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Transcript
PROTEIN
SYNTHESIS
Teacher's Guide
The Series
Protein Synthesis
Teacher's Guide
The programs are broadcast by TVOntario, the
television service of The Ontario Educational
Communications Authority. For broadcast dates
consult the appropriate TVOntario schedule. The
programs are available on videotape. Ordering
i nformation for videotapes and this publication
appears on page 25.
Canadian Cataloguing in Publication Data
Rosenberg, Barbara L.
Protein synthesis
To be used with the television program, Protein
synthesis.
Bibliography: p.
I SBN 0-88944-080-8
1. Protein synthesis (Television program) 2. Protein
biosynthesis. 3. Protein biosynthesis - Study and
teaching (Secondary)
QP551.R671985
574.19'296
C85-093027-8
©Copyright 1985 by The Ontario Educational
Communications Authority
All rights reserved.
Printed in Canada.
Producer/ Director: David Chamberlain
Writer: Alan Ritchie
Narrator. James Moriarty
Consultant: Hubert Dollar
Animation: Animation Drouin Inc.
The Guide
Project leader: David Chamberlain
Writer: Barbara L. Rosenberg
Editor: Loralee Case
Designer: John Randle
Consultant: H. Murray Lang
Contents
Program 1:
Protein: The Stuff of Life .................... 1
Program 2:
DNA: The Molecule of Heredity .............. 8
Program 3:
DNA Replication: The Repeating Formula ...... 12
Program 4:
RNA Synthesis: The Genetic Messenger ....... 15
Program 5:
Transfer RNA: The Genetic Messenger ........ 19
Program 6:
Ribosomal RNA: The Protein Maker ........... 22
Protein: The Stuff of Life
Objectives
Students should be able to:
1. Recognize that each species manufactures its own unique
proteins.
2. Cite an example of protein specificity in the human body.
3. List the different functions of proteins and cite examples for each.
4. Describe the basic components found in protein.
5. Draw the molecular structure of some common amino acids.
6. Illustrate the formation of a peptide bond between two amino
acids.
7. Explain why, an infinite variety of protein molecules is possible.
8. Develop the concept of the significance of protein for the survival
of the cell and the organism as a whole.
Program Description
Each cell contains hundreds of different proteins, and each kind of
cell contains some proteins that are unique to it. Not only does every
different species of plant and animal synthesize protein peculiar to
that species, but every individual likely produces some protein
molecules slightly different from those of every other individual. The
degree of difference in the proteins of two species depends upon
the evolutionary relationship of the forms involved. Organisms less
closely related by evolution have proteins that differ more markedly
than those of closely related forms.
The protein specificity of an individual is particularly evident in the
i mmunological reactions of animals, by which foreign proteins are
prevented from remaining permanently in the body of the organism.
The body manufactures specific protein -antibodies. The surface
features of the antibody complement the surface irregularities of the
foreign protein. As a result the antibody binds to the foreign protein
and disposes of it.
The types and functions of proteins are extremely varied. Proteins
are found as enzymes - catalysts that make chemical reactions of
living matter possible. An example is amylase, which begins to break
down starch in the mouth. Muscles, responsible for the movement of
living organisms, are largely composed of ordered protein molecules,
actin and myosin. Transport proteins are responsible for carrying
many materials through the circulatory system. Haemoglobin
transports oxygen and carbon dioxide in the blood. Interaction of a
number of different proteins results in the clotting of blood.
Antibodies can recognize and inactivate virtually any foreign
substance that gains access to the body. Hormones, which regulate
and coordinate bodily functions, are proteins. Major structural and
protective material in animals is made up of protein. Collagen and
elastin provide the strength and resilience of connective tissues
such as skin and ligaments. Keratin i s a major protein found in hair
and nails. Lastly, proteins also serve as food reserves, as in
ovalburain (egg white) and casein, a major protein found in milk.
Proteins are macromolecules (polypeptides) containing atoms of
carbon, hydrogen, oxygen, nitrogen, and often sulfur. Trace amounts
of phosphorus, copper, and zinc may also be found. Proteins are
composed of a linear arrangement of amino acids, the building
blocks of protein, bonded together in long chains varying in length
from less than 100 to more than 50 000 amino acids. An amino acid
contains a carbon atom which is attached by covalent bonds: a
hydrogen atom; an amino (NH2) group; a carboxyl group (COOH); and
"something else" - an "R" group that establishes which particular
amino acid it is. Amino acids are joined together in peptide linkages
to form polypeptides. Peptide bond formation involves the elimination of a water molecule between each pair of amino acid groups.
Not only do different proteins vary in length, they also vary in
shape because each type of protein contains specific numbers and
kinds of amino acids arranged in particular sequences. Consequentl y, each type of protein molecule has a unique chain of amino acids
and a unique shape. For example, the molecules of fibrous proteins
are long and threadlike. They have special functions, such as binding
body parts together or forming the fine threads of a blood clot.
Globular proteins are coiled or folded into compact masses;
enzymes are usually of this type. Considering that there are 20
commonly occurring amino acids and that proteins may consist of
thousands of these units, the possible number of different
sequences of amino acid units, and hence the number of different
protein molecules, is staggeringly large. An almost infinite variety of
protein molecules is possible.
1
2
Although the difference between two proteins may be as slight as
the replacement of a single amino acid in a chain of several hundred,
it may lead to profound consequences within the organism. Incorrect
sequencing of the protein insulin results in the condition known as
diabetes mellitus.
The synthesis of protein is perhaps the most significant synthesis
carried on by the cell. Since enzymes are proteins, protein synthesis
controls the nature of the enzymes produced. Enzymes in turn
determine the reactions the cell can perform. These reactions
control the substances that can be synthesized or degraded, and the
substances produced and stored determine the structure and
function of the cell. The structure and function of the cells in turn
control the nature of the entire organism.
Drawing a small portion of the amino acid
sequence for insulin with the peptide
linkages illustrates a growing
polypeptide (Figure 2).
Figure 2
Before Viewing
Program 1 may stand alone as an introduction to proteins in a
biochemistry unit. When used as part of the unit on protein
synthesis, the structure of amino acids should be reviewed as well
as the formation of peptide bonds. Functions of proteins should also
be reviewed at this time.
You may want to perform Activity 1 now to introduce students to
the fact that many of the foods we eat contain protein. You may also
choose to do similar experiments from any of the resource material
listed in Activity 1.
After Viewing
The vital role of proteins in the maintenance and continuation of
living organisms should be stressed. Examples of specific amino
acids may be drawn on the board to illustrate basic differences
among them. Common examples to use are glycine, alanine, and
l eucine. Refer to Figure 1 for their chemical compositions.
Figure 1
The amino acid sequence for insulin (a protein that causes a
decrease in the level of blood sugar) could also be placed on an
overhead to explain that an incorrect sequencing could be
disastrous (Figure 3). This figure shows a sequence of amino acids
i n the insulin molecule. The molecule consists of two polypeptide
chains held together by two disulfide bridges.
anemia. Figure 4 shows the N-terminal portions of normal
haemogiobin and sickle-cell haemoglobin.
