Chapter 5
Cell Activities
5.1
The Concept of Metabolism
5.2
The Roles of Enzymes in Metabolism
5.3
Factors Affecting Enzyme Activities
5.4
Transporting Materials in and out of Cells
What is metabolism?
• All chemical activities taking place inside the body cells of
an organism are collectively called metabolism
Examples:
• Bone and wood:
 contain living cells  carry out metabolism
• Finger nails and shells:
 truly dead materials  no metabolism
Catabolism
complex molecules
simpler molecules
Examples:
• Digestion
complex food molecules  simpler molecules
enhance absorption
• Respiration
glucose  carbon dioxide & water
release energy
Anabolism
simpler molecules
complex molecules
Examples:
• Photosynthesis
- combination of simple inorganic molecules
(carbon dioxide & water)
 complex organic molecules
(glucose)
Metabolic rate = the overall speed of the chemical
reactions in the organisms
Low metabolic
rate at rest
High metabolic rate
when active
Why are enzymes so important?
• Catalysts: substances that speed up chemical
reactions
• Enzymes: catalysts which speed up chemical
reactions inside the cells
Enzymes = biological catalysts
• Enzymes
– maintain metabolic rate
– may work inside or outside of the cells
(e.g. digestive enzymes in the gut)
What are the properties of enzymes?
•
•
•
•
•
•
Made up of proteins
Speed up the rates of chemical reactions
Are regenerated / reused after reaction
Effective in very small quantities
Easily affected by changes in temperature and pH
Very specific in the reactions they catalyse
How do enzymes work?
Based on experimental evidence, scientists have
proposed a hypothesis to explain how enzymes work
and why enzymes are so specific in their actions:
Lock-and-key hypothesis
Substrate : the reacting molecule in an enzymecatalyzed reaction
Active site : the particular site of the enzyme for
reaction to occur; specific for each type of enzyme
Enzyme-substrate complex : an unstable structure
formed by the enzyme and substrate molecules
during reaction
Lock
(substrate)
Key
(enzyme)
Enzymesubstrate
complex
Opened lock
(product)
Key (enzyme,
remain unaffected)
The lock-and-key hypothesis of enzyme action
Enzyme action in a catabolic reaction
Enzyme action in an anabolic reaction
Practical 5.1
Investigating the presence of catalase in various
plant and animal tissues
Introduction
Catalase is an enzyme that speeds up the breakdown
of hydrogen peroxide into water and oxygen. It is an
example of catabolic reaction.
Procedure
1. Prepare six boiling tubes in a holder.
2. Add 5 cm3 hydrogen peroxide solution into each tube.
3. Add a small piece of potato, apple, meat, liver and
boiled liver of similar sizes into five of the tubes
respectively. Do not add any tissues into the last tube.
4. Note whether any
gas is evolved in
each tube. Test the
gas with a glowing
splint.
5. Measure the highest
point reached by the
bubbles in each
tube.
Analysis
1. What is the purpose of testing boiled liver?
Ans: To show the effect of boiling on the activity
of the enzyme catalyse in liver
2. What is the purpose of doing the test in a test tube
without any tissue?
Ans: It acts as a control set-up
3. Compare the results obtained. Give a conclusion for
this experiment.
Ans: Different tissues may contain different
amounts of catalase so that they have different
rates for the same reaction
~ End of Practical 5.1 ~
Factors Affecting Enzyme
Property of Enzyme:
• Protein has an active site with a particular threedimensional shape maintained by weak
bondings
Under certain conditions:
Breaking of weak bondings
Change in shape of the active site
Affect enzyme activity
How does temperature affect
enzyme activity?
Enzyme activity = the rate of substrate utilization
or product formation
From 0 – 40/50C
• Rise in temperature
 increase the rate of an enzyme-catalyzed
reaction
Effect of temperature on enzyme activity
Reasons
at higher temperature, the substrate molecules
move faster
higher chance for collision with the active sites
higher chance to form enzyme-substrate
complexes
greater amount of product molecules
Above 50C
• Reaction rate drops rapidly
 stops completely at 70 - 80C
Effect of temperature on enzyme activity
Reasons
at high temperature
breaking of weak bondings
change in shape of the enzymes and their active sites
lose the catalytic activity
enzyme is denatured
Around 0C
• Very low reaction rate
Reasons
movement of the substrate molecules slow down at
low temperature
there is less chance for the substrate molecules to
collide with enzyme molecules
enzyme is inactivated
Optimum temperature
• The temperature at which an enzyme-catalyzed
reaction occurs most rapidly
• Specific to different enzymes
• Lies around 45C for most enzymes
Effect of temperature on enzyme activity
How does pH affect enzyme activity?
