The immune system is a vast collection of tissues and cells that work together to keep the body free of infection. In doing so, the system has to balance between hypoactivity, which allows infection to reign, and hyperactivity, which causes allergies and autoimmune conditions. A complex system of local and long-distance signals allow for the system to respond appropriately. Long distance signals place the body and the immune system on alert. Local signals tell arriving immune cells that they are entering the war zone, and to draw their weapons (keeping the weapons holstered until they arrive at the battle field reduces friendly-fire incidents)
There is also the problem of recognition - immune cells are worthless if they don't know what to attack, since they ignore pathogens (bacteria, etc) and attack host cells (self). The system by which immune cells are schooled requires specialized environments and signaling processes.
This site shows animations that explain many of the intricacies of the immune process. It's much more than you need for the MCAT, but very useful for those of you already into med school.
Saturday, September 26, 2009
Thursday, September 24, 2009
The War of 1812 and the US Navy
I'm going to give you a hypothetical situation to think about: what if an Iranian frigate took - destroyed - a U.S. frigate?
Now, I'm aware of the attack on the U.S. Cole several years ago - and this is not what I'm driving at. Some crewmen died in that attack, but the Cole survived, was repaired, and returned to service in the U.S. Navy. As shocking as the attack was, the Cole wasn't destroyed, and she wasn't taken. But what if an Iranian frigate took a U.S. one?
In 1812, the United States of America declared war on England. The reasons for this are long and complex (as is so often the case on war) and are beyond the point of this post, but the outcome of the war arguably marked the entrance of the U.S. onto the world stage.
Before the war, the U.S. was merely a loose group of former colonies - a third-rate nation at best. They possessed little in the way of a navy, with 19 vessels, of which 16 were actually in service. Seven of these were frigates, with the remainder being smaller vessels such as brigs and sloops. England's navy (the Royal Navy) possessed over 600 in-service vessels, of which about 175 were ships of the line - a class of ships that would eventually come be known as battleships, and which were larger and heavier than the frigates that formed the largest ships in the American navy. So on paper, there was no contest: the American navy would be lucky to capture a few British merchantmen before being captured itself, or at best bottled up by Royal Navy blockade.(1)
The course of history also seemed to be against the Americans. For the past 20 years, the Royal Navy had routinely routed every enemy it had faced. Nelson's victory at Trafalgar(2) had been notable only for the scale of the victory; the Royal Navy simply won and won, even when outmanned and outgunned. It was a foregone conclusion that the war at sea would be swiftly over, with England victorious.
It was with supreme confidence, therefore, that Captain Dacres of the HMS Guerriere met the USS Constitution (Captain Hull) on August 19th, 1812. He addressed his men, saying that he exepcted them to beat the Constitution in 30 minutes, and that he would be "offended with them if they did not do their business in that time." Dacres was not too far off in the length of the battle (Constitution ceased firing less than 25 minutes after she opened fire at 6:05pm) but he was wrong in his prediction of its outcome: Constitution destroyed the Guerriere, so badly shattering her that she was worthless as a prize and had to be burned so as not to be a menace to navigation. Besides their frigate, the British lost 23 killed, plus another 56 wounded. American casualties were seven killed, and seven wounded.
Let me pause here to see if I can put this in modern terms. England no longer rules the waves - if anyone does, I suppose it is America. So again, what would we think if, say, an Iranian frigate engaged a U.S. frigate - and destroyed her in less than half an hour?
Of course, this only begins to approach the reality of what happened in the War of 1812, because the U.S. Navy hasn't spent twenty years defeating every other armed nation on earth. If the U.S. Navy were to tomorrow take on, say, the combined English and German navies, I don't know who would win. And, of course, not only did the Constitution take the Guerriere on August 19th, but a little over two months later the USS United States took the HMS Macedonian. And then on December 29th, Constitution met and took the HMS Java. The United States, an infant nation with an insignificant navy, met and smashed the forces of the most powerful international force in the world. The world took notice.
[EDIT, 1 OCTOBER 2009: A friend of mine recently pointed out that if Iran were to successfully attack an American warship in any meaningful way, their joy would be short-lived: "I think Iran would regret their victory. The 19th century Royal Navy, for all its immense power had nothing like a B-52 or, heaven forbid, the U.S.S. Tennessee." I think he's correct, and that's part of my point, since England in 1812 felt similarly confident about any naval clash they had with the U.S. So my point is this: England in late 1812 was shocked by the American successes, as shocked as America would now be if its navy repeatedly lost to the Iranians.]
