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While methanol is significantly smaller than propanol the hemolysis time in methanol was 30% longer than propanol. Can you explain why?
Although methanol is smaller than propanol, it's hemolysis time was much larger than that of propanol because propanol's lipid/water partition coefficient (1.4x10-1)was much larger than that of methanol (9.5x10-3). This is significant because it means that when put in a solvent, in this case the erythrocytes, the solvent is more likely to absorb the substance with a higher lipid/water partition coefficient.
In real life situations, how does this fundamental process of diffusion feature in the functioning of the following vital systems:
Within the digestive system, diffusion follows the principle of equalizing a difference in concentration gradients. The subject of interest, nutrients, arrive in the digestive tract and are absorbed into the bloodstream by the cells lining the tract. The nutrients travel through the blood stream and are transported into cells because there is a lack of nutrients inside the cell which creates the concentration gradient between the inside and outside and causes the nutrients to move down the gradient through the blood vessel membrane, past the target cell membrane and into the cell.
Ingested medicines are diffused into cells in a similar manner because they are arriving with the digested food and nutrients. If the medicine is in pill form then the pill capsule must be digested to release it's contents. The medicine follows the nutrients path and is absorbed by cells in the digestive tract, into the bloodstream, then delivered to cells in need of the medicine that first recognize the medicine, approve of it and allow it to be absorbed into the cell.
Transdermal patches allow medications to be absorbed by the layers of cells of the skin, directly to the bloodstream by the process of diffusion. This process may take longer than diffusion in the digestive system because the medicine must pass multiple layers of cells.
Explain why molecules move down their concentration gradients.
Molecules move down there concentration gradient by the continuous random movements of molecules due to the kinetic energy that all molecules and ions possess. If a difference in concentration exists, then random movements will move molecules from higher concentrations to areas of lower concentrations, or down their concentration gradients. By doing so, the distribution of particles levels out and becomes even on either side of the membrane. This comes from the fundamental principle of homeostasis, where an organism strives to maintain internal stability.
Why is the diffusion of substance A influenced only by the concentration gradient of A, and not also by the concentration gradients of other substances (B, C, D., etc.)?
The diffusion of substance A is influenced only by the concentration of its own gradient because other substances are dealing with their own concentration gradients. The cell works in an organized manner and so only deals with the concentrations of one substance at a time. Even if A were to be bound to a non-permeating solute B on one side of the membrane, and A existed alone on the other side, the concentration of A would determine the passive transport of the bound AB molecule.
What factors determine the net direction of movement of water across the plasma membrane?
The factors that determine the net direction of movement of water across the plasma membrane are concentration and osmotic pressure. Water is a substance that will move down a concentration gradient in order to dilute the more concentrated area to even out the concentration of both sides of the membrane. Osmotic pressure will ensure that the membrane will strive to achieve a concentration of solutions on each side of a membrane that is the same, and is also important in maintaining the volume of a cell.
Is diffusion via facilitated transport endothermic? Please explain your answer.
Facilitated diffusion is a form of protein-aided transport of molecules across a membrane down its concentration gradient at a rate greater than that obtained by passive diffusion. If this process were endothermic, this would mean the reaction that would take place would absorb heat. However, this is not true because facilitated diffusion is a form of passive transport that does not require ATP. Because no energy is required in the process, no energy as heat would be released or absorbed.
Many spontaneous reactions occur slowly. Why don’t they all occur instantly? In biological processes, how is this rate sped up?
Factors that effect the rate of reactions are: concentration of reactants, temperature, viscocity and the presence of an enzyme or catalyst. Not all reactions occur instantly because ideal conditions of concentration, temperature, viscocity and presence of an enzyme or catalyst for a fast reaction aren't always possible to exist. In biological processes this rate is sped up by the catalytic enzymes that enhance the reaction and make it more efficient.
Beta-oxidation is the process by which fatty acids are cut up two carbons at a time to begin the process of generating ATP from lipids. Would you expect that some enzyme in that process might be down regulated by the six carbon molecule citric acid? Why or why not?