Figure 4
Figure 3
The amino acid residue in one of the protein chains of haemoglobin
i s glutamic acid. Certain individuals inherit a gene that results in the
replacement of this glutamic acid by another amino acid, valine. As a
result, the individual suffers from a chronic disease known as
sickle-cell anemia. Figure 4 shows the N-terminal portions of normal
haemoglobin and sickle-cell haemoglobin.
Activities 2, 3, and 4 can be performed at this time.
Although not mentioned in the program, you may wish to discuss
the classic example of haemoglobin as well (Figure 4). The amino
acid residue in one of the protein chains of haemoglobin is glutamic
acid. Certain individuals inherit a gene that results in the replacement of this glutamic acid by another amino acid, valine. As a result,
the individual suffers from a chronic disease known as sickle-cell
3
4
Activities
Activity 1(a): Identification
of Proteins
The purpose of this activity is to inform
students that various foods we eat contain
proteins.
Apparatus
10% solution egg albumin
1 % solution gelatin
Milk
Chicken bouillon
Distilled water
0.02M copper sulfate solution
6M sodium hydroxide
Concentrated nitric acid
Test tubes
Test tube rack
10 mL graduated cylinder
Eye dropper
Method
Biuret Test
1. Label the test tubes with the names of the
solutions to be tested; include distilled water
as a control sample.
2. To 2 mL of each solution add 2 mL of 6M
sodium hydroxide and four drops of 0.02M
copper sulfate solution. Gently shake the
tubes to mix the solutions.
3. The appearance of a violet or violet-pink color
i ndicates the presence of proteins and
specifically the presence of peptide bonds
between the amino acids.
Xanthoproteic Test
1. Label the test tubes with the names of the
solutions to be tested.
2. To 1 mL of each solution add five to ten drops
of concentrated nitric acid. CAUTION: Nitric
acid is corrosive.
3. The appearance of a yellow color indicates
the presence of proteins and specifically the
presence of the benzene nucleus.
Observations
Record the results in chart form.
Type of test:
Sample
Observations
1.
I nterpretation
2.
3.
4.
5.
Discussion
1. Explain which test i s the better test for
proteins.
2. Compare and contrast coagulation and
denaturation and name several agents
i nvolved in these reactions.
3. Describe the functions of proteins.
4. What is a polymer?
Other experiments in the identification of
proteins may be found in McCormack et al,
Biology Laboratory Manual, p.193; Feldman,
Experiments in Biological Design, p. 52; and
Abramoff and Thomson, Investigations of Cells
and Organisms, p. 49. (Refer to Further
Reading.)
Activity 1(b): Identification
of Proteins
Activity 2: What Is the Nature of
the Protein Molecule?
For a very simple and quick experiment
1. Put 3 cm of water in a test tube. Add 1/8
teaspoon of gelatin powder. Shake well.
2. Add six drops of Biuret solution. Notice how
the color of the Biuret solution changes from
blue to violet. (A violet color indicates the
presence of proteins.)
3. As an alternative, use 1/8 teaspoon of dried
egg white instead of gelatin, or finely
chopped hard-boiled egg white. If using
hard-boiled egg white the solution should be
heated.
If the equipment is available, try this
experiment. The chemical extraction procedures are particularly clear cut and easy.
Proteins as well as nucleic acids are obtained.
(The nucleic acids can be used later following
Program 4.) Students will also be introduced to
centrifugation and chromatography, procedures
that are widely used in biochemical research.
The organism involved is Saccharomyces
cerevisiae, obtained commercially as baker's
yeast.
Apparatus (for each group)
10 g moist baker's yeast
30 mL trichloroacetic acid (TCA) 5% solution
30 mL sodium chloride 10% solution
100 mL ethyl alcohol 95%
5 mL pancreatic enzyme in buffer solution
5 mL phosphate buffer solution without enzyme
Two crystals thymol
35 mL isopropanol
5 mL formic acid
20 g fine sand
Eight capillary tubes or toothpicks
Three centrifuge tubes
Two test tubes
One mortar and pestle
One sheet of Whatman No. 1 chromatographic
paper (15 cm x 15 cm)
One 1 L jar with lid.
Apparatus (for class)
111 L jars with lids
One or two centrifuges
Cellophane tape
Solutions of amino acids: alanine, aspartic acid,
histidine, lysine, methionine
1 L ninhydrin reagent
Water bath
I ce bath
Refrigerator
Oven or electric iron
CAUTION: Ninhydrin must only be used by
teachers with a knowledge of carcinogenic
materials and the proper use, safe storage,
and safe disposal of the chemical.
Method
Refer to Figure 4.
Figure 4
5
6
Step 1
Disrupting the yeast cells and separating
glycogen from protein and nucleic acid
Flow Chart
1. To 10 g of washed baker's yeast in a mortar, add twice as much sand; grind thoroughly for five
minutes. Add 30 mL of 5% TCA and grind two to three minutes more. Let the sand settle. Pour
off the milky suspension into a 50 mL plastic centrifuge tube. Centrifuge for five minutes.
Precipitate
(Nucleic acids, proteins)
Step 2
Separating protein and nucleic acid
2. Add the precipitate to 15 mL of 10% sodium chloride solution. Stir and heat in a boiling water
bath for 10 minutes. Store in a refrigerator until the next class.
3. Centrifuge three minutes.
Precipitate
(coagulated proteins)
Step 3
Hydrolyzing protein
Supernatant
(glycogen - discard)
4. Measure out small portions of the precipitate
(about the size of a small pea) into each of
two labelled test tubes. To one test tube add
4 mL of a solution of pancreatic enzyme in
phosphate buffer and a crystal of thymol.
Swirl the tube to suspend its contents. To the
protein in the other tube add 4 ml- of buffer
without enzyme and add thymol. This tube
serves as a control. Label both tubes and
store until the next class.
Hydrolyzed
(proteins)
Supernatant
(nucleic acids)
5. Pour supernatant into another centrifuge
tube. Precipitate the nucleic acids by adding
30 mL of ethyl alcohol; stir and cool several
minutes in an ice bath. Store in the
refrigerator for use after viewing Program 4.
Unhydrolyzed
(proteins)
Step 4
Preparing chromatographic equipment for
analysis of amino acid
6. Touch the paper only along the edges. Using a sharp pencil, draw a fine line parallel to one side
of the filter paper about 1.5 cm from the edge to serve as the bottom of the chromatogram.
Allow a margin of about 1.5 cm on each side of the paper, and place eight pencil dots evenly
spaced between the margins along the line.
