Chapter 13: Biotechnology

Transcription

1 Chapter Review 1. Explain why the brewing of beer is considered to be biotechnology. The United Nations defines biotechnology as any technological application that uses biological system, living organism, or derivatives thereof to make or modify products or processes. Since the brewing of beer involves a man-made fermentation process that manipulates living organisms (especially yeast), it meets the broad definition of biotechnology. 2. Restriction enzymes are sometimes called the immune system of bacteria. Explain how restriction enzymes protect bacteria from viruses. Restriction enzymes cut double-stranded DNA molecules, such as those injected by a bacteriophage, into smaller, noninfectious fragments, thus helping to protect the bacterial genome from DNA insertion. 3. How does a bacterium protect its own DNA from being cleaved by restriction enzymes? A bacterium protects itself by modifying the restriction sites in its own DNA. Specifically, enzymes called methaylases add methyl groups to certain bases at the restriction sites after the chromosome has been replicated, and the restriction enzymes do not recognize or cut the methylated sites in the host s DNA. Nonmethylated phage DNA is efficiently recognized and cleaved, thus protecting the bacterium. 4. Explain how sticky ends are be used to join DNA from two different organisms. Restriction enzymes can cut two DNA strands in a manner that leaves a short sequence of singe-stranded DNA at each cut end. These overhangs are called sticky ends because they bond with sequences that are made up of the complementary bases on other DNA molecules, including those from other organisms. 5. Describe the type of atomic interaction that hold together two fragments of DNA after they have been joined with sticky ends. What characteristics can strengthen this interaction? The sticky ends first hydrogen bond to complementary sticky ends from another DNA fragment, and the resulting recombinant DNA can then be sealed with DNA ligase. This enzyme creates covalent phosphodiester bonds between the strands and is the same enzyme that joins Okazaki fragments during DNA replication. 6. Describe how DNA is rearranged by meiosis. Compare that description to a description of how recombinant DNA is produced. DNA rearrangement can occur during meiosis by crossing over where genetic material is exchanged between two homologous chromosomes during prophase I and independent assortment of the chromosomes. No cloning vector or second organism is involved in the process; it is simply the rearrangement of DNA that already exists within the genome of a cell. Recombinant DNA is produced by combining the DNA of two organisms, the host and a cloning vector, such as a virus or bacterium. Restriction enzymes are utilized to cut the DNA fragments, and DNA ligase catalyzes newly joined fragments to create the recombinant stand of DNA. 7. Plasmids are small circular pieces of DNA. A particular plasmid (puc 19) is cut with the AvaII restriction endonuclease (enzyme). The puc 19 plasmid has a size of 26,867 base pairs. The Ava II cuts the plasmid at two locations: 1837 bp and 2059 bp.

2 a. The circle below represents a plasmid. Draw a map of this plasmid showing the location of these cuts and the size of the fragments. b. On the gel below, sketch the results that you would expect to see if the fragments were run using gel electrophoresis. Label which end is positive, and which is negative. 8. A DNA molecule of 12,000 bp (12 kilobase, or kb) is cut by restriction enzymes as follows: SIZES OF CONDITION FRAGMENTS (KB) Enzyme A 2, 10 Enzyme B 2, 10 Enzymes A+B 2, 8 a. Sketch the results of the three cuts on the electrophoresis gel below. b. Indicate on a linear map where each enzyme cuts the DNA.

3 9. What is the advantage of using yeast over bacteria as a model organism for the introduction of recombinant DNA? The advantages of using yeast in recombinant DNA work include a rapid life cycle (completion in 2-8 hours), ease of growth in the laboratory, and a relatively small genome size. Additionally, yeast is an eukaryotic organism unlike the prokaryotic bacteria. Also, yeasts have many of the characteristics, including protein interactions, of interest in studies of other eukaryotes. 10. Explain the benefits of using a viral vector in place of a plasmid vector. Constraints on plasmid replication limit the size of the new DNA that can be inserted into a plasmid to about 10,000 bp. This works for many prokaryotic genes, but many eukaryotic genes of interest are much larger and require a vector that can accommodate larger DNA inserts. In addition, since viruses infect cells naturally, they offer a great advantage over plasmids, which often require artificial means to enter host cells. 11. What is a reporter gene? Give an example. A reporter gene is a genetic marker included in recombinant DNA to indicate the presence of the recombinant DNA in a host cell. It can also be attached to a promoter to study how promoters function under different conditions or in different tissues of a transgenic multicellular organism. An example is the luciferase gene, the protein from which is luminescent and therefore visible evidence of gene insertion. 12. Outline the steps that a researcher would need to follow if he or she wished to insert a foreign piece of DNA into a plasmid. Foreign DNA can become part of a replicon within a host cell by simple insertion into a random host chromosome or by introduction of a DNA sequence via a vector that can integrate into the host chromosome. To obtain the DNA fragments for insertion, restriction enzymes are used to make cuts in double-stranded DNA of the source organisms. These fragments are then be separated by size using gel electrophoresis. DNA fragments from different sources can then be used to create recombinant DNA by splicing them together using DNA ligase. Bacterial plasmids are commonly used as hosts for the recombinant DNA. Insertion of the recombinant DNA into plasmids can be accomplished by treating the host cells chemically to make their outer membranes more permeable to allow entrance by diffusion, or by electric shock treatment (electroporation) to create temporary pores in the membranes through which the DNA can enter. A reporter gene should be used so that the researcher can detect which cells have incorporated the new DNA molecules into their genome. 13. A researcher has a small gene that she wants to insert into E. coli for expression. This gene has been cut on both ends with the EcoRI restriction enzyme. The researcher obtained a plasmid vector that includes two functional genes: the β-galactosidase gene (lacz) that codes for an enzyme that can convert the colorless substrate X-gal into a bright blue product, and the gene for resistance to the antibiotic ampicillin (amp). The EcoRI restriction enzyme site on the lacz gene was cut to insert the small gene of interest. The researcher grew E. coli bacteria with the following treatments: with no plasmid added with plasmid alone with recombinant plasmid (gene added inside the lacz gene site).

