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UNIT 5:TRANSPORT ACROSS THE CELL MEMBRANE
Key unit competence: Explain the physiological processes by which materials move in and out of cells and the significance of these processes in the life of organisms.
Introductory activity 5
Observe the figure below and answer the following questions. Different molecules are moving as you see them on this figure.
1. Given that the molecules are on both sides of the cell membrane, why K and L are happening?
2. The origin of arrow L has ATP while the origin of K has no ATP, explain why.
5.1 Types of transport of substances across the cell membrane
Activity 5.1
a. Identify all types of molecules transported across the cell membrane.
b. Discuss different types of transport of substances across the cell membrane
Different types of transport of substances across the cell membrane exist. Their main types are active transport which requires energy stored in ATP (adenosine triphosphate) and passive transport which does not require energy. The internal environment of a cell is maintained differently from that of its external environment by a thin surface membrane called the cell membrane or plasma membrane or plasmalemma. The plasma membrane is a phospholipid bilayer with phosphate heads and fatty acid tails. It governs the entry and exit of molecules and ions. This property of the plasma membrane that regulates the exchange occurring between the cell and its medium is referred to as “cell permeability”. A cell membrane, therefore, determines which substances can enter the cell (comprising those which may be important for the vital activities within the cell) and also regulates the outflow of substances (consisting of excretory waste and water).
This feature of the membrane not only maintains difference in the composition of intracellular and extracellular fluid, but also establishes a balance in their osmotic pressure. Therefore, a membrane may be permeable to some substances while impermeable to others. This is called “selective permeability”. The lipid bilayer (phospholipid bilayer) of the membrane is permeable to non-polar and uncharged molecules such as O2, CO2 and steroids but is impermeable to charged or polar molecules and ions like glucose. It may be slightly permeable to water and urea though being polar molecules due to their smaller size.
A concentration gradient refers to difference in the concentration of a substance from one region to another across a plasma membrane. In such a case, the solute will have a tendency to move from a region of high to that of low concentration.
Due to the distribution of positively and negatively charged ions, the inner surface of plasma membrane is more negatively charged than the outer. This difference in the electrical charge between two regions creates an electrical potential and since this is established across a plasma membrane, it is termed as membrane potential.
Table 5.1: Types of transport of substances across the cell membrane.
Passive transport involves the movement of molecules along the electrochemical gradient or concentration gradient without the use of ATP (Downhill transport). It occurs by diffusion or osmosis.
Active Transport drives the molecules against their electrochemical gradient by hydrolysis of ATP (Uphill transport).
Application activity 5.1
1. Use the following list of terms to complete these statements
Terms list: active transport, shrinks, diffusion, osmosis, Turgor pressure, higher, exocytosis.
Statements:
i. In hypertension solution, a cell ................................. .
ii. In lungs, CO2 diffuses out of blood by the process of .............................
iii.Reabsorption of water in kidney tubule takes place by the process of .................. .
iv. The pressure exerted by plants’ cells on cell wall is ................................
v. The larger the surface area of membrane, the ........................ is the rate of diffusion.
2. During the dry seasons, most plants wilt and finally die. Relate this drying process to the transport of substances across the cell membrane.
5.2 Diffusion and factors affecting the process of diffusion
Activity 5.2
a. Materials Required: Visking tubing with capillary, Beaker with water, Sucrose solution (10%), pieces of beetroot.
b. Illustration:
Procedure:
1. Fill the visking tubing with sucrose and starch solution.
2. Fill a beaker with water and glucose. Now, put some pieces of beetroot into the visking tubing.
3. Partly submerge the visking tubing into the beaker.
4. Observe the change in the level of liquid in the capillary tube attached to the visking tubing.
5. Observe the movement of red pigment from a region of high concentration in the vacuoles to a region of low concentration in the solution outside the pieces of beetroot.
Tasks:
1. Identify the type of this movement or transport.
2. Discuss factors affecting the process of diffusion and its significance in living organism
5.2.1 Simple and facilitated diffusion
Diffusion is the movement of molecules such as O2, CO2 from their higher to their lower concentration that is, down their concentration gradient until equilibrium is achieved and they are distributed equally.
It is included in passive transport which does not require energy.
