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Is plasma membrane permeable to sucrose


I searched a lot in the net but don't found a clear answer. I just want to know if plasma membrane is permeable to sucrose.


No, but yes.

Sucrose is a large polar solute. Because it is polar, it cannot easily pass the hydrophobic core of the membrane. So, if the lipids of the plasma membrane are mostly impermeable to sucrose, how do cells take in sucrose?

Plants

In biology, membrane-bound proteins are used for efficient transport across the membrane (Brian, 2011 from a review of Suc transport in plant cells). The specific group of membrane proteins used for sucrose transport are unimaginatively called Suc transporters (SUT).

So in a plant cell, sucrose can be moved across the membrane.

Humans

In mammals, there are no SUTs. Instead, sucrose is broken down into fructose and glucose by the enzyme sucrase.

From there, glucose transporters GLUT proteins carry glucose and fructose across the plasma membrane.


A Physical Barrier

The plasma membrane surrounds all cells and physically separates the cytoplasm, which is the material that makes up the cell, from the extracellular fluid outside the cell. This protects all the components of the cell from the outside environment and allows separate activities to occur inside and outside the cell.

The plasma membrane provides structural support to the cell. It tethers the cytoskeleton, which is a network of protein filaments inside the cell that hold all the parts of the cell in place. This gives the cell its shape. Certain organisms such as plants and fungi have a cell wall in addition to the membrane. The cell wall is composed of molecules such as cellulose. It provides additional support to the cell, and it is why plant cells do not burst like animal cells do if too much water diffuses into them.

Selective Permeability

Plasma membranes are selectively permeable (or semi-permeable), meaning that only certain molecules can pass through them. Water, oxygen, and carbon dioxide can easily travel through the membrane. Generally, ions (e.g. sodium, potassium) and polar molecules cannot pass through the membrane they must go through specific channels or pores in the membrane instead of freely diffusing through. This way, the membrane can control the rate at which certain molecules can enter and exit the cell.

Endocytosis and Exocytosis

Endocytosis is when a cell ingests relatively larger contents than the single ions or molecules that pass through channels. Through endocytosis, a cell can take in large quantities of molecules or even whole bacteria from the extracellular fluid. Exocytosis is when the cell releases these materials. The cell membrane plays an important role in both of these processes. The shape of the membrane itself changes to allow molecules to enter or exit the cell. It also forms vacuoles, small bubbles of membrane that can transport many molecules at once, in order to transport materials to different places in the cell.

Cell Signaling

Phospholipids

The membrane is partially made up of molecules called phospholipids, which spontaneously arrange themselves into a double layer with hydrophilic (“water loving”) heads on the outside and hydrophobic (“water hating”) tails on the inside. These interactions with water are what allow plasma membranes to form.

Proteins

Proteins are wedged between the lipids that make up the membrane, and these transmembrane proteins allow molecules that couldn’t enter the cell otherwise to pass through by forming channels, pores or gates. In this way, the cell controls the flow of these molecules as they enter and exit. Proteins in the cell membrane play a role in many other functions, such as cell signaling, cell recognition, and enzyme activity.

Carbohydrates

Carbohydrates are also found in the plasma membrane specifically, most carbohydrates in the membrane are part of glycoproteins, which are formed when a carbohydrate attaches to a protein. Glycoproteins play a role in the interactions between cells, including cell adhesion, the process by which cells attach to each other.

Fluid Mosaic Model

Technically, the cell membrane is a liquid. At room temperature, it has about the same consistency as vegetable oil. Lipids, proteins, and carbohydrates in the plasma membrane can diffuse freely throughout the cell membrane they are essentially floating across its surface. This is known as the fluid mosaic model, which was coined by S.J. Singer and G.L. Nicolson in 1972.


Osmosis Experiment using Potato Strips

Osmosis refers to a special form of transport of water molecules across the plasma membrane. It involves water molecules movement through a selectively permeable membrane from a region having high water potential to a region having lower water potential. Therefore, water molecules moves through a semi-permeable membrane to a region containing higher concentration of solute from a region containing lower concentration of solute. The ability of water molecules to freely move in a particular system or environment is referred to as water potential. This can be facilitated by osmosis, framework impacts, gravity, or mechanical weight (Sperelakis, 1995). Normally, in defining osmosis, three kinds of solutions are mentioned these include hypertonic, isotonic, and hypertonic solutions. A solution that has the same concentration of solutes both outside and inside of the cell is called isotonic solution. A hypertonic solution is that solution that has high solute concentration compared to the cells neighboring it thus more solvent molecules move from the cells towards it. Lastly, the hypotonic solution is characterized by the low concentration of solute molecules and solvent molecules move from this solution to other regions that have high solute concentrations (Sperelakis, 2011).

The Aim of the Experiment

To determine the relationship between an increase in sucrose concentration and the resulting effect on the weight of potato strips.

Variables

For this experiment, the varying sucrose concentration solution is the independent variable. The final weight of potato strips at the end of the experiment is the dependent variable. Similarly, the original weight of the potato strip is the control variable.

Hypothesis

An increase in the concentration of sucrose solution results in a subsequent decrease in the weight of potato strips.

Methods

Equipment

  • Cork borer
  • Balance
  • Ruler
  • Stirrer
  • Pipette
  • Measuring cylinder
  • Razor blade
  • Tile
  • Flat wooden scapula
  • Tissue roll
  • Weighing boat
  • Marker and pencil
  • 6 beakers (200ml)
  • Cling film
  • Potatoes
  • Sucrose solution
  • Lab coat

Procedure

Preparing sucrose solutions of different concentrations

The beakers are first labeled appropriately as per the sucrose concentration they will contain. To prepare 0.1M sucrose solution, 68.40 grams of was weighed. Then 200ml of water was measured using a measuring cylinder and a pipette. The sucrose and water were then mixed thoroughly in the in the beaker till the sucrose dissolved. The same process was repeated for 0.2M, 0.4M, 0.8M, and 1.0M with 13.68g, 27.36g, 54.72g, and 68.40g respectively.

Preparing the strips

The cork borer was hard-pressed through the potato to obtain the strip. The strip was then pushed out onto the tile using a pencil. At both sides, the skin was removed. potato strips of 4cm in length was obtained using razor blade after accurate measurement using the ruler

Placing the potato cylinders in the different sucrose concentration solutions

The potato strips were then carefully bloated using tissue paper without squeezing them to remove water from the external part. Their respective weights were measured and noted. Subsequently, they were placed in the prepared sucrose solutions. After forty-five minutes the strips were removed, bloated, and reweighed.

Results

Table 1: The mean deviation in the weights of potato strips after forty-five-minute exposure in sucrose solution.

Concentration required (M)Average initial mass (g)Average final mass (g)Percentage change (%)
0.03.463.614.34
0.13.233.21-0.62
0.23.093.5414.56
0.43.463.17-8.38
0.83.253.04-6.46
1.03.563.25-8.71

Figure 1: Graph showing percentage change against sucrose concentration.

Discussion and Conclusion

From the results in table 1, it is evident that there is an overall decrease in the mass of potato cylinder as the concentration is increased except for the 0.2M and 0.4M concentrations. Figure 1 further confirms the overall trend and the correlation between the sucrose concentration and the mass of the potato strips. The general trend indicates that the mass of the strips decreases as the sucrose solution concentration increases. This observation can be clearly explained that the process of osmosis took place. As the sucrose concentration increases, the solution becomes hypertonic. Therefore, water moves from the cells of the potato to the surrounding hypertonic solution in the beaker through osmosis (Kurzweil & Walker, 2009). The movement of water through the process of osmosis into the hypertonic solution results in the decrease in the mass of the potato strips after 45 minutes. An increase in the solute concentration makes the solution in the beaker hypertonic compared to the cytoplasmic water concentration which is hypotonic. Water then naturally moves along the concentration gradient through the process of osmosis to the surrounding solution (Lenart & Flink, 1984).

However, there was some experimental error that affected the results obtained for the 0.2M and 0.4M sucrose concentrations. This has resulted in the skewed line that is observed in figure 1. Generally, the negative correlation that exists between the concentration and the percentage change in mass makes the mass of potatoes trips to reduce as the sucrose concentration increases (hypertonic) (Kurzweil & Walker, 2009). Though the data obtained is biased due to the experimental error in the 0.2M and 0.4M concentration, I accept the hypothesis for the experiment that states an increase in the concentration of sucrose solution results in a subsequent decrease in the weight of potato strips. The variation could be as a result of external factors that affect the osmosis rate such as humidity and temperature. The high temperature and humidity in the laboratory could be the possible cause of faster rate of osmosis observed in the 0.2M and 0.4M concentrations. It is noteworthy noting that the cling film used in the experiment prevented evaporation of the contents of the beaker. Also, measurement errors involved when cutting the potato strips is another potential of the experimental inaccuracy.

Moreover, experimental errors could emerge in the blotting process. There is a possibility that some potato strips were squeezed unconsciously. Despite the few errors involve I can conclude that the experiment was carried out successfully in the laboratory. To reduce the errors involved like the one for the 0.2M and 0.4M, the experiment should be conducted in a controlled condition in future. The laboratory setting for the experiment should be controlled from the external factors and the solutions that contain the potato strips should be kept at constant humidity and temperature to avoid errors like the one observed in this experiment.

References

Kurzweil, A., & Walker, R. (2009). Potato Chip Science: Book & Stuff. New York: Workman Pub.

Lenart, A., & Flink, J.M. (1984). Osmotic Concentration of Potato. International Journal of Food Science and Technology, 19(1), 65-89.

Sperelakis, N. (1995). Cell Physiology Source Book. Burlington: Elsevier Science.

Sperelakis, N. (2011). Cell Physiology Sourcebook: Essentials of Membrane Biophysics. San Diego, CA, USA: Elsevier Science.


Is plasma membrane permeable to sucrose - Biology

The following exercises are designed to help you understand the cell as a functional unit.

Diffusion (Canadian Campbell 2nd ed - Concept 7.3)

Molecules are in a constant state of motion. If, for example, NaCl is dissolved in water so that the NaCl molecules have a completely even and random distribution throughout the water, then there will be no net movement of NaCl in any direction. If however, the NaCl concentration is initially higher in one part of the water than another, diffusion will occur so that there is a net movement of NaCl from the area of high concentration to an area of lower concentration.

-What factors would influence the rate of diffusion?
-Where in an organism do you think diffusion would be an important process?
-What type of substances would enter and leave cells by simple diffusion?

Osmosis (Canadian Campbell 2nd ed - Concept 7.3)

The membrane of the cell or of various subcellular organelles (chloroplasts, mitochondria) serves as a regulatory structure by controlling the entry or exit of substances into and out of the cell. The membrane may be permeable, impermeable or partially permeable to a given substance. In general, membranes are freely permeable to water molecules. In cells, water movement through the cell membrane is determined by the process of osmosis.

During lab 3, three lengths of dialysis tubing were filled with a 2.0 M sucrose solution. A knot was then tied in the other end so that as much air as possible is excluded. The tubes were then rinsed in distilled water. The tubes were weighed and one tube each was placed in the following solutions.

At 15 minute intervals the sacs were removed weighed

  • isosmotic (isotonic) to the solution in the dialysis tube?
  • hypoosmotic (hypotonic) to the solution in the dialysis tube?
  • hyperosmotic (hypertonic) to the solution in the dialysis tube?
  • How does the dialysis tubing simulate a cell membrane?
  • What osmotic problems does an algal, bacterial or protozoan cell face in a fresh water environment?
    • In a sea water environment?
    • in freshwater
    • then in the ocean,
    • and then again as an adult returning to fresh water?

    Plant Cells and Osmosis

    Recall the slide you made of the onion during lab 1 (click to view). Note the position and shape of the cell wall, cytoplasm and vacuole. Now view an onion section which has had a 10% NaCl solution added (click to view).

    - What is the positional relationship of the cell wall to the protoplast?

    • What happens to the water that was inside the cell?
    • What two characteristics of cell membrane are necessary in order for osmosis to occur?
    • Why wouldn't the cell burst if placed in distilled water, as an animal's cell would?
    • How is osmotic pressure important in the normal structure of a plant?

    Cell Permeability

    This refers to the ability of substances to move into or out of the cell by crossing the plasma membrane. Some substances cross the membrane very easily and the membrane is said to be very permeable to these substances others move across with difficulty, while others are excluded completely. In the latter case the cell membrane is impermeable to these substances.

    • What happens to the membrane when the cell is killed by boiling?
    • Does it become leaky?
    • For what types of substances is chloroform a solvent?
    • How does chloroform affect cell membranes?

    One feature you have noticed in plant cells which is absent from animal cells is a cell wall. This wall protects the protoplast and the membrane - plasmalemma.

    First published Sept 95: Modified June 2020
    Copyright © Michael Shaw 2019 (Images and Text)


    Is plasma membrane permeable to sucrose - Biology

    61 notecards = 16 pages ( 4 cards per page)

    Campbell Biology 10th edition Chapter 7

    1) For a protein to be an integral membrane protein, it would have to be _____.

    C) amphipathic, with at least one hydrophobic region

    D) exposed on only one surface of the membrane

    C) amphipathic, with at least one hydrophobic region

    You have a planar bilayer with equal amounts of saturated and unsaturated phospholipids. After testing the permeability of this membrane to glucose, you increase the proportion of unsaturated phospholipids in the bilayer. What will happen to the membrane's permeability to glucose?

    A) Permeability to glucose will increase.

    B) Permeability to glucose will decrease.

    C) Permeability to glucose will stay the same.

    D) You cannot predict the outcome. You simply have to make the measurement.

    A) Permeability to glucose will increase.

    According to the fluid mosaic model of cell membranes, phospholipids _____.

    A) can move laterally along the plane of the membrane

    B) frequently flip-flop from one side of the membrane to the other

    C) occur in an uninterrupted bilayer, with membrane proteins restricted to the surface of the membrane

    D) have hydrophilic tails in the interior of the membrane

    A) can move laterally along the plane of the membrane

    The membranes of winter wheat are able to remain fluid when it is extremely cold by _____.

    A) increasing the percentage of unsaturated phospholipids in the membrane

    B) increasing the percentage of cholesterol molecules in the membrane

    C) decreasing the number of hydrophobic proteins in the membrane

    D) cotransport of glucose and hydrogen

    A) increasing the percentage of unsaturated phospholipids in the membrane

    Some regions of the plasma membrane, called lipid rafts, have a higher concentration of cholesterol molecules. At higher temperatures, these regions _____.

    A) are more fluid than the surrounding membrane

    B) are less fluid than the surrounding membrane

    C) detach from the plasma membrane and clog arteries

    D) have higher rates of lateral diffusion of lipids and proteins into and out of these regions

    B) are less fluid than the surrounding membrane

    Singer and Nicolson's fluid mosaic model of the membrane proposed that membranes_____.

    A) are a phospholipid bilayer between two layers of hydrophilic proteins

    B) are a single layer of phospholipids and proteins

    C) consist of protein molecules embedded in a fluid bilayer of phospholipids

    D) consist of a mosaic of polysaccharides and proteins

    C) consist of protein molecules embedded in a fluid bilayer of phospholipids

    An animal cell lacking oligosaccharides on the external surface of its plasma membrane would likely be impaired in which function?

    A) transporting ions against an electrochemical gradient

    C) attaching the plasma membrane to the cytoskeleton

    D) establishing a diffusion barrier to charged molecules

    Which of these are NOT embedded in the hydrophobic portion of the lipid bilayer at all?

    D) All of these are embedded in the hydrophobic portion of the lipid bilayer.

    Why are lipids and proteins free to move laterally in membranes?

    A) The interior of the membrane is filled with liquid water.

    B) Lipids and proteins repulse each other in the membrane.

    C) Hydrophilic portions of the lipids are in the interior of the membrane.

    D) There are only weak hydrophobic interactions in the interior of the membrane.

    D) There are only weak hydrophobic interactions in the interior of the membrane.

    Cell membranes are asymmetrical. Which of the following statements is the most likely explanation for the membrane's asymmetrical nature?

    A) Since the cell membrane forms a border between one cell and another in tightly packed tissues such as epithelium, the membrane must be asymmetrical

    B) Since cell membranes communicate signals from one organism to another, the cell membranes must be asymmetrical.

    C) The two sides of a cell membrane face different environments and carry out different functions.

    D) Proteins only function on the cytoplasmic side of the cell membrane, which results in the membrane's asymmetrical nature.

    C) The two sides of a cell membrane face different environments and carry out different functions.

    In what way do the membranes of a eukaryotic cell vary?

    A) Phospholipids are found only in certain membranes.

    B) Certain proteins are unique to each membrane.

    C) Only certain membranes of the cell are selectively permeable.

    D) Some membranes have hydrophobic surfaces exposed to the cytoplasm, while others have hydrophilic surfaces facing the cytoplasm.

    B) Certain proteins are unique to each membrane.

    Which of the following is a reasonable explanation for why unsaturated fatty acids help keep a membrane more fluid at lower temperatures?

    A) The double bonds form kinks in the fatty acid tails, preventing adjacent lipids from packing tightly.

    B) Unsaturated fatty acids have a higher cholesterol content and, therefore, more cholesterol in membranes.

    C) Unsaturated fatty acids are more polar than saturated fatty acids.

    D) The double bonds block interaction among the hydrophilic head groups of the lipids.

    A) The double bonds form kinks in the fatty acid tails, preventing adjacent lipids from packing tightly.

    What kinds of molecules pass through a cell membrane most easily?

    Which of the following most accurately describes selective permeability?

    A) An input of energy is required for transport.

    B) Lipid-soluble molecules pass through a membrane.

    C) There must be a concentration gradient for molecules to pass through a membrane.

    D) Only certain molecules can cross a cell membrane.

    D) Only certain molecules can cross a cell membrane.

    Which of the following is a characteristic feature of a carrier protein in a plasma membrane?

    A) It exhibits a specificity for a particular type of molecule.

    B) It requires the expenditure of cellular energy to function.

    C) It works against diffusion.

    D) It has no hydrophobic regions.

    A) It exhibits a specificity for a particular type of molecule.

    Which of the following would likely move through the lipid bilayer of a plasma membrane most rapidly?

    Which of the following allows water to move much faster across cell membranes?

    A) the sodium-potassium pump

    You are working on a team that is designing a new drug. For this drug to work, it must enter the cytoplasm of specific target cells. Which of the following would be a factor that determines whether the molecule selectively enters the target cells?