• pH
 the degree of acidity or alkalinity of the reaction
medium
• Small changes in pH
 greatly affect the enzyme activities
• Each enzyme has its optimum pH
• Most enzymes work best at pH 5 to 9, but some work
best at an extreme pH
When pH  optimum pH
change in enzyme structure
decrease in activity
Effect of pH on enzyme activity
When there is drastic changes in pH
 denaturation of enzyme
 a great decrease in catalyzing ability
Effect of pH on enzyme activity
Critical Thinking 5.1
The optimum pHs of some enzymes
The graph below shows the effect of pH on the
activities of three different enzymes.
Questions
1. Which enzyme works best in alkaline conditions?
Ans: Trypsin
2. Which enzyme works best in acidic conditions?
Ans: Pepsin
3. Which enzyme works best in neutral conditions?
Ans: Salivary amylase
4. What is the range of pH at which each enzyme is
active?
Ans:
Pepsin: pH 1-3; salivary amylase: pH 5.5-8.5;
trypsin: pH 7.5-10.5
Practical 5.3
Investigating the effect of temperature on the activity
of amylase
Introduction
Amylase is a type of digestive enzyme found in saliva
and the gut. It speeds up the breakdown of starch.
In this experiment, we are going to test the effect of
temperature on the activity of amylase.
Procedure
1. Prepare six water baths with the following
temperatures:
- 0C
- room temperature
- 40C
- 60C
- 80C
- 100C
2. Put 1 cm3 of amylase solution into each of six test
tubes, and place them into the suitable water baths.
3. Put 4 cm3 of starch solution into each of six other test
tubes, and place them into the suitable water baths.
4. After 5 minutes, pour the amylase solution from a
test tube into the starch solution in the test tube in
the same water bath.
5. After 10 minutes, pour 5 cm2 of Benedict’s solution
to each tube and shake gently. Boil for 5 minutes.
6. Shake the tubes at 1-minutes intervals. Note the
colour changes of the mixtures. Compare the
amount of brick-red precipitate formed in the six
tubes.
Analysis
1. What is the purpose to keep the test tubes containing
the starch solution and amylase solution in each water
baths for about five minutes before mixing?
Ans: To ensure that both starch and amylase solutions
have reached the desired temperatures before mixing
2. What is the optimum temperature for the action of
amylase? Ans: At 60C
4. Explain the effect of temperature on the activity of
amylase. Ans: Amylase is inactive at 0C. The rate of
reaction increases steadily with temperature, and
reaches the maximum at the optimum temperature of
60C. Beyond the optimum temperature, the reaction
rate drops rapidly and finally stops completely due to
denaturation of enzyme
Practical 5.4
Investigating the effect of pH on the activity of amylase
Introduction
In this experiment, we are going to test the effect of pH
on enzyme activity. Amylase is extracted from mung
beans for the test.
Procedure
1. Soak 10 mung beans in
water for two days.
2. Remove the seed coat.
Grind them with 2 cm3 of
distilled water.
3. Filter the mixture
and collect the
filtrate in a test
tube.
4. Prepare and label six cavities of a spot tile as follows:
A: Amylase extract + 1 drop of distilled water
B: Dilute hydrochloric acid (HCl)
C: Amylase extract + 1 drop of dilute HCl
D: Dilute sodium bicarbonate solution (NaHCO3)
E: Amylase extract + 1 drop of dilute NaHCO3
A
F: Distilled water
D
E
B
C
F
5. Prepare six small
filter paper discs.
6. Put one filter paper
disc into each of
cavities A to F.
A B
C
D E
F
7. Drain away excess
liquid from the
paper discs, and put
them onto a starchagar plate.
8. Incubate the starchagar plate at 40C
for half an hour.
9. Remove the paper
discs from the plate.