Example: Diffusion of oxygen from air sacs (alveoli) to the blood in lung capillaries depends on the concentration of oxygen in alveoli; gaseous exchanges (O2 and CO2 ) in plant occurs also by diffusion.
Simple diffusion is the simplest mechanism in which a molecule dissolves in the phospholipid bilayer, diffuses across it and then dissolves in the aqueous solution present on the other side of the cell membrane. It neither requires ATP nor any protein. The direction of movement is determined by the concentration gradient (i.e., molecules flow from a region of higher concentration to a region of lower concentration) or electrical gradient. Therefore, any molecule that is soluble in the phospholipid layer is capable of crossing the plasma membrane. This is the reason why only small, relatively hydrophobic (water repelling) molecules (example - benzene), gases (O2 , CO2 ) and even small polar, uncharged molecules diffuse easily across the plasma membrane while other larger molecules are restricted. Diffusion is also regarded as the random mixing of particles in which the particles continue to move down their concentration gradient until an equilibrium is reached and the particles are evenly distributed throughout the solution.
Facilitated diffusion (also known as facilitated transport or passive-mediated transport) is the process of spontaneous passive transport of molecules or ions across a biological membrane via specific transmembrane integral proteins.
5.2.2 Factors affecting the rate of diffusion
1. Concentration gradient: The greater the concentration difference between the two sides of the membrane, the faster is the rate of diffusion.
2. Temperature: As the temperature increases, the rate of diffusion increases.
3. Size of diffusing molecules: Smaller molecules have faster rate of diffusion while the ones with larger mass, diffuse slowly.
4. Surface area to volume ratio: The larger the surface area of membrane available for diffusion, the higher is the rate of diffusion.
5. Thickness the membrane: The greater the distance across which diffusion is to occur, the longer it takes for molecules to pass through.
5.2.3 Significance of diffusion
Diffusion plays an important role in living systems. Below are a few examples where its diverse significance can be understood.
1. In the human body, nutrients (in the form of ions and small molecules) are absorbed from the food by the surrounding blood cells in the vessels by way of diffusion.
2. In the lungs, CO2diffuses out of blood in alveolar sacs whereas O2 (present in high concentration in the inhaled air) diffuses into the cells in the blood vessels (with low O2 concentration).
3. Cutaneous respiration (through skin) is the most common mode of respiration in lower non-chordates wherein gases directly diffuse through the air into the surface epithelium of the organisms.
4. The eyes lack a great number of blood vessels (which carry oxygen) and therefore, needs an extra supply of oxygen. The atmosphere provides this extra needed oxygen, which is taken up by the eye through direct diffusion of O2 into the cornea, the hardouter covering on the eye. In absence of diffusion, the eyes would dry out.
5. For medicines taken orally as pills, the medicine must somehow find its way into the bloodstream. Once in the stomach, the medicine from the pill is absorbed into the lining of the stomach and then into the bloodstream, both by the process of diffusion.
6. Gaseous exchange during the process of respiration and photosynthesis takes place with the help of diffusion.
7. Transpiration or loss of water from the aerial parts of the plants involves the process of diffusion.
8. Diffusion is involved in passive uptake of mineral salts.
9. Odour molecules of the flowers to attract the pollinating animals, spreads in the air by diffusion.
10. Diffusion plays an important role in imbibition (special type of diffusion involving the absorption of water molecules by solids such as absorption of water by wood or seeds) and osmosis.
Application activity 5.2
1. Choose from term listed below, use them to complete the following statements:
List of terms: diffusion, osmosis, facilitated diffusion, imbibition.
Statements:
i. Transpiration or loss of water from the aerial parts of the plants involves the process of .......
ii. Diffusion which involves the carrier protein is called....
2. Use the following data.
a. Plot those data on a graph placing the thickness of the membranes on X- axis and rates of diffusion on Y- axis.
b. Interpret your graph.
c. Apart from distance, identify other possible factors affecting diffusion.
5.3 Process of osmosis
Activity 5.3
Aim: To investigate the process of osmosis.
Materials: Irish or sweet potatoes, knife, water, moist salt or moist sugar, Petri dish or beaker.
