1. An argument of some creationists states that evolution is impossible because it
defies the second law of thermodynamics. They state that molecules should not
form the ordered structures seen in organisms. Instead, the second law of
thermodynamics dictates that the molecules on earth should become more
disordered. Based on your knowledge of the second law, what is the flaw in this
argument?
This argument is flawed because the system has not been properly defined. Earth is not a
closed, isolated system. Energy is always coming in to the earth in the form of solar
radiation. This energy allows for certain components of the earth to become more
ordered. Thus, the earth is only a small part of the system as a whole, with that system
being the universe. Certain parts of a system can become more ordered as long as the
system as a whole is becoming more disordered. The universe as a whole is increasing in
disorder (entropy).
2. Outline the flow and eventual fates of energy in this animal by labeling this
diagram, similar to figure 6.2 in HWA. Is this animal an open or a closed system?
Ingested:
Chemical bond energy
Inside body:
Biosynthesis of compounds used for growth or storage
Maintenance(tissue repair, circulation, respiration, etc.)
Chemical bond energy in the feces
Chemical bond energy in exported matter (hair, skin cells, etc)
Mechanical energy through external work
Heat
Egested:
3. What is the difference between direct and indirect calorimetry?
Direct calorimetry measures an animal’s heat production, which is used to determine
metabolic rate, whereas indirect calorimetry measures metabolic rate by measuring gas
exchange or material balance, then using these data to quantify heat production.
4.
What is the overall reaction for the complete oxidation of glucose?
C6H12O6 + 6 O2  6CO2 + 6H20 + 2820 kJ/mol
5. What is the respiratory quotient (RQ), and why is it different for different
foodstuffs?
The RQ is the ratio of CO2 produced per mole of O2 consumed per unit time. It is
different depending on foodstuffs because different foodstuffs produce different amounts
of CO2 per unit oxygen. The stoichiometry of the equations are different.
6. If an animal exhibits an RQ value of 0.83, what is it likely oxidizing? How do we
know (or not know!)?
It could be oxidizing proteins, or it could be oxidizing a mixture of foodstuffs. We don’t
know, because this number is between the RQ values for carbohydrates and lipids. We
must interpret this value carefully.
7. Discuss the differences between basal, standard, and resting metabolic rate.
Basal metabolic rate is measured in endothermic animals that are resting, fasting, and in
their thermoneutral zone. Resting indicates that the measurements are taken during the
inactive phase of the animals’ circadian rhythms, at night for a diurnal animal, during
the day for a nocturnal animal.
Resting metabolic rate in endothermic animals stipulates only that animals are fasting
and thermoneutral. The definition of resting is relaxed to only mean that the animals are
not engaged in excessive activity.
Standard metabolic rate is measured in poikilothermic animals, and stipulates that the
animals must be resting and fasting. For a single animal, a different resting metabolic
rate may be calculated at many different temperatures.
8. As a new student in a physiology lab, you are studying metabolism in birds. You
arrive in the lab around 10:00 a.m. and feed the birds. As you are feeding the
birds, you notice that it is quite chilly in the room, causing you to shiver slightly.
As soon as you finish this task, you tell your advisor that you plan to measure the
basal metabolic rate of these birds immediately. Your advisor tells you that there
are at least 3 flaws in your experimental setup. What are these flaws, and how
might you correct them?
1. It is daytime. Birds are diurnal. To measure BMR, you must measure them
during their resting phase at night. You might correct this by measuring the birds
during the night, or, if you are conducting a long-term acclimation experiment,
you might manipulate the lights such that it is dark during the day, and light at
night. You might also simply give up on measuring BMR and measure resting
metabolic rate instead.
2. You have just fed the birds. The birds will have an elevated metabolic rate
because of specific dynamic action as they digest their food. You need to wait a
few hours before these birds can be termed as “fasting”.
3. It is cold in the room, causing you to shiver. If you are cold in the room, there is
a chance that the birds are also cold, which could mean that their metabolic rate
is elevated. You should either raise the temperature in the room to about 30 oC,
or, if your lab has such a setup, control the temperature of the metabolic chamber
to ensure that the birds are in their thermoneutral zone.
Metabolic Rate
9. Place the following labels on the graph below: thermoneutral zone, upper critical
temperature, lower critical temperature. Define each term.
Temperature
The point at which the downward sloping line meets the horizontal line is the lower
critical temperature. This temperature represents the lowest temperature at which an
animal does not need to elevate its metabolic rate to warm itself and maintain a
normal body temperature.
