Gene Isolation and Manipulation

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1 10 Gene Isolation and Manipulation WORKING WITH THE FIGURES 1. Figure 10-1 shows that specific DNA fragments can be synthesized in vitro prior to cloning. What are two ways to synthesize DNA inserts for recombinant DNA in vitro? DNA can be amplified from genomic sequences in vitro using the polymerase chain reaction (PCR) or by copying the mrna sequences into cdna using reverse transcriptase. 2. In Figure 10-4, why is cdna made only from mrna and not also from trnas and ribosomal RNAs? cdna is made from mrna and not from trnas or rrnas because polyt primers are used to prime the first DNA strand synthesis. Only the polyadenylated mrnas will anneal to the primers. 3. Redraw Figure 10-6 with the goal of adding one EcoRI end and one XhoI end. Below is the Xhol recognition sequence. Recognition sequence:... C T C G A G G A G C T C... After cut:... C T C G A G G A G C T C...

2 Chapter Ten Redraw Figure 10-7 so that the cdna can insert into an XhoI site of a vector rather than into an EcoRI site as shown. 5. In Figure 10-11, determine approximately how many BAC clones are needed to provide 1 coverage of

3 302 Chapter Ten a. the yeast genome (12 Mbp). b. the E. coli genome (4.1 Mbp). c. the fruit-fly genome (130 Mbp). a. The yeast genome is approximately 12 Mb, if the average insert size is kb, then BAC clones will give 1X coverage. b. For E. coli, only 21 to 41 clones will be required. c. For fruitfly, clones will be required. 6. In Figure 10-15, why does DNA migrate to the anode (+ pole)? DNA moves toward the positive pole in agarose gel electrophoresis because it is negatively charged. 7. In Figure 10-18(a), why are DNA fragments of different length and all ending in an A residue synthesized? As DNA is synthesized in the sequencing reaction, the enzyme will randomly insert either datp or ddatp across from T residues. If ddatp is selected, then the chain will terminate, as there is no 3 OH available for addition of the next nucleotide. As a result, the reaction will include fragments that have terminated at every position where an A is required. 8. As you will see in Chapter 15, most of the genomes of higher eukaryotes (plants and animals) are filled with DNA sequences that are present in hundreds, even thousands, of copies throughout the chromosomes. In the chromosome-walking procedure shown in Figure 10-20, how would the experimenter know whether the fragment he or she is using to walk to the next BAC or phage is repetitive? Can repetitive DNA be used in a chromosome walk? The researcher would find that the fragment he or she was using as a probe to identify adjacent sequences would hybridize to many non-overlapping clones. Such repetitive sequences could not be used in a chromosome walk, since the repeated sequence would be present in many different genomic locations and it would be impossible to determine which was the correct one. 9. Redraw Figure to include the positions of the single and double crossovers.

4 Chapter Ten In Figure 10-26, why do only plant cells that have T-DNA inserts in their chromosomes grow on the agar plates? Do all of the cells of a transgenic plant grown from one clump of cells contain T-DNA? Justify your answer. The T-DNA carries the kanamycin resistance gene; therefore, only cells that have acquired T-DNA inserts will grow on the agar plates containing the drug that selects for Kan R. 11. In Figure 10-28, what is the difference between extra-chromosomal DNA and integrated arrays of DNA? Are the latter ectopic? What is distinctive about the syncitial region that makes it a good place to inject DNA? Extrachromosomal arrays are maintained independently of the C. elegans chromosomes, while the integrated arrays become incorporated into the genome. The integrated arrays are ectopic, as they do not integrate into the homologous sequences in their normal chromosomal locus. The syncitial regio is a good place to inject DNA because there are a large number of nuclei in shared cytoplasm, any of which can take up the injected DNA. In addition, these cells will become egg or sperm, so the introduced genes will be passed on to individuals in the next generation. BASIC PROBLEMS 12. From this chapter, make a list of all the examples of

5 304 Chapter Ten (a) the hybridization of single-stranded DNAs and (b) proteins that bind to DNA and then act on it. a. The following are examples of hybridization of single-stranded DNAs discussed in this chapter: use of primers (DNA sequencing, PCR), hybridization of sticky ends in cloning, probe hybridization (Southern blotting, Northern blotting, diagnosis of mutations, identification of clones, and in situ hybridization). b. The following are examples of proteins that bind to and act on DNA discussed in this chapter: restriction enzymes, DNA polymerase (Taq polymerase), reverse transcriptase, and DNA ligase. 13. Compare and contrast the use of the word recombinant as used in the phrases (a) recombinant DNA and (b) recombinant frequency. a. Recombinant DNA is the term used to describe techniques used to generate hybrid molecules consisting of DNA from two different sources. In this use, recombinant refers to the artificial joining of DNA from different sources. b. Recombinant frequency is used to describe the statistical likelihood that two loci in the same organism will assort together or separately during meiosis. Here, recombinant refers to the combinations of loci that are different from the parent. 14. Why is ligase needed to make recombinant DNA? What would be the immediate consequence in the cloning process if someone forgot to add it? Ligase is an essential enzyme within all cells that seals breaks in the phosphate-sugar backbone of DNA. During DNA replication it joins Okazaki fragments to create a continuous strand, and in cloning, it is used to join the various DNA fragments with the vector. If it was not added, the vector and cloned DNA would simply fall apart. 15. In the PCR process, if we assume that each cycle takes 5 minutes, how manyfold amplification would be accomplished in 1 hour? Each cycle takes 5 minutes and doubles the DNA. In 1 hour there would be 12 cycles, so the DNA would be amplified 2 12 = 4096-fold amplification.

