Unit 2 Transport Across the Cell Membrane
Key Unit Competence
To be able to explain the physiological processes by which materials move in and out ofcells and the significance of these processes in the life of organisms.
LEARNING OBJECTIVES
At the end of this unit, the learner will be able to:
• Describe and explain the processes and significance of movement in and out of the cell
mentioned in the content.
• Recall that the increasing size of organisms is constrained by its ability to obtain resources
through diffusion across the cell surface and its ability to move substances out of cells.
• Explain the movement of water between cells and solutions with different water potentials
and explain the effects on plant and animal cells.
• Carry out an investigation on simple diffusion using plant tissues and non-living materials.
• Apply the knowledge of hypertonic environments in food preservation by salting.
• Research adaptations of plants and animals to salty habitats.
• Interpret and present data in graphic and table form on the effects of varying concentrations
of : e.g., sugar and salt on plant and animal tissues.
• Appreciate the importance of movement of substances across cells.
• Show concern when exposing living organisms to concentrated media.• Distinguish between endocytosis and exocytosis.
INTRODUCTORY ACTIVITY
A cell is like a fenced factory that produces goods. Apart from goods produced, there are waste
products that should be rid of the factory and on the other hand there is permanent need of
raw materials that should be used by the factory. Therefore, though fenced, the factory is not an
isolated system. It needs exchange with the surrounding environment.
1) In the cell context, what does mimic the fence?
2) State any two materials which should enter the cell
3) State any two material which the cell should get rid of4) Tell your peer how the cell exchanges materials with its environment.
2.1 GRADIENTS ACROSS THE PLASMA MEMBRANE
ACTIVITY 2 .1
1) Use scientific reasoning to prove that the old man is right but the young man is wrong.
2) Make research and bring diagrams that illustrate the following concepts:
(a) Isotonic solutions(b) Hypertonic vs hypertonic solution.
The internal environment of a cell is maintained differently from that of its external environment
by a thin surface membrane called the cell or plasma membrane. The plasma membrane is a
lipid 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 of the membrane is permeable to non-polar and uncharged molecules such as O2,
CO2, 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 plasmamembrane, it is termed as membrane potential (Figure 2.1).
Figure 2.1: Movement of particles or ions with respect to concentration and
electrical gradient across a semi-permeable membrane
The occurrence of concentration gradient and membrane potential helps in the movement
of substances across plasma membrane. Substances tend to move down the concentration
gradient (downhill movement) known as passive movement, i.e., from high to low
concentration. Similarly, a negatively charged molecule will tend to move to positively
charged region and vice-versa. Therefore, these two parameters greatly affect the ion
movements across the membrane (by diffusion as explained in the following section) and
together constitute the electrochemical gradient.
2.2 MOVEMENT OF SUBSTANCES ACROSS PLASMA MEMBRANE
ACTIVITY 2 .2
1. From O-level you studied many types of transport across the cell membrane. Use available
resources to fin
d out short note on each of the following processes:
Simple diffusion – facilitated d
iffusion – active transport – endocytosis – Phagocytosis
2. Carry out the following experiment and thereafter answer questions that follow:
A. Put a crystal of KMnO4 in a beaker containing 200ml of water. Write your observation
after 5 – 10 – 20 minutes.
B. Take the activity 3 from the SBC. Take the activity 5 from the SB
As mentioned before, plasma membrane is selectively permeable. Figure 2.2 enlists all theprocesses involved in movement of substances across cells.
Figure 2.2: Processes involved in the transport of molecules across plasma membrane
Transport mechanisms can be broadly classified into two types:
Passive Transport—It involves the movement of molecules along the electrochemical gradientwithout the use of ATP (Downhill transport). Occurs by diffusion or osmosis.
Active Transport—It drives the molecules against their electrochemical gradient by hydrolysisof ATP (Uphill transport).
2.3 PASSIVE TRANSPORT
Below is an account of different means of transport across the plasma membrane:
2.3.1 Simple Diffusion
It 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 areevenly distributed throughout the solution (refer to Figure 2.1).
ACTIVITY 2.3
Aim: To investigate simple diffusion using plant tissues and visking tubing.
Materials Required: Visking tubing with capillary, Beaker with water, Sucrose solution (10%),
pieces of beetroot.
Background: Visking tubing is a partially permeable
artificial membrane that mimics the cell membrane,
i.e., it allows
only smaller molecules (e.g., water
molecules, glucose etc.) to pass through while restricts
the larger ones (e.g., starch). In simple diffusion,
the direction of movement
is determined by the
concentration gradient and the molecules
flow from a
region of its higher concentration to a region of lower
concentration (or along electrical gradient).
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 diffusion 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.
Discussion:
1. Compare the change in water level (in the capillary) with what you have studied in theory.
2. Discuss your observation.
3. What would you expect if instead of water you had taken 20% sucrose solution in the beakerand performed the same experiment?