(1) This disparity is lessened by the fact that England was then also embroiled in the Napoleanic wars, which placed great demands on her navy, but the fact remains that the Royal Navy was much more powerful than the U.S. Navy, with larger, heavier ships and greater reserves of men and materiel.
(2) Nelson, with 27 ships of the line, trounced a combined Franco-Spanish fleet of 33 ships of the line, sinking one and capturing 17 while losing none of his own.
Sources:
* Battle of Trafalgar: Grant, R. G. Battle at Sea: 3,000 Years of Naval Warfare. DK Publishing, New York. 2008 @ pp 188-189.
* War of 1812:
- relative strength of the Royal and American Navies: Toll, Ian W. Six Frigates: The Epic History of the Founding of the U.S. Navy. Norton, New York. 2006. @ pp 331-333.
- Constitution:Guerriere engagement: Toll (ibid) @ pp 347-354.
- Constitution:Java engagement: Toll (ibid) @ pp 375-380.
- United States:Macedonian engagement: Toll (ibid) @ pp 360-365.
Cross-posted on main page
Now, I'm aware of the attack on the U.S. Cole several years ago - and this is not what I'm driving at. Some crewmen died in that attack, but the Cole survived, was repaired, and returned to service in the U.S. Navy. As shocking as the attack was, the Cole wasn't destroyed, and she wasn't taken. But what if an Iranian frigate took a U.S. one?
In 1812, the United States of America declared war on England. The reasons for this are long and complex (as is so often the case on war) and are beyond the point of this post, but the outcome of the war arguably marked the entrance of the U.S. onto the world stage.
Before the war, the U.S. was merely a loose group of former colonies - a third-rate nation at best. They possessed little in the way of a navy, with 19 vessels, of which 16 were actually in service. Seven of these were frigates, with the remainder being smaller vessels such as brigs and sloops. England's navy (the Royal Navy) possessed over 600 in-service vessels, of which about 175 were ships of the line - a class of ships that would eventually come be known as battleships, and which were larger and heavier than the frigates that formed the largest ships in the American navy. So on paper, there was no contest: the American navy would be lucky to capture a few British merchantmen before being captured itself, or at best bottled up by Royal Navy blockade.(1)
The course of history also seemed to be against the Americans. For the past 20 years, the Royal Navy had routinely routed every enemy it had faced. Nelson's victory at Trafalgar(2) had been notable only for the scale of the victory; the Royal Navy simply won and won, even when outmanned and outgunned. It was a foregone conclusion that the war at sea would be swiftly over, with England victorious.
It was with supreme confidence, therefore, that Captain Dacres of the HMS Guerriere met the USS Constitution (Captain Hull) on August 19th, 1812. He addressed his men, saying that he exepcted them to beat the Constitution in 30 minutes, and that he would be "offended with them if they did not do their business in that time." Dacres was not too far off in the length of the battle (Constitution ceased firing less than 25 minutes after she opened fire at 6:05pm) but he was wrong in his prediction of its outcome: Constitution destroyed the Guerriere, so badly shattering her that she was worthless as a prize and had to be burned so as not to be a menace to navigation. Besides their frigate, the British lost 23 killed, plus another 56 wounded. American casualties were seven killed, and seven wounded.
Let me pause here to see if I can put this in modern terms. England no longer rules the waves - if anyone does, I suppose it is America. So again, what would we think if, say, an Iranian frigate engaged a U.S. frigate - and destroyed her in less than half an hour?
Of course, this only begins to approach the reality of what happened in the War of 1812, because the U.S. Navy hasn't spent twenty years defeating every other armed nation on earth. If the U.S. Navy were to tomorrow take on, say, the combined English and German navies, I don't know who would win. And, of course, not only did the Constitution take the Guerriere on August 19th, but a little over two months later the USS United States took the HMS Macedonian. And then on December 29th, Constitution met and took the HMS Java. The United States, an infant nation with an insignificant navy, met and smashed the forces of the most powerful international force in the world. The world took notice.
[EDIT, 1 OCTOBER 2009: A friend of mine recently pointed out that if Iran were to successfully attack an American warship in any meaningful way, their joy would be short-lived: "I think Iran would regret their victory. The 19th century Royal Navy, for all its immense power had nothing like a B-52 or, heaven forbid, the U.S.S. Tennessee." I think he's correct, and that's part of my point, since England in 1812 felt similarly confident about any naval clash they had with the U.S. So my point is this: England in late 1812 was shocked by the American successes, as shocked as America would now be if its navy repeatedly lost to the Iranians.]