Flood it with iodine
solution for 1 minute
and then flush gently
with tap water.
10.Observe the colour
changes.
Analysis
1. Why do we carry out the tests for the liquids in the
cavities A, C and E in the spot tile?
Ans: To investigate the amylase activity at neutral,
acidic and alkaline conditions respectively
2. Why do we carry out the tests for the liquids in cavities B,
D and F in the spot tile?
Ans: They act as control set-ups
3. Why do we incubate the agar plate at 40C instead of
room temperature?
Ans: It provides an optimum temperature for most
enzyme-catalyzed reactions
4. In which medium does amylase work best?
Ans: In an alkaline medium
5. From the results, give a conclusion for this experiment.
Ans: Amylase works best in an alkaline medium
~ End of Practical 5.4 ~
Practical 5.5
Comparing the protease activities in different fruit juices
Introduction
In some fruit juices,
there are enzymes that
can break down proteins.
We will use a milk-agar
plate to compare the
protease activities of
different types of fruits.
Procedure
1. Prepare a milk-agar plate.
2. Prepare a small amount of juice from each of
several types of fruits (e.g. pineapple, kiwi fruit,
papaya, guava and mango) by using a mortar and
a pestle.
3. Collect the filtrate and dilute with distilled water
(2cm3 of filtrate with 8cm3 distilled water).
4. Remove 6 circular discs of agar from the agar plate.
5. Fill a well in the milk-agar with two drops of a fruit
juice.
6. Incubate the plate at 40C for 1 to 2 hours.
7. Note and measure the diameter of the clear zone
around each well.
Analysis
1. Suggest a sample for the control experiment.
Ans: Distilled water
2. How are the clear zones formed in the milk agar?
Ans: In some fruits, there are enzymes (proteases)
that can break down proteins. If the milk protein in
the milk-agar plate is digested, a clear zone will
appear in the agar plate
3. What is the relationship between the diameter of the
clear zone and protease activity?
Ans: The higher the protease activities, the larger
the diameter of the clear zone
Further Investigation
You may do a similar experiment to compare or
study the protease activities of the following
solutions:
1. biological washing powder,
2. meat tenderizer,
3. protein-removal tablets for contact lenses.
~ End of Practical 5.5 ~
Making use of enzymes in everyday life
Questions for Discussion
1. Why are commercial enzymes mainly obtained
from microorganisms?
Ans: It is because the reproductive rate of
microorganisms is fast and the yield of
commercial enzymes is high
2.Why is it possible to obtain maltose from starch?
Ans: Starch is converted to maltose by amylase
3. Why can proteases
be used to make
meat more tender?
Ans: Proteases partially break down proteins
in the meat tissues that are originally
responsible for its tough nature
4. Why can proteases be used to remove egg
stains and blood stains from clothes?
Ans: Proteases can break down the proteins
present in egg stains and blood stains
~ End of Enrichment Reading 5.1 ~
Activities of the cell
Various supplied
to
substances
Cells
removed
from
Waste
product
Carry out various life processes, such as
- releasing energy
- building up cellular structures
Cell membrane
• Has many tiny pores
• Selectively permeable
• Allows some particles to pass through but not others
Processes involved in transport of materials
• Diffusion
• Osmosis
• Active transport
What is diffusion?
• The net movement of the particles of a substance
from a region of higher concentration to a region of
lower concentration
• The net movement continues until the concentration
of the substance in the two regions are the same
• It is a passive process
Concentration gradient
• The difference in concentration between two regions
where diffusion occurs
• Determines the rate and the direction of diffusion
Critical Thinking 5.2
The diffusion of ink in water
A small amount of blue ink was put in a beaker
containing water. The photographs below show
the colour change of the water immediately, 10
minutes later and 20 minutes later.
Questions
1. How does the colour of the water change
within 20 minutes?
Ans: The blue colour spreads out gradually
in water
2. Explain the colour change.
Ans: Ink particles spread from a region
of high concentration to a region of low
concentration by diffusion
Questions
3. What do you expect to see if the solution is
allowed to stand for a long time? Are the
ink particles still moving at this state?
Ans: The water will become blue evenly;
ink particles are still moving randomly
4. Will the result be different if the demonstration
is carried out at a higher temperature? Why?
