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Dialysis and Creatinine Clearance According to the National Kidney Foundation, in 1997 more than 260,000 Americanssuffered from chronic kidney failure and needed an artificial kidney (dialysis) to stay alive. (Source: The National Kidney Foundation, wriv.kidney.org.) When the kidneys fail, toxic waste products such as creatinine and urea build up in the blood. One way to remove these wastes is to use a process known as peritoneal dialysis, in which the paticnt's peritoneum, or lining of the abdomen, is used as a filter. When the abdominal cavity is filled with a certain dialysate solution, the waste products in the blood filter through the peritoneum into the solution. After a waiting period of several hours, the dialysate solution is drained out of the body along with the waste products.

In one dialysis session, the abdomen of a patient with an elevated concentration of creatinine in the blood equal to 110 grams per liter was filled with two liters of a dialysate (containing no creatinine). Let $f(t)$ denote the concentration of creatinine in the dialysate at time $t .$ The rate of change of $f(t)$ is proportional to the difference between 110 (the maximum concentration that can be attained in the dialysate) and $f(t)$ Thus, $f(t)$ satisfies the differential equation
$$
y^{\prime}=k(110-y)
$$
(a) Suppose that, at the end of a 4 -hour dialysis session, the concentration in the dialysate was 75 grams per liter and it was rising at the rate of 10 grams per liter per hour. Find $k.$
(b) What is the rate of change of the concentration at the beginning of the dialysis session? By comparing with the rate at the end of the session, can you give a (simplistic) justification for draining and replacing the dialysate with a fresh solution after 4 hours of dialysis? [Hint: You do not need to solve the differential equation.]

Those question explains a simple procedure. An experiment. Um, and they were just looking at the fraction of cells and my toes is over time and were given a graph for this. So promotion, eh? It asks what all cells be accepted are expected to contain the radioactive DNA. Um, after they are labeled in the experiment and on lee the cells that are in the S phase of their cell cycle, Um, and the ass phasing that cells are making DNA. Um, during the 30 minute labeling period may contain any radioactive unite number. Question be it asks. Initially, there are no my tonic cells that contain radioactive DNA. We want to know why. This is, um so initially might have excels contained no radioactive Dina, because these cells, um, we're not engaged And, uh, Deena synthesis during living period. It takes about two hours for the 1st 8 will make hot Excel with here. Next, we went to explain the rise and fall, and then the rise again of this curve that we see and just for reference, the curve looks something like this. Besides, what? This question is your friend. So the initial rise of the curve corresponds to cells that were just finishing DNA replication. When the radioactive diamond in was added, a curve rises, more is more labelled. Cells entered my toe, sis, and the peak corresponds to those cells. I just started the s phase in the radioactive gravity was added. So this first arise is, um so starting ass fades. Whether that was they were just finishing it when this, I mean, it was at or they were just starting it when it was added, the labelled cells, then exit for my toes is replaced by unable. Night hot excels which were not in the s phase during the labeling period. So cells then exit my closest and that's the fall and last after 20 hours occur. So it's rising again because it labeled toes. Enter second, my cases and lastly, want to estimate the length of the S phase from this graph. So the initial two hour lag before any label might have excels appear corresponds to the G T phase. Um, which is the time between the end of the S phase and beginning of my toe sis. The first labelled cells seen in my toes is where those just finishing s phase when they radioactive gabardine was added. So in summary, the, um again, the first table cells were those that were just finishing the s phase eso The initial two hour lag corresponds to the G two phase.

All right. So part a asked us to explain each term in the equilibrium or in the system of differential equations here. So if we notice here, these two terms are the same in both equations. Because, um, here, see, is in the blood represents the amount in the blood. So if you look at a P first, this is the our rate of transfer rate of Yuria. Uh, this is from the blood. Sorry. Not from the blood from the pool to blood. Okay. And then BC is gonna be the rate of Yuria. Um, be going the other direction from blood, uh, to the pool. Now, the first term here. Now, this is the hemodialysis part. Okay, the K over VC, this is the rate of hemodialysis. You know, part be asked us to construct the face plot, so let's do so by first looking at the, um no clients. So for the sea, no client, we have negative K v. Um, plus a P minus specie is equal to zero. So we have that Pete is going to be able to be over a time C plus que over V also time, see, Or we could back out the sea and it becomes, um, k over V all that time. See, you know, for the wine no client we just have here. P is going to be equal to, um or sorry, the negative AP first plus B c is equal to zero. Or we get that. Um uh, he is gonna be be over a time. See here this slope, we'll always be greater than this slope. So when we construct that face plain and everything should always be positive. So we have. See here, pee here. Okay, we have this, um, and then this should be slightly less slow. Okay, um, now, for let's say that we have the 0.1 For example, if we plug in 01 as in Ah, see, zero. What? The test point here. Now we should get, um Let's put that into our c prime. So we get negative K over V time. Zero plus a times one minus B time. Zero. That's equal to a which is greater than zero. So we'll be going to the right over here. And since we're linear, we'll also be going to the left in this region. Now let's play the inn for our next one. So if you have negative eight times one plus B times zero isn't me negative A which is less than zero. So in this region, we're gonna be going down like so and then up in this region here, our equilibrium point is just going to be at the origin. 00 Okay, so then we'll have this. So if you see here that all of our, um, all of our solutions, they will eventually go towards the, um 00 here. So for a part, be what happened to the Yuria concentrations as he goes to infinity. Ah, they the Yuria concentrations, they both go to zero concentrations, uh, go towards the equilibrium. So go to zero. Right, and then you're done.