(1) This disparity is lessened by the fact that England was then also embroiled in the Napoleanic wars, which placed great demands on her navy, but the fact remains that the Royal Navy was much more powerful than the U.S. Navy, with larger, heavier ships and greater reserves of men and materiel.
(2) Nelson, with 27 ships of the line, trounced a combined Franco-Spanish fleet of 33 ships of the line, sinking one and capturing 17 while losing none of his own.
Sources:
* Battle of Trafalgar: Grant, R. G. Battle at Sea: 3,000 Years of Naval Warfare. DK Publishing, New York. 2008 @ pp 188-189.
* War of 1812:
- relative strength of the Royal and American Navies: Toll, Ian W. Six Frigates: The Epic History of the Founding of the U.S. Navy. Norton, New York. 2006. @ pp 331-333.
- Constitution:Guerriere engagement: Toll (ibid) @ pp 347-354.
- Constitution:Java engagement: Toll (ibid) @ pp 375-380.
- United States:Macedonian engagement: Toll (ibid) @ pp 360-365.
Cross-posted on main page
Monday, September 21, 2009
Essays: keep your eyes on the prize
NOTE: The MCAT no longer includes an essay.
I once had an SAT student who wrote a brilliant essay in which she evaluated both sides of an issue before deciding that there was merit to both sides, and that we could determine which side to go with based on a particular criterion, which she listed and discussed. As I said, it was brilliant, and it was better than many of the essays I receive from MCAT students. Unfortunately, she failed to adequately address the question that had been asked of her, so I couldn't give her a good grade.
When you're asked to write an essay, the first step is to be sure that you understand what you're assignment is. the second step is to be sure that you actually write to address that assignment. Both steps are important.
The first step doesn't take much: you just have to take the time to read the assignment and make sure that you understand it. Try to paraphrase the assignment question (put it into your own words) to be sure that you understand it. If you taking an exam where you're allowed to do so, and you're uncertain about the assignment, ask your instructor.
The second can be trickier, since it can be tempting to use the assignment as a jumping-off point for an essay that ultimately charges off into other territories, or to only answer part of a more complex assignment. Once we're writing, our thoughts may focus on the what's in front of us - is this fact correct, is my grammar ok - and we can lose sight of where we're actually supposed to be going.
The solution to keeping on track is to plan out the essay before writing it. Take a few minutes to sketch out the points you want to make, with their examples and/or reasoning. Look at the completed sketch to make sure that it actually matches the assignment. And then write the essay, keeping to the sketched-out plan: if a new example comes to us as we write, don't add it unless there is the time to go back and rework the original plan to include it (in other words, it's okay to add examples to a take-home assignment, but not for an in-class exam).
When you're asked to write an essay, the first step is to be sure that you understand what you're assignment is. the second step is to be sure that you actually write to address that assignment. Both steps are important.
The first step doesn't take much: you just have to take the time to read the assignment and make sure that you understand it. Try to paraphrase the assignment question (put it into your own words) to be sure that you understand it. If you taking an exam where you're allowed to do so, and you're uncertain about the assignment, ask your instructor.
The second can be trickier, since it can be tempting to use the assignment as a jumping-off point for an essay that ultimately charges off into other territories, or to only answer part of a more complex assignment. Once we're writing, our thoughts may focus on the what's in front of us - is this fact correct, is my grammar ok - and we can lose sight of where we're actually supposed to be going.
The solution to keeping on track is to plan out the essay before writing it. Take a few minutes to sketch out the points you want to make, with their examples and/or reasoning. Look at the completed sketch to make sure that it actually matches the assignment. And then write the essay, keeping to the sketched-out plan: if a new example comes to us as we write, don't add it unless there is the time to go back and rework the original plan to include it (in other words, it's okay to add examples to a take-home assignment, but not for an in-class exam).
If f (x) = 3x + 4
Function problems, like f (x), g (x), h (x), etc, leave a lot of students confused. This is unfortunate, because function problems are nothing more than substitution problems. I'll give you an example:
f (x) = 3x + 4
f (2) = ?
We note that in left half of the question, 2 has been plugged in where x was. To solve the problem, then, all we need to do is substitute (plug in) 2 for x in the right half of the equation:
f (2) = 3(2) + 4 = 6 + 4 = 10
Similarly,
f (3) = 3(3) + 4 = 9 + 4 = 13
f (4) = 3(4) + 4 = 12 + 4 = 16
f (5) = 3(5) + 4 = 15 + 4 = 19
f (6) = 3(6) + 4 = 18 + 4 = 22
f (7) = 3(7) + 4 = 21 + 4 = 25
f (8) = 3(8) + 4 = 24 + 4 = 28
f (9) = 3(9) + 4 = 27 + 4 = 31
f (10) = 3(10) + 4 = 30 + 4 = 34
f (y) = 3(y) + 4 = 3y + 4
f (z) = 3(z) + 4 = 3z + 4
etc.