Procedure
i. Cut a potato into 2 pieces using a knife.
ii. Use a knife to shape each piece in a way allowing it to be stable in a Petri dishes.
iii. Label Petri dish A and B.
iv. Make a small depression in each potato piece.
v. Fill a half of Petri dish A with distilled water.
vi. Place a moist sugar or salt in the depression of potato piece of Petri dish A.
vii. Place the potato piece A in Petri dish A.
viii. Fill a half of Petri dish B with 60 % sugar or salt solution.
ix. Place distilled water in the depression of potato piece of Petri dish B.
x. Place the potato piece B in Petri dish B.
xi. Observe during at least 30 minutes.
Tasks
1. Identify the investigated process.
2. Interpret your observations.
3. Discuss factors affecting this process.
4. Discuss the significance of the process in living organisms.
Osmosis is the movement of water molecules from molecules from hypotonic solution (less concentrated solution having many molecules of water and few molecules of solutes ) to hypertonic solution ( highly concentrated solution containing few molecules of water and many molecules of solutes) through a differential or selective or semi- permeable membrane due to concentration gradient.
In osmosis, the movement of water (solvent) occurs due to the difference of chemical potential (water potential in case of water) on the two sides of the cell membrane.
The kinetic energy or free energy possessed by the molecules of a substance is called chemical potential. The chemical potential of water is called water potential. The chemical potential of pure water (solvent) is higher than that of the same water in a solution. Presence of solute particles decreases the chemical potential (free energy) of water. The lowering of chemical potential (free energy) is due to attraction and collision between solvent (water) and solute molecules. Thus, in terms of thermodynamics, ‘Osmosis is the movement of water or solvent molecules from the region of their higher chemical potential (free energy) to the region of their lower chemical potential (free energy) across a semipermeable membrane’.
When a cell is placed in a solution containing a solute (e.g., salt or sugar) dissolved in water, any of the three conditions may arise:
i. If the medium is hypotonic with respect to the cell, i.e., if it has solute concentration lower than the cell interior, water will tend to move into the cell. This may lead to swelling of the cell and even cause it to burst if it is animal cell. The plant cell is termed turgid. For example, red blood cells when placed in a hypotonic solution, there is hemolysis.
ii. If the medium is hypertonic with respect to the cell, i.e., has high solute concentration than the cell interior, then water will tend to move out of the cell into the medium. This would cause cell to shrink in size. For example, a plant cell when placed in hypertonic solution, shows plasmolysis in which the plasma membrane shrinks and becomes widely separated from the cell wall.
iii.If the medium is isotonic with respect to the cell, i.e., the solute concentration is equal to that in the cell, the net movement of water across the membrane would be zero. The cell size and concentration would remain constant. The cell is termed flaccid. For example, 0.9% solution of NaCl is nearly isotonic to blood serum.
Difference between turgidity, flaccidity and plasmolysis
Food preservation by salting using hypertonic solutions
When a cell is placed in a hypertonic solution, water actually flows out of the cell into the surrounding solution thereby causing the cells to shrink and lose its turgidity. Hypertonic solutions are used for antimicrobial control.Salt and sugar are used to create hypertonic environment for microorganisms and are commonly used as food preservatives.
“Salting is the preservation of food with dry edible salt. It is related to pickling (preparing food with brine, i.e., salty water). It is one of the oldest methods of preserving food. Salting is used because most bacteria, fungi and other potentially pathogenic organisms cannot survive in a highly salty environment, due to the hypertonic nature of salt. Any living cell in such an environment will become dehydrated through osmosis and die or become temporarily inactivated.”
Salting methods
Cut your vegetables up in pieces before you put them into the saltwater to preserve food by salt-curing. As you chop a vegetable and put it into the saltwater, it makes its own juice. Nowadays, you might want to use a smaller container. Just make sure the water has plenty of salt added. Let the vegetables stand in the saltwater for at least 10 days in order to “pickle.” Pickle simply means preserve in brine. Then cover tightly with a lid. Preserve meats by saltcuring. Rub meat completely with salt pellets and allow it to cure for 4 to 8 weeks. At the end of this time, the meat will be almost dry. It can be stored this way for a long time. This method is called “dry-curing.”
Soak meat in a solution of brine for a period of 3 to 4 weeks. It will be ready to eat, but it won’t last long this way. You can also use a syringe to inject brine into the muscles of the meat in order to preserve the food by salt-curing. It will be ready to eat in 2 to 3 weeks. Just remember that these wet methods of saltcuring meat do not preserve it as long as the dry method does.