Step 5
Running the chromatogram
7. Spot one solution per dot of the amino acids, the hydrolyzed protein, and the unhydrolyzed
control. (See Figure 4). The remaining place is for an unknown amino acid selected by the
teacher. Using a different, capillary tube (or a toothpick) for each solution, place a drop (about
3 mm) on the paper. Allow it to dry, and repeat the process twice for each of the amino acid
Step 6
Analyzing the chromatogram
solutions and four times for the hydrolyzed and unhydrolyzed protein samples. Roll the paper as
shown in Figure 4 and lower it into the solvent (10 parts formic acid, 70 parts isopropanol, 20
parts water). (There should be enough solvent to cover the bottom of the jar.) Allow the solvent
to run up the paper to within a centimetre of the top. Remove the paper and make a pencil line
at the solvent front. Place the paper in a safe place to dry.
8. Wearing plastic gloves, dip the paper into the ninhydrin reagent and let it dry again. (CAUTION:
Ninhydrin reagent is a poisonous carcinogen. Use only in a fume cupboard.) Heat the paper in a
warm oven (30°C) or place between two sheets of clean filter paper and heat with an electric
i ron for a few minutes until colored spots appear. Ninhydrin reacts with amino acids, forming
blue or pinkish-blue spots. Outline the spots with a pencil. Determine the R F value of each
amino acid and record it.
For further details about this experiment, refer
to BSCS, Laboratory Block: The Molecular
Basis of Metabolism, pp. 6-16. (See Further
Reading).
Discussion
The ratio of the distance the solvent has moved
i s called the R F. The R F of a substance under
particular conditions is important information in
identifying the substance. Two substances with
the same R F i n a number of solvents are
probably 'identical. The R F can be expressed as
follows:
distance of flow of known or
R F = unknown compound
distance of flow of solvent
The R F i s usually written as a decimal and is
always less than one. Explain to students the
principle involved in paper chromatography.
Activity 3: Amino Acid
Composition of an Unknown
This lab exercise investigates the occurrence
of amino acids as an indication of protein
metabolism within the body. The presence of
amino acids is once again tested for by using
paper chromatography. The experiment can be
found in BSCS, Biological Science, An Inquiry
into Life, Student Laboratory Guide, (2nd ed.),
I nquiry 6-2. (See Further Reading.)
Activity 4: Review
1. Describe an example of protein specificity in
the human body.
2. Describe the different roles that proteins play
within the body. Give examples.
3. List the different elements found in all
proteins. What other elements may be found
i n proteins?
4. Draw the molecular structures for glycine
and alanine. What is the basic difference
between the two?
5. How does a peptide linkage occur'? What
substance is formed during this process?
6. There are millions of different proteins in the
living world. How is this possible?
Further Reading
Abramoff, P and R. Thomson. Investigations of
Cells and Organisms: A Laboratory Study in
Biology. Englewood Cliffs, New Jersey:
Prentice Hall, 1968.
Benson, G.D. et al. Investigations in Biology.
Toronto: Addison Wesley,1977.
BSCS. Laboratory Block: The Molecular Basis
of Metabolism. Colorado: Raytheon Education Co., 1968.
BSCS. Biological Science: An Inquiry Into Life.
Student Laboratory Guide. 2nd ed. New York:
Harcourt, Brace and World, 1968.
Feldman. Experiments in Biological Design.
New York: Holt, Rinehart and Winston,1965.
I ngram,,P.J. Biosynthesis of Macromolecules.
2nd ed. Menlo Park, California:
WA. Benjamin, 1972.
Kimball, J. Biology. 5th ed. Reading,
Massachusetts: Addison-Wesley, 1983.
McCormack, J. et al. Biology Laboratory
Manual. Glenview, Ilinois: Scott, Foresman,
1980.
7
8
DNA: The Molecule of Heredity
Objectives
Students should be able to:
1. State where the "blueprint," or genetic information, is stored.
2. Describe the general structure of a DNA molecule.
3. Describe the composition of a nucleotide.
4. Differentiate between purines and pyrimidines.
5. State the specific base pairing that occurs within the DNA
molecule.
6. State what constitutes the genetic code.
7. Recognize that DNA carries a code that specifies the primary
structure for all proteins in the body of an organism.
Program Description
We resemble our parents because we inherit "genetic traits" from
them. What is actually passed on is genetic information - instructions for carrying out life processes. This genetic information, or
"blueprint," is located in DNA molecules found in the genes
located in the chromosomes inside the nucleus of the parents' sex
cells and is then passed from cell to cell by mitosis as the child
develops. The "blueprints" direct the developing cells to construct
specific protein molecules, which in turn function as structural
materials, enzymes, or other vital substances.
To find out how DNA could serve its genetic functions - storing
and replicating information - a model was built to explain all the
i nformation known about DNA. DNA is an exceedingly long
molecule, very thin, yet rather rigid. It is composed of two strands of
polynucleotides coiled around each other in a helical manner and
held together by hydrogen bonds between pairs of nucleotide bases.
A nucleotide consists of a 5-carbon sugar, deoxyribose; a
phosphate group attached to the 5-carbon atom of the sugar; and a
nitrogen-containing ring structure called a base. The base of a DNA
nucleotide can be one of four kinds: adenine (A) and guanine (G),
which are purines, and thymine (T) and cytosine (C), which are
pyrimidines. The backbone of each of these two chains is composed
of alternating deoxyribose sugar residues and phosphoric acid
molecules; it is uniform throughout the enormous length of the
molecule and apparently carries no genetic information. The purine
or pyrimidine base bonded to each pentose sugar projects in
towards the axis of the helix. One purine base and one pyrimidine
base bond together and hold the chains together. The resulting
structure is something like a ladder or zipper in which the outer rails
represent the sugar and phosphate backbones of the two strands
and the crossbars, or "teeth of the zipper," represent the organic
bases. In addition, the molecular ladder is twisted to form a double.
helix, or corkscrew, effect.
The DNA strand consists of nucleotides arranged in a particular
sequence. Moreover, the nucleotides of one strand are paired in a
special way with those of the other strand. Only certain bases will fit
and bond their nucleotides together. One purine and one pyrimidine
just fill the available space between the two chains. Specifically,
adenine will only bond with a thymine and a cytosine will only bond
with a guanine. As a consequence of such base pairing, a DNA
strand possessing the base sequence A, G, C, T would have to be
bonded to a second strand with the complementary base sequence
T, C, G, A. It is the particular sequence of base pairs that encodes the
genetic information held in the DNA molecule. A single DNA
molecule may contain from hundreds of thousands to close to 100
million base pairs in endless possible sequences. The order of
these sequences constitutes the genetic code for the construction
of protein.
Before Viewing
As an introduction you might ask students to consider why humans
give birth to humans, cats give birth to cats, and birds give birth to
birds. What makes it possible for characteristics to be carried on
from one generation to the next?
You may wish to complete Activity 1 at this time. In this exercise
students will use a specific staining procedure to determine where
DNA is located in the cells.
After Viewing
The molecular structures of adenine, guanine, thymine, and cytosine
could be drawn on the board, making note of the differences
between the purines and pyrimidines. Refer to Figure 1. Complete
Activities 2, 3, 4, 5, and 6.
Figure 1
Activities
Method
Refer to Figure 2.
