BIO112 - LAB 7
FACTORS THAT AFFECT ENZYME ACTIVITY
Objectives:
1. Have a basic understanding of enzymes: what they are, how they function, and their role in cell metabolism
2. understand the importance of enzyme 3-D structure for their function
3. know how temperature and pH affect enzyme function
4. know the relationship between enzyme concentration and reaction rate
5. know the relationship between substrate concentration and reaction rate
6. understand the action of the enzyme amylase: its substrate, its products, and the conditions under which it
functions
7. understand and be able to explain the procedures that were used in this lab to demonstrate amylase activity
8. know the source of amylase used in this lab
9. be able to graph your results and explain what they mean
10. know all terms that are in boldface type.
Materials:
Student bench:
1. Iodine (IKI)
2. 1% starch solution
3. 1% pancreatin
Side bench:
1. Buffer at pH 3, 5, 7, 10
2. Water baths at 5, 20, 35, 600C
Procedures:
Each group will do three experiments in order to determine how incubation conditions affect the rate of an
enzymatic reaction. In each experiment you will vary temperature or pH or enzyme concentration and measure
the rate at which pancreatic amylase digests starch.
Readings:
Sadava: pp. 125-130, 134-135
45
Introduction
Proteins serve many functions in cells. The function of a protein is determined by its three-dimensional shape.
One category of protein (based on their function) is the enzyme. Enzymes are proteins that can catalyze (speed
up) the rate of a chemical reaction. An enzyme can catalyze only one specific chemical reaction. Because
enzymes are specific for one type of reaction, there must be a different enzyme for every chemical reaction that
a cell carries out. Each enzyme has one or more active sites. Active sites are depressions in the protein where
the enzyme binds temporarily to its substrate(s). Substrates are the reactants in the chemical reaction that the
enzyme catalyzes. The substrate’s shape and chemical properties match the active site in a way that is similar to
the way in which a key fits into a lock.
Recall that a protein’s three-dimensional shape is the result of the way in which its amino acids interact with
each other and their environment. A change in the environment (temperature, pH, solute concentration, etc.) can
cause a protein to fold in a different way. Since a protein’s function is a result of its three-dimensional shape
and its shape is influenced by the environment, then it should be clear that a protein can only function properly
under certain optimal conditions; changing the conditions will alter its shape and its ability to function. Extreme
conditions of pH or temperature (such as boiling) will cause a protein’s to become so badly distorted that it can
not function at all. We say that the protein has been denatured.
The rate at which enzymes work is strongly affected by pH and temperature. Although most enzymes work
when the pH is near 7, some will only function when conditions are very acidic or alkaline. For example, the
enzymes in the human stomach work best when the pH is about 2; it stops completely if the pH is above 7.
In this lab you will look at digestion (hydrolysis) of starch by the enzyme Amylase. The enzyme amylase is
produced by the salivary glands and pancreas. Amylase breaks the starch molecule into smaller chains of
variable length called Dextrins. Amylase then breaks down the Dextrins to Maltose (a disaccharide) and
Glucose (a monosaccharide).
Starch
Amylase
>
Dextrins
Amylase
> Maltose + Glucose
How can you tell if the enzyme Amylase is breaking down starch to maltose and glucose? You look for either
(1) the appearance of glucose and maltose or (2) the disappearance of starch. In previous labs you learned to use
the Benedict's test to detect reducing sugars and the Iodine test to detect starch. You will use these same tests in
today's experiments. The iodine test can also be used to test for the formation of Dextrins since Dextrin in the
presence of iodine turns a reddish color.
You can use the Iodine test to follow the progress of the enzyme at breaking down the starch to dextrins and
then maltose and glucose.
Sample:
Starch
Iodine Test: Black
Amylase
>
Dextrins
Reddish
Amylase
>
Maltose + Glucose
Yellow (iodine color)
TIME
Our real interest is to study the rate at which Amylase hydrolyzes starch under different experimental
conditions. You will look at the effect of (1) denaturing the enzyme, (2) temperature while incubating, (3) pH,
and (4) enzyme concentration on the rate of the reaction. The source of amylase is Pancreatin, an extract made
from pancreas tissue.
46
Consider the following thought experiment
Procedure:
1. Prepare 3 test tubes with the materials shown in the table. Mix.
2. Allow the tubes to stand (a.k.a. Incubate) for 10 minutes at room temperature.
3. Carry out the Iodine test on a small sample (several drops) from each tube.
4. Carry out the Benedict's test on a sample from each of the tubes.
1
2
3
Tube
1 ml Starch Solution
1 ml water
1 ml Starch Solution
1 ml Pancreatin Solution
1 ml Starch Solution
1 ml Boiled-Pancreatin Solution
Iodine Test Result (+/-)
Benedicts Test Result (+/-)
+
-
-
+
+
-
⋅
What is the purpose of tube #1?
⋅
Write the equation for this reaction.
⋅
What do these results tell you about the importance of tertiary structure for protein function?
Overview of the Experiments
In the following experiments you will measure the rate at which pancreatic amylase digests starch. You will
mix pancreatic amylase with starch to start the reaction. Then, at one minute intervals, you will do an iodine
test on a sample taken from the reaction mixture. In this way, you can determine the time needed to convert the
starch to dextrins and the time needed to convert the starch to maltose. In each experiment, you will look at the
impact of varying either temperature or pH or enzyme concentration on the reaction rate.
Iodine Test Procedure:
‚ Remove a sample (4 drops) from each of the tubes and place them on the spot plate.
‚ Add 1 drop of Iodine to each of the samples. Record the result in the table.
Black = starch present
Reddish = dextrins present
Some practical advice: You will follow the reaction under four
different conditions (of temperature, pH or enzyme
concentration) at the same time. You have only 1 minute to take
each of the 4 samples and add the iodine. Proceeding in an
orderly fashion will make this possible. Use 1 row in the spot
plate for the samples collected at each 1 minute interval as
shown in the diagram. Use the same order each time.
Do NOT use the same pipette to sample from all 4 test tubes; this
would transfer material from one tube to another. If this
happens, your results are meaningless. Use a different pipette
for each of the 4 tubes. It may be re-used for the different times.
Yellow = no starches or dextrins present
Time
50C
200C
350C 550C
0
1
2
:
:
15 min
Record your results on paper. Arrange your data table in the same way as the spot plate. Record the color change
after doing each iodine test: Black, Red, Yellow.
Do NOT clean the spot plates until the experiment is over and everyone has seen the results. The color in the spot
plate wells will be stable for hours.
47
Experiment 1 - Temperature
Goal: To determine the effect of temperature on the rate of starch hydrolysis by Amylase.
Approach: You will incubate starch-amylase mixtures at 4 temperatures and determine (1) the time when
dextrins first appear and (2) time when digestion is complete, i.e. when no starch remains.
