Biotechnology: Manipulating and Cloning DNA PDF

8.1 Manipulating and Cloning DNA diabetes mellitus a disease in which the Diabetes mellitus is a disease caused by a deficiency in the production of insulin, resulting blood glucose level is too high because of in elevated blood glucose levels. Approxi...

8.1 Manipulating and Cloning DNA diabetes mellitus a disease in which the Diabetes mellitus is a disease caused by a deficiency in the production of insulin, resulting blood glucose level is too high because of in elevated blood glucose levels. Approximately 5 % of all deaths are caused by diabetes. the insufficient production or activity of the Type 1 diabetes results when the body fails to produce insulin. Many people with type 1 hormone insulin diabetes require insulin by mouth or by injection to prevent the disease from seriously type 1 diabetes a type of diabetes damaging their body. Type 2 diabetes results when the body does not produce enough caused by an inability to produce insulin insulin or when it cannot use the insulin produced. The two main risk factors associ- ated with type 2 diabetes are a genetic predisposition and being overweight. In North type 2 diabetes a type of diabetes caused America, type 2 diabetes is on the rise. Canadians eat more refined foods and sugar, and by low insulin or an inability to use insulin less natural foods and fibre, than ever before. As a consequence, more people are suf- fering from insulin resistance, obesity, and type 2 diabetes. Diabetes can lead to many health complications, including blindness, organ problems, limb amputations, and early death. You will learn about the differences between type 1 and type 2 diabetes in Unit 4. Insulin was isolated by two researchers, Dr. Frederick Banting and Dr. Charles Best, at the University of Toronto in 1922. Helen Free, born a year after the discovery of insulin, invented a method to analyze the blood sugar by dip-test urinalysis. (Before this, doctors tasted a patient’s urine. Sweet urine was an indicator of dia- betes.) This allowed people with diabetes to monitor their blood sugar level easily at home. These two innovations have vastly improved the lives of diabetics. Insulin was initially collected from the pancreases of pigs and cows. Some people, however, had allergic reactions to this insulin, even though it was necessary to keep them alive. Scientists attempted to mass-produce actual human insulin. They discov- ered that they could insert the human insulin gene into bacteria, and the bacteria genetic engineering the intentional would make human insulin. Genetic engineering is the intentional alteration of a production of new genes and alteration genome by substituting or introducing new genetic material into the genome. It is of genomes by the substitution or now used to mass-produce human insulin economically and safely. introduction of new genetic material Bacteria are versatile tools for genetic engineers because they reproduce quickly and often, are relatively inexpensive to maintain, and contain plasmids (small cir- cular pieces of DNA that replicate independently of the bacteria’s chromosome). The bacteria Escherichia coli (commonly known as E. coli) are very common in the human body and are used to produce biosynthetic human insulin. The human gene that codes for insulin is inserted into an E. coli plasmid. The E. coli transcribes and translates the piece of human DNA to make the human protein insulin, which is then recombinant DNA a DNA strand that is harvested from the bacteria. Safflowers are also used to produce human insulin. After created using DNA pieces from two or the human insulin gene in the plasmid is inserted into the E. coli or the safflower, the more sources bacterial or plant DNA contains genes of two species and is called recombinant DNA. Restriction Enzymes The first step in genetic recombination is to isolate, or cut out, a DNA fragment that restriction enzyme an enzyme that contains the desired gene. Scientists use restriction enzymes (also called restriction cuts DNA at a specific location in a endonucleases), which occur naturally in prokaryotic cells. A restriction enzyme acts base sequence; also called restriction like molecular scissors, cutting a DNA molecule at specific locations. Each restriction endonuclease enzyme recognizes a specific sequence of nucleotides on a DNA strand. This is known recognition site a sequence of bases as the recognition site for that particular enzyme. When the restriction enzyme cuts the on a DNA strand that restriction enzymes DNA molecule, the pieces it creates are known as restriction fragments. Hopefully, one of bind to the restriction fragments will contain the entire target gene, such as the gene for insulin. Each restriction enzyme cuts (or digests) at only one recognition site and in only one restriction fragment a fragment that is direction. For example, the enzyme EcoRI binds to a recognition site with the base-pair produced when a DNA strand is cut by a sequence 59-GAATTC-39 (Table 1 and Figure 1, next page). EcoRI then cuts the phospho- restriction enzyme diester bond in the DNA backbone between the G and the A. Notice that the recognition sites on the DNA are palindromic when you consider both strands (Figure 1). Another EcoRI enzyme makes the same cut in the complementary DNA strand. This leaves only a small number of the hydrogen bonds holding the DNA molecule together, allowing the DNA molecule to be easily separated, resulting in complementary “sticky” ends. 366   Chapter 8 Genetic Technologies NEL 8159_Bio_Ch08_pp364-421.indd 366 3/27/12 6:02 PM  Table 1 Examples of Restriction Enzymes Enzyme Outcome after restriction name Recognition site End type enzyme digestion EcoRI 59-GAATTC-39 sticky 59-G AATTC-39 39-CTTAAG-59 39-CTTAA G-59 XhoI 59-CTCGAG-39 sticky 59-G TCGAG-39 3'-GAGCTC-59 39-GAGCT C-59 HindIII 59-AAGCTT-39 sticky 59-A TCGAC-39 39-TTCGAA-59 39-CAGGT G-59 SmaI 59-ACCCGGGT-39 blunt 59-ACCC GGGT-39 39-TGGGCCCA-59 39-TGGG CCCA-59 EcoRI 5 - A T T C G A C G G A A T T C T T A A C G C G -3 1 EcoRI locates the 5 -GAATTC-3 recognition site in the top strands and cuts the DNA 3 - T A A G C T G C C T T A A G A A T T G C G C -5 backbone between the G and A bases. 5 to 3 direction of top strand 5 - A T T C G A C G G A A T T C T T A A C G C G -3 2 Top strand has been cut by the restriction enzyme but the two strands remain together. 3 - T A A G C T G C C T T A A G A A T T G C G C -5 5 to 3 direction of bottom strand 5 - A T T C G A C G G A A T T C T T A A C G C G -3 3 EcoRI locates a 5 -GAATTC-3 recognition site in the bottom strand and cuts the phosphodiester 3 - T A A G C T G C C T T A A G A A T T G C G C -5 bond (backbone), again between the G and A bases. EcoRI 5- A T T C G A C G G A A T T C T T A A C G C G -3 4 Both strands have been cut by restriction enzymes. The DNA separates into two 3- T A A G C T G C C T T A A G A A T T G C G C -5 fragments with “sticky ends.” Figure 1 How the restriction enzyme EcoRI cuts a DNA sequence Two possible outcomes result from a restriction enzyme cutting a DNA molecule. blunt end the end that remains after If cuts are made straight across the strand, blunt ends are created. If cuts are made in a restriction enzymes cut straight across a zigzag, sticky ends are created. For example, EcoRI produces sticky ends, whereas SmaI DNA strand; a blunt end is more difficult produces blunt ends (Table 1). Molecular biologists prefer to work with restriction than a sticky end to recombine with enzymes that produce sticky ends because the DNA fragments that are created are another strand easier to join to any other DNA strand that has been cut by the same enzyme. sticky end the end that remains after In 1970, Dr. Hamilton Smith, of Johns Hopkins University School of Medicine, was restriction enzymes cut on a zigzag across wondering how some bacteria resisted viral infections when he accidentally discovered a DNA strand; a sticky end of a DNA restriction enzymes. Since then, many such enzymes have been catalogued. At least fragment can form hydrogen bonds with 2500 restriction enzymes, with specificity for about 200 target sequences, have been iso- a complementary sticky end on any other lated, mostly from prokaryotic cells. About 200 of the most helpful restriction enzymes DNA molecule that has been cut by the are available commercially for use in molecular biology laboratories. WEB LINK same enzyme NEL 8.1 Manipulating and Cloning DNA   367 8159_Bio_Ch08_pp364-421.indd 367 3/27/12 6:02 PM  The biological function of restriction enzymes is to protect the cells in which they are found. When a virus injects its own DNA or RNA into a bacterial cell, the bacterial restric- tion enzymes destroy the viral nucleic acid by cutting it in pieces. The fact that scientists can use this function of restriction enzymes for DNA technology is indeed fortunate! Restriction enzymes are highly specific. Most have recognition sites for four to eight base