1. Place the root tips in the beaker and cover
the beaker with cheesecloth. Wash the root
Place several onions in water a week prior to
tips in running tap water for five to ten
performing this lab exercise. Before the lab cut
minutes to remove the acid.
off 1 cm of the root tips that grew out from the
2.
Pour off the excess water and then add
onions avid fix them in acetic acid, rinse in
enough
of the Schiff's reagent to cover the
distilled water, and hydrolyze in hydrochloric acid.
root tips.
Apparatus
3. Leave the root tips in the dye for 20 minutes.
Onion root tips - fixed
They should turn a purple color. If they don't,
l et them stand until the tips are stained, then
Schiff's reagent
remove all excess dye by washing the tips
Sodium bisulfite
under tap water for three to five minutes.
Acetic acid
4.
Add enough sodium bisulfite to cover the
Beakers
root tips and leave them in this solution for
Forceps
one to two minutes. This bleaches all parts of
Microscope
the cell that do not contain DNA.
Slides
5. After bleaching, remove the root tips from
Cover slips
the beaker and place them on a microscope
Medicine droppers
slide. Add a drop of acetic acid.
Cheesecloth
6. Add a cover slip, and with a pencil gently roll
Rubber band
the preparation to squash the root tips and
Figure 2
Activity 1: Where Is DNA Located?
Schiff's reagent
Cheesecloth
Rubber band
9
10
separate the cells.
7. Examine the slide under the microscope, first
with low power and then with high power.
The dye stains the DNA in the cell purple.
The longer staining period and the washing
provide a better delineation of the DNA.
Discussion
1. In what part of the cell is DNA located?
2. What is the color of the cytoplasm?
3. Can you identify individual chromosomes in
any of the cells?
4. Are there any cells i n the process of
dividing?
Activity 2: Structure of DNA
Apparatus
Tracing paper
Pencil
Scissors
Tape that can be written on
Method
Make copies of each of the nucleotides shown
here and have students build the DNA
molecule.
Discussion
1. How can the models be arranged so that
the width of the total structure is always the
same?
2. What nucleotides will "fit" next to the
thymine nucleotide? To the guanine
nucleotide?
3. What "fits" are possible for the adenine
nucleotide? For the cystine nucleotide?
4. What characteristics of the models determined your answers in 2 and 3?
5. From observations of shape and size alone,
you know that adenine could pair with
either thymine or cytosine. But does it pair
with one or the other, or both, in a real DNA
molecule?
6. If adenine always pairs with cytosine, how
should the relative amounts of adenine and
cytosine compare in a DNA molecule?
7. If adenine always pairs with thymine, what
relative amounts of these bases would you
expect to find?
8. If adenine sometimes pairs with cytosine
and sometimes with thymine, what relative
amounts of adenine, compared to the
amount of the other two, would you expect
to find?
9. What do you notice about the percentage of
any single base in the different kinds of
cells? Consider the following data.
Amount of DNA Found in Human Cells
Tissue
Adenine Guanine Thymine Cytosine
Thymus cells
30.9
19.9
29.4
Spleen cells
21.0
29.4
20.4
29.2
Liver cells
30.3
19.5
30.3
19.9
Sperm cells
30.7
19.3
31.2
18.8
10. (a) Which bases pair together in human
DNA? Use examples from the data to
explain your answer.
(b) Would the same pattern of pairing hold
true for all organisms? Explain.
Activity 3: What Is the Molecular
Basis of Heredity?
This dry l ab can be found in Abramoff and
Thomson, Investigations of Cells and Organisms, Exercise 60. (Refer to Further Reading.)
You may wish to use it to reinforce the
structure of the DNA molecule and as a lead-in
to Program 3.
19.8
Activity 4: Extracting DNA from
Cells
If time and equipment allows, you may wish to
t ry this experiment in McCormack et al, Biology
Laboratory Manual, p. 39. This is a biochemical
procedure illustrating how biochemists extract
and study chemical substances found in living
organisms. It involves a 24-hour nutrient broth
culture of E, coli. After centrifugation the DNA
strands are lifted out of the test tube and
observed under blue light. A wet amount of the
extracted DNA can then be prepared and
viewed under the microscope.
Activity 5: Reports
Many people were involved in the research
l eading to the structural model of DNA
proposed by Watson and Crick. Have students
i nvestigate the contributions of one of the
following and write a brief report: Friedrich
Meescher (1844-1895); Albrecht Kossel (18351927); Phoebus Aaron Levene (1869-1940);
Alexander Todd (1907-); Oswald Avery (18771955); Heinz Fraenkel-Conrat (1910-); and
Rosalind Franklin (1920-1958).
Activity 6: Review
1. Where are the master instructions for protein
synthesis located in a cell?
2. DNA is described as a zipper and a
corkscrew. Unwound it l ooks like a ladder. Of
what substances are the rails and rungs
composed?
3. What is the composition of each of the four
nucleotides?
4. What is the basic difference between a
purine and a pyrimidine? Draw the structures.
5. What are the four possible combinations in
the base pairing?
6. What is the significance of the complementary base pairing?
7. What is the function of DNA in the cell?
Further Reading
Abramoff, P and R. Thomson. Investigations
of Cells and Organisms. Englewood Cliffs,
New Jersey: Prentice-Hall, 1968.
"Biopolymer Models of Nucleic Acids." Journal
of Chemical Education. March 1979. Vol. 56.
p. 168.
Frankel, E. DNA: The Ladder of Life. 2nd ed.
New York: McGraw-Hill, 1979.
Lessing, L. DNA: At the Core of Life Itself. New
York: Macmillan, 1967.
McCormack, J. et al. Biology Laboratory
Manual. Glenview, Illinois: Scott, Foresman,
1980.
Watson, J. The Double Helix. New York:
Atheneum, 1969.
11
12
DNA Replication: The Repeating Formula
Objectives
Students should be able to:
1. Describe the structure of a DNA molecule.
2. State what is meant by compiementarity
3. Draw the molecular structures for the four bases and explain the
i mportance of hydrogen bonds in DNA.
4. State at what point during cell division DNA replication occurs.
5. Describe the basic steps involved in DNA replication.
Program Description
The rails of the DNA molecule are made up of sugar molecules
bonded to phosphoric acid molecules; the rungs are made up of
nitrogenous bases - a purine (adenine or guanine) bonded with a
pyrimidine (thymine or cytosine). The specific sequencing of the
base pairs builds the genetic code, the "blueprint" necessary for the
synthesis of protein.
An important question now arises: How is the genetic code read,
or how is the DNA decoded to make an exact copy of itself? The key
to the copying is in the architecture of the DNA molecule. The
; nucleotides of one strand are paired in a special way with those of
the other strand. Only certain bases will fit and bond their nucleotides together. One purine and one pyrimidine just fill the available
space between the two chains. The shape of the cytosine molecule
i s such that a stable set of three hydrogen bonds could only be
formed between it and guanine, and vice versa. Similarly, the shape
of the thymine molecule is such that the most stable set of
hydrogen bonds (two) would be formed between it and adenine, and
vice versa. This complementarity is essential to the accurate
replication of DNA.