    A) hydrophobicity of the drug molecule

    B) lack of charge on the drug molecule

    C) similarity of the drug molecule to other molecules transported by the target cells

    D) lipid composition of the target cells' plasma membrane

    C) similarity of the drug molecule to other molecules transported by the target cells

    A) is very rapid over long distances

    B) requires an expenditure of energy by the cell

    C) is a passive process in which molecules move from a region of higher concentration to a region of lower concentration

    D) requires integral proteins in the cell membrane

    C) is a passive process in which molecules move from a region of higher concentration to a region of lower concentration

    Which of the following processes includes all others?

    D) transport of an ion down its electrochemical gradient

    When a cell is in equilibrium with its environment, which of the following occurs for substances that can diffuse through the cell?

    A) There is random movement of substances into and out of the cell.

    B) There is directed movement of substances into and out of the cell.

    C) There is no movement of substances into and out of the cell.

    D) All movement of molecules is directed by active transport.

    A) There is random movement of substances into and out of the cell.

    Which of the following is true of osmosis?

    A) Osmosis only takes place in red blood cells.

    B) Osmosis is an energy-demanding or "active" process.

    C) In osmosis, water moves across a membrane from areas of lower solute concentration to areas of higher solute concentration.

    D) In osmosis, solutes move across a membrane from areas of lower water concentration to areas of higher water concentration.

    C) In osmosis, water moves across a membrane from areas of lower solute concentration to areas of higher solute concentration.

    Which component is a peripheral protein?

    Which component is cholesterol?

    Which component is a protein fiber of the extracellular matrix?

    Which component is a microfilament (actin filament) of the cytoskeleton?

    Which component is a glycolipid?

    The solutions in the two arms of this U-tube are separated by a membrane that is permeable to water and glucose but not to sucrose. Side A is half-filled with a solution of 2 M sucrose and 1 M glucose. Side B is half-filled with 1 M sucrose and 2 M glucose. Initially, the liquid levels on both sides are equal.

    Question: Refer to the figure. Initially, in terms of tonicity, the solution in side A with respect to the solution in side B is _____.

    The solutions in the two arms of this U-tube are separated by a membrane that is permeable to water and glucose but not to sucrose. Side A is half-filled with a solution of 2 M sucrose and 1 M glucose. Side B is half-filled with 1 M sucrose and 2 M glucose. Initially, the liquid levels on both sides are equal.

    Question: Refer to the figure. After the system reaches equilibrium, what changes are observed?

    A) The molarity of sucrose is higher than that of glucose on side A.

    B) The water level is higher in side A than in side B.

    C) The water level is unchanged.

    D) The water level is higher in side B than in side A.

    B) The water level is higher in side A than in side B.

    A patient was involved a serious accident and lost a large quantity of blood. In an attempt to replenish body fluids, distilled water—equal to the volume of blood lost—is added to the blood directly via one of his veins. What will be the most probable result of this transfusion?

    A) The patient's red blood cells will shrivel up because the blood has become hypotonic compared to the cells.

    B) The patient's red blood cells will swell and possibly burst because the blood has become hypotonic compared to the cells.

    C) The patient's red blood cells will shrivel up because the blood has become hypertonic compared to the cells.

    D) The patient's red blood cells will burst because the blood has become hypertonic compared to the cells.

    B) The patient's red blood cells will swell and possibly burst because the blood has become hypotonic compared to the cells.

    The solutions in the arms of a U-tube are separated at the bottom of the tube by a selectively permeable membrane. The membrane is permeable to sodium chloride but not to glucose. Side A is filled with a solution of 0.4 M glucose and 0.5 M sodium chloride (NaCl), and side B is filled with a solution containing 0.8 M glucose and 0.4 M sodium chloride. Initially, the volume in both arms is the same..

    Question: Refer to the figure. At the beginning of the experiment,

    A) side A is hypertonic to side B.

    B) side A is hypotonic to side B.

    C) side A is hypertonic to side B with respect to glucose.

    D) side A is hypotonic to side B with respect to NaCl.

    B) side A is hypotonic to side B.

    The solutions in the arms of a U-tube are separated at the bottom of the tube by a selectively permeable membrane. The membrane is permeable to sodium chloride but not to glucose. Side A is filled with a solution of 0.4 M glucose and 0.5 M sodium chloride (NaCl), and side B is filled with a solution containing 0.8 M glucose and 0.4 M sodium chloride. Initially, the volume in both arms is the same..

    Question: Refer to the figure. If you examine side A after three days, you should find _____.

    A) a decrease in the concentration of NaCl and glucose and an increase in the water level

    B) a decrease in the concentration of NaCl, an increase in water level, and no change in the concentration of glucose

    C) a decrease in the concentration of NaCl and a decrease in the water level

    D) no change in the concentration of NaCl and glucose and an increase in the water level

    C) a decrease in the concentration of NaCl and a decrease in the water level

    Five dialysis bags constructed of membrane, which is permeable to water and impermeable to sucrose, were filled with various concentrations of sucrose and then placed in separate beakers containing an initial concentration of 0.6 M sucrose solution. At 10-minute intervals, the bags were massed (weighed) and the percent change in mass of each bag was graphed.

    Question: Which line in the graph represents the bag that contained a solution isotonic to the 0.6 M solution at the beginning of the experiment?

    Five dialysis bags constructed of membrane, which is permeable to water and impermeable to sucrose, were filled with various concentrations of sucrose and then placed in separate beakers containing an initial concentration of 0.6 M sucrose solution. At 10-minute intervals, the bags were massed (weighed) and the percent change in mass of each bag was graphed.

    Question: Which line in the graph represents the bag with the highest initial concentration of sucrose?

    Five dialysis bags constructed of membrane, which is permeable to water and impermeable to sucrose, were filled with various concentrations of sucrose and then placed in separate beakers containing an initial concentration of 0.6 M sucrose solution. At 10-minute intervals, the bags were massed (weighed) and the percent change in mass of each bag was graphed.

    Question: Which line or lines in the graph represent(s) bags that contain a solution that is hypertonic at 50 minutes?

    Celery stalks that are immersed in fresh water for several hours become stiff. Similar stalks left in a 0.15 M salt solution become limp. From this we can deduce that the fresh water_____.

    A) and the salt solution are both hypertonic to the cells of the celery stalks

    B) is hypotonic and the salt solution is hypertonic to the cells of the celery stalks

    C) is hypertonic and the salt solution is hypotonic to the cells of the celery stalks

    D) is isotonic and the salt solution is hypertonic to the cells of the celery stalks

    B) is hypotonic and the salt solution is hypertonic to the cells of the celery stalks

    What will happen to a red blood cell (RBC), which has an internal ion concentration of about 0.9 percent, if it is placed into a beaker of pure water?

    A) The cell would shrink because the water in the beaker is hypotonic relative to the cytoplasm of the RBC.

    B) The cell would shrink because the water in the beaker is hypertonic relative to the cytoplasm of the RBC.

    C) The cell would swell because the water in the beaker is hypotonic relative to the cytoplasm of the RBC.

    D) The cell will remain the same size because the solution outside the cell is isotonic.

    C) The cell would swell because the water in the beaker is hypotonic relative to the cytoplasm of the RBC.

    Which of the following statements correctly describes the normal tonicity conditions for typical plant and animal cells? The animal cell is in _____.

    A) a hypotonic solution, and the plant cell is in an isotonic solution

    B) an isotonic solution, and the plant cell is in a hypertonic solution

    C) a hypertonic solution, and the plant cell is in an isotonic solution

    D) an isotonic solution, and the plant cell is in a hypotonic solution

    D) an isotonic solution, and the plant cell is in a hypotonic solution

    In which of the following would there be the greatest need for osmoregulation?

    A) an animal connective tissue cell bathed in isotonic body fluid

    B) a salmon moving from a river into an ocean

    C) a red blood cell surrounded by plasma

    D) a plant being grown hydroponically in a watery mixture of designated nutrients

    B) a salmon moving from a river into an ocean

    When a plant cell, such as one from a rose stem, is submerged in a very hypotonic solution, what is likely to occur?

    B) Plasmolysis will shrink the interior.

    C) The cell will become flaccid.

    D) The cell will become turgid.

    D) The cell will become turgid.

    A sodium-potassium pump _____.

    A) moves three potassium ions out of a cell and two sodium ions into a cell while producing an ATP for each cycle

    B) move three sodium ions out of a cell and two potassium ions into a cell while consuming an ATP for each cycle

    C) moves three potassium ions out of a cell and two sodium ions into a cell while consuming 2 ATP in each cycle

    D) move three sodium ions out of a cell and two potassium ions into a cell and generates an ATP in each cycle

    B) move three sodium ions out of a cell and two potassium ions into a cell while consuming an ATP for each cycle

    The sodium-potassium pump is called an electrogenic pump because it _____.

    A) pumps equal quantities of Na+ and K+ across the membrane

    B) contributes to the membrane potential

    C) ionizes sodium and potassium atoms

    D) is used to drive the transport of other molecules against a concentration gradient

    B) contributes to the membrane potential

    Which of the following membrane activities requires energy from ATP?

    A) facilitated diffusion of chloride ions across the membrane through a chloride channel

    B) movement of Na+ ions from a lower concentration in a mammalian cell to a higher concentration in the extracellular fluid

    C) movement of glucose molecules into a bacterial cell from a medium containing a higher concentration of glucose than inside the cell

    D) movement of carbon dioxide out of a paramecium

    B) movement of Na+ ions from a lower concentration in a mammalian cell to a higher concentration in the extracellular fluid

    The voltage across a membrane is called the _____.

    D) electrochemical gradient

    Ions diffuse across membranes through specific ion channels down _____.

