Biology 155 Laboratory Supplement: 1
Biology 155 - General Biology I
Laboratory Supplement
David L. Schultz
cite this document as:
Schultz, D. L. 2006. Biology 155 General Biology I Laboratory Supplement. 78 pp.
Biology 155 Laboratory Supplement: 2
Table of Contents
Enzyme Action: Catalytic Properties of Catechol Oxidase
3
Intracellular Localization of Respiration and Glycolysis
22
Genetic Transformation of Acinetobacter
30
Pedigree Analysis
36
Evolution by Genetic Drift and Natural Selection
46
An Introduction to Systematics Using the Caminalcules
64
Writing Scientific Reports
75
Biology 155 Laboratory Supplement: 3
Enzyme Action:
Catalytic Properties of Catechol Oxidase
Synopsis -- This experiment investigates the activity of the enzyme catechol oxidase in
homogenate of potato. The course of the enzyme-catalyzed reaction is followed using a
spectrophotometer. Using variations on the basic protocol we will observe the effects of several
factors on the reaction including (a) change in enzyme concentration; (b) change in substrate
concentration; (c) temperature variation; and (d) enzyme inhibitors. The results obtained will allow
conclusions to be made about the specific properties of this enzyme and the general properties of
enzymes.
Objectives
• To learn the basis of several techniques:
cellular homogenization
centrifugation
spectrophotometry
graphing
calculation of slopes from graphs
pippetting
serial dilution
• To learn the basics of experimentation, especially what is meant by control and treatment.
• To observe the behavior of an enzyme-catalyzed reaction under various conditions
varying amounts of enzyme
varying amounts of substrate
varying physical conditions
presence of inhibitors
Upon completion of this laboratory you should be able to:
• Plan and carry out a serial dilution of a substance.
• Construct a graph of absorbance units versus time from data.
• Determine the relative rate of the reaction from the graph and data.
• Construct a graph of reaction rate versus time from data.
• Use the results of several experiments to describe the general effects of a variable on reaction
rate.
Introduction
Enzymatic Reactions -- All living cells carry out a great variety of chemical reactions essential
for maintenance of life. Chemical reactions in living cells are different from chemical reactions in
most other systems in that (1) they are much faster (reactions that are two slow to measure in nonliving systems can occur very rapidly in cells); (2) they are highly ordered, regulated, and balanced,
with hundreds of interdependent reactions proceeding simultaneously. The source of both of these
differences is that the chemistry of living things is carried out by specific biological catalysts,
called enzymes, that accelerate and regulate biochemical processes. An understanding the
properties of enzymes is essential for an understanding of the chemical basis of life.
Biology 155 Laboratory Supplement: 4
Enzymes are proteins and like all proteins, their properties are determined by their structure.
The structure of proteins is ultimately determined by the sequence of amino acids because the
amino acid sequence determines the three-dimensional conformation. Associations of amino acids
in distant parts of the polypeptide chain cause the chain to twist and fold upon itself giving it a 3-D
shape. The 3-D conformation creates regions or sites that can associate with other chemicals and
catalyze their conversion to some other chemical entity. The catalytic or active site of an enzyme is
very specific. The amino acids that are found there can only associate with a limited class of
chemicals. Because of this, a given enzyme can only catalyze a very limited number of reactions.
Most often, an enzyme will catalyze only one specific reaction. In addition to the active site, other
sites are found on the enzyme that can associate with other chemicals. Thus, the presence of other
chemicals can influence the catalytic properties of enzymes. One important class of chemicals that
can influence enzyme action is cofactors. Cofactors are enzyme specific and highly diverse,
ranging from simple ions to large molecules. They usually associate with the enzyme at or near the
active site and can greatly enhance the catalytic properties of the enzyme. In part of the experiment
that follows you will investigate the role of cofactors.
Usually, any change in the conformation of an enzyme will affect the catalytic activity of
the enzyme. High temperature or extremes of pH, can cause an otherwise stable enzyme to unfold,
and loose its conformation and properties. More subtle changes in pH, temperature, or the
concentration of specific chemicals, can reduce or enhance an enzyme's activity to lesser degrees.
Cells actively maintain conditions that allow enzymes to function at rates appropriate to their
current needs.
One of the important properties of enzymes is that although they can cause chemical
changes in other molecules, they are themselves unchanged by the reaction. As a result, a single
enzyme can catalyze the transformation of many molecules.
Enzymes can only promote chemical transformations that are favorable, and happening as a
slow rate anyway. They can't make the impossible happen, but they can greatly enhance the natural
tendency of chemicals to associate, combine, dissociate or decompose. They do this by lower the
activation energy of the reactants.
The reactants of any enzyme catalyzed reaction are called substrates and the results of the
reaction are called products. At temperatures favorable for living things many substrates are very
stable, but they can become unstable if they are heated or if another source of energy is provided.
The amount of energy that must be added to a chemical system to the promote the conversion of
substrates to products is called the activation energy. Enzymes interact with substrates in a manner
that lowers the activation energy, and thus promotes the reaction. The enzyme may change the
physical shape of the substrates, or change in the charge distribution on the substrate and
destabilize chemical bonds. The way enzymes promote chemical reactions is summarized by the
"enzyme-substrate complex" model:
ENZYME + SUBSTRATE(S) ↔ ENZYME-SUBSTRATE COMPLEX ↔ ENZYME + PRODUCT(S)
According to this model the enzyme can only bind a single set of substrates at a time,
forming an enzyme-substrate complex. While they are bound the reaction occurs. Next, the
enzyme dissociates from (releases) the product and then can bind another set of substrates. Because
the enzyme is not consumed, and because enzymes can make reactions happen extremely quickly,
often only small amounts of enzyme are needed for the rapid production of products. In some
cases, enzyme concentrations are so low that they can only be detected through their ability to
catalyze a specific reaction.
Enzyme Activity -- Enzyme activity is measured by determining the rate at which it catalyzes a
reaction. Rate is expressed as a change in concentration of either a substrate or a product over a
period of time. The concentration of enzyme, substrate and product can all influence the reaction
rate.
Biology 155 Laboratory Supplement: 5
Enzyme activity can also be influenced by the presence of other substances, such as
cofactors (as mentioned above) and inhibitors. Inhibitors are defined as chemicals that decrease
enzyme activity. The effect of an inhibitor measured as the percentage decrease in reaction rate.
Percent inhibition is calculated as:
Percent Inhibition =
Rate without inhibitor − rate with inhibitor
× 100
rate without inhibitor
Percent inhibition can range from 0 to 100%. Percent inhibition depends upon the
concentration of the inhibitor, concentration of the substrate, and concentration of the enzyme. To
measure percentage inhibition you must measure reaction rate for a given concentration of substrate
and enzyme, and then repeat all measurements with the addition of an inhibitor.
Inhibitors influence the rate of reaction by interacting with the enzyme. If the inhibitor is
chemically similar to the substrate, it may interact with the enzyme at the active site and thus
restrict access of substrate molecules to the active site. If such an inhibitor interacts with the active
site in a reversible manner, in other words it is capable of associating (binding) an dissociating
(unbinding) from the active site, it is called a competitive inhibitor. It is called competitive because
it will compete with the substrate for access to the active site. While the inhibitor is bound, the
substrate can't bind, but when the inhibitor dissociates the substrate can bind and be converted to
product. The percentage inhibition of a competitive inhibitor varies with the concentration of
substrate. High substrate concentration relative to inhibitor concentration will result in lower
percent inhibition than will low substrate concentration.
In contrast, the percent inhibition produced by noncompetitive inhibitors is not influenced
by substrate concentration. Noncompetitive inhibitors do not bind to the enzyme reversibly,
usually do not interact with the active site, and thus they do not compete with the substrate for
access to the active site. Some noncompetitive inhibitors bind to the enzyme and change its
conformation, and thus change its ability to interact with the substrate. Others bind to cofactors and
prevent them from interacting with the enzyme. Either mode of noncompetitive inhibition results
in a fractional decrease in the activity of a solution of enzymes. Thus, percentage inhibition with a
noncompetitive inhibitor will be constant at any substrate concentration.
Biology 155 Laboratory Supplement: 6
Biological Significance of Catechol Oxidase
Catechol oxidase if found widely in both plants and animals. Catechol oxidase is also
called polyphenoloxidase and in animals it is called tyrosinase. In this laboratory we will use a
form of this enzyme that is found in potatoes.
Potatoes are members of the family Solanaceae (so are tomatoes, jimson weed, and deadly
nightshade). Solanaceous plants originated in the new world, and the potato and its relatives were
first introduced to Europe in the 16th century. Potatoes and tomatoes have since become a standard
part of the diets of people worldwide.
The edible part of a potato, the tuber, is derived from the potato stem. It is intended to be
used by the plant for vegetative (asexual) reproduction. It is rich is starch and has some protein just
under the skin. These nutrients are stored in the potato and are intended by the plant to be used for
supplying energy and raw materials for new plant growth. However, the starch and protein are also
a good source of nutrition for many other organisms, including microbes, fungi, insects, and
mammals. Although potatoes are readily exploited by many organisms, they are not defenseless.
They have a relatively tough skin and they can produce chemicals that some organisms find
noxious. One class of chemicals that is noxious to many microbes and insects is the quinones.
Catechol oxidase, the enzyme that will be used in this laboratory exercise, catalyzes the production
of a quinone.
Catechol oxidase catalyzes the oxidation of catechol to ortho-quinone and water. The
reaction is diagrammed below. While catechol is colorless, ortho-quinone has a brown color. The
familiar browning of cut apples and potatoes to due to the production of ortho-quinone from
catechol. The brown colored product of this reaction can be used to measure the reaction rate using
a spectrophotometer, a device that allows quantification of the amount of a colored substance in
solution.
There is evidence that insect attacks on potatoes causes increased production of orthoquinone through increased enzyme activity, possibly due to increased enzyme concentration.
Quinones promote healing of the potato through inhibition of further insect attacks and inhibition of
the growth of microbes.
Quinones belong to a highly variable group of compounds called "secondary plant
compounds". One thing all secondary plant compounds have in common is that they appear to be
intended to be used to influence other organisms to interact or not interact with the plant. Flower
pigments and odors attract pollinators. Strong spicy-smelling oils or bitter-tasting tannins inhibit
many herbivores from eating the plants. Some secondary compounds, like pyrethrins and rotenone,
are extremely toxic to insects and are now being used by humans as insecticides. In all cases, these
compounds appear to play no other role in the biology of the plant than to promote its survival and
reproductive potential through enhancing or inhibiting the actions of other organisms.
References
Cruikshank, I. A. M. and D. R. Perrin 1964. Pathological function of phenolic compounds in
plants. In: Biochemistry of Phenolic Compounds, J. B. Harborne, ed., pp. 511-544.
Academic Press.
Levin, D. A. 1971. Plant phenolics: an ecological perspective. American Naturalist 105:157-181.
Whittaker, R. H. 1970. Biochemical ecology of higher plants. In: Chemical Ecology, E.
Sondheimer and J. B. Simeone, eds., pp. 32-70. Academic Press.
Biology 155 Laboratory Supplement: 7
Techniques used in this experiment.
In this experiment you will prepare a suspension of broken cells that contain the enzyme.
The process of preparing the suspension, or homogenate, is called homogenization. To prepare the
homogenate, you will take the source of the enzyme, the potato, cut it into small pieces and blend
it, using a standard kitchen blender. Blending ruptures cell membranes, the vesicles, and any other
intracellular membrane bound organelles. In the process it releases the enzyme and cofactors
necessary for this laboratory exercise.
After preparing the homogenate, you will centrifuge the homogenate. Centrifugation forces
insoluble particles to the bottom of the tube. The insoluble mass at the bottom is called a pellet.
The liquid portion of the homogenate is called the supernatant. The enzyme we want to work with
in this experiment is found in the supernatant.
Because ortho-quinone has a brownish color we can measure its concentration with a
spectrophotometer. Spectrophotometry is based on the relationship between concentration of a
colored substance and the amount of light it absorbs. Spectrophotometry can be used to measure
the increase in a colored product or the disappearance of a colored substrate. In this experiment
you will measure the light absorbance of a reaction mixture at regular time intervals and then by
plot absorbance versus time. From this plot you will estimate reaction rate as the slope of the
relationship between absorbance and time.
In this experiment you will also need to make serial dilutions. This is a simple and widely
used technique for making a series of solutions of decreasing concentration of a solute. In this lab,
you will make two fold dilutions, each solution in a series will have one-half the concentration of
the previous one. Thus the first will be full strength, the second will be 1/2 strength, the third will
be 1/4 strength, and so on to 1/8, 1/16, 1/32, and 1/64 strength. We can make a serial dilution as
follows. Start with twice the volume of full strength solution as you wish to have in a test tube or
some other container. If you need 3 ml for your experiment start with 6 ml. Transfer one-half (3
ml) to a second test tube and then dilute this with an equal volume of dilutent (usually distilled
water). Mix this new solution thoroughly. The concentration of the new solution will be 1/2 that of
the full strength solution in the first tube. To make 1/4 strength solution, transfer one-half (3 ml) of
the solution in the second tube to a third tube and add an equal volume of dilutent. Mix thoroughly.
The solution in the third tube has 1/2 the concentration of the solution in the second tube and 1/4
the concentration of the full strength solution in the first tube. We can repeat this procedure
indefinitely. Table 1 shows the steps in making a 2 fold dilution series. Dilution series are one
way of making an accurate dilution of a chemical in solution. If you are careful, it can be much
more accurate than measuring out a known quantity of dry chemicals to make each concentration in
the series.
Table 1. How to make a two-fold dilution series. Start with twice the volume of a stock (full
strength) solution as will be needed, in this case 3 ml will be needed so start with 6 ml.
Step
1
2
3
4
5
6
Volume to transfer
to a new tube
3 ml
3 ml
3 ml
3 ml
3 ml
3 ml
Source
stock
tube1
tube2
tube3
tube4
tube5
Amount of dilutent to
add to the new tube
3 ml
3 ml
3 ml
3 ml
3 ml
3 ml
Final concentration in
the new tube
1/2 strength
1/4 strength
1/8 strength
1/16 strength
1/32 strength
1/64 strength
Biology 155 Laboratory Supplement: 8
Protocol
In this lab you will do several experiments with catechol oxidase. Each experiment is different,
but there are many things that are the same in each experiment. Become familiar with the
similarities between each experiment.
•
•
•
•
All of the experiments will involve the enzyme and the substrate. The enzyme is catechol
oxidase and it is present in the supernatant of the homogenate. The homogenate is full strength
enzyme solution. In most of the experiments you work with dilutions of this homogenate. The
substrate is catechol. Catechol will be available in the lab in different concentrations and also
available in combination with a buffer, in which case it will be called buffered substrate. The
buffer simply acts to stabilize the pH.
The reaction produces a colored product. As the number of product molecules in the reaction
mixture increases so will the intensity of the color in a direct fashion. Thus, we can measure
the relative amount of product in a reaction mixture by measuring the amount of light that is
absorbed by the mixture at a specific wavelength using a spectrophotometer.
Each of the experiments tests the effect of a single variable. All experiments will consist of a
comparison of the reaction rate of a standard, or control, tube with a tube that has had one
variable altered. This tube is called the treatment tube. If there is a difference between the two
tubes, you will be able to conclude that the variable changed in the treatment had and effect.
Each experiment will be conducted in the same manner. You will run your experiment in
special tubes, made to fit into the spectrophotometer, and made from a special glass. The last
step in the set-up for each experiment will be to add the enzyme to a substrate mixture, or to
add the substrate to an enzyme mixture, and mix thoroughly. At this point the reaction begins
and it is very important to begin recording data immediately. After mixing, put the tube in the
spectrophotometer, take a reading, and record the time, to the nearest second. Remove the tube
from the machine. Wait one minute. Take another reading and record the time again. Repeat
this for 10 minutes, or until there is little change in the absorbance for 3 consecutive minutes.
Preparation for Lab
„ Review the properties of enzymes as described here, and in your lecture text.
„ Read through all of the experiments described below and develop predictions for the outcome
of each experiment.
„ Become familiar with the experimental procedure. Rewrite the instructions in a flow chart that
you will be able to follow easily.
„ Prepare your lab notebook before lab with tables for data. You will need a separate table for
each reaction that you run. Each table should have three columns: one for the time that a
reading is taken, one for the absorbance measurement, and another for the change in absorbance
between readings. At the top of each table you should record the enzyme concentration,
substrate concentration, and any special treatment. Leave space to graph the data.
„ For practice, graph the data in Table 2 and estimate the initial reaction rate. The initial reaction
rate is the slope of the "best-fit" line through the first few data points. Bring this graph to lab
for discussion.
„ Read the instructions on the use of the Spec 20 Spectrophotometer.
Biology 155 Laboratory Supplement: 9
Table 2. Sample data for practice.
Time
Absorbance Change in
(minutes)
Units
Absorbance
0
0.10
1
0.30
0.2
2
0.60
0.3
3
0.90
0.3
4
1.20
0.3
5
1.40
0.2
6
1.60
0.2
7
1.70
0.1
8
1.80
0.1
9
1.85
0.05
10
1.90
0.05
Preliminary Preparation
You will use potato as a source of enzyme. At the beginning of this lab you will prepare a
homogenate. Throughout this preparation and in all experiments it is very important to avoid
contaminating the solutions with tap water, dirt from unclean glassware or hands.
Procedure:
1. Wash and weigh the potato. Cut the potato into small cubes about 1/4" on each side. Blend the
potato with an equal mass of water (1 ml of water weighs 1 g) thoroughly for about 2 minutes.
2. After washing your hands, rinse cheesecloth with distilled water. Fold it double, and filter the
homogenate through it into a 400 ml beaker. Squeeze the cheesecloth to extract most of the
liquid (filtrate). Note the color of the filtrate and continue to watch it throughout the
laboratory period.
3. Pour the filtrate into two centrifuge tubes. Make sure there is an equal volume of filtrate in
each tube. Place these on opposite sides of the centrifuge and spin at 2000 rpm for 5 minutes.
4. Decant the supernatant and discard the pellet. The supernatant will be used as the source of
enzyme. Cover the tube containing the supernatant with Saran wrap and keep it on ice
throughout the lab.
Biology 155 Laboratory Supplement: 10
Part I - Trial measurement of reaction rate.
In this exercise you will become familiar with the experimental procedure and analysis of
data. In this experiment you will determine the effect of addition of substrate on the reaction. You
will also use this data to determine what dilution of supernatant (enzyme solution) should be used
for the other experiments. Each group will use a different dilution of supernatant.
Step 1. Get your Spec 20 ready. See Appendix I for instructions. Use 5 ml of potassium
phosphate buffer as a blank for zeroing the spec.
Step 2. If you have not been assigned a dilution of enzyme (supernatant), ask your instructor, and
record that dilution in your notebook.
Step 3. Prepare two Spec tubes. Into one put 5 ml of buffered catechol (tube 1) and into the other
put 5 ml of distilled water (tube 2).
Step 4. When you are ready to proceed, add 1 ml of the assigned dilution of the enzyme solution to
tube 1. Mix thoroughly and immediately put the tube into the Spec 20 and take an
absorbance reading. Record the time and absorbance. As soon as you remove the tube
from the Spec 20 add the assigned volume of enzyme to tube 2, mix and take readings.
Step 5. 60 seconds after you took your first reading of tube 1, take another. Record the time to the
second. Do the same for tube 2. Repeat this step for 10 minutes, or until three successive
readings are nearly the same.
Step 6. Construct a graph of absorbance versus time. Plot your data from both tubes on this graph.
Calculate the initial reaction rate for each tube as the average change in absorbance per
minute over the first three minutes. See Appendix II.
Step 7. Report your results to your instructor.
Your instructor will use the data from each group to determine what dilution of enzyme
solution will allow data from the remaining experiments to be graphed most easily. Throughout the
remainder of this experiment this will be called the "full strength enzyme."
Biology 155 Laboratory Supplement: 11
Part II - Effect of Enzyme Concentration on Reaction Rate
In this experiment you will examine the effect of decreasing amounts of enzyme on the rate
of reaction. Substrate concentration will be kept constant. Use the dilution of enzyme decided in
Part I throughout this experiment. Four separate reaction tubes will be used in this experiment.
ecause you may not be able to handle all four tubes at once, you will probably want to run just two
tubes at a time.
Procedure:
1. Starting with 6 ml of enzyme solution, prepare a serial dilution of enzyme, following the
diagram in figure 1. Use regular test tubes to hold the dilutions. Make dilutions to give you
four concentrations: undiluted (full strength), 1/2, 1/4, and 1/8.
2. Label four Spec tubes for the four enzyme concentrations. Add 5 ml of buffered catechol to
each tube.
3. Add the 1 ml of enzyme from the first tube of the dilution series to the appropriately labelled
Spec tube, mix, and take a reading. Repeat with the second tube using the appropriate tube from
this dilution series. (You will reserve tubes 3 and 4 at this time, and take readings from them
after you are finished with tubes 1 and 2.)
4. Continue taking readings from the tubes at 60 second intervals for 5 minutes, or until the
readings no longer changde. (When you are finished taking readings from one set of tubes
repeat steps 3 and 4 with the other tubes)
5. Determine reaction rates as average change in absorbance per minute during the first three
minutes.
6. Plot a graph of rate (vertical axis) versus concentration of enzyme (horizontal axis).
Figure 1. Protocol for Part II Effect of Enzyme
Concentration on Reaction
Rate
Using four regular test tubes,
prepare a serial dilution of
enzyme. Then …
… add 5 ml of buffered catechol
to each of 4 Spec tubes. When
ready to begin recording
absorbance readings, add 1ml of
one of the enzyme dilutions to the
tube. Begin recording absorbance
readings immediately.
* Add enzyme to the Spec tube
only when you are ready to begin
taking absorbance readings!
Biology 155 Laboratory Supplement: 12
Part III - Effect of Substrate Concentration on Reaction Rate
In this experiment, you will investigate the effect of decreasing amounts of substrate on the
rate of reaction. Throughout this experiment enzyme concentration will be kept constant. Use 1 ml
of full strength enzyme throughout this experiment. You will use 5 different substrate
concentrations. Thus you must take readings from 5 reaction tubes. You should do this in two
runs, using 2 tubes in the first run and 3 tubes in the second run.
Procedure:
1. Prepare a serial dilution of substrate (see figure 2). Start with 6 ml of 0.024M catechol as the
full strength (stock) solution. Make your dilution series using 3 ml of the catechol solution and
3 ml of distilled water as the dilutent. Make dilutions to give you 5 concentrations: full
strength, 1/2, 1/4, 1/8, and 1/16.
2. Label 5 Spec tubes for the five dilutions and into each tube put:
1.0 ml potassium phosphate buffer
3.0 ml distilled water
1.0 ml catechol from appropriate tubes in the dilution series
3. Decide how many tubes you want to run at once. (Try three at most.) In the first tube you are
going to run, add 1 ml of full strength enzyme, mix, and immediately take an absorbance
reading. Record the reading and time. Do the same with the other tubes that you are going to
run. Continue to take readings and record the time at 60 second intervals for 5 minutes. (When
you are finished taking readings from the first set of tubes, start another set as described in step
3.)
4. Calculate the reaction rate as the average change in absorbance per minute over the first three
minutes.
5. Plot a graph of rate (vertical axis) versus substrate concentration (horizontal axis)
Figure 2. Protocol for Part III Effect of Substrate
Concentration on Reaction Rate
Using five regular test tubes,
prepare a serial dilution of 0.024M
catechol. Then …
… prepare 5 Spec tubes with
water, potassium phosphate buffer,
and 1ml of one of your catechol
dilutions.
Add 1 ml of full strength enzyme
to the Spec tube when you are
ready to begin recording
absorbance readings.
NOTE: Add enzyme to the Spec
tube only when you are ready to
begin taking absorbance readings!
Biology 155 Laboratory Supplement: 13
Part IV. Effects of Inhibitors on Enzyme Activity
In this experiment you will investigate the effects of adding inhibitors to the reaction
mixture. Using the procedure described below you will be able to decide if an inhibitor acts as a
competitive or as a noncompetitive inhibitor. Noncompetitive inhibitors are not influenced by
variation in substrate concentration; although reaction rate increases as substrate concentration
increases, percentage inhibition does not. Competitive inhibitors are influenced by variation in
substrate concentration; reaction rate increases as substrate concentration increases, and percentage
inhibition decreases. If you don't understand why this is so, review the relevant portion of the
introduction to this laboratory and discuss it with your instructor.
Inhibitors can also provide information about the molecular structure of the enzyme. Some
inhibitors are known to interact with metal ions, and thus if such an inhibitor influences an enzyme
you can conclude that a metal ion is probably important to the enzyme. Other inhibitors are known
to interact with sulfhydryl (-SH) groups. Sulfhydryl groups are important in maintaining the
conformation of enzymes. If an inhibitor that interacts with sulfhydryl groups influences an
enzyme you can conclude that the enzyme likely has a conformation that is maintained, in part, by
sulfhydryl groups.
Note: all the inhibitors used in this experiment are also POISONS. Be very careful when
handling these solutions. If you spill the enzyme on yourself or the benchtop clean it up with large
volumes of tap water. Any leftover inhibitor should be disposed of in specially marked containers
for each inhibitor. Ask you instructor where these are found.
Table 3. Properties of inhibitors used in this experiment.
Inhibitor
Potassium cyanide
Potassium arsenite
Parahydroxybenzoic acid
Phenolthiourea
Action
Reacts with iron or copper, if present
Reacts with sulfhydryl groups, if present
Resembles catechol (see figure 2)
Reacts with copper, if present.
Figure 3. Molecular structure of catechol, parahydroxybenzoic acid, and phenolthiourea.
In this experiment you will determine the effect of an inhibitor by comparing the percentage
inhibition of an inhibitor at different substrate concentrations. In this experiment you will work
with each inhibitor in turn, but ask your instructor which inhibitor to start with.
Biology 155 Laboratory Supplement: 14
Procedure:
1. As in part III, Prepare a serial dilution of catechol. Start with 10 ml of 0.024M catechol and use
distilled water as dilutent. Use regular test tubes for this dilution series. Make dilutions to give
four concentrations: full strength, 1/2, 1/4, 1/8, and 1/16. Set the dilutions aside. You will use
aliquots from each tube several times during the experiment.
2. Prepare 5 Spec tubes. To each tube add:
1 ml of one of the catechol dilutions
1.0 ml potassium phosphate buffer
2.5 ml distilled water
0.5 ml of the inhibitor (consult with your instructor about which inhibitor to use)
3. You can run 2 tubes simultaneously. Decide which two to use. When you are ready to begin
taking absorbance readings add 1.0 ml of full strength enzyme solution, mix thoroughly and
begin recording data. Take readings and record time every 60 seconds for 5 minutes. Repeat
step 3 until all 5 tubes are completed.
4. Calculate reaction rate for each tube as the average change in absorbance per minute over the
first three minutes.
5. Repeat steps 2, 3 and 4 for another inhibitor.
6. Repeat steps 2, 3 and 4, but substitute 0.5 ml distilled water for the inhibitor. This is a distilled
water control.
7. Calculate percentage inhibition for each inhibitor at each substrate concentration:
8. Graph percentage inhibition versus substrate concentration for each inhibitor.
Figure 3. Protocol for Part
IV - Effect of Inhibitors on
Enzyme Activity
Using five regular test tubes,
prepare a serial dilution of
0.024M catechol. Then …
… prepare 5 Spec tubes
with water, inhibitor,
potassium phosphate buffer,
and 1ml of one of your
catechol dilutions.
Add 1 ml of full strength
enzyme to the Spec tube
when you are ready to begin
recording absorbance
readings.
NOTE: Add enzyme to the
Spec tube only when you
are ready to begin taking
absorbance readings!
Biology 155 Laboratory Supplement: 15
Part V - Effect of Temperature on Enzyme Activity
You and the other members of your group should devise a protocol for examining the effect
of temperature on enzymes and reaction rates.
One experiment you could consider is measurement of reaction rate at different
temperatures. To measure reaction rates at different temperatures you should prepare tubes as you
did in Part II (5 ml of buffered catechol in each tube). Bring the tube to the temperature at which
you want the reaction to run by immersing it in a water bath at the appropriate temperature. After
giving the tubes time to equilibrate to the temperature of the bath, add the working volume of
enzyme and begin taking readings. Between readings keep the tube in the water bath. Make sure
to wipe the tube before putting it in the Spec 20.
Another experiment you could consider is exposing the enzyme to different temperatures
before beginning the experiment. You could boil the enzyme, freeze the enzyme, etc. After
exposure, allow the enzyme solution to return to room temperature before adding it to reaction
mixture. Add 5 ml of buffered catechol to the Spec tube and then add 1 ml of exposed enzyme to
the tube.
If you would like to try another experiment discuss it with your instructor first. When you
are completed with your experiment, plot absorbance versus time, determine initial reaction rate,
and then graph reaction rate versus your experimental treatment.
Biology 155 Laboratory Supplement: 16
Questions for discussion
1. How did the color of the homogenate change during the course of the laboratory? Why did this
occur? Do you think this color change might be sufficient to influence the results of your
experiments?
2. Why can't you assume that the absorbance of your reaction tube is 0 at the beginning of the
experiment?
3. You probably found that reaction rate increased in proportion to enzyme concentration in Part
II. How is this result consistent with the enzyme-substrate complex model? How might this
relationship be important in living cells?
4. How is the relationship between reaction rate and substrate concentration consistent with the
enzyme-substrate complex model? How might this relationship be important in living cells.
Could compartmentalization of substrates be an important means of regulating reaction rate?
5. How does temperature influence enzyme activity? Should all enzymes from all organisms
behave in a similar fashion?
6. If an enzyme is not inhibited by arsenite, could we conclude that it has no sulfhydryl groups?
7. When an enzyme's conformation is destroyed by unfolding, it is said to have been denatured. If
an enzyme is noncompetitively inhibited is it denatured?
8. If you were attempting to predict what chemical or chemicals could act as competitive
inhibitors of an enzyme, what chemical could you use as a basis for your predictions?
9. The effects of some noncompetitive inhibitors can be reduced through the addition of other
chemicals, but others appear to inhibit in a permanent fashion. How are both behaviors
possible?
Biology 155 Laboratory Supplement: 17
Appendix I: Instructions for using the Spec 20 and Spec 20D Spectrophotometer.
The Spec 20 an Spec 20D are equivalent devices for taking spectrophotometric data. The
Spec 20D has a digital readout. The Spec 20 has an analog (gauge with a needle and scale)
readout. The two machines will give equivalent data in this experiment. The procedure for setting
up the machines is similar but not identical, so they are presented separately below.
Spec 20:
1. Turn on your Spec by turning the power switch (left front) clockwise. Allow the Spec to warm
up for at least 15 minutes before proceeding. Note: the power switch is also called the zero
control knob.
2. Set the wavelength to 525 nm with the wavelength control knob (top right). Make sure the
sample compartment is empty and the lid is closed.
3. Adjust the meter needle to 0% transmittance (absorbance at infinity) by the using the zero
control knob (left front).
4. Place a clean Spec 20 tube (not a regular test tube) filled with water or some other solvent
reagent blank) in the sample holder (top left). Push the tube all the way in. Turn the tube so that
the index line on the tube matches the index line on the sample holder. Close the lid.
5. Adjust the meter needle to 100% transmittance by using the light control knob (right front).
6. Remove the blank and place samples for absorbance readings in the sample holder.
7. Record absorbance and any other relevant data such as time. Do not leave the tube in the
compartment any longer than necessary to avoid heating the sample.
8. Repeat steps 6 and 7 for all samples.
Spec 20D:
1. Turn on your Spec by turning the power switch (left front) clockwise. Allow the Spec to warm
up for at least 15 minutes before proceeding. Note: the power switch is also called the zero
control knob.
2. Set the wavelength to 525 nm with the wavelength control knob (top right). Make sure the
sample compartment is empty and the lid is closed.
3. Set the display mode to FACTOR by pressing the MODE control key until the red light beside
'factor' is lit. If the display does not read 1.00, press the INCREASE or DECREASE key until
the display reads 1.00.
4. Set the display mode to TRANSMITTANCE by pressing the MODE control key until the light
beside 'transmittance' is lit. Adjust the display to 0.0 transmittance with the zero control knob
(left front).
5. Place a clean Spec 20 tube (not a regular test tube) filled with water or some other solvent
reagent blank) in the sample holder (top left). Push the tube all the way in. Turn the tube so that
the index line on the tube matches the index line on the sample holder. Close the lid.
6. Adjust the display to 100.0 transmittance by using the light control knob (right front). The last
decimal place does not need to be 0.
7. Set the display mode to ABSORBANCE by pressing the MODE control switch until the light
beside 'absorbance' is lit. If you have done everything correctly to this point, the display should
read 0.000. Remove the spec tube and place in a test tube rack. The display should read 1.999
and be flashing when there is no tube in the sample compartment.
8. Place samples to be read into the sample compartment.
9. Record absorbance and any other relevant data such as time. Do not leave the tube in the
compartment any longer than necessary to avoid heating the sample.
10. Repeat steps 8 and 9 for all samples.
Biology 155 Laboratory Supplement: 18
Appendix II - Drawing and Reading Graphs
There are many ways of displaying data that consists of paired values. One way is a table
like the following:
Table AII-1. Paired variables.
r
1
2
5
3
7
8
v
1
4
25
9
49
64
Another way is to determine the mathematical relationship between the two variables
and express this relationship as an equation:
v = r2
A graph is a third way to represent the relationship between the variables. Graphs are very
appealing because they allow us to visualize, and often more quickly comprehend, the relationship
between variables. Paired data are commonly plotted on a two-dimensional graph. A two
dimensional graph has two axes, a horizontal axis that is also called the abscissa or x-axis, and a
vertical axis that is also called the ordinate or y-axis. The scale of each axis is chosen as
appropriate for the data to be plotted. The two axes need not intersect at 0,0 but they may. Figure
AII-1 is a graph of the data in table AII-1.
Figure AII-1. Graph of v versus r.
By convention, the horizontal axis represents the independent variable and the vertical axis
represents the dependent variable. The independent variable is most commonly the variable that is
varied in a controlled manner in an experiment. The dependent variable is the variable that we
expect to change in response to change in independent variable. Whenever we expect that change
in one variable will cause a change in another variable, the former is the independent variable and
the latter is the dependent variable. In the graph above, we use r as the independent variable and v
as the dependent variable.
Biology 155 Laboratory Supplement: 19
One of the values of graphing data is that it allows us to mentally "fill in the blanks" in the
data. For example, the paired data in Table AII-1 did not have a data pair for which r equalled 4.
However, if we wish to assume that there is a regular and predictable relationship between r and v,
we can graphically predict what the value of v would be if we had an observation when r equalled
4. Further, we can fill in the blanks for all missing values by connecting the data points in a series
with a line. The line connecting all points is a prediction of what we would find if we measured v
at an infinite number of values of r spanning the whole range of values of r on the horizontal axis.
Determining Rates from Data and Graphs
A rate is a measure of the change in the dependent variable when the independent variable
changes by one unit. In the data above, when the independent variable (r) changes from 2 to 3, the
dependent variable (v) changes from 4 to 9. Thus, over the interval r = 2 to 3, the rate can be
calculated as:
change in v ∆v 9 − 4 5
= =5
rate =
=
=
change in r ∆r 2 − 1 1
The rate of change in this interval is 5 units of v for per unit r. The rate of change over the interval
r = 5 to 7 is:
∆v (49 − 25) 24
=
= 12
=
∆r
(7 − 5)
2
Thus over the interval r = 5 to 7 the rate of change v per unit r is 12, much higher than in the
previous interval. The increase in rate is reflected in the graph shown in figure AII-1. The
slope of the curve changes and becomes more steep as r increases.
In your experiments you will have to calculate reaction rates as the change in
absorbance per unit time. In the example that follows the unit of time is seconds. The data
that you will collect might look something like this:
Table AII - 2. Data obtained
from a spectrophotometer.
Time is recorded in seconds.
Time
15
74
130
187
248
300
360
430
480
And when you graph this data it will look like this:
Absorbance
1.10
1.64
2.16
2.51
3.02
3.47
3.65
3.84
3.92
Figure AII-2. A graph of absorbance versus time
Notice that the absorbance increases relatively steadily with time, but there are some slight
jumps and dips in the line connecting the data points. Slight deviations are not uncommon in data
obtained from biological experiments and may reflect inaccuracies in readings, peculiarities of this
Biology 155 Laboratory Supplement: 20
particular run, or some other unknown factor. Greater experience with the protocol of the
experiment can often reduce inaccuracy, but many times experiments continue to yield data that is
less than ideal. However, data like that shown here can still be useful for determining many things
about enzymes or other biological phenomena.
We can use the data in Table AII-2 to calculate reaction rate over each time interval of the
experiment. The following table shows the original data in columns 1 and 2, the change in time in
column 3, the change in absorbance in column 4, and the rate in column 5 (calculated as change in
absorbance divided by change in time).
Table AII - 3. Table for determining reaction rate. t represents time, A represents
absorbance, ∆t is change in time, ∆A is change in absorbance, and rate is ∆A/∆t.
t
15
74
130
187
248
300
360
430
480
A
1.10
1.64
2.16
2.51
3.02
3.47
3.65
3.84
3.92
∆t
-59
56
57
61
52
60
70
50
∆A
---0.54
0.52
0.35
0.51
0.45
0.18
0.19
0.08
rate
-----0.0092
0.0093
0.0061
0.0084
0.0087
0.0052
0.0027
0.0016
Using the data from column 1 and 5 a graph of rate versus time can be constructed:
Figure AII-3. Reaction rate (∆A/∆t) versus time.