Procedure:
1. Mark 4 test tubes with the temperature at which it will be incubated: 50, 200, 350, and 600C
2. Add 4 ml of 1% starch solution to each of 4 test tubes (Shake the flask before using). This is the Substrate.
3. Place each tube in the appropriate water bath (5, 20, 35, and 600C) for 10 minutes. Mix occasionally.
This gives the starch solution time to reach the proper temperature. Leave the tubes in the water baths as
you do the following steps.
4. Add 1 ml Pancreatin solution to each of the 4 test tubes. Mix. Note the time; this is Time = 0.
Pancreatin will now start digesting starch.
5. Immediately take a sample from each tube and put
it on the spot plate. Do the Iodine test.
This is your Time = 0 reading.
6. Repeat the Iodine test at one minute intervals.
Be sure to shake the tube before taking the sample.
Stop when all 4 samples give a negative Iodine
test or at 15 minutes.
Record your data in the table.
Black = starch present
Reddish = dextrins present
Yellow = no starches or dextrins present
7. Examine your data and find the time required for
(1) Dextrins to appear, and
(2) complete Starch digestion.
Record your results in the table.
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0
5
Temperature
200
350
600
Incubation Temperature
0
5C
200C
350C
600C
Time for Dextrins
to appear
Time for complete
starch digestion
Relative Enzyme
Rate (see Step 9)
48
8. Draw graphs of your results to show the relationship between temperature and time required to completely
digest the starch. Place the Temperature on the x-axis and Time on the y-axis.
‚ Draw a plot of the time required for Dextrins to form.
‚ Draw a second plot of the time required for starch to disappear completely.
Ö What is the relationship between incubation temperature and the time required to digest starch?
9. Use your data to find the relationship between temperature and the rate of enzymatic reaction.
‚ You did not measure the actual rate of enzymatic reaction, but you can compare the times required to
digest the starch under different temperature conditions to each other and calculate the Relative Enzyme
Rate for each temperature. Here is the procedure.
1. Look at the times required to completely digest the starch at each temperature. Select the shortest time.
2. Divide the shortest time by each of the other times. These fractions are the Relative Enzyme Rates.
Notice that the shortest time will give a value of 1; each of the other times will be a fraction of 1.
Record the Relative Enzyme Rates in the table above.
For example, if the times were 3, 6, 9, and 12 minutes, then the shortest time was 3 minutes.
The Relative Enzyme Rates would be 3÷3 = 1.0; 3÷6 = 0.5; 3÷9 = 0.33; and 3 ÷ 12 = 0.25. The
enzyme that took 12 minutes to complete digestion worked at a rate that was ¼ (0.25) that of the
enzyme that took only 3 minutes to complete digestion.
‚ Draw a graph to show the relationship between temperature and the enzyme rate. Place Temperature on
the x-axis and Relative Enzyme Rate on the y-axis.
Ö What is the relationship between incubation temperature and Rate of Enzymatic action?
49
Experiment 2 - pH
Goal: Determine the effect of pH on the rate of starch hydrolysis by Amylase.
Approach: You will incubate starch-amylase mixtures under four different pH conditions and determine when
dextrins first appear and when digestion is complete, i.e. no starch remains.
Procedure:
1. Label 4 test tubes with the pH at which the reaction will be carried out: pH3, pH5, pH7, pH10
2. Add 5 ml of the appropriate phosphate buffer (pH3, pH5, pH7, pH10) to each tube.
3. Add 3 ml of 1% Starch Solution (shake the flask before using). Mix.
4. Add 1 ml Pancreatin Solution to each of the 4 test tubes. Mix. Note the time; this is Time = 0.
Pancreatin will now start digesting starch.
5. Immediately take a sample from each tube and put
it on the spot plate. Do the Iodine test.
This is your Time = 0 reading.
6. Repeat the Iodine test at one minute intervals.
Be sure to shake the tube before taking the sample.
Stop when all 4 samples give a negative Iodine
test or at 15 minutes.
Record your data in the table.
Black = starch present
Reddish = dextrins present
Yellow = no starches or dextrins present
7. Examine your data and find the time required for
(1) Dextrins to appear, and
(2) complete Starch digestion.
Record your results in the table.
pH
Time
0
1
2
3
5
7
10
3
4
5
6
7
8
9
10
11
12
13
14
15
Incubation pH
3
5
7
10
Time for Dextrins
to appear
Time for complete
starch digestion
Relative Enzyme
Rate (see Step 9)
50
8. Draw graphs of your results to show the relationship between pH and time required to completely digest the
starch. Place the pH on the x-axis and Time on the y-axis.
‚ Draw a plot of the time required for Dextrins to form.
‚ Draw a second plot of the time required for starch to disappear completely.
Ö What is the relationship between pH and the time required to digest starch?
9. Use your data to find the relationship between pH and the rate of enzymatic reaction.
‚ You did not measure the actual rate of enzymatic reaction, but you can compare the times required to
digest the starch under different pH conditions to each other and calculate the Relative Enzyme Rate for
each pH. Here is the procedure.
1. Look at the times required to completely digest the starch at each pH. Select the shortest time.
2. Divide the shortest time by each of the other times. These fractions are the Relative Enzyme Rates.
Notice that the shortest time will give a value of 1; each of the other times will be a fraction of 1.
Record the Relative Enzyme Rates in the table above.
For example, if the times were 3, 6, 9, and 12 minutes, then the shortest time was 3 minutes.
The Relative Enzyme Rates would be 3÷3 = 1.0; 3÷6 = 0.5; 3÷9 = 0.33; and 3 ÷ 12 = 0.25. The
enzyme that took 12 minutes to complete digestion worked at a rate that was ¼ (0.25) that of the
enzyme that took only 3 minutes to complete digestion.
‚ Draw a graph to show the relationship between temperature and the Relative Enzyme Rate. Place pH on
the x-axis and Relative Enzyme Rate on the y-axis.
Ö What is the relationship between incubation pH and Rate of Enzymatic action?
Ö Will the optimal pH be the same for every enzyme?
51
Experiment 3 – Enzyme Concentration
Goal: Determine whether the concentration of enzyme has an effect on the rate of starch hydrolysis.
Approach: Set up a series of tubes with different concentrations of Amylase then use the Iodine test to determine
when hydrolysis is complete in each tube. You will make SERIAL DILUTIONS of the Pancreatin stock solution
in order to get solutions of varying concentrations. See steps 1 - 5 below.
Procedure:
1. Label 4 tubes with following final dilutions:
1:4, 1:16, 1:64, 1:256
2. Add 6 ml distilled water to each of the tubes.
3. Add 2 ml stock Pancreatin solution to the tube labeled "1:4". Mix. This gives a final dilution of 1:4.
4. Transfer 2 ml of this solution to the tube labeled "1:16". Mix. This gives a final dilution of 1:16.
5. Repeat Step #4 for each of the remaining tubes; each time transferring 2 mls from one tube to the next tube
in the series. Each tube will have a final concentration which is 1/4 the concentration of the previous tube in
the series.
1:4 -----> 1:16 -----> 1:64 -----> 1:256
6. Remove 2 mls of solution from the tube labeled"1:256" and discard the 2 mls. This will leave 6 ml of
Pancreatin in each of the 5 tubes.
