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From what I understand, the larger the molecular size, the slower the rate of diffusion and vice versa. However, after doing a lab (similar to this one) about diffusion through a cellulose membrane, my instructor said that despite substances having different molecular sizes, the rate of diffusion (thru the membrane) will be the same. Could someone further explain or did I hear incorrectly?
I think that your instructor was wrong.
For the type of experiment that you describe the dialysis tubing (cellulose acetate with pores) is acting as a semi-permeable membrane which can block the diffusion of a polymer (starch) but will allow the passage of a small molecule (glucose). The molecular weight cut-off of the tubing was probably 10 kilodaltons or greater, so glucose (180 daltons) would diffuse through the pores without hindrance, as would sucrose (342 daltons). But would the monosaccharide and the disaccharide move through the membrane at different rates?
This table gives values for the diffusion coefficient (D) of these two sugars as:
glucose 6.7 cm² s⁻¹ × 10⁻⁶
sucrose 5.2 cm² s⁻¹ × 10⁻⁶
The time τ to travel distance L by diffusive movement is given by τ = L²/6D
Here all that matters is that τ is inversely proportional to D, so we can estimate that glucose 'diffuses' approximately 30% faster than sucrose.
Bulk diffusion is, of course, driven by a concentration gradient: if you were to start with, say, 0.1 M each of glucose and sucrose in a dialysis tube the glucose should reach equilibrium with the surrounding liquid faster than the sucrose.
- Contributed by CK-12: Biology Concepts
- Sourced from CK-12 Foundation
What will eventually happen to these dyes?
They will all blend together. The dyes will move through the water until an even distribution is achieved. The process of moving from areas of high amounts to areas of low amounts is called diffusion.
Determining the Rate of Osmosis Through a Semi-Permeable Membrane
Determining the Rate of Osmosis through a Semi-Permeable Membrane Schneider, Justin* Sec 16 and Thomas Hudson
Water is the principle solvent in cells. There are three conditions that a cell may be subjected to in the cells physical environment. The cells may be isotonic, hypotonic, and hypertonic. Isotonic is when the cell and is environment have equal concentrations of solutes. Therefore cells in an isotonic solution do not experience osmosis. When the cell is hypotonic it has a lower concentration of solutes so water will move out of the cell. When the cell is hypertonic, it will have a higher concentration of solutes and water will move into the cell to attempt to reach an isotonic state. The primary purpose of this exercise is to measure the rate and direction of osmosis under different concentrations. The greater the concentration of the solute (sucrose) the faster the solution (water) will pass through the membrane trying to achieve an isotonic state (Vodopich and Moore. 2011.) Materials and Methods
To start the experiment 8 pieces of string were gathered approximately 2 inches in length, sewing thread were used for this experiment. Four water soaked pieces of dialysis tubing that were 15 cm long were also used. A small beaker for bag A was filled with a solution of 25% sucrose, was labeled to keep separate from the other beaker. A larger beaker was used for bags B, C, and D. It was filled with a 1% solution of sucrose, and was labeled for good practice. Each end of the tubes was sealed by folding and tying it tightly with one of the pieces of thread. The other end of the tube was soaked with water and opened gently rolling it between the index and thumb fingers. Once one end of each tube was sealed and the ends were opened each tube was filled with a solution. Before filling each tube with its selected solution a small piece of paper with a letter A, B, C, or D written in pencil were inserted so they could easily be.
References: Raven, Johnson et al. 2011. Biology.9th edition. McGraw Hill. New York, NY.
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The author would like to thank Miss Kendra Hill for the education, and instruction. He would also like to thank Roel Rabara for his guidance and thorough instruction in the Biology 151 lab that has helped him understand how the scientific methods work as well as helping him improve his scientific research writing. Last but not least his lab partner Thomas Hudson who has helped find the results of their experiments and tests.
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1) Fick’s First Law gives rise to the formulae:
P is the permeability, an experimentally determined membrane ‘conductance’ for a given gas at a given temperature
c2-c1 is the difference in concentration of the gas across the membrane for the direction of flow from c1 to c2
2) Applicable to explain diffusion for two miscible liquids when they are brought in contact since the diffusion occurs at a macroscopic level.
3) Fick’s first law is also important in radiation transfer equations. However, in this context, it becomes inaccurate when the diffusion constant is low, and the radiation becomes limited by the speed of light rather than by the resistance of the material the radiation is flowing through. In such situation a flux limiter is used.