Factors Affecting the Rate of Diffusion
1. The greater the concentration difference between the two sides of the membrane, the
faster is the rate of diffusion.
2. As the temperature increases, the rate of diffusion increases.
3. Smaller molecules have faster rate of diffusion while the ones with larger mass, diffuse
slowly.
4. The larger the surface area of membrane available for diffusion, the higher is the rate of
diffusion.
5. The greater the distance across which diffusion is to occur, the longer it takes for molecules
to pass through.
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, CO2 diffuses 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 O2concentration).
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 epitheliumof 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 hard outer 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 of the flowers to attract the pollinating animals, spreads in the air by diffusion.10. Diffusion plays an important role in imbibition and osmosis.
2.3.2 Osmosis
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. 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 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 potentialto the region of their lower chemical potential 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 (Figure 2.3):
• 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. The cell
is termed turgid. For example, red blood cells
when placed in a hypotonic solution, causehemolysis
Figure 2.3: Movement of water molecules when a cell is placed in three types of solution—
Hypotonic, Hypertonic and Isotonic with respect to the cell
• 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 the 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.
• If the medium is hypotonic with respect to the cell, i.e. the concentration of solutes in
the cytosol is higher than that of the solution. In this condition, water diffuses into the
cell due to osmotic pressure and the cell becomes turgid, or bloated.
• 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, and two
historically significant such foods are dried and salted cod (usually referred to as salt fish) and
salt-cured meat. 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 orbecome temporarily inactivated.”
Salting Methods
• Cut your vegetables up in pieces before you put them into the salt water to preserve food
by salt-curing. As you chop a vegetable and put it into the salt water, 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 salt water for at least 10 days in
order to “pickle.” Pickle simply means preserved in brine. Then cover tightly with a lid.
• Preserve meats by salt-curing. 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 thisway 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 salt-curing meat do not preserve it aslong as the dry method does.
ACTIVITY 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, 0.8Mand 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.
5. Seal the tube with an aluminium foil paper to prevent water loss by
evaporation.
6. Leave the tube rack aside for 1h.
7. Measure the length of the cylinders in each tube. This is the final length (Lf).
8. Calculate the % change in length for each using the formula -
% change in length = [(Lf – Li)/ Li] × 100.
9. Calculate the average of the three readings obtained for each tube.
10. Plot a graph of mean % change in length versus the sucrose concentration used.
11. Draw the best fit line for all the points obtained.
12. 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).
Discussion: Discuss the following questions after observing the results drawn.
• What happened to the potato tissue cylinders? Did they swell or shrink?• Which process do you think brought the change?
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 the 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. Water molecules (or molecules in gaseous state) show random movement
as they possess kinetic energy. Therefore, as the concentration of water in a solution increases,
the kinetic energy or its water potential increases. Hence, when two solutions are kept in close
contact, water molecules with higher kinetic energy tend to move towards the one with lower
kinetic energy. 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 signalled 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 directionto 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, andbody 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
c ells 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 salt water 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 haveto constantly regulate the movement of water out or into their body (called osmoregulation).
1) Complete the sentence with the correct word
(a) In hypotonic solution, a cell ................................. .
(b) In lungs, CO2 diffuses out of blood by the process of ............................ .
(c) Purification of blood by kidneys takes place by the process of .................. .
(d) The pressure exerted by plants’ cells on the cell wall is ............................. .
(e) The larger the surface area of the membrane, the ........................ is the rate of diffusion.
2) (a) what is the key difference between active and passive transports?(b) Explain briefly how salting is an efficient way of conserving meat
2.4 ACTIVE TRANSPORT
ACTIVITY 4
Active transport is the movement of dissolved molecules into or out of a cell through the cell
membrane, from a region of lower concentration to a region of higher concentration. The
particles move against the concentration gradient, using energy released during respiration.
Use the search engine to find out simulation that illustrates the following:
a) The membrane proteins / Carrier proteins are involved
b) ATP is usedc) Molecules or ions move against their concentration gradient
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 isprovided either by another coupled reaction or by direct hydrolysis of ATP.
2.4.1 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 (Figure 2.4); 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 pumpsin 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 transportof glucose against concentration gradient from the intestinal lumen to the cell.
Figure 2.4: Working of Na+ – K+ pump. 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, and
(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
favourable 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+ (Figure 2.5).
(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 (Figure 2.5). Another example is
Na+–H+ , which transports Na+ into the cell with the export of H+, thereby removingexcess of H+ and preventing acidification of cell cytoplasm.
Figure 2.5: 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 thatthe two molecules are transported in opposite direction
2.4.2 Factors Affecting the Process of Active Transport
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.
2.4.3 Significance of Active Transport in Organisms
(i) In the intestinal lining, glucose is absorbed by active transport from a lower concentrationto a higher concentration in the cells lining the intestine.
ACTIVITY 4
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) was
taken in different boiling tubes. 1 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 fromMf. When the data was 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: What concentration of salt is isotonic to the potato tissue? Why?