Ans: The blue colour will spread out much
faster, because the ink particles possess
more kinetic energy
What factors affect the rate of
diffusion?
1. Molecular size
– Small particles diffuse faster than larger particles
2. Concentration gradient / difference
– Diffusion occurs more rapidly if there is a larger
difference in concentration between two regions
3. Distance of diffusion
– Diffusion occurs more slowly if the distance
betweem the two regions is longer
4. Surface area
– For substances diffusing through a membrane,
a larger surface area promotes a faster rate of
diffusion
5. Temperature
– Diffusion occurs more rapidly with a rise in
temperature (because the kinetic energy of particles
increases)
Why is diffusion important to
organisms?
It is important in the exchange of useful and wasteful
substances between:
• cells and body fluids
• the organism and its surroundings
Examples:
• Gas exchange in the breathing organs of animals
• Gas exchange on the leaf and stem surfaces of
plants
• Absorption of digested food in the small intestine
What is osmosis?
Osmosis refers to the diffusion of water from
the dilute solution to the concentrated solution
across a selectively permeable membrane
Why does osmosis occur?
What will happen when a
concentrated sucrose solution is
separated from a dilute sucrose
solution by a selectively permeable
membrane?
Note:
The membrane has numerous tiny pores that allow small
particles (e.g. water molecules) to pass through, but not
larger particles (e.g. sucrose molecules)
• Region B contains a
higher proportion of
sucrose molecules and
a lower proportion of
freely moving water
molecules
• In a sucrose solution,
the sucrose molecules
and water molecules
are in continuous,
random motion
• Only water molecules
can pass through the
membrane in both
directions and more
pass from region A to
region B
• There is a net
movement of water
molecules from dilute
to concentrated
solution across the
membrane
• Osmosis is a special
type of diffusion
• Takes place only when a
selectively permeable
membrane is present
• Definition: the diffusion
of water molecules from
a dilute solution to a
more concentrated
solution across a
selectively permeable
membrane
What is water potential?
• Water potential is the total energy for the water
molecules in a solution to move about
• Net movement of water molecules: from a region
of higher water potential to a region of lower water
potential
• Pure water has the highest water potential, its
water potential is defined as zero
• When substances are dissolved in water, water
potential become less than zero
• Concentrated sucrose solution has a lower water
potential than dilute sucrose solution
How does osmosis affect organisms?
Water molecules may move into or out of the cell by
osmosis depending on the concentration of substances
dissolved in the fluid around a cell
3 terms to describe the relative concentration of solutions:
• Hypertonic
• Hypotonic
• Isotonic
Hypertonic solution
• A solution with a water potential lower than that of
another solution, e.g. the water potential of the
solution in the cell
• When animal cells are placed in a hypertonic solution:
 water leaves the cell by osmosis
 shrinking
• When plant cells are placed in hypertonic solution:
 water leaves the cell by osmosis
 shrinking of cytoplasm
cytoplasm becomes separated from cell wall (i.e.
cell is plasmolysed)
plant tissue becomes flaccid
Red blood cells
shrink (4 000)
Plant cells
become
plasmolysed
Hypotonic solution
• A solution with a water potential higher than that of
another solution, e.g. the water potential of the
solution in the cell
• When animal cells are placed in hypotonic solution:
 water enters the cell by osmosis
 expand or even burst
• When plant cells are placed in hypotonic solution:
 water enters the cell by osmosis
 expand, but swelling is limited by the cell wall
 become turgid
Red blood cells
swell (4 000)
Plant cells
become
turgid
Isotonic solution
• A solution with the same water potential as that of
another solutions, e.g. the water potential of the
solution in the cell
• Important to maintain a suitable water potential of
body fluids (e.g. the blood)
No net movement of
water into or out of
the cell
Water
leaves
the cell
Immersed in
isotonic solution
Immersed in
hypertonic
solution
The cell
shrinks and
becomes
plamolysed
Water
enters
the cell
Immersed in
hypotonic
solution
A plant cell
The cell
swells and
becomes
turgid
The cell
shrinks
An animal cell
Immersed in
hypertonic
solution
Water
leaves
the cell
Immersed in
isotonic solution
The cell
swells and
eventually
burst
Immersed in
hypotonic
solution
Water
enters
the cell
No net movement of
water into or out of
the cell
Importance of osmosis to plants
• For water absorption
- water potential of the root epidermal cells is lower
than that of soil water
- water moves from the soil (higher water potential)
into the root epidermal cells (lower water potential)
• For support
- young, non-woody plants depend entirely on
turgidity for support
- if water loss is significant, plant tissue will become
flaccid and the plant will wilt
Practical 5.6
Studying osmosis by using a dialysis tubing
Introduction
The phenomenon of
osmosis can be
demonstrated by:
• using a selectively
permeable membrane
• using two solutions of
different concentrations
Procedure
1. Soak a dialysis tubing in water.
2. Open up and tie a knot close to one end of the
dialysis tubing.
3. Fill the dialysis tubing with 20% sucrose solution.
4. Tie the other end of the dialysis tubing to a 1 cm3
pipette.
5. Wash the dialysis tubing with distilled water.
6. Place the dialysis tubing into a beaker of distilled
water.
7. Measure & record the change in liquid level.
8. Set up a control.
Analysis
1. Why do we rinse the outer surface of the dialysis tubing?
Ans: To avoid introducing sucrose solution to the
surrounding medium
2. How does the volume of sucrose solution in the dialysis
tubing change with time?
Ans: It rises with time, indicating that the volume
of sucrose solution in the dialysis tubing increases
3. Explain the result of this demonstration.
Ans: There is a net movement of water molecules into
the sucrose solution from surrounding through the
selectively permeable membrane by osmosis
~ End of Practical 5.6 ~
Practical 5.7
Studying osmosis in a living plant tissue
Introduction
Instead of using a dialysis tubing, a living tissue
(potatoes) can be used to demonstrate osmosis.
Procedure
1. Peel off the skins of potatoes and make 3
potato cups.
2. Place the potato cups into three petri dishes
containing distilled water, and label as:
- A (unboiled, cavity filled with distilled water)
- B (unboiled, cavity filled with sucrose solution)
- C (boiled, cavity filled with sucrose solution)
3. Observe the change in the level of the solution in
each potato cup 24 hours later.
Analysis
1. Describe your observation.
Ans:
There is an increase in the liquid level in B.
The liquid level remains more or less unchanged in A.
The level of the sucrose solution in C drops until it is
equal to the level of distilled water outside.
2. Explain the changes in liquid level in A, B and C.
Ans:
A: no osmosis occurs, because there is no
concentration difference between distilled water in
the cavity and that in the petri dish;
B: osmosis occurs, because distilled water has a
higher water potential than that of the sucrose
solution. Water molecules diffuse across potato
cells into the cavity;
C: no osmosis occurs, because strong heat kills the
potato cells and destroys their cell membrane which
become freely permeable. The sucrose solution
moves out of the cavity until its level is equal to the
level of water outside
3. At the end of the experiment, in which set-up, B or C,
may we find sucrose in the water outside the potato cup?
Explain briefly.
Ans: We may find sucrose in the water outside the
potato cup in set-up C because the cell membrane of
potato cells in C become freely permeable after
strong heating. Thus, the sucrose molecules move out
of the cell into the distilled water outside.
~ End of Practical 5.7 ~
What is active transport?
Active transport
• It is need, for instance, to transport substances
against a concentration gradient
• It is a process of transporting substances across
the cell membrane with the use of energy
• It can only take place in living cells
• It involves carriers (special protein molecules)
located on the cell membrane
Movement of
substances
against the
concentration
gradient
Low concentration
outside the cell
Substance to be transported
Carrier molecule
Cell membrane
High concentration
inside the cell
1. The substance enters
the carrier molecule
The action
of the
carrier
protein in
active
transport
2. The carrier molecule
combines with the
substance and changes
its shape. The energy
needed for this comes
from respiration
3. The change of shape of
the carrier molecule
releases the substance
into the cell
Why is active transport important
to organisms?
• Enables a cell to take in useful substances / remove
waste substances against the concentration gradient
• Allows the transport of substances across the
membrane when:
- diffusion is too slow
- there is no concentration gradient
Examples:
In the human body:
• Absorption of glucose by the cells lining the small
intestine
• Absorption of minerals by the cells lining the kidney
tubules
In plants:
• Absorption of minerals from the soil