4) The exchange rate of oxygen and carbon dioxide in the lungs across the alveolar membrane can be determined using the Fick’s first law and Graham’s law.
1) Used to model transport processes in foods, neurons, biopolymers, pharmaceuticals, porous soils, population dynamics, nuclear materials, plasma physics, and semiconductor doping processes.
2) Integrated circuit fabrication technologies, model processes like CVD, thermal oxidation, wet oxidation, and doping, use diffusion equations obtained from Fick’s laws.
Factors that Influence Membrane Diffusion
There are several factors that can influence the diffusion of molecules both within and across a membrane physical barriers, electrostatic attraction or repulsion nodes, and partitioning phenomena (3), are just a few examples.
Physical obstacles can become considerably crowded, which can obstruct the free passage of molecules. Several adaptor proteins (alpha-actinin, talin, vinculin, etc.) can attach the cortical cytoskeleton to the long cytosolic tails of transmembrane proteins within the plasma membrane and act as a fence, which restricts the movement of molecules across the membrane (4,5). Such obstruction by the cortical cytoskeleton has lead to the observation that molecules undergo a &ldquohop diffusion&rdquo in which they &ldquohop&rdquo intermittently between confinement zones (4,5) in order to diffuse (Figure7).
Figure (PageIndex<7>): An image of the mesh actin cytoskeleton (light purple) attaching to transmembrane proteins (dark purple) and tightly associated lipids to form a barrier that limits the diffusion of molecules (3). The red arrows indicate limited diffusion.
Membrane-matrix junctions can also contribute to a physical obstruction that impedes membrane diffusion. This is due to the binding between integrin receptors located in the plasma membrane and the extracellular matrix (fibrous network of proteins that cells attach to) (Figure (PageIndex<8>)). If this interaction is increasingly accumulated at a high enough density, then diffusion is limited. A recent study showed that this type of physical barrier blocked the diffusion of membrane molecules whose dimensions exceeded the width of the integrin-matrix interaction (6).
Figure (PageIndex<8>): An image of the extracellular matrix (yellow) linking to membrane receptors (purple) to generate a mesh that limits diffusion (3). The red arrows indicate limited diffusion.
Aqueous ethyl cellulose dispersions containing plasticizers of different water solubility and hydroxypropyl methylcellulose as coating material for diffusion pellets. I. Drug release rates from coated pellets
The present work investigates release mechanisms of theophylline pellets coated with an aqueous ethyl cellulose (EC) dispersion containing plasticizers and hydroxypropyl methylcellulose (HPMC) as a water soluble pore former. Three different drug release mechanisms from coated pellets can be determined as a function of the water solubility of the plasticizers and the ionic strength of the release medium. Coated pellets with the addition of more hydrophilic plasticizers such as triethyl citrate (TEC) or diethyl phthalate (DEP) show an approximate zero-order-release rate. In contrast, two-phase release profiles can be observed from pellets coated with dispersions containing hardly soluble plasticizers such as dibutyl phthalate (DBP) or dibutyl sebacate (DBS). Only in a release medium of high ionic strength the water soluble pore former will remain in the coating. Thus the drug diffuses through a hydrated swollen membrane containing EC, HPMC and insoluble plasticizer. The release mechanisms depend on the glass transition temperature of the ethyl cellulose and therefore on the migration of the plasticizers and the pore former. This was shown by investigation of the migration of the additives and the influence of the temperature of the release medium on the release. Additionally, the study investigates the effect of curing and storage conditions of coated pellets on the drug release rate.
Adult kidneys, of which there are two, are approximately 3 centimeters thick, 6 centimeters wide, and 12 centimeters long 1 1 . The kidneys lie in the back abdominal wall. Their major function is to maintain homeostasis within body and they accomplish this by filtering the blood 1 2 . The end process of kidney function is urine, which is produced to maintain the internal body environment through the regulation of certain solutes, such as potassium and sodium ions 1 3 .
The kidneys are arguably one of the most complex of all the organs of the human body. There is an overwhelming amount of information regarding the many different mechanisms of kidney function, including how it regulates solute concentration in urine, how it adjusts itself according to changes in the internal environment, and the physiology that explains how it all works. However, not all of this information is necessary for the proposed curriculum unit. In order to help focus on only what is necessary to teach osmosis and diffusion, only urine formation and the transport of salt, water and glucose will be discussed in detail.
