Part IV


Applications of Biotechnology:

Genetic Engineering




REFERENCES


Genetic Engineering


Below is a list of references you may find helpful in understanding the topic. If you need a copy of the item, call your local library or state university library. Michigan State University telephone number is (517) 353-8700. They can arrange for Interlibrary Loan if the item is not in their collection.

 

Drlica, K. (1984). Understanding DNA and gene cloning: A guide for the curious. New York: John Wiley and Sons.

 

Genetic Engineering of Plants. Natl. Research Council. Washington, DC: National Academy Press. 1984.

 

Genetic Engineering in Food and Agriculture. Ames, IA: Council for Agricultural Science and Technology (CAST), September 1986.

 

Genetic Engineering: A Natural Science. St. Louis, MO: Monsanto Company, 1985.

 

Hyde, M.O., & Hyde, L.E. (1984). Cloning and the new genetics. New Jersey: Enslow Publishers.

 

Jacbosson, S., Jamison, A., & Rothman, H. (1986) The biotechnical challenge. New York: Cambridge University Press.

 

Jacksin, D.A. & Sitch, S.P. (eds.). (1979). The recombinant DNA debate. Englewood, NJ: Prentice Hall.

 

Judson, H.F. (1979). The eighth day of creation: Maker of the revolution in biology. New York: Simon and Schuster.

 

Karp, L. E. (1976). Genetic engineering: Threat or promise? Chicago: Nelson Hall.

 

Keiffer, G.H. (1987). Biotechnology, genetic engineering and society: Monograph III. Reston, VA: National Association of Biology Teachers.

 

Kieffer, George H. Biotechnology, Genetic Engineering and Society. University of Illinois at Urbana-Champaign, IL: National Association of Biology Teachers, 1987.

 

Mertens, T.R. (ed). (1975). Human genetics: Reading on the implications of genetic engineering. New York: John Wiley & Sons.


 

Of Earth, Genetic Engineering. Monsanto will provide 40 copies of each publication to educators for classroom use. The booklets cover the topics of plant and animal biotechnology. Monsanto, 800 N. Lindbergh Boulevard, St. Louis, MO 63167. 314/694-7233(Gary Burton)

 

Wheelis, M. & Gonick, L. (1983). The cartoon guide to genetics. New York: Harper and Row.

 

Witt, S.C. (1985). Briefbook: Biotechnology and genetic diversity. San Francisco: California Agricultural Lands Project.


DNA


Below is a list of references you may find helpful in understanding the topic. If you need a copy of the item, call your local library or state university library. Michigan State University telephone number is (517) 353-8700. They can arrange for Interlibrary Loan if the item is not in their collection.

 

Biotechnology Kit:DNA Extraction. Catalog #21-1100. Available from Carolina Biological Supply Company, 2700 York Rd., Burlington, NC 27215; 800-334- 5551.

 

Dixion, L. Teaching recombinant DNA technology in high school biology courses. (1988, September). The American Biology Teacher, 50(6), pp. 368-373.

 

DNA plant products won't be marketed nationally by Kraft. (DNA Plant Technology Corp.).(1987, June 29). Wall Street Journal, p.22(W), p. 28(E) col 4.

 

DNA Extraction Kit #21-1100. Available from Cabisco Biotechnology, 2700 York Rd., Burlington, NC 27215; 800-334-5551

 

Guilfoile, P. A simple DNA isolation technique using halophilic bacteria. (1989, March). The American Biology Teacher, 51(3), pp. 170-171.

 

Micklos, D.A. & Freyer, G.A. (1990). DNA Science: A first course in recombinant DNA Technology. Cold Spring Harbor Laboratory Press and Carolina Biological Supply Company. 477 pp.

 

Myers, R. Recombinant DNA technology in high school biology laboratory. (1988, January). the American Biology Teacher, 591(1), pp. 43-45.

 

Myers, R. A method for isolating DNA from E. coli. (1985, September). The American Biology Teacher, 47(6), pp. 362-364.

 

Pebbles, P. & Leonard, W.H. A hands-on approach to teaching about DNA structure and function. (1987, October). The American Biology Teacher, 49(7), pp. 436- 438.

 

Pines, M. (1087). Mapping the human genome. Bethseda, MD: Howard Hughes Medical Institute.

 

Sigsmondi, L.A. A paper model of DNA structure and replication. (1989, October). The American Biology Teacher, 51(7), pp.422- 423.

 

"Spooling Chromosomal DNA." Kit #107. Available from EDVOTEK, Inc.,P.O. Box 1232, West Bethesda, MD 20817; 800-EDVOTEK.

 

Thomson, R.G. Recombinant DNA made easy I: "Jumping genes". (1988, February). The American Biology Teacher, 50(2), pp. 101-106.


Gene Splicing


Below is a list of references you may find helpful in understanding the topic. If you need a copy of the item, call your local library or state university library. Michigan State University telephone number is (517) 353-8700. They can arrange for Interlibrary Loan if the item is not in their collection.

 

Advances in Genetic Technology-teacher's edition. (1989).BSCS, D.C. Health and Company.

 

Gene   and surroundings. (1983). Colorado Springs,CO: Kendall/Hunt-Biological Sciences Curriculum Study.


Grady, D. (1987, June). The ticking of a time bomb in the genes. Discover, pp.26.

 

Hotz, R.L. (1987, May 12). Secrets of life, part 3: Genetic testing could give employers inside view of workers. Palm Beach Post. pp. 8A.

 

Hotz, R.L. (1987, May 11). Secrets of life, part 2: Couple fearful of diseased baby, hunting genetically clean egg. Palm Beach Post, pp. 2A+.

 

Hotz, R.L. (1987, May 10). Secrets of life, part 1: Awesome questions for mankind created for test-tube advances. Palm Beach Post, p. 1A+

 

Lyon, J. & Gorner, P. (1086, March 2-9). Altered fates: Gene therapy altars future. Chicago Tribune, Tempo, pp. 1+.