Note that this graph is very different from the graph of absorbance versus time. Note also that
reaction rate appears to fall over time. How is this consistent with the enzyme substrate
complex model?
Biology 155 Laboratory Supplement: 21
Appendix III: Spectrophotometry
Spectrophotometry, the measurement of light intensity at particular wavelengths, is
important for the study of many biological phenomena. Wherever light is used or absorbed,
spectrophotometry can be a useful tool for studying the process. It is particularly useful for
studying the course of biological reactions where a colored product is produced or a colored
substrate is consumed. Spectrophotometric methods depend on several general principles that are
outlined below.
Consider a solution of a colored compound, a dye or a plant pigment, for example. When
white light passes through the solution certain wavelengths are absorbed partially or completely by
the pigment. The color we perceive is the remaining, unabsorbed light. In many circumstances, one
is interested in knowing what proportion of the light that strikes the solution is transmitted or
absorbed. Transmittance and absorbance are related measures. Transmittance (T) is the ratio of the
intensity of light that passes through the solution (I = transmitted light) to the ratio of the light
striking the solution (I0 = incident light):
I
I0
The proportion of light transmitted depends on two factors relevant to this discussion: (1)
the concentration of the dissolved substances and (2) the length of the optical path. For a given
wavelength, the transmittance is related to these factors as:
T=
I
= 10- kcl
I0
where k is a constant, c is concentration, and l is the length of the optical path. This equation can
be rewritten in logarithmic form to give
T=
I
log10 T = log10   = -kcl
 I0 
Since the logarithm of a ratio is equal to the negative logarithm of the reciprocal ratio, this equation
can also be written as
I 
- log10 T = log10  0  = kcl = A
I
Here, the relationship between transmittance (T) and absorbance (A) can be seen. Transmittance is
the ratio (I/I0). Absorbance is the logarithm of the ratio (I0/I). Note that absorbance is directly
proportional to concentration, while transmittance is exponentially related to concentration. Given
this, why is absorbance the preferred measure for following the production of a colored product in a
biochemical reaction?
The equations above are only accurate for monochromatic light (of a single wavelength).
Spectrophotometers can create light that is nearly monochromatic by passing white light through a
prism or diffraction grating to separate the light into its component wavelengths. The light can then
be passed through a small slit, so that only a very narrow range of wavelengths can be sampled.
The wavelength of spectrophotometers is continuously variable and may extend into the invisible,
ultraviolet or infrared, regions of the spectrum.
Spectrophotometry can be used to the study of many important biological phenomena
including the determination of absorption spectra (the relationship between absorbance and
wavelength) for a particular molecule or set of molecules, and the determination of concentration of
a colored substance in a solution.
Biology 155 Laboratory Supplement: 22
INTRACELLULAR LOCALIZATION
OF RESPIRATION AND GLYCOLYSIS
Synopsis -- This experiment utilizes the method of differential centrifugation, together with
enzyme assays, to separate mitochondria from a homogenate and establish where the key processes
of respiration and glycolosis are localized in eukaryotic cells.
Objectives –
• Learn the basis and use of differential centrifugation for cell fractionation.
• Learn the use of indicator dyes (in this case methylene blue) in chemical experiments.
• Deduce the subcellular location of glycolysis and respiration from the results of the experiment.
INTRODUCTION
Cell Fractionation
Two major types of evidence have provided our present concept of the compartmentation of
intracellular function: (1) inferences from microscopic observation, and (2) evidence from actual
physical separation and biochemical analysis of intracellular constituents. This experiment will
provide evidence for the localization of cellular respiration in mitochondria, and that of glycolysis
in the soluble portion of the cytoplasm. The purpose of the experiment is not merely to verify that
a given function resides in a particular cell part, but rather to introduce you to an extraordinarily
important method of studying living matter and to lead you into some of the mental processes
required to infer how things work at the subcellular level.
Note that in this experiment we will be concerned with the separation of cell organelles.
Our separation will not be complete; that is to say, we will not obtain pure mitochondria, but will
exploit a rapid procedure which lends itself to easy execution while illustrating more complex
procedures used in the research laboratory. Our procedure yields a rich harvest of mitochondria
which are functional as judged by the most sensitive biochemical criterion available: the ability to
carry out ATP synthesis coupled to electron transport.
We will actually use two methods: homogenization and centrifugation. You are already
familiar with homogenization from the experiment on the enzyme action. Remember that
homogenization produces a solution of soluble cell constituents and a suspension of insoluble
constituents. The latter include intracellular organelles such as mitochondria. These may be
separated from the other organelles and soluble material by means of centrifugation. In a
centrifuge, material is spun on an axis of rotation and thereby subjected to a force (centrifugal
force) directed outward from the rotational axis. This force causes the suspended cell organelles to
move away from the axis of rotation down the length of the centrifuge tube. If the centrifuge runs
fast and long enough, the organelles will eventually become sedimented at the bottom of the tube
(pellet).
The rate of sedimentation varies for different kinds of suspended particles. At a given speed
of rotation, heavier and larger particles move faster than lighter, smaller ones. With elaborate
centrifugation procedures, nuclei, which are relatively large and dense, may be separated from
"microsomes" (bits and pieces of endoplasmic reticulum with any attached ribosomes).
Our choice of experimental material derives from the need for high glycolytic and
respiratory metabolism and the easiest possible separability of mitochondria from the soluble
fraction. We therefore selected insect flight muscle, specifically that of flesh flies (Sarcophaga).
As in many other flies, these have extraordinary contractile performance, causing wingbeat
frequencies of many hundreds to over a thousand cycles per second. This is energized by
glycolysis and respiration with glucose as a fuel.
Biology 155 Laboratory Supplement: 23
The most striking features of flight muscle are (1) the enlarged size of the muscle fibrils
whose contractions are responsible for the performance mentioned and (2) the equally striking giant
mitochondria. These features lend flight muscle to our present purposes for two reasons: (1) the
activity of glycolytic and respiratory enzymes is very high, permitting easy detection; and (2)
owing to the abundance and large size of the mitochondria, they can be isolated in substantial
quantity at relatively low centrifugal forces.
Recommended Reading
Baker & Allen: The Study of Biology, 3rd ed., pp. 195-232.
Keeton: Biological Science, 3rd ed., pp. 137-138, 163-180.
Loewy & Siekevitz: Cell structure and Function, 2nd ed., pp.310-314.
Raven & Johnson: Biology 4th ed., pp. 193-205.
Biology 155 Laboratory Supplement: 24
PROTOCOL
Special Preparations Before Coming To Lab
Carefully study the logic of the experiment (see table) in order to understand why each tube
is necessary and why each component is used. Know the function of the substrates glucose and
succinate in cellular metabolism.
Assay for Glycolysis and Respiration
To understand the assay method, it is crucial to know the meaning of the terms
glycolysis and respiration. Glycolysis is best defined as the conversion of glucose into two
molecules of pyruvic acid (pyruvate). This conversion does not consume oxygen, and
accomplishes only a partial chemical degradation of a glucose molecule, thus affording the cell
only a partial utilization of glucose's potential as a metabolic fuel. In contrast to glycolysis,
cell respiration consumes oxygen, and pyruvic acid is completely oxidized to carbon dioxide
and water. As you should expect, both glycolysis and cell respiration are accomplished
enzymatically; the former is carried out by the glycolytic enzyme system, and the latter by the
enzymes of the Krebs cycle and electron-transport chain. NOTE: The usage of the term
respiration in some texts embraces both glycolysis and the conversion of pyruvate to carbon
dioxide and water, which requires water. The usage here conforms to that of the scientists
who established this field of biochemistry and is more convenient for a variety of reasons.
As a consequence of the points stated in the preceding paragraph, a homogenate which
is glycolyzing but not respiring will not use oxygen. If oxygen is utilized, its concentration in
solution will fall. We will use the dye, Methylene Blue to indicate when the oxygen is used
up as a result of respiration. When the concentration of dissolved oxygen is very low, the dye
becomes colorless. This happens as a result of chemical reduction of the dye. Reducing
equivalents from metabolic intermediates in the Krebs cycle are passed to NAD (or in the case
of succinate, FAD) and from these nucleotides on via the electron-transport chain. The dye
intercepts this passage and gets reduced. The dye color disappears when it is reduced.
Therefore, if we supply glucose to a homogenate in the presence of Methylene Blue
and the dye becomes colorless after a period of time, we have evidence for both glycolysis and
respiration. If instead of glucose we supply a substrate in the Krebs cycle derived from
pyruvate (we will use succinate for this purpose) and the dye bleaches, we have evidence for
respiration but not glycolysis. If we supply glucose to a homogenate capable only of
glycolysis, the dye would not be reduced and any inference as to the occurrence of glycolysis
would have to be based on other data. Be sure you understand the reasoning behind all these
points.
The strategy of the experiment is designed to reveal which part of the muscle
homogenate carries out glycolysis and which part carries out respiration. The procedure
breaks down into four sections: (1) preparation of the homogenate; (2) fractionation of the
homogenate by centrifugation; (3) biochemical analysis of the enzymatic capabilities of the
fractions obtained; and (4) microscope observation of the fractions (optional). Students should
work in pairs. Because the preparation of the homogenate is a relatively complicated
procedure, a flow sheet has been prepared for you (see Figure 1). Read the instructions very
carefully.
Biology 155 Laboratory Supplement: 25
Materials:
• 1 stoppered shell vial or small Erlenmeyer flask for flies
• 70 flesh flies, Sarcophaga bullata (ice-cold to anesthetize them)
• 1 wire test-tube rack (small mesh)
• a few paper towels to cut the flies on
• 1 razor blade
• 1 enamel pan
• 1 thermometer
• 1 grease pencil
• 1 piece of plastic wrap
• 1 large plastic beaker filled with crushed ice
• 7 large glass test tubes to hold reagents, homogenate, and centrifuged fractions
• 7 glass reaction tubes (SMALL size, about 3 inches tall)
• 7 plastic pipettes, 1 ml and controls
Procedure:
Part A- Preparation of homogenate
1. The instructor will assign one team of two students to prepare the homogenizer. This team
should obtain a clean homogenizer. Chill it for 5-10 minutes before use in an ice bath.
2. Meanwhile, the instructor will distribute 60 flies among the remaining teams. If the flies are
kept in a closed plastic tube on ice, they will be immobilized. Cold causes anesthesia in insects.
3. Using the razor blades, QUICKLY (to avoid warming) cut off the wings, legs, heads and
abdomens. Save the thorax of each fly. The thorax is the segment of the body where the wings
and legs were attached. The entire class should do this at the same time, as rapidly as possible.
Finally, each team should cut each thorax in half (to facilitate grinding tissue). The thoraces
should be put into the chilled glass homogenizer tube as it sits in ice.
4. Add 15.0 ml of ice-cold homogenizing medium to the homogenizer tube (0.32M mannitol
containing 0.02M phosphate buffer, pH 7.4).
5. The team designated in Step 1 above should now make the homogenate for the entire class.
Run the homogenizer up and down into the mix of medium and thoraces until the mixture
becomes thick (like a milk-shake). During this process keep the homogenizing tube on ice.
6. A different pair of students, assigned in advance, should prepare a filtering device. This
consists of a 5 inch diameter circle of cheesecloth, two layers thick, wetted with homogenizing
medium (but not dripping), placed in a short-stemmed glass funnel. The center of the
cheesecloth should be pushed down as far as the beginning of the stem. The stem, in turn,
should be fitted into a 50 ml graduated cylinder. It helps to put the cylinder in a beaker of ice.
7. Transfer the homogenate to the cheesecloth. It should start to filter through. If it does not,
shake the funnel slightly to start filtration. If it still does not, squeeze it through by hand,
making sure your hands are clean before doing so.
8. Add an additional 10.0 ml of ice-cold medium to the homogenizing vessel and run the
homogenizer to to suspend any residual tissue debris. Transfer the 10.0 ml to the cheesecloth in
the funnel, using this 10 ml to wash down the material trapped on the cheesecloth. Repeat the
washing of the homogenizer and the cheesecloth with an additional 5.0 ml of ice-cold medium.
Finally, with clean hands, squeeze out the cheesecloth bag into the funnel. This filtration
should provide about 30 ml of homogenate in the cylinder. Most of the material held back in
the cheesecloth includes pieces of thoracic integument and large muscle fibers.
Biology 155 Laboratory Supplement: 26
9. Mix the filtered homogenate thoroughly. This is crucial. Measure its total volume, and record
this value to the nearest mililiter. Transfer 15.0 ml of the homogenate to a clean tube marked H
(for whole homogenate). Keep this tube on ice at all times.
10. Transfer the remaining 15 ml homogenate to a clean centrifuge tube and place the tube in a
beaker of crushed ice.
11. Prepare a balance tube by putting 15 ml distilled water into a new centrifuge tube like the one
used in the previous step.
12. Place both tubes in the refrigerated centrifuge, on opposite sides of the rotor. Centrifuge at
5000 rpm for 20 minutes.
13. Another crucial step: immediately after the centrifuge stops, retrieve the tube containing the
homogenate, carefully holding it at the same angle at which it lay in the centrifuge. Pour all the
supernatant into a clean (rinsed with distilled water and shaken dry) 25 ml graduated cylinder.
Do not pour out any of the pellet (the pellet is whitish in color); the point is to achieve a
"clean" separation. Restore the volume of the supernatant to 15 ml with the homogenizing
medium; shake well to mix contents. Transfer to a clean tube, mark it S, and keep it on ice.
The volume of S must exactly equal that in step 10 above.
14. Add ice-cold homogenizing medium to the pellet. The amount of medium added should be just
enough to make the final volume of resuspended pellet exactly equal to the original volume
centrifuged (15.0 ml). This is crucial. Stopper the tube and shake it to resuspend the pellet
thoroughly. When the pellet is resuspend, label it P.
15. You should now have three labled tubes on ice: uncentrifuged homogenate (H), resuspended
pellet (P), and supernatant (S). The pellet contains nuclei, glycogen (polysaccharide) granules,
mitochondria, and bits of the muscle's contractile apparatus. The supernatant contains most
soluble muscle constituents, including glycolytic enzymes and some membranous material
(reticulum) which is too small to centrifuge out at the speeds used.
16. Each individual team of students should now obtain 1 ml of H, l ml of S, and 1 ml of P in
labeled tubes. Keep them in a beaker of crushed ice.
Biology 155 Laboratory Supplement: 27
FLOW CHART
Cut
100 flies, ON ICE
Distribute 15 flies/team for preparation of thoraces
Students return halved thoraces (wings, legs, head, and abdomen removed)
Grind
Pool thoraces in homogenizer, ON ICE
Add 15.0 ml homogenizing buffer
Homogenize for 2 minutes, rheostat set at 50
Filter
Through cheesecloth, into a 50 ml graduated cylinder
Wash &
Rinse homogenizer with 10 ml homogenizing buffer; pour through
cheescloth
Repeat
Rinse with another 5 ml of homogenizing buffer; squeeze out
cheesecloth
Divide
Pour off 15 ml of homogenate into a labeled tube
Centrifuge
The remaining 15.0 ml homogenate for 20 minutes at 5000 rpm.
Separate
from
Supernatant (S): Restore to 15.0 ml with homogenizing buffer; mix well
Pellet (P):Restore to 15.0 ml with homogenizing buffer; mix well
Distribute
1 ml of H/team, 1 ml of P/team and 1 ml of S/team
Part B- Biochemical analysis of fractionated homogenate
1. Each team of two students should obtain the following materials in the quantities indicated.
Keep them in clean, dry, labeled test tubes in a rack at room temperature, not on ice.
Substance
Concentration
Quantity
Mannitol buffer*
0.32M
3ml
Buffer-cofactor-dye mixture
**
5ml
Glucose
0.015M
3ml
Succinate
0.2M
3ml
* Mannitol buffer = homogenizing medium
** Potassium phosphate buffer, pH 7.4 (0.2M); ATP (0.0125M); MgCl2(0.005M); Methylene blue (0.5
mg/ml).
2. Obtain 1 ml plastic pipettes for these solutions, and three pipettes for dealing with tubes H, S,
and P. Each pipette should be marked to avoid cross-contamination of solutions and should be
left in the tube rather than placed on the table top.
3. Obtain seven clean, small glass test tubes to be used as reaction vessels in the next step. Be
sure these test tubes are identical in size. Number the tubes clearly 1-7, using a grease pencil.
4. Obtain a pan containing water adjusted to 35°C, about two inches deep.
Biology 155 Laboratory Supplement: 28
5. The following table shows what to add to each of the seven reaction tubes. You will work
faster if you add the first ingredient (mannitol) to all the required tubes, then the buffer mix, and
so on down the line. Do not add homogenate, pellet, or supernatant until last, and only when
you are really ready for the reactions to start. The exact time of adding H, P, and S should be
recorded carefully. Be sure tubes H, P, and S are thoroughly agitated before pipetting from
them) their contents may settle). All volumes shown in the following table are in milliliters.
Ingredient
Mannitol
Buffer mix
Glucose
Succinate
Whole Homogenate (H)
Pellet (P)
Supernatant (S)
1
0.25
0.45
0.20
0.25
-
2
0.25
0.45
0.20
0.25
-
Reaction Tube Number
3
4
5
0.25
0.25
0.45
0.45
0.45
0.20
0.20
0.20
0.25
0.25
0.25
0.25
-
6
0.25
0.45
0.20
0.25
7
0.45
0.45
0.25
-
Immediately after adding H, P, or S, each tube should be rapidly and thoroughly mixed. Gentle
sloshing is no good. The best index of complete mixing is uniformly distributed dye color
(blue) in every tube. Immediately go on to step 6.
6. Place the rack of reaction tubes in the pan of water at 35 °C, noting the exact time. Do not
agitate the tubes at all from this point on. The water must be deep enough to come to a level
above the reaction mixture in the seven tubes. During the course of the experiment it will be
necessary to add hot water to maintain the temperature at 35 °C. When you do this, add it away
from the tube rack and mix the water in the pan thoroughly without any disturbance, even
slight, to the position of the tubes.
7. As noted earlier, the indicator of reaction in the tube is dye color. When oxygen is exhausted,
methylene blue turns from bright blue to colorless. The tube then takes on whatever color its
other contents have, in this case homogenate color. Note the exact time at which individual
tubes lose blue color and record this. The time is not necessarily the same for all tubes and
some may "never" bleach, i.e., in more than one hour. Note: the bleaching of a tube depends
on metabolism using oxygen faster than it can diffuse back in from the air. When a tube
bleaches, you will still see a bluish upper margin at the surface in contact with air. This can be
disregarded for present purposes. It may take 10 minutes or longer before tubes bleach.