Let's answer the following questions concerning the cell membrane for a determine the capacitance of the membrane for the typical cell described to calculate, the capacitance were given the formula just e notes. E comes a over the, um over l. Andi. So eh, not were given these constants Um, 8.854 times, 10 to the negative, 12 on We've got the area. Andi e not would be or the die electric constant here you not e three times on. So let's plug in our values. E is going to be three times 8.854 times 10 to the negative 12. Yeah, two looms squared, Yeah, over en meters squared. And then the area here would be one times 10 to the negative. Six centimeters squared but will convert this to meters. I think so. Don't put a conversion write in the equation and then l is equal to one times 10 to the negative six sent 2 m but will convert this 2 m 100 centimeters in 1 m and solving this will yield a capacitance of 2.66 times 10 to the negative 13 C squared meter. We want to convert this to Faraday. So one Faraday over one C squared meter. And so our capacities and Faraday's is 2.66 times 10 to the negative. 13. Faraday. So there's our capacitance for the cell membrane. Given the values for B, what is the net charge required to maintain the observed membrane potential? So, since the capacitance, um, is, um, in far day Um, hmm. One. Faraday is equal to, um one colon per volume, therefore 2.66 times 10 to the negative. 13. Faraday's but cool um per column per volts are not going to lump revolt times. A voltage of 0.85 volts will yield 2.26 times 10 to the negative. 14. Cool. Um, So there's the charge, uh, required to maintain the observed potential, as were given the uh, self potential is 0.85 volts for see how maney potassium mines must flow through the cell membrane to produce the membrane potential. So this'll be Q over E. On the charge is 2.26 times 10 to the negative 14 Q loans, and this would be 1.602 times 10 to the negative 19 Kellems per iron, which will yield 1.41 times 10 to the fifth potassium ions that would be required. So 1.41 times 10 of the fifth for D. How maney potassium ions air in the typical cell. So, um, we have the mill arat e here. So 6.22 times 10 to the 23rd irons Permal times 155 times 10 to the negative, three moles per leader. And then we've got 1000 centimeters cube for leader and this'd be times one times 10 to the negative eight centimeters cube solving This would yield 93 times 10 to the 11th irons. Uh, that would be the number of potassium mines in a typical cell. And then, lastly, for E the fraction of intracellular potassium mines transferred to produce the member remembering potential so that it does not change the potassium lines in the cell. So the fraction would be 14 times 10 to the fifth irons or answer from part. See over 93 times 10 to the 11 irons or answered from part D on. This is equal to 1.5 times 10 to the negative seven. So the concentration, um, would remain, uh, the concentration of potassium ions in the south remains constant at 155 million. Mueller, um, this would be the fraction of mines involved here in part.

For this question. We're looking at the role of CD four on the surface of T cells, so first off CD is going to refer to what we call a cluster of differentiation. This is just a fancy term for saying it's a protein on the surface of a cell that we can use to identify the cell and tell it apart from others, we use this CD sort of abbreviation. So that way we don't have to memorize many different names for different proteins on the surface of many, many different cells. So for our T cells, we typically have two or three types. You may have heard of these, but we have to help ourselves Teesside or toxic cells, and we have natural killer cells. So when we talk about CD four, these are most typically related to our T helper cells. Archie's side or toxic cells are more related to SETI eight, and natural killer cells are related to C D 16 and CD 56. As you may know, T helper cells have a more helping role are they allow for stimulation of cells to under grow immune responses. R C site A toxic cells are for interest. Cellular infections such as viruses and can have some role in cancer fighting and natural killer cells is most specifically for fighting. Cancers are cancerous cells. So for this question, we're looking at the actual role of CD four and measuring its response. So if you remember how T cells are going to interact and undergo their response, this experiment is displaying this process in two different ways. So on this left side, we'll have our T cell, and this is going to express our CD four marker, which means these are going to be our t help ourselves. Whereas this cell in our right will be our professional antigen presenting cells. This will be, say, you're gentry, ticks cells or, say a B cell. And these will express at MHC two molecule. And this just means it's some sort of extra cellular auntie gin. So say, a bacterial protein or some sort of protein found in the blood or tissue that is not typically produced inside of a cell. And, of course, that MHC two molecule is going to hold some form of antigen to be recognized. So for this experiment, you have a regular interaction between the T cell and the antigen presenting cells, and they're going to do an antibody interaction. So for that antibody interaction, they're going to bind an antibody to that CDP CD four molecule and prevent its interaction with the MHC complex. So from there we have two different graphs. We have our antibody graph where you can see it starts off really low until it gets a sudden jump and an immune response. So on our Y axis, we have the way they're measuring their response, which is going to be through calcium uptake. And from this graph, you can interpret that it has a slow start. So you have to have a lot of this MHC complex in order to get a response and stimulation of the antigen presenting cells. Whereas in the regular model you can see that it has a very quick response due to the interaction of CD four and the MHC too complex. So here the role of the CD four complex is going to be to stimulate that antigen presenting cells, which means it's going to allow for the more uptake of androgen. This would be saved through optimization and through the engulfing of foreign material, but they're also cause additional interaction in a cell media aided response as well or through the extra secretion of antibodies, say by B cells. However, this experiment demonstrates that this T cell CD four response is to quicken this response and stimulation process to make it more ready and available for smaller cases our concentrations of MHC androgen complexes.


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