7. Add 2 mls 1% Starch solution to all 5 tubes. Mix.
Note the time; this is Time = 0.
The Pancreatin will now start digesting the starch.
8. Immediately take a sample from each tube and put
it on the spot plate. Do the Iodine test.
This is your Time = 0 reading.
9. Repeat the Iodine test at one minute intervals.
Time
0
1:4
Dilution Factor
1:16
1:64
1:256
1
2
3
4
Be sure to shake the tube before taking the sample.
Stop when all 4 samples give a negative Iodine
test or at 15 minutes.
5
Record your data in the table.
Black = starch present
Reddish = dextrins present
Yellow = no starches or dextrins present
8
6
7
9
10
11
12
13
14
15
10. Examine your data and find the time required for
(1) Dextrins to appear, and
(2) complete Starch digestion.
Record your results in the table.
Dilution Factor =
Enzyme Concentration
1:4
1:16
1:64
1:256
Time for Dextrins
to appear
Time for complete
starch digestion
Relative Enzyme
Rate (see Step 12)
52
11. Draw graphs of your results to show the relationship between enzyme concentration and time required to
completely digest the starch. Place the Enzyme Concentration (expressed as dilution) on the x-axis and
Time on the y-axis.
‚ Draw a plot of the time required for Dextrins to form.
‚ Draw a second plot of the time required for starch to disappear completely.
Ö What is the relationship between Enzyme Concentration and the time required to digest starch?
12. Use your data to find the relationship between Enzyme Concentration and the rate of enzymatic reaction.
‚ You did not measure the actual rate of enzymatic reaction, but you can compare the times required to
digest the starch using different enzyme concentrations to each other and calculate the Relative Enzyme
Rate for each enzyme concentration. Here is the procedure.
1. Look at the times required to digest the starch at each Enzyme concentration. Select the shortest time.
2. Divide the shortest time by each of the other times. These fractions are the Relative Enzyme Rates.
Notice that the shortest time will give a value of 1; each of the other times will be a fraction of 1.
Record the Relative Enzyme Rates in the table above.
For example, if the times were 3, 6, 9, and 12 minutes, then the shortest time was 3 minutes.
The Relative Enzyme Rates would be 3÷3 = 1.0; 3÷6 = 0.5; 3÷9 = 0.33; and 3 ÷ 12 = 0.25. The
enzyme that took 12 minutes to complete digestion worked at a rate that was ¼ (0.25) that of the
enzyme that took only 3 minutes to complete digestion.
‚ Draw a graph to show the relationship between temperature and the enzymatic rate. Place Enzyme
Concentration on the x-axis and Relative Enzyme Rate on the y-axis.
Ö What is the relationship between Enzyme Concentration and Rate of Enzymatic action?
53
54
BIO112 - LAB 8
DIFFUSION
Objectives:
By the end of this lab you will
1. be able to describe the fluid mosaic model of the membrane
2. know how the fluid mosaic model accounts for selective permeability
3. know the difference between active and passive transport
4. know the meaning of diffusion and osmosis
5. know how temperature, particle size & concentration gradient affect the rate of diffusion
6. know the meaning of osmolarity and tonicity
7. know how cells will respond to isotonic, hypertonic & hypotonic solutions
8. know the meaning of crenation and hemolysis
9. know the relationship between lipid solubility and permeability
10. know how ionic compounds and covalent compounds differ in their affect on the osmolarity of a solution
11. know all terms that are in boldface type
Materials:
Student Benches:
1. Electronic balance
2. Petri dish with Agar
3. 0.9% NaCl (saline)
4. Powdered Carmine dye
5. Methylene blue
6. Potassium Permanganate
7. Cork borer
8. thread
Side Benches:
1. Demo of diffusion of gases – observe results only
2. Blood Collection: lancets, alcohol swabs, biohazard container, Sharps container
3. Beaker with dialysis tubing soaking in water
4. Sucrose Solutions: 15%, 30%, 45%
5. Methanol, Ethanol, Propanol
6. NaCl Solutions – Dilution Series: 1/6 M, 1/8 M, 1/10 M, 1/12 M, 1/14 M, 1/16 M
7. Glucose Solutions – Dilution Series: 1/6 M, 1/8 M, 1/10 M, 1/12 M, 1/14 M, 1/16 M
Procedures:
You will do a series of experiments to illustrate Brownian motion, diffusion of gases and liquids, and the effect
of concentration gradient on the rate of osmosis across a synthetic membrane. You will also do two experiments
with blood cell membranes. One looks at the effect of lipid solubility on membrane structure. The other looks
at the af
Readings:
Sadava pp. 105 - 111
55
Introduction
The plasma membrane forms an extremely thin boundary between the cell’s interior and its surrounding
environment. The membrane is described as selectively permeable because it allows only a few types of
particles to move across the membrane while excluding most other particles. In addition to acting as a barrier,
the membrane can actively transport some particles in or out of the cell. Thus, the membrane creates a chemical
environment inside the cell that differs from its external environment. The cell must maintain the proper
internal chemical environment in order to live. If the membrane is ruptured, the composition of the cell’s
interior will change and the cell may die.
It is the membrane’s chemical structure that determines how the membrane alters the movement of materials in
or out of the cell. Scientists now know a great deal about the biochemical structure of the membrane. The most
widely accepted description of the membrane is the “fluid mosaic” model that was proposed by Singer in the
1970’s. According to this model, the membrane is formed by an array of lipid and protein molecules that is only
a few molecules thick. Each class of molecules has an impact on the properties of the membrane.
The lipid component is made primarily from phospholipids and cholesterol. The phospholipids are arranged
into two layers that are commonly referred to as a “Phospholipid Bilayer”. Why do they form a bilayer?
Recall that one end of a phospholipid molecule is non-polar (the fatty acid “tails”) and therefore hydrophobic,
but the other end is polar (the phosphate-containing “head”) and therefore hydrophilic. The molecules line up so
that their polar heads face towards either the cell’s cytoplasm or the cell’s exterior. The non-polar fatty acid
tails face towards the interior of the membrane forming a hydrophobic region at the center of the membrane.
This arrangement places the polar heads in contact with water and orients the tails so that they do not come in
contact with water. The only force holding the molecules into place is the hydrophilic attraction of the heads to
water. As a result, the phospholipids may shift within the bilayer, but they cannot move out of the bilayer.
Cholesterol, which is located within the bilayer, helps to stabilize the membrane.
Many different proteins are found in the membrane. They differ in their locations and functions. Proteins that
are located on the surface of the membrane are called peripheral proteins; proteins that are located within the
phospholipid bilayer are called integral proteins. Membrane proteins may function as enzymes, as receptors,
or for the transport of particles across the membrane. The transport proteins are integral proteins that span the
membrane. There are several types of transport proteins. Channel proteins form a tube that crosses through
the hydrophobic region of the membrane and allows for passage of particles. Carrier proteins bind to particles
on one side of the membrane then change shape in order to transport the particle to the opposite side of the
membrane.
