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Tutorial #6:

Gases, temperature and metabolism: poikilothermic and homeothermic responses.
by Syreeta Williams on 16 November 2012

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Transcript of Tutorial #6:

Gases, temperature and metabolism: Poikilothermic and Homeothermic responses Tutorial #6 Define and Explain the following terms : Poikilotherms
A poikilotherm refers to an organism in which their internal temperature varies continuously with their surrounding temperatures. Organisms that are poikilotherms are classified as cold blooded organisms as well. Cold blooded animals are considered to be any organism that requires the environment to provide their bodies with body heat, due to their inability to control their own internal body temperatures. 4a) A balance equation A balance equation is input = output.
- the products are equal to the equal to the reactants 2b. You conduct an experiment to measure gas exchange. A 17 g mouse consumes 4.8 ml of oxygen and produces 3.89 ml carbon dioxide (both volumes are corrected to standard temperature and pressure).
Determine its respiratory quotient (RQ).

What is the respiratory quotient? The respiratory quotient refers to the ratio of the volume of carbon dioxide released to the volume of oxygen consumed by a body tissue or an organism in a given period. Respiratory Quotient equation
RQ = CO2 eliminated / O2 consumed 3.98ml Carbon Dioxide / 4.8 ml Oxygen
= 0.8219 Homeothermy
Homeotherms maintains its body temperature at a constant level that is usually ...? a. Below the environmental temperature b. Above the environmental temperature c. Same temperature of the environmental temperature Homeothermy refers to the study of organisms that are capable of maintaining a stable internal body environment. Homeotherms are warm blooded animals. Warm blooded animals convert the food they consume into energy. Warm blooded animals have to eat an ample amount of food in order to retain the sufficient amount of energy needed by the animal. Multiple choice : Group E Avena , Daniel , Helen, Syreeta Fun Fact: Quantitatively, RQ is usually between .7 and 1. When it is closer to 1, the metabolism is one of carbohydrates and when it is closer to .7 it is one of fats. Intermediate numbers are harder to specify :) Explain what would happen in each case if an animal were in i) negative or ii) positive balance. Give examples. Negative water balance occurs when the amount of water excreted is greater than the amount of water ingested, also known as dehydration. Possible causes include diarrhea, vomiting, and high perspiration levels, especially when combined with low water consumption. Positive Energy balance: The body would gain more energy than it needs, and would compensate by storing that energy as fat, leading to weight gain. If the animal is lethargic, and does not exercise it cannot lose the weight and may become obese. Its heart, and other vital organs, will need to work harder to compensate for the increased weight. ENERGY BALANCE Negative Energy Balance: More energy is lost than gained, perhaps due to extreme exercise, dieting, or starvation. The body is burning fat storages to produce the needed energy. Positive water balance occurs when the amount of water excreted is less than the amount of water ingested. The animal would certainly become bloated from excess water in their body. If circumstances become severe, Edema may develop and excess water surrounding systemic cells could damage the internal cells. Thermal Balance Negative Thermal Balance: Here, more energy is lost than gained. Usually this occurs when the external temperature fall below the thermoneutral zone. Mammals decrease internal temperature by shivering, while ectotherms would decrease their metabolic rates. Positive Thermal balance: In this scenario, the animal gains more energy than it loses. When temperatures exceed the critical temperature, the body must increase its metabolic rate to decrease its internal temperature. An ectotherm would change its behavior by perhaps moving to a cooler spot. More developed species can increase their metabolic rates. Endotherms increase their metabolic rate by sweating. Organismic: Environmental 4b) A gecko consumes 1.3g/day of insects containing 75% water and an energy content of 25kcal per gm of oven-dried insect. 30% of the energy is lost in feces and 4% lost in urine. Assuming that the gecko is in energy balance: Determine the average daily metabolic rate and the digestive efficiency. What are the components of.... i) energy Remember: energy is never created or destroyed, it can only be converted between various types of energy including : chemical, kinetic, potential, and themrmal. - food (chemical energy)
- caloric energy including the sun, basal metabolic rate in which ATP is exerted, specific dynamic action and activities such as thermoregulation and alertness that result in the production of energy, body products that require energy such as reproduction and growth, and heat. Water ii) Water: Water balance esists on an organismic and enviornmental level.
ORGANSMS:
Water gained from: food, drink, metabolic processes, respiratory surfaces
Water Lost in: feces, urine, sweat, respiratory processes.

ENVIORNMENT:
run off, evapotranspiration, change in H20, storage (soil/bedrock) VS. precipitation Thermal: Heat gained: from the environment
Heat lost in: energy, feces, urine, metabolic processes such as metabolic activity and brain activity. heat is transfered through through conduction, convection, radiation and evaporation 4c) Input and output eqations: i.) Energy: food = ATP + BMR + Body activities + reproduction + growth + heat

ii.) Water: Food + Drink + Metabolism = Feces + Urine + Sweat + Respiration + Metabolic Processes

iii.) Thermal Balance: Environment Heat + Muscle heat production = Energy + Feces + Urine + Metabolic Processes. Water weight : 0.75 x 1.3 g/day = 0.975 g/ day
Total Dry Energy weight = 25 kcal/g x (1.3 g/day – 0.975 g/ day) = 25 kCal/g x 0.325 g/day
Input = 8.125 kCal/dayOutput:
a.) Feces: 8.125 kCal/day x 0.30 = 2.4375 kCal/day b.) Urine: 8.125 kCal/day x 0.04 = 0.325 kCal/dayMetabolic Rate = Input – output à 8.125 – (2.4375 + 0.325) = 5.3625 kCal / day

Digestive Efficiency = [(energy input – energy lost) / food intake] x 100
Digestive Efficiency = [(8.125 – (2.4375 + 0.325)) / 8.125] x 100Digestive Efficiency = [(8.125 – 2.7625) / 8.125] x 100
Digestive Efficiency = (5.3625 / 8.125) x 100
Digestive Efficiency = (0.66) x 100 = 66 % 4d) Explain how all three regulatory systems (energy, water, heat) are dependent upon each other. The body needs energy which it obtains from food in order to carry out the other processes. ->
From this energy water is generated within the body and taken in from the external environment. ->
This energy thus allows the body to regulate its water balance and create heat. ->
With the proper internal heat the animal is able to acquire energy efficiently and process the sufficient amount of water to not dehydrate.->
Without water the other regulatory systems will not be able to function because water is vital
for the cells in which these regulations occur. 2d. What is the heat production? The average amount of heat produced for each liter of oxygen consumed in metabolism is about 4.8 kCal or 20.1 kJoules. Heat production can be calculated by multiplying the volume of oxygen that the mouse consumes, by the heat produced per unit of oxygen consumed.


If Carbohydrates are Oxidized = (4.8 ml x 21.1 J/ml) = 101.28 Joules
If Lipids are Oxidized = (4.8 ml x 19.8 J/ml) = 95.04 Joules
If Proteins are Oxidized (4.8ml x 18.7 J/ml) = 89.76 (Assume a caloric equivalent of 4.8 Kcal/L) The mouse weights 17 grams. Also, the mouse consumes 4.8 ml of oxygen and produces 3.89 ml carbon dioxide. Remember !
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