Sometimes, we might find multiple functions used together. When this is the case, we just follow the usual rules of math to untangle the question:
if f (x) = 3x + 4, and g(x) = 5x - 7
f (4) - g (2) = ?
As before, we merely substitute in. Let's work with each function separately, then put them together, being sure to keep straight that 4 was given to us for the f function and 2 was given to us for the g function:
f (4) = 3(4) + 4 = 12 + 4 = 16
g (2) = 5(2) - 7 = 10 - 7 = 3
Taking the original equation and then substituting in these values, we have:
f (4) - g (2) = ?
16 - 3 = ?
and of course that equals 13.
We may also find cases where functions are nested within parentheses:
if f (x) = 3x + 4, and g(x) = 5x - 7
f (g (10)) = ?
Note that I've defined a new function for g. These are solved in the same way: by following the usual rules of math. g (10) is found inside a set of parentheses, so we start with that:
g (10) = 5(10) - 7 = 50 - 7 = 43
We then substitute this value in for g (10):
f (g (10)) = f (43)
And then we solve f (43):
f (43) = 3(43) + 4 = 129 + 4 = 134
PS: If you had to reach for your calculator to do any of that math, then you're relying on your calculator too much.
f (x) = 3x + 4
f (2) = ?
We note that in left half of the question, 2 has been plugged in where x was. To solve the problem, then, all we need to do is substitute (plug in) 2 for x in the right half of the equation:
f (2) = 3(2) + 4 = 6 + 4 = 10
Similarly,
f (3) = 3(3) + 4 = 9 + 4 = 13
f (4) = 3(4) + 4 = 12 + 4 = 16
f (5) = 3(5) + 4 = 15 + 4 = 19
f (6) = 3(6) + 4 = 18 + 4 = 22
f (7) = 3(7) + 4 = 21 + 4 = 25
f (8) = 3(8) + 4 = 24 + 4 = 28
f (9) = 3(9) + 4 = 27 + 4 = 31
f (10) = 3(10) + 4 = 30 + 4 = 34
f (y) = 3(y) + 4 = 3y + 4
f (z) = 3(z) + 4 = 3z + 4
etc.
Sometimes, we might find multiple functions used together. When this is the case, we just follow the usual rules of math to untangle the question:
if f (x) = 3x + 4, and g(x) = 5x - 7
f (4) - g (2) = ?
As before, we merely substitute in. Let's work with each function separately, then put them together, being sure to keep straight that 4 was given to us for the f function and 2 was given to us for the g function:
f (4) = 3(4) + 4 = 12 + 4 = 16
g (2) = 5(2) - 7 = 10 - 7 = 3
Taking the original equation and then substituting in these values, we have:
f (4) - g (2) = ?
16 - 3 = ?
and of course that equals 13.
We may also find cases where functions are nested within parentheses:
if f (x) = 3x + 4, and g(x) = 5x - 7
f (g (10)) = ?
Note that I've defined a new function for g. These are solved in the same way: by following the usual rules of math. g (10) is found inside a set of parentheses, so we start with that:
g (10) = 5(10) - 7 = 50 - 7 = 43
We then substitute this value in for g (10):
f (g (10)) = f (43)
And then we solve f (43):
f (43) = 3(43) + 4 = 129 + 4 = 134
PS: If you had to reach for your calculator to do any of that math, then you're relying on your calculator too much.
Saturday, July 26, 2008
The Action Potential
Action Potential is the term describing the sending of a signal from one end of an axon (the axon hillock) to the other end of the axon (by the synapse). What follows is the basics of this phenomenon.
Remember that in almost every cell, including neurons, the Na/K pump is continuously running in the background. This pump, of course, pumps three Na+ out for every two K+ it pumps into the cell. In doing so, it creates several gradients:*
Because of these gradients, if we were to open a Na+ channel, Na+ would be drawn into the cell for two reasons:
If we were to open a K+ channel, K+ would be torn between two impulses:
An action potential involves channels for both Na+ and K+. Both channels are voltage-gated (which means that they are triggered to open by changes in membrane potential). Na+ channels open quickly, but after being open for a brief time they lock closed. K+ channels don’t open as fast as the Na+ channels do. K+ channels do not lock closed. Threshold potenital is the membrane potential at which the channels are triggered.
