemical methods [28].
Thus, I measured the resting energy expenditure and respiratory quotient in passerine and non-passerine species of during the winter and summer. The measurements were obtained at different temperatures ranging from +5 to +35 °C.
Using a third series of measurements, I determined the oxygen consumption and carbonic gas exhalation in birds using a FoxBox C flow-through respirometer (Sable Systems Inc.). Simultaneously the rate of air passage through the chamber, temperature in the chamber, and the concentration of carbonic gas and oxygen were recorded. The intensity of ventilation of the respirometer chamber (passage rate) was set within 600–850 mL/min. Air was continuously blown through the hermetic respiration chamber in which the bird was located. The rate of oxygen consumption and carbonic gas exhalation was calculated as the difference between the concentration of these gases at the output of the respiration chamber containing the bird and the output of a similar empty chamber, multiplied by the rate of air passage through the chamber. The concentration of carbonic gas and oxygen in the respiration chamber containing the bird and a similar empty chamber was successively measured for 24–30 and 6–10 min, respectively. The readings were obtained once every 10 s. Measurements were obtained at night in a darkened chamber at different ambient temperatures. The measurements were obtained at 1–2 h after the birds fell asleep and the chamber temperature stabilized. The respiratory quotient was determined during these experiments. The energy metabolism was continuously calculated based on the calculated values of the respiratory quotient at a given moment of time. This series of experiments were performed at the start of the experiment, and after 25 species were measured, the results concerning one of these species were published [29].
The values for oxygen consumption were obtained using an updated respirometer, and these values were corrected to standard pressure and temperature according to the equations of Depocas and Hart [30].
The stoichiometric approach to calculation of total evaporative water loss and relationship between evaporative and non-evaporative heat loss
Simultaneous determination of the energy expenditure and mass loss in resting birds was used to estimate their energetic equivalent of body mass loss (q). During the measurements of q, the birds were in the post-absorptive state, and because measurements were performed during the non-breeding, non-molting, non-migrating seasons of the year, no production (growth or reproduction) occurred during these measurements. The birds did not have drinking water, and they metabolized previously ingested food. I performed preliminary experiments to determine the time of a day (in 24-h cycle) when the loss of body mass of the bird was minimal and bird’s rate of metabolism was constant. My data for both species were consistent with the results of previous experiments [3, 31–34]. That is mass losses resulting from cloacal excretions and products of nitrogen metabolism remain very small 4 hours after feeding and they may be ignored. Therefore, the main variable affecting the q value is the pulmonary and cutaneous evaporation of water. This interrelationship is expressed as follows:
$$q = C_{\text{s}} { \cdot }D_{\text{W}} /D_{\text{m}}$$
(1)
where C
s is the energetic content (kJ/g) of the oxidized ingredient (fat = 39.7 kJ/g; carbohydrate = 17.6 kJ/g; protein = 18.4 kJ/g), D
W is the mass of food oxidized during energy metabolism (in further calculations D
W = 1 g), and D
m is the loss of body mass comprising the following components:
$$D_{\text{m}} = \left( {C_{\text{s}} { \cdot }W{ \cdot }\% H_{\text{e}} } \right)/\left( { 2. 4 2 {\kern 1pt} \times {\kern 1pt} 100} \right) + {\text{cloacae discharge}} + D_{{({\text{CO}}_{2} - {\text{O}_{2}}}} )$$
(2)
where C
s·W/2.42 × 100 is the amount of water (g) necessary for the evaporative removal 1 % heat generated through the oxidation of W g compound of known energetic content (C
s), %H
e is the evaporative heat loss expressed as a percentage of total heat produced during the oxidation of any amount of this compound in the organism, cloacae discharge is the loss of body mass as excreted urine and feces, and \(D(CO_{2} - O_{2} )\) is the difference in body mass gained from the oxygen consumed and the mass lost from the CO2 released.
The q value, representing the energy equivalent of body mass lost, is associated with the level of respiratory evaporation. For example, at higher ambient temperatures, evaporation rapidly increases relative to the metabolic rate observed when birds pant. The relationship between the energy equivalent and the level of evaporation is readily quantified [25, 26, 31, 32, 35].
During the oxidation of 1 g of fat, 39.7 kJ of heat is produced, and the mass of oxygen consumed is 0.07 g greater than the mass of carbon dioxide released, yielding 1.07 g of metabolic (oxidation) water.
During the oxidation of 1 g of fat, the mass of oxygen consumed is 0.07 g greater than the mass of carbon dioxide released, yielding 1.07 g of metabolic (oxidation) water, and 39.7 kJ of heat is also produced. Thus, in the absence of evaporation, the bird gains 0.07 g of body mass for each gram of fat oxidized as a result of metabolic water. The heat of water vaporization is 2.42 kJ/g. When a portion of the heat produced during metabolism is lost via evaporation, the loss of body mass (D
m) per 1 g of the oxidized substrate (fat) resulting from evaporative heat loss (H
e) is:
$$D_{\text{m}} = 3 9. 7 \times \% H_{\text{e}} /\left( { 2. 4 2 \times 100} \right) - 0.0 7$$
where %H
e is the percent of total heat produced lost through evaporation. Considering the constant addition of mass resulting from excess metabolic water, which, with respect to loss of body mass, is a negative value, q is dependent on the proportion of body heat production lost via evaporation at rest during the oxidation of fat, determined as
$$q = 3 9. 7/\left( { 3 9. 7/\left( { 2. 4 2 \times 100} \right)} \right) \times \% H_{\text{e}} - 0.0 7 = 3 9. 7/0. 1 6 4 \times \% H_{\text{e}} - 0.0 7$$
(3)
There are two unknown variables in this equation: q and %H
e; however, q is an experimentally determined value obtained at different ambient temperatures. Thus, %H
e can be calculated as
$$\% H_{\text{e}} = ( 3 9. 7/q + 0.0 7)/0. 1 6 4$$
(4)
Upon the oxidation of 1 g of carbohydrate, 17.6 kJ of heat is released and 0.56 g of metabolic water is produced, and because the mass of released carbon dioxide is 0.44 g more than the mass of consumed oxygen, this variable is not associated with the loss of body mass through evaporation. Correspondingly, the dependence of body mass loss and q on the level of evaporative heat loss during carbohydrate oxidation is represented as
$$D_{\text{m}} = \left( { 1 7. 6/ 2. 4 2 \times 100} \right) \times \% H_{\text{e}} + 0. 4 4 = 0.0 7 2 \times \% H_{\text{e}} + 0. 4 4$$
$$q = 1 7. 6/D_{\text{m}} = 1 7. 6/(0.0 7 2 \times \% H_{\text{e}} + 0. 4 4)$$
(5)
The oxidation of 1 g of protein yields 18.4 kJ of heat, 0.49 g of metabolic water and 0.47 g of nitrogen metabolism products. Because the weight of released CO2 is 0.04 g more than the mass of consumed oxygen and because nitrogen metabolism products are excreted during prolonged experiments, the loss of body mass at %H
e = 0 is 0.47 g + 0.04 g = 0.51 g per 1 g of oxidized protein, determined as
$$D_{\text{m}} = \left( { 1 8. 4/ 2. 4 2 \times 100} \right) \times \% H_{\text{e}} + 0. 5 1 = 0.0 7 6 \times \% H_{\text{e}} + 0. 5 1$$
$$q = 1 8. 4/D_{\text{m}} = 1 8. 4/(0.0 7 6 \times \% H_{\text{e}} + 0. 5 1)$$
(6)
Equations (4), (5) and (6) can be transformed into an exponential form. For fat metabolism, Eq. (4) is represented as
$$\% H_{\text{e}} = 2 3 8. 3\,q^{ - 0. 9 8} ,\,\,{\text{ where}}\;q\;{\text{is in kJ}}/{\text{g, or }} \% H_{\text{e}} = 2. 7 6q^{ - 0. 9 8}, \,\,{\text{ where}}\;q\;{\text{is in W}}/{\text{g}}$$
(7)
Equation (7) facilitates the calculation of %H
e for specific q values for fat metabolism, determined during the winter.