Molecular replication occurs during the process of cell division
(mitosis) and enables the "blueprints" for creating proteins to be
passed on from organism to organism. DNA replication begins with
an "unzipping" of the "parent" molecule. The hydrogen bonds
between the base pairs are broken by a special enzyme and the two
halves of the molecule unwind. The exposed strands now bond with
complementary free-floating nucleotides in the nucleus. The growing
chain of nucleotides is then linked together by the bonding of the
sugar and adjacent phosphoric acid molecules. At the end of the
process there are two DNA molecules; half of each is derived from
the parent molecule; the other half is the new DNA. The two new
DNA molecules will exactly .resemble the original one. The
duplicated double helix molecules can then be distributed, one to
each of the products of cell division. In this way cell continuity as
well as the continuity of the organism as a whole is ensured.
Before Viewing
The structure of the DNA molecule should be reviewed. It is
i mportant to stress that it is the "blueprint" necessary for the
continuation of the organism. Since students were exposed to the
structures of the nitrogenous bases in Program 2, you might take the
ti me now to review the structures and discuss the hydrogen bonding
that occurs between specific base pairs. Refer to Figure 1.
Figure 1
Thymine
After Viewing
Cytosine
It is important for students to understand that the molecular shape
of the DNA molecule is of immense consequence. Consider a single
polynucleotide strand, a small portion of the DNA molecule. In the
course of random molecule motion, the free-floating nucleotides
within the nucleus would impinge against the base molecules of the
DNA strand. Sooner or later a thymine residue would come into
opposition with an adenine residue in the chain. A pair of hydrogen
bonds would be formed, and this thymine nucleotide would be
bound to the DNA strand. In the same way the free-floating cytosine
nucleotide would be bound to the guanine residue, an adenine
nucleotide to the thymine residue, and a guanine nucleotide to the
cytosine residue. The sequence of nucleotides bound to the DNA
strand would be in the proper configuration to allow for ester
linkages to form, between the sugar and phosphate residues of
adjacent molecules. In this way, in spite of the essential randomness
of molecular collisions and the reactions resulting from them, a
polynucleotide strand of specific structure would be built. Activities
1, 2, and 3 should be completed now.
Activities
Activity 1: DNA - How Does It
Make Copies of Itself?
1. Use the patterns shown here to make parts
of a DNA molecule. Trace and cut out the
following number of parts: 16 Ss, 12 Ps, 4 Gs,
4 Ts, 4 As, 4 Cs.
2. Build a model of a segment of a DNA
molecule. The segment should contain five
rungs. Some pieces will be left over; this is
the way it might be in a real cell. There would
be a strand of DNA in the nucleus and many
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"spare parts" circulating in the cell. Refer to
Figure 2.
3. Once the model has been made, carefully
separate it down the centre so that there are
two parts. Refer to Figure 3. Use the "spare
parts" to make a new strand of DNA by
matching the extra pieces to each of the two
halves.
Discussion
1. Compare the two strands of DNA. How
closely alike are they?
2. Explain how DNA makes copies of itself.
Activity 2: Bacteria, Pneumonia,
and DNA
Approach, I nvestigation 8-A. (Refer to Further
Reading.)
This is a "dry" lab in which students are led
step by step to a conclusion as they interpret
data from experiments where mice are injected
with pneumococcus cells. Students should
answer the discussion questions at the end of
each experiment before reading about the next
experiment. From these experiments students
should understand that DNA is the source of
hereditary instructions. The activity.i s found in
BSCS, Biological Science - A Molecular
Activity 3: Review
Figure 2
1. What is the importance of hydrogen bonds in
the explanation of the DNA model?
2. Define complementanty and its significance
with respect to DNA.
3. Illustrate the bonding patterns that occur
between specific base pairs.
4. At what point during cell division does DNA
replicate? Why?
5. Discuss the basic steps involved in DNA
replication.
Further Reading
Figure 3
BSCS. Biological Science - A Molecular
Approach. 4th ed. Lexington, Massachusetts:
D.C. Heath, 1980.
"Genetic Repair." Sciquest. January 1981.
I ngram, D.J. Biosynthesis of Macromolecules.
2nd ed. Menlo Park, California:
W.A. Benjamin, 1972.
Lessing, L. DNA: At the Core of Life Itself.
New York: Macmillan, 1967.
Sagre, A. Rosalind Franklin and DNA. New York:
Norton, 1975.
"Selfish DNA." Sciquest. January 1981.
"The Teaching of DNA Replication in Schools:
Thirty Years on, Thirty Years Out of Date?"
Journal of Biological Education. Vol. 18.
September 1984. p. 25.
Watson, J. The Double Helix: A Personal
Account of the Discovery of the Structure of
DNA. New York: Atheneum,1968.
RNA Synthesis: The Genetic Messenger
Objectives
Students should be able to:
1. List seven differences between DNA and RNA.
2. List and state the functions of the three types of RNA.
3. Explain how a messenger RNA (mRNA) molecule is made from
DNA.
4. Identify mRNA as a complementary copy of DNA.
5. Define the terms codon, terminator codon, and initiator codon.
6. State the significance of the codon AUG.
7. State the significance of the "poly-A" tail.
8. State where protein synthesis occurs.
Program Description
Although DNA molecules are located in the chromosomes of a cell's
nucleus, protein synthesis occurs in the cytoplasm. Therefore the
genetic information must be transferred from the nucleus into the
cytoplasm. This transfer of information is the function of mRNA. The
differences between DNA and RNA are described in the program.
The ribonucleic acids are classified into three groups:
(a) mRNA, the template for protein synthesis:
(b) tRNA, whose function is to act as the amino acid-adaptor
molecule carrying specific amino acids into their specific places
on the protein synthesizing template; and
(c) rRNA, which controls the manufacturing process.
As mRNA is produced, a double-stranded section of a DNA
molecule seems to unwind and pull apart. A molecule of mRNA is
then formed of nucleotides that are complementary to those
arranged along the exposed strand of DNA. In this way the mRNA
molecule that contains the information for arranging the amino acid
of a protein molecule in the sequence dictated by the DNA "master
i nformation" is synthesized. Once formed, mRNA molecules can
move out of the nucleus into the cytoplasm. Once the genetic code
has been transcribed to mRNA, the message can then be read three
nucleotides at a time. The triplet code with three adjacent nucleotide
bases is termed a codon and is responsible for a specific amino
acid. For example, GCU codes for alanine, GGA for glycine, and CAC
for histidine.
The minimum coding relationship between nucleotides and amino
acid is three nucleotides per amino acid. Four different kinds of
nucleotides taken three at a time provide for 64 combinations
(43 -64). All but three of the 64 combinations code for one or another
amino acid, and as many as six different nucleotide triplets may
specify the same amino acid. The three codons which are not
specific for any amino acid are called terminators. Terminator triplets
signal the end of the polypeptide chain and cause the protein chain
to become detached from the ribosome..UAG is an example of a
terminator codon. The nucleotide triplet AUG is also unique. It is the
initiator codon as well as the codon for the amino acid methionine.