    A) their chemical gradients

    B) their concentration gradients

    C) the electrical gradients

    D) their electrochemical gradients

    D) their electrochemical gradients

    Which of the following would increase the electrochemical gradient across a membrane?

    A) a sucrose-proton cotransporter

    D) both a proton pump and a potassium channel

    The phosphate transport system in bacteria imports phosphate into the cell even when the concentration of phosphate outside the cell is much lower than the cytoplasmic phosphate concentration. Phosphate import depends on a pH gradient across the membrane—more acidic outside the cell than inside the cell. Phosphate transport is an example of _____.

    In some cells, there are many ion electrochemical gradients across the plasma membrane even though there are usually only one or two proton pumps present in the membrane. The gradients of the other ions are most likely accounted for by _____.

    C) pores in the plasma membrane

    D) passive diffusion across the plasma membrane

    Which of the following is most likely true of a protein that cotransports glucose and sodium ions into the intestinal cells of an animal?

    A) Sodium and glucose compete for the same binding site in the cotransporter.

    B) Glucose entering the cell down its concentration gradient provides energy for uptake of sodium ions against the electrochemical gradient.

    C) Sodium ions can move down their electrochemical gradient through the cotransporter whether or not glucose is present outside the cell.

    D) A substance that blocks sodium ions from binding to the cotransport protein will also block the transport of glucose.

    D) A substance that blocks sodium ions from binding to the cotransport protein will also block the transport of glucose.

    Proton pumps are used in various ways by members of every domain of organisms: Bacteria, Archaea, and Eukarya. What does this most probably mean?

    A) Proton gradients across a membrane were used by cells that were the common ancestor of all three domains of life.

    B) The high concentration of protons in the ancient atmosphere must have necessitated a pump mechanism.

    C) Cells of each domain evolved proton pumps independently when oceans became more acidic.

    D) Proton pumps are necessary to all cell membranes.

    A) Proton gradients across a membrane were used by cells that were the common ancestor of all three domains of life.

    Several epidemic microbial diseases of earlier centuries incurred high death rates because they resulted in severe dehydration due to vomiting and diarrhea. Today they are usually not fatal because we have developed which of the following?

    A) antiviral medications that are efficient and work well with most viruses

    B) intravenous feeding techniques

    C) medications to slow blood loss

    D) hydrating drinks with high concentrations of salts and glucose

    D) hydrating drinks with high concentrations of salts and glucose

    The force driving simple diffusion is _____, while the energy source for active transport is _____.

    A) the concentration gradient ADP

    B) the concentration gradient ATP

    C) transmembrane pumps electron transport

    D) phosphorylated protein carriers ATP

    B) the concentration gradient ATP

    An organism with a cell wall would most likely be unable to take in materials through _____.

    White blood cells engulf bacteria using _____.

    D) receptor-mediated exocytosis

    Familial hypercholesterolemia is characterized by _____.

    A) defective LDL receptors on the cell membranes

    B) poor attachment of the cholesterol to the extracellular matrix of cells

    C) a poorly formed lipid bilayer that cannot incorporate cholesterol into cell membranes

    D) inhibition of the cholesterol active transport system in red blood cells

    A) defective LDL receptors on the cell membranes

    The difference between pinocytosis and receptor-mediated endocytosis is that _____.

    A) pinocytosis brings only water molecules into the cell, but receptor-mediated endocytosis brings in other molecules as well.

    B) pinocytosis increases the surface area of the plasma membrane, whereas receptor-mediated endocytosis decreases the plasma membrane surface area.

    C) pinocytosis is nonselective in the molecules it brings into the cell, whereas receptor-mediated endocytosis offers more selectivity.

    D) pinocytosis can concentrate substances from the extracellular fluid, but receptor-mediated endocytosis cannot.

    C) pinocytosis is nonselective in the molecules it brings into the cell, whereas receptor-mediated endocytosis offers more selectivity.

    In receptor-mediated endocytosis, receptor molecules initially project to the outside of the cell. Where do they end up after endocytosis?

    A) on the outside of vesicles

    B) on the inside surface of the cell membrane

    C) on the inside surface of the vesicle

    D) on the outer surface of the nucleus

    C) on the inside surface of the vesicle

    A bacterium engulfed by a white blood cell through phagocytosis will be digested by enzymes contained in _____.

    Use the paragraph and accompanying figure to answer the following questions.

    Human immunodeficiency virus (HIV) infects cells that have both CD4 and CCR5 cell surface molecules. The viral nucleic acid molecules are enclosed in a protein capsid, and the protein capsid is itself contained inside an envelope consisting of a lipid bilayer membrane and viral glycoproteins. One hypothesis for viral entry into cells is that binding of HIV membrane glycoproteins to CD4 and CCR5 initiates fusion of the HIV membrane with the plasma membrane, releasing the viral capsid into the cytoplasm. An alternative hypothesis is that HIV gains entry into the cell via receptor-mediated endocytosis, and membrane fusion occurs in the endocytotic vesicle. To test these alternative hypotheses for HIV entry, researchers labeled the lipids on the HIV membrane with a red fluorescent dye.

    Question: In an HIV-infected cell producing HIV virus particles, the viral glycoprotein is expressed on the plasma membrane. How do the viral glycoproteins get to the plasma membrane? They are synthesized _____.

    A) on ribosomes on the plasma membrane

    B) by ribosomes in the rough ER and arrive at the plasma membrane in the membrane of secretory vesicles

    C) on free cytoplasmic ribosomes and then inserted into the plasma membrane

    D) by ribosomes in the rough ER, secreted from the cell, and inserted into the plasma membrane from the outside

    B) by ribosomes in the rough ER and arrive at the plasma membrane in the membrane of secretory vesicles

    Use the paragraph and accompanying figure to answer the following questions.

    Human immunodeficiency virus (HIV) infects cells that have both CD4 and CCR5 cell surface molecules. The viral nucleic acid molecules are enclosed in a protein capsid, and the protein capsid is itself contained inside an envelope consisting of a lipid bilayer membrane and viral glycoproteins. One hypothesis for viral entry into cells is that binding of HIV membrane glycoproteins to CD4 and CCR5 initiates fusion of the HIV membrane with the plasma membrane, releasing the viral capsid into the cytoplasm. An alternative hypothesis is that HIV gains entry into the cell via receptor-mediated endocytosis, and membrane fusion occurs in the endocytotic vesicle. To test these alternative hypotheses for HIV entry, researchers labeled the lipids on the HIV membrane with a red fluorescent dye.

    Question: What would be observed by live-cell fluorescence microscopy immediately after HIV entry if HIV is endocytosed first, and then later fuses with the endocytotic vesicle membrane?

    A) A spot of red fluorescence will be visible on the infected cell's plasma membrane, marking the site of membrane fusion and HIV entry.

    B) The red fluorescent dye-labeled lipids will appear in the infected cell's interior.

    C) A spot of red fluorescence will diffuse in the infected cell's cytoplasm.

    D) A spot of red fluorescence will remain outside the cell after delivering the viral capsid.

    B) The red fluorescent dye-labeled lipids will appear in the infected cell's interior.


    5 Conclusion

    This assessment of the movement of water and cryoprotectants in oocytes and embryos shows that the pattern of the movement is more stage-specific than species-specific, although species specificity exists in some cases. Therefore, the protocols that have been developed for the cryopreservation of oocytes and embryos in one species generally would be applicable to other species at a similar stage, in terms of permeability.

    In order to design protocols for the vitrification of mammalian oocytes and embryos, it would be important to consider the pathway of the movement of water and cryoprotectants for each stage. In vitrification, the time of exposure and temperature of the vitrification solution are important because the vitrification solution contains a high concentration of cryoprotectant and thus is highly toxic to oocytes and embryos. When water and cryoprotectants move through oocytes and embryos principally by simple diffusion, the temperature and time of exposure to the vitrification solution are important because temperature affects the permeability to water and cryoprotectants. When water and cryoprotectants move through embryos principally by facilitated diffusion via channels, the amount of time of exposure to the vitrification solution is more important because the permeability is less affected by temperature. However, the exposure of oocytes and embryos to the vitrification solution at a high temperature should be avoided because cryoprotectants are more toxic at higher temperatures.


    Carbon Fixation Is Catalyzed by Ribulose Bisphosphate Carboxylase

    We have seen earlier in this chapter how cells produce ATP by using the large amount of free energy released when carbohydrates are oxidized to CO2 and H2O. Clearly, therefore, the reverse reaction, in which CO2 and H2O combine to make carbohydrate, must be a very unfavorable one that can only occur if it is coupled to other, very favorable reactions that drive it.

    The central reaction of carbon fixation, in which an atom of inorganic carbon is converted to organic carbon, is illustrated in Figure 14-38: CO2 from the atmosphere combines with the five-carbon compound ribulose 1,5-bisphosphate plus water to yield two molecules of the three-carbon compound 3-phosphoglycerate. This �rbon-fixing” reaction, which was discovered in 1948, is catalyzed in the chloroplast stroma by a large enzyme called ribulose bisphosphate carboxylase. Since each molecule of the complex works sluggishly (processing only about 3 molecules of substrate per second compared to 1000 molecules per second for a typical enzyme), many enzyme molecules are needed. Ribulose bisphosphate carboxylase often constitutes more than 50% of the total chloroplast protein, and it is thought to be the most abundant protein on Earth.