8. (Optional for the curious: when a tube has bleached, slight agitation will re-aerate it and turn it
blue again. This can be demonstrated easily and the tube will bleach all over again. Why does
it bleach? Why is the time different from step 7?
9. Tube 7 deserves special attention. Among cell biologists, this is loosely termed a "no-substrate
control tube". This tube was set up to test a specific question, namely: does the action in tube 1
depend on the addition of glucose? What happened to the dye in Tube 7, if anything? What
conclusions does this permit you to draw? Is there any way in which the results which this tube
provides can still be consistent with the idea that glucose is the initial substrate for glycolysis
and respiration in vitro and in vivo?
Biology 155 Laboratory Supplement: 29
QUESTIONS FOR DISCUSSION
1. To start a meaningful discussion of cell respiration and glycolysis, you have to understand what
the terms mean and what the general features of the underlying processes are. Refer to your
text for a summary of both processes. Be sure you understand the three overall states in the
breakdown of sugar (glucose):
• glycolysis: the sequential, multi-step conversion of pyruvate, the associated utilization and
production of ATP, the production of two reduced NAD;
• Kreb's cycle: the sequential, multi-step conversion of pyruvate, via acetyl coenzyme A, to
carbon dioxide with the removal of hydrogens and formation of reduced NAD;
• electron-transport cycle: the sequential, multi-step passage of electrons to oxygen,
accomplishing the oxidation of hydrogens (removed in step b) to water, with the energy
yield used to drive ATP synthesis.
An understanding of these processes does not require learning the names of many intermediate
chemical compounds and enzymes, (but be sure you know succinate). Rather, concentrate on
where carbon chains are broken, where hydrogens come from, where ATP is produced, and
how the processes are connected to each other.
2. Be sure that you understand the purpose of every chemical ingredient in the reaction mixtures.
You should be able to figure out all of them except mannitol. This "sugar alcohol" is an inert
substance used to maintain the integrity of the mitochondria during the preparative procedure
(they would burst in distilled water and clump together in saline solution). Mannitol also helps
maintain similar reaction conditions in tubes 1 to 7. Why is it important to keep the conditions
constant?
3. In light of the purpose of each chemical ingredient of tubes 1 to 7, explain what the mixture in
each individual tube is designed to test for. Be sure to ascertain why tube 4 and tube 7 are run;
what do these two control for? What additional tubes might have been controls? What
alternative substrates could have been used?
4. Does this experimental design permit you to make any inference about the intracellular
localization of (1) glycolysis, (2), the Kreb's cycle, (3) the electron-transport chain? How
detailed an inference can you make and still defend you views in a logical manner? In the event
you do not feel you can make a totally definitive conclusion, be prepared to present and defend
the kind of additional evidence that you would seek, using any method feasible in the research
lab.
5. The organization of flight muscle is unusual and for many years the mitochondria were known
as sarcosomes. How can one prove they are in fact mitochondria? Again, utilize any method
available in a research lab.
6. Try to avoid illogical reasoning in the preceding questions: i.e., if the presence of X is taken as
evidence for Y, is it valid then to use the presence of Y as evidence for X?
7. What is the functional significance of the actual intracellular localization of glycolysis, Kreb's
cycle, and the electron-transport chain? What advantageous functional features are conferred
upon the cell by this arrangement? What possible evolutionary events have led to this
arrangement?
Biology 155 Laboratory Supplement: 30
GENETIC TRANSFORMATION OF ACINETOBACTER
Synopsis -- A crude extract containing DNA from a streptomycin-resistant strain of
Acinetobacteris used to transform wild-type (sensitive) cells to a resistant phenotype. The
experimentdemonstrates the use of bacteria, sterile technique, and the selective methods in
bacterialgenetics that helped establish DNA as genetic material.
Objectives
• Learn the basic procedure of sterile technique.
• Demonstrate the chemical nature of inheritance (and note the similarity of this experiment with
those of Griffith and Avery, MacLeod and McCarthy).
INTRODUCTION
Although F. Miescher discovered the nucleic acids in living cells in 1871, their central role
in inheritance was not apparent for nearly 70 years. During this time, however, a series of
important experiments were performed that directly led to the identification of DNA as the genetic
material.
The first of these experiments was conducted by a British bacteriologist, Fred Griffith, in
1928. Griffith worked with Streptococcus pneumoniae (also called Pneumococcus), a bacterium
known to cause pneumonia in both humans and mice. Pneumococcus can exist in two different
forms with respect to colony appearance. Cells of one form, called smooth (S), have an outer
capsule of polysaccharide. The presence of the capsule causes the colonies resulting from these
cells to appear smooth and glistening and also protects the cells from the defense mechanisms of a
host organism. When a mouse, for example, is injected with pneumococci of this type, it develops
pneumonia and usually dies. The S phenotype is therefore virulent. The other type of
Pneumococcus is called rough (R) and has no capsule. Mice injected with R cells generally do not
contract pneumonia and continue living, and hence, R is avirulent. Studies prior to 1928 had
demonstrated that the presence or absence of the polysaccharide capsule was an inherited
characteristic in these bacteria. Moreover, it was known that the encapsulated S type occurred in
several distinct forms or strains, differing from one another in the chemical composition of the
capsule. For his experiments, Griffith worked with two S strains denoted IIS and IIIS, and their
corresponding rough phenotypes, IIR and IIIR.
In an important experiment, Griffith injected several mice with living Pneumococcus of the
IIR type. Normally these mice would show no ill effects. However, Griffith also injected these
mice at the same time with a large number of heat-killed IIIS type bacteria. To his surprise, many
of the mice died. When samples of blood from the infected mice were examined, Griffith found
not only the R type cells living the host, but virulent, encapsulated S type as well. Moreover, the
encapsulated cells were type IIIS. These cells could not have come from the IIR cells by mutation
since that would only produce IIS bacteria. There was something in the heat-treated IIIS inoculum,
he reasoned, causing a transformation of the bacteria from IIR to IIIS. Griffith was unable to
specify the chemical composition of this substance, which came to be called the transforming
principle.
It was not until 1944 that the transforming principle was identified. By that time,
experimenters had perfected techniques for performing transformation experiments in test tubes (in
vitro) as well as in mice (in vivo). At that time, three experimenters (O. T. Avery, C. M. MacLeod,
and M. McCarthy, 1944) found that extracts from heat-killed IIIS bacteria containing almost pure
DNA were capable of transforming IIR bacteria in a test tube culture. Moreover, when this purified
transforming principle was treated first with deoxyribonuclease (an enzyme which breaks down
DNA) and then mixed with cultures of IIR, no transformants resulted. This experiment, then,
provided the first direct evidence that DNA is the chemical basis of heredity.
Biology 155 Laboratory Supplement: 31
Since 1944, similar experiments have revealed transformation in a few other species
ofbacteria (Haemophilus influenzae, Bacillus subtilis, Neisseria gonorrhoeae, and Acinetobacter
calcoaceticus). The majority of bacteria cannot be transformed naturally, primarily because the
cell membrane is impermeable to DNA but also because of nucleases which break down
unprotected DNA. However, using special laboratory procedures which alter the permeability of
the cell membrane and protect against nuclease activity, many other species can be transformed,
including the important experimental species, Escherichia coli. Although sporadic claims for
transformation have been made for certain eukaryotic organisms, transformation has not been
conclusively established in organisms other than bacteria. What possible reasons can you suggest?
Techniques used this experiment
In this experiment, you will be working with a bacterium known to be competent, i.e.,
capable of transformation. This species, Acinetobacter calcoaceticus, was first described by the
famous Dutch microbiologist Beijerinck in 1911 after isolation from a quinate enrichment.
Acinetobacter can almost invariably be isolated from soil or water supplemented with acetate,
and is an important decomposer of organic matter.
Just as in Griffith's experiment, you will be working with two distinct phenotypes of one
species. But where Griffith's bacteria differed with respect to virulence (the smooth and rough
types). the strains of Acinetobacter available to you differ in their resistance to the antibiotic
Streptomycin. You will be provided with a strain of wild-type Acinetobacter BD413 (hereafter
called StrS)which is naturally sensitive to streptomycin and will not grow on the medium containing
the antibiotic. A culture of a mutant strain of BD413 (hereafter called StrR) which is streptomycin
resistant will also be available. In the experiment you will be looking to see if the sensitive
phenotype can be transformed to resistant in much the same way that Griffith's Pneumococcus was
transformed from avirulent to virulent.
References
Avery, D.T., C.M. MacLeod, and M. McCarthy. 1944. Studies on the chemical nature of the
substance inducing transformation of pneumococcal types. Induction of transformation by
deoxyribonucleic acid fraction isolated from Pneumococcus Type III. Jour. Exp.
Med.79:137-158. (Reprinted in Gabriel, M.L. and S. Fogel, Great Experiments in Biology,
and in Peter, J.A., Classic Papers on Genetics).
Baumann, A., M. Doudoroff, and R. Stanier. 1968. A study of the Moraxella group II.Oxidativenegative species (Genus Acinetobacter). J. Bacteriol. 95:1520.
Fox, M.S. and M.K. Allen. 1964. On the mechanism of deoxyribonucleate integration in
pneumococcal transformation. PNAS. 52:412-419.
Hotchkiss, R. and E. Weiss. 1956. Transformed bacteria. Reprinted in Facets of Genetics,A.M. Srb,
R.S. Edgar, R.D. Owen, eds., W.H. Freeman & Co.
Juni, E. and A. Janik. 1969. Transformation of Acinetobacter calcoaceticus (Bacterium anitratum).
J. Bacteriol. 98:281-288.
Leonard, C.G. and R. M. Cole. 1972. Purification and properties of streptococcal competence factor
isolated from chemically defined medium. J. Bacteriol. 110:273-280.
Stent, G.S. 1971. Molecular Genetics, Ch. 7, "Transformations", pp.175-193.
Tomasz, A. 1965. Control of the competent state in Pneumococcus by a hormone-like cell product:
an example for a new type of regulatory mechanism in bacteria. Nature. 208:155-159.
Tomasz, A. and J.L. Messer. 1966. On the nature of the pneumococcal activator substance.PNAS.
55:58-66.
Biology 155 Laboratory Supplement: 32
PROTOCOL
This experiment has three parts, described as follows: one for preparing the initial cultures
for the transformation, a second for transferring bacteria to a selective medium, and a third for
observing the subsequent growth of the cultures. Before the first laboratory period, be sure to
acquaint yourself with sterile techniques as explained in Appendix 1 at the end of this lab exercise.
Because microorganisms are normally found in the air, in your mouth and breath, on your skin,
etc., it is essential that you follow these techniques to avoid contamination of the sterile material
and strains you will be using. If necessary, ask your lab instructor to demonstrate the technique for
you.
Procedure:
First Lab Period
1. Obtain a Petri dish of the streptomycin-resistant bacterium, StrR. Open the dish and with a
sterilized loop scrape off a generous loopful of the bacteria from the surface of the plate as
shown by your instructor. (Do not push the loop into the agar. The bacteria are only on the
surface.) Replace the cover and transfer the bacteria from the loop into a sterile lysate vile
containing 1 ml detergent solution (0.15 M NaCl, 0.015 M sodium citrate and 2% sodium
dodecyl sulfate). These are the tubes with the soapy solution with caps. Be careful that the
bacteria do not scrape off on the sides of the tube on the way down. Twirl the loop in the
solution until the bacteria are fully dispersed; if you see lumps in the fluid, the bacteria are not
dispersed enough. Twirl again! Cap the tube and return the used Petri dish.
2. Set the tube in a rack in a 60°C water bath containing 2 inches of water and leave it for 60
minutes. Be sure to time this carefully. The combined action of the detergent (sodium dodecyl
sulfate) and the heat will lyse the cells, releasing DNA. The solution also contains sodium
citrate and NaCl which help to stabilize DNA. Hopefully, no cells will survive this treatment.
How would you test this? Note: Do not let any water from the bath get into the tube, or
bacteria which are resistant to the treatment may enter.
3. Obtain a sterile nutrient agar plate and, keeping it closed, divide it into four sectors as shown
below in Figure 1 by marking the underside of the plate with a marker. Mark your name and
section number on the top lid of the plate at this time. Sectors a and c should be inoculated with
strain StrS; and sector b with strain StrR. To inoculate, use a sterile but cool inoculating loop.
Transfer one small loop of bacteria from the stock culture to the surface of your plate in the
sector desired. Enough cells will be transferred if you just touch the center of the appropriate
section of the growing cells. Sterilize the loop between inoculations.
IMPORTANT; Do not load on too many cells. You should only be able to just detect where
the cells have been streaked. If there are more cells, the results will be less distinct.
Biology 155 Laboratory Supplement: 33
Figure 1: Inoculating the soy agar plate.
4. Sterilize the loop once again and transfer a loopful of your lysate from the heated tube to the
patch you have already made on sector c only. Spread it around carefully. This sector now
contains a mixture of cells of strain StrS and DNA from strain StrR.
5. Sterilize the loop again and transfer a loopful of lysate to sector d. This sector should contain
only lysed (dead) cells and their contents. Make sure you understand the purpose of this sector.
If, after a week of incubation, you find growth on this sector, what does it mean? Give your
plate to your instructor; it will be incubated and ready next class to make another set of
transfers.
Second Lab Period
1. You should obtain a sterile plate of nutrient agar containing streptomycin (200 mg/ml agar); the
plate will be marked "strep". This time mark it into 3 sectors on the underside as shown in
Figure 2. Examine your plate from the first period and record the amount of growth in each
sector with a sketch in your laboratory notebook. Then transfer a loop of cells from the center
of each patch (omitting d) to the appropriate sector of your new plate.
IMPORTANT; Transfer only a very few cells to this plate. It is of critical importance not to
overload the cells here. Why?
Once again, mark your name and section on the lid of your plate at this time. Give your plate to
your instructor. It will be incubated until the next lab period.
2. When you are done with your first Petri dish from last week, ask your instructor where to put it
for disposal.
Figure 2: Inoculating the strep plate.
Biology 155 Laboratory Supplement: 34
Third Lab Period
1. Record the amount of growth in each sector once again. Do you see any convincing evidence
for transformation? If you did not, why might there have been an error?
2. How could you show that the apparent transformants which you observed, if any, really resulted
from transfer of DNA from one strain to the other?
3. It has been shown in Pneumococcus that one strand of the incoming DNA in transformation
forms a hybrid region with one strand of the recipient DNA (Fox and Allen, 1964). StrR is
recessive to StrS. The reason StrS is dominant is not well understood (the mutation alters a
polypeptide found in the 3OS ribosomal subunit). In view of these facts, how do you interpret a
change in phenotype from sensitive to resistant? Why do you plate the cells and the DNA of
sector c on a nutrient medium and then transfer the culture to a nutrient medium augmented
with streptomycin?
4. Recent studies (Tomasz, 1965; Leonard and Cole, 1972) have demonstrated that the recipient
cells are not able to take up DNA at all times, but only during a special period of growth when
they become competent. This seems to hold for all transformable bacteria and it appears to be
related to changes in cell wall synthesis. Under optimal conditions, as many as 10 to 20% of
the cells in a culture may be transformed for a single genetic marker. With this in mind, can
you expand your answer to Question 3?
5. These experiments have also shown that there is a protein-like compound released from
competent cells (an activator substance) which rapidly stimulates neighboring cells in a culture
to become competent, in turn also producing an activator. Subsequently, an inhibitor appears in
the culture that blocks activator substances and renders the cells no longer transformable. What
chemical control mechanism in vertebrates does this remind you of? Do you agree with
Tomasz that this constitutes "differentiation" in a population of bacteria?
6. What do you think is the significance of gene transfer to bacteria?
Biology 155 Laboratory Supplement: 35
APPENDIX: STERILE TECHNIQUES
A common problem in microbiology is how to keep unwanted microorganisms from
contaminating media and cultures used in experiments. As you will see in this exercise, the
solution to this problem is relatively simple, but the techniques involved require careful attention
and practice to be successful. Basically, the idea is first to sterilize the medium to be used, by one
of several techniques, and then to work with it in a manner that prevents all but the desired
organisms from having access to it.
Sterilization -- In the experiment, you will be provided with already sterile Petri plates of nutrient
agar. Glass Petri dishes and other glassware are usually sterilized by heating at 170°C for 1 hour.
Heat sensitive materials such as the plastic Petri dishes you will be using in this lab are often
sterilized with the gas ethylene oxide which lethally damages DNA. The nutrient agar in the
bottom of your Petri dishes was sterilized by still another method used for liquids:autoclaving.
Briefly, autoclaving consists of heating the materials to temperatures that are lethal for all
organisms present (generally this means heating to 121°C for 15 minutes, though some bacterial
and fungal spores can survive this treatment). The autoclave works something like a pressure
cooker: by limiting an increase in volume, it allows water to be heated to temperatures above its
boiling point. After the hot agar has been poured into the sterile Petri plates, it will remain sterile
virtually indefinitely if the plates are not opened.
Plating -- During experimentation, great care must be taken when opening the dishes to avoid
contamination. The following guidelines are suggested to aid you in reducing the chances of
contamination while you are plating your cultures:
• Open the plates as infrequently as possible.
• Open the plates for as short a time as possible.
• Do not touch the inside except with sterile tools. Try not to breathe directly into the plates.
When the Petri dish must be open (during an inoculation) lift the lid until there is just enough
space to maneuver in. Do not remove the lid completely.
• Carefully sterilize the inoculating loops in a Bunsen burner (or alcohol flame) until the loop and
several inches of wire above it are red-hot. The first point of contact with the flame should be
two inches above the loop. Now draw the loop into the flame. Be sure to allow a flamed loop
to cool to room temperature before executing a transfer, otherwise your bacteria will be killed.
After each inoculation, flame the loop to avoid contaminating other materials in the lab.
• You can substantially reduce the likelihood of contamination if you first wipe down your work
space on the lab bench with alcohol or detergent.
• These bacteria have a rapid doubling time and will grow very fast. Streak lightly to get optimal
results.
One last point. The agar in your Petri plates is nutrient-supplemented polysaccharide extract
originally from seaweed. While sterile, and before it became cool enough to solidify (roughly
45°C), the agar was poured into the Petri dishes. While the agar was cooling, moisture from the
agar solution probably condensed on the underside of the top of the Petri dish. Unless the plates
are always kept stored upside-down, the condensed water may "rain" onto the surface of the
medium, causing smears and runs in your cultures. For this reason, always keep your Petri dishes
inverted when not in use, especially when your cultures are to be stored for some time.
These simple precautions should help you obtain the contamination-free cultures that you will need
in you bioassay for transformation. However, reading about sterile technique is just not enough:
you should practice these simple precautions in a few dry runs using extra Petri dishes (not those
for your actual cultures) that are available in you lab (perhaps try plating samples from the mucus
lining of your mouth). Practice and patience will pay off.