6 Chapter Ten The position of the gene for the protein actin in the haploid fungus Neurospora is known from the complete genome sequence. If you had a slow-growing mutant that you suspected of being an actin mutant and you wanted to verify that it was one, would you (a) clone the mutant by using convenient restriction sites flanking the actin gene and then sequence it or (b) amplify the mutant sequence by using PCR and then sequence it? The great advantage of PCR is that fewer procedures are necessary compared with cloning. However, it requires that the sequence be known, and that the primers are each present once in the genome and are sufficiently close (less than 2 kb). If these conditions are met, the primers determine the specificity of the DNA segment amplified and thus would be most efficient. 17. You obtain the DNA sequence of a mutant of a 2-kb gene in which you are interested and it shows base differences at three positions, all in different codons. One is a silent change, but the other two are missense changes (they encode new amino acids). How would you demonstrate that these changes are real mutations and not sequencing errors? (Assume that sequencing is about 99.9 percent accurate.) Resequencing the relevant gene should be done, as this will tell you if the original sequence was correct. You can then use the mutant sequence in a gene replacement experiment (depending on the organism) to see if the mutant phenotype is actually the result of the sequence variation you detected. 18. In a T-DNA transformation of a plant with a transgene from a fungus (not found in plants), the presumptive transgenic plant does not express the expected phenotype of the transgene. How would you demonstrate that the transgene is in fact present? How would you demonstrate that the transgene was expressed? You could isolate DNA from the suspected transgenic plant and probe for the presence of the transgene by Southern hybridization. 19. How would you produce a mouse that is homozygous for a rat growth- hormone transgene? Inject the rat growth hormone gene (RGH) into fertilized eggs of mice. These eggs are then implanted into a surrogate mother and any resulting offspring mated to test their offspring to see if any are transgenic for RGH. Because RGH will be inherited in a dominant fashion, any large and transgenic offspring will be heterozygous at this point. The transgenic siblings will need to

7 306 Chapter Ten be mated to each other in order to generate mice that are homozygous for the RGH transgene. 20. Why was cdna and not genomic DNA used in the commercial cloning of the human insulin gene? The commercial cloning of insulin was into bacteria. Bacteria are not capable of processing introns. Genomic DNA would include the introns, while cdna is a copy of processed (and thus intron-free) mrna. 21. After Drosophila DNA has been treated with a restriction enzyme, the fragments are inserted into plasmids and selected as clones in E. coli. With the use of this shotgun technique, every DNA sequence of Drosophila in a library can be recovered. a. How would you identify a clone that contains DNA encoding the protein actin, whose amino acid sequence is known? b. How would you identify a clone encoding a specific trna? a. Because the actin protein sequence is known, a probe could be synthesized by guessing the DNA sequence based on the amino acid sequence. (This works best if there is a region of amino acids that can be coded with minimal redundancy.) Alternatively, the gene for actin cloned in another species can be used as a probe to find the homologous gene in Drosophila. If an expression vector was used, it might also be possible to detect a clone coding for actin by screening with actin antibodies. b. Hybridization using the specific trna as a probe could identify a clone coding for itself. 22. In any particular transformed eukaryotic cell (say, of Saccharomyces cerevisiae), how could you tell if the transforming DNA (carried on a circular bacterial vector) a. replaced the resident gene of the recipient by double crossing over or single crossing over? b. was inserted ectopically? a. The transformed phenotype would map to the same locus. If gene replacement occurred by a double crossing-over event, the transformed cells would not contain vector DNA. If a single crossing-over took place, the entire vector would now be part of the linear Saccharomyces chromosome.

8 Chapter Ten 307 b. The transformed phenotype would map to a different locus than that of the auxotroph if the transforming gene was inserted ectopically (i.e., at another location). 23. In an electrophoretic gel across which is applied a powerful electrical alternating pulsed field, the DNA of the haploid fungus Neurospora crassa (n = 7) moves slowly but eventually forms seven bands, which represent DNA fractions that are of different sizes and hence have moved at different speeds. These bands are presumed to be the seven chromosomes. How would you show which band corresponds to which chromosome? Size, translocations between known chromosomes, and hybridization to probes of known location can all be useful in identifying which band on a PFGE gel corresponds to a particular chromosome. 24. The protein encoded by the cystic-fibrosis gene is 1480 amino acids long, yet the gene spans 250 kb. How is this difference possible? Conservatively, the amount of DNA necessary to encode this protein of 445 amino acids is = 1335 base pairs. When compared with the actual amount of DNA used, 60 kb, the gene appears to be roughly 45 times larger than necessary. This extra DNA mostly represents the introns that must be correctly spliced out of the primary transcript during RNA processing for correct translation. (There are also comparatively very small amounts of both 5 and 3 untranslated regions of the final mrna that are necessary for correct translation encoded by this 60-kb of DNA.) 25. In yeast, you have sequenced a piece of wild-type DNA and it clearly contains a gene, but you do not know what gene it is. Therefore, to investigate further, you would like to find out its mutant phenotype. How would you use the cloned wild-type gene to do so? Show your experimental steps clearly. The typical procedure is to knock out the gene in question and then see if there is any observable phenotype. One methodology to do this is described in the companion text. A recombinant vector carrying a selectable gene within the gene of interest is used to transform yeast cells. Grown under appropriate conditions, yeast that have incorporated the marker gene will be selected. Many of these will have the gene of interest disrupted by the selectable gene. The phenotype of these cells would then be assessed to determine gene function.

9 308 Chapter Ten CHALLENGING PROBLEMS 26. Prototrophy is often the phenotype selