ENERGY METABOLISM
I. Energy Metabolism - introduction
A. C6H12O6 + 6 O2 --->---> 6 H2O + 6 CO2 + energy (686 Kcal/mole or 38 ATP)
(2867 KJ/mole)
B. It is logical for an animal physiologist to ask how much energy an animal must use. But problems develop:
1. Definitions
2. How do you measure energy utilization?
3. How do you reasonably compare one animal to another?
II. Definitions:
A. The broad use of the word metabolism means ³The sum total of the chemical reactions that occur in an organism, or some subset of that total² (Purvis). Unless otherwise specified, in this course metabolism will be used in the sense of energy metabolism, which means the chemical reactions involved in energy usage in an animal. ATP is typically the goal in energy metabolism.
B. Basal metabolism - rate of energy utilization under minimal stress. The very fact that you are taking the measurement means the animal is under stress. On the other hand you do not want the animal abnormally relaxed.
III. Measurements:
A. Various factors affect the energy utilization rate and must be considered (degree of activity, environment, O2 concentration and temperature, animal size).
B. Calorimetry - accurate but difficult, slow, and often not practical.
1. Calorie= amount of heat to raise 1 gram of water 1 degree Celsius.
2. 1 calorie = 4.18 joules.
3. ³Around 60% of the energy that is used to recharge ADP appears as heat Hence, metabolic activity ALWAYS generates heat " So, even 'cold blooded animals' produce heat!
C. Measure energy in minus energy out: Energy of food taken in minus energy in feces (or whatever) left over.
D. Measure O2 uptake.
1. If carbohydrates are the primary oxidized material, then O2 uptake can be converted into Kcal utilization.
2. If proteins are used (protein + O2 ----> CO2 products + energy) the value is 4.5 Kcal/L of O2 (18.8 KJ/ L O2)
3. Lipids usage yields 4.7 Kcal/L O2 (19.6 KJ/ L O2)
4. How does one know actually what the animal is consuming? Respiratory quotient gives a good idea (RQ = CO2 produced/O2 consumed)
RQ = 1 for carbohydrate
RQ = .8 for proteins
RQ = .71 for fats
"Protein used" may be directly determined by the nitrogen excreted.
TOTAL RQ = 1.0 (C) + 0.8 (P) + 0.71 (F)
1 = C + P + F
where C, P and F are the proportions of carbohydrate, protein and fat used. If one of the variables is known (e.g. protein) then the other two can be determined.
5. An animal may not immediately intake O2 for the energy it is utilizing. Using anaerobic respiration acts as a buffer system or delay system (O2 debt) for O2 consumption. (Oxygen debt) (in some cases the animal may never 'repay')
D. CO2 measurements.
IV. Energy Storage.
A. Fat is a good form of energy storage since it has the least weight per unit of energy (9.4 kilocalories per gram compared to sugar at 4.2 kilocalories per gram). BUT fat is difficult to digest (the most difficult food source to digest) because it is water insoluble and the lipases used to digest it are water soluble. In the digestive tract fat typically has to be emulsified for digestion (emulsification by bile breaks fat into small globules so that there is more surface area for the lipases to act upon.)
B. Glycogen (a common animal carbohydrate polymer) is very easily broken down (thus also faster) but does weigh more per unit of energy. Additionally, being water soluble, about 3 g of water for every gram of glycogen is needed to store glycogen!
C. Glycogen also has the advantage that it can be used in anaerobic conditions (broken down to lactic acid) to make ATP while fat can not.
D. Fat is particularly useful for animals ³on the go² (less weight) while glycogen is more useful for sessile animals. In reality all animals have some of both.
V. Metabolism as a function of oxygen concentration:
A. For terrestrial animals, the rate
of energy metabolism is often relatively independent of the oxygen
concentration of the air. Aquatic environments however can be more challenging.
Understand what figures 5.1, and 5.2 of Schmidt-Nielsen
are telling us. (saturated fresh water at atmospheric pressure has 7.2 ml of
oxygen per liter or 159 mm Hg). Obviously energy metabolism is also affected by
temperature, an issue that will be more directly addressed in a subsequent
chapter.
Appreciate the concept that earth started without oxygen and that oxygen was probably toxic to the first organisms. Mechanisms had to be developed t overcome this toxicity.
B. Acclimation to low oxygen
concentration . Understand figures 5.3, of Schmidt-Nielsen.
C. Anaerobic metabolism:
1. glucose -> 2 lactic acid +
2ATP. Glycogen is good for this.
2. Gold fish, very tolerant of low oxygen situations (carp is related) generate ethanol and CO2 (and of course ATP) as the end product of anaerobic glycolysis. Sounds like a potential special project to me!
D. Just appreciate that there can be
other anaerobic pathways. Appreciate there is an oxygen minimum layer in the
ocean and that many animals live there.
VI. Diving of air breathing animals.
A. Animals most studied for diving
skills: Waddell seal, elephant seal, whale, loon.
Sperm Whale: 2500 meters (sonar) (1
atm for each 10 m of depth)
Elephant seal: 1529 meters
2
hours
B. Major problems:
1. Bends: Gas coming out of solution as pressure drops. Nitrogen is usually the gas and
blockage in the circulatory system by bubbles is the problem.
Oxygen is
typically not a problem because it is not being used. Also less soluble in
water (i.e. blood).
Note that swim
bladders of fish use oxygen rather than air.
2. Maintaining metabolism under
water without oxygen.