If we focus on the cell body of the neuron, and on its dendrites, we find several ion channels in it. These channels open and close in response to incoming signals, and let positive and negative ions enter and leave the cell. If enough positive ions enter the cell, then the voltage across the membrane (the membrane potential) reaches the threshold potential and triggers those Na+ and K+ channels on the axon hillock. These channels then open.
DEPOLARIZATION:
The Na+ channels open first. Na+ pours into the cell, bringing its positive charge with it. This positive charge continues to grow, and eventually spreads as far as the next voltage-gated Na+ channel. When the positive charge at this next Na+ channel grows to a large enough size, it triggers that Na+ channel (remember, the Na+ channels are triggered to open by the membrane potential). Sodium flows in through this Na+ channel, and the positive charge spreads down to a third Na+ channel, triggering it to open. This process continues, opening the Na+ channels one at a time, with each Na+ channel allowing in the Na+ ions that trigger the next Na+ channel to open. This advancing wave of positive charge is the signal that is passed down from one end of the cell to the other. We call this phenomenon depolarization because the initial polarity across the cell membrane is lost as the positively charged Na+ ions enter the cell. (In fact, we actually end up with the cell interior being slightly positive.)
REPOLARIZATION:
If all we had was Na+ channels, we could send a signal, as described above, but then positive charge would have filled the axon, and there would be no way to send a second signal. So, we have to have a way to reset this membrane potential. This resetting is the job of the K+ channels. Recall that the same membrane voltage changes that trigger the Na+ channels also trigger the K+ channels, but that the K+ channels are slower to react. Eventually, however, the K+ channels do react, and do open, and K+ rushes out of the cell. K+’s exit moves positive charge out of the cell, returning the cell membrane voltage to its normal value of slightly negative inside the cell, and frankly overshooting a bit, so we end up a bit too negative. The Na/K pump helps return the cell from this overshoot (hyperpolarization) to normal membrane potential. We have regained our normal, resting membrane voltage, so we call this repolarization.
DEPALARIZATION + REPOLARIZATION = ACTION POTENTIAL:
The wave of depolarization, followed by its wave of repolarization, is known as the action potential.
REFRACTORY PERIODS:
Above, I noted that the Na+ channels, after being open for a brief moment of time, lock closed. Obviously, as long as they are locked closed, they cannot open, and so cannot participate in an action potential. This period during which the Na+ channels are locked closed is known as the absolute refractory period, since no amount of stimulation can cause another action potential to pass down the neuron.
Following the absolute refractory period is the relative refractory period. Remember how the K+ channels cause us to overshoot our target membrane voltage? Until the Na/K pump has returned the membrane to its proper resting voltage, the cell membrane is too negative, and a larger than normal force is required to change the cell membrane voltage enough to trigger an action potential.
* The pump also creates an osmotic gradient, where the number of particles outside the cell is greater than those inside the cell. The osmotic gradient is not used in the action potential, and is further complicated by the fact that running the pump splits ATP (a single osmotic particle) into ADP and Pi (two osmotic particles), which would tend to cancel the osmotic effect of pumping three ions out of the cell for each two ions pumped in. However, the ADP and Pi are generally swiftly recycled back into ATP, etc, etc, and at this point we're getting to be more complicated than we need to be. Focus on the three gradients already mentioned, and you should be fine.
Remember that in almost every cell, including neurons, the Na/K pump is continuously running in the background. This pump, of course, pumps three Na+ out for every two K+ it pumps into the cell. In doing so, it creates several gradients:*
- a Na+ gradient, where Na+ is greater on the outside of the cell than on the inside
- a K+ gradient, where K+ is greater on the inside of the cell than on the outside
- an electrical gradient, where the outside of the cell has more positive charges (is more positive than) the inside of the cell. Thus, the inside of the cell is negative in comparison to the outside of the cell, and a polarity exists across the cell membrane. This electrical gradient is known as the membrane potential, and it’s normal value (inside perhaps -70 mV with respect to the outside) is referred to as the resting (membrane) voltage, or resting potential.
Because of these gradients, if we were to open a Na+ channel, Na+ would be drawn into the cell for two reasons:
- there is more Na+ outside of the cell than inside of it
- the inside of the cell is negative with respect to the outside of the cell, attracting the positively charged Na+ ions
If we were to open a K+ channel, K+ would be torn between two impulses:
- the K+ gradient, which would push K+ out of the cell
- the electrical gradient, which would pull the positively charged K+ ions into the relatively negative interior of the cell
An action potential involves channels for both Na+ and K+. Both channels are voltage-gated (which means that they are triggered to open by changes in membrane potential). Na+ channels open quickly, but after being open for a brief time they lock closed. K+ channels don’t open as fast as the Na+ channels do. K+ channels do not lock closed. Threshold potenital is the membrane potential at which the channels are triggered.