How the correct AUG is selected for initiation is not known, but
once the correct AUG has been located the subsequent nucleotides
are read in groups of three.
Many RNA molecules are synthesized within the nucleus. Those
i nvolved in protein synthesis will have a long string of adenine
residues (100-200 poly-A) attached to the end of the mRNA molecule.
This poly-A tail is retained as the molecule enters the cytoplasm. The
l onger the tail, the more stable the molecule. Perhaps the length of
the tail in some way determines how many times that particular
mRNA will be translated.
Once in the cytoplasm mRNA links up with a ribosome. It is here
that the genetic instructions carried by mRNA are received and, with
the assistance of tRNA, translated into functional proteins.
Before Viewing
Emphasize once again the fact that DNA is a huge macromolecule
formed by thousands of nucleotide units. There are four different
nucleotides depending on the nitrogenous base each contains, and
these four nucleotides are linked in a specific sequence within each
DNA molecule. The essential feature of DNA is that it is a selfduplicating molecule; it is the basis of life.
Evidence clearly indicates that the four different nucleotides act
as the letters of the alphabet and are used to encode the information
necessary for the synthesis of protein and hence for control of the
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cell. Since the DNA molecule can build exact replicas of itself, the
genetic code can be copied and recopied so that each cell arising
from the original parent cell may possess a copy of instructions that
determine its nature.
After Viewing
Discuss the basic differences between DNA and RNA and have
students complete the following chart.
The other point to be stressed is that if the sequence of the DNA
bases is, for example, A, T, G, C, G, T, A, A, C, then the complementary bases in the developing mRNA molecule would be U, A, C, G, C,
A, U, U, G. Thus the sequence allows for the genetic information to
be carried to the ribosome and to be translated by tRNA, which will
be introduced in Program 5. Activities 1, 2, and 3 may be completed
at this time.
Activities
Activity 1:
Extraction of DNA and RNA
This exercise involves the extraction of both
DNA and RNA from plant tissues and the identification of three of the four nitrogen bases by
means of paper chromatography. It is similar to
Activity 2 except that geranium leaves are used
instead of yeast. The technique for chromatography is also less involved but the overall
experiment is not as effective. The exercise can
be found in Benson et al, Investigations in
Biology, Investigation 9. A similar activity can be
found in BSCS, Biological Science. An inquiry
into Life, I nquiry 8-1. (Refer to Further Reading.)
Pentose sugar
Structure
Size
Amount
Where
Kinds
Nitrogenous bases
Deoxyribose
Double helix
Large molecule
(can be thousands of
nucleotides long)
Few molecules in cell
Nucleus'
One type
Adenine, guanine,
thymine, cytosine
Ribose
Single helix
Smaller molecule
(transcribes only a section
of the total DNA molecule)
Many molecules in cell
Nucleus and cytoplasm
3 types - mRNA, tRNA,
rRNA
Adenine, guanine, uracil .
cytosine
"The mitochondria and chloroplasts have their own DNA separate from the DNA of
the nucleus.
Activity 2: What Is the
Nature of Nucleic Acids?
The yeast extract contains both RNA and DNA,
but there is much more RNA. That is why this
experiment is done in this program rather than
i n Program 2: DNA.
Apparatus (for each group)
5 mL 1M sulfuric acid
5 mL barium hydroxide
l mL bromthymol blue
One sheet Whatman No. 1 chromatogram
paper (15 cm x 15 cm)
One 1 L jar and lid
Six capillary tubes or toothpicks
10 mL acetic acid
30 mL butanol
Cellophane tape
I ndividual solutions of adenine, guanine,
cytidine monophosphate (CMP), and uridine
monophosphate (UMP)
A mixture of adenine, guanine, cytidine
monophosphate, and uridine monophosphate
One or two clinical-type centrifuges
Refrigerator
Water bath
FLOW CHART
Step 1
Hydrolyzing nucleic acid
1. Centrifuge the alcohol-nucleic acids from Program 1, Activity 2, Step 2 (4) for three minutes.
Precipitate (nucleic acids)
Supernatant (discard)
2. Dissolve the precipitate in 2 mL of sulfuric acid, CAUTION: Sulfuric acid is a strong oxidant.
Transfer half the solution to a small test tube and heat in a boiling water bath for an hour (or as
near an hour as convenient - 30 to 60 minutes.) During heating, maintain the volume by adding
water. Transfer the unheated half of the sulfuric acid solution to another test tube. Label both.
Neutralize both portions of the solution with barium hydroxide, using a drop of bromthymol blue
as an indicator. Add barium hydroxide to the acid solution drop by drop until the indicator turns
blue. Do not add more barium hydroxide than is required to produce a color change. Barium
sulfate precipitates; ignore it. The sample that is not heated should be neutralized as soon as
the precipitate is dissolved. Store the test tubes in the refrigerator for the next period.
Step 2
Preparing chromatographic equipment and a
chromatogram sheet
Step 3
Running the chromatogram
Step 4
Analyzing the chromatogram
3. Refer to Program 1, Activity 2, Step 4 (6) and Step 5 (7) and to Figure 1 in that program. Prepare
the jar and sheet as before. Put on at least five superimposed spots of each of the test sub
stances. These should include adenine, guanine, CMP, UMP, a mixture of these four, the nucleic
acid hydrolysate, and the unhydrolyzed sample. Pour enough of the solvent to cover the bottom
of the jar. (The solvent should be 15 parts acetic acid, 60 parts butanol, and 25 parts water.) Let
the chromatogram develop as before. Remove the paper when the solvent nears the top and let
it dry.
4. When the paper is dry examine it under ultraviolet light in a dark room. The spots containing the
bases will appear dark against a blue background. CAUTION: Do not look at ultraviolet light or
its reflections unless you are wearing protective glasses. Serious damage can be done to your
eyes.
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Discussion
1. Use the R F values of the known bases to
identify the bases of the hydrolyzed nucleic
acids. Record the R F values of all
substances.
2. What are the chemical, structural, and
functional differences between DNA and
RNA?
3. Explain the principle involved in paper
chromatography.
Activity 3: Review
1. List seven differences between DNA and
RNA.
2. What are the three kinds of RNA and what
are their functions?
3. How does mRNA become a pattern for
protein synthesis?
4. An "unzipped" DNA strand exposes the
following nucleotides: ATGGCATTGAC.
What mRNA sequence will form?
5. Define the following: codon, terminator
codon, initiator codon.
6. Where in the cell does protein synthesis
occur?
Further Reading
Benson, G.D. et al. Investigations in Biology.
Toronto: Addison-Wesley, 1977.
BSCS. Laboratory Block: The Molecular
Basis of Metabolism. Colorado: Raytheon
Education Co., 1968.
BSCS. Biological Science: An Inquiry into Life.
Student Laboratory Guide. 2nd ed. New York:
Harcourt, Brace and World, 1968.