Biology 155 Laboratory Supplement: 36
Pedigree Analysis
Synopsis – This exercise introduces the analysis of pedigrees as a means of determining the most
likely mode of inheritance of a trait. The logic of pedigree analysis allows the formulation of clear
testable hypotheses for the inheritance of a trait. This exercise emphasizes the power of hypothesis
rejection in pedigree analysis and in science.
Objectives
• To understand the most common patterns of inheritance of mammalian traits and how to
distinguish between, autosomal dominant, autosomal recessive, sex-linked dominant, and sex
linked recessive traits using data that can be obtained from pedigrees.
• To learn the logic of hypothesis rejection in pedigree analysis.
Introduction
In his classic experiments with peas, Gregor Mendel discovered that inheritance is
particulate. Particulate inheritance has several interesting and important properties:
• Traits are passed from parents to offspring through the transfer of discrete packets of
information that we now call genes.
• Each individual carries two copies of genes for each trait.
• Genes come in alternate forms. We now call alternate forms of a gene alleles.
• Within an individual, one allele of a gene may mask the expression of another. This property is
called "dominance".
Mendel worked with garden peas, but the principles of particulate inheritance are the same for
all plants, animals, and fungi. Mendel observed only a single type of particulate inheritance in his
peas. Today we know there are several other common modes of particulate inheritance. In this
laboratory you will gain experience in determining the mode of inheritance of a trait using data
obtained from human pedigrees.
To determine the mode of inheritance of a trait in peas, fruit flies, or almost any other plant or
animal, one can do a series of controlled crosses, much like Mendel did. However, in humans and
some other organisms, controlled crosses are not possible, so one has to rely on data that can be
obtained from pedigrees. A pedigree is a graphical description of the genetic history of a family.
Data derived from pedigrees is one way of determining the mode of inheritance of a trait.
Basic Principles -- In humans, many other animals, and a few plants, the sex of an individual is
determined by its chromosomal constitution. Special chromosomes, called sex chromosomes,
determine the sex of an individual. In humans there are two sex chromosomes, the X chromosome
and the Y chromosome. Normal human females possess two X chromosomes. Normal human
males possess one X chromosome and one Y chromosome. The remaining 44 chromosomes of
humans are called the autosomes. Traits that are determined by genes carried on the autosomes are
called autosomal traits. Traits that are determined by genes on the X chromosome are called sex
linked or X-linked. Traits that are determined by genes carried on the Y chromosome are called Ylinked or holandric. Holandric traits are very rare and will not be discussed further in this exercise.
Both sex linked and autosomal genes can have dominant and recessive alleles, which allows
traits to be divided into four common classes, autosomal dominant, autosomal recessive, sex linked
dominant, and sex linked recessive. Each mode of inheritance can produce specific patterns of
genetic expression from one generation to the next, and each mode can't produce other patterns of
genetic expression. Understanding what can and can't be produced by each mode of inheritance is
the key to understanding how to analyze pedigrees. In the exercises below you will analyze
pedigrees to determine if a particular mode of inheritance is possible. In these exercises you will
Biology 155 Laboratory Supplement: 37
learn an important principle of science, that conclusions about the likely nature of any phenomenon
are made by rejecting alternatives.
Autosomal recessive traits -- The expression of an autosomal trait is not normally influenced by
the sex of the individual. So, for the moment we can ignore sex and look at what is possible and
what is not in the inheritance of such traits. If there are two alleles in a population, a dominant and
a recessive, then we will see only two phenotypes, the dominant phenotype and the recessive
phenotype. Using the letter "A" to designate the dominant allele and the letter "a" to designate the
recessive allele, the following table shows the phenotype of each of the three possible genotypes. If
the trait in question caused a malady or disease, we could call all individuals that express the trait
"affected" and all that don't express the trait "unaffected".
Genotype
AA
Aa
aa
Phenotype
unaffected (dominant)
unaffected (dominant)
affected (recessive)
Using the information in this table, we can make conclusions about what we might and might not
see in a pedigree. But first we need to learn more about pedigrees.
Symbols used in pedigrees. -- In all pedigrees, the sex of an individual is represented by a symbol.
Females are represented by circles, and males are represented by squares. Individuals whose sex is
unknown, such as an unborn child, are represented by diamonds. Affected individuals, those that
exhibit the trait in question, are shown as shaded symbols. Unaffected individuals are shown as
unshaded symbols.
The pattern of relationship between individuals is shown by lines connecting them. A mating
between a male and a female is shown by a horizontal line connecting the symbols
The children of such a mating shown below the mating and connected to it with a vertical line. A
series of children from a mating are shown as shown below.
Biology 155 Laboratory Supplement: 38
Exercise 1. Consider the following pattern and ask yourself if it could be produced by the
inheritance of an autosomal recessive allele. In this exercise and all that follow, first assume (or
hypothesize) the trait is inherited according to the mode in question, then try to determine the
genotypes of all relevant individuals. Discuss your conclusions with your lab partners. You should
try do determine if the mode of inheritance can account for the pattern seen. If it can't, then you
can reject that mode of inheritance as a possibility. Do this before reading the next paragraph.
Conclusions. -- The data presented in figure 4 should have allowed you to conclude that the trait
could be inherited as an autosomal recessive. Under this hypothesis the affected child must have
the genotype "aa", and it must have inherited one "a" allele from each parent. The mother is also
affected, so she must also have the genotype "aa". The father is unaffected, but under the
hypothesis of autosomal recessive inheritance, he must be a heterozygote, "Aa", in order for his
child to be "aa". You should also have concluded that the unaffected child has genotype "Aa".
Exercise 2. Consider another pattern. Can this pattern of inheritance be interpreted as the product
of an autosomal recessive allele? Proceed as you did in the last exercise. Make all the conclusions
that you can before reading the next paragraph.
Conclusions. Again, assuming the trait is inherited as an autosomal recessive, the affected
individual must be "aa". For this to be the case the unaffected parents must both be "Aa". The
unaffected child could be "AA" or "Aa". This type of uncertainty is fine in pedigree analysis, and
such individuals are often designated as "A-".
Exercise 3. Consider a third pattern. Can this pattern of inheritance be interpreted as the product of
an autosomal recessive allele? Proceed as you did in the last exercise. Make all the conclusions
that you can before reading the next paragraph.
Conclusions. Under the hypothesis that an autosomal recessive allele is responsible for the
inheritance of this trait, all affected individuals must be have the genotype "aa". The affected ("aa")
parents could clearly produce an affected child, but they could not produce an unaffected ("A-")
child. The inheritance pattern shown in this exercise clearly allows you to reject the possibility that
this trait is the product of an autosomal recessive allele.
Biology 155 Laboratory Supplement: 39
Autosomal dominant traits -- Like autosomal recessive traits, autosomal dominant traits are not
normally influenced by the sex of the individual. The dominant allele that produces the trait can be
designated with an upper case letter, like "A", and the recessive allele, the one that does result in
expression of the trait, can be designated with a lower case letter, like "a". The following table
shows the phenotype of the three possible genotypes:
Genotype
AA
Aa
aa
Phenotype
affected
affected
unaffected
Using this information we can attempt to interpret the data available in a pedigree. As in the last
set of exercises, start by hypothesizing that the trait is inherited as an autosomal dominant and
attempt to make pattern fit with this hypothesis.
Exercise 4. Consider the following pedigree and attempt to evaluate the hypothesis that this trait is
the product of an autosomal dominant allele. Assign genotypes to all individuals to the degree
possible. Discuss your assignments with your lab partners. Do this before reading the next
paragraph.
Conclusions This pattern in consistent with the hypothesis that this trait is the product of an
autosomal dominant allele. All affected individuals can be assigned the genotype "A-". However,
because the unaffected child must be genotype "aa", the parents must both have the genotype "Aa".
Exercise 5. Consider another pedigree and attempt to evaluate the hypothesis that this trait is
the product of an autosomal dominant allele. Attempt to interpret this pattern as being due to
the inheritance of an autosomal dominant allele. Assign genotypes to all individuals to the
degree possible. Discuss your assignments with your lab partners. Do this before reading the
next paragraph.
Conclusions -- If this trait were the product of an autosomal dominant allele, then all
unaffected individuals would have the genotype "aa" and all affected individuals would have at
least one "A" in their genotype, (i.e. "A-). Clearly this pattern is not consistent with the
hypothesis. The affected child, under this hypothesis has genotype "Aa" and both his affected
parents have must have genotype "aa". Thus, if the trait were the product of an autosomal
dominant allele the unaffected parents could not have produced an affected offspring. This
pedigree allows you to reject the hypothesis that this trait is the product of an autosomal
dominant allele.
Biology 155 Laboratory Supplement: 40
Sex-linked recessive traits. -- A sex linked trait is carried on the X chromosome. Because females
have two X chromosomes and males have only one the pattern of inheritance of a sex linked trait is
different for males and females. To make it clear that traits are sex linked the alleles are often
designated as superscripts on the the letter "X". So in this case the alleles would be XA and Xa. All
females have two X chromosomes, but males have only one. However, males also have Y
chromosome so the genotype of a male often includes the Y. Because males can have only one
copy of an allele they can't be called homozygous or heterozygous. Instead they are called
hemizygous. Males always have a Y chromosome that they received from their father, and an X
chromosome that they received from their mother. Females receive an X chromosome from each
parent. The following table shows the phenotypes of each of the possible genotypes for a trait that
is the product a sex linked recessive allele.
Female Genotypes
XAXA
XAXa
XaXa
Phenotypes
unaffected
unaffected
affected
Male Genotypes
XAY
XaY
unaffected
affected
Exercise 6. Consider the following pedigree and evaluate the hypothesis that the trait is the product
of a sex linked recessive allele. Again, attempt to assign genotypes to all individuals as fully as
possible. Discuss you assignments with your lab partners. When you finish you may read the next
paragraph.
Conclusions. -- The affected father has to have the genotype XaY and the affected daughter has to
have the genotype XaXa. The daughter had to have received an Xa allele from both parents, so the
unaffected mother is genotype XAXa. The unaffected son has genotype XAY. He received the Y
from his father and the XA from his mother. Clearly, the pedigree supports the hypothesis that this
trait is the product of a sex linked recessive allele.
Exercise 7. Consider the following pedigree and evaluate the hypothesis that this trait is the product
of a sex linked recessive allele. Again, attempt to assign genotypes to all individuals as fully as
possible. Discuss you assignments with your lab partners. When you finish you may read the next
paragraph.
Conclusions. -- Under the hypothesis that the trait is the product of a sex linked recessive allele,
the affected daughter must be XaXa and the affected male child and father must both be XAY. If
this trait were a sex linked recessive trait, the affected daughter would have to have received a Xa
from both parents. It is possible that she could have received a Xa allele from her mother, if her
Biology 155 Laboratory Supplement: 41
mother had the genotype XAXa, but she could not have received the Xa allele from her father. Thus,
this pedigree is not consistent with the hypothesis that this trait is the product of a sex linked
recessive allele. Before proceeding, evaluate the same pedigree as with the hypothesis that this trait
is the product of an autosomal recessive allele. Discuss your conclusions with your lab partners.
Sex linked dominant traits. -- Sex linked dominant traits are also carried on the X chromosome.
In this case a single dominant allele, designated XA is sufficient to produce an affected individual.
The table below shows the phenotypes of each of the possible genotypes for a trait that is the
product of a sex linked dominant allele.
Female Genotypes
XAXA
XAXa
XaXa
Phenotypes
affected
affected
unaffected
Male Genotypes
XAY
XaY
affected
unaffected
Exercise 8. Consider the following pedigree and attempt to evaluate the hypothesis that this trait is
the product of a sex linked dominant allele. Assign genotypes to all individuals as completely as
possible. Discuss your assignments with your lab partners before reading the next paragraph.
Conclusions. -- The affected father must have genotype XAY. The unaffected mother must have
genotype XaXa. The unaffected male child must have genotype XaY and he could have received
the recessive allele from his mother. The affected daughter can be assigned the genotype XAXa
under this hypothesis. She received the XA allele from her father and the Xa allele from her mother.
Clearly, this pedigree is consistent with the original hypothesis.
Exercise 9. Consider the following pedigree and attempt to evaluate the hypothesis that this trait is
the product of a sex linked dominant allele. Assign genotypes to all individuals as completely as
possible. Discuss you assignments with your lab partners before reading the next paragraph.
Conclusions. -- Under the hypothesis that this trait is the product of a sex linked dominant allele,
the affected father must have the genotype XAY, the unaffected mother must have the genotype
XaXa, the unaffected son must have the genotype XaY, and the unaffected daughter must have the
genotype XaXa. Clearly, this pedigree is not consistent with the hypothesis because under this
hypothesis the daughter would had to have received the XA allele from her father. So, there is no
Biology 155 Laboratory Supplement: 42
way we could expect to see an unaffected daughter when the father is affected if the trait is the
product of a sex linked dominant allele.
Complete Analysis of Pedigrees
Normally, pedigree analysis consists of evaluating a pedigree for a number of different modes of
inheritance and rejecting all modes that are not consistent with the data. Often there is too little
data to reject all but one mode. When this is the case, researchers have to look for other pedigrees
that may provide just the right information for ruling out one or more possible modes of
inheritance. It is by rejection of the alternatives that a single mode of inheritance is assigned as the
most likely mode of inheritance.
Exercise 10. In this exercise, you are to evaluate the pedigree shown below relative to the four
common modes of inheritance, autosomal recessive, autosomal dominant, sex linked recessive, and
sex linked dominant. If you are working in groups, you can divide the testing of modes among
members of your group. Unlike the previous exercises, assign genotypes only so far as is needed to
reject a mode of inheritance. If a mode of inheritance can be rejected, record it as rejected. Discuss
your conclusions with your lab partners before reading the next paragraph.
Conclusions: -- Both dominant modes of inheritance can be ruled out because the affected female
generation III could not have received a dominant allele from either parent. The hypothesis of sex
linked recessive can be rejected. The affected female in generation III would have to have received
the allele from both parents and since her father is unaffected he could not have the allele in his
genotype. The hypothesis of autosomal recessive is the only hypothesis that can't be rejected.
Discussion questions:
1. What does the complete analysis of a pedigree prove or disprove?
2. When you reject all but one mode of inheritance does this prove that this mode of inheritance is
the correct one? (Could other modes of inheritance be possible?)
3. Can your conclusions about proof and rejection in pedigree analysis be extended to science in
general? Can you think of any specific cases in science where conclusions were made after a
hypothesis was rejected?
Biology 155 Laboratory Supplement: 43
Exercises 11-15. The next several pages have a series of pedigrees for you and other members of
your group to evaluate. Each of you should start your evaluation independently of other members
of the group. Whenever you think you can reject one mode of inheritance tell other members of
your group that you can, and stop and discuss it with them. Reject as many modes as possible.
Come to a conclusion about the most likely mode or modes of inheritance. For several of the
following pedigrees you will conclude that more than one mode of inheritance is possible. For
these, you should discuss what other offspring, from a mating that is shown, or a from a mating in a
subsequent generation might allow you to reject one or more of the alternatives.
Exercise 11. Consider the following trait that has only been observed in males from one family.
Exercise 12. The following pedigree is for congenital deafness. Shaded symbols represent deaf
individuals.
Biology 155 Laboratory Supplement: 44
Exercise 13. Consider the following trait that has only been observed in females in one family.
Exercise 14. In this exercise, assume that this is a very rare trait in the general population. Such an
assumption allows you to conclude that none of the individuals marrying into this family are likely
carriers of the trait.
Biology 155 Laboratory Supplement: 45
Exercise 15. Although you will probably conclude that two modes of inheritance could have
produced this pedigree, you may be able to conclude that one is much more likely based on the data
in the pedigree.
More pedigrees: -- For more information and practice with pedigrees see the CD ROM that
accompanies your text.
Glossary
affected - exhibiting the trait
autosomal - a gene or trait that is carried on a chromosome that is not a sex chromosome
carrier - an individual that carries a single copy of a recessive allele. The allele is not seen because
it is masked by a dominant allele.
dominant - an allele that can mask the expression of another.
genotype - the genetic consititution of an individual
hemizygous - the state of having only one copy of a sex linked gene. This is the normal state of
males.
holandric - a gene or trait that is carried on the Y chromosome.
hypothesis - an educated guess, or a working guess about the nature of a phenomenon.
phenotype - the outward appearance of an individual
recessive - an allele whose expression can be masked by another.
sex linked - a gene or trait that is carried on the X chromosome.
unaffected - not exhibiting a trait.
x-linked - see sex linked.
y-linked - see holandric.
Biology 155 Laboratory Supplement: 46
Evolution by Genetic Drift and Natural Selection
Synopsis. -- The exercise uses computer simulations to investigate two of the most important
evolutionary processes: genetic drift and natural selection. One of the foundations of population
genetics, the Hardy-Weinberg Law, forms the basis for these simulations. Genetic drift will be
investigated by varying the simulated population size and allowing reproduction to continue for a
number of generations. Natural selection will be investigated by assuming that the reproducing
population is large, assigning starting genotypic frequencies, assigning fitness values to each
genotype, and allowing reproduction for a number of generations. The simulations will be used to
investigate effects of selection against a recessive allele, selection favoring a dominant allele, and
selection favoring a heterozygote. You will be able to extend these investigations to alternative
models of evolution.
Objective. -- To investigate and compare evolution resulting from natural selection and from
genetic drift using computer simulations.
Introduction
Evolution. -- Evolution is a change in allelic frequencies within a population. Evolution can be
brought about by a number of factors. Because mutation can introduce new alleles into a
population, and thus change the frequency of existing alleles, it is a potential cause of evolution.
Similarly, immigration or emigration can alter the frequency of alleles. However, the best-known,
and probably most important, cause of evolution is natural selection. Natural selection is the
differential perpetuation of alleles that results from differences in survival or reproduction of
individuals of different genotypes. Differential perpetuation of alleles means that some alleles
increase in frequency while others decrease in frequency from one generation to the next. An
important part of the definition of natural selection is that the different abilities of individuals is the
product of the alleles they possess. These alleles and abilities they confer can be passed from
parents to offspring.
Natural selection contrasts with another process that may result in differential perpetuation
of alleles: genetic drift. Genetic drift can be defined as differential perpetuation of alleles due to
chance. Chance events result in some alleles being perpetuated even though they do not contribute
to greater survival or reproductive ability. Because chance events are unpredictable, the alleles that
happen to increase due to genetic drift do not convey greater abilities for survival in the future.
Thus, we cannot expect the bearers of alleles favored by genetic drift in one generation to have
greater abilities for survival or reproduction in subsequent generations.