The citric acid cycle's purpose is to generate energy by a series of oxidation reactions to form CO2 and ATP. Since the citric acid and beta-oxidation are both trying to generate more ATP and to down regulate means to decrease a quantity, it is unlikely an enzyme in the beta-oxidation process would be down regulated by the six carbon molecule citric acid.
In the text Figure 8.1, which was also shown as a slide in lecture, the speed of ATP production at the start of demand was described for the 4 mechanisms of ATP production. Aerobic production was slowest and the other three mechanisms were fastest. Propose a hypothesis for why this speed differential might be the case.
Aerobic production may be slowest because the transport of oxygen may take up a lot of time. Then the oxidation of substances like NADH and the break down of substances like glycogen, take up a fair amount of time as well. It would follow that anaerobic production during glycolysis would be faster because the body is not utilizing any outside substance and therefore nothing has to wait for a delivery of oxygen to carry out the reaction.
Which of the following statements regarding maximal rate of oxygen consumption (VO2dot max) is false? Give an example that illustrates why it is false.
a. It is the same for individuals within a species, but is different between species.
Maximal oxygen consumption depends on many factors, and therefore cannot be the same for individuals within a species, although the values may be similar. Among these factors are an individuals body build, size and weight, gender, presence of favorable genes and age. Even the external environment and level of oxygen in the air would affect the levels found in an individual's body.
Where exactly inside a cell is the oxygen consumed in aerobic metabolism?
Oxygen is consumed in the mitochondria in cells in aerobic metabolism.
We inhale O2 and exhale CO2.
a. Does the inhaled oxygen atoms leave our bodies in the CO2 that we exhale?
Inhaled oxygen does not leave our bodies in the CO2 that we exhale. The oxygen atoms in the O2 molecules that we breath in get back out of our bodies after they go through the process of aerobic metabolism and leave the body in the form of water. The carbon and oxygen atoms that are in the CO2 we exhale are waste products of cellular aerobic respiration as well, that originate from food that we have eaten.
Explain why in Figs. 8.10 and 8.11, the label for the y-axis reads “Rate of O2 demand or supply” rather than “Rate of O2 consumption or supply.”
The figures say demand instead of consumption because these values are not interchangeable. The amount of oxygen that a process requires is not the same amount of oxygen that is used during the consumption. In certain cases, ATP that is made by processes like anaerobic glycolysis can be used, instead of ATP that is made with the use of oxygen.
Ornithologists have described birds that are about to undertake long-distance migratory flights as “butter balls” because they have large fat stores. Why have these birds stored up energy as fats (lipids) rather than carbohydrates?
Carbohydrates would be a heavier load to carry for birds. This is because the energy density of fats is higher than the energy density of carbohydrates, so less fats would carry the same amount of energy as more carbohydrates would.
Lactate dehydrogenase causes a build-up of lactic acid in the muscles of animals. Although current research indicates that lactate does not cause cramps, it can be an indicator of reduced muscle performance and fatigue. Since this clearly hinders performance, what is the advantage to having this enzyme in muscles?
The advantage of having lactate dehydrogenase in muscles is that it would provide lactic acid that could be used, anaerobically, by the process of gluconeogenesis, to form glucose. In cases where not enough ATP is being delivered, this would allow cells to produce ATP without oxygen.
You are thinking of trying out for the Cornell rowing crew. You weigh 65 kg and you’ve just had your VO2dot max measured at the energetics facility in Martha Van Rensselaer Hall. You’ve been told that your VO2dot max is exactly 5 liters of O2 per minute.
a) What is your maximum sustained power output, expressed in terms of watts (=joules per second)? Note: Assume that in aerobic catabolism the consumption of 1 liter of oxygen makes available to you 20 kilojoules of energy.
[ 1 L = 20 kJ ]
[ 1 W = 1 J/sec ]
[ VO2dot max = 5 L/minute ]
5 L/minute * 20 kJ/L = 100 kJ/minute
100 kJ/minute / 60 seconds/minute = 1.67 kJ/second
1.67 kJ/second = 1670 J/sec = 1670 W
b) Do you have an Olympic level VO2dot max?
[Hint: you may want to check Fig. 9.10 or Table 9.4]
Based on above, I do not. An Olympic elite human athlete has a VO2max of about 1.5 mLO2/kgs, so 90 mLO2/kgmin which would require 5.85 LO2/min, where I only use 5LO2/minute.