    Figure 14-38

    The initial reaction in carbon fixation. This reaction, in which carbon dioxide is converted into organic carbon, is catalyzed in the chloroplast stroma by the abundant enzyme ribulose bisphosphate carboxylase. The product is 3-phosphoglycerate, which (more. )


    Lab 1 Osmosis

    Cells have kinetic energy. This causes the molecules of the cell to move around and bump into each other. Diffusion is one result of this molecular movement. Diffusion is the random movement of molecules from an area of higher concentration to areas of lower concentration. Osmosis is a special kind of diffusion where water moves through a selectively permeable membrane (a membrane that only allows certain molecules to diffuse though). Diffusion or osmosis occurs until dynamic equilibrium has been reached. This is the point where the concentrations in both areas are equal and no net movement will occur from one area to another.
    If two solutions have the same solute concentration, the solutions are said to be isotonic. If the solutions differ in concentration, the area with the higher solute concentration is hypertonic and the area with the lower solute concentration is hypotonic. Since a hypotonic solution contains a higher level of solute, it has a high solute potential and low water potential. This is because water potential and solute potential are inversely proportional. A hypotonic solution would have a high water potential and a low solute potential. An isotonic solution would have equal solute and water potentials. Water potential (y) is composed of two main things, a physical pressure component, pressure potential (yp), and the effects of solutes, solute potential (ys). A formula to show this relationship is y = yp + ys. Water will always move from areas of high water potential to areas of low water potential.
    The force of water in a cell against its plasma membrane causes the cell to have turgor pressure, which helps maintain the shape of the cell. When water moves out of a cell, the cell will loose turgor pressure along with water potential. Turgor pressure of a plant cell is usually attained while in a hypotonic solution. The loss of water and turgor pressure while a cell is in a hypertonic solution is called plasmolysis.
    Hypothesis:
    During these experiments, it will be proven that diffusion and osmosis occur between solutions of different concentrations until dynamic equilibrium is reached, affecting the cell by causing plasmolysis or increased turgor pressure during the process.
    Materials:
    Lab 1A – To begin Lab 1A, first collect the desired equipment. The materials needed are dialysis tubing, Iodine Potassium Iodide (IKI) solution, 15% glucose/ 1% starch solution, glucose Testape or Lugol’s solution, distilled water, and a 250-mL beaker.
    Lab 1B – For Lab 1B you will need to collect six presoaked dialysis tubing strips, distilled water 0.2M, 0.4M, 0.6M, 0.8M, and a 1.0M sucrose solution six 250-mL beakers or cups, and a scale.
    Lab 1C – Lab 1C these items are needed: a potato, knife, potato core borers, six different solutions, and a scale.
    Lab 1D – During Lab 1D, only paper, pencil, and a calculator will be needed to make the calculations.
    Lab 1E – n Lab 1E these items are needed: a microscope slide, cover slip, onion cells, light microscope, and a 15% NaCl solution.
    Procedures:
    Lab 1A – After gathering the materials, pour glucose/starch solution into dialysis tubing and close the bag. Test the solution for presence of glucose. Test the beaker of distilled water and IKI for presence of glucose. Put the dialysis bag into the beaker and let stand for 30 minutes. When time is up test both the bag and the beaker for presence of glucose. Record all data in table.
    Lab 1B – Obtain the six strips of dialysis tubing and fill each with a solution of a different molarity. Mass each bag. Put each bag into a beaker of distilled water and let stand for half and hour. After 30 minutes is up, remove each bag and determine its mass. Record all data in its appropriate table.
    Lab 1C – sing the potato core borer, obtain 24 cylindrical slices of potato, four for each cup. Determine the mass of the four cylinders. Immerse four cylinders into each of the six beakers or cups. Let stand overnight. After time is up, remove the cores from the sucrose solutions and mass them. Record all data in its appropriate table.
    Lab 1D – Using the paper, pencil, and calculator collected, determine solute potentials of the solutions and answer the questions asked to better understand this particular part of the lab.
    Lab 1E – Using the materials gathers, prepare a wet mount slide of the epidermis of an onion. Draw what you see of the onion cell under the microscope. Add several drops of the NaCl solution to the slide. Now draw the appearance of the cell.
    Data:
    Lab 1A – Table 1.1

    Contents Initial Color Final Color Initial Presence of Glucose Final Presence of Glucose
    Bag 15% Glucose/ 1% Starch Solution clear Dark blue + +
    Beaker H2O+IKI Orange to brown Orange to brown _ +

    Lab 1A Questions
    1) Glucose is leaving the bag and Iodine-Potassium-Iodide is entering the bag. The change in color of the contents of the bag and the presence of glucose in the bag prove this.
    2) In the results, the IKI moved from the beaker to the bag, this caused the change in the color of the bag. The IKI moved into the bag to make the concentrations outside the bag equal to inside the bag. The glucose solution moved out of the bag making glucose present in the beaker. The glucose moved to make the solute concentration inside and out of the bag equal.
    3) If the initial and final percent concentration of glucose and IKI for in the bag and the beaker were given, they would show the differences and prove the movement of these substances to reach dynamic equilibrium.
    4) Based on my observations, the smallest substance was the IKI molecule, then the glucose molecules, water molecules, membrane pore, and then the starch molecules being the largest.
    5) If the experiment started with glucose and IKI inside the bag and starch in the beaker, the glucose and IKI would move out of the bag to make the concentrations equal, but the starch could not move into the bag because its molecules are too big to pass through the semipermeable membrane.

    Lab 1B Table 1.2 Dialysis Bag Results


    Tetrahymena Thermophila

    Alejandro D. Nusblat , . Aaron P. Turkewitz , in Methods in Cell Biology , 2012

    B Sterol Metabolism

    Sterols affect membrane fluidity and permeability ( Ohvo-Rekila et al., 2002 ). In addition, they are essential components of the “lipid rafts” that have been characterized principally in animal cells, which are currently understood as membrane microdomains whose formation depends upon the affinity of sterols for sphingolipids. The partitioning of proteins in lipid rafts may be important for regulation of signal transduction pathways ( Simons and Toomre, 2000 ). Sterols also serve as precursors of bile salts and steroid hormones in mammals, brassinosteroids in plants, and fungi and ecdysteroids in arthropods.

    Eukaryotic organisms can satisfy their sterol requirement by de novo synthesis in vertebrates (cholesterol), plants (stigmasterol, sitosterol, and campesterol), and fungi (ergosterol), or by obtaining them from food. Sterol auxotrophs include invertebrates (nematodes and arthropods), some ciliates (Paramecium tetraurelia), apicomplexans (Plasmodium falciparum), and some flagellated parasites (Giardia intestinalis and Trichomonas vaginalis). T. thermophila is unusual in this regard, having no detectable sterols in its membranes and, accordingly, no sterol requirement. Instead, it synthesizes tetrahymanol, a compound similar to hopanoids found in bacteria, which acts as a surrogate sterol. However, when sterols are added to the growth medium, tetrahymanol synthesis is suppressed and T. thermophila incorporates the exogenous sterol, either with or without modifications( Conner et al., 1968 ). In particular, the ciliate desaturates sterols at positions C5(6), C7(8), and C22(23) and removes the C24 ethyl group in C29 sterols (phytosterols) ( Mallory and Conner, 1971 ). By the activity of these three sterol desaturases (C-5, C-7, and C-22 sterol desaturases) and C-24 sterol deethylation, the ciliate modifies exogenous sterols and accumulates the tri-unsaturated products in its membrane.

    C-22 sterol desaturases have been characterized in other eukaryotes. In T. thermophila, the C-7 and C-22 sterol-desaturating activities, found mainly in a microsomal fraction, require cytochrome b5 as shown by their inhibition with azide and cyanide ( Nusblat et al., 2005 Valcarce et al., 2000 ). This cytochrome b5 dependence is not characteristic of the C22 desaturases of plants and fungi, which require cytochrome P450. The difference is underscored by the insensitivity of the ciliate C22 desaturase to azole, a compound that strongly inhibits the corresponding plant and fungal activities. Moreover, no clear orthologs can be found in the T. thermophila genome for known C-22 sterol desaturases ( Morikawa et al., 2006 ). These observations suggest that the T. thermophila enzyme represents a new class of C-22 sterol desaturases.

    The C-5 sterol desaturase present in most eukaryotic cells belongs to the fatty acid hydroxylase (FAH) superfamily of integral membrane proteins that bind an iron cofactor via a 3-histidine motif. The C-5 sterol desaturase in T. thermophila, DES5A, was identified by characterizing the phenotype resulting from deletion of a putative FAH gene ( Nusblat et al., 2009 ). The deletion mutant, which was fully viable, showed strongly diminished C-5 sterol desaturase activity, while C-7(8) and C-22(23) desaturase activities were unaffected.

    The gene involved in C-24 sterol deethylation, DES24, was similarly confirmed by the disruption of putative FAH genes ( Tomazic et al., 2011 ), resulting in a strain unable to eliminate the C-24 ethyl group from different phytosterols, and probably defective at the first step in dealkylation. Interestingly, the mutant strain was highly sensitive to phytosterols in the culture media, showing defects in growth and morphology and altered tetrahymanol biosynthesis. This observation suggests that C29 sterols can impair the normal growth of Tetrahymena. While C-24 sterol deethylation activity has been characterized in other organisms including nematodes, arthropods, and green algae, the Tetrahymena enzyme represents the first molecular characterization. However, DES24 clusters phylogenetically with bacterial FAH sequences of unknown function, with no obvious orthologs in other eukaryotes, and may therefore have been acquired by lateral transfer. A variety of other observations, including substrate specificity and inhibitor studies, are also consistent with the hypothesis that the mechanism of T. thermophila C-24 deethylation differs from that in other eukaryotes.