If we focus on the cell body of the neuron, and on its dendrites, we find several ion channels in it. These channels open and close in response to incoming signals, and let positive and negative ions enter and leave the cell. If enough positive ions enter the cell, then the voltage across the membrane (the membrane potential) reaches the threshold potential and triggers those Na+ and K+ channels on the axon hillock. These channels then open.
DEPOLARIZATION:
The Na+ channels open first. Na+ pours into the cell, bringing its positive charge with it. This positive charge continues to grow, and eventually spreads as far as the next voltage-gated Na+ channel. When the positive charge at this next Na+ channel grows to a large enough size, it triggers that Na+ channel (remember, the Na+ channels are triggered to open by the membrane potential). Sodium flows in through this Na+ channel, and the positive charge spreads down to a third Na+ channel, triggering it to open. This process continues, opening the Na+ channels one at a time, with each Na+ channel allowing in the Na+ ions that trigger the next Na+ channel to open. This advancing wave of positive charge is the signal that is passed down from one end of the cell to the other. We call this phenomenon depolarization because the initial polarity across the cell membrane is lost as the positively charged Na+ ions enter the cell. (In fact, we actually end up with the cell interior being slightly positive.)
REPOLARIZATION:
If all we had was Na+ channels, we could send a signal, as described above, but then positive charge would have filled the axon, and there would be no way to send a second signal. So, we have to have a way to reset this membrane potential. This resetting is the job of the K+ channels. Recall that the same membrane voltage changes that trigger the Na+ channels also trigger the K+ channels, but that the K+ channels are slower to react. Eventually, however, the K+ channels do react, and do open, and K+ rushes out of the cell. K+’s exit moves positive charge out of the cell, returning the cell membrane voltage to its normal value of slightly negative inside the cell, and frankly overshooting a bit, so we end up a bit too negative. The Na/K pump helps return the cell from this overshoot (hyperpolarization) to normal membrane potential. We have regained our normal, resting membrane voltage, so we call this repolarization.
DEPALARIZATION + REPOLARIZATION = ACTION POTENTIAL:
The wave of depolarization, followed by its wave of repolarization, is known as the action potential.
REFRACTORY PERIODS:
Above, I noted that the Na+ channels, after being open for a brief moment of time, lock closed. Obviously, as long as they are locked closed, they cannot open, and so cannot participate in an action potential. This period during which the Na+ channels are locked closed is known as the absolute refractory period, since no amount of stimulation can cause another action potential to pass down the neuron.
Following the absolute refractory period is the relative refractory period. Remember how the K+ channels cause us to overshoot our target membrane voltage? Until the Na/K pump has returned the membrane to its proper resting voltage, the cell membrane is too negative, and a larger than normal force is required to change the cell membrane voltage enough to trigger an action potential.
* The pump also creates an osmotic gradient, where the number of particles outside the cell is greater than those inside the cell. The osmotic gradient is not used in the action potential, and is further complicated by the fact that running the pump splits ATP (a single osmotic particle) into ADP and Pi (two osmotic particles), which would tend to cancel the osmotic effect of pumping three ions out of the cell for each two ions pumped in. However, the ADP and Pi are generally swiftly recycled back into ATP, etc, etc, and at this point we're getting to be more complicated than we need to be. Focus on the three gradients already mentioned, and you should be fine.
Sunday, May 25, 2008
Dalton's Law of Partial Pressures: HW problems
1. Consider Dalton’s Law of Partial Pressures:
Ina. Restate the law (put it into your own words) Answer
Inb. Consider a closed container with 7 atm of gas inside. There are 2 mol of gas X, 5 mol of gas Y, and 7 mol of gas Z. The gasses do not interact with each other, and thus Dalton’s Law of Partial Pressures applies. What are the partial pressures of each of the three gases? Answer
Inc. Consider another closed container. Inside, there are 2 mol gas A. When we add 3 mol of gas B, we see that the total pressure inside the cylinder increases by 6 atm. If gas A does not interact with gas B, what was the original pressure in the cylinder? Hint: Hint 2 Hint 3 Answer
HINTS:
1.c.
Hint 1: Another way of looking at Dalton’s Law of Partial Pressures tells us that as long as the gases in a container don’t interact, their pressures are independent. In other words, as long as the pressures don’t interact, we can apply PV=nRT to them independently.