[ 90 mLO2/kgmin * 65 kg * 1/1000 ] = 5.85 LO2/min
Is the Km of an enzyme that is the rate limiting step in a series of reactions, such as glycolysis, less than or greater than the Km of the other enzymes in the series? Why?
The Km of an enzyme that is the rate limiting step, is greater than the Km of the other enzymes in the series. The rate limiting step is the slowest step of the rate of reaction. A lower Km is associated with a greater enzyme-substrate affinity and therefore a greater rate, and because the rate limiting step is the slowest step, the Km would be higher because the rate would be slower.
Km = substrate concentration so that the rate of reaction = ½ the Vmax
What is meant by feedback in a series of enzymatic reactions?
Enzymatic reactions are controlled by many regulating enzymes that can either inhibit or activate the enzyme. One method of this is feedback mechanisms, found at allosteric sites, that are able to control allosteric enzymes. They do so by inhibiting an allosteric site, which changes the actual active sit of the enzyme. If the product of this inhibiting reaction builds up, then the enzyme will stop reacting. This works as a sort of feedback mechanism that controls how much product is made by the enzyme.
Positive or negative feedback
It is possible to have position inhibition (context dependent)
Statement: “Lactic acid is a waste product of anaerobic glycolysis and cells try to get rid of it.” State whether you think this statement is true or false and provide a few sentences to defend your position.
Energy storage product in central nervous system
You have been studying hard at Cornell and decide to kick back and just chill for a day.
My mass specific BMR will be 0.254 LO2/h [ 0.676 * 50 kg^(-0.25) ] so for 24 hours [ 0.254 LO2/h * 24 h ] I will need 6.096 LO2. If I were to burn just lipids or just carbs, then I would use 121.92 kJ/kg [ 6.096 LO2 * 20 kJ/LO2 ].
*must use regular BMR because you are at rest, so:
[ 398 * 50 kg^(.75) ] = 7483.6 mLO2/h = 7.484 LO2/h
[ 7.484 LO2/h * 24 ] = 179.616 LO2
[ 179.6 * 20 kJ ] = 3592 kJ
*use mBMr for comparisons (differences in mass as well)
b. How much fat could you burn to supply those needs?
[ 3592 kJ/ 4.18 kJ ] = 859 kcal
[ 859 kcal / 9.4 kcal ] = 91.38 grams lipid
c. How much carbohydrate ?
[ 3592 kJ / 4.18 kJ ] = 859 kcal
[ 859 kcal / 4.0 kcal ] = 214.75 grams carbohydrate
Compare your value for required kJ (1a above) with that of one of your lab mates. Does their value scale proportionally with yours, e.g., if they weigh 20% less than you do, is their value 20% less?
[to be done in discussion I hope?] Theoretically, if my labmate weighs 75 kg, then their mass specific BMR will be 0.229 LO2/h [ 0.676 * 75^(-0.25) ] and they will use 109.92 kJ over 24 hours.
[ 121.92 kJ/ 109.92 kJ ] = 1.11
The value does not scale proportionally. This is because of the negative exponent in the mass specific BMR equation, which makes the equation exponential.
Professor Gilbert tips the scales at 127kg, probably more than most of you. Nevertheless, he has a lower mass specific metabolic rate than you do. If he decides to chill for a day, will he require fewer kcal? If not, why not?
*as mass increases, metablic rate increases, but on a logarithmic scale
Professor Gilbert when using the regular BMR equation, he will require more kcal because his mass is more than my own.
Usain Bolt currently holds the human sprint speed record 10.4 m/s, running the 100m dash in 10.58s. The cheetah can cover 200m in shorter time, 6.9s, for a burst speed of 28.9 m/s. If we wanted to measure metabolic rate during such displays of human or carnivore prowess, why would measuring VO2dot not be a reasonable approach.
*Initially, the body uses anaerobic metabolism, so for such a short time, measuring the volume of oxygen using will not apply entirely.
If your 70 lb dog uses 20 kJ running for 1 min is it working very hard?