The chemical properties of the membrane determine which types of particle can pass through the membrane and
the route that they may take. The phospholipid bilayer allows small, non-polar molecules to cross since they are
soluble in the hydrophobic region of the membrane. Lipids and the gases O2 and CO2 cross via this route. On
the other hand, most particles are not lipid-soluble, and they must cross with the assistance of transport proteins.
For example, ions and small polar molecules such as glucose and amino acids require the aid of proteins for
transport. Each transport protein is specific about which particles it can assist in crossing the membrane. The
types of proteins in the membrane determine which particles can enter or exit the cell.
Movement of particles across the membrane is the result of either passive or active processes. Active processes
are those in which the cell must use energy in the form of ATP. Active processes always involve proteins. The
ATP energy is used to cause a change in protein shape. Passive processes do not require the use of ATP;
energy is supplied by other sources. In today’s lab you will study several passive processes including diffusion,
osmosis, dialysis, and filtration. A description of each process precedes each experiment.
For most of the experiments, you will study passive processes in non-living semi-permeable membranes.
Although the properties of the synthetic membranes are not the same as those of cell membranes, many of the
same basic principles still apply. These synthetic membranes behave much like a kitchen strainer that separates
particles on the basis of size.
56
DIFFUSION
Diffusion is the movement of particles from a region in which they are present at a high concentration
to an area of lower concentration. Diffusion may occur in mixtures of gases or liquids. For example,
perfume diffuses through the air, and a drop of food coloring will diffuse throughout a cup of water.
Since the source of energy for this movement is the kinetic energy (motion due to heat) of the particles,
it is a passive process. Kinetic energy causes particles to move about randomly. When the particles
collide with each other, they ricochet, moving farther apart. Since collisions are more common when
the particles are closer together, particles tend to move out of a region of high particle concentration
into regions with lower concentrations. This net flow of particles down their concentration gradient
(from high to low concentration) is diffusion. Diffusion will continue as long as there is a
concentration gradient. When the concentration of particles is uniform throughout the mixture, the
particles will continue to move, but there will be no net movement.
Diffusion can cause particles to move across a membrane if that membrane is permeable to the
particle. Those molecules which are able to either dissolve in the lipid portion of a membrane, or are
small enough to pass through its pores, will diffuse across the membrane in both directions. If a
concentration gradient exists, the net movement of the molecules will be from the area of higher
concentration to the area of lower concentration.
Experiment # 1
When large particles are mixed with water, they move about randomly as a result of collisions with the
water molecules. This random movement is called Brownian motion. You can observe Brownian
motion under the microscope by observing the movement of powdered dye particles suspended in
water.
Materials
1. Microscope slide & coverslip
2. dropper bottle with water
3. powdered Carmine dye
Procedure
1. Place a small drop of water on a microscope slide.
2. Use a toothpick to pick up a tiny quantity of Carmine dye and mix it with the drop of water.
3. Place a coverslip on the water-carmine droplet
4. Observe with a compound microscope under high power (about 450X). Look for movement by the
small particles.
Ö Do the particles move in random directions? _____________
Ö Do particles of different size move at the same or different speeds? _______________________
Ö What happens to the speed of movement if the temperature increases? You can check this by
leaving the slide on the microscope stage for a few minutes; the microscope lamp will warm the
slide.
57
Experiment # 2
The rate at which particles move is dependent upon their molecular weight. In the following
experiment you will determine the relationship between molecular weight and rate of diffusion. You
will compare the rate of diffusion of potassium permanganate (MW = 158) with methylene blue (MW
= 320) through an agar gel. The agar gel is similar to gelatin, consists of 98.5% water, and is freely
permeable to the particles used here.
Materials
1. Petri dish with agar
2. Cork borer
3. Wax pencil
4. Ruler marked in millimeters
5. 0.1 M Potassium permanganate in a dropper
bottle
6. 0.1 M Methylene blue in a dropper bottle
Procedure
1. Use the cork borer to punch two holes approximately 3 centimeters apart in the agar plate. You
will be able to easily remove the agar if you seal the end of the cork borer with you finger before
pulling away from the agar. Press down on the agar around the holes so that it will seal to the petri
plate.
2. Mark the bottom of the petri dish with the wax pencil so you will know which dye goes into which
hole.
3. Partially fill one of the holes with potassium permanganate solution and the other with methylene
blue. Work carefully so that you do not spill or overfill the holes.
4. Record the time.
5. Cover the dish and set it aside where it will not be disturbed.
6. Use a ruler to measure the distance (in millimeters) that each dye diffused after approximately 30,
60, and 90 minutes. Record your results in the table below.
Time
Distance traveled by methylene
blue
Distance traveled by
potassium permanganate
20 minutes
40 minutes
60 minutes
Ö What can you conclude about the relationship between molecular weight and the speed of diffusion
based upon these results?
________________________________________________________
Experiment #3
You will measure the rate of diffusion by two substances of different molecular weight that are in a
gaseous state. You will observe the diffusion of ammonium hydroxide (NH4OH) and hydrochloric
acid (HCl) in a glass tube. Your instructor will place a small quantity of each substance at opposite
ends of a glass tube. NH4+ and Cl- ions will diffuse towards the center of the tube. When they meet,
they will form a precipitate (NH4Cl) that will coat the inside of the tube. The distance traveled by each
can then be measured. The molecular weight is 18 for NH4+ and 35.5 for Cl- ions.
Ö Which particle diffuses faster?
Ö Is this result consistent with your observations of the rate of diffusion in the agar plate?
58
OSMOSIS
The diffusion of a solvent1, such as water, through a semi-permeable membrane is called osmosis. In the
human body, water is the solvent in the majority of cases.
Osmosis occurs when a semi-permeable membrane separates a solution with low solute concentration (high
water concentration) from a solution with high solute concentration (low water concentration). Water will
diffuse down its own concentration gradient moving from the solution with low solute concentration into the
solution with high solute concentration. The movement of water across the membrane during osmosis, causes
pressure to build on the side of the membrane that has the higher solute concentration. This is called Osmotic
Pressure. The osmotic pressure can be measured. It is equal to the amount of pressure needed to stop the
movement of water across the membrane.
Organisms must deal with osmotic pressure all the time. Plant cells use osmotic pressure to their advantage. As
water enters the cell, the cell expands and presses against the cell wall. The cell wall limits expansion by the
cell and the cell becomes turgid (stiff) and this helps support the plant. The plants also take advantage of
osmotic pressure to help move water throughout the plant body. You will study this in Biology 111. In contrast
to plants, many animal cells may be destroyed by osmotic pressure since these cells have no cell walls to resist
expansion of the cell. As water enters the cell it may cause the cell to swell and burst (this is called cell lysis).
Many animals solve the problem by regulating the concentration of the solutes in the fluids that surround the
cells. For example, animals such as humans use their kidneys to keep the solute concentration of their blood and
extracellular fluids the same as the solute concentration inside their cells.