Hint 2: Since the pressures of A and B are independent, the partial pressure of gas A doesn’t change when we add gas B.
Hint 3: Since the partial pressure of gas A doesn’t change when we add gas B, we can calculate A’s partial pressure after we add gas B, as the pressure will be the same.
ANSWERS:
Question 1
Part a
Dalton tells us that, if we have more than one gas in an enclosed container, and those gases don't react, then the ratio of each gas's partial pressure to the total of the pressure of all of the gases equals the ratio of the moles of each gas present to the total number of moles of gas present.
Part b
gas X: 1 atm; gas Y: 2.5 atm; gas Z: 3.5 atm
Explanation: Initially, we have:
We can easily fill in the total number of moles:
Now, we can determine the ratio of moles to atmospheres: 14 moles to 7 atm, which reduces to 2 mol to 1 atm, or 1 mol to 0.5 atm. We then apply this ratio to the individual gases:
Note that the pressure each gas individually adds up to the total pressure in the container.
Part c
4 atm
Explanation: Dalton tells us that, if gasses don't interact with each other, for a given volume and temperature the ratio of moles to presure is fixed. When we added three moles of gas B, we saw that it generated 6 atm, telling us that the ratio of moles:atm was 3:6, or 1:2. Since there are two moles of gas A, these moles exert 4 atm.
Last edited July 26, 2008
Ina. Restate the law (put it into your own words) Answer
Inb. Consider a closed container with 7 atm of gas inside. There are 2 mol of gas X, 5 mol of gas Y, and 7 mol of gas Z. The gasses do not interact with each other, and thus Dalton’s Law of Partial Pressures applies. What are the partial pressures of each of the three gases? Answer
Inc. Consider another closed container. Inside, there are 2 mol gas A. When we add 3 mol of gas B, we see that the total pressure inside the cylinder increases by 6 atm. If gas A does not interact with gas B, what was the original pressure in the cylinder? Hint: Hint 2 Hint 3 Answer
HINTS:
1.c.
Hint 1: Another way of looking at Dalton’s Law of Partial Pressures tells us that as long as the gases in a container don’t interact, their pressures are independent. In other words, as long as the pressures don’t interact, we can apply PV=nRT to them independently.
Hint 2: Since the pressures of A and B are independent, the partial pressure of gas A doesn’t change when we add gas B.
Hint 3: Since the partial pressure of gas A doesn’t change when we add gas B, we can calculate A’s partial pressure after we add gas B, as the pressure will be the same.
ANSWERS:
Question 1
Part a
Dalton tells us that, if we have more than one gas in an enclosed container, and those gases don't react, then the ratio of each gas's partial pressure to the total of the pressure of all of the gases equals the ratio of the moles of each gas present to the total number of moles of gas present.
Part b
gas X: 1 atm; gas Y: 2.5 atm; gas Z: 3.5 atm
Explanation: Initially, we have:
| 2 mol gas X | = ? atm |
| + 5 mol gas Y | = ? atm |
| + 7 mol gas Z | = ? atm |
| = ? mol | = 7 atm |
We can easily fill in the total number of moles:
| 2 mol gas X | = ? atm |
| + 5 mol gas Y | = ? atm |
| + 7 mol gas Z | = ? atm |
| = 14 mol | = 7 atm |
Now, we can determine the ratio of moles to atmospheres: 14 moles to 7 atm, which reduces to 2 mol to 1 atm, or 1 mol to 0.5 atm. We then apply this ratio to the individual gases:
| 2 mol gas X | = 1 atm |
| + 5 mol gas Y | = 2.5 atm |
| + 7 mol gas Z | = 3.5 atm |
| = 14 mol | = 7 atm |
Note that the pressure each gas individually adds up to the total pressure in the container.
Part c
4 atm
Explanation: Dalton tells us that, if gasses don't interact with each other, for a given volume and temperature the ratio of moles to presure is fixed. When we added three moles of gas B, we saw that it generated 6 atm, telling us that the ratio of moles:atm was 3:6, or 1:2. Since there are two moles of gas A, these moles exert 4 atm.
Last edited July 26, 2008
Tuesday, April 1, 2008
Electrochemistry
ELECTROCHEMISTRY seems to be a subject that lends its self to being confusingly taught. Do we add the half-cells together, or do we subtract one from the other? How do we know which direction the reactions run? Is this set-up voltaic, or electrolytic? Or perhaps it’s galvanic? And how many volts are really involved here?
Let's start with a galvanic cell, also known as a voltaic cell.