Question 3: Which of the salt concentrations are hypotonic and hypertonic with respect to
potato tissue?
Question 4: What is the effect on the size and weight of the tissue when it is placed in a
hypotonic, hypertonic and isotonic solution?Discussion: Discuss your interpretation
APPLICATION 2.2
1. Define in your own term active transport
2. Differentiate between facilitated diffusion and active transport3. Discuss the significance of active transport in animals.
2.5 ENDOCYTOSIS AND EXOCYTOSIS
ACTIVITY 2.5
Conduct the culture of Paramecium using decaying leaves (e.g. cabbage leave).
After three days, mount on the glass slide a drop sample from the top layer.
Observe under the microscope the movement and feeding of paramecia.
Share your observation.
1) Explain how Paramecia feed2) What name would you give to the feeding process of paramecia?
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 evenintact cells.
Pinocytosis - Also called “cell drinking”. It involves uptake of fluids by the cell.
2.5.1 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 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 systemthat show phagocytosis and are therefore also called “professional phagocytes” (Figure 2.6).
Figure 2.6: Process of phagocytosis showing the ingestion of a bacteria by a cell
2.5.2 Pinocytosis
It 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 broken down
substances. This non-specificity in the ingested substance distinguishes it from phagocytosis
which takes up specific substances from the exterior. Also, phagocytosis involves breakdownof the particle which is lacking in case of pinocytosis (Figure 2.7).
Figure 2.7: Process of pinocytosis showing the uptake of fluid by a cell
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 allits contents into the surrounding medium. (Figure 2.8)
Table 2.1: Comparision of Endocytosis and Exocytosis
APPLICATION 2.3
1. Complete the sentence with the correct word
(a) The process of cell drinking is known as ..................... .
(b) Ca2+ ions are required for .............................. .
(c) When two molecules are transported in opposite direction, it is .................... .
(d) ............................ involves ingestion
2. The Amoeba is a single-celled organism that lives in water. Describe how it engulfs particlesof food by endocytosis.
2.6 SUMMARY
• Every cell is surrounded by cell or plasma membrane which regulates the movement or
exchange of ions or molecules between the cell and its medium. This property of cell is
called cell permeability.
• The presence of concentration and membrane potential (together called electrochemical
gradient) helps in the movement of substances across the membrane.
• Plasma membrane mediates transport of smaller molecules by passive and active
transport whereas larger molecules are transported by endocytosis.
• In passive transport, ions/molecules move from higher concentration to lower
concentration which includes diffusion and osmosis and there is no utilization of energy.
• Simple diffusion is the movement of small hydrophobic molecules from higher
concentration
to lower concentration by dissolving in phospholipid bilayer till equilibrium
is reached.
• Osmosis is a movement of water molecules from low solute concentration to high solute
concentration side (or from higher solvent concentration to lower solvent concentration).
• Active transport is the movement of ions/molecules from lower concentration to higher
concentration. It is of two types: Primary and Secondary active transport. The former
involves direct utilization of energy in the form of ATP hydrolysis while the later involves
movement of molecules against concentration gradient but coupled with the movement
of a second molecule in an energetically favourable direction without direct utilization
of ATP. The movement may be symport, uniport or antiport.
• Endocytosis is the ingestion of large particles such as bacteria, macromolecules and fluids
into the cell in the form of small vesicles. It is further of two types, viz., phagocytosis
(cell eating, engulfing of solid particles) and pinocytosis (cell drinking, uptake of liquidfluids).
2.7 GLOSSARY
• Active transport: Transport of moleculer across cell membrane utilizing energy in the
form of ATP.
• Diffusion: It is the movement of solutes from higher concentration to lower concentration
through semi-permeable membrane.
• Endocytosis: Cells engulf molecules which cannot pass through the plasma membrane.
• Exocytosis: Cells expel molecules which cannot pass through the plasma membrane
by exocytosis.
• Osmosis: It is a movement of water molecules from ion solute concentration to higher
solute concentration.
• Osmotic potential: The potential of water molecules to move from a hypotonic solution
to a hypertonic solution across a semi-permeable membrane.
• Passive transport: Transport of ions and molecules through cell membrane without
utilization of energy.
• Phagocytosis: It is a food capturing process of bacteria and protozoans. In eukaryotes,
this process is used for remolving aged cells or those which have stopped functioning.
• Pinocytosis: It is a non-specific process which involves engulfing of either pre-dissolved
or already broken down substances.
• Plasmolysis: It is the process in which cells lose water in a hypertonic solution. Plasmolysis
can lead to cell’s death.
• Wall pressure: The pressure exerted by water inside the cell against the cell wall. It is
also called Turgor pressure.
• Water potential: It is the measure of potential energy in water. It drives the movement
of water through plants.
END UNIT ASSESSMENT 2
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) Passive transport (b) Active transport
(c) Endocytosis (d) 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. Distinguish between endocytosis and exocytosis giving suitable examples.