The Nephron and Urine Production
The basic functional unit of the kidney is the nephron. The nephron is a tubular structure that is lined with one layer of epithelial cells. On average, each kidney is comprised of approximately one million nephrons. Blood first enters the kidneys by way of the renal artery. A large volume (1300 mL) of blood enters the kidney every minute. Upon entering the renal artery, blood then travels through the afferent arteriole, which leads into an individual nephron 1 4 .
The afferent arteriole directs the blood into a capillary structure called the glomerulus. The glomerulus functions as a high pressure filter. As blood passes through the glomerulus, proteins and blood cells are separated from the plasma. The filtrate, a protein-free plasma solution, ends up in the Bowman's capsule of the nephron. The Bowman's capsule is a membrane bound sac that surrounds the glomerulus. The blood cells and proteins that were not filtered into the Bowman's capsule flow out of the nephron through the efferent arteriole, which carries blood out of the nephron. At this point, the fluid that has been collected in the nephron is referred to as filtrate and is the precursor to what will eventually become urine 1 5 .
The glomerulus is essentially a capillary bed, but it has unique properties that make it possible for simple filtration to occur without the use of any energy (other than the energy that is consumed in the heart to create pressure in the blood). The first is that the pressure in the glomerular capillaries is much higher than an average capillary bed and the second is that the capillaries are much more permeable than most capillary beds 1 6 .
The filtrate that is produced in the glomerulus is protein-free plasma that contains glucose, amino acids, vitamins, minerals and any other solutes that are contained in the blood. If the nephron's structure allowed for only simple filtration, as occurs in the glomerulus, all of these important chemicals necessary in the body for healthy functioning would be excreted in the urine and would not be available for use. If these molecules were lost, they would have to be continually replaced in the diet. Fortunately, the nephron is able to recover most of these useful, filtered molecules. As the filtrate moves through the remainder of the nephron, these materials are reabsorbed into the body, since a healthy kidney does not produce urine that contains glucose or many of the other solutes found in the filtrate at this early stage of urine production.
The next region of the nephron is the proximal tubule, which receives about 120 mL of filtrate per minute. The proximal tubule is the region of the nephron responsible for the reabsorption of certain solutes, especially glucose. To understand how glucose is moved out of the fluid in the tubule, and eventually reabsorbed into the blood stream by way of the peritubular capillary bed that surrounds the nephrons, it is necessary to follow the movement of sodium ions that are also present at high concentration in the filtrate. There is a high concentration of sodium ions inside of the proximal tubule because of the initial filtration process. But sodium concentrations are kept relatively low in the epithelial cells that line the proximal tubules. Because of the concentration gradient, sodium ions will move from inside the tubule, through the cells and into the interstitial space. Of course, this requires a protein to transport them out of the cell, since ions cannot pass through the lipid bilayer. Sodium ions are transported into the cells and then pumped from the cells to the interstitial space outside of the cells. The peritubular capillary bed that surrounds the nephron has a low salt concentration as well, since salt was originally removed in the glomerulus. Therefore, the sodium ions diffuse into the capillary bed and are reabsorbed. As salt concentrations increase in the interstitial space outside of the proximal tubule, osmotic pressure increases and water moves outside of the proximal tubule as well. Ultimately, this water is also reabsorbed by the peritubular capillaries 1 7 .
Reabsorption of Glucose
Glucose, like most molecules, is not capable of simply diffusing across the cell membrane when there is a concentration gradient. In order for glucose to be reabsorbed by the body after being filtered out of the blood, glucose transport is actually coupled with the sodium ion transport, as described above. Transport proteins in the cell membrane are activated when both glucose and Na + are available. The transporters require both substances to be present and they operate without the use of any energy. As previously stated, Na + concentrations are kept low in the cells that line the proximal tubules. Therefore, sodium will continue to move down its gradient, so long as there is glucose available in order to activate the transport protein. In this mechanism, sodium is transported from the lumen of the proximal tubule into the cells that line the tubule. Once inside, the sodium and the glucose molecules are still at a higher concentration than they are in the surrounding interstitial space and the peritubular capillaries. There are specialized transport proteins on the side of the nephron cells closest to the peritubular capillaries that are capable of transporting glucose separately from hydrogen. Since glucose is moving down its concentration gradient, as is sodium, this movement of molecules is passive and requires no energy 1 8 .