The flat portion of the graph is the thermoneutral zone. This zone is the range of
temperatures over which the metabolic rate does not need to be raised to maintain a
normal body temperature. An animal must be in this zone to measure basal or
standard metabolic rate.
The point at which the horizontal line meets the upward sloping line of the graph is
the upper critical temperature. This is the highest temperature at which an animal
need not raise its metabolic rate to cool itself and maintain a normal body
temperature.
10.
a. Your lab does research using meadow voles as test subjects. As the
undergraduate in the lab, you are charged with taking care of these voles. You feed
the 30g voles about 175g of food per week. Your advisor gets curious about other
animals as test subjects, and traps some meadow jumping mice. These mice weigh
approximately 15g. You feed the jumping mice 87.5g of food per week, exactly half
of what you fed the voles. Will the jumping mice be healthy on this diet? Why or
why not?
The mice will not be healthy, because metabolic rate is not a linear function, but an
allometric function. The mice will actually require more than 87.5 g of food, because
their weight-specific metabolic rate is higher than the meadow voles’.
b. Using figure 6.9 from your book, calculate the expected mass specific
metabolic rate of the meadow vole and the meadow jumping mouse.
For all mammals, weight specific metabolic rate can be calculated using the following
equation:
M/W = 4.46W-0.30
This equation is identical to equation 6.4 in HWA (M/W = aW(b-1)), with the constants a
and b derived from empirical data for mammals.
Meadow Vole:
M/W = 4.46*30g-0.30 = 1.61 mL O2/g*hr
Jumping Mouse:
M/W = 4.46*15g-0.30 = 1.98 mL O2/g*hr
11. Dune Larks reside in the Namib Desert, one of the driest regions on earth. For Dune
Larks with an average body mass of 27.3 g, BMR equals 74.7 mL O2/h or 36.0 kJ/d.
Below 30 C, total evaporative water loss (TEWL) varied little with ambient temperature
(Ta) and averaged 66.0 mg H2O/h, or 1.58 g/day. Above 30 C, TEWL was described by
the equation mg
H2O/h = e-0.164 + 0.161(Ta). Between -3.9 and 35 C, body temperature did not vary and
averaged 40.8 C. Above 35 C, Tb increased linearly with Ta and was described by the
equation
Tb = 35.1 + 0.2 (Ta).
Below the lower critical temperature, oxygen consumption of Dune Larks was
described by ml O2/h = 201.95 - 4.56(Ta). Above the thermal neutral zone, oxygen
consumption was described as ml O2/h = -118.1 + 5.5(Ta).
a. Construct a graph of metabolism vs. Ta and TEWL vs Ta. Notice that there are three
lines that you can graph. 1. A line for metabolism below the lower critical temperature, a
line for BMR, and a line above the upper critical temperature. Note that the intersection
of these lines will provide information about Tlc and Tuc.
b. What is the upper and lower critical temperature of Dune Larks?
Upper critical temp. = 35.05 oC (74.7= -118.1 + 5.5Ta)
Lower critical temp. = 27.91oC (74.7=201.95 – 4.56Ta)
c. Calculate the heat transfer coefficient H at 15 C, at 30 C, and at 45 C. When Ta > Tb,
should the bird minimize h or maximize h? Recall that 20.08 J of heat are produced per
ml oxygen consumed and that 2.4 J of heat are lost per mg water lost as evaporation.
At 15oC:
Ta = 15oC
Tb=40.8oC
Hm=201.95 – 4.56*Ta (Bird is below TNZ) * 20.08J
=201.95 – 4.56*15 = 133.55 mL O2/h * 20.08 = 2681.68 J/hr
He=66.0 mg H2O/hr * 2.4 J = 158.4 J
h=(Hm-He)/(Tb-Ta)
=(2681.68 – 158.4)/(40.8 – 15)
97.80 J/oC
The bird should minimize h, to ensure that it doesn’t lose excessive heat to its
surroundings at cool temperatures
At 30oC:
Ta = 30oC
Tb=40.8oC
Hm=74.7 * 20.08 = 1499.98 J/hr
He=e-0.164+0.161*30 = 106.27 mg H2O/hr * 2.4 J = 255.05 J
h=(Hm-He)/(Tb-Ta)
=(1499.98 – 255.05)/(40.8 – 30)
115.27 J/oC
At 45oC:
Ta = 45oC
Tb=35.1 + 0.2*Ta = 35.1 + 0.2*45 = 44.1oC = HOTTER THAN BEJESUS
Hm= -118.1 + 5.5*Ta * 20.08 (Bird is above TNZ)
=( -118.1 + 5.5*45) * 20.08 = 2598.35 J/hr
He=e-0.164+0.161*45 = 1189.16 mg H2O/hr * 2.4 J = 2853.98 J
h=(Hm-He)/(Tb-Ta)
=(2598.35 – 2853.98)/(44.1 – 45)
284.033 J/oC
The bird should attempt to push its value of h to a minimal value, to reduce the heat
influx from the environment. However, the bird does not appear able to do this. It’s
heat flux is h*(Tb-Ta) = -255.63 J, meaning that 255.63 J are flowing into the bird!
