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Up-to-date Wth security updates fixes, and improvements, choose Check ior Updates:BACKGROUND: Many molecules and even single atom ions can't diffuse easily thr...

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Up-to-date Wth security updates fixes, and improvements, choose Check ior Updates:BACKGROUND: Many molecules and even single atom ions can't diffuse easily through the cell membrane. How do they get into and out of cells? Until now you have only observed oxygen and carbon dioxide diffusing through a cell membrane_ The way those two molecules directly pass through the cell membrane is actually quite uncommon: Most molecules are unable to pass through the membrane, So some kind of hole must b

up-to-date Wth security updates fixes, and improvements, choose Check ior Updates: BACKGROUND: Many molecules and even single atom ions can't diffuse easily through the cell membrane. How do they get into and out of cells? Until now you have only observed oxygen and carbon dioxide diffusing through a cell membrane_ The way those two molecules directly pass through the cell membrane is actually quite uncommon: Most molecules are unable to pass through the membrane, So some kind of hole must be made to allow them to diffuse through. This is done using one Or more proteins to create pore in the membrane that allows only specific molecules or ions t0 pass through: 10. Which are the only types of molecules that pass easily through a cell membrane without a protein channel? Positive ions Negative ions Small non-polar molecules Small polar molecules Medium sized polar = molecules What is true" of most naturally occurring pores? They allow only ions to pass through They allow most molecules to pass through They let anything that is the right size through D. They are specific t0 one kind of molecule 699 words 0 English (United States) Focus



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Cells are bounded by membranes composed of phospholipids. A phospholipid consists of a pair of fatty acids that may or may not have carbon-carbon double bonds, fused at the carboxylic acid with a three-carbon glycerol that is terminated by a phosphate, as shown in the figure below. Most cell membranes comprise two phospholipid layers with the hydrophilic phosphate ends of each molecule in the outer and inner surfaces. The hydrophobic chains of carbon atoms extend into the space between these two surfaces. The exchange of matter between the interior of the cell and the environment is mediated by this membrane with selective permeability. A. Pose questions that identify the important characteristics of this lipid bilayer structure the molecules that must be acquired from the environment and eliminated from the cell relationships between the structures of these molecules and the structure of the bilayer Because the plasma cell membrane has both hydrophilic and hydrophobic properties, few types of molecules possess structures that allow them to pass between the interior of the cell and the environment through passive diffusion. The fluidity of the membrane affects passive transport, and the incorporation of other molecules in the membrane, in particular cholesterols, has a strong effect on its fluidity. Fluidity is also affected by temperature.
Measurements of the speed of movement of oxygen molecules, $\mathrm{O}_{2},$ through three types of membranes were made (Widomska et al, Biochimica et Biophysica Acta, $1,768,2007$ and compared with the speed of movement of $\mathrm{O}_{2}$ through water. These measurements were carried out at four different temperatures. One type of membrane was obtained from the cells in the eyeball of a calf (lens lipid). Synthetic membranes composed of palmitic acid with cholesterol (POPC/CHOL) and without cholesterol $($ POPC) were also used. The results from these experiments are shown in the table below. B. Represent these data graphically. The axes should be
labeled, and different symbols should be used to plot data for each material. C. Analyze the data by comparing transport of oxygen through the biological membrane, water, and the synthetic membranes. Consider both membrane composition and temperature in your analysis. The plasma membrane separates the interior and the exterior of the cell. A potential to do work is established by defining regions inside and outside the cell with different concentrations of key molecules and net charge. In addition to the membrane defining the cell boundary, eukaryotic cells have internal membranes. D. Explain how internal membranes significantly increase the functional capacity of the cells of eukaryotes relative to those of prokaryotes.

Hi, everyone In this problem, it is given the distance. Pick things up. Rates of the kind described to be, Oh, 7.59 awaited. That is 7 25 and to 10 to the department's 9 m. We have to find cap pistons per unit area. It is to be absolutely not. By deep. That is 8.85 10 to the parliament. Spread upon. 7.5 into can go to the bar when it's nine. So it is to be 1.18 10 24 minus six ferret per meter square. Okay, be part electric field will be be upon ding. Yeah, Crazy. 55 into 10 to the par minus three upon 7.5 and 2 10 to the power when that's nine. So yeah, 1.1 33 10 to the power six. Both per meter. That's all. Thanks for about

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.