    T. thermophila, which is exposed in its environment to phytoplankton, higher plants and algae, may have developed the ability to metabolize otherwise-harmful phytosterols upon acquisition of DES24 from bacteria. Interestingly, however, Paramecium tetraurelia does not have C-24 dealkylation activity ( Conner et al., 1971 ) and requires phytosterols ( Whitaker and Nelson, 1987 ). Overall, sterol metabolism in T. thermophila seems to be the evolutionary product of a fascinating combination of gene losses (e.g., typical eukaryotic genes involved in sterol biosynthesis) combined with acquisition of bacterial genes to allow for synthesis of unusual compounds, with potentially novel mechanisms of sterol modification. This evolutionary history may be illuminated by interrogating the genomes of other Tetrahymena species as these are sequenced. In addition, further studies of the sterol pathways in T. thermophila may yield more information about lipid diversity and function.


    Process of Diffusion in Plant Cell (With Diagrams)

    The movement of various substances into a plant, usually from the soil, out of which the green plant synthesises the numerous complex organic compounds, is accomplished, principally through the agency of the process known generally as diffusion.

    In some cases, however, the operation of the diffusion phenomena is complicated by other factors.

    We know that in higher green plants, some substances enter the living cells through the aerial organs—the diffusion of CO2 and O2 from the atmosphere into the plants is principally through stomata. From soil, water and ions of simple inorganic salts pass into the plants through the root cells by a process which is basically diffusion, though greatly modified by other factors.

    Similarly, the loss of large quantities of water as vapour from leaves and other aerial organs into the atmosphere is also accomplished by diffu­sion. O2 evolved in photosynthesis and the respiratory CO2 also diffuse out of the plant into the atmosphere. The exhaled CO2 in respiration of the root cells diffuses out into the soil.

    The movement of substance within a plant, i.e., from one part of a plant to another, is mainly by diffusion. The gases move through the intercellular spaces and the water and solutes through the dead and living vessels and also from cell to cell by diffusion, each following its own independent diffusion pressure or concentration gradient.

    There are very few, if any, physiological processes occurring in plants which do not directly or indirectly involve the diffusion phenomena. What is diffusion then? According to the kinetic theory, the molecules or ions of all substances (gases, liquids, solutes or solids) are in constant motion in all directions, which is due to their individual kinetic activity.

    This is called diffusion. In a mixture of several substances separated by a membrane, the molecules or ions of each substance will diffuse in that direction which is from a region of its own higher concentration or pressure to its own lower con­centration or pressure or in other words from a region of its own higher activity to one of its own lesser activity of the particular substance concerned.

    And the direction of this diffusion is independent of the direction of diffusions of ions or molecules of other sub­stances present. From plants, the outward diffusing water vapour molecules in trans­piration or oxygen evolved in photosynthesis do not carry respiratory CO2 with them, but each gas diffuses independently of the other, depending only on the difference of its own concentration or pressure, inside and outside the plant.

    It is well known that when a gas is allowed entry in a room, the molecules of the gas, owing to their kinetic energy, become evenly distributed throughout the room. If a bottle of chloroform or ether is opened indoors, the distinctive odour can be perceived in all parts of the room within a short time.

    This is diffusion of gases and the rate of diffusion is dependent upon the concentration of the particular gas, the temperature, the density and also to some extent on the presence of other gases.

    Diffusion of Dissolved Substances (Solutes):

    If a blue crystal of CuSO4 is dropped in a beaker containing water, the diffusion of the dissociated ions of Cu++ and SO4 = or entire molecules of CuSO4 can be observed by the slow change in the colour of water.

    This rate of diffusion of solute molecules or ions through the solvent is extremely slow and may principally be due to the fact that the densely packed particles of water (solvent) considerably impede the diffusion of particles of the solute CuSO4.

    The direction of diffusion of a particular Solute ion of molecule follows the same general pattern, i.e., from higher concentration or pressure to its lower concentration or pressure, regardless of diffusion, of other solute particles, if any, in the same system. The rates of diffusion are also generally controlled by principles which are essentially similar to those governing diffusion of gases.

    Diffusion Through a Membrane:

    Diffusion takes place also through membranes. Three types of membranes are possible, permeable, impermeable and semipermeable. Permeable membranes are those which allow diffusion of both solvent and solute molecules or ions through them. Ordinary cellulose cell walls of young non-vacuolated cells are examples of completely permeable membranes.

    Lignified cell walls are also quite freely permeable to water and when wet, to solute ions and molecules as well. Impermeable membranes are those which prohibit the entry of both solvent and solute particles.

    Suberised or heavily cutinised cell walls in plants are practically impermeable as regards the passage of water and dissolved substances. Semipermeable or differentially permeable membranes are those which allow the diffusion only of the solvent particles through them while restricting the entrance of the solute particles, i.e., the membrane is permeable to one substance but impermeable to the other.

    In this connection, the only solvent we need to consider is water, for water is the only important liquid in living organisms which moves from cell to cell and hence all our further discussions of this phenomenon will naturally be in terms of water and aqueous solutions.

    If a U tube (Fig. 663A) is divided by a permeable membrane, filled on one arm with pure water and the other with a solution of sucrose, the molecules of sucrose will diffuse through the permeable membrane towards water because the diffusion pressure and concentration of sucrose molecules are higher on the right-hand than on the left- hand side and the water molecules will move in the opposite direction for its diffusion pressure or concentration is higher in the left portion than in the right.

    Thus there will be a thorough mixing of both sucrose and water molecules in the two arms of the tube. An equilibrium is eventually established between the two arms of the tube when equal quantities of sucrose and water molecules are present on the two sides of the partition.

    If the membrane is more rapidly penetrated by one of the two substances that substance will at first pass more quickly than the other but ultimately equalisation of the concentration and the pressure of the solutions in the two arms must result.

    The case will be very different, however, when the partition, separating the two liquids is semipermeable, i.e., permeable only to the molecules of water while molecules of sucrose are quite unable to pass through it (Fig. 663C).

    The result will be that owing to greater diffusion pressure (or concentration) of water in the left arm of the U tube, water molecules will pass through the membrane towards the right arm. Volume for volume, the number of molecules of water is certainly higher in the left arm than in the right arm since in the solution of sucrose, a large proportion of molecules of water are replaced by sucrose molecules.

    Thus there will be a greater accumulation of water molecules on the side of the membrane containing sucrose solution and the level of solution in the right arm will consequently increase considerably.

    This diffusion of water molecules through a semipermeable or differentially permeable membrane from the region of higher concentration and higher diffusion pressure of water to a region where it is lesser is known as osmosis and the pressure developed as a result of this is known as osmotic pressure.

    It must be pointed out here, however, that semipermeable membranes which completely restrict the movement of solute particles or will comple­tely restrict the movement of water molecules from the right arm to the left arm of the U tube, are only theoretically possible.

    In practice, however, most semipermeable membranes are really only differentially permeable and allow passage to some sucrose molecules or even to some solvent (water) molecules from the right to the left of the tube. In U tube B, where the membrane is impermeable, there is theoretically no net diffusion of either the solvent or the solute molecules.

    Obscurities have sometimes resulted from the application of the term osmosis to the movement of solutes through the membranes. The use of the term in this sense has not only led to much misunderstanding but the concepts of many students have remained vague as regards the process of absorption of water and absorption of solutes which two terms, they are apt to consider synonymous.

    The movement of solutes through a mem­brane, though certainly primarily a process of diffusion, may even be simple diffusion, but complicated enormously by a number of physico-chemical factors which do not play any part in the process of diffusion we know of.

    According to the principles of diffusion, the molecules or ions of all substances in solution, gases or even dispersed particles in a colloidal system, tend to attain an equal distribution in terms of concentration through all parts of a system. The osmotic move­ment of water is in response to the same general principles equal distribution is, however, seldom attained in living organisms.

    Semipermeable membranes and the process of osmosis play the primary role in the absorption of water by vegetable cells. In studying the phenomena and the laws govern­ing them, consideration of natural vegetable cells may be deferred for a while, and the study of the operation of the osmotic phenomena and the effects produced in an artificial cell with an artificially prepared synthetic membrane containing an osmotically active substance, might give us a clearer understanding of the process that actually takes place in a living cell.

    An artificial osmotic chamber with a true semipermeable membrane can be prepared easily by taking a clay cell of galvanic elements coated on the inside with copper ferrocyanide. Copper ferrocyanide is prepared by mixing CuSO4 with potassium ferrocyanide.

    Another type of artificial semipermeable membrane can be obtained by mixing a solution of gelatine with tannic acid.

    The naturally occurring semipermeable membranes include bladders of fishes and animals, the thin white membrane in eggs when the hard outer calcium coatings are removed by dissolving in acids and the most important of all, the plasma membranes of living vacuolated mature cells of the vegetable tissue. The plasma membrane is the semipermeable membrane in plant cells.

    It is the presence of this differentially permeable membrane which gives to osmosis its distinctive aspect as compared with other diffusion processes.

    How can you distinguish between a living and a non-living cell? Sometimes, the streaming movements of the protoplasm show us that a cell is living. Another method is by staining a cell with a dye such as neutral red or methylene blue.