Osmolarity is a measure of the relative solute concentration of two solutions. Two solutions with the same
osmolarity are isoosmotic. If two solutions have unequal solute concentration, one is hyperosmotic and the
other is hypoosmotic relative to the other. The tonicity of a solution is defined by the response of a cell to
being immersed in the solution. Tonicity reflects both the concentration of the solute particles and the particle’s
ability to cross the membrane. When a cell is placed into an isotonic solution there will be no change in its size
due to osmosis. When a cell is placed in a hypotonic solution, water will enter the cell and the cell will swell.
Red blood cells (RBCs), which have no mechanism to oppose this swelling, will swell until they burst. This is
called hemolysis. When a cell is placed into a hypertonic solution, water will exit the cell and the cell will
shrink. This is called crenation. An RBC will appear to collapse around its cytoskeleton and appear bumpy.
Experiment #4
In this experiment you will see how the relative osmolarity of the solutions on either side of a membrane affects
the rate of osmosis. You will use a synthetic membrane made from cellophane. The properties of this semipermeable membrane are different from those of a cell membrane since it limits movement of particles only on
the basis of their size. Large particles cannot cross the membrane. For example, glucose is small enough to
cross the membrane, but sucrose is too large to cross. In this experiment, you will use commercially produced
cellophane tubing (dialysis tubing) to create bags. Each bag will be filled with a solution of sucrose, sealed shut,
and then immersed in either water or a sucrose solution. You will measure the movement of water by measuring
changes in the weight of each bag.
Materials:
1. 5 pieces of 15 cm cellophane tubing, pre-soaked
2. thread
3. 5 - 250 ml beakers
4. Sucrose solutions: 20%, 40%, 60%
1
A solution is a homogeneous mixture of two or more substances; generally a liquid and some other solid or
gaseous substance. In a solution the solvent and solute always remain mixed. The most abundant substance in
the mixture (usually a liquid) is the solvent. The other substances are the solutes. In most biological systems,
water is the solvent and salts, sugars, and other small polar molecules that dissolve in the water are the solutes.
Suspensions are heterogeneous mixtures in which the particles tend to separate. Flour and water form a
suspension.
59
Procedure:
1. Fill 5 beakers with water or sucrose solution as indicated in the table below.
2. Prepare 5 bags as follows.
⋅ Tie a knot in the end of a piece of cellophane tubing so that it forms a bag about 5 inches long.
Keep the cellophane tubing wet while working with it. See Figure 1.
⋅ Fill each bag with about 10 ml of the solution indicated in the table below. It should be about
half full.
⋅ Use thread to close the end of the bag. Tie the knot near the end of the bag so that there is slack
in the bag. This provides room for water to enter the bag.
⋅ Rinse the bag under tap water to remove any sucrose from the bag’s surface. Pat dry on paper
towel.
⋅ Weigh each bag on the electronic balance. Record the mass in the table in the row Time = 0
3. After all 5 bags have been weighed, place each bag in the appropriate beaker
4. Visually check for signs that the bags are leaking. If any are leaking, start over.
5. Measure the mass of each bag at 15 minute intervals. Record your results in the table below.
6. Calculate the change in mass (∆ Mass) of the bag since Time = 0.
7. For each bag, plot the ∆ Mass (y axis) against the time (x axis).
Figure 1
Select a piece of
tubing and form a
knot at one end
Fill the bag with about 10
ml of Glucose. The bag
should be about ½ full.
1
Time
(min.)
2
Bag 1 = dH2O
Beaker = dH20
Mass
(g)
Squeeze gently to remove air.
Fold the end over and tie
securely with a piece of thread.
∆ Mass (g)
Bag 2 = 15% sucrose
Beaker = dH20
Mass
(g)
∆ Mass (g)
Rinse bag
under tap
water
3
Bag 3 = 30% sucrose
Beaker = dH20
Mass
(g)
∆ Mass (g)
Place the bag in a beaker of
distilled water. Check for leaks.
4
Bag 4 = 45% sucrose
Beaker = dH20
Mass
(g)
∆ Mass (g)
5
Bag 5 = dH2O
Beaker = 45% sucrose
Mass
(g)
∆ Mass (g)
0
15
30
45
60
Demonstration
Your instructor will set up two dialysis bags attached to capillary tubes. Each bag will be filled with
different concentrations of sucrose then immersed in a beaker of water. Observe the movement of
water in the tubes. Compare these results to your results in the previous experiment.
60
BLOOD CELLS CAN BE USED TO STUDY MEMBRANE BEHAVIOR
Human red blood cells are very sensitive to changes in osmotic pressure; when placed in a hypotonic
solution they undergo hemolysis. You can see if hemolysis has occurred by (1) looking at a drop of
blood under the microscope; all cells will be ruptured and only the membrane fragments are present, or
(2) looking at the appearance of a suspension of blood with the naked eye. A suspension of intact
blood cells (i.e. cells in either an isotonic or hypertonic solution) is cloudy and distorts the passage of
light through it. A suspension of hemolyzed blood cells (i.e. cells in a hypotonic solution) is clear and
allows light to pass through. This difference between the appearance of a suspension of normal and
hemolyzed blood cells provides a simple way to determine whether blood cell membranes are intact or
damaged. You will use this “Hemolysis Test” as a tool in the last two experiments.
Hemolysis Test
Hold up the test tube containing the blood suspension in front of a page of printed text.
‚
‚
If the text is not legible (solution cloudy), the blood cells are intact :. the solution is either isotonic or hypertonic.
If the text is
legible, (solution clear), hemolysis has occurred :. the solution is hypotonic.
Preparation of Blood Cell Suspension = your source of blood for the last two experiments.
œ Work carefully so that you do not expose anyone to your blood. Work on top of 2 sheets of paper towel.
1. Add 3 ml of 0.9% saline to a small test tube.
2. Clean the fingertip of 1 student volunteer with an alcohol swab.
3. Use the sterile lance to draw blood.
4. Add 5 drops of blood to the test tube. Stretch parafilm over the opening of the tube and mix. Use
this suspension as your source of blood for the rest of the procedures.
œ Discard the Lance in the Sharps container. Discard any blood-contaminated paper materials in the
biohazard bucket. Report any spilled blood to your instructor. The blood should be handled only by the
student that contributed the blood.
Does A Molecule’s Lipid Solubility Affect its Ability to Diffuse Through the Plasma Membrane?
Based upon the structure of a membrane, we might expect that lipid soluble substances should be able to diffuse
across a membrane. If this is true, then substances that are very lipid soluble should diffuse across the
membrane faster than substances that are less soluble in lipid.
The Partition Coefficient is a measure of the relative solubility of a substance in oil and water. The higher the
partition coefficient is, the higher the lipid solubility.
Solubility in lipid
Partition Coefficient = Solubility in water
The chart below shows the partition coefficient for three alcohols. Notice that the partition coefficient increases
as the length of the carbon chain in each alcohol molecule increases.
Experiment #5
Goal: In this experiment, you will compare the rate at which these three alcohols diffuse across the membrane.