So this problem is about how cell membranes could be modeled. His capacitors basically with di electrics inserted. So here's our capacitor or a cell membrane. And let's go ahead and get down some of the givens in the problem. So it says that the membranes are a seven and 1/2 nanometers thick, so we can say D is 7.5 times 10 to the minus nine meters. Okay. And we also know that the dye electric constant is 10. So cafa is equal to 10. And first we want to get the capacitance per square centimetre of the wall. So we want to use capacitance is equal to Absalon. Not a overdue. You let me just double check the room, actually modeling it as a parallel plate. Yeah, Parallel plate capacitor. Okay, great. So that's up. Absolutely not a divided by D. So? So the sea Per eh Um oh, we need to be concerned with units. So this But if we use s i units for capacitance and, uh absolutely not area D. This is gonna be in units of, um Oh, what are the units of capacitance? Let me double check that. Well, regardless, the area will be in units of meters squared, so it'll be whatever. The units of capacitance are divided by meters squared, so we need to convert that to centimeters squared. So we can do that. We'll need to multiply our answer by ah 100 centimeters or excuse me. We go back. You know what? I'm just gonna rewind a little bit. I think I've kind of gotten down a confusion whole. So I'm just going to calculate Seper a and then I'm gonna convert it. I think to me make it a little bit less confusing. So that's just gonna be absolutely not divided by D. And so I'm gonna go ahead and Oh, and then we also need Kappa. So, um and this is just a general formula that you can multiply by Kapo. So I'm gonna go ahead and plug that into a calculator. It crane 85 times. 10 to the minus 12. Kappa rose 10 and then D is seven and 1/2 nanometers. Seven points, Right. First time from light. It's night. Great. And so I got there. The capacitance per unit area is 0.118 Well, capacitance units is fair and I don't know where I forgot that. It's kind of funny. I've been writing it all day, but I blanked 0.118 fads per meter squared, and we want to get that two centimeters squared. So there was 100 centimeters in one meter, and then we have to go out twice. Multiply that twice to get, uh, you can get rid of the meter squared and converted two centimeters squared. So let's see. It's it's gonna be divided by 10 to the minus for so when So 10 to the minus four. Right now it's already 10 to the minus two. So then that will be 1.18 times 10 to the minus six fare ads per centimeter squared. Great. And next, we want to get the electric field inside the membrane. So we given that it has an 85 mobile voltage. So this is the answer for a So be we know V is equal toe 85 Mila Bolds. What is E? So I guess there's a few ways toe Answer this, um see, well, he is equal to deem my derivative of the with respect toe acts. So, uh, the magnitude of e could be said to be Delta bi over a delta eggs. So our delta acts is just the distance. So Alcoa had it right distance here on DSO. We just wanted to buy this 85. Miller holds by 7.5 on the animators. I'm gonna go ahead and do that eight point pry Brady of Fire 85 millet balls. Try to buy some one point front minus nine, huh? It's very small. I'm smarter than the wavelength of light. Wow, that seems really big, I guess because the distance is so small. I'm just gonna double check that. This was question was copied over, right? Because these numbers don't seem realistic to me, So I'll go ahead and pause the video while I do that. Hey, I double check the question. It really is Nana meters. So this tells us that the my calculation tells us that the electric field is, um, 11 million volts per meter. That seems crazy. 11 times 10 to the minus are tied to a plus six volts per meter. 88 5 seven in a hot Yeah,

Everybody. So, for this one we need find was the capsules per square centimeter of cell wall. So get a equals C over area, which is K absalon, not de very So get plan Times eight point E five times ton to negative 12 c squared in, um, squared of a divided by seven point five times 10 to negative nine m. Okay. And this has been equal to 1.18 times 10 to your negative. Six, uh, centimeters squared. Or you could say it's 1.18 tie 1.18 new centimeters square. Okay. And now or be one you know is defined in the normal Western states. Cell's potential difference of 85 u of years was left field inside the membrane. So much of field in side the membrane. We have 85 times 10 at native, three bowls divided by 7.5 times, 10 to negative nine meters and we get one point one at three times 10 to 7 votes meters. Okay. I mean,


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