The Null Model: Hardy-Weinberg Equilibrium
Before we can understand natural selection and genetic drift we should consider the fate of
alleles in a population that is not evolving. The population we will consider has the following
characteristics: it is large; the organisms are diploid and reproduce sexually; individuals mate
randomly (i.e., every genotype mates with every other in proportion to their frequencies);
generations do not overlap (i.e., one generation dies as the next begins); and there is no mutation,
migration, or natural selection. These assumptions are listed in Table 1.
Table 1. Assumptions required for a population to be in Hardy-Weinberg Equilibrium.
Large population size
Generations do not overlap
Diploid organism
No mutation
Sexual reproduction
No immigration or emigration
Random mating
No natural selection
Biology 155 Laboratory Supplement: 47
Consider a gene in this population with two alleles, A and a. We will designate the
frequency of the AA genotype D (for dominant homozygote), the frequency of the Aa genotype H
(for heterozygote), and the frequency of the aa genotype R (for recessive homozygote). Since these
are the only genotypes possible, the sum of the three frequencies must equal 1. In summary:
frequency of AA = D,
frequency of Aa = H,
frequency of aa = R
D+H+R=1
Given any set of genotypic frequencies, the frequencies of the alleles can be calculated as:
ƒ(A) = (2D + H)/2 = D + (H/2)
ƒ(a) = (2R + H)/2 = R + (H/2)
Here, and in what follows, ƒ is used to indicate frequency. The terms 2D and 2R in the
above equations reflect the fact that AA and aa individuals carry two copies of each allele, while
heterozygotes carry only one. We divide by 2 because each genotype consists of two alleles. By
convention the frequency of the A allele is assigned the letter p, and the frequency of the a allele is
assigned the letter q, so that:
ƒ(A) = p
ƒ(a) = q,
and p + q = 1.
Assuming that individuals choose their mates at random, we can calculate the frequency of
each mating combination from the frequency of each genotype (see Table 2). For example,
because AA individuals make up a proportion of the population, D, we expect AA individuals to
choose other AA individuals as mates DxD = D2 of the time. Similarly, we expect AA individuals
to mate with Aa individuals 2DH of the time (2DH since an AA x Aa mating can occur between an
AA male and Aa female or an AA female and an Aa male). The expected frequencies of the six
different mating combinations in this population are gathered from column two and shown in the
second column of Table 3.
Table 2. Expected frequencies of different mating combinations between genotypes if
genotypes mate randomly.
Male
D (AA)
H (Aa)
R (aa)
D (AA)
Female
H (Aa)
R (aa)
D2
DH
DR
DH
H2
HR
DR
HR
R2
To see how alleles behave in a randomly mating population we use the mating frequencies to
calculate the frequency of each genotype in the next generation. The AA x AA mating occurs with
frequency D2 and produces only AA offspring. Thus, in the next generation there will be D2 AA
offspring that result from AA x AA matings. Similarly, the mating AA x Aa occurs with frequency
2DH, and produces ½ AA offspring and ½ Aa offspring. Therefore, in the next generation there
will be DH AA offspring and DH Aa offspring from AA x Aa matings. We can do this type of
calculation for each of the six possible mating combinations in order to find the expected
frequencies of all genotypes among offspring of the next generation. The results are shown in
Table 3.
Biology 155 Laboratory Supplement: 48
Table 3. Expected mating frequencies and offspring genotypic frequencies for a randomly
mating population.
F2 Genotypic Frequency
Mating
Mating
Combination
Frequency
AA
Aa
aa
2
2
AA x AA
D
D
AA x Aa
2DH
DH
DH
AA x aa
2DR
2DR
Aa x Aa
H2
H2/4
H2/2
H2/4
Aa x aa
2HR
HR
HR
2
aa x aa
R
R2
If we total the columns for each offspring genotype and call the new genotypic frequencies D',
H', and R' we get:
D' = D2 + DH + H2/4
H' = DH + 2DR + H2/2 + HR
R' = H2/4 + HR + R2
If these expressions are rewritten, factored and simplified we get:
D' = D2 + DH + H2/4
= (D + H/2)2
= p2
H' = DH + 2DR + H2/2 + HR = 2(D + H/2)(R + H/2)
= 2pq
R' = H2/4 + HR + R2
= q2
= (R + H/2)2
One important result of all of these calculations and manipulations is this: After one
generation of random mating the genotypic frequencies will have a binomial distribution
D' = ƒ(AA) = p2, H' = ƒ(Aa) = 2pq, R' = ƒ(aa) = q2. (Notice that we made no assumptions about
the initial values of D, H, or R.) This result is the Hardy-Weinberg Law, named after its two
discoverers. Further, once this genotypic frequency distribution is reached, it will not change,
provided the population continues to mate randomly and all of the other assumptions of Table 1 are
true. The condition of stable genotypic frequencies that result from random mating is called the
Hardy-Weinberg Equilibrium.
Another way of looking at the genotypic frequencies under Hardy-Weinberg Equilibrium is
shown in Table 4. Under HWE, AA individuals represent p2 of the population because random
mating results in A bearing sperm (whose frequency is p) fertilizing A bearing eggs (whose
frequency is also p) p2 of the time. Similarly, we expect fertilizations that produce heterozygotes to
occur 2pq of the time and fertilization to produce aa homozygotes to occur q2 of the time.
Table 4. Genotypic combinations and frequencies
with random pairing of gametes.
Female gamete
Male gamete
A (p)
a (q)
2
A (p)
AA (p )
Aa (pq)
a (q)
Aa (pq)
aa (q2)
The Hardy-Weinberg Equilibrium yields another very important conclusion: given the
assumptions made in Table 1, genotypic frequencies will never change; a population will not
evolve. So, if we see a change in allelic or genotypic frequencies, then one of the assumptions of
Table 1 has been violated.
Biology 155 Laboratory Supplement: 49
Genetic Drift: The Effect of Finite Population Size
An important assumption that was made in the derivation of the Hardy-Weinberg Law is
that the population is very large. Without this assumption there is the possibility that chance
factors may influence allelic frequencies. This is so because, in any sampling process, large
samples are more likely to be representative of the population from which they came than are small
samples. The smaller the sample, the greater is the likelihood that the sample will be different in
some way from the population from which it came just by chance alone.
Consider a simple life history of a diploid organism:
F0 Adult → Gamete → F1 Zygote → F1 Adult
If the assumptions necessary for HWE hold, then allelic frequencies will not change between any
two stages in this life history. In other words, ƒ(A) will be the same in F0 Adults, their gametes,
their zygotes, and in F1 Adults. However, the assumption about large population size is a very
important one and if violated there could be substantial changes in allelic frequencies. For
example, if the frequency of allele A is 0.5, we should expect that half the gametes produced by a
population of 1 million oysters to carry A. However, if were are considering a much smaller
population, such as the last surviving female Dodo bird, who is for example a heterozygote Aa, and
she produces only one egg, that egg must carry either the A or the a allele. Thus, ƒ(A) for her
gamete must be either 0 or 1, not 0.5. Even small scale chance events can have an effect. For
example, use Table 4 to see how much change in genotypic frequencies could happen if female
gametes are not produced in frequencies p and q, but (p - .01) and (q + .01).
At the next phase in the life cycle, from zygote to adult, the same sort of arguments are true.
If there are many offspring, the accidental deaths are likely to be randomly distributed among all
genotypes. For example, if 5% of all zygotes die before becoming adults as a result of being struck
by lightning, then we expect by chance alone, that 5% of all individuals of each genotype, AA, Aa,
or aa, to be struck by lightning. Among 1 million individuals we would not expect lightning or any
other chance event to prefer AA individuals over Aa or aa. However, if the population has only 20
members, 5% is just one individual. That individual would have just one of the three genotypes,
and loss of that individual will alter allelic frequencies. If that individual carries a rare allele, that
allele could be eliminated from the population.
Changes in the allelic frequencies of a population due to chance events is called genetic
drift, or random genetic drift, to emphasize the difference between changes due to random factors
and those due to selection or other nonrandom factors. Because the effective population size (the
number of reproducing adults) is never infinite, there can always be genetic drift. However, the
potential for substantial genetic drift is much greater in small populations than in large populations.
Just as the frequency of heads in a series of coin tosses will often differ from 0.5 when few tosses
are made, the allele frequency in an offspring generation may differ greatly from that of the parent
generation when the effective population size is small. You will have an opportunity to observe the
effects of population size on allelic frequencies from one generation to the next in the laboratory
exercises that follow.
Biology 155 Laboratory Supplement: 50
Natural Selection: The effect of differential survival and reproduction
Natural selection was defined earlier as differential perpetuation of alleles due to different
survival and reproductive abilities of individuals of different genotypes. The ability of an
individual to survive and reproduce is a measure of its fitness. The fitness of individuals of any
genotype can be measured in a absolute sense or in comparison to other individuals. For example, if
70% of all AA individuals survive and reproduce, 70% of all Aa individuals do likewise, but only
35% of aa individuals survive and reproduce, the absolute fitness of the genotypes are 0.70, 0.70,
and 0.35, respectively. If the fitness of each genotype is to be compared to the others we could
divide the absolute fitness of each genotype by the highest fitness value. This is a genotype's
relative fitness. In this case, 0.7/0.7 = 1.0, 0.7/0.7 = 1.0, and 0.35/0.70 = 0.5, respectively. Relative
fitness measures are often symbolized with the letter W. In this example, WAA = 1.0, WAa = 1.0,
and Waa = 0.5. Selection is given the symbol S, and is the complement of relative fitness: S = 1 W. This information is summarized in Table 5.
Table 5. Summary of the relationship between survival, fitness, and selection.
Absolute
Relative
Selection
Genotype
Survival
Fitness
Fitness (W)
(1-W)
AA
70%
0.70
1.0
0.0
Aa
70%
0.70
1.0
0.0
aa
35%
0.35
0.5
0.5
Fitness values are valuable because they can be use to predict how allelic frequencies will
change from one generation to the next. In the case above, one homozygote and one heterozygote
had equal fitness. With respect to the effects of selection, the A allele is dominant and masks the
effect of the a allele. This is a case of selection against a recessive phenotype, or selection favoring
a dominant phenotype. We can look at how selection affects the frequency of two alleles over the
course of one generation as follows.
In the case above, if the starting alleles frequencies are
ƒ(A) = p = 0.5, and ƒ(a) = q = 0.5
and the adults of the former generation mated randomly then the genotypic frequencies of new
formed zygotes are
ƒ(AA) = p2 = 0.25,
ƒ(Aa) = 2pq = 0.50, ƒ(aa) = q2 = 0.25
If selection acts before the zygotes reach reproductive age then we will see a reduction in the
relative numbers of some genotypes. To do this calculation fitness values are multiplied by the
genotypic frequencies, then the sum of all values is used to adjust the genotypic frequencies so they
sum to 1.
Biology 155 Laboratory Supplement: 51
Table 6. A numerical example of the effect of selection
Genotypic x relative
preliminary
Genotype frequency
fitness
=
value
AA
0.25
x
1.0
=
0.25
Aa
0.50
x
1.0
=
0.50
aa
0.25
x
0.5
=
0.125
sum = 0.875
The genotypic frequencies of adults after selection are
ƒ'(AA) = 0.25/0.875 = 0.29
ƒ'(Aa) = 0.50/0.875 = 0.57
ƒ'(aa) = 0.125/0.875 = 0.14
The allele frequencies among the adults are then
ƒ'(A) = D + H/2 = 0.29 + (0.57/2) = 0.575
ƒ'(a) = R + H/2 = 0.14 + (0.57/2) = 0.425
Thus, as a result of one generation of selection there has been a substantial change in the frequency
of the two alleles. In the case of allele A the frequency has changed from 0.5 to 0.575, and the
frequency of allele a has changed from 0.5 to 0.425. These allele frequencies could be used to
calculate the frequencies of zygotes in the next generation, and the above process repeated for
many generations to study how selection changes allele frequencies. The computer simulations of
natural selection in this laboratory exercise will allow you to study natural selection without having
to make these laborious calculations.
References
Ayala, F. J. 1982. Population and evolutionary genetics. A primer. Benjamin Cummings,
Reading, Massachusetts
Hartl, D. L. 1981. A primer of population genetics. Sinauer Associates, Inc. Sunderland,
Massachusetts.
Wilson, E. O. and W. H. Bossert 1971. A primer of population biology. Sinauer Associates,
Sunderland, Massachusetts.
Inc.
Biology 155 Laboratory Supplement: 52
Simulation of Genetic Drift
You will use a computer model to observe the effects of population size on changes in the
frequency of alleles in the population. The program simulates the process of reproduction in
randomly mating population but only a limited number of the offspring survive to reach
reproductive maturity. The number that survives is the effective population size. Every offspring
has an equal chance of surviving, so there are no fitness differences, and the offspring that survive
are chosen randomly by the computer. In this simulation, just as in nature, chance can result in
some genotypes surviving more often than you would expect and this will result in a change in
allele frequencies from one generation to the next. At thebeginning of each simulation the
population contains only two alleles, each with a frequency of 0.5.
Figure 1. A sample plot of a simulation of genetic drift.
Figure 1 shows the appearance of the monitor's screen at the end of one simulation. The xaxis is the number of generations and is given in scientific notation. The y-axis shows the
frequency of the A allele. Below the x-axis are two labels. The first reports the current generation
and the second reports the frequency of the A allele in that generation. Because there are only two
alleles in this population, the frequency of the a allele is always 1 - ƒ(A). The figure shows one
simulation of genetic drift with a population size of 100 over the course of 100 generations. With
this population size, the frequency of each allele changes slightly, either increasing or decreasing,
from one generation to the next soley due to chance.
Biology 155 Laboratory Supplement: 53
Procedure:
Refer to the appendix of this lab for specific instructions about running the computer
program. After you have completed a run, you will see that it is quite easy to use. Follow through
and answer the questions on the following pages.
1. Simulate a population that has an effective population size of 10 over the course of 100
generations. Record the frequency of the A allele at the end of the simulation. If the simulation
stopped before 100 generations, record the frequency of the A allele when it stopped and the
generation in which it stopped.
2. Take data on 10 simulations of an effective population size of 10 over 100 generations, and 10
simulations of an effective size 100 over 100 generations. Record the data in the table below.
Trial
Population size = 10
number of
generations to
conclusion
ƒ(A)
at end
Trial
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
Number of extinctions of A =
Number of extinctions of a =
Total number of extinctions =
Population size = 100
number of
generations to
conclusion
ƒ(A)
at end
Number of extinctions of A =
Number of extinctions of a =
Total number of extinctions =
3. Does this simulation allow you to make any conclusions about the fate of alleles or the retention
of genetic diversity in populations of different sizes?
Biology 155 Laboratory Supplement: 54
4. Clear the results of the previous simulation and begin a new series of simulations. Each
simulation will be for 100 generations. Run one simulation for each of the following effective
population sizes 10, 100, 1000, and 10,000. Sketch the results on the following graph. Label
the curves from each simulation.
5. What is the general effect of population size on genetic drift?
6. Is genetic drift evolution? Why or why not?
7. Is genetic drift a predictable process? What sorts of predictions can be made?
8. In the each of the simulations you ran, the starting allele frequencies were ƒ(A) = 0.5 and
ƒ(a) = 0.5. How might the results have differed if the initial frequencies were ƒ(A) = 0.95 and
ƒ(a) = 0.05? What is the most likely fate of rare alleles in small populations?
Biology 155 Laboratory Supplement: 55
Simulation of Natural Selection
To determine the fate of an allele, and to observe changes in the genetic structure of
populations under different types of selection, you will use a computer simulation of natural
selection. Your simulations will be of a single gene with two alleles. At the beginning of each
simulation, you will be prompted to assign fitness values to each genotype, to assign starting
frequencies to the AA and Aa genotype, and to choose the number of generations over which the
simulation will run. Since the sum of the frequencies of the three genotypes must add to 1, the
frequency of the aa genotype will be determined by subtraction.
Table 6 above gives an example of the type of calculations the computer simulation will
use. Fitness values will remain constant throughout the time of the simulation. The model assumes
that adults mate randomly and selection acts in the time between the formation of zygotes and the
maturation of adults. The process of zygote formation, maturation, and reproduction is repeated in
each generation and the results are shown graphically on the computer monitor. The model
assumes that the population is very large (so there is no genetic drift) and that there is no
immigration, emigration, or mutation.
During a simulation the monitor screen will appear as in Figure 2. The left graph shows the
frequency of allele A through time. The current generation is shown at the bottom and the current
frequency of allele A is shown at the top. The right graph is a histogram that shows the frequency
of genotypes in the current generation. Each genotype is represented by two bars. The left bar for
each genotype is the frequency zygotes of that genotype. The right bar for each genotype is the
frequency of reproducing adults for that genotype. Thus, the paired left and right bars represent the
frequencies of a genotype before and after selection. At the top of the right graph is the average
fitness of individuals in this population.
Figure 2. A sample plot of a simulation of natural selection.
Biology 155 Laboratory Supplement: 56
Procedure:
There are a number of questions below that you will be able to answer after you have
completed all of the experimental simulations that follow. Read these questions before you do the
experiments so that you will have them in mind while you are doing the experiments. At the start
of each simulation, record the fitness values you are assigning and initial genotypic frequencies.
During the simulation, note how the average fitness and the genotypic structure of the population
changes. At the end of the simulation, make a sketch of the left graph on the screen (the graph of
f(A) versus time) so you can compare the results of different simulations. Feel free to develop
hypotheses and test them with further simulations. Also, feel free to work in groups and discuss
your results with others. Four experiments are suggested that will help you answer the following
questions.
Questions:
1. What effect does changing the starting genotypic frequencies have on the eventual fate of an
allele for different types of selection?
2. Does the disadvantageous allele persist in the population longer when there is selection against
the dominant phenotype or when there is selection against the recessive phenotype?
3. When a heterozygote has a higher fitness than either homozygote, how does the equilibrium
frequency of the two alleles relate to the relative fitness of the two homozygotes?
4. How does the average fitness of a population change over time when there is selection? Does
this value appear to reach an equilibrium?
5. Is the course of change in a population under natural selection predictable? What information
would you need to make predictions about the genetic structure and eventual genetic fate of a
population? What assumptions would your predictions depend upon?
6. How do natural selection and genetic drift differ? If either were occurring in a natural or
laboratory population how could you tell? How could you tell which was occurring?
Biology 155 Laboratory Supplement: 57
Experiment 1: The effect of selection against a recessive phenotype.
If selection acts only against a recessive phenotype, the dominant homozygote and the
heterozygote each have a relative fitness of 1. For this experiment begin by assigning fitness
values as follows: WAA = 1, WAa = 1, and Waa = 0.5. Use the following starting genotypic
frequencies: ƒ(AA) = 0.0 and ƒ(Aa) = 0.1. (Thus ƒ(aa) = 0.9.) Observe selection through 200
generations. This set of starting conditions simulates a population that has a rare dominant allele
that confers high fitness. At the end of the simulation be sure to record the results shown on the left
graph. Initial values for more simulations of this type of selection are shown next to the graphs that
follow. Additional graphs are provided for you to use for choosing your own initial conditions.
