Read Part A first, and answer all the questions asked in this part of the exercise (20 points). Once you have done this part go to Part B and answer all questions (30 points). Finally read the paper identified for you in part C and write a report according to the instructions given (50 points). You should hand in the journal paper along with your report. All reports and answers have to be typed (12 size characters). Part A.The Gene cloning technique permits biologists to produce a sufficient quantity of a single gene so that they can study the gene’ detail DNA sequence, as well as the gene function(s). Cells containing cloned genes can be induced to produce large quantities of the protein coded for by the cloned gene. By having the protein in large amounts, one can study the protein in detail or use it for its commercial value. For example, scientists can produce large quantities of proteins that are used as enzymes in manufacturing and processing food commercially. Other proteins produced by gene cloning, such as human growth hormone (HGH), are medically useful since children deficient in HGH that would have remained short (dwarfs) can now be treated and reach normal height. The simplest way to clone a gene is to induce bacteria to take up the gene and replicate the “foreign” DNA when they replicate their own DNA. Because bacteria can replicate so rapidly, this generates huge numbers of copies of the foreign gene quickly and easily. Vectors are tools used by molecular biologists to insert pieces of foreign DNA into host cells. Two naturally occurring vectors are bacteriophages and plasmids. Bacteriophages (or phages, for short) are viruses that infect bacterial cells by injecting their genetic material into the bacterial cell. Plasmids are small pieces of circular DNA that are sometimes present in bacteria in addition to the main bacterial genome of circular DNA, which is larger. Many bacteria readily absorb plasmids from the environment. Like the main bacterial genome, plasmid DNA contains genes that code for specific proteins, which normally are not necessary for the bacterium’s survival. However, these proteins may confer an advantage under certain conditions. For example, genes for antibiotic resistance are often found in plasmids. One way scientists can insert foreign DNA into host cells is by combining the “foreign” DNA with the DNA of a vector. The resulting DNA is called recombinant DNA. If the recombinant DNA gets inside a host cell (bacterial or eukaryotic cell), it can replicate along with the DNA of the host cell and thus the recombinant DNA is amplified with each cell division. The phages and plasmids used for gene cloning are usually artificial constructs, which often contain additional DNA that has been added to the original DNA of the vector. For example, some plasmids contain pieces of phage DNA that make them especially versatile. These plasmids are called phagemids. Another type of plasmid vector with both plasmid and phage components is the cosmid, which allows the cloning of especially large stretches of DNA. The following procedure describes how scientists can combine phage DNA with a plasmid called pUC18 to form a recombinant vector. It also explains one way scientists can screen E. coli cells (a bacterium commonly used in genetic engineering experiments) to determine which cells have absorbed the recombinant vector: 1. The plasmid pUC18 includes 2 genes. One gene confers resistance to the antibiotic ampicillin, which enables bacteria to grow in the presence of ampicillin, while bacteria that lack this gene are not. The second gene is called the lacZ gene. The lacZ gene codes for the enzyme β–galactosidase, which is normally used by E. coli cells to digest lactose. When bacteria with this gene are grown in a medium containing a synthetic analog of lactose called Xgal, the enzyme digests Xgal and releases compound “X”, which is blue. Therefore, bacteria with a functional lacZ gene will produce a blue color when grown in the presence of Xgal. 2. A restriction enzyme is used to linearize the pUC18 plasmids. This means each circular plasmid is cut at a single location to produce a linear piece of DNA. The restriction enzyme that is used is called EcoRI. This enzyme will cut the plasmid at a single point, somewhere within the lacZ gene. The same restriction enzyme is used to cut the phage DNA into fragments: Note that only combination “C” contains recombinant DNA made by joining DNA from 2 different sources, in this case, the plasmid and the phage. All the other combinations (A, B, D, and E) contain non-recombinant DNA (DNA from one source only.)4. E. coli cells are incubated with the DNA vectors. During incubation, some E. coli cells will absorb no DNA, some will absorb non-recombinant DNA (either plasmid DNA alone or phage DNA alone), and some will absorb a recombinant plasmid. How do scientists know which bacterial cells absorbed a recombinant plasmid? 5. To screen for bacterial cells that contain a recombinant plasmid, the bacteria are placed on growth medium that contains ampicillin and Xgal. There are 3 possible outcomes:a. E. coli cells that absorbed no DNA at all, as well as those that absorbed only phage DNA (B or E above) will lack the gene for ampicillin resistance (present only on the plasmid.) Therefore, these bacteria will not grow and no bacterial colonies will be formed.b. E. coli cells that absorbed only plasmid DNA (A or D above) will have the gene for ampicillin resistance and the lacZ gene. The gene for ampicillin resistance allows them to grow and form colonies. The lacZ gene produces β–galactosidase which digests Xgal to releases compound “X”, which is blue. Therefore, these E. coli cells will produce blue colonies.c. E. coli cells that absorbed a recombinant plasmid (C above) will have the gene for ampicillin resistance (present on the plasmid.) Therefore they will grow and produce colonies. However, the lacZ gene on the plasmid has been cut into 2 pieces by the restriction enzyme and a piece of phage DNA has been inserted between the 2 parts of the lacZ gene. This inactivates the lacZ gene, meaning no β–galactosidase will be produced. Therefore, Xgal is not digested, no blue color is produced, and the colonies remain white. How efficient is gene cloning for producing many copies of a gene?Under ideal conditions, E. coli can undergo binary fission (divide) every 20 minutes. Imagine that one E. coli cell with a single plasmid containing a foreign gene was inoculated into a flask of nutrient broth and incubated for 12 hours.1.Q. How many copies of the foreign gene could it produce? How are restriction enzymes and DNA ligase used to combine foreign DNA with a plasmid?In order to use a plasmid to insert “foreign” DNA into a bacterial cell, two steps are required:1) The “foreign” DNA must be combined with a plasmid. The resulting DNA is called recombinant DNA and the procedure used to make recombinant DNA is called recombinant DNA technology.2) A bacterial cell must absorb the recombinant plasmid For the first step, restriction enzymes and DNA ligase are used. Restriction enzymes are naturally occurring enzymes that cut DNA. Many restriction enzymes have been isolated from bacteria, providing a valuable tool for molecular biologists. Each restriction enzyme cuts DNA only where a specific sequence of base pairs occurs.Because these 2 fragments have unpaired bases which can potentially pair with complementary bases, we say they have “sticky ends.” By cutting DNA from two different sources with the same restriction enzyme, DNA fragments with complementary sticky ends are formed. When these fragments are mixed together, the complementary singlestranded ends may join together to form double stranded DNA by hydrogen bonding between the bases. Hydrogen-bonding between complementary bases is not sufficient to completely rejoin the DNA fragments. The broken bonds between the deoxyribose and phosphate groups that form the “side-rails” of the DNA double helix (the phosphodiester linkages) must also be repaired. DNA ligase is the enzyme, also isolated from bacteria, that catalyzes this reaction.
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