You can measure the rate of diffusion by placing blood cells into each alcohol and measuring the time until the
cells undergo hemolysis. See the description of the Hemolysis Test above.
Tube
Alcohol
Formula
Partition
Coefficient
1
Methyl alcohol (methanol)
CH3OH
0.010
2
Ethyl alcohol (ethanol)
C2H5OH
0.036
3
Propyl alcohol (propanol)
C3H7OH
0.156
Time until hemolysis
(Seconds)
61
Procedure
1. Add 2 ml of each alcohol to a small test tube.
2. Add 2 drops of your blood suspension to the first test tube.
3. Record the number of seconds until hemolysis occurs (suspension becomes clear).
Be ready. The times may be very short (a few seconds).
4. Repeat for each of the other two alcohols.
Ö Which alcohol diffuses across the membrane fastest?
Ö Do your results support the expected relationship between lipid solubility and membrane
permeability?
Ö Based upon your results, will Hexanol cross the membrane faster, or slower, than propanol?
Ionic Compounds and Covalent Compounds differ in their impact on osmotic pressure.
Ionic Compounds and Covalent Compounds differ in the number of solute particles that they form in a
solution. This affects their impact on the osmotic pressure of the solution since each solute particle in
a solution contributes to the osmotic pressure of the solution. Covalent compounds do not ionize in
water, so their impact on a solution’s osmotic pressure is directly related to their molar concentration2
in the solution. In contrast, ionic compounds can ionize in solution creating 2 or more particles that
contribute to osmotic pressure. Ionic compounds differ in the extent to which they ionize. Thus, the
impact of an ionic compound on osmotic pressure depends on both its molar concentration in the
solution and its tendency to ionize in solution.
In this experiment, you will (1) determine the concentration of glucose (a covalent compound) that
yields an isotonic solution, (2) determine the concentration of NaCl (an ionic compound) that yields an
isotonic solution, and (3) calculate the proportion NaCl that is ionized. This is an optional activity.
How can you determine if a solution is isotonic? You will use the Hemolysis Test. You will prepare a
set of 7 tubes that contain a concentration series for one solute. You will then add blood to each tube
and perform the hemolysis test (as described above). If hemolysis occurs, then you know that the
solution is hypotonic. If hemolysis does not occur, then the solution is either hypertonic or isotonic.
We will assume that the tube with the lowest solute concentration that is not hypotonic is isotonic.
Experiment #6
Procedure:
1. Label 7 test tubes for the glucose series and 7 tubes for the NaCl series.
The following concentrations will be used: Tube 1 = 1/6 M, Tube 2 = 1/8 M, Tube 3 = 1/10 M,
Tube 4 = 1/12 M, Tube 5 = 1/14 M, Tube 6 = 1/16 M, and Tube 7 = 1/18 M.
2. Add 2 mls of the appropriate glucose or NaCl solution to each of the tubes.
3. Add 2 drops of the Blood Cell Suspension to each of the tubes and mix.
4. Let the tubes stand for 15 – 20 minutes.
5. Examine the tubes for hemolysis. See the description of the Hemolysis Test above if you have forgotten
what you should see. Record your results in the table.
‚ Indicate on the table which tubes have hemolysis and which tubes have no hemolysis.
‚ Based on your results, indicate on the table which tubes contain hypotonic solution and which contain
isotonic or hypertonic solution. We will assume that the tube with the lowest solute concentration that
is not hypotonic is isotonic.
2
Molar Concentration. The mole is a measure of the number of molecules of a substance. One mole of a molecule
contains 6.02 x 1023 molecules. A 1M (=1 molar) solution has 1 mole of solute/liter of solution. A 2M solution contains 2
moles/liter of solution. Molar concentration differs from a percent concentration which is based on mass/volume.
62
Tube Number and Molarity of the solution
Decreasing solute concentration
Solution
Tube 1
1
/6 M
Tube 2
1
/8 M
Tube 3
Tube 4
Tube 5
Tube 6
Tube 7
1
1
1
1
1
/10 M
/12 M
/14 M
/16 M
/18 M
Glucose
NaCl
Ö Which isotonic solution (Glucose or NaCl) has the higher solute concentration?
Ö Explain this result.
Ö If you had to prepare an isotonic solution of KCl, what molar concentration should it have based
upon your results?
---------------------------------------------- Optional Activity --------------------------------------------------Isotonic Coefficient
The Isotonic Coefficient provides a measure of the number of solute particles formed by an ionic
compound relative to a covalent compound such as glucose.
Isotonic Concentration of glucose (M)
Isotonic Coefficient = Isotonic Concentration for the ionic solute (M)
Ö The more a compound ionizes in solution, the larger the coefficient will be. Examine the equation
and your results to see why.
Ö Based on your results, what is the isotonic coefficient for NaCl? __________________________
You can calculate the percentage of NaCl particles that have ionized with the following equation.
i
a = 100 X
1 + (k-1)
where i = the Isotonic Coefficient
k = the number of ions produced by ionization of the compound. K = 2 for NaCl.
a = the percentage of the solute compound that is ionized
⋅
Suppose that you want to make solutions of MgCl2 and NaCl that are isotonic. Which solution
(MgCl2 or NaCl) would need to have the higher molar concentration?
---------------------/ /-----------------------
63
64
BIO112 - LAB 9
CELLULAR RESPIRATION
Objectives:
1. know the meaning of anaerobic and aerobic
2. be able to name and describe (starting materials/end products/yields) the anaerobic metabolic pathways
3. know how we monitor the anaerobic metabolism in yeast
4. be able to name and describe (starting materials/end products/yields) the aerobic pathways
5. know how to use the respirometer to monitor cell respiration
6. know what Redox Reactions are
7. know the action of dehydrogenases, oxidases, catalases, and peroxidases
8. be able to explain how we were able to monitor the action of turnip peroxidase
Materials:
Yeast + sucrose suspension
Yeast suspension
Germinating peas
Turnip root
Guaiacol
Glass beads
KOH pellets
Manometer =Side-arm flask with clamp and stopper with pipette
Respirometer = large test tube and stopper with capillary tubing
Dye to use as marker in the respirometer
Spec 20
Procedures:
Each group will use manometers to measure fermentation in yeast. You will use a respirometer to measure
respiratory rate in peas at 2 temperatures. You will also use the Spec 20 to monitor the action of peroxidase
enzymes that are found in the turnip root.
Readings:
Sadava, Chapter 7
65
Cells metabolize glucose and other organic molecules in order to acquire energy and the materials
needed for synthesis of macromolecules. The energy that is released during the breakdown of glucose
is used to produce ATP. Most cell machines (enzymes, pumps, motor proteins) use ATP as fuel.
Glucose metabolism requires the actions of many enzymes to break down glucose and transfer the
energy to ATP. Each enzyme performs a specific action; each requires a specific substrate molecule
and produces a specific product. Since the product of one enzymatic reaction becomes the substrate
for the next enzymatic reaction, cell biologists refer to these sequences of enzymatic reactions as
enzymatic pathways. Several major pathways are involved in the break down of glucose. The
enzymes for each pathway are found in a specific location within the cell. Each has a name. The
enzymatic pathways are very similar in most organisms. In this lab you will study these major
pathways.