pairs. The fewer base pairs that they need for the recognition sequence, however, the more cuts that are made in a DNA strand. In a random DNA sequence, the probability of 1 finding a particular four-pair sequence is one in 44 (256 or 0.4 %). The chance of finding 1 the six-pair sequence that EcoRI needs to perform a cut is one in 46 (4096 or 0.02 %). A restriction enzyme is named after its cell of origin, plus a Roman numeral if more than one restriction enzyme has been isolated from this species. For example, EcoRI (pronounced “eco-R-one”) was the first restriction enzyme isolated from E. coli, XhoI was the first enzyme isolated from Xanthomonas holcicola, and HindII and HindIII were the second and third enzymes isolated from Haemophilus influenzae. DNA Ligase DNA ligase is the enzyme that is used to join cut strands of DNA (Section 6.4). DNA ligase works best with sticky ends of DNA, but a second form, T4 DNA ligase, works well with blunt ends. The two DNA fragments must have overlapping complementary portions or be blunt ends that are properly aligned end to end. They will be comple- mentary if they were both generated using the same restriction enzyme, since each restriction enzyme cuts specifically at one sequence of base pairs. Hydrogen bonds form between the complementary bases, but this is not a stable arrangement. The DNA is not fully linked until phosphodiester bonds form between the backbones of the double strands. DNA ligase is the enzyme that makes this happen. DNA ligase works, as many organic reactions do, by promoting a dehy- dration reaction. Thus, water molecules are released when DNA ligase joins the DNA strands. Plasmids The next tools that are needed for recombinant DNA techniques are plasmids, small circular pieces of DNA that are found in bacteria (Figure 2). Plasmids replicate inde- pendently of the chromosomal DNA. They often contain genes that code for specific proteins, such as proteins that provide resistance to antibiotics such as ampicillin or protect from the toxic effects of certain heavy metals. In fact, when you hear about Figure 2 This electron micrograph bacteria that mutate quickly so that diseases become difficult to treat, it is often the shows plasmids inside E. coli. plasmid DNA that is mutating. A cell that is able to take up foreign (often plasmid) DNA, such as a healthy competent cell a cell that is able to take E. coli cell, is called a competent cell. A plasmid that has been designed to be a vehicle up foreign DNA from its surroundings for transferring foreign genetic materials into a cell is called a vector. The plasmid copy number, the number of copies of a plasmid within a bacterial cell, is variable and vector a DNA molecule that is used as a vehicle to transfer foreign genetic material is characteristic of a particular plasmid. If a plasmid with a high copy number has into a cell, for example, a plasmid been engineered to produce insulin, more insulin will be produced per cell because more copies of the insulin-producing gene are present. copy number the number of plasmids of If a fragment containing a target gene is created using the same restriction enzyme a specific type within a cell that a plasmid is cut with, the fragment will possess the same complementary ends as the linear plasmid. The foreign fragments and the plasmid fragments can be placed in the same solution, where they anneal because of the complementary sticky host cell a cell that has taken up a foreign ends. DNA ligase is then added to re-form the phosphodiester bonds between the plasmid or virus and has used its cellular fragments, resulting in a new circular piece of DNA that carries the foreign gene machinery to express the foreign DNA fragment. The plasmid is now recombinant DNA, a combination of the original cloned gene an identical copy of an plasmid DNA and the foreign DNA. The plasmid can be introduced into a host cell original target gene that can be made by (often a bacterial cell), where it will replicate to form many copies within the cell. The introducing the target gene into a host cell gene has been cloned because, as the plasmid replicates, many copies of the plasmid and having it copied with recombinant DNA will be produced. Cloned genes are identical multiple copies 368   Chapter 8 Genetic Technologies NEL 8159_Bio_Ch08_pp364-421.indd 368 3/27/12 6:02 PM  of a gene. Each copy of the plasmid will include a copy of the original inserted gene. The gene can now begin to express its function. For example, if the insulin gene is in the plasmid, the host cell will begin to produce insulin. The procedure that is used by researchers to verify that bacteria are carrying a target gene is outlined in Figure 3. Refer back to this figure as you continue. Later in this sec- tion, you will learn how the bacteria with the target gene are identified (hybridization). cell recombinant DNA molecules bacterium bacterial chromosome progeny gene of bacteria interest 1 Cut the DNA into fragments using restriction enzymes; one fragment should include the target gene. plasmid from bacterium inserted DNA fragment 2 Use an artificial engineered 3 Insert a gene fragment 4 Transform bacterial cells with recombinant 5 Identify the plasmid that is antibiotic resistant into each plasmid to make DNA plasmids. Each bacterium receives a bacterium containing and has a multiple cloning site. recombinant DNA. The different plasmid. As the bacteria divide, the plasmid with the Cut the circular plasmid to make recombinant DNA molecule the recombinant plasmids replicate, amplifying target gene inserted it linear. is the recombinant plasmid. the piece of DNA inserted into the plasmid. into it. Figure 3 An overview of cloning DNA fragments in a bacterial plasmid Restriction Maps Plasmid mapping has revolutionized molecular biology. A restriction map is a diagram restriction map a diagram that shows that shows the relative locations of all the known restriction enzyme recognition sites restriction enzyme recognition sites and on a particular plasmid and the distances, in base pairs (bp), between the sites. This the distances, measured in base pairs (bp), technique allows molecular biologists to determine which plasmids might be most between the sites suitable for a particular recombinant DNA procedure and to evaluate quickly the suc- cess of the cloning experiments. EcoRI Scientists often use a restriction map to determine which restriction enzyme they 200 bp HindIII will use to cut the plasmid. Figure 4 shows a restriction map that illustrates two rec- 500 bp ognition sites for the restriction enzyme EcoRI and one site for HindIII. This map also shows the number of base pairs between the cut locations of each enzyme. When EcoRI added up, we can see that the plasmid has a total length of 1800 base pairs. Figure 5 shows the results of exposing the plasmid in Figure 4 to the two restric- tion enzymes on their own and in combination. EcoRI cuts the plasmid in two 1100 bp plasmid (DNA loop) 200 bp Figure 4 A simple example of a 500 bp 500 bp restriction map 1800 bp 1300 bp 1100 bp (a) 1 fragment: 1800 bp (b) 2 fragments: 500 bp, (c) 3 fragments: 200 bp, 1300 bp 500 bp, 1100 bp Figure 5 Plasmid digestion with (a) HindIII only, (b) EcoRI only, and (c) EcoRI and HindIII NEL 8.1 Manipulating and Cloning DNA   369 8159_Bio_Ch08_pp364-421.indd 369 3/27/12 6:03 PM  locations and produces two fragments, while HindIII cuts the plasmid in only one location and produces one fragment. When the plasmid is exposed to both restriction enzymes, three fragments are produced. Because plasmids are circular, the number of fragments is always equal to the number of cuts. In the following tutorial, you will construct restriction maps from the given information. Tutorial 1 Constructing Restriction Maps A restriction map shows restriction enzyme recognition sites and the distances, measured in base pairs (bp), between the sites. You have learned how a restriction map can be used to predict the fragment lengths that will be produced by different digestions. In this tutorial, you will learn how geneticists use restriction enzymes to gather data and construct a restriction map. Sample Problem 1: Constructing a Restriction Map to Show Recognition Sites Construct a restriction map for Plasmid X to show the recognition sites for two commonly used restriction enzymes: EcoRI and BamHI. Step 1. First, two separate restriction enzyme digestions are performed on different samples of Plasmid X (two single digestions). Then a double digestion is performed on the plasmid with both restriction enzymes. After the digestions are performed, the fragment lengths are determined (Table 2). (This is done using gel electrophoresis, which you will learn about later in the chapter.) The length of the undigested plasmid is also determined. Notice that the sum of the fragments is always equal to the total length of the plasmid. Table 2 Results of Plasmid X Digested with EcoRI and BamHI Plasmid X Plasmid X Plasmid X Plasmid X digested digested digested with undigested with EcoRI with BamHI EcoRI and BamHI 1400 bp 1400 bp 600 bp 100 bp 800 bp 600 bp 700 bp Step 2. Draw Plasmid X restriction maps for the individual restriction enzymes (Figure 6). EcoRI Plasmid X Map BamHI Plasmid X Map 1400 bp 800 bp 600 bp 1 fragment: 1400 bp 2 fragments: 600 bp, 800 bp Figure 6 Step 3. Draw a restriction map for all the restriction sites. We do not know the location of the EcoRI recognition site relative to the BamHI sites. However, we do know the fragment lengths for the combined digestion (column 4 of Table 2). We can use this information to determine the relative locations of the restriction sites. First, we redraw a restriction map for one of the single digestions. It is easier to begin with the digestion that produced the greater number of fragments, so we redraw the BamHI map (Figure 7(a), next page). 370   Chapter 8 Genetic Technologies NEL 8159_Bio_Ch08_pp364-421.indd 370 3/27/12 6:03 PM  Next, we consider the fragment lengths produced by the double digestion. Note the following: There are two new fragment lengths: 100 bp and 700 bp One fragment length is unchanged: 600 bp Since we now have a fragment of only 100 bp in length, we know that the EcoRI site must be within 100 bp of a BamHI cut. Such a cut could occur within the 600 bp or 800 bp fragment (Figure 7(b)). However, because we also know that the remaining two fragments are 600 bp and 700 bp, we can infer that the EcoRI site is located within the 800 bp fragment. The final restriction map is shown in Figure 7(c). BamHI Plasmid X Map Locating EcoRI Site Final Plasmid X Map 100 bp 100 bp EcoRI BamHI ? ? 100 bp 600 bp 600 bp 600 bp 800 bp 800 bp 700 bp (a) (b) (c) BamHI Figure 7 Sample Problem 2: Constructing a Restriction Map Using Digestion Fragment Length Data Construct a restriction map, beginning with data already obtained from the digestion of Plasmid Y using the restriction enzymes EcoRI and BamHI. The data are provided in Table 3. Table 3 Results of a Restriction Fragment Digestion of Plasmid Y EcoRI BamHI EcoRI + BamHI 600 bp 600 bp 200 bp 1200 bp 1200 bp 200 bp 400 bp 1000 bp Step 1. Draw Plasmid Y restriction maps for the individual restriction enzymes. The results of the first two digestions are shown in Figure 8. EcoRI Plasmid Y Map BamHI Plasmid Y Map EcoRI BamHI 600 bp 600 bp 1200 bp 1200 bp EcoRI BamHI 2 fragments: 600 bp, 2 fragments: 600 bp, 1200 bp 1200 bp Figure 8 Although the fragment lengths produced by both digestions are the same, we know that each restriction enzyme recognizes different base-pair sequences. Therefore, these sites are not at the same positions on the plasmid. NEL 8.1 Manipulating and Cloning DNA   371 8159_Bio_Ch08_pp364-421.indd 371 3/27/12 6:03 PM  Step 2. Draw a restriction map of all the restriction sites. First, we redraw one of the restriction maps for a single digestion (Figure 9(a)). Next, we consider the fragment lengths that are produced by the double digestion: 200 bp (2), 400 bp, and 1000 bp. Using logic, we can infer the following:  One BamHI site must be within the 600 bp EcoRI fragment, and the other must be within the 1200 bp EcoRI fragment. (There are no 600 bp or 1200 bp fragments remaining after the double digestion.)  Both BamHI cuts produce a fragment of 200 bp, and they are 600 bp apart. (Since there is no 400 bp fragment to begin with, we know that a single cut cannot produce both 200 bp fragments.) Based on this logic, we can infer that one cut is 200 bp from an EcoRI site and within the 600 bp fragment (Figure 9(b)). This cut creates a 200 bp fragment and a 400 bp fragment. The second BamHI site must be located 600 bp away from the first site, but closer to the second EcoRI site in order to produce a second 200 bp fragment. The final restriction map is shown in Figure 9(c). EcoRI Plasmid Y Map Locating of BamHI Sites Final Plasmid Y Map EcoRI EcoRI EcoRI BamHI BamHI 200 bp 200 bp 600 bp 1000 bp 400 bp 1000 bp 400 bp 1200 bp 200 bp 200 bp EcoRI EcoRI EcoRI BamHI BamHI (a) (b) (c) Figure 9 Practice 1. Plasmid Z has yet to be mapped. It was digested with two enzymes: EcoRI and PstI. The following data were obtained (Table 4). Table 4 Results of a Restriction Enzyme Digestion of Plasmid Z Plasmid Z Plasmid Z cut Plasmid Z Plasmid Z cut with uncut with EcoRI cut with PstI EcoRI and PstI 1500 bp 500 bp 1500 bp 300 bp 1000 bp 500 bp 700 bp (a) In how many locations did EcoRI cut Plasmid Z? Did PstI cut Plasmid