The simplest way to look at electrochemistry is to realize that the universe is a brutal place: might makes right. When we couple two reactions, (for instance, two half cells), whichever is stronger (has the largest magnitude (absolute value) for Eº) runs the show, which means that it gets to go the direction it wants to go. And like any chemical reaction, it wants to go in the direction that it's spontaneous.
Now, if we consider the equation
Let's start with a galvanic cell, also known as a voltaic cell.
The simplest way to look at electrochemistry is to realize that the universe is a brutal place: might makes right. When we couple two reactions, (for instance, two half cells), whichever is stronger (has the largest magnitude (absolute value) for Eº) runs the show, which means that it gets to go the direction it wants to go. And like any chemical reaction, it wants to go in the direction that it's spontaneous.
Now, if we consider the equation
∆Gº = -nFEº
we note that it shows us the relationship between Eº and ∆Gº. F represents Faraday’s number, which is about 10,000, and n indicates the number of electrons involved. F and n will always be positive numbers, so a negative ∆Gº value (indicating a spontaneous reaction) corresponds to a positive Eº value. In other words, a positive Eº value indicates a spontaneous electrochemical reaction.
So, putting these items together, since the biggest Eº value goes the direction it wants to (i.e. it is spontaneous) we give it a positive value. If the Eº value that we’re given is positive, then we're good - the half-cell reaction runs as it is shown. If the value we’re given is negative, then the reaction is written backwards.
Of course, this is only one half of our cell. The half-cell with the big Eº is either generating electrons (electrons are a product) or absorbing them (electrons are a reactant). If it's generating them, then the second half-cell MUST absorb them. If the equation given for the second half-cell shows it absorbing electrons, then we can use the Eº value given. If the equation shows the half-cell generating electrons, then we have to run that reaction backwards, which means we keep the magnitude of the Eº value, but reverse its sign.
If the half-cell with the bigger Eº is absorbing electrons, then the second half-cell MUST generate those electrons. If its equation shows it generating electrons, we use its Eº as noted; if its equation shows it absorbing electrons then we have to reverse the sign of the Eº, but keep its magnitude.
RECAP
1. Biggest Eº runs the show - it goes in the spontaneous direction (direction of positive Eº value)
2. The direction of the second half-cell is determined by whether it has to absorb the electrons generated by the first half-cell, or supply electrons to be absorbed by the first half cell. If its equation is written in the correct direction, then we take the Eº value given - otherwise we reverse its sign.
3. The total Eº is the sum of the two Eº values. The Eº value of the first half-cell is positive. The sign of the second half-cell is determined by the direction it must run in order to cater to the electron needs of the first half-cell.
EXAMPLE
Create a galvanic/voltaic cell using the following 2 half cells:
SO4 + 4 H + 2e -> H2SO3 + H2O Eº = + 0.170
ZnS + 2e -> Zn + S Eº = - 1.440
Which half-cell is the anode, and which the cathode?
Which electrode is positive, and which is negative?
How many volts do we get out of the cell?
(Assume 1M concentrations and 25 degrees C)
NOTE: this all describes a galvanic/voltaic cell. Electrolytic cells are exactly the opposite.
The answers
1. The half-cell with ZnS has a larger absolute value (greater magnitude) for its value of Eº, so it runs in its spontaneous direction.
2. As written, this half-cell yields a negative Eº value; its spontaneous direction is therefore the reverse of how it's written, so we actually have:
Zn + S -> ZnS + 2 e Eº = + 1.440 Volts
3. In its spontaneous direction, the first half-cell generates electrons; the second half-cell MUST therefor absorb them (i.e. use them as a reactant)
4. As written, the second half-cell does have electrons as a reactant, so it's written the correct way:
SO4 + 4 H + 2 e -> H2SO4 + H20 Eº = + 0.170
5. Total voltage equals the voltage of half-cell 1 plus the voltage of half-cell 2:
1.440 + 0.170 = 1.610 Volts
6. OIL RIG: Oxidation Is Loss of electrons, Reduction is gain of electrons - electrons are lost at the Zn half-cell, so this is where oxidation occurs. Reduction occurs at the SO4 half-cell.
7. An Ox; Red Cat: The anode is the site of oxidation, while reduction occurs at the Cathode. The Zn half-cell is the site of Oxidation, and thus is the Anode. The SO4 half-cell is the site of reduction, and thus is the Cathode
8. The Zn half-cell is running the show. It generates electrons and sends them away (repels them). Electrons are repelled by a negative charge, so Zn is the negative electrode. SO4 is absorbing (attracting) electrons; it must have a positive charge to do so.
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