I ngram, D.J. Biosynthesis of Macromolecules.
2nd ed. Menlo Park, California:
W.A. Benjamin, 1972.
Temin, H.M. "RNA-Directed DNA Synthesis."
Scientific American. January 1972. p. 24.
Travers, A.A. Transcription of DNA.
Oxford Biology Reader. 2nd ed. London:
Oxford University Press, 1977.
Transfer RNA: The' Genetic Messenger
Objectives
Students should be able to:
1. Explain the role of the endoplasmic reticulum.
2. Name the building blocks of proteins.
3. Describe the structure of the tRNA molecule.
4. State the function of tRNA.
5. Define the terms anticodon and acceptor codon.
6. Identify how a particular tRNA attaches to a particular place on the
mRNA.
7. State how a triplet of bases in the mRNA determines the specific
amino acid.
8. Discuss the role of the ribosome in protein synthesis.
Program Description
Most cells contain an extensive system of tubules with thin
membranes called endoplasmic reticulum. Associated with the
endoplasmic reticulum are ribosomes, and it is here that protein
synthesis occurs. Before a protein molecule can be synthesized, the
correct amino acids must be present in the cytoplasm to serve as
building blocks. Furthermore, these amino acids must be positioned
i n the proper locations along a strand of mRNA. The positioning of
the amino acid molecules is the function of transfer RNA (tRNA).
Since at least 20 different amino acids are involved in protein
synthesis, there must be at least 20 different kinds of tRNA
molecules to serve as guides. Each kind of tRNA molecule
recognizes and binds to one kind of amino acid.
Transfer RNA molecules are polynucleotide chains of some 75 to
85 nucleotides. Many of these nucleotides containing the regular
nitrogenous bases link two portions of the chain in a double helix
similar to that of the DNA molecule. Because of this several loops
are formed in the chain. The nucleotide sequences for the tRNA
molecules vary widely, but all molecules of tRNA have an unlinked
codon at one end - the acceptor end of the tRNA molecule. The
amino acid is attached at this end. In each case, the amino acid is
activated and attached to the appropriate tRNA molecule by means
of an activating enzyme specific for that amino acid. Energy of
activation is supplied by ATP The tRNA then ferries the amino acid to
the ribosome. One portion of the nucleotide sequence in the tRNA
represents the anticodon, an unpaired nucleotide-triplet complementary to the codon in mRNA that specifies that amino acid. The
nucleotides on the anticodon bond to the complementary nucleotides on the mRNA strand. In this way the tRNA carries its amino
acid to a correct position on a mRNA strand. The ribosome then
moves onto the next codon and the tRNA for the next amino acid .
links up with the specific codon on the mRNA. The preceding amino
acid is linked to the incoming one by a peptide bond, and its tRNA is
released and is available to pick up another amino acid. As a result,
amino acids are placed in the sequence needed to form a particular
protein molecule. Once the polypeptide chain is completed, the protein is then released and becomes a separate functional molecule.
By knowing the nucleotide sequence of a particular protein one
can trace its protein structure back to its DNA "blueprint" and can
reconstruct an exact copy of the DNA molecule.
The role of the ribosome is to provide the proper orientation of the
amino acid-transfer RNA, the messenger RNA, and the growing polypeptide chain so that the genetic code on the template, or mRNA,
can be read accurately to ensure that the correct protein is formed.
Before Viewing
It may be wise at this time to introduce the structure of the tRNA
molecule. The program depicts the molecule as an "L" shape.
However, most texts illustrate tRNA as a "clover leaf." An acetate
outline such as Figure 1, which shows methionine - tRNA isolated
from E. coli, may be used to point out exactly where the anticodon
and the acceptor codon are located. There may be a slight discrepancy between the guide and the program as to the description
of the acceptor codon. All tRNAs end in the unpaired nucleotides
ACC, and it is the arrangement of the nucleotides within the
molecule in combination with a specific enzyme that recognizes the
specific amino acid that will attach to the specific tRNA. The
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program suggests that the triplet codon at the acceptor end is
specific for a particular amino acid. Specific base pairings can also
be mentioned, pointing out to students the formation of a double
helix similar to the DNA molecule.
After Viewing
Stress once again the fact that the function of the tRNA molecule is
to act as an acceptor for an activated amino acid and as an adaptor
for carrying the amino acid to the site of protein synthesis in the
mRNA template in the ribosome, ensuring that the correct amino
acid is placed on the correct coding site. Activities 1, 2, and 3 should
be completed now
AC ivi'ties
Further Reading
Activity 1: Reconstruction of the
DNA "Blueprint"
Asimov, I. The Genetic Code. New York: New
American Library, 1962.
This activity reinforces the concepts of
transcription and translation. Refer to Figure 3
i n Program 1 for the amino acid sequence of
i nsulin. A chart of the nucleotide triplets for the
specific amino acids can be found in Kimball's
Biology, p. 256. (See Further Reading.) Have
students trace part of the protein structure of
i nsulin back to its DNA "blueprint."
Beadle, G. and M. Beadle. Language of Life:
An Introduction to the Science of Genetics.
New York: Doubleday,1966.
Bouk, E. The Code of Life. New York: Columbia
University Press, 1965.
BSCS. Biological Science - A Molecular
Approach. 4th ed. Lexington, Massachusetts:
D.C. Heath, 1980.
I ngram, D.J. Biosynthesis of Macromolecules.
2nd ed. Menlo Park, California:
WA. Benjamin, 1972.
Kimball, J.W. Biology. 5th ed. Reading,
Massachusetts: Addison-Wesley, 1983.
Activity 2: What Controls the
Synthesis of Large Molecules in a
Living Cell?
This investigation is presented as a "dry" lab.
The discussion questions concerning the data
should l ecid students to the conclusion that for
each enzyme (protein) in Neurospora present
and functioning in the synthesis of amino acids
or vitamins, there is a particular gene or group
of genes responsible for its formation. This lab
i s extremely helpful to the students' understanding of the gene-enzyme relationship: that
DNA directs the synthesis of the enzyme, a
protein necessary for the formation of specific
substances in the organism. The investigation
i s found in BSCS, Biological Science - A
Molecular Approach, I nvestigation 9-C. (Refer to
Further Reading.)
Activity 3: Review
1. What is the function of the endoplasmic
reticulum?
2. Name the building blocks of protein.
3. Draw a simplified model of the tRNA
molecule.
4. What is the function of tRNA?
5. Define anticodon and acceptor codon.
6. How many different kinds of tRNA must
exist in a cell? Explain.
7. (a) Distinguish between transcription and
translation.
(b) Where do each occur?
"The Teaching of Protein Synthesis - A
Microcomputer Based Method" Journal
of Biological Education. Vol. 17. Fall 1983.
pp. 222-24.
21
22
Ribosomal RNA: The Protein Maker
Objectives
Students should be able to:
1. Describe the structure and state the function of the ribosome.
2. State what factors must be present in order for peptide initiation
to occur.