 

Mertens, T.R. & Pursifull, J. The gene scene: A human genetics game. (1986, February) The American Biology Teacher, 48(2), pp. 104-108.

 

Miller, J.A. (1986, January 25). Fight continues over gene-splice release (field test by Advanced Genetic Science Inc.). Science News, 129,56.

 

Schmeck,H.M. (1985, November 10). The promises of gene therapy. New Times Magazine, pp.114+.

 

Schwartz, R. (1985, September). Designer genes at the dinner table. FDA Consumer, pp. 31-33.

 

Scientists set to unravel 80,000 element sequence of human genetic code.(1986, September 7). Baltimore Sun, p11A.

 

Seidel-Rogol, B.L. An exercise in gene mapping: Using hybridization analysis to detect a restriction site polymorphism in the rDNA gene of Neurospora crassa. (1990, February). The American Biology Teacher, 52(2), pp. 103-106.



TITLE OF BIOTECHNOLOGY ACTIVITY:


A.    Gene Splicing

 

OBJECTIVE(S):

 

Upon completion of this activity, students should be able to:

 

                        1.         Describe in their own words, using a model, the basic outline of gene splicing.

 

                        2.         Identify the agents involved in gene splicing; plasmid, gene of interest, restriction enzyme, and ligase.

 

                        3.         Tell what each of the agents does.

 

MATERIALS AND/OR EQUIPMENT:

 

Sandwich bag filled for each student as follows:

 

                                  plasmid duplicated on colored paper and cut by teacher

                                  4 blue dots with sticky backs, approximately 1 cm in diameter (restriction enzyme)

                                  4 red dots (ligase)

                                  a strip of chromosomal DNA which can be in a different bag to represent a different cell.

 

TIME REQUIRED FOR ACTIVITY:

 

One class period

 

BACKGROUND MATERIAL:

 

Gene splicing resulted from molecular biological research, and its application has developed the science of biotechnology. Certain naturally occurring restriction enzymes cut DNA at specific sites. Plasmids, tiny DNA rings, were discovered in bacteria. Some plasmids carry genes which convey resistance to antibiotics. Restriction enzymes cut at predictable sites. The discovery of enzymes, ligases, which help the attachment of the new genes into the plasmid ring allowed the development of gene splicing or the science of recombinant DNA.

 

Beyond the realm of health-care for humans and animals, biotechnology is having a wide spread effect on agriculture. Genetic engineering has great potential in the area of breeding in both plants and animals. Plants are being introduced with genes that produce characteristics that act as herbicides and others genes that produce frost resistance. Plants that are resistant to drought and acceptable to the use of salt water for irrigation are also being tested. Another plant with the characteristic ability to fix or use atmospheric nitrogen is being tested. This will reduce the reliance on nitrogen fertilizers that require expenditure of large amounts of energy to produce but are necessary for plant growth in most parts of the world. Continued changes in the genetics of crop plants may be imperative to feed the world's population.

 

Another important area of plant biotechnology is the culture of plant cells which carry out many complex metabolic processes that yield useful products like rubber, pain killers, and other chemical substances. Some plants are difficult to grow from seed but are being cloned from single cells from which the whole plant can be raised. One of the earliest applications of biotechnology was the biological control of natural predators of certain plants by introducing organisms into them that would provide protection from the predator. The organisms remain in the plants and provide an indefinite amount of protection reducing the need for chemical pesticides and they are non-polluting.

 


PROCEDURE:

 

                        1.         Each student will open a baggie (bacteria cell) and extract a paper plasmid.

 

                        2.         The plasmid model has a gene which is held in position by colored dots. These dots represent a specific restriction enzyme and are removed by the students.

 

                        3.         The gene is removed from the plasmid and allows the splicing of a "gene of interest."

 

                        4.         The student removes a piece of DNA from another bag. This gene has the same base pairing ends as the plasmid with the same color restriction enzyme dots.

 

                        5.         The student detaches the gene from the DNA.

 

                        6.         The "gene of interest" is then moved to the plasmid where it is spliced by the ligase, dots of a different color. The "gene of interest" is now part of the plasmid and it will be duplicated as the cell divides and expresses itself if selected for that purpose.

 

DATA AND SUMMARY ANALYSIS:

 

This hands-on activity attempts to make concrete a very abstract concept, gene splicing. Students are given a paper model of a plasmid from which a gene is cut at specific sites by a restriction enzyme, represented by self-stick dots. The same restriction enzyme is used to remove a "gene of interest" from chromosomal DNA. This is then inserted in the plasmid with another colored dot representing ligase, the chemical glue.

 

QUESTION(S) FOR INVESTIGATION:

 

                        1.         What impact does gene splicing have on biotechnology?

 

IDEA(S) FOR ADDITIONAL ACTIVITIES:

 

                        1.         Have students write a proposal of how they would like to change an organism through genetic engineering and explain how they would go about making the change.









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| G C G T C A T T C C G G T C A C A T T C A |

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| C G C A G T A A G G C C A G T G T A A G T |

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REFERENCE(S):

 

Genetic Engineering: A Natural Science, St. Louis: Monsanto Company, 1987.

 

Of the Earth: Agriculture and the New Biology, St. Louis: Monsanto Company, 1987.

 

Introduction to Plant Biotechnology, Urbana: Vocational Agriculture Service, College of Agriculture, University of Illinois.

 

Recognizing the Impact of Technology on Agriculture: Biotechnology, Urbana: Illinois Agricultural Core Curriculum, Agricultural Education, University of Illinois.


 



TITLE OF BIOTECHNOLOGY ACTIVITY:


B.    The Size of the Genome

                  (Adapted from "A Sourcebook of Biotechnology Activities", North Carolina Biotechnology Center.)

 

OBJECTIVE(S):

 

Upon completion of this activity, students should be able to:

 

                        1.         Describe the relative sizes of bacterial cells, chromosomes, genes and plasmids.

 

                        2.         Explain that strands of DNA are extremely thin and must be very tightly coiled; an immense length of DNA fits into a relatively tiny cell volume.