Z? (b) Draw the restriction map for Plasmid Z. 2. Plasmid W has yet to be mapped. It was cut with two enzymes: HindIII and BamHI. The following data were obtained (Table 5). Table 5 Results of a Restriction Enzyme Digestion of Plasmid W Plasmid W Plasmid W cut Plasmid W cut Plasmid W cut with uncut with HindIII with BamHI HindIII and BamHI 1000 bp 450 bp 400 bp 150 bp 550 bp 600 bp 250 bp 300 bp 300 bp (a) In how many locations did HindIII cut Plasmid W? Did BamHI cut Plasmid W? (b) Draw the restriction map for Plasmid W. 372   Chapter 8 Genetic Technologies NEL 8159_Bio_Ch08_pp364-421.indd 372 3/27/12 6:03 PM  3. Plasmid Q is known to contain recognition sites for three restriction enzymes. To prepare a restriction map for this plasmid, researchers performed seven different digestions, including single, double, and triple digestions. Use the data in Table 6 to generate a final restriction map, showing the relative locations of all the restriction enzyme recognition sites. Table 6 Results of a Restriction Fragment Digestion of Plasmid Q EcoRI + EcoRI + EcoRI + BamHI + BamHI + EcoRI BamHI SmaI BamHI SmaI SmaI SmaI 800 bp 250 bp 900 bp 200 bp 350 bp 150 bp 150 bp 1500 bp 700 bp 1400 bp 250 bp 450 bp 250 bp 200 bp 1350 bp 300 bp 450 bp 550 bp 250 bp 500 bp 1050 bp 600 bp 300 bp 1050 bp 750 bp 350 bp 450 bp 600 bp Transformation Under specific conditions, plasmids enter bacterial cells, multiply, and express the foreign gene that has been inserted into the plasmid genome. The successful intro- duction of DNA from another source is called transformation (Section 6.2). The cell that has received the DNA is said to be transformed. If a bacterial cell is not able to take up a plasmid that contains foreign DNA, it can sometimes be made competent in a laboratory. One way to do this is to place the bacteria in a solution that contains calcium chloride and recombinant plasmids in an ice-water bath. As the bacteria cool, the calcium ions stabilize the negative phosphate ions on the phospholipid bilayer of the cell membrane. The CaCl2 solution is then heated quickly and re-cooled. The sudden change from cold to hot momentarily disrupts the membrane, allowing the plasmid to enter (Figure 10). The cells are then kept at 37 °C for a period of time to stabilize and grow.   plasmid DNA            calcium ions              negatively charged       phospholipids     inner cell membrane                             Investigation 8.1.1  phospholipid bilayer  Transformation of Bacteria with   the Gene for Green Fluorescent          Protein (p. 399) Now that you have learned what opening in cell membrane outer cell membrane cytosol genetic engineers do to produce genetically modified bacteria, you Figure 10 Calcium ions neutralize the negative charge from the phosphate group on the plasmid can perform Investigation 8.1.1 and DNA and on the phospholipids found in the cell membrane. This minimizes the repelling effect of genetically engineer bacteria yourself. like charges. DNA can then enter the bacterial cell more easily. NEL 8.1 Manipulating and Cloning DNA   373 8159_Bio_Ch08_pp364-421.indd 373 3/27/12 6:03 PM  After the cells have stabilized and grown, they are tested for ampicillin resist­ ance. Those that contain a plasmid with the gene for ampicillin resistance grow in a medium that contains ampicillin. This means that they were successfully made com- petent and transformed. If they do not grow and replicate, they do not contain the plasmid with the ampicillin-resistant gene, so the process was unsuccessful. Hybridization: Identifying Bacterial Clones with Target Genes The technique of DNA hybridization is used to identify the cells that contain the introduced plasmids with the desired gene. This gene can be identified by its unique DNA sequence because it will pair with a short, single-stranded complementary hybridization probe a fragment of DNA DNA molecule, called a hybridization probe. For the gene with the recombinant DNA that is used to detect the presence of for insulin production, the hybridization probe is a piece of the DNA that codes for complementary nucleotide sequences insulin, ranging in size from 15 to 2500 bases. Figure 11 shows the steps taken to identify a target DNA sequence during DNA hybridization. After the presence of the DNA for insulin production is confirmed, the bacteria are grown in huge quantities, enough to produce insulin on a commercial scale. culture medium containing ampicillin 1 Bacteria containing plasmids are