Evaporative heat loss for any combination of oxidized compounds can be calculated using this method. Our early work [36–38], based on the change of the diurnal variations of body composition in finches and house sparrows during the annual cycle, suggested that during the summer period the ratio of oxidizable substrates at night was close to 0.7 for fat, 0.2 for carbohydrates and 0.1 for proteins. With such ratio of oxidizable substrates RQ must be equal to 0.7 × 0.7 + 0.2 × 1 + 0.1 × 0.82 = 0.77.
Oxidation of a molecule of Carbohydrate \(6\,{ {\rm O}_{2}}+\text{C}_{6}\text{H}_{12}\text{O}_{6} \to 6\,{ {\rm CO}_{2}}+6\,{ {\rm H}_{2}\text{O}}+38\,\text{ATP RER} = \text{V}_{{\rm CO}_{2}}/\text{V}_{{\rm O}_{2}} = 6\,{ {\rm CO}_{2}}/6\, { {\rm O}_{2}}=1.0.\) Oxidation of a molecule of Fatty Acid \(23\,{ {\rm O}_{2}}+\text{C}_{16}\text{H}_{32}\text{O}_{2} \to 16\,{ {\rm CO}_{2}}+16\,{ {\rm H}_{2}\text{O}}+129\,\text{ATP RER} = \text{V}_{{\rm CO}_{2}}/\text{V}_{{\rm O}_{2}} = 16\,{ {\rm CO}_{2}}/23\, { {\rm O}_{2}}=0.7.\) Oxidation of a molecule of albumin \(63\,{ {\rm CO}_{2}}+38\,{ {\rm H}_{2}\text{O}}+\text{SO}_{3}+9\,{\text{CO}(\text{NH}_{2})_{2}}+36\,\text{ATP RER} = \text{V}_{{\rm CO}_{2}}/\text{V}_{{\rm O}_{2}} = 63\,{ {\rm CO}_{2}}/77\, { {\rm O}_{2}}.\) Calculation of the RQ for the same ratio of molecules oxidized substrates gives 0.7(16 CO2/23 O2) + 0.2(6 CO2/6 O2) + 0.1 (63 CO2/77 O2) = 11.2 CO2/16.1 O2 +1.2 CO2/1.2 O2 + 6.3 CO2/7.7 O2 = 18.7 CO2/25.0 O2 = 0.748. But 1 g of fat, carbohydrates or proteins contains different amounts of molecules. Substrates themselves are not made of pure glucose, albumin and different ratios of fatty acids, so the meanings are slightly different (0.77 and 0.748). Therefore, it is permissible to take the proportion of oxidized substrates equal to 0.7 for fat, 0.2 for carbohydrates and 0.1 for proteins for the summer period at night.
A combination of oxidized substrates in the bird species during the summer, including fats, carbohydrates and proteins at 0.7:0.1:0.2, respectively, was used. The total q comprises 0.7 q for fats, 0.1 q for carbohydrates and 0.2 q for protein, represented as
$$\begin{aligned} q &= 0. 7 \times 3 9. 7/(0. 1 6 4 \times \% H_{\text{e}} - 0.0 7) + 0. 1 \times 1 7. 6/(0.0 7 2 \times \% H_{\text{e}} + 0. 4 4) \\ &\quad + 0. 2 \times 8. 4/(0.0 7 6 \times \% H_{\text{e}} + 0. 5 1)\end{aligned}$$
(8)
For the exponential form, this equation is represented as
$$\% H_{\text{e}} = 2 3 9. 3 { }q^{ - 1.0 5} ,\,\,{\text{ where}}\;q\;{\text{is in kJ}}/{\text{g, or}}\;\% H_{\text{e}} = 2. 7 7q^{ - 1.0 5},\quad q\;{\text{in W}}/{\text{g}}$$
(9)
The regressions (7) and (9) were fitted using the least-squares method of linear regression.