Glycolysis - The enzymes for glycolysis are located in the cytoplasm. Glycolysis splits glucose into
two 3-carbon molecules called G3P. This requires ATP energy. G3P is then converted into Pyruvate.
Energy that is released in this process is used to convert ADP+ Pi into ATP and reduce NAD+ to
NADH.
1 Glucose + 2ATP + 2NAD+ + 4(ADP+Pi) ―2G3P→ 2Pyruvate + 2(ADP+Pi) + 4ATP + 2NADH
[6C]
[3C]
[3C]
As a result of glycolysis, the amount of pyruvate in the cell increases and the amount of NAD+
decreases. Cells must have a way to (1) remove the pyruvate as it forms, and (2) get more NAD+ or
else glycolysis will stop. Cells differ in the methods used. Some types of cells use fermentation
reactions which do not require oxygen. Other cells use aerobic pathways which do require oxygen.
Fermentation Reactions & Anaerobic Glucose Metabolism
Fermentation reactions give cells a way to remove pyruvate and to regenerate NAD+ (from NADH
produced in glycolysis) so that glycolysis can continue. Fermentation does not produce any ATP, but
it is necessary so that glycolysis can continue.
Fermentation can occur in several ways.
1. Yeast cells use alcoholic fermentation:
2Pyruvate [3C] + 2NADH →→ 2Ethanol [2C] + 2CO2 + 2NAD+
2. Certain bacteria and human skeletal muscle produce lactate:
2Pyruvate [3C] + 2NADH → 2Lactate [3C] + 2NAD+
Cells that metabolize glucose by using glycolysis followed by fermentation do not require oxygen.
These cells are anaerobic. Anaerobic glucose metabolism captures only a small fraction of the
energy that is contained in the glucose.
Aerobic Respiration
Many cells can capture more energy from pyruvate by using a pathway that requires oxygen. These
cells are described as aerobic, and this oxygen-requiring process is called aerobic respiration. The
enzymes that carry out aerobic respiration are located in the mitochondria. Aerobic cells use
glycolysis followed by three additional processes to completely breakdown glucose to CO2 and capture
energy as 38 ATP. The overall reaction is shown here.
C6H12O6 + 6O2 + 38(ADP+Pi) ―→ 6CO2 + 6H2O + 38ATP
66
The following aerobic processes occur inside the Mitochondria:
Oxidation of Pyruvate
This is a single enzymatic step which converts the pyruvate that is made by glycolysis into acetate and
carbon dioxide. Hydrogen is transferred from pyruvate to NADH.
2Pyruvate[3C] + 2NAD+ + 2 CoA → 2Acetyl CoA[2C] + 2CO2 + 2NADH
Krebs Cycle
This pathway completes the breakdown of acetyl CoA to CO2. Hydrogen is transferred from acetyl
CoA to NADH and FADH2. Some energy is used to make ATP.
2AcetylCoA[2C] + 6NAD+ + 2FAD + 2(ADP + Pi) → 4CO2 + 6NADH + 2FADH2 + 2ATP
Electron Transport System (ETS)
The ETS is the process that uses oxygen. The ETS removes the Hydrogen from NADH and FADH2
and transfers it to Oxygen forming water. The large amount of energy that this releases is used to
produce ATP through chemiosmotic phosphorylation.
NADH + H+ + ½O2 + 3(ADP + Pi) → NAD+ + H2O + 3ATP
FADH2 + H+ + ½O2 + 2(ADP + Pi) → NAD+ + H2O + 2ATP
In this lab you will demonstrate fermentation in yeast, respiration in peas, and the action of the
enzymes that carry out reactions similar to those seen in the ETS.
A. ANAEROBIC GLUCOSE METABOLISM:
The alcoholic fermentation of glucose by yeast cells has great commercial value. Fermentation of
grapes and other fruits yields the alcohol found in wine and other alcoholic beverages. Bakers use
yeast to make raised breads. The general equation for fermentation is:
glucose [C6H12O6] → pyruvate → 2 ethyl alcohol[2C] + 2CO2 + 2ATP
The rate of this process can be determined by measuring the rate at which CO2, a gas, is produced.
Experiment # 1
1. Make up these two samples in a beaker:
sample 1 = 60 ml yeast suspension + 60 ml water + 60 ml
sucrose solution
sample 2 = 60 ml yeast suspension + 120 ml water
2. Pour enough of sample 1 into a flask to completely fill it. Mark
this flask #1.
Pour enough of sample 2 into a second flask to completely fill it.
Mark this flask #2.
3. Seal each flask with a rubber stopper that holds an inverted 10 ml pipette. The solution will rise up
inside the pipette as the stopper is pressed in. Be careful, it may even overflow. Open the clamp
67
attached to the plastic tubing on the side arm of the flask. Lower the level of sample in the pipette
drop to the 0.0 mark. You will need to drain this liquid into a beaker or sink. Close the clamp.
4. Record the level of the liquid in the pipette every 10 minutes for 60-70 minutes.
TIME (minutes)
Flask #1 Pipette (ml)
Flask #2 Pipette (ml)
0
10
20
30
40
50
60
70
⋅
Note the accumulation of gas in the flask. What is this gas?
⋅
Open the flask and smell the aroma. What is the smell?
⋅
What is the function of Flask #2? What does it measure?
⋅
Why does liquid rise up the pipette? Why does the liquid in Flask #1 rise faster?
B. MEASURING RESPIRATION IN PEAS WITH A RESPIROMETER
Cells that are carrying out aerobic respiration consume O2, so we should be able to measure the rate of
respiration by measuring the loss of O2 from a chamber in which an organism is respiring.
Unfortunately, respiration also produces CO2. If we are measuring changes in gas volume, any
decrease in gas volume due to O2 consumption will be masked by CO2 production. What is needed is a
way to remove CO2 as it is produced so that it will not alter the gas volume. This can be done by
creating a chemical “trap” to remove the CO2.
We can use a device called a respirometer to measure changes in the volume of the gas around an
organism that is respiring. The respirometer is simply a closed chamber that can change volume as the
organism produces/consumes gases. KOH pellets will act as a trap for CO2. The changes in gas
volume due to respiration will be very small so it must be capable of measuring tiny changes in
volume. Respirometers are very sensitive to changes in volume due to temperature change, so you will
set up two respirometers. One respirometer chamber will have the organisms inside. The second
respirometer chamber will have no organisms, and it will serve as a control which reflects changes in
volume as a result of temperature change. This is called the compensation chamber. The true
volume of oxygen consumed by the organism is equal to the difference between the volume change in
the chamber with the organism and the volume change in the compensation chamber.
68
Experiment #2
1. Mount 2 large test tubes (20 cm x 20 mm) on a ring stand.
2. Fill one tube to a depth of 12 cm with germinating peas.
Fill the other tube with an equal volume of glass beads.
| | | | | | | | | | | | |
Treat each tube in the same way for the remaining steps
millimeter
of the procedure.