[ 70 lbs / 2.2 lbs ] = 31.81 kg
VO2dot max (mlO2/min) = 118.2 * (31.81 kg)^0.87 = 2398.278 mlO2/min
[ 2398.278 mlO2/min * 1 min ] = 2398.278 mlO2
1 L = 1000 ml
[ 2398.278 mlO2 / 1000 ml ] = 2.398 LO2
1 LO2 burning lipids OR carbs = 20 kJ
[ 2.398 LO2 * 20 kJ ] = 47.96 kJ
Based on the calculations above, the maximum metabolic rate for a dog weighing 70 lbs over a minute would be 47.96 kJ. Since the dog is only using 20 kJ, less than half of the maximum energy needed, the dog is not working very hard.
What elements of the heat exchange equation affect an animal’s body temperature?
Tbody = Tambient + Tmetabolic +/- Tradiation +/- Tconduction +/- Tconvection – Tevaporation
According to the heat exchange equation, the temperature of the environment, the energy used during metabolism, and radiative, conductive, convective and evaporative cooling affects an animal's body temperature.
Some of the elements can cause gain or loss of heat, but two can only influence Tb in one
way. What are those elements and why can they only affect Tb asymmetrically?
Tb can only be influenced in one way by Tmetabolism and Tevaporation. This is because Tmetabolism represents the energy that is being used by the body, so the process of metabolism will always generate heat, causing this value to always be positive. Tevaporation is a representation of the input of energy that takes away heat from the temperature of the body Tb, and so can only cool the body, causing this value to always be negative.
Many parasites have bizarre life cycles. Some have two hosts, an invertebrate animal, such as
a mosquito, and a mammal such as a person. The parasite needs to function in both hosts.
Would you expect the parasite to have narrow or broad thermal tuning of its enzymatic
Although most ectotherms are thermal conformers, some are not. Give an example and
suggest a mechanism by which a specific ectotherm is able to regulate and maintain Tb above
Many creatures engage in behavioral thermoregulation, whether they are ectotherms or
endotherms. Provide a couple of examples of behaviors that contribute of modification of an
What is the thermal neutral zone (TNZ)?
How is an animal able to maintain BMR constant even though ΔT changes in the TNZ ?
How are the mitochondria different in brown fat compared to white fat ? (facilitated transport)
The mitochondria in BAT are larger in size, and have an abundant supply of yellow cytochrome pigments which give the cells their color. The inner membranes of mitochondria also hold the proton-transport protein uncoupling protein 1 that allows the tissue to undergo uncoupling. The process: Norepinephrine released in BAT binds to beta-adrenergic receptors in cell membranes of BAT which activates G proteins in cell membranes and leads to intracellular production of second messenger cyclic AMP, receptors are G protein-coupled receptors, cyclic AMP activates intracellular lipase enzyme that rapidly hydrolyzes triacylglycerols stored in cells to release free-fatty-acid fuels for mitochondrial oxidation. Molecules of the uncoupling protein are activated, mitochondria carry lipid oxidation in uncoupled state.
Write the heat exchange equation and describe why some terms are only added or subtracted.
M is animal's metabolic rate and equals animal's rate of heat loss
C is animal's thermal conductance
Tb is body temperature
Ta is ambient temperature
Ta is subtracted because this means that heat is transferring out of an animal's body by dry heat transfer.
If Tb was subtracted, then that would mean that heat was moving into the body.
Tb = Ta + Tmet +/- Tradiation +/- Tconduction +/- Tconvection - Tevaporation
Metabolism will always generate heat
Evaporation can only cool, input of energy, takes away heat
Radiation, conduction, convection can be all three depending on situation
Define the conditions under which Evaporative Cooling will be the major feature used to maintain homeostasis in core body temperature.
When subjected to Hyperthermic conditions explain how the principles of Convection and Conduction could be employed to regain a core temperature set point of between 36.5 and 37.5 (°C)?
Conduction is the transfer of heat through a material substance that is motionless. Convection is the transfer of heat through a material substance that is moving. To regain a certain core temperature set point after being exposed to a hyperthermic environment, convection could cool the body if an animal would find a cooler mass to touch, for example a lizard hiding under a cool rock, and conduction would be a cool breeze blowing.
Convection could be cold shower, conduction could be cold compress.
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