    These dyes only stain the cell contents in the vacuole and may accumulate there. Non-living cells, or living cells when killed never show this type of staining. A third characteristic of the living cells is that their cell contents can be shrunk or plasmolysed away from the cell wall when placed in a solution whose concentration is greater than that of the cell contents.

    The cell contents thus can be separated from its surrounding cell wall. Obviously then, even the living plant cell consists of a non-living part (cell wall) and a living part (protoplasm). The non-living cell wall, however, has pronounced physiological signi­ficance since it markedly affects the protoplasm, particularly its water content and growth.

    The most important part of a cell is this living protoplasm for here are synthesised and broken down all the organic chemical substances from which life itself is regene­rated. Energy is absorbed in synthesis and released in the breakdown.

    We are concerned here with the activities of living cells which certainly depend upon the properties of the living protoplasm. The nature of the protoplasm can, how­ever, be only properly understood if the chemistry and the physics of the protoplasm are known to us.

    A chemical analysis of protoplasm obtained from some lower plants such as slime molds, shows that as percentage of the fresh weight, water content varies from 80- 90%, proteins 7-10%, lipid substances 1-2%, other organic and inorganic materials, just over 1%.

    The analysis is gross, and it certainly does not mean that new protoplasm could be formed just by mixing the above substances in the proportion indicated above. It only gives us some insight into the essential physical and chemical properties of proto­plasm.

    Among the many kinds of proteins in the protoplasm are the nucleoproteins. The nucleoproteins are perhaps the most fundamental substance of life as indicated by the fact that the smallest living entity (?) virus, consists almost entirely of nucleoproteins which are the bearers of heredity in all organisms—plants or animals.

    The phospholipids (e.g., lecithin) are the main fatty substances of the protoplasm being mostly confined to the structural components of protoplasm, e.g., plastids, mito­chondria, etc. They are also associated with the cell membrane.

    All these structural components are suspended in the ground substance of the protoplasm which is made up predominantly of proteinaceous sol or gel. The ground substance is usually considerably viscous than the cell sap of the vacuoles and since viscosity and gel formations as we know are characteristics of a colloidal system, the protoplasm is obviously a colloid.

    The non-living part of the cell wall also shows colloidal properties. The vacuolar cell sap also possess enough colloidal materials, e.g., tannins which hold dyes such as methylene blue. Optically, the cell sap, however, appears as homogeneous. The cell walls are made of colloidal gels of cellulose and other organic substances, e.g., pectin.

    All the components of a living cell, the cell wall, vacuole (both non-living) and protoplasm contain large amounts of water which pass readily from the living to the non-living parts and vice versa.

    The planes separating the three components of a living cell—the protoplasm, cell wall and the vacuole are known as interfaces. At each interface, special, physical forces such as surface tension, operate freely. The substances near the interfaces tend to be concentrated sometimes producing the phenomenon of adsorption.

    Water movement across the cell membrane has usually been expressed in terms of diffusion pressures and their differences and osmotic pressure of solutions. However, it is more reasonable to explain this phenomenon in thermodynamic terms, since movement takes place only when energy is available in some form.

    All molecules at any temperature above absolute zero have a free energy with which it has a tendency to move about in the surrounding space. As the number of molecules per unit space increases, the total free energy also increases the free energy of Avo-gadro’s number (6 . 03 x 10 23 ) of a chemical species present in 1 gram mole is termed the chemical potential of that species—whether in the molecular or in the ionic form.

    In solutions both solute and solvent particles have chemical potentials. When we talk of chemical potential of water, we refer to it simply as water potential, usually designated by the Greek letter ¥ (psi).

    When two solutions are separated by a permeable membrane both solute and solvent molecules diffuse across the membrane until the chemical potentials of both on two sides of the membrane are equal. Movement is always from a higher to a lower potential.

    There is diffusion even after equilibrium is attained, but the rate of diffusion in both directions is equal. When the membrane is semi­permeable, movement of only solute or solvent particles is permitted by the membrane. In the case of plant membranes only the solvent water molecules are permitted to pass.

    Water molecules in pure water have the maximum chemical potential however, when solute molecules are dissolved in it, this potential is decreased, the greater the number of such solute molecules, that is greater the concentration, less is the water potential.

    As in the case of the Celsius (centigrade) scale of temperature measurement, where the freezing point of water is arbitrarily fixed to be zero, water potential of pure water is arbitrarily taken to be zero. Consequently the water potential- of any solution containing solute particles is negative.

    The water potential of a 0 . 1 M solution is higher than a 0 . 2 M solution, because it is more dilute and its water potential is less negative. Water potential is usually expressed in terms of pressure, i.e., bars on atmospheres. Increase in temperature or pressure increases water potential. Water potential is also affected by substances in the environment which absorbs water.

    Movement of water across membranes practically impermeable to solute particles’ is called osmosis. Osmotic pressure of a solution is a function of its solute concentration at a particular temperature according to the equation

    where π is the osmotic pressure, C the molar concentration (since pressure is the result of the total activity of the number of molecules), R is the Gas constant (0 . 082 litre atmospheres/mole degree or 0 . 0357 litre cal. per degree centigrade) and T is the absolute temperature. It is usually expressed in litre atmospheres.

    Although osmotic pressure is a positive quantity, osmotic potential is always negative, since it is the value by which water potential is lowered when the solute is dissolved in pure water.

    Two major components of water potential of a solution are pressure potential and osmotic potential. At a particular temperature water potential is the sum of pressure potential (ψр) and osmotic potential (ψπ) Pressure potential is the effect of atmos­pheric pressure or any other pressure applied.

    Thus, pressure potential is the difference of water potential and osmotic potential. It may be positive or negative accordingly as pressure is applied or the cells are in a state of tension. Plant cells also contain many hydrophilic, colloidal gel-like and porous surfaces, e.g., the cell wall, which absorb and retain water with some tenacity the water potential of this matrix is referred to as matrix potential, ψm.

    If the colloidal matrix and the cell solution are two distinct phases in equilibrium with each other, then osmotic potential of the cell solution may well be equal and either value added to the pressure potential should constitute the water potential of the system. However, several workers consider them as separate entities and the total effect is believed to be additive according to the equation

    Water movement in plant cells is along a water potential gradient. Although it is primarily a diffusion process, the rates observed are much higher than what may be provided by a pure diffusion process.

    There is evidence that there is a bulk move­ment of water. Presumably, the gradient in the transit region of the membrane surface is quite steep, so that the rate is considerably accelerated and water molecules are pulled in a bulk, somewhat as in the case of a siphon.

    In most earlier text books water potential was considered as diffusion pressure of water and the difference in the water potentials of the external solution and cell solu­tion as the diffusion pressure deficit or the suction pressure.

    We shall now explain the process of osmosis particularly with reference to an osmometer using some of this terminology to familiarise the student with these concepts.

    If an osmometer, as the artificial cell mentioned before is called (or simply osmotic chamber), containing a strong solution of sucrose is immersed in a vessel containing pure water and a mercury manometer is attached to it as shown in Fig. 668 , the rise of column of mercury in the manometer tube due to inward diffusion of water will denote the measure of the pressure of the solution inside the cell.The artificial membrane is here rigid and as this membrane is permeable only to water, the inwardly directed diffusion of water from outside into the cell will cause a pressure to develop inside the membrane. The maximum pressure thus developed is quantitatively equal to the pressure on the solution to prevent any increase in its volume due to entrance of water.

    The inflow of water from outside (endosmosis) due to higher water potential there, causes a hydrostatic pressure to develop inside the cell. This hydrostatic pressure is directed against the cell wall and this actual pressure developed within the cell as a result of osmosis, is known as turgour pressure.

    Actual pressure or turgour pressure develop­ed during osmosis is seldom equal to osmotic pressure. If a solution, having an osmotic pressure of 10 atm. be immersed in a solution, having an osmotic pressure of 6 atm., water will diffuse inwards until at equilibrium, i.e., when the water potentials of the external and internal solutions are equal, the actual pressure developed, can at its maximum be only 4 atm.

    Even if the external liquid be pure water, the actual pressure or the turgour pressure developed in the internal solution would still not be quite equal to its original osmotic pressure unless the membrane is completely inelastic. Thus in an osmometer, the maximum turgour pressure developed can only equal the osmotic pressure but can never exceed it.

    When a solution is confined within a semipermeable membrane and the membrane is immersed in water, the passage of water through the membrane results in the deve­lopment of turgour pressure on the solution side of the membrane.

    But the maximum potential osmotic pressure which can develop in the solution must be equal to the excess of diffusion pressure of pure solvent over the diffusion pressure of water in the solution, but only when the solution is not under any pressure except, of course, atmospheric pressure.

    In plant cells, other pressures must be taken into account in order to determine the suction pressure or diffusion pressure deficit (DPD) of the cell sap. Suction pressure is water absorbing capacity of the solution in the membrane.

    For example, if a solution has an osmotic pressure of 10 atmospheres, it certainly means that the diffusion pressure of water in the solution is less than the diffusion pressure of pure water by 10 atm. or in other words, the suction pressure of such a solution is 10 atm.

    Hence if a cell with cell sap solution of 10 atm. (suction pressure of also 10 atm.) is immersed in a solution of 4 atm. (suction pressure also 4 atm.) and is separated by a membrane permeable only to water, water will diffuse towards the region of its lesser diffusion pressure or greater diffusion pressure deficit (suction pressure), i.e., towards the solution whose osmotic pressure and suction pressure are both 10 atm.