3. Loosen the fibers of ½ ball of cotton and place it on top
of the peas and glass beads. This will separate the
KOH from the peas.
KOH
4. Add KOH to a depth of about 1 cm in each tube.
cotton
Peas or
5. Get the stoppers that will be used to seal the test tube.
Glass Beads
It will have a section of glass tubing bent at a right
angle and a second piece of tubing with a short piece of
rubber tubing and a clamp. Attach a millimeter ruler to
the glass tubing so that the millimeter rule lies parallel to the tube. Remove the tubing clamp.
Inspect the rubber tubing for cracks. Then put the stopper in place on each test tube.
6. Lower the 2 tubes into the fishbowl filled with water that is at room temperature. Let the
temperature in the tubes stabilize for 10 minutes.
7. Add a drop of dye to the tip of the glass tube with a pipette. Let the temperature stabilize for a few
minutes. The dye can move freely in the glass tube in response to any change in gas pressure in
the test tube.
8. Put the clamp on the rubber tubing.
9. Record the position of the dye on the millimeter ruler.
10. Record the position of the dye on the ruler at 1 minute intervals for 5 minutes.
11. When you finish taking your readings at room temperature, remove the tubing clamp, then repeat
the measurements (steps 6 through 10) at a different temperature by placing the respirometers in a
water bath.
12. Plot both sets of data. Place time on the X-axis. Place “volume” on the Y-axis. The Volume is the
Net Movement of the dye (= movement in the Respiring Peas tube – movement in the
Compensation Chamber), expressed in millimeters. Indicate the temperature used on each graph.
Distance Moved by Dye (mm)
Time
elapsed
(minutes)
Room
Temperature
=
0
C
0
Respiring
Peas
Ruler
Reading
Compensation
Chamber
P = Total Distance
Moved (mm)
0
Ruler
Reading
C = Total Distance
Moved (mm)
0
1
2
3
4
5
69
Net Movement
= Peas – Compensation
(mm)
0
Distance Moved by Dye (mm)
Time
elapsed
(minutes)
Temperature
=
0
C
0
Respiring
Peas
Ruler
Reading
Compensation
Chamber
P = Total Distance
Moved (mm)
0
Ruler
Reading
C = Total Distance
Net Movement
= Peas – Compensation
(mm)
Moved (mm)
0
0
1
2
3
4
5
⇒ At what temperature was Oxygen consumed faster?
⇒ Obtain the results of other students for different temperatures. What is the relationship between
temperature and the rate of glucose metabolism?
70
C. REDOX REACTIONS
The ETS consists of several membrane-bound proteins which transfer H atoms (electrons and protons)
from NADH and FADH2 to O2 through a series of redox (oxidation-reduction) reactions. These
redox reactions release energy which is used to drive H+ pumps. The H+ pumps create concentration
gradients that are used by the enzyme ATP synthase to make ATP.
Redox reactions involve the transfer of electrons from one molecule to another molecule. The
molecule that receives the electrons is “reduced” by the reaction. The molecule that supplies the
electrons is “oxidized”. The electrons are commonly transferred as an atom of H (i.e. 1 electron + 1
proton). We can describe the overall result of the ETS as being the oxidation of NADH and the
reduction of oxygen to form H2O. Oxygen is the final electron acceptor in the respiration of glucose.
Dehydrogenases are enzymes that transfer H from one molecule to another molecule.
AH2 + B ―→ A + BH2
An Oxidase is a type of dehydrogenase that transfers H from a substrate to Oxygen forming H2O2.
AH2 + O2 ―→ A + H2O2
H2O2 is Hydrogen Peroxide. It is toxic and must be broken down by other enzymes. This can be done
several ways.
Catalase is an enzyme that converts 2 molecules of hydrogen peroxide to water and oxygen:
2H2O2 ―→ 2H2O + O2
Peroxidase is an enzyme that removes H from hydrogen peroxide and transfers it to another molecule.
This molecule is reduced.
H2O2 + B ―→ BH2 + O2
Demonstration of Peroxidase Activity
In this experiment you will study a reaction that is catalyzed by peroxidase. The peroxidase will be
obtained from turnips. The peroxidase will break apart H2O2. releasing H2 and O2. The H2 will reduce
O (gas) to make H2O. The O2 that was released will be reduced by Oxidase using H2 obtained from a
molecule called Guaiacol that we will add to the reaction mixture. Guaiacol-H2 (in its reduced form)
is colorless. Guaiacol-- (in its oxidized form) has color. We will use this color change in guaiacol as a
way to indirectly follow the activity of the peroxidase reaction.
H2O2 + O2 ―peroxidase → 2H2O + O2
Guaiacol-H2 ――oxidase→ Guaiacol-- + 2H2O
(Reduced, no color)
(oxidized, color)
The overall reaction can be summarized by:
Guaiacol-H2 + H2O2 + O2 ―peroxidase→ Guaiacol-- + 2H2O
(Reduced, no color)
(oxidized, color)
Reduced Guaiacol is colorless. Oxidized Guaiacol is brown. We can use the Spectrophotometer to
measure the color change that is produced by the reaction.
71
Experiment #3
Your instructor will prepare the Turnip Extract as follows:
‚ Homogenize 27 g of turnip flesh + 50 ml water in the blender.
‚ Filter the turnip pulp through cheese cloth into a beaker.
‚ Dilute the filtrate with 200 ml water
1. Turn on the Spec 20 and set the filter to 500 nm. Let if warm up for 20 minutes.
2. Label 4 large test tubes (A-D) and prepare as follows:
⋅ Tube A = 4.0 ml water + 1.0 ml boiled Turnip Extract (boil for 2 minutes in a water bath)
⋅ Tube B = 4.5 ml water + 0.5 ml Turnip Extract
⋅ Tube C = 4.0 ml water + 1.0 ml Turnip Extract
⋅ Tube D = 3.0 ml water + 2.0 ml Turnip Extract
3. Prepare your “blanks”. Fill 4 small test tubes with the following:
5 ml H2O + 1 drop Guaiacol solution + 2 drops H2O2
4. Do the following for Tube A, then repeat for each of the remaining tubes.
a. Set the mode on the Spec 20 to “% Transmittance”.
b. Insert one of the “blanks” (step 3 above), into the Spec 20. Adjust the % Transmittance to 100%.
c. Remove the “blank” tube from the Spec20 and quickly pour its contents into Tube A (step 2).
Quickly mix them together, then, pour the mixture back into the small test tube.
d. Wipe the small tube clean and insert it into the Spec 20. Immediately read the % Transmittance
and record.
e. Leave the tube in the Spec 20 and take a reading every 20 seconds for 2 minutes.
5. Plot your data for each tube so that you can see how % Transmittance changes as a function of
time. Plot Time on the x-axis and %Transmittance on the y-axis.
% Transmittance
Time
Tube A
Tube B
0
20 seconds
40 seconds
60 seconds
80 seconds
100 seconds
120 seconds
72
Tube C
Tube D
73