 

                        3.         Explain the differences between bacterial and human gnomes.

 

                        4.         Recognize and understand these words: DNA, genome, plasmid, bacterium, chromosome, gene, supercoiled DNA, prokaryote.

 

                        5.         (Optional) Use scientific notation to confirm the dimensions for bacterial DNA.

 

MATERIALS AND/OR EQUIPMENT

 

For entire class

 

▾ felt tip pen

▾ scissors

▾ glue

▾ meter stick or metric ruler

 

For each lab team of four students, a letter-size envelope containing:

 

▾ a 2 cm (20 mm) gelatin capsule (No. 0"; these capsules can be purchased at pharmacies or health food stores)

▾ 10 meters of white thread; 10 meters of colored thread

▾ 220 mm strands of colored yarn or twine; red and blue or choose your school colors.

▾ 2 blank 3"x5" index cards


 

TIME REQUIRED FOR ACTIVITY:

 

Two class periods

 

BACKGROUND MATERIAL:

 

Teaching about genes as "things" requires special measures. Students have difficulty grasping the incredibly small scale of biology at the molecular level, so they need ways to imagine what they cannot see. How big are bacteria? How long are their strands of DNA? How big is a gene? What is a plasmid? How big are these things in relation to each other? The answers require more than words and numbers.

 

Scientists use bacteria as a model system for studying the molecular genetics of more complex multicellular organisms. Bacterial DNA is easier to study because it is not surrounded by a nuclear membrane; instead, the bacterial chromosome is a circle of DNA that is anchored in the cell membrane. In addition, the amount of DNA found in bacteria is much less than the amount found in multicellular organisms. The biotechnology techniques related to genetic engineering, monoclonal antibodies, DNA probes and cell/tissue culture all require an understanding of the size relationship among DNA, genes, plasmids and chromosomes.

 

This activity demonstrates various size relationships among genes, bacterial chromosomes, plasmids and bacterial cells. Students will also learn about some of the physical properties of DNA and the sizes of bacterial gnomes compared to the human genome.

 

The Genome

 

All organisms, no matter how simple they may seen to us, require a substantial amount of DNA to carry the instructions necessary to live and reproduce. This DNA is referred to as the "genome." Because DNA is microscopic, it's hard for humans to visualize the actual size of the genome. This is especially difficult when the organism itself is tiny, such as the bacterium that is about a millionth of a meter (1-2 microns) long. Scientists who study genes have used E. coli in many of their experiments, so it is a good organism for us to use as a model today.

 

The Size of E. coli

 

In this activity you will make a model of an E. coli cell that will be 10,000X bigger than actual size. (This is comparable to inflating an ant to the size of a tractor trailer.)

 

                                  An E. coli bacterium expanded 10,000X would be approximately 20 millimeters long, and look like a small pill.

 

The Size of E. coli's DNA

 

In bacteria, the DNA is a large circle. It accounts for 2-3% of the cell's weight and occupies about 10% of its volume. The remarkable features of DNA are its length and slenderness. If we enlarged the DNA of an E. coli 10,000X, its total length would go around a large inflatable swimming pool. If you cut the loop and straightened it into a line, the strand would be two car lengths long. Imagine how slender the 10-meter strand must be to fit in only 10% of the space in the bacterium model?

 

The Size of E. coli's Plasmids

 

In many recombinant DNA experiments, biologists use a tiny circlet of bacterial DNA that is much smaller than the length of chromosomal DNA. This circlet is called a plasmid. Plasmids occur naturally and are small enough to pass through living bacterial cell membranes. As one might expect, plasmids contain only a few genes, none of which is essential to the bacteria. (In contrast, E. coli's loop of DNA contains about 2,000 genes). Molecular biologists add desired genes to plasmids, then insert the new plasmid with the added gene into a living bacterium. Plasmids are considered to be "gene taxis," and are often called vectors.

 

An E. coli plasmid enlarged 10,000X would be about 10 millimeters in circumference (a circle 3.2mm in diameter). If you cut the plasmid and straightened it into a line, the strand would be about one-half the length of a bacterium. The plasmid would also be extremely thin. Floating in the bacterial medium, the plasmid can collapse and coil like a very fine rubber band.


 

The Size of E. coli's Genes

 

Genes are lengths of DNA that code for a specific protein. A gene can be a continuous length of DNA, or it can be many different pieces of DNA interspersed along a chromosome. Researchers in the early 1970's began the age of recombinant DNA (rDNA) by identifying and isolating genes that they wanted to express in E. coli. Once the genes were identified and isolated, they were packaged into plasmids. The altered plasmids were inserted into living bacterial cells, and the proteins are coded for the "new" genes and expressed by the bacteria. A bacterial gene enlarged 10,000X would be approximately 4mm in length.

 

PROCEDURE:

 

Teacher Preparation

 

Make a demonstration kit for yourself and one for each lab team.

 

                        1.         Place one gelatine capsule into each envelope.

 

                        2.         Use a meter stick or ruler and scissors to cut the two pieces of colored yarn or twine and the sewing thread to the following lengths:

 

▾ 1 length of white thread and 1 length of colored thread - 10 meters each

▾ 1 piece of blue yarn or twine - 4 millimeters long (This will be just a piece of blue"fuzz" if you use yarn.)

▾ 1 piece of red yarn or twine - 10 millimeters long

 

                        3.         Put the two 10-meter lengths of thread together and coil around your hand. Each length of thread represents on side of the DNA double helix.

 

                        4.         Glue the 4-millimeter piece of blue yarn or twine to a 3"x5" card and label it "Length of DNA in a single bacterial gene."

 

                        5.         Glue the 10 mm piece of red yarn to a 3"x5" card and label it. "Bacterial plasmid." If you use thread or string, glue it into a circle on the card to stimulate the circular nature of plasmids.