4 a. Draw a diagram of the uncut and the recombinant plasmid below showing how the different genes will look. b. Complete the table below with the results the researcher will see after growing the E coli bacteria on media containing X-gal and the antibiotic ampicillin. Treatment Grown with no plasmid Transformed with plasmid Transformed with recombinant plasmid Selection by ampicillin (colonies present or absent) Absent Present Present Color of colonies if present (blue or white) N/A Blue White 14. Use the figure at the right to explain how knock-out genes are produced. Label FIVE parts of the figure (A, B, C, D and E) as you refer to them in your discussion. A: The targeted gene is inactivated by insertion of the reporter gene. B: The vector is inserted into a stem cell. C: Then the targeted genes on the vector and genome of the host cell line up via homologous sequence recognition. D: Recombination occurs. The inactivated gene Is now in the host genome, and the vector is lost during cell division. E: The stem cell is transplanted into an early embryo of the host organism, where it replaces most of the embryo s cells during development. 15. How does antisense RNA regulate the expression of DNA? Gene expression can be controlled in nature by the production of short, single-stranded RNA molecules (micrornas) that are complementary to specific mrna sequences. This type of complementary molecule is called antisense RNA because it binds by base pairing to the sense bases on the mrna. Even though the

5 gene continues to be transcribed, the resulting partially double-stranded RNA hybrid effectively halts the translation of the mrna bound to microrna. 16. Identify three reasons why biotechnology is advantageous over more traditional plant breeding techniques. The ability to identify specific genes the use of genetic markers allows breeders to select for specific desirable genes, making the breeding process more precise and rapid. The ability to introduce any gene from any organism into a plant or animal species this ability, combined with mutagenesis techniques, greatly expands the range of possible new traits. The ability to generate new organisms quickly manipulation of cells in the laboratory and regenerating a whole plant by cloning is much fast than traditional breeding. 17. Many people oppose the insertion genes from one organism onto another to produce genetically modified organisms (GMOs). Describe THREE concerns about GMOs. Genetic manipulation is an unnatural interference with nature while it can be argued that this is true, it is also true that all crop are unnatural because they come from artificially bred plants growing in a manipulated environment. Genetically altered foods are unsafe to eat advocates argue that only single genes with specific known functions are added, but evidence may not be convincing to many. Genetically altered crop plants are dangerous to the environment there are concerns about the transfer of altered genes from crops to native species, adversely affecting ecosystem balance. Science Practices & Inquiry 18. A fellow student tells you that organic chemists frequently will try to substitute one functional group for another on a large molecule to try and make minor changes to the molecule. Cocaine, for example, has four functional groups that can be substituted for or deleted to make different pharmaceutical drugs of different potency. So if chemists can make new drugs this way, why can t biologists simply change one amino acid in a protein to try and make new proteins? a. Explain why changing one amino acid in a protein is not as easy as changing a functional group on a large molecule. The principal determinant of a protein s function is its shape, determined by its sequence of amino acids. The primary structure of a protein is its amino acid sequence, determined by the sequence of nucleotides in its gene (DNA). Although the protein s tertiary structure is set up by the amino acid sequence, it may be difficult for the newly formed polypeptide molecule to retain its shape in its cytoplasmic environment where it may interact with hydrophobic amino acids. Instead of associating with its own hydrophobic amino acids of its own, it may instead associate with other nearby proteins. This could lead to alteration of its intended tertiary structure, affecting the function of the protein. To avoid this situation, cells contain molecular chaperones that stabilize newly-formed polypeptides as they fold into their correct structures. In complex proteins, special chaperones called chaperonins are required to help proteins retain their prescribed shapes. Chaperonins are hollow cylinders into which new proteins fit as they fold. They are lined with hydrophobic molecules that bind with the protein to direct proper folding and encapsulate it within the chaperonin. Therefore, changing one amino acid in any given protein is fraught with the potential to alter the function of the entire protein by changing its tertiary structure. In addition, molecular binding of the amino acids in the protein to the chaperonin would make removal of an amino acid difficult. Other large molecules, such as carbohydrates, are easier to manipulate, as they are simpler chains that do not have the three-dimensional issues nor the chaperonin binding that make proteins more complex.

6 b. If it were possible to change an amino acid in a protein, identify two biotechnology techniques that could be used to determine if a change had occurred or to determine the properties of the new molecule. One method to detect a change in an amino acid for a protein is the use of a reporter gene, a selectable marker gene whose expression is easily assayed. Some reporter genes can be detected visually, such as those from plasmid vectors with the gene for green fluorescent protein (GFP), which emits green light when exposed to ultraviolet light. Another method to mark recombinant DNA is to inactivate a gene. Selectable marker genes are used to select for bacteria that have taken up a plasmid. A second reporter gene allows for the identification of bacteria harboring the recombinant plasmid. The resulting phenotype(s) reveal the presence of the recombinant DNA. This method is used to manipulate the DNA in cells with antibiotic resistance genes. Only cells carrying the antibiotic resistance gene can grow in the presence of that antibiotic. During the process of making recombinant DNA, a second antibiotic resistance gene can be used to identify cells that carry the vector with the desired insert. If the second antibiotic resistant gene is inactivated by the insertion of additional DNA, then cells carrying copies of the vector with the inserted DNA can be identified by their sensitivity to that antibiotic.