Total evaporative water loss is represented as
$${\text{TEWL}} = ({\text{MR}} \times \% H_{\text{e}} / 100)/ 2. 4 2$$
where MR is the total heat produced at any T
A, and %H
e is the percent of total heat lost through evaporation at this T
A.
Therefore, a definite correlation between the energy equivalent and level of evaporative heat loss for any ambient temperature was observed. Because there is also a specific level of heat production for any temperature, it was possible to calculate the heat dissipated through evaporation at any ambient temperature and the amount of water spent to dissipate that amount of heat, as the evaporative heat of 1 g of water is equal to 2.42 kJ [39].
The correlation of evaporative and non-evaporative heat loss was obtained as follows: starting with q, I calculated %H
e for specific ambient temperatures from Eqs. (7) and (9), then H
e was calculated from the total level of heat loss at specific ambient temperatures (0 °C for SMR, T
lc for BMR, T
uc for BMR, 25 °C for SRM25°) as follows: H
e = SMR × %H
e/100 or H
e = BMR × %H
e/100 or H
e = SRM25° × %H
e/100. The evaporative water loss (TEWL, in g/day) was calculated from the following equation: TEWL = H
e/2.42. Deducting H
e from the total heat loss (SMR or BMR), heat loss through conduction, convection and radiation (or non-evaporative heat loss, H
s) was as follows: H
s = SMR − H
e or H
s = BMR − H
e.
The TEWL can be used to determine the nonevaporative thermal conductance in birds. The determination of non-evaporative heat loss facilitated the calculation of heat loss through conduction, convection and radiation via body insulation, i.e., nonevaporative thermal conductance at different ambient temperatures (h
min and h
max) was calculated according to the following equations.
Non-evaporative thermal conductance at low ambient temperatures (at T
A < T
lc, when the bird minimizes heat loss, h
min):
$$\begin{aligned}h_{\text{min} } &= [({\text{SMR}} - {\text{SMR}} \times \% H_{\text{e1}} / 100) \\ & \quad- ({\text{BMR}} - {\text{BMR}} \times \% H_{\text{e2}} / 100]/(T_{\text{lc}} - T_{\text{A}} ) \end{aligned}$$
(10)
where SMR is the standard metabolic rate at 0 °C, %H
e1 is the percentage of evaporative heat loss at this temperature, BMR is basal metabolic rate, %H
e2 is the percentage of evaporative heat loss at T
A = T
lc, T
lc is the lower critical temperature and T
A is the ambient temperature at which SMR is measured (in this case 0 °C).
Non-evaporative thermal conductance at high ambient temperatures (at T
A = T
uc, when the bird “undresses itself” to maximize heat loss, h
max):
$$h_{ \text{max} } = ({\text{BMR}} - {\text{BMR}} \times \% H_{\text{e3}} / 100)/(T_{\text{B}} - T_{\text{uc}} )$$
(11)
where %H
e3 is the percentage of evaporative heat loss at T
uc, T
uc is the upper critical temperature, and T
B is the body temperature.
Calculations and statistical processing of the results were performed with the Statgraphics program package. All data are expressed as mean ± SD. Figures (including curve fits and correlation coefficients) were produced using the Harvard Graphic 3.0 software package. Linear curve fits are plotted in Figs. 2 and 3, whilst the lines for H
e and %H
e are polynomial curve fits. The relationships in this study were estimated by analysis of variance (ANOVA), and significance was determined by t test, as appropriate. The following abbreviations associated with statistics are used in this paper: n, sample size; p, statistical significance; t-test for independent samples: Variables for Summer and Winter were treated as independent samples r, Pearson’s linear correlation; SD, standard deviation.