 

                        6.         Put all materials from Steps 2-5 into each envelope.

 

Teaching Tips

 

                        1.         If time allows, have your students prepare the "gene" and "plasmid" cards, and measure out the lengths of thread (Teacher Prep, steps 2-5). This enables them to practice metric measurement.

 

                        2.         The student questions involving scientific notation should be assigned relative to the math abilities of your students.

 

Teaching Procedure

 

                        1.         Have students do background reading. (You may want to go over it with them in class.)

 

                        2.         Students split into groups of four and follow the procedure on the activity sheet. (In step #7, the best way to fit the thread in the capsule is to fold the lengths of thread in half several times until they have a "wad" of thread approximately 3 inches long. Insert one end of the folded thread into the longer section of capsule. Push the "wad" into the capsule with a twisting motion.)

 

                        3.         Students do the question sheet. (This can be in class or as homework.)

 

Student Procedure

 

                        1.         Remove the gelatin capsule from the envelope. It represents a single E. coli bacterium that has been enlarged 10,000X.

 

                        2.         Remove the two 3" x 5" cards from the envelope. The lengths of yarn on the card represent the lengths (and not the diameter) of an E. coli gene and plasmid magnified 10,000X.

 

                        3.         Remove and stretch out the two lengths of thread in the envelope. These represent the bacterial chromosome.

 

                        4.         Two people should make a circle with the two threads. This represents the bacterial chromosome which is simply a circle of DNA that is attached to the cell membrane.

 

                        5.         A third person now holds up the 3"x5" card labeled "Length of DNA in a single bacterial gene" next to the DNA in a single bacterial gene" next to the DNA "loop." The 4 mm long piece of yarn glued on the card represents the length of a single gene. Remember, a gene is composed of all the segments of DNA that instruct the cell to make a single protein, whether these segments are continuous or not.

 

                        6.         With the white thread loop and the card labeled "Bacterial gene" still in view, ask the fourth person to hold up his/her 3"x5" card labeled "Bacterial plasmid." The 10 mm piece of yarn glued on this card represents a typical bacterial plasmid. Plasmids carry several genes necessary for their own replication and stability, but you can see that the plasmid is tiny in comparison with the bacterial chromosome. Plasmids can also carry genes that add important characteristics (such as antibiotic resistance) to a bacterium.

 

                        7.         Now that you have compared the sizes of the chromosomes, plasmids and genes, try to construct the bacterium. This means that the long lengths of thread representing the chromosome have to go inside the capsule. It's not easy!

 

QUESTION(S) FOR INVESTIGATION:

 

                        1.         Bacteria are difficult to study because of their extremely small size. One thousand average-sized bacteria linked together like a strand of beads would only be as big as the period at the end of this sentence. How can one study bacteria if they are so small? Propose two solutions to this question.

 

                        2.         Discuss how the following bacterial structures relate to each other: bacteria, DNA, chromosome, gene, plasmid.

 

                        3.         Why does it take two lengths of thread to represent the chromosome?

 

                        4.         How many bacterial genes would fit on your DNA circle (formed in Step #4)?

 

                        5.         Is the thread that you tried to stuff in the capsule to thick to represent the DNA's actual thickness? Hint: what % of bacterial cell volume is your thread using, and what is the actual volume that DNA occupies in bacteria?

 

                        6.         In this activity, how many meters did it take to represent the E. coli genome? If the human genome is 1,000X greater than the E. coli genome, how many meters would it take to represent the human genome? How many miles to represent the human genome?

 

Mathematical calculations of bacterial and human gnomes:

 

▾ The distance between DNA base pairs is 3.4 x 1010m.

▾ The DNA chromosome in E. coli has about 3 million base pairs; the average gene has 1,2000 base pairs; and the average plasmid has about 3,000 base pairs.

 

Using the information above, calculate the following:

 

                        7.         How long (in meters) is the DNA strand in E. coli?

 

                        8.         How long (in meters) is the average E. coli gene?

 

                        9.         What is the circumference (in meters) of an average-sized E. coli plasmid?

 

                        10.       If E. coli is magnified 10,000X normal size, how long (in millimeters ) is its DNA strand?

 

                        11.       If E. coli is magnified 10,000X normal size, how long (in millimeters) is a gene?

 

                        12.       If E. coli is magnified 10,000X normal size, how long (in millimeters) is a plasmid?

 

The human genome can be related to a length of railroad track. Imagine that you have uncoiled and untwisted the DNA molecule in a single human cell. The ties will represent base pairs and the rails will represent the deoxyribose sugar and phosphate sides of the DNA molecule. The railroad ties are approximately two feet apart.

 

                        13.       Remember that there are 3 x 109 base pairs in the human genome. How many miles (or kilometers) of track will it take to represent a human genome?

 

                        14.       How many miles (or kilometers) of track would wrap around the earth at the equatorial circumference = 24,000 miles)?

 

                        15.       Your textbook can be used to relate the number of characters (letters and spaces) on a page to the number of base pairs (3 x 109) in the human genome. How many books like your textbook would be needed to represent the human genome?

 

▾ Choose a page in your textbook that is mostly printed.

▾ Count the number of characters on five randomly selected lines.

Calculate the average number of characters per line (C).

▾ Count the number of lines on the page (L). Multiply the average number of characters on a line times the number of lines on the page. (CxL)

▾ Multiply the number of characters for a page by two (2). This will give you the approximate number of characters for a full page (front and back). (CxLx2)

▾ Divide the total genome of a human cell by the number of characters on a full page in the textbook. This will give you the number of pages that will be needed to represent the human genome. (3 x 109) divided by (C x L x 2) = (P).

▾ Divide the number of pages needed for a human genome by the number of pages in your textbook (T). This will tell you number of books (N) that would be needed to record the information in the genome of one human cell. (P) divided by (T) = (N)!

 

ANSWERS TO QUESTIONS FOR INVESTIGATION:

 

                        1.         Use a microscope or make a model you can see.

 

                        2.         Bacteria contain DNA just like other living cells. The main portion of bacterial DNA is contained in its singular chromosome. The DNA segments of the chromosome that code for a specific protein are called genes. Bacteria also contain circular, non-chromosomal pieces of DNA called plasmids.