Osmosis in animal cells
When there is more water outside an animal cell than inside or animal cell is kept in hypertonic solution, more water particles will enter the cell than leave. This will lead to swelling of the cytoplasm and push it outwards. Consequently, the cell membrane will stretch and finally the cell will burst. On the other hand, when there is less water outside the cell (hypotonic solution) in comparison to inside, more water molecules move out of the cell and finally cell will shrink. Therefore, both the conditions are harmful for animal cells, so, the water concentration surrounding the animal cell must be kept constant for the cells to carry out their functions normally.
Process of osmosis in plant cells Unlike the animal cells, the plant cells do not have the ability to osmoregulate the water that enters the cells. Therefore, the water tends to move into the cells continuously due to the water potential. Water Potential is a key concept for understanding the movement of water, i.e., the plant-water relation.
Pure water has the highest water potential which at room temperature in absence of any pressure is zero. If solutes are added to water, its water potential decreases because the number of water molecules with kinetic energy would tend to be low. This magnitude of decline in water potential due to presence of solutes is referred to as the solute potential.
The continuous uptake of water by the plant cells causes the cells to swell to the limit when the hydrostatic pressure within the cell prevents any more water to get in. This pressure is known as Osmotic pressure and the cells are said to be turgid. One of the critical functions of plant cell walls is thus to prevent cell swelling as a result of this osmotic pressure. In contrast to animal cells, plant cells do not maintain an osmotic balance between their cytosol and extracellular fluids. Consequently, osmotic pressure continually drives the flow of water into the cell. This water influx is tolerated by plant cells because their rigid cell walls prevent swelling and bursting. Instead, an internal hydrostatic pressure called Turgor pressure builds up within the cell, eventually equalizing the osmotic pressure and preventing the further influx of water.
Turgor pressure is responsible for much of the rigidity of plant tissues. In addition, turgor pressure provides the basis for a form of cell growth that is unique to plants. In particular, plant cells frequently expand by taking up water without synthesizing new cytoplasmic components. Cell expansion by this mechanism is signaled by plant hormones (auxins) that activate certain proteins which allow turgor pressure to drive the expansion of the cell in a particular direction. As this occurs, the water that flows into the cell accumulates within a large central vacuole, so the cell expands without increasing the volume of its cytosol. Such expansion can result in a ten to hundred fold increase in the size of plant cells during development. The pressure exerted on the contents of a plant cell by the cell wall that is equal in force and opposite in direction to the turgor pressure is known as wall pressure.
Adaptations of plants and animals to salty conditions
Plants in salty areas take up more salt from the soil resulting in an increase in salt concentration in the cells and thus maintaining a water potential that is more negative than that of the soil.
The difference in osmotic potential between plant cells and soil water leads to the movement of water into the cells through the cell membrane via osmosis. Water is evaporated from the leaves. This also helps the movement of water from the roots up the stem to the leaves. Some plants restrict the opening of stomata to conserve their water in salty conditions and some turn down leaves to decrease surface area of evaporation. Plants have glands to store salt which burst when concentration of salt increases and causes the release of salt to the soil again. Some plants regulate salt levels by transporting sodium and chloride ions into the central vacuole. High salt concentration in the vacuole causes more water uptake and swelling.
Some plants avoid salt stress by releasing leaves in which excess sodium chloride accumulates in petioles.
Animals adapt to the salty conditions very well as plants. For example, fishes in salt water intake a lot of water and reduce the loss of water by excreting less amount of urine by having a kidney with relatively few small glomeruli. Fishes also have chloride secretory cells on their gills which actively transport salts from the blood to the surroundings. Salt glands are also found in other animals inhabiting salty conditions. Therefore, specially developed kidneys, gills, and body functions help equalizing salt concentrations across membranes through osmosis.
Significance of osmosis
Listed below are a few examples that illustrate the importance of osmosis:
1. Osmosis is of prime importance in living organism, where it influences the distribution of nutrients and the release of metabolic waste products. Living cells of both plants and animals are enclosed by a partially permeable membrane called the cell membrane, which regulates the flow of liquids and of dissolved solids and gases into and out of the cell.
2. It helps maintain the pressure on either side of the cell membrane thereby preventing the cells to become turgid and burst or to become flaccid.