d. At what temperature does Ta = Tb?
Set both Ta and Tb as x, since the value is the same, and solve for x
x = 35.1 + 0.2 * x
0 = 35.1 – 0.8*x
35.1 = 0.8*x
x = 43.88 oC
12. Why is it important to use isotopes for both hydrogen and oxygen in the doubly
labeled water method?
The loss of doubly labeled water in an animal is a function of both its CO2 production
and its rate of water loss. Oxygen isotopes will be lost through both water loss and CO2
production, and therefore by themselves can not provide an accurate index of metabolic
rate. Hydrogen isotopes will only be lost through water loss. Therefore, a scientist can
calculate the amount of water lost from the hydrogen isotopes lost, and then subtract this
value from the amount of oxygen isotopes lost, and calculate the amount of oxygen
isotopes lost through CO2 production. In this way, the metabolic rate can be accurately
calculated.
13. Suppose that under the conditions sufficient to measure basal metabolic rate, a
kangaroo rat consumes 2 mL O2/g*hr. You force this kangaroo rat to run on a
wheel until it is exhausted, and find that its maximum O2 consumption (VO2 Max)
was 14 mL O2/g*hr. What is the aerobic scope for this animal? What is its
aerobic expansibility?
Aerobic scope:
VO2 max – BMR O2 Consumption = 14 – 2 =12 mL O2/g*hr
Aerobic expansibility:
VO2 max / BMR O2 Consumption = 14/2 = 7 mL O2/g*hr
14. Bar-tailed Godwits are currently recognized as the record holders among birds for
the longest non-stop flight during migration. This species takes an oceanic route from its
breeding grounds in Alaska and Russia to its wintering grounds in New Zealand. One
individual was tracked as traveling 11,026 km (6,851 miles) over the course of 9 days!
A closely related species, the Hudsonian Godwit, also breeds in Alaska, but takes an
overland route on its trip south, making stops to feed along the way. How might we
expect the optimal speed of travel to differ between these two species during migration?
We might expect the Hudsonian Godwit to fly faster, because it is able to feed during its
journey. The Bar-tailed Godwit will need to minimize energy cost per unit distance, as it
must cover a great distance without feeding. The Hudsonian Godwit does not have this
concern, as it can easily seek a “pit stop” if its energy stores begin to run low.
15. Suppose the heat of metabolism is 2,681.7 J/h, and the heat lost by evaporation is
160.2 J/h. At a body temperature of 40.8 C and an ambient temperature of 15 C, what is
the heat transfer coefficient?
h=(Hm-He)/(Tb-Ta)
(2,681.7 – 160.2)/(40.8-15)
97.73 J/oC
16. What physiological parameters might we expect to change in a person who
undergoes endurance training?
Higher VO2 Max, Higher levels of muscle citrate synthase activity, higher levels of
muscle cytochrome oxidase activity, higher levels of testerone in males.
17. Many people who feed birds in their backyard will put out suet (beef fat) in
addition to seeds for the birds to eat. During the winter months, the birds
consume the suet much more quickly than they do during other times of the year.
Why might this be so in terms of ecological energetics?
During the winter, the temperature is very low, and these low temperatures require that
birds expend a lot of energy simply to keep themselves warm, primarily through
shivering. These energetic costs must be balanced by increased energetic intake if a bird
is to survive. The act of foraging itself may expose the bird to harsher
microenvironments, thus it is most beneficial for the bird to minimize the time spent
foraging. To do this, the bird should choose foodstuffs that provide the most energy per
unit weight. Fats are extremely high in energy compared with seeds, or other items that
might be available to birds in the winter. Therefore, birds will preferentially seek suet to
minimize energetic expenditure and maximize energetic gain.