 

                        3.         Each length is one side of the DNA helix.

 

                        4.         If each gene is 4 mm long there would be 250 genes in a meter, or 2,500 in 10 meters. Actually, some genomic DNA is "nonsense." Most bacterial gnomes contain only 2,000 genes that code for proteins.

 

                        5.         This activity reinforces the supercoiling nature of DNA and the extreme smallness in diameter relative to its length that DNA must possess.

 

Have your students fold the lengths of thread in half several times until they have a "wad" of thread approximately 3 inches long. Insert one end of the folded thread into the longer section of capsule. Push the "wad" into the capsule with a twisting motion. The thread will fit into the capsule, but obviously takes up more (approx. 90%) than 10% of the volume.

 

                        6.         10.2m to represent the E. coli genome.10.2 x 1,000 = 10,200m to represent the human genome 10,200m divided by 1,609 m/mile = 6.3 miles to represent the human genome in miles.

 

Note: Remember that we have expanded linear lengths by 10,000X. An extension of this activity would be for the students to calculate real linear lengths of DNA. Merely divide values above by 10,000.

 

                        7.         3.0 x 106 x 3.4 x 10-10m = 10.2 x 10-4 = 1.02 x 10-3m.

 

                        8.         1.2 x 103 x 3.4 x 10-10m = 4.08 x 10-7m.

 

                        9.         3.0 x 103 x 3.4 x 20-10m = 10.2 x 10-7 = 10.2 x 10-6.

 

                        10.       1.02 x 10-3m x 1.0 x 104 = 4.08 x 103m =10.2mm.

 

                        11.       4.08 x 10-7m x 1.0 x 104 = 4.08 x 103m = 4.08mm.

 

                        12.       1.02 x 10-6m x 1.0 x 104 = 1.02 x 10-2m = 10.2mm

 

                        13.       3 x 109 ties x 2 ft per tie = 6 x 109 ft.

 

                        14.       The length (1,136,363.6 miles) of the genome divided by the equatorial circumference (24,000 miles) will equal 47.3 circuits around the earth.

 

15.Will vary depending on the size of yours students' textbook.



TITLE OF BIOTECHNOLOGY ACTIVITY:



C.    How Genes Make Proteins

                  (Adapted from "A Sourcebook of Biotechnology Activities", North Carolina Biotechnology Center.)

 

OBJECTIVES:

 

Upon completion of this activity, students should be able to:

 

                        1.         Name five types of proteins and describe their functional roles.

 

                        2.         Define and describe the two major processes of protein synthesis, including:

● where these processes occur and the major cellular structures involved.

● the roles of DNA, mRNA, tRNA and amino acids

 

                        3.         Use any two sequences of DNA and mRNA, or mRNA and tRNA to demonstrate the concept of complementary base-pairing.

 

                        4.         Recognize and understand these words:

protein, synthesis, template, complementary base pairing, amino acids, nucleotides, genetic code, transcription, translation, DNA, mRNA, TRNA, codon, anticodon.

 

MATERIALS AND/OR EQUIPMENT:

 

                                  photocopy masters (4 DNA, 4 MRNA, 4 TRNA and 2 amino acid)

                                  colored paper (card stock if possible), four colors, two pages of each color

                                  scissors

                                  letter-sized envelope

 

TIME REQUIRED FOR ACTIVITY:

 

One or Two class periods

 

BACKGROUND:

 

Naturally occurring mutations have created new genetic combinations since the origin of life. For centuries, humans have developed and used selective crossbreeding to improve organisms used for food, clothing, transportation, etc. Since the early 1970s, genetic engineers have developed molecular techniques to alter the genetic make-up of organisms. All three of these mechanisms involve changing the genetic make-up of organisms. How do these changes in an organism's genetic make-up (genotype) affect the trait that is expressed (phenotype)?

 

Proteins provide the structural and functional basis of life. They play a part in every conceivable life function:

 

                                  A structural protein called collagen helps make up cartilage and tendons. Another protein (keratin) is found in our hair and fingernails.

                                  Hemoglobin is a transport protein that carries oxygen through the body.

                                  Plant hormones such as auxins and gibberellins are proteins that enhance or regulate biochemical messages. Insulin stimulates our blood to remove sugar.

                                  Proteins that catalyze chemical reactions in organisms are called enzymes. Amylase helps us to digest starches, and RNA polymerase assist with the transcription process that you will learn about in this activity.

 

Proteins have different functions, activities, shapes and chemical nature. Proteins are long, chainlike molecules that assume twisted 3-dimensional shapes. Each link of a protein chain is a simple organic unit called an amino acid. There are 20 amino acids that are used to form protein chains. The proteins we eat are broken down and then rearranged into the proteins we need.

 

DNA, a type of nucleic acid, is a long, double-stranded molecule made up of units called nucleotides. One nucleotide consists of deoxyribose sugar, a phosphate group and one nitrogenous base. The sequence of nucleotides contains information necessary for making a chain of amino acids - a protein chain. That sequence of nucleotides is called a gene. Sometimes several DNA sequences work together to make a protein; a gene is not always one continuous stretch of DNA.

 

Protein synthesis involves two basic processes, transcription and translation, that make use of another nucleic acid, RNA. RNA, like DNA, is made up of a chain of nucleotides. In transcription, enzymes catalyze the transfer of DNA's information to messenger RNA (MRNA) molecules. The MRNA molecules then move out of the nucleus to the ribosomes, where protein synthesis occurs.

 

Translation is the process of decoding the transcribed DNA message contained in mRNA. A second type of RNA, a cloverleaf shaped molecule called transfer RNA (tRNA) is involved. In the cytoplasm of the cell, specific tRNAs attach to their particular amino acid. At the base of each tRNA molecule is a sequence of three nucleotides (anticodon) that will recognize a complementary set of three nucleotides on the mRNA molecule (codon). The tRNAs, bonded to their amino acids, move to the ribosome where the mRNA is attached. The tRNA's and mRNA's bond together, as do the amino acids. The mRNA's and tRNA's release from each other, and the sequence of amino acids that is left is what defines the protein.