3. Plant roots absorb water and minerals from soil and take it upwards to the leaves and other plant parts which are essential for plant growth.
4. Purification of blood by kidneys also involves osmosis. Osmosis maintains the balance of inter- and intracellular fluids. 5. Reverse osmosis is used to purify water.
6. Plants wilt when watered with saltwater or provided too much of fertilizer as this makes the soil hypertonic than the plant roots and disrupts water uptake.
But osmosis may also be harmful, especially, in case of marine and freshwater fishes which have to constantly regulate the movement of water out or into their body (called osmoregulation).
Application activity 5.3
I. Associate the column A and B
5.4 Water potential, osmotic potential, wall pressure
Activity 5.4
Aim: To investigate and present the effects of immersing plant tissue in solutions of different water potentials and using the results to estimate the water potential of tissue.
Materials required: Potatoes (plant tissue), Cork borer, Measurement scale, Knife/Blade, Boiling tubes, Aluminium foil, Graph paper, Sucrose solution (0.0M, 0.2M, 0.4M, 0.6M, 0.8M and 1M), Water.
Procedure:
1. Using a cork borer, cut cylinders of potato tissue.
2. Slice the cylinders into 4 cm length each. This is the initial length (Li).
3. Take sucrose solutions of different concentration in boiling tubes and label them.
4. Immediately add 3 potato tissue cylinders in each boiling tube. Seal the tube with an aluminium foil paper to prevent water loss by evaporation.
5. Leave the tube rack aside for 1h.
6. Measure the length of the cylinders in each tube. This is the final length (Lf).
7. Calculate the % change in length for each using the formula: % change in length = [(Lf– Li)/ Li] × 100.
8. Calculate the average of the three readings obtained for each tube.
9. Plot a graph of mean % change in length versus the sucrose concentration used.
10. Draw the best fit line for all the points obtained.
11. Using the graph, determine the sucrose concentration at which the tissue showed no change in its length. This is the water potential of the potato tissue used (in terms of molarity).
Questions: Discuss the following questions after observing the results drawn.
1. What happened to the potato tissue cylinders? Did they swell or shrink?
2. Which process do you think brought the change?
Water potential (Ψ) is a measure of the tendency or capacity of water molecules to move from one place to another. In other words, it is the potential energy of water
The Ψ of pure water is 0 and addition of solutes makes Ψ negative (-). So, Ψ is generally comprised between 0 and except in glomerulus of kidney nephron where Ψ is greater than 0.
Units of measuring water potentials
Water potential is measured in kPa (KiloPascal). Confusingly, pure water – which has the highest water potential – has a value of 0 kPa Dissolving solutes in water reduces water potential. So, water potential values go down from 0 to negative figures. The larger the negative figure the more solute dissolved and the lower the water potential. Below, there is a ladder showing water potential, free water molecules and kPa values.
Table 5.2: Decrease of water potential as solute concentration increases.
Osmotic (solute) potential: ΨS
Osmotic (solute) potential is the potential of a solution to cause water movement into the cell across a partially permeable membrane as a result of dissolved solutes. The amount that the solute molecules lower the Ψ shows the solute potential.
ΨS: is always negative
Pressure potential: ΨP.
Pressure potential is the capacity of pressure to contribute to water movement. The more the pressure inside the cell, the more the tendency will be for water to leave it
ΨP is generally positive except in plant xylem in which there is a tension (negative pressure) and so negative ΨP
Calculation of water potential
Ψ = ΨS + ΨP
Where:
Ψc or Ψ: water potential of the cell
ΨS: osmotic (solute) potential
ΨP: ¬pressure potential
Wall pressure is the pressure that the plant cell wall exerts against the internal cell content. This pressure prevents the bursting of plant cells placed in hypotonic solutions.
Application activity 5.4
1. If 10 grams of salts are added pure water, does its water potential increase or decreases? Justify your answer.
2. If the solute potential of beetroot cells is -1400 kPa and their water potential is -950 kPa, what is their mean pressure potential?
Show the formula and calculations.
5.5 Process of active transport
Activity 5.5
Aim: To interpret data on movement of solvents and ions in and out of the cell in the given graph.
Materials required: Given data and the plotted graph.