One thing you should try is changing the initial genotypic frequencies.
WAA
WAa
Waa
= 1.0
= 1.0
= 0.5
ƒ(AA) = 0.0
ƒ(Aa) = 0.1
ƒ(aa) = 0.9
number of
generations = 200
final ƒ(A) =
final average fitness =
WAA
WAa
Waa
= 1.0
= 1.0
= 0.3
ƒ(AA) = 0.0
ƒ(Aa) = 0.1
ƒ(aa) = 0.9
number of
generations = 200
final ƒ(A) =
final average fitness =
Biology 155 Laboratory Supplement: 58
WAA
WAa
Waa
= 1.0
= 1.0
= 0.9
ƒ(AA) = 0.0
ƒ(Aa) = 0.1
ƒ(aa) = 0.9
number of
generations = 200
final ƒ(A) =
final average fitness =
WAA
WAa
Waa
=
=
=
ƒ(AA) =
ƒ(Aa) =
ƒ(aa) =
number of
generations =
final ƒ(A) =
final average fitness =
WAA
WAa
Waa
=
=
=
ƒ(AA) =
ƒ(Aa) =
ƒ(aa) =
number of
generations =
final ƒ(A) =
final average fitness =
Biology 155 Laboratory Supplement: 59
Experiment 2. The effect of selection against a dominant phenotype.
If selection acts only against a dominant phenotype, the dominant homozygote and the
heterozygote have equal fitness and their relative fitness is less than 1. The recessive homozygote
has a fitness of 1. For this simulation assign fitness values as follows: WAA = 0.5, WAa = 0.5 and
Waa = 1.0. Use the following starting genotypic frequencies: ƒ(AA) = 0.9 and ƒ(Aa) = 0.1. (Thus
ƒ(aa) = 0). Observe selection through 200 generations. These starting conditions simulate a
population with a rare recessive allele that confers high fitness. At the end of the simulation be sure
to record the results shown on the left graph. Initial values for more simulations of this type of
selection are shown next to the graphs that follow. Additional graphs are provided for you to use
for choosing your own initial conditions. One thing you should try is changing the initial genotypic
frequencies
WAA
WAa
Waa
= 0.5
= 0.5
= 1.0
ƒ(AA) = 0.9
ƒ(Aa) = 0.1
ƒ(aa) = 0.0
number of
generations = 200
final ƒ(A) =
final average fitness =
WAA
WAa
Waa
= 0.3
= 0.3
= 1.0
ƒ(AA) = 0.9
ƒ(Aa) = 0.1
ƒ(aa) = 0.0
number of
generations = 200
final ƒ(A) =
final average fitness =
Biology 155 Laboratory Supplement: 60
WAA
WAa
Waa
= 0.9
= 0.9
= 1.0
ƒ(AA) = 0.9
ƒ(Aa) = 0.1
ƒ(aa) = 0.0
number of
generations = 200
final ƒ(A) =
final average fitness =
WAA
WAa
Waa
=
=
=
ƒ(AA) =
ƒ(Aa) =
ƒ(aa) =
number of
generations =
final ƒ(A) =
final average fitness =
WAA
WAa
Waa
=
=
=
ƒ(AA) =
ƒ(Aa) =
ƒ(aa) =
number of
generations =
final ƒ(A) =
final average fitness =
Biology 155 Laboratory Supplement: 61
Experiment 3. The effect of selection favoring the heterozygote.
If selection favors the heterozygote, then the heterozygote has a relative fitness of 1 and the
fitness of each homozygote is less than 1. For this simulation assign fitness values as follows
WAA = 0.5, WAa = 1.0, and Waa = 0.5.
Use the following starting genotypic frequencies:
ƒ(AA) = 0.9 and ƒ(Aa) = 0.1. (Thus ƒ(aa)) = 0.) Observe selection through 200 generations.
These initial conditions simulate selection in a population where the heterozygote has greater
survival and reproductive potential than either homozygote and the recessive allele is rare. At the
end of the simulation be sure to record the results shown on the left graph. Initial values for more
simulations of this type of selection are shown next to the graphs that follow. Additional graphs are
provided for you to use for choosing your own initial conditions. One thing you should try is
changing the initial genotypic frequencies.
WAA
WAa
Waa
= 0.5
= 1.0
= 0.5
ƒ(AA) = 0.9
ƒ(Aa) = 0.1
ƒ(aa) = 0.0
number of
generations = 200
final ƒ(A) =
final average fitness =
WAA
WAa
Waa
= 0.5
= 1.0
= 0.9
ƒ(AA) = 0.9
ƒ(Aa) = 0.1
ƒ(aa) = 0.0
number of
generations = 200
final ƒ(A) =
final average fitness =
Biology 155 Laboratory Supplement: 62
WAA
WAa
Waa
= 0.9
= 1.0
= 0.5
ƒ(AA) = 0.9
ƒ(Aa) = 0.1
ƒ(aa) = 0.0
number of
generations = 200
final ƒ(A) =
final average fitness =
WAA
WAa
Waa
= 0.9
= 1.0
= 0.9
ƒ(AA) = 0.9
ƒ(Aa) = 0.1
ƒ(aa) = 0.0
number of
generations = 200
final ƒ(A) =
final average fitness =
WAA
WAa
Waa
=
=
=
ƒ(AA) =
ƒ(Aa) =
ƒ(aa) =
number of
generations =
final ƒ(A) =
final average fitness =
Biology 155 Laboratory Supplement: 63
Appendix
Using the Population Genetics Simulation Software
The population genetics simulation software (PGEN) can be accessed on the Biology 155
homepage: http://www.nicholls.edu/biol-ds/biol155/default.htm. Follow the instructions there for
starting the program.
The initial screen of the population genetics software will give you some background on
the simulations. Press a key to get to the menu.
From this point on the ESC key will back you out of whatever activity you are attempting.
At the menu, the ESC key will allow you to quit the program. While the simulation is running, you
can pause the activity by pressing any letter key or the space bar.
From the MENU, press 1 [enter] to simulate genetic drift or 2 [enter] to simulate natural
selection, or ESC to quit.
Genetic Drift -- Under "Enter Parameters" you will be asked to enter the number of generations to
simulate, then the size of the population. If the number isn't legal (e.g. a population size of 0), you
will be prompted to enter another number. You have to press the ENTER or RETURN key after
you have typed the number. During the simulation you will see a graph like figure 1 in the lab
write-up. The vertical axis is the frequency of the allele A. This number varies from 0 to 1. A
frequency of 0 means the allele is no longer in the population, and a frequency of 1 means that all
individuals are homozygous for the allele (i.e. the other allele is no longer in the population). The
horizontal axis indicates the generation number. The horizontal line through the middle of the
graph shows the starting frequency of 0.5. The simulation will proceed for the number of
generations you chose, or it will stop when the frequency of the allele reaches 0 or 1. Because the
model does not allow for reintroduction of an extinct allele, there can be no further change in the
allele frequency and the program stops. At the end of a simulation you will be prompted to enter
the population size for another simulation. You can enter a number, or just press enter to use the
same number as was used in the last simulation.
The length of time each simulation takes to run to completion depends upon the number of
generations that you choose to observe and the size of the population. Larger population sizes and
generation numbers will increase the length of time that it takes a simulation to run.
If the graph becomes too cluttered, press ESC and restart the simulation.
Natural Selection -- More parameters must be entered to start a simulation of natural selection.
Under "Enter Parameters" you must first enter the fitness of each genotype. Although you can enter
very large numbers, the program will scale them all to relative fitness values, so that the most fit
genotype will have a fitness of 1.0 and less fit genotypes will be scaled proportionately.
NOTE: The program uses two alleles A and a, which combine to create three genotypes,
AA, Aa, and aa, but there is no necessary dominance relationship between the alleles. You create
selection against a recessive allele by giving the aa genotype a lower fitness than the other two
genotypes. You create selection against a dominant allele by giving the genotypes AA and Aa
lower fitness than the aa genotype. If you give Aa and aa a higher fitness than AA then you are
implying that the a allele has an effect on the fitness of the heterozygote and is not fully recessive.
Next, you will be asked to enter the initial frequency of the AA and Aa genotypes. These
numbers must sum to a number less than or equal to 1. You do not need to enter the frequency of
the aa genotype, the program will calculate its frequency so that the sum of the frequencies of the
three genotypes is exactly 1.
At this point you will be shown a summary of the parameters you entered. If they are not
correct, press ESC and enter them again. Next, enter the number of generations you will to see
simulated. This number must be an integer. The screen for the natural selection simulation looks
like that shown in figure 2 of the lab write-up. Review that section to familiarize yourself with the
information on the screen.
At the end of the simulation, you must go back to the menu and choose Drift or Selection again.
Biology 155 Laboratory Supplement: 64
AN INTRODUCTION TO SYSTEMATICS USING THE CAMINALCULES
Synopsis -- In this laboratory exercise you will learn how to use and construct a dichotomous key,
how to use outgroup comparison to develop a hypothesis of the direction of evolution of a
character, how to use derived character states to develop hypotheses of evolutionary relationship.
In this laboratory exercise you will also become familiar with the phylogenetic relationships of the
major phyla of metazoan animals.
Objectives
• Learn how to use a dichotomous key.
• Learn how to construct a dichotomous key.
• Learn how to determine the direction of evolution of a character.
• Learn how to develop an hypothesis of evolutionary relationship.
• Learn how to resolve a phylogenetic tree.
INTRODUCTION
Systematics is the study of the relationships of organisms. A systematist uses knowledge of
relationships to classify organisms. In this exercise you will use some of the tools of systematists
for identifying and determining relationships among organisms using the Caminalcules as model
organisms.
The Caminalcules are a fanciful group of organisms that are the product of the imagination
of the late Joseph R. Camin. Camin was a systematist who was interested in both the theoretical
and practical issues of systematics and developed the Caminalcules as a means of illustrating and
testing models of evolution and systematics. Caminalcules are an ideal group for student use
because they can illustrate several principles of systematic zoology, introduce the character as a
unit of information in systematics, illustrate the need for careful observation in systematics, and
allow hypotheses of relationship to be developed and tested. The word Caminalcule is obviously
derived from Camin's name, the suffix "cule" is from the Latin for sack. Caminalcules are
"Camin's sack animals".
Dichotomous keys
A dichotomous key is a tool for the identification of organisms. It is used by both
professional and amateur biologists to identify unfamiliar organisms. A dichotomous key is a
series of paired questions, or couplets, that split a group of organisms two smaller groups. Each
couplet generally focuses on one or a few alternate states of a single character. A character is some
feature of the morphology, biochemistry, physiology, development, or behavior of an organism.
Dichotomous keys use characters that have alternate states, such as presence or absence, big or
little, etc. Classically, one half the couplet will apply to organisms with one or more states of a
character and the other half will apply to all other organisms in the group in question. As one
progresses through a key, the questions become applicable to a smaller and smaller group of
organisms until a definite identification is reached. Often it takes a series of couplets to arrive at a
definite identification. Each couplet will either provide a definite identification or direct the user to
another couplet.
Biology 155 Laboratory Supplement: 65
Biology 155 Laboratory Supplement: 66
Biology 155 Laboratory Supplement: 67
Exercise 1: Cut out all of the Caminalcules from the sheets provided (include their numbers in the
cut-out). Use the following key to the genera of Caminalcules to identify all members of each
genus of Caminalcule.
Key to the Genera of Caminalcules
1A. Front appendages elongate or tentacle-like
B. Front appendages not elongate or tentacle-like
2
3
2A. Abdomen enlarged, rear appendage with two small pads
B. Abdomen not enlarged, rear appendage is broad or flipper-like
Abdoculea
Caudiculea
3A. Front appendages subdivided with clear toe-like digits
B. Front appendages not subdivided
Pediculea
4
4A. Front appendages broadly flattened dorsoventrally resembling wings Pteroculea
B. Front appendages absent, or if present only as small buds
Bathyculea
How to construct a dichotomous key
There are various styles for dichotomous keys. The style given above is commonly used by
zoologists. Botanists often use a slightly different, but functionally equivalent, style key.
Regardless of the type of organism, the job of constructing a dichotomous key consists of finding
characteristics that will divide a group into two groups, and then within each of these subgroups
find characteristics that will divide them into two groups again. This process is repeated until all
taxa are identified unambiguously. When choosing characters it is important to use features that
can be determined without comparison to other taxa whenever possible. For example, character
states like large or small can only be determined through comparison, but character states like
single or double can be determined without comparison. If a character that requires comparison is
used, like enlarged abdomen, then other supporting characters that are less ambiguous should also
be included in the couplet. Often the couplets can be constructed as the characters are chosen. To
construct the text of the key, work out all the subgroups of one group fully, constructing as many
couplets as necessary, and leave blanks for the couplet numbers of subgroups that you will return to
later.
For practice, start by laying out all of the Caminalcules in one pile and then split it into two
piles using the key above. This will give you one group with tentacle-like appendages and one
without tentacle-like appendages. Now, split the first group into two groups following the second
couplet; those with enlarged abdomens and those without. Continue splitting until there are no
directions to another couplet and then backtrack to the couplet for a group that has not been
classified fully. If you do this neatly on a bench or table top you will have a series of piles and
subpiles that correspond exactly to the key above and how it was constructed.
Exercise 2. Choose one genus of Caminalcule and construct a dichotomous key that can be used to
identify each member of the group. Assign a name to each species in the genus before you start.
Give your key to another group and let them evaluate the effectiveness of your key.
Exercise 3. Evaluate the effectiveness of another group's key to the species of a genus of
Caminalcule. Provide feedback (positive and negative) to the group on the choice of characters and
character states.
Biology 155 Laboratory Supplement: 68
Evolution, Phylogenetic Inference, and Reconstruction of Phylogenies
Evolution has two distinct components that are equally important in the creation of
biological diversity. Anagenesis, or change in genetic lineage through time, and cladogenesis, the
splitting of genetic lineages. Anagenesis is change from generation to generation and is responsible
for the origin of new states of a character. Cladogenesis is the splitting of genetic lineages by
speciation. When a single species becomes two, new independently evolving genetic lineages are
created. Within each of the new lineages, both anagenesis and further cladogenesis can occur for
the creation of ever more diversity.
With both anagenesis and cladogenesis as components of the evolutionary process, we can
envision the history of evolution of a group of organisms, or of all organisms, as having a
branching or tree-like pattern. This representation of the evolution of a group of organisms is
called a phylogeny. An important part of the study of any group of organisms, and of the study of
evolution in general, is the study of the phylogeny of that group of organisms. Reconstruction of
phylogenetic histories is the endeavor of the systematist, and forms the basis of groupings in animal
classifications. The goal of the systematist is to reconstruct the phylogeny of a group of organisms
for classification, and for making further discoveries about evolutionary processes.
The principles of phylogenetic reconstruction are simple in theory, but often difficult in
practice. What a systematist would like to do is identify character states that are indicative of
shared ancestry. Using the Caminalcules as an example, if we knew that toes on rear appendages
evolved within the genus Pediculea, then we could hypothesize that the group of species with such
toes probably are the descendants of the species that first evolved toes. If they are all descendants
of the original toed ancestor, then they are members of a single branch of the evolutionary tree of
the Caminalcules, and form a subbranch of the branch that is the Pediculea. A group that consists
of all of the descendants of a common ancestor is called a monophyletic group. Once a
monophyletic group (a branch on the tree) has been identified, subbranches within that group can
be identified using other characters until a complete hypothesis of evolutionary relationships is
constructed.
A problem that makes phylogenetic reconstruction difficult is that not all shared character
states are indicators of shared ancestry. A systematist must determine the likely direction of
evolution of a character. For example, within the Caminalcules there are species with rear
appendages with toes and those with rear appendages without toes. A systematist has to decide
which of these states is derived (or advanced) and which is ancestral (or primitive). The derived
state can be used as evidence of shared ancestry, but the ancestral state cannot.
There are several tools for deciding which state of a character is derived. The first, and
most important of these, is called "outgroup comparison". To use outgroup comparison, a
systematist must first identify a likely monophyletic group. This is a group that shares one or more
derived character states. For example, the Pediculea can be defined as the group that has toes on its
front appendages. For the sake of example at this point, we must assume that the Pediculea is a
monophyletic group; all species in this group are descended from a single ancestor that also had
front appendages with toes. The following figure shows a phylogeny that groups all the members
of the Pediculea together, but does not provide any information about how they are related. The
character "toes on front appendages" is the basis for our estimate that they form a monophyletic
group and is designated with a labeled bar under the group.
Biology 155 Laboratory Supplement: 69
Figure 1. A completely unresolved phylogeny of the Pediculea.
Now, we can look at another character, such as the presence of toes on rear appendages, and
ask "Is the presence of toes on the rear appendages a derived or ancestral state?". To determine
which is the ancestral state we can look at other closely related Caminalcules outside the Pediculea.
The Pediculea are the "ingroup" and other Caminalcules are the "outgroup". We look at the
outgroup because members of the outgroup share a common ancestor with the ingroup, but a more
distant or ancient relationship. Thus, the descendants of this more distant ancestor (the outgroup)
are likely to possess the ancestral state of the character. If members of the outgroup have toes, then
we can reasonably conclude that the ancestral state is the presence of toes, and the absence of toes
is the derived state and this state originated within the Pediculea. If most members of the outgroup
do not have toes, then we can conclude that the absence of toes is the ancestral state, and the
presence of toes is the derived state; that toes originated within the Pediculea and that all of the
members of the Pediculea with toes form a monophyletic group. When we look outside the
Pediculea, we find no species with toes on the rear appendages. Thus, the absence of toes can be
hypothesized to be the ancestral condition and the presence of toes can be hypothesized to be the
derived condition. This information can then be used to construct an hypothesis of phylogenetic
relationship within the Pediculea. This hypothesis is shown in figure 2.
Biology 155 Laboratory Supplement: 70
Figure 2. A refinement of the phylogeny of the Pediculea using the character state
"presence of toes on the rear appendages".
A hypothesis of phylogenetic relationship is a educated guess about shared ancestry. The
character used in the example above, "presence of toes on the rear appendages", suggests that
species 2, 12, and 22 shared a relatively recent common ancestor and are more closely related to
each other than they are to any other species of Pediculea.
We can resolve the relationship of species 2, 12, and 22 further by looking for character
state variation within this group. Among these three species we find that species 2 and 12 have a
crest, or dorsal protruberance, on their head, while species 22 does not. We can now use the other
members of the Pediculea as the outgroup and ask if a crest is ancestral or derived. When we look
at the other members of the Pediculea we find that no species have a crest. Thus, we can
hypothesize that the absence of a crest is the ancestral state, and the presence of a crest is the
derived state. This allows further refinement of our hypothesis of evolutionary relationship:
Biology 155 Laboratory Supplement: 71
Figure 3. A refinement of the phylogeny of Pediculea using the character state "presence of
a crest on the head".