 

In this activity you will act out the steps of transcription and translation in protein synthesis.

 

PROCEDURE:

 

Teacher Preparation

 

                        1.         Use the photocopy masters and colored paper to make the sheets from which the cards can be cut. The master sheets need to be photocopied as two-sided pages. The first two master sheets make one page, sheet 3-4 make the second page, etc. You should have two pages of DNA, mRNA and tRNA, and one page of amino acids when you finish.

 

                        2.         Cut out the code cards from the sheets (or have your students cut them out in class). Arrange the cutouts into four stacks: DNA, mRNA, tRNA and amino acids. Keep the "start" sequence (TAC) on top of the DNA stack, and the stop sequence (ATC) on the bottom.

 

                        3.         Store the cards in an envelope.

 

Teaching Tips

 

                        1.         Laminate the four stacks of cards for use with your other classes.

 

                        2.         Multiple copies of the cards can be used for quizzes, remedial flash cards or make-up challenges.

 

Teaching Procedure

 

                        1.         Have students do Background Reading. You may wish to discuss this as a class.

 

                        2.         Have students follow the procedure of the Activity Sheet. They set up the classroom, then act out Transcription and Translation.

 

                        3.         Have students complete Question Sheet (can be a homework assignment or done as a class).

 

Student Procedure

 

In this activity you will act out the steps of transcription and translation in protein synthesis.

 

A.Setting the scene.

 

▾ The classroom's floor, walls and ceiling are analogous to a cell membrane. The windows and doors represent the membrane's selectively permeable pores because they regulate the size of the objects that can enter or leave the "cell." (It would be hard for an elephant to come into the room!)

▾ Designate one area of the class as the "nucleus," where transcription occurs. Enclose it with chairs and desks to represent the nuclear membrane. Spaces between chairs and desks simulate nuclear ports that regulate the movement of MRNA in and out of the nucleus.

▾ All other non-nucleus areas in the room are "cytoplasm."

▾ In the cytoplasmic region designate and area of the class as the "ribosome," where translation occurs.

 

                        1.         The teacher will distribute the DNA sequences and their complementary MRNA codes to two groups of students. The cards are arranged in groups of three letters because the nitrogenous bases of the genetic code function as triplet-base units. The large letters on the cards refer to first letter of the nucleotide bases (A=Adenine, C=Cytosine, G+guanine, T=thymine and U=uracil). The nucleotide base "thymine" found in DNA is replaced by the base "uracil" in all RNA molecules. Do not distribute the tRNA and amino acids cards until step #8.

 

                        2.         Review DNA structure and the concept of complementary base-pairing. Recall that the informational part of DNA is within the sequence of nitrogenous bases. Protein synthesis, just like DNA replication, does not begin until a stretch of DNA gets the signal to "unzip" and expose the nitrogenous bases.

 

                        B.        Transcription: The DNA message is transcribed into mRNA by the enzyme RNA polymerase.

 

                        1.         Assume that a strand of DNA has unzipped, exposing DNA's bases. In reality, one of the two strands in "active," while the other acts as a "dummy." You will be working with the active strand in this activity.

 

                        2.         Students with DNA cards should line up in the classroom area designated "nucleus." The student with the DNA card labeled "TAC" (the start sequence) should be on the left as the class sees him/her, and the "ATC" (the stop sequence) card should be on the right. All other DNA cards can be arranged in any order.

 

                        3.         RNA polymerase catalyzes the pairing of DNA's exposed bases with complementary RNA bases. (Remember, only one of the two DNA strands is active.) Students with the RNA cards should match their 3-letter sequence with the 3-letter sequence of the DNA cards. The 3-base mRNA sequence is called a codon.

▾ RNA cytosine always pairs with DNA guanine.

▾ RNA uracil (Remember: "U" substitutes for "T") always pairs with DNA adenine.

▾ RNA adenine always pairs with DNA thymine.

▾ RNA guanine always pairs with DNA cytosine.

 

                        4.         After everyone matches up their cards, use the following table to check the DNA/RNA pairs. Notice that of the nine pairs, the stop and start pairs are the only ones that must be in a certain position.



DNA

TAC

GGC

TTA

CAG

CTC

GAT

AGG

CCG

ATC

Transcription

mRNA

AUG

start

CCG

AAU

GUC

GAG

CUA

UCC

GGC

UAG

stop

 

 

                        5.         Students with the DNA cards can sit down, leaving a chain of RNA sequences. You have simulated the process of transcription as it happens in protein synthesis. Notice you have made a very short (shorter than in real life) complementary section of RNA that almost reflects the exact opposite of the DNA code.

 

                        C.        Translation: The mRNA message is translated into a chain of amino acids called a protein via enzymes and tRNA.

 

Notice that the tRNA cards are also arranged in groups of three letters. The three-base sequence of tRNA is called an anticodon. In addition, each tRNA card had a 3-letter abbreviation (in the "arrow" part of the card) for one of the 20 amino acids. This activity includes only seven of the 20 protein-building blocks called amino acids.

 

                        1.         If there are enough students, distribute the tRNA and amino acid cards to two new groups of students. Students with either type of card should be randomly scattered in the "cytoplasm."

 

                        2.         Students with the tRNA cards should find their specific amino acids. For example, the tRNA anticodon card "GGC" with the letters PRO should find the amino acid PROLINE. Simultaneously, students with mRNA cards should walk out the "nucleus" through the rows of desks/chairs that represent the nuclear pores and stop in the area designated "ribosome."

 

                        3.         After the tRNA students find the students with amino acid cards, both students should proceed over to the mRNA card that matches the tRNA card. Use the rules of complementary base-pairing:

▾ RNA cytosine always pairs with the RNA guanine.

▾ RNA uracil always pairs with the RNA adenine.