Figure 2. A. Movement of solutes and water from the tubular fluid to the blood is regulated by tubular epithelial cells. B. Tubular epithelial cells transport sodium ions from the luminal fluid to the interstitium. C. Tubular cells have co-transporters that allow glucose to be reabsorbed together with sodium. Reproduced with permission from ref. 2 (Saltzman, 2009).
The Loop of Henle and Reabsorption of Water
At this point in the journey to create urine from blood filtrate, the glucose absorbed in the filtrate has been reabsorbed by the body, as has some salt and some water. However, it is in the body's best interest for the kidney to recover as much water as possible from the filtrate, creating the most concentrated urine. In order to get water to leave the filtrate through diffusion, the area surrounding the nephron must have a high salt concentration. A high salt concentration in the interstitial fluid outside of the nephron will provide a driving force for osmosis, allowing water to be recovered from the filtrate.
The reabsorption of water occurs in many places in the nephron, but especially in the collecting duct, which is the final segment of tubule in the nephron. To allow for the reabsorption of water, the nephron needs a mechanism for creating high solute concentrations (i.e. high osmotic forces) in the fluid outside the collecting duct. This is accomplished by the Loop of Henle. If you could imagine a tube shaped like a U, this is how the Loop of Henle looks. In the descending side of the loop, the cells are water permeable. In the ascending portion of the loop, the cells are not permeable to water. The cells in the ascending tubule also contain pumps that use energy to transport salt into the interstitial fluid surrounding the Loop and collecting duct. These pumps that remove salt from the ascending loop are important because they pump NaCl into the interstitial fluid that surrounds the Loop of Henle. This helps keep the concentration of salt higher in the interstitial fluid outside of the descending loop 1 9 . Because of the presence of high salt concentrations in the interstitial fluid, as the filtrate travels down the collecting duct&mdashon the way out of the nephron&mdashthere is a strong driving force for water to diffuse through the cells that line the collecting duct. In these last stages of flow through the tubule, the filtrate can become extremely concentrated 2 0 .
As osmosis continues and more water moves into the interstitial fluid surrounding the loop, the filtrate becomes much more concentrated. This is the instance of osmosis that the curriculum unit will focus on, and rather than get into specifics of osmotic pressure and the role of the ascending loop, the "story" will be simplified and students will simply be informed that the body uses energy&mdashwithin the Loop of Henle&mdashto keep the area surrounding the Loop of Henle and collecting duct high in salt concentration, in order to encourage osmosis and the reabsorption of water by the body. Of course, this eliminates much of the more subtle detail involved in this process. If this were to be taught in a high school class, the role of the ascending loop would be appropriate to include and could also be used to further explain active transport, as opposed to the passive transport that happens more frequently in the kidney.
The final stage of urine formation removes certain waste products from the blood that could be toxic if they were allowed to accumulate. Tubular secretions are how metabolic wastes, certain drugs, hydrogen and potassium ions and other materials end up in the urine 2 1 . The collection of these materials occurs in the distal tubule. The filtrate now flows into the collecting duct of the nephron. As the filtrate travels through the collecting duct, certain materials may once again be reabsorbed, depending on the current conditions of the body. The mechanism for this response to the internal environment is key to understanding how the kidney maintains homeostasis 2 2 . It is not important to this unit however, and will not be described in detail.
The Collecting Duct and Excretion of Urine
The collecting duct contains the filtrate, which is now referred to as urine. The process described above is occurring in every one of the one million nephrons in each kidney. Each collecting duct at the end of each nephron will funnel into a larger tubule. This happens repeatedly until all of the urine is funneled into the ureter 2 3 . The ureter is a tubular structure that carries the urine from the renal pelvis to the bladder. Once urine volume in the bladder reaches a specific volume, approximately 200- 400 mL, a receptor in the wall lining of the bladder is stimulated, creating the feeling that the bladder needs to be emptied. From the bladder, urine leaves the body through the urethra 2 4 .
Graham's Law Formula
Graham's law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass. See this law in equation form below.
r(M) ½ = constant
In these equations, r = rate of diffusion or effusion and M = molar mass.
Generally, this law is used to compare the difference in diffusion and effusion rates between gases, often denoted as Gas A and Gas B. It assumes that temperature and pressure are constant and equivalent between the two gases. When Graham's law is used for such a comparison, the formula is written as follows:
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