We can refine our estimate of the relationships among the Pediculea further by looking at other
characters. One character that might also be useful is the degree of separation of rear appendages.
Some species have two completely separate rear appendages (species 2, 3, 4, 5, 12, and 22) and
others have the two appendages that are joined at their base (species 18 and 23). If we look outside
the Pediculea we find that any species that have rear appendages have a single appendage or
appendages that are joined at their base. Thus, from this we can hypothesize that rear appendages
joined at their base is the ancestral state and separate rear appendages is the derived state. This
allows a further refinement of the relationships of the members of the Pediculea:
Figure 4. A refinement of the phylogeny of Pediculea using the character state "rear
appendages separate".
Biology 155 Laboratory Supplement: 72
Exercise 4. For each of the following characters, decide which state is ancestral and which is
derived and use this information to refine the phylogeny of the Pediculea.
Character
front appendage with elbows
horn on front of head
state
present
present
state
absent
absent
Exercise 5: Use outgroup comparison to decide which of the following character states is ancestral
and derived in each of the following groups of Caminalcules. After identifying the derived
character state, use it to construct, and where possible revise, a hypothesis of evolutionary
relationship.
Group
Pteroculea
Bathyculea
Bathyculea
Caudiculea
Character
rear appendage
neck
front appendages
rear appendage
state
presence
straight
present
flipper-like
state
absence
folded
absent
pad-like
Other tools for deciding the direction of evolution are also available to the systematist. If a
fossil record for a group of organisms exists, then the direction of evolution may be readily
apparent. Older, and usually deeper, sedimentary strata have fossil remains of species that existed
before species that were fossilized in younger strata. Thus, where two character states are found,
one in older fossils and one in younger fossils, the state found in the older fossils can be
hypothesized to be the ancestral state and used for estimating phylogenetic relationships.
Developmental sequences are also valuable for deciding the direction of evolution. It is
common for organisms to possess ancestral states of a character during early stages of development
and to develop derived states later in development. For example, if we had a developmental
sequence of those members of the Abdoculea that have a single eye, we could decide if a single eye
is the ancestral or derived state. If development begins with two separate eyes that later fuse, then
we can conclude that separate eyes is ancestral and fused eyes derived.
Reconstruction of the Phylogeny of the Metazoans
The most difficult aspect of phylogenetic reconstruction, and the aspect that requires the
most knowledge and expertise, is recognizing and choosing characteristics that are likely to be
informative about evolutionary relationships. The difficulty of choosing characters and interpreting
derived and ancestral states results in continual revision and refinement of estimates of
evolutionary history. Today, few phylogenies are universally agreed upon. However, the major
features of the phylogeny of many groups are generally agreed upon. One phylogeny that is fairly
well accepted is that of the Metazoa, the members of the animal kingdom that have true tissues.
The major metazoan phyla are the Cnidaria (jellyfish, corals,anemonaes), Platyhelminthes
(flatworms), Nematoda (roundworms), Rotifera (rotifers), Annelida (segmented worms), Mollusca
(molluscs: snails, clams, octopus), Arthropoda (shrimp, crabs, barnacles, spiders, scorpions,
insects), Echinodermata (starfish, sea urchins), and Chordata (tunicates, sea squirts, vertebrates).
The following characteristics are thought to be indicators of shared ancestry in the in the metazoan
animals.
One character that is used in the construction of the current phylogeny of animals is the
symmetry plan of the body, especially during the embryonic stage. There are two states, radial
symmetry or bilateral symmetry. Radially symmetric animals have a body plan that is disk- or
wheel-like. The Cnidaria is the only example of an animal phylum with radially symmetric
embryos that never become bilaterally symmetrical. Most other animals become bilaterally
Biology 155 Laboratory Supplement: 73
symmetrical during early development and remain so into adulthood. An exception to this is the
Echinoderms. Echinoderms pass through an embryonic stage that is bilaterally symmetrical and
then become radially symmetrical as adults. Thus, it appears that Echinoderms have a new type of
radial symmetry. All metazoans pass through a radially symmetrical embryonic stage, the blastula,
and thus it is thought that radial symmetry is the ancestral state and bilateral symmetry is the
derived state.
Another character that is important in the construction of phylogenies is the number of
tissue layers in the body plan. The Cnidarians all have a two tissue layer body plan; they are
diploblastic. All other metazoans are triploblastic. The mesoderm is the third tissue layer to form
during development, and thus it is thought that the ancestral state is diploblastic and the derived
state is triploblastic.
Another character that is used in the estimate of the evolutionary history of the metazoa is
the presence or absence of a body cavity and if present what type of cavity exists. The body cavity
or coelom is present in all phyla except the Cnidaria and the Platyhelminthes. Again
developmental sequences suggest that absence of a coelom (acoelomate) is ancestral and the
presence (coelomate) is derived. Although some would argue that the flatworms may be descended
from a species that had a coelom, their development does not support this suggestion. Among the
coelomate phyla, two different body plans exist, the pseudocoelomate and the eucoelomate plan.
Pseudocoelomate animals (Nematoda and Rotifera) do not have mesoderm associated with their
gut, while the Eucoelomate animals (Annelida, Mollusca, Arthopoda, Echinodermata, and
Chordata) have mesoderm associated with their gut. There is little evidence to suggest that the
pseudocoelomate plan is ancestral to the eucoelomate plan, or vice versa. It thus appears that both
originated independently from acoelomate ancestors; both are derived states that evolved from the
ancestral state.
There is an important difference among the metazoans in the fate of the blastopore. The
blastopore is the point on the blastula at which invagination begins. The invagination continues
and ultimately forms the archenteron, the primitive gut. Thus, the blastopore represents the first
opening to the primitive gut. In some metazoans the mouth forms at, or in the vicinity of, the
blastopore (Cnidaria, Platyhelminthes, Nematoda, Rotifera, Annelida, Mollusca, Arthropoda). In
other metazoans the anus forms at, or in the vicinity of, the blastopore, with the mouth forming
later at the other end of the primitive gut (Echinodermata, Chordata). The former group can be
called protostomes ("first mouth"). The latter are called deuterostomes ("second mouth"). Because
some metazoans have no anus (Cnidaria, Platyhelminthes) this suggests that the first opening to the
primitive gut was the mouth and the deuterostome condition is derived.
Among eucoelomates (Annelida, Mollusca, Arthropoda, Echinodermata, Chordata) there is
a difference in the mode of coelom formation. In the Annelida, Mollusca and Arthropoda the
coelom forms through splitting of blocks of mesoderm. This mode is called schizocoelous
formation of the coelom. In the Echinodermata and the primitive members of the Chordata the
coelom forms through outpocketing from archenteron. This mode is called enterocoelous coelom
formation. Since no other metazoans have a coelom that is completely enclosed within mesoderm,
there is no basis for suggesting that one mode is ancestral to theother. So, both conditions should
be considered derived from some unknown ancestral eucoelomate condition.
Other characters are correlated with the characters described above. However, the mode of
direction of the characters is not clear, so they will not be dealt with here. Consult your text after
you have completed the following exercise for other characteristics that may be relevant to your
phylogeny. Table 1 summarizes the characters and character states discussed above.
Biology 155 Laboratory Supplement: 74
Table 1. Characteristics used to reconstruct the phylogeny of the metazoans. The characters
are described more fully in the text. Ancestral states are designate with an (a). Derived states
are designated with a (d).
Character
Phylum
symmetry
body plan
coelom
coelom type
blastopore
fate
coelom
formation
Cnidaria
radial
(a)
bilateral
(d)
bilateral
(d)
bilateral
(d)
bilateral
(d)
bilateral
(d)
bilateral
(d)
bilateral
(d)
bilateral
(d)
diploblastic
(a)
triploblasti
(d)
triploblastic
(d)
triploblastic
(d)
triploblastic
(d)
triploblastic
(d)
triploblastic
(d)
triploblastic
(d)
triploblastic
(d)
absent
(a)
absent
(a)
present
(d)
present
(d)
present
(d)
present
(d)
present
(d)
present
(d)
present
(d)
none
(a)
none
(a)
pseudocoelomate
(d)
pseudocoelomate
(d)
eucoelomate
(d)
eucoelomate
(d)
eucoelomate
(d)
eucoelomate
(d)
eucoelomate
(d)
mouth
(a)
mouth
(a)
mouth
(a)
mouth
(a)
mouth
(a)
mouth
(a)
mouth
(a)
anus
(d)
anus
(d)
none
(a)
none
(a)
none
(a)
none
(a)
schizocoelous
(d)
schizocoelous
(d)
schizocoelous
(d)
enterocoelous
(d)
enterocoelous
(d)
Platyhelminthes
Nematoda
Rotifera
Annelida
Mollusca
Arthropoda
Echinodermata
Chordata
Exercise 6. Use the characters given in Table 1 to develop an estimate of the evolutionary
relationships of the metazoans.
Biology 155 Laboratory Supplement: 75
WRITING SCIENTIFIC REPORTS
Please follow the instructions given below when writing lab reports for this course. Don't hesitate to
ask if you have questions about form or content. Above all, remember to write with precision,
clarity, and economy.
Writing- Your writing should be in full sentences and easily understood. It should conform to the
conventions of standard written English (sentence form, grammar, spelling, etc.). Good writing is
as important in science as it is in other disciplines because one's ideas have little impact, no matter
how important they may be, if they are not well communicated. While style is mostly an individual
characteristic, everyone should strive for presentations that are easily understandable as well as
grammatically correct. One reason for emphasizing clarity is that writing and thinking are closely
related; as many people have said, "fuzzy writing reflects fuzzy thinking." When people have
difficulty translating their ideas into words, they generally do not know the material as well as they
think.
Style- Scientific writing is usually in the past tense because one reports on experiments that have
been completed. Opinions vary on whether scientific reports should be written in active voice (e.g.
“the dog fetched the ball”) or passive voice (e.g. “the ball was fetched by the dog”), but the current
trend in most journals is to prefer the active voice. Generally active voice is more natural, easier to
read, and write, so, use active voice as much as possible. It is fine to use “I” or “we” when
describing what you did, but repetitive use of either is tedious to read. Remember: past tense,
active voice.
Presentation - The first page of a lab report should be a title page with the title of the report, your
name, the date, the course (e.g., Biology 155), and your lab partners. There should then follow text
that is a minimum of three pages and a maximum of ten double-spaced, typewritten pages in length
(tables, figures, and references do not count in this total). All writing should be on only one side of
the page, and the reports should be stapled in the upper left-hand corner. Do not put your report in a
folder or binder. The best length is shorter than the maximum, so don't expand a shorter report to
reach a ten page limit. It is important to write concisely. The report must be typed or wordprocessed. If you don’t know how to use a word processor, now is the time to learn. Go to the
Writing Across the Curriculum Laboratory and get started. Neatness and clarity of presentation are
almost as important as clarity of thought.
Audience- Write the report as if you were writing to other students who are taking a similar course
but have not done this experiment. Assume that they have some familiarity with the subject matter
but no expertise. Give your reader enough information to repeat the experiment, but do not give
details on mundane tasks like pipetting, setting up equipment, or preparing dilutions. Do not write
specifically for the instructor.
Collaboration- You may talk with others about how to write the report. However, you may not
turn in a report that “borrows” sentences, phrases, graphs, tables, etc., from anyone else’s report
from this semester or previous semesters. You are encouraged to have a WAC lab tutor read and
critique your report. The WAC lab is located on the ground floor of the library. You may consult
with your instructor on how to write your report. However, you must write your own report.
References- Likely references to use in your report are your lab manual, lab supplement, and
lecture text. Many other sources are available in the library. For all reports in this course you must
use at least 3 references. Cite your sources in the body of the report and list the references in a
literature cited section. Citations should be made with a standard scientific format (not by
footnotes); cite the author and date of publication only, so that a quick look at the Literature Cited
can provide the reader with all necessary information. When there are more than two authors,
simply list the first author and et al., along with the date. You should not use direct quotations from
Biology 155 Laboratory Supplement: 76
the references; paraphrase information and give credit to the source of the idea. The following are
sample citations:
"Garrett (1989) showed that a gene in yeast ..."
"... is found in the urinary bladder of the turtle (Gapp et al., 1990)."
"... as reported recently (Miller, 1986; Pfitsch & Pearcy, 1989)."
You should list a reference for every idea not your own. Plagiarism is more than copying material
word for word; it is also using someone else's ideas or phraseology without giving reference to the
other work or other person. Fortunately, the reference format is so simple that it is very easy to
include references to all the work that one has used (Williams, 1983). If the idea is not published
but is provided by a lab partner or someone else, give the reference as a personal communication
(N. Cutler, pers. comm.). Be aware of the difficulties that arise when one uses material from
another source and changes only a word here or there without acknowledging the source. Such
actions are plagiarism, even though the statement may not be word-for-word the same as in the
original. Just remember the basic rule: list a reference for every idea or statement not your own.
Format- There are five fundamental sections to a scientific report, with acknowledgments,
literature cited, and appendices being additional sections. Each section should begin with the title of
that section (Introduction, Materials and Methods, etc.).
Abstract – The abstract is a summary of your entire report. Usually the presentation of material in
the abstract parallels the presentation in each section of the report. The abstract should summarize
the problem being investigated, the methods used in the investigation, the major results,
conclusions, and the significance of the findings. Since the abstract is a summary of the report it
should be written last. The most common error that students make in writing the abstract is to
include new information or to make statements that are unsupported by the information in the
report. If information is not contained in your report it should not be included in your abstract.
Introduction – The introduction should (1) give sufficient background for the reader to understand
the problem being investigated., (2) clearly describe the problem being investigated, (3) describe
the organism or system used in the investigation, (4) describe the purpose and goals of the
investigation and (5) state any hypotheses or predictions that will be tested. Start your introduction
with broad statements, including enough background information (with references to outside
sources) to set the stage for your experiment. Then narrow down to your particular study,
explaining why it is of interest. Specify the objectives of the experiment, and make your hypotheses
clear. An introduction consisting of two to four paragraphs is usually sufficient. Do not regurgitate
the lab handout; write your own introduction.
Materials and Methods - Summarize the entire process that was followed and the materials that
were used in enough detail so that a student with abilities and knowledge comparable to your own
could repeat the experiment. Do not present your materials in a list. Describe the materials needed
for a procedure as part of your description of the procedure. Be sure to explain the purpose of
procedures that you used. For example: “The reaction rate was measured with and without the
inhibitor so that the effect of the inhibitor could be determined.” Describe any manipulations of the
raw data that you did. For example: “Reaction rate was calculated as the average change in
absorbance over the first three minutes.” Do not include mundane details. Do not include results
here.
Results - The data and results are given here as summary tables or graphs. Do not provide all the
raw data. Also, describe your results for the reader in a narrative form. Students often omit text
from their results section, but it is a necessity. Tell the reader what he or she should see in your
data but do not simply repeat the information that is presented in tables or graphs. Also, do not
simply direct the reader to the table or graph with a sentence like “Figure 1 shows the relationship
between enzyme concentration and reaction rate.” Instead, provide a summary of what the reader
Biology 155 Laboratory Supplement: 77
should see when they look at the figure. For example: “As enzyme concentration increased
reaction rate increased (figure 1).” All tables and figures should be numbered sequentially. All
graphs are called figures. In graphs be sure to clearly label the axes and units. Each table should
be given a descriptive title that is placed above the table. Each figure should have a descriptive
caption that is placed below the figure. For example “Figure 1. The relationship between enzyme
concentration and reaction rate.” Sample tables and figures are given below. Do not include tables
and figures in the body of your report. Instead, place them at the end of your report (after the
literature cited section), tables first and then figures.
Discussion - This section should develop the major conclusions of your paper. Interpret your
results and explain their significance. Be sure to relate what you have to say about your results to
the goals, hypotheses, and predictions that you presented in your introduction. Also, develop the
broader significance of your results; how they relate to the world beyond the laboratory.
Begin the discussion by interpreting your specific results and end it more broadly by placing your
results in context. Don't declare the experiment a success or failure; evaluate the results in view of
the purpose of the experiment. If your feel your results were erroneous discuss the results you
expected as well as those you received. You may also compare methods or discuss difficulties, but
if you list sources of error, you should estimate how important each source of error may be. If you
were to do the experiment again, what if anything, would you do differently? It is inappropriate to
include statements such as "I learned a lot from this experiment...". The discussion is a very
important section. It is your chance to show how well you understand the ideas and techniques
involved and to relate your results to the ideas expressed in outside sources (the literature cited).
Acknowledgments - The acknowledgments section is optional. If you wish to thank someone, such
as a lab partner or a tutor at the WAC lab, for help in understanding the experiment or in organizing
the report, you do so here. Scientists regularly acknowledge others for helping with experiments or
commenting on written drafts.
Literature Cited - List any publications referred to in your paper alphabetically by first author; do
not number them. Every item in your bibliography should be cited in the body of your paper, or it
shouldn't be listed at all. If you use information from an intermediary source, you should list the
original reference but should also note the intermediary: "...cited in...". Use the following standard
forms (some journals use a variation of this form).
an article with one author:
Reynolds, P.D. 1992. Mantle-mediated shell decollation increases posterior aperture size in
Dentalium rectius Carpenter 1864 (Scaphopoda: Dentaliida). Veliger 35:26-35.
an article with more than one author:
Gapp, D.A., R.N. Taranto, E F. Walsh, P.J. Favorito, and Y. Zhang. 1990. Insulin cells are found in
the main and accessory urinary bladders of the painted turtle, Chrysemys picta. J. Exp. Zool.
254:332-337.
a book:
Stokes, D., L. Stokes, and E. Williams. 1991. The Butterfly Book. Little, Brown and Co., Boston.
96 pp.
a chapter from an edited volume:
Pearcy, R.W., and W.A. Pfitsch. 1994. Consequences of sunflecks for photosynthesis and growth
of forest understory plants. Pages 343-359 in E.D. Schulze and M.M. Caldwell, editors.
Ecophysiology of Photosynthesis. Springer Verlag, New York.
Final Check The last thing to do before turning a report in is to read it. Correct all typographical
errors and other mistakes, and ensure that you have said what you wanted to say!
Biology 155 Laboratory Supplement: 78
Sample Table:
--------------------------------------------------------------------------------------------Table 1- Frequency of ABO blood types in Biology 155. The results are
given as both the number and percentage of students with different blood
types, and they are reported for both Monday's lab and for the entire course.
--------------------------------------------------------------------------------------------Monday Lab
All Sections
Blood type
number
percentage
number
percentage
A
6
23.1
47
32.9
B
4
15.4
26
18.2
AB
0
0.0
7
4.9
O
16
61.5
63
44.0
Totals
26
100.0
143
100.0
Sample Figure:
Figure 1. The relationship between absorbance and time in experiment 1.