    If we dis­regard dilution for the present and the volume changes, the solution inside the mem­brane will exert a turgour pressure of 6 atm. and this will be the actual hydrostatic pressure developed within the cell.

    As each action has an equal and opposite reaction, imposition of a pressure on the wall of the membrane will result in an out wardly directed wall pressure of 6 atm. The wall pressure evidently is always equal in magnitude to turgour pressure but acts in the opposite direction.

    The wall pressure will necessarily increase the diffusion pressure of the enclosed solution of 10 atm. by 6 atm. Since the initial diffusion pressure deficit of the enclosed solution was also 10 atm., the development of a wall pressure of 6 atm., raises the diffu­sion pressure of water in the internal solution by 6 atm.

    While at the same time reduces the diffusion pressure deficit of water in the internal solution to 4 atm. (10—6 = 4) which is the suction pressure or diffusion pressure deficit in the external solution.

    Osmotic pressure is proportional to the number of particles of solutes per given volume of water under standard conditions of temperature and atmospheric pressure and irrespective of kind, weight or size of the particles.

    If one membrane contains a solution with twice as many particles—molecules or ions—as that in the second, the first, evidently will have twice as much osmotic pressure as the second.

    A solution, such as found inside the vacuoles of a mature plant cell containing a mixture of many solute molecules (or ions) will have the same osmotic pressure as a solution with a single salt if the ratio of the total number of dissolved molecules and ions to the number of water molecules is the same in each case.

    In order to determine this proportional number of particles it is necessary to know the proportional weights of the particles. Since molecular weights give the proportional weights of the molecules, it is customary to use molecular solutions.

    Molar solutions are solutions in which as many grams of solutes as its molecular weight are added to water to a final volume of 1,000 ml. Sucrose with an empirical formula of C12H22O11 has a molecular weight of 342 g the molecular weight of glucose is 180 g, etc.

    These amounts dissolved in 1,000 ml of water will give molar solutions (M) of sucrose and glucose respectively. A0.1 M solution of glucose will develop the same osmotic pressure as 0 . 1 M solution of sucrose as these two solutions will contain the same number of particles of molecules. Equimolecular solutions of non- electrolytes are thus iso-osmotic or isotonic.

    A colloidal system should also possess an osmotic pressure, for osmotic pressure is determined by the proportion of the solute particles to solvent molecules and is independent of the kind and the size of particles.

    The particles of the dispersed phase of a colloidal system theoretically then will have the same effect on the magnitude of the osmotic pressure as smaller molecules or ions. Large molecules or molecular aggregates theoretically have the same effect as smaller molecules or ions. Thus osmotic pressure, in a colloidal system, will depend upon the concentration of the particles of the dispersed phase.

    As the total concentrations of the particles of the dispersed phase are relatively much smaller compared to particles in a true solution, all colloidal systems have very low osmotic pressure values, seldom exceeding a fraction of an atmosphere.

    Sugars and complex organic substances are, as we know, non-electrolytes whereas the inorganic salts are electrically active, i.e., certain percentage of inorganic salts is dissociated when the salts are in solution. This increases the number of particles in a particular volume of water.

    At a concentration 0.1 M NaCl, 80% of NaCl molecules are dissociated and therefore for every 100 NaCl molecules, 20 molecules will remain undissociated whereas 80 molecules are ionised into equal number of Na+ and Cl ions.

    There will, therefore, be 180 particles either as molecules or ions, instead of 100 particles and the osmotic pressure would be 1.8 times that of a substance of the same molar concentration with only undissociated molecules. A 0 . 1 M solution of NaCl, therefore, will have an osmotic pressure approximately equivalent to 0.1 X 1.8 or 0.18 M solution of cane sugar or glucose or any other non-electrolyte.

    The number which expresses by how much the osmotic concentration of an electrolyte is greater than an equimolecular solution of a non-electrolyte, is termed isotonic coefficient for that parti­cular concentration for 0.1 M NaCl, as we have seen, it is 1 .8.

    The conditions affecting osmotic pressure are very similar to those which affect gas pressure and within certain limits are the same. General laws of vapour pressure also hold good for osmosis.

    According to Avogadro’s hypothesis:

    Equal volumes of all gases under identical condition of temperature and pressure, contain same number of mole­cules or particles. One molar weight of any gas at standard conditions (0°C. and 1 atm.) occupies 22.4 litres. If this gas be compressed to a volume of 1 litre, it will certainly exert a pressure of 22.4 atm. at 0°C. (From Boyle’s law) The osmotic pressure of a solution varies directly with concentration of the solution and inversely with the volume.

    This theoretical value of 22.4 atm. sometimes slightly differs from actually measured pressure due to various reasons—24.83 atm. instead of 22.4. The theoretical value 22𔈂 atm. can be corrected for any desired temperature and pressure. (At 25°C., the osmotic pressure of 1 M solution is aproximately 27 atm.)

    We shall again consider the case of an artificial rigid osmotic chamber with a membrane, permeable only to water (semipermeable membrane) and a long vertical glass tube of very small diameter attached at the open end as in Fig. 664.

    Theoretically such a membrane, either inorganic or organic, should allow only the solvent mole­cules to pass through it, but as a matter of fact all artificial inorganic membranes— the same is true for organic as well—allow certain solute molecules to pass through them when certain conditions are fulfilled.

    The osmotic chamber containing a 0.45 M solution of sucrose is immersed in pure water as in Fig. 664(A). Net movement of water inwards will take place by osmosis resulting in raising the column of water in the tube. At equilibrium, the suction pressure of water in the chamber will be zero, because of the imposition of a hydrostatic pressure (turgour pressure) of 10 atm. on the internal solution.

    When equilibrium will be reached between osmotic chamber and the sur­rounding water, the level of solution in the vertical tube will remain stationary. What will be the maximum height of solution attained in the vertical tube at that condition? At this stage, the osmotic pressure of the solution in the chamber is equal to the actual pressure developed due to incoming water, i.e., turgour, which is also equal to 10 atm. at its maximum.

    1 atmospheric pressure is equivalent to 76 cm of mercury and the specific gravity of mercury is 13 . 6. Thus the maximum height attained by the solution in the vertical tube attached to the osmotic chamber will be 76 x 10 x 13 . 6 cm or approximately 103 m. (We disregard here the dilution of the solution in the chamber and the changes in the volume.)

    If the osmotic chamber is now taken out of water and then immersed in a solution of 0 . 25 M concentration as in Fig. 664(B), the level of column of water in the tube will certainly drop until the actual pressure developed on the membrane and the wall of the chamber is determined by the difference in concentration of the solutions inside and outside the chamber, i.e., approximately 4.48 atm. (22 . 4 x 0 . 2 =4 . 48 atm.).

    The actual pressure developed (turgour) within the chamber can thus only attain a value of 4.48 atm. The suction pressure of the water in the solution inside the chamber (which was zero in pure water) will increase for it will now be subjected to a pressure of only 4.48 atm.

    Imposition of a turgour pressure of only 4-48 atm. now will make the suction pressure equal to 10—4 . 48, i.e., 5 . 52 atm., equivalent to 0 . 25 M, which is exactly the concentration of the external solution. Thus the water level in the tube will be stationary at a height of 4 . 48 x 76 X 13 . 6 =46 m (this height is due to the actual hydrostatic pressure or turgour developed within the chamber, keeping the water column at that height).

    If the chamber with its attached tube is again removed and transferred to an external solution of 0 . 15 M concentration, water will again start entering the chamber by endos- mosis and the turgour pressure will further rise and at its maximum will now be equal to 0 . 30 M (or 6 . 72 atm.).

    The level of water in the tube will show a further rise until it reaches 6.72 X 76 x 13.6=69 m. Imposition of a turgour pressure of 6 . 72 atm. will College botany reduce the suction pressure of the solution inside to 3 . 28 atm. which is exactly the suction pressure of water in the external solution (0 . 15 M=3 . 28 atm.).

    If the chamber is now finally transferred to an external solution, having the same concentration as that in the chamber (0.45 M), water will come out again by exosmosis until the levels of water column in the tube and in the external solution are exactly the same.

    The turgour pressure will be progressively reduced to zero, and all the hydrostatic pressure developed on the walls of the chamber is thus released due to water exit.

    And as the suction pressure of water in the solution inside the chamber is progressively released from the imposed turgour pressure, the suction pressure value progressively increases until it attains a maximum of 10 atm. (0.45 M) equal to the suction pressure of the external solution.

    There is no net movement of water into the chamber when the chamber is thus brought into equilibrium with external solution—movement of water molecules certainly there is, but the rates of exosmosis and endosmosis are equal.

    Dialysis:

    Dialysis is essentially a sort of filtration in which the pore size of the membrane permits the diffusion of smaller particles from inside outside and vice versa, but larger molecules are retained.

    Thus coenzymes can be separated from apoenzymes by dialysis. If the enzyme solution is taken inside a dialysis bag (collodion or other similar membranous structures) and the contents equilibrated with a buffer solution or water the coenzyme molecules and metal ions which are small, pass out of the long but the larger protein molecules are retained within the bag.

    Several changes of the outside solution are required, so that the diffusion of the smaller particles is complete or almost complete. If the outside solution is not changed, after sometime equilibrium is attained and the small particles are present both inside and Outside the Dialysis Bag or Tubing. Kidney Of Man And Animals Is An Efficient Dialyser.


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