▾ RNA adenine always pairs with RNA uracil.

▾ RNA guanine always pairs with RNA cytosine.

 

                        4.         Messenger RNA and tRNA anticodon sequences are given below based on the original DNA sequence given in step #7. Except for the start and stop cards, student cards will not necessarily be in this order across the row, but they should match up vertically (e.g., mRNA'S "CCG" will pair with tRNA'S "GGC"). Use the following table to check the mRNA/tRNA pairs:



mRNA

AUG

CCG

AAU

GUC

GAG

CUA

UCC

GGC

UAG

Translation

tRNA

and

amino

acid

UAC

met

GGC

pro

UUA

asn

CAG

val

CUC

glu

GAU

leu

AGG

ser

CCG

gly

AUG

stop

 

 

                        5.         As each tRNA anticodon finds its corresponding codon on the mRNA strand, the tRNAs detach from the amino acids. The amino acids remain at the ribosome and form a peptide bond with the amino acid brought by the previous tRNA. Two or more amino acids linked in this way are called polypeptides. Translation is complete when a sequence of mRNA information translates into a polypeptide. A protein is one or more polypeptide chains linked together.

 

QUESTION(S) FOR INVESTIGATION:

 

                        1.         Name five types of proteins and describe their functional roles.

 

a.

 

b.

 

c.

 

d.

 

e.

 

                        2.         Compare and contrast:

 

▾ mutations, selective breeding and genetic engineering



 

▾ DNA reliction and protein synthesis



 

▾ transcription and translation



 

                        3a.       What is the difference between a DNA sequence of codons and an RNA sequence of codons?




 

                         b.         Why do we say that DNA determines the structural arrangement of proteins?




 

                        4.         Suppose an individual has a nutrient deficiency due to poor diet and is missing a particular amino acid. Which process of protein synthesis would be more affected? Why?



 

                        5a.       What is the relationship of the "active" and "dummy" stands of DNA? What do they look like?







 

                         b.         Would there be a problem if the dummy strand were used to make a protein? Why?







 

                        6a.       Given below are some tRNA anticodons/amino acid relationships and a stretch of imaginary DNA. Fill in the missing boxes in the chart below by writing the correct mRNA codons, tRNA anticodons and amino acids.



Use the following

tRNA/amino acid relationships

GGC

pro

UUA

asn

CAG

val

CUC

glu

GAU

leu

AGG

ser

CCG

gly

DNA:

TAC

AGG

GGC

CTC

TTA

CAG

CTC

GAT

AGG

GAT

ATC

mRNA

 

 

 

 

 

 

 

 

 

 

 

tRNA

 

 

 

 

 

 

 

 

 

 

 

Amino

Acid

met

start

 

 

 

 

 

 

 

 

 

stop

 

                         b.         What are the similarities between the DNA sequence and the tRNA sequence?





 

                        7.         A new and exciting branch of biotechnology is called protein engineering. To engineer proteins, molecular biologists work backward through the protein synthesis process. They first determine the exact sequence of the polypeptides they want, and then create a DNA sequence to produce it. Use the rules of transcription and translation to "engineer" the peptide sequence below. Fill in the rows for tRNA anticodons, mRNA codons and DNA.





Use the following

tRNA/amino acid relationships

GGC

pro

UUA

asn

CAG

val

CUC

glu

GAU

leu

AGG

ser

CCG

gly

DNA:

 

 

 

 

 

 

 

 

 

 

 

 

mRNA

 

 

 

 

 

 

 

 

 

 

 

 

tRNA

 

 

 

 

 

 

 

 

 

 

 

 

Amino

Acid

met

start

leu

val

pro

gly

asn

ser

glu

glu

pro

val



stop

 

ANSWERS FOR QUESTIONS FOR INVESTIGATION:

 

                        1.         Collagen, keratin - structural Hemoglobin, serum albumin -transport Hormones (auxins, gibberellins, insulin) -chemical messengers Enzymes (RNA polymerase, amylase) - catalyze biochemical reactions Actin, myosin -contractile, found in muscles Antibodies - fight foreign substances in the body

 

                        2.         ▾ mutations, selective breeding and genetic engineering: Compare: All are mechanisms of genetically modifying organisms. Contrast: Mutations occur naturally (or can be induced), as a change in a single base pair, gene or whole chromosome; selective breeding is man-made and occurs at the organism level; does not change genetic make-up of organisms involved in breeding event, but alters the genetic makeup of the population or gene pool; genetic engineering is man-made and occurs at the base pair or gene level. ▾ DNA replication and protein synthesis: Compare: Both require and unzipping of the DNA molecule. Contrast: The function of DNA replication is to make more DNA, occurs in the nucleus; the function of protein synthesis is to make proteins, occurs partly in the nucleus and partly in the cytoplasm. ▾ transcription and translation: Compare: Both are major processes of protein synthesis. Contrast: Transcription occurs in the nucleus, involves DNA and mRNA; translation occurs in the cytoplasm, involves mRNA, tRNA and amino acids.

 

                        3a.       Uracil in RNA substitutes for thymine in DNA.

                         b.         The order of nucleotides in DNA dictates the order of nucleotides in mRNA. The linear sequence of codons in the mRNA dictates the order of tRNAs. Since specific amino acids are attached to specific tRNAs, DNA also dictates the order of amino acids. Amino acids are the building blocks of proteins.

 

                        4.         Translation. Because transcription involves DNA and mRNA, and can occur without the presence of a particular amino acid. Translation depends on the presence of a particular amino acid. Translation depends on the presence of amino acids to build proteins.

 

                        5a.       They are exact opposites of each other.

                         b.         Because the opposite message would be transcribed and translated and the desired protein not produced.

 

6a.