3. Pressure collapsing gaseous filled cavities.
4. Carbon dioxide buildup. (Drop in pH) (only a problem if actually using
oxygen)
5. Lactic acid buildup
B. Exhaling before diving and small
lungs are common approaches for
divers.
C. Higher blood amounts and higher
carrying capacity of oxygen in the blood are usually characteristics of divers.
D. Diving bradycardia (slowing of the heart rate) (also known as diving
reflex) and general slowing of metabolism
is common. Brain and central nervous system remains well supplied, but other
circulation to other organs is greatly reduced and correspondingly the
metabolism of those organs is reduced. (Almost all animals – even
non-divers- demonstrate a slowing of the heart when their face is put in water.).
Note that blood pressure is maintained.
E. Spleen in Weddell Seals stores an
unusual amount of oxygenated blood that can be released upon a dive. Turtles
and frogs can draw oxygen from the water via the skin.
F. Anaerobic metabolism is common
among divers. 2 ATP per glucose and a buildup of lactic acid. It is the muscles that mostly generate
lactate and since there is little blood flow to the muscles, little enters the
blood until surfacing. There is also more
of aa tolerance to lactic acid in these animals. More correctly,
probably a tolerance to low pH.
G. Pressure, per se, on the surface
of an animal is not a problem because pressure in will equal pressure out.
Collapse of gaseous spaces can be a problem. Change in chemical reaction rates
(as well as tertiary protein shape and ionizations) due to pressure may also be
a problem.
H. A sperm whale can hold four tons
of spermaceti oil and it is thought that it can change its density (sink to
depths or float to top) by changing the temperature of this oil and thus its
density.
VII. How does the metabolic rate of an animal change with respect to the size of the animal? (allometry). There advantages to being large and advantages to being small. : The larger the animal the more it can . The smaller the animal the more it can .
(side note: the largest living earth animal EVER is alive today: blue whale).
A. Simply dividing the energy rate by the mass of the animal tells us if a given "mass" of animal uses the same amount of energy from animal to animal. (Will a liver cell of an elephant use energy at the same rate as a liver cell of a mouse?) The answer is found that a large animal uses less energy per body weight than a smaller animal. (all else equal)
(Note: The word ³specific² is often used to mean ³divided by body mass². So specific metabolic rate means metabolic rate per unit mass.)
B. Is it based on a body surface area value? No, it turns out a larger animal has a higher rate per unit of body surface area.
C. Surface area is not normally measured, but rather inferred from the mass. Given the mass of a cube of water, could you determine the surface area.
1 gram ====> 1 cm3 ====> 1 cm/side ====> 6(1 cm)(1 cm) surface area
8 gram ====> 8 cm3 ====> 2 cm/side ====> 6(2 cm)(2 cm) surface area
In general for any given shape:
Surface area = constant
* M2/3
Length = constant * M1/3
Volume = constant * M1
(In this case we are somewhat familiar with the constant - the inverse of it is what we call density (mass per unit volume))
D. Met rate = aMb
log (met rate) = log(Mb) + log(a)
log (met rate) = b log(M) + log(a)
y = m x + b
E. Such a plot reveals a slope of 0.75, not 1 and not .667
F. The value of knowing the constants "a" and "b" is the capability of prediction and finding animals that deviate from the norm.
G. Metabolic rate = aMb where b = 0.75 for any animal (or plant or unicells!).
H. The constant "a" would be different for endotherms, ectotherm, unicells, and plants. For two animals of the same size, a would be larger if it had a higer metabolic rate (Thus, endotherms have a higher ³a² value (intercept) than ectotherms.)
I. Almost any function that you want to compare to the size of an animal body will in fact follow the relationship of:
(rate of any physiological function)
= const Mconst.
J. We often measure specific metabolism or specific activity which means that the function divided by the mass of the animal. One must watch the units carefully.
1. So a 250 g rat that uses 50 ml of O2 per minute has a specific metabolism of
(50 ml of O2*min-1) / 250 g = 0.20 ml of O2*min-1g-1
2. Divide by mass through the equation of Met rate = aMb to get the equation for specific metabolism (metabolism per unit weight):
Since b is equal to 0.75 then the plot of log (metabolism per unit weight) versus log(mass) would have a slope of b-1 = -0.25.
V. Energy metabolism as a function of activity. (go over text figures and text discussion).
A. How does one determine the most economical rate and method of locomotion? Want to minimize the Kcal*Km-1 (or Liters O2*Km-1). Walking , running, swimming, flying. Which are most economical, effective. How does the media (land, air, water) affect metabolism of movement.
B. Water is good (assuming the animal is neutrally buoyant) because no effort is needed in supporting the body. Water is bad because it is viscous. Streamlining is important. The larger the animal the more important is the streamlining.
C. Animals will naturally switch from one form of motion to the other when it is economically advantageous.
D. Larger animals can move farther and faster. Larger animals use less energy per km PER KG. (of course their basal metabolism is less PER KG too.)
E. The actual cost does not seem to vary much depending on the method of terrestrial locomotion or endothermy or ectothermy (fig 5.20)!
F. Note the U shaped curve in energy expenditure versus speed for the bird in figure 5.21. If you want to consider the most distance per unit on energy, you need to see where the tangent line through the origin touches the curve. This gives you the smallest O2 kg-1hr-1.
G. Fig 5,22 shows flying as more economical than running (up to a certain body size) and swimming as being even more economical!