3. Identify how a particular tRNA attaches to a particular place on
mRNA.
4. Describe the initiation process.
5. Describe how a chain of amino acids attached to tRNA molecules
becomes linked together.
6. Explain the cause and effect of a mutation.
Program Description
The machinery for the assemblage of a protein is in the ribosome,
the key building block being ribosomal RNA (rRNA). The ribosome is
composed of two subunits - a smaller subunit containing one RNA
molecule and 21 proteins and a larger subunit containing two RNA
molecules and 34 proteins.
Peptide initiation includes the formation of a complex between
the smaller subunit, mRNA, tRNA, GTP (energy necessary for
synthesis), and three distinct initiation factors, F 1, F2, and F3. The
next step appears to be the attachment of the larger subunit.
Transfer RNA, with its specific amino acid, now binds to a specific
site on the ribosome, the "R" site. At this point the anticodon of
tRNA binds to the codon on mRNA (usually AUG) that starts every
message. Stresses in the molecular bonds between the codon and
the anticodon cause the tRNA to "flip over." This ensures proper
binding. The tRNA is then translocated to a second site, the "P" site,
on the ribosome. Now there is the addition of the next amino
acid-tRNA added to the first site on the ribosome, which was
recently vacated. In other words, the new tRNA is capable of base
pairing with the next mRNA codon at the "R" site. This tRNA now
flips sideways and the two amino acids are aligned. The preceding
amino acid is linked to the incoming one by a peptide bond. The
product is located at this point on the "R" site. The tRNA on the "P"
site is now released and a shift of the new peptidyl-tRNA from the
"R" site to the "P" site on the ribosome occurs. Also at this point,
the ribosome has moved along the mRNA by three nucleotides, or
one codon. Everything is "now ready for the addition of a further
tRNA, and the cycle repeats itself. This process is repeated again
and again until the protein is completed.
The program ends by briefly summarizing the role of the ribosome
i n engineering the correct link-up between mRNA and tRNA. There is
a brief mention of mutations, alterations in the DNA code that may
eventually result in the evolution of a new organism.
Before Viewing
Have students review the structure of the tRNA molecule and the
role played by the ribosome in the synthesis of protein molecules.
After Viewing
The program describes the building of the polypeptide in general
terms. It might be easier for students to understand the concepts of
polypeptide formation if the process is divided into three phases:
i nitiation, elongation, and termination. This is described on page 254
of Kimball's Biology. (Refer to Further Reading.)
There is a slight discrepancy dealing with the initiation process.
The program states that the tRNA initially binds to the "R" site. In
most texts this is listed as the "A" site. The tRNA is then transl ocated to the "P" site. To put things into perspective, you may
want to use a chalkboard or acetate outline of the formation of polypeptides similar to Figure 1.
It may be of interest at this time to discuss mutations. Although
most mutations seem to be harmful to the cell or organism,
occasionally one occurs that has a slight new advantage over others
of its species. By reproduction the new mutant organism may pass
the advantageous gene into the population of that particular species.
This can open a pathway toward evolution into a new species. This
idea can then be followed up in Activities 1 and 2.
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s
Activities
Activity 1: How Can Mutant
Strains of Bacteria Be Isolated?
Because of the importance of bacterial
research in the areas of biochemistry, genetics,
and medicine, students should be introduced
to the basic bacteriological techniques of
transferring and culturing. The work with
bacteria introduces the idea that mutations are
changes in the biochemistry of an organism
and do not necessarily produce a visible
change. Students innoculate agar media with
bacteria and observe the growth pattern of the
bacteria in the presence of antibiotics. Upon
completion of this activity students should be
able to relate the meaning of mutations to the
changes that occur to the code of the genetic
material within the cell. This investigation can
be found in BSCS, Biological Science - A
Molecular Approach, I nvestigation 9-A. (Refer to
Further Reading.)
Activity 2: Effects of Radiation on
Microorganisms
Understanding some of the useful, as well as
the harmful, effects of radiation is important in
our modern age. Cancer treatment and food
preservation by radiation are possible because
of the lethal effect of the rays on the exposed
cells. In this activity, students will have
experimental evidence of the killing effect of
ultraviolet light on yeast cells. The investigation
can be found in BSCS, Biological Science - A
Molecular Approach, I nvestigation 9-B. (Refer to
Further Reading.)
Activity 3: Genetic Engineering
Considerations as to the handling of this
bioethical issue are in order. Once the transfer
of genetic information is fully understood, there
i s no technical reason why humans cannot
affect the process. It might soon be possible,
for example, to correct the genetic code for an
i ndividual who has a genetic disease, such as
sickle-cell anemia or phenylketonuria. Tamperi ng with the transfer of genetic information is
called genetic engineering. Have students form
a debate on this topic. Questions to consider
are: Do scientists think it is possible to
eventually "build babies to order"? Do you
think this is a good idea? Why or why not?
What are the arguments for and against genetic
engineering?
Activity 4: Research Paper
Have students prepare a report on a hereditary
disease such as PKU, cystic fibrosis, diabetes,
Huntington's Chorea, Down's syndrome,
TaySachs disease, sickle-cell anemia, or cleft
palate. Their reports should discuss the causes,
symptoms, detection, and treatment of the
disease.
Activity 5: Review
1. Describe the structure and state the function
of the ribosome.
2. What factors are necessary in order to
i nitiate polypeptide formation?
3. Describe where the initiator tRNA binds to
the ribosome. How does the tRNA bind to
mRNA?
4. Describe the elongation process and the
formation of the polypeptide.
5. Describe the role of the DNA code, codons,
and anticodons relative to RNA, ribosomes,
and amino acids.
6. Explain the result of errors in the transfer of
genetic information.
Further Reading
Brown, D.D. "The Isolation of Genes." Scientific
American. July 1975. p. 24.
BSCS. Biological Science - A Molecular
Approach. 4th ed. Lexington, Massachusetts:
D.C. Heath, 1980.
I ngram, D.J. Biosynthesis of Macromolecules.
2nd ed. Menlo Park, California:
W.A. Benjamin, 1972.
Kimball, J. Biology. 5th ed. Reading,
Massachusetts: Addison Wesley,1983.
Kornberg, A. DNA Synthesis. San Francisco:
W.H. Freeman, 1974.
Patt, D.A., and G.R. Patt. An Introduction to
Modern Genetics. Reading, Massachusetts:
Addison-Wesley, 1975.
Pines, M. Inside the Cell. The New Frontier of
Medical Science. U.S. Dept. of Health,
Education and Welfare. DHEW Publication
No. (NIH) 78-1051.
Ordering
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Videotape
Program 1: Protein:
The Stuff of Life
BPN
248901
Program 2:DNA: The Molecule 248902
of Heredity
248903
Program 3: DNA Replication:
The Repeating Formula
248904
Program 4: RNA Synthesis:
The Genetic Messenger
248905
Program 5: Transfer RNA:
The Genetic Messenger
248906
Program 6: Ribosomal RNA:
The Protein Maker
25