Use the following tRNA/amino

acid relation

GGC

pro

UUA

asn

CAG

val

CUC

glu

GAU

leu

AGG

ser

CCG

gly

 

DNA:

TAC

AGG

GGC

CTC

TTA

CAG

CTC

GAT

AGG

CCG

GAT

ATC

mRNA

AUG

UCC

CCG

GAG

AAU

GUC

GAG

CUA

UCC

GGC

CUA

UAG

tRNA

UAC

AGG

GGC

CUC

UUA

CAG

CUC

GAU

AGG

CCG

GAU

AUC

Amino

Acid

met

start

ser

pro

glu

asn

val

glu

leu

ser

gly

leu

stop

 

                            b.         They are the same except for "T" and "U".

 

 

 

 

 

 

7.

 

Use the following

tRNA/amino acid

relationships:

GGC

pro

UUA

asn

CAG

val

CUC

glu

GAU

leu

AGG

ser

CCG

gly

DNA:

TAC

GAT

CAG

GGC

CCG

TTA

AGG

CTC

CTC

GGC

CAG

ATC

mRNA

AUG

CUA

GUC

CCG

GGC

AAU

UCC

GAG

GAG

CCG

GUC

UAG

tRNA

UAC

GAU

CAG

GGC

CCG

UUA

AGG

CUC

CUC

GGC

CAG

AUG

Amino

Acid

met

start

leu

val

pro

gly

asn

ser

glu

glu

pro

val

stop

 

 

RESOURCES:

 

Kits

Demonstration kit: DNA Made Easy

Catalog #17-1040. Carolina Biological

Supply Company (2700 York Rd., Burlington, NC 27215).

 

Audio/Visuals

6-part videotape: Protein Synthesis.

Catalog #84-2489. TV Ontario (TV Ontario, 143 W. Franklin St.,

Suite 206, Chapel Hill, NC 27514).

 

 

 


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TITLE OF BIOTECHNOLOGY ACTIVITY:

 

D.    DNA MOLECULE MODEL

                  (Adapted from "Biotechnology in Agriculture", MAVCC, Stillwater, OK.)

 

OBJECTIVES:

 

Upon completion of this activity, students should be able to:

 

                        1.         Create a DNA molecule model.

 

                        2.         Explain how the DNA molecule model works.

 

MATERIALS AND/OR EQUIPMENT:

 

                                  1 -- 6-inch square block of 2" x 6" x 6" wood

                                  1 -- 24-inch long 1/4" diameter dowel

                                  2 -- sheets of tabs (included)

                                  2 -- approx. 4-foot long strips of computer paper edging

                                  Tacky glue or glue stick

                                  Scissors

                                  Protractor

                                  Electric drill and 1/4" drill bit

 

TIME REQUIRED FOR ACTIVITY:

 

One or Two class periods

 

BACKGROUND MATERIAL:

 

The unique structural arrangement of DNA allows for storage and transmission of all genetic information from one generation to another. This model will be a visual representation of DNA's structure.

 

Directions: Build a model of a DNA molecule by using the following steps. Refer to the photograph of the completed model at the end of the procedure steps for a better idea of what you are trying to build.

 

PROCEDURE:

 

            A.        Build the support stand.

 

                        1.         Drill a 1/4" hole in the center of the wood block.

 

                        2.         Insert the dowel in the hole and stand the assembly upright.

 

                        Figure 1 -- Stand support

 

ole.gif

                 

 

B.Attach the support tabs to the dowel.

 

                        1.         Cut 26 support tabs labeled "S" from the sheet provided. Cut on the solid lines surrounding the "S." Do not cut on the center line.

 

                        Figure 2 -- Support tabs 

                       

ole1.gif

                        2.         Align the dashed "S" line with the long axis of the dowel and glue the first tab in place at the bottom of the dowel.

 

                        3.         Use a protractor to rotate the second tab 36 degrees from the first and glue it in place just above the first tab. This 36-degree rotation mimics the actual rotation of a DNA molecule where 10 nucleotide base pairs make one full rotation - 36º x 10 = 360º (one full rotation).

 

Note: A copy of a modified protractor is included on the last page of this assignment. Use it for measuring the 36-degree rotation if a protractor is not available.

 

                        4.         Continue attaching all 26 of the tabs rotating each 36 degrees from the one below it.

 

Attach the nucleotide bases in the correct sequence.

 

                        1.         Cut out 17 "T-A" tabs from the sheets provided. Do not cut on the dashed line.

 

                        2.         Cut out 9 "C-G" tabs from the sheets.

 

                        3.         Starting at the bottom, attach the nucleotide tabs to the support tabs in the exact arrangement shown in the diagram below.

 

            Note:   The diagram is linear but your model will rotate. Keep track of left and right as you move up the model. The easiest way to do this is to rotate the stand as you move up.

 

Attach the deoxyribose-phosphate "backbone" to the ends of the nucleotide bases.

 

                        1.         Bend each of the nucleotide base tabs back 90 degrees at the dashed line.

 

                        2.         Glue one strip of computer paper edging to each tab end on the right

                                    side. This will coil upward as you progress.

 

                        3.         Repeat step 2 with the other computer strip for the left side.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This model represents a piece of DNA showing the nucleotide bonds and the deoxyribose-phosphate backbone. The double helix coiled structure is a reasonably accurate representation of the actual structure of a DNA molecule.

 

                        ┌-T      >         A-| 

                        |-C       (          G-|

                        |-T       >         A-|

                        |-T       >         A-|

                        |-C       (          G-|

                        |-G       )          C-|

                        |-T       >         A-|

                        |-A       <         T-|

                        |-C       (          G-|

                        |-T       >         A-|

                        |-T       >         A-|

                        |-A       <         T-|

                        |-A       <         T-|

                        |-G       )          C-|

                        |-C       (          G-|

                        |-A       <         T-|

                        |-T       >         A-|

                        |-A       <         T-|

                        |-A       <         T-|

                        |-C       (          G-|

                        |-G       )          C-|

                        |-T       >         A-|

                        |-T       >         A-|

                        |-A       <         T-|

                        |-A       <         T-|

                        |-C       (          G-|

 

 


DNA model - Your model will look similar to this structure

dna.gif

 


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