grown on a culture medium containing ampicillin. Only those bacterial bacteria that picked up a plasmid colony with the ampicillin-resistant gene will survive and grow into a colony. filter paper replica of 3 The filter paper is treated to bacterial cause the bacteria to break open. 2 A special filter paper laid on colonies The double-stranded DNA is the bacterial plate picks up cells denatured to single-stranded DNA, from each bacterial colony. which remains stuck to the filter A replica of the plate is made. paper in the same position as the colony it came from. labelled probe plasmid DNA labelled single- (single stranded) (single stranded) stranded DNA probe for the gene of interest 4 A radioactive-labelled probe complementary to the target gene (DNA or RNA) is incubated bag with the single-stranded DNA on the filter paper. The radioactive probe hybridizes (forms base pairs) filter Hybridization has occurred between only with the target gene. Excess probe is washed off. the labelled probe and the plasmids released from the bacteria in this colony. The hybridization is detected in subsequent steps. corresponds to 5 The filter is placed against film, which is exposed developed one colony on wherever the radioactive probe has hybridized. photographic master plate The position of any radioactive spot is correlated to the original plate. This colony is isolated and used film to produce large quantities of the target gene. original master plate Figure 11 DNA hybridization to identify the presence of a target gene 374   Chapter 8 Genetic Technologies NEL 8159_Bio_Ch08_pp364-421.indd 374 3/27/12 6:03 PM  8.1 Review Summary Insulin can be produced by genetically engineering bacterial cells and then cloning them. Appropriate restriction enzymes are used to remove and create a relatively small DNA fragment containing the target gene from the original DNA. To construct a recombinant DNA molecule, the target gene is inserted into a DNA vector. Plasmids are one example of a vector. Bacterial host cells can be manipulated to take up foreign DNA using a calcium chloride transformation. A restriction map is a diagram that shows distances (in base pairs) between restriction enzyme recognition sites. Cells that have been successfully transformed with recombinant DNA are isolated based on their ability to resist an antibiotic. Those that contain plasmids with the target gene are identified using hybridization techniques. Questions 1. Create a flow chart to show how a piece of DNA 8. The plasmid in Figure 12 was digested using containing a target gene is excised, combined different restriction enzymes whose sites have with a plasmid, and then expressed by a host been mapped. The plasmid is 7896 base pairs cell. K/U C long and contains a gene (shown in green) 2. HindIII and SmaI are used to cut up samples of that confers antibiotic resistance to tetracycline. a large piece of DNA (Table 1, page 367). Which T/I A enzyme would you expect to produce the larger number of fragments? Explain your reasoning using HindIII EcoRI (7524 bp) EcoRI mathematics. T/I C (0/7896 (1236 bp) HindIII 3. Why are restriction enzymes that produce DNA (7095 bp) bp) fragments with sticky ends more valuable to geneticists than restriction enzymes that produce DNA fragments with blunt ends? K/U BamHI HindIII (3569 bp) 4. What are the benefits of a bacterial cell that contains a plasmid? K/U (6024 bp) BamHI (4968 bp) 5. Why is a test performed to determine whether Figure 12 transformed cells are antibiotic resistant before they are isolated and grown? K/U (a) Determine the size and number of fragments that would be produced if the plasmid was 6. A fragment of DNA is 120 000 base pairs long and digested with the following enzymes: is cut by a restriction enzyme with a recognition site that is five base pairs long. How many cuts would (i)   EcoRI you expect? T/I (ii)   BamHI (iii) HindIII 7. Using the Internet or other sources, find an actual restriction map. Copy the restriction map into your (iv)  EcoRI and HindIII notebook. Locate and label the antibiotic-resistant (v)    EcoRI, HindIII, and BamHI gene and the multiple cloning site, if it exists. (b) Which cut would inactivate the antibiotic Identify one restriction enzyme that would create resistance? Why? a single fragment and, if applicable, one restriction WEB LINK enzyme that would cut the fragment into two pieces. T/I C NEL 8.1 Manipulating and Cloning DNA   375 8159_Bio_Ch08_pp364-421.indd 375 3/27/12 6:03 PM