Background experiment of the given graph: Using a cork borer, cylinders of potato tissue were cut. The cylinders were sliced into 5 cm length each and weighed. This is the initial mass (Mi). Different concentrations of salt solutions (0%, 0.2%, 0.4%, 0.6%, 0.8% and 1% NaCl) were taken in different boiling tubes. One potato tissue cylinder was added to each boiling tube. The tube was sealed and left aside for 24h. Next day, the weight of each cylinder was measured to obtain the final mass (Mf). The change in weight was then calculated by subtracting Mi from Mf. When the data were plotted on a graph paper, it gave the below shown result.
Procedure
1. Read and understand the background experimental details and the graph thoroughly.
2. Based on your understanding, interpret the result in terms of answering the below mentioned questions:
Question 1: What pattern do you observe with respect to some potatoes gaining water and some losing water? Why?
Question 2: Which of the salt concentrations are hypotonic and hypertonic with respect to potato tissue?
Active transport is the movement of ions or molecules from a region of lower concentration to higher concentration across the plasma membrane (Uphill transport). For this, the energy is provided either by another coupled reaction or by direct hydrolysis of ATP.
5.5.1 Process of active transport
i. Primary active transport: It involves direct hydrolysis of ATP. Example includes ion pumps, for example, Na+ –K+ pump (Na+ –K+ ATPase), responsible for maintaining gradients of ions across the plasma membrane; Ca2+ ions are actively transported across the plasma membrane with the help of Ca2+ pump which is powered by ATP hydrolysis, and; H+ ions are actively transported out of the cells by ion pumps in plasma membranes of bacteria, yeasts and plant cells.
ii. Secondary active transport (Active Transport Driven by Ion Gradients): Molecules are transported against the concentration gradient not using energy derived directly from ATP hydrolysis but coupled with the movement of second molecule in an energetically favourable direction, i.e., from higher concentration to lower concentration. For example, glucose is transported with the coupled transport of Na+ ions. Na+ gradient is responsible for transport of glucose against concentration gradient from the intestinal lumen to the cell.
The concentration of Na+ is more outside than inside while that of K+
ions is more inside. Steps involved:
1. 3 Na+ ions bind to the pump facing the cytoplasm,
2. Binding of Na+ promotes ATP hydrolysis and thus phosphorylation of pump,
3. Conformational change in transporter causes it to face outwards and low binding affinity for Na+, so 3 Na+ released outside,
4. High binding affinity for K+, so 2 K+ ions bind to pump,
5. Binding of K+ promotes dephosphorylation and therefore, conformational change in pump,
6. Low affinity for K+, so 2 K+ ions are released into the cytoplasm.
The transporter simultaneously binds to one molecule of glucose and two ions of Na+ . Energetically favorable movement of Na+ drives the uptake of glucose against its concentration gradient.
The coordinated uptake of molecules may be symport, uniport and antiport.
a. Symport: When two molecules transport in the same direction, e.g., coordinated uptake of glucose and Na+
b. Uniport: Transport of only a single molecule, e.g., diffusion of glucose.
c. Antiport: When two molecules are transported in the opposite direction, e.g., Na+ – Ca2+ antiporter transports Na+ into the cell and Ca2+ out. Another example is Na+ –H+ , which transports Na+ into the cell with the export of H+ , thereby removing excess of H+ and preventing acidification of cell cytoplasm.
A (1 and 2) - Symport of 1 molecule of glucose with 2 molecules of Na+ ions by secondary active transport. Note that the two molecules are transported in the same direction. B- Antiport of 1 molecule of Na+ into the cell and 1 molecule of Ca2+ out of the cell. Note that the two molecules are transported in opposite direction
5.5.2 Factors affecting the process of active transport
Any factor that affects energy production (during cell respiration) affects the rate of active transport. Factors affecting rate of active transport include the following:
i. Oxygen concentration: Oxygen is used to break down food to release energy. When the concentration of oxygen is low, less energy is produced; hence rate of active transport is slow. The higher the oxygen concentration, the more energy is released and therefore the rate of active transport is increased.
ii. Temperature: Increase in temperature up to optimum levels increases rate of chemical reactions that release energy in the cell. Increase in energy enhances the rate of active transport. Temperatures above optimum levels denature enzymes that speed up chemical reactions. This results in low energy production and therefore rate of active transport is slowed down. Low temperatures inactivate enzymes hence less energy is produced. This slows down active transport.
iii. Enzyme inhibitors: Enzyme inhibitors are substances that slow down the rate of enzyme activity. Presence of enzyme inhibitors slows down the rate of active transport. These block the enzyme active sites which makes it hard for the enzymes to bind and react with the substrates.
iv. Cofactors and coenzymes: Cofactors and coenzymes are substances that activate enzymes. Their presence increases the rate of chemical reactions leading to more energy production. This increases rate of active transport. A cofactor is nonprotein adjunct required by an enzyeme in order to function;many cofactors are metal ions (such as Mg2+,Magnesium ions), others are coenzymes. A coenzyme is a nonprotein organic molecule that aids the action of the enzyme to which it is loosely bound.Examples of coenzymes are Nicotinamide adenine dinucleotide (NAD), Flavine adenine dinucleotide (FAD).
v. pH : This is the acidity or alkalinity of a solution. Some enzymes function best in acidic, alkaline or neutral pH. If the pH of a chemical reaction is altered, enzyme activity will be slowed down or stopped. This will slow down or stop energy production. Consequently, active transport will be slowed down or may stop.
It is known that active transport is carried out with the help of pumps. There are two factors which importantly affect the active transport, including the rate of transport by individual active transporters and the number of active transporters present in the membrane or in another term the surface area.
Furthermore, the rate of transport by individual transporter in turn depends upon the nature of transporter, electrochemical gradient or electrochemical driving force on either side of the membrane, and the conditions under which a transporter must operate.
5.5.3 Significance of active transport in organisms
i. In the intestinal lining, glucose is absorbed by active transport from a lower concentration to a higher concentration in the cells lining the intestine.
ii. Na+ and K+ gradients established by the Na+ –K+pump is required for the propagation of electric signals in nerves and muscles.
iii. Ca2+ ions are actively transported by Ca2+ pump which is required for muscle contraction.
i. H+ ions are actively pumped out of the cell lining the stomach which results in the acidity of gastric fluids which help in the digestion. H+ ions are actively transported into the endosomes and lysosomes with the help of pumps. Active transport is also important for the transport of nutrients, including ions, sugars, amino acids into the cells and transport of toxic substances out of the cell (e.g., ABC transporters in bacteria and eukaryotic cells).
Application activity 5.5
The following data is provided.
Concentration of oxygen in a cell (mol /L) Rates of active transport (molecules/minute)
10 10
12 11
13 12
14 13
a. Plot those data on a graph placing the thickness of the membranes on abscissa axis and rates of active transport on ordinates axis.
b. Interpret your graph.
c. Suggest other possible factors affecting active transport.
5.6 Endocytosis and exocytosis
Activity 5.6
Discuss on the process of endocytosis and exocytosis and draw the diagram showing the process of endocytosis and exocytosis. Present your findings to your classmates.
5.6.1 Endocytosis
Christian de Duve (1963) coined the term “endocytosis” which is responsible for ingestion of large particles (such as bacteria), macromolecules and fluids into the cell in the form of small vesicles. Unlike all the above mentioned processes involved in transport molecules, endocytosis is the only means by which large molecules or particles can be taken up by the cell, especially in eukaryotes. It is further categorized into:
Phagocytosis, also called “cell eating”. It involves ingestion of bacteria, cell debris or even intact cells.
Pinocytosis, also called “cell drinking”. It involves uptake of fluids by the cell.
5.6.1 Phagocytosis
This serves as a means of food capturing by bacteria and many protozoans while in eukaryotic cells it serves as a defense mechanism to fight against harmful 167 microorganisms and even to get rid of the cells that have stopped functioning normally or are aged. In mammals (such as man), macrophages (of spleen and liver) and neutrophils are key components of the immune system that show phagocytosis and are therefore also called “professional phagocytes”.
5.6.2 PinocytosisIt is also called “fluid endocytosis” and is used primarily for the uptake of extracellular fluids. It is a non-specific process which involves engulfing of either pre-dissolved or already brokendown substances. This non-specificity in the ingested substance distinguishes it from phagocytosis which takes up specific substances from the exterior. Also, phagocytosis involves breakdown of the particle which is lacking in case of pinocytosis
5.6.3 Exocytosis
Unlike endocytosis in which macromolecules or fluids are taken into the cell, exocytosis results in secretion or release of substances out of the cell. It also involves membrane enclosed secretory vesicles which are formed within the cell and fuse with the plasma membrane to drain off all its contents into the surrounding medium.
Table 5.2: Comparison of Endocytosis and Exocytosis
Application activity 5.6
1. Use the following terms listed below to complete the following statements
Terms list: Osmosis, Pinocytosis, active transport, plasmolysis, antiport, Phagocytosis, Christian de Duve.
Statements:
i. The process of cell drinking is known as ..................... .
ii. Ca2+ ions are required for .............................. .
iii.When two molecules are transported in opposite direction, it is .................... .
iv. ............................ involves ingestion of bacteria.
v. (v) Endocytosis is discovered by ……………..
2. Draw and interpret the phagocytosis of Entamoeba histolytica by a macrophage.
Skills lab 5
Aim: Meat preservation by salting method and money generation.
Suppose that there is no refrigerator at your home or butcher. Practise the salting method studied in class to preserve your meat which will be eaten or sold for earning money.
Materials: Container such as a bucket, salt, string, knife, fish/meat, suspension metal.
Procedure
i. Using a knife, make a hole in the meat.
ii. Insert the string in those holes.
iii. Place many salt molecules on that meat.
iv. Suspend that meat on the suspension metal for its preservation.
v. Sell the preserved meat to earn money.
4. Portfolio Report:
i. Write your skills lab project implementation report focusing on how this skill lab has helped you to get money and new biological skills, submit it to your teacher.
ii. Bring a product (preserved meat /fish) and present it to the whole class.
Note: Invent or discover other skill labs related to this unit.
End unit assessment 5
I. Choose whether the given statements are True (T) or False (F)
1. Passive transport occurs by diffusion or osmosis.
2. Simple diffusion involves uphill transport of ions or molecules.
3. Osmosis is the movement of water or solvent molecules from the region of their higher chemical potential (free energy) to the region of their lower chemical potential (free energy) across a semipermeable membrane.
4. Not all transport mechanisms occurring across a cell membrane require ATP utilization.
5. Molecules or substances that are large in size are transported across the membrane by active transport.
6. In the human body, nutrients (in the form of ions and small molecules) are absorbed from the food by the surrounding blood cells in the vessels by way of osmosis.
7. Purification of blood by kidneys involves diffusion.
8. Reverse osmosis is used to purify water.
9. In the intestinal lining, glucose is absorbed by active transport from a lower concentration to a higher concentration in the cells lining the intestine.
10. Salting is one of the oldest methods of preserving food.
II. Multiple choice questions
1. ………………….. is the movement of ions or molecules from a region of lower concentration to higher concentration across the plasma membrane.
a. Active transport b. Passive transport
c. Pinocytosis d. Exocytosis
2. In the absence of ………….. eyes would dry out.
a. osmosis b diffusion
c endocytosis d exocytosis
3. Gaseous exchange during the process of respiration and photosynthesis takes place with the help of
a. osmosis b. diffusion
c. endocytosis d. exocytosis
4. Transpiration involves the process of
a. osmosis b. diffusion
c. endocytosis d. exocytosis
5. ……………………… is important for the transport of nutrients into the cells and toxic substances out of the cell.
a. Active transport b. Passive transport
c. Pinocytosis d. Exocytosis
6. For transport by simple diffusion
a. Particles should be small in size
b. Particles should be soluble in lipid
c. Both of the above
d. None of the above
7. Which of the following transport mechanisms describes the process by
which a macrophage engulfs bacteria?
a. Active transport
b. Endocytosis
c. Transcytosis
III. Long answer type questions
1. In your own words, explain the processes by which materials move in and out of cells.
2. Give four examples, showing significance of diffusion in living systems.
3. Give four examples, showing the importance of osmosis in living systems.
4. In your own words, explain the significance of Active transport in living organisms.
5. With examples, explain how can you apply the knowledge of hypertonic environments in food preservation by salting?
6. How do plants and animals adapt to salty conditions?
7. Justify the statement:
The interplay between HIV and the plasma membrane has much to offer in terms of understanding viral tropism and pathogenicity and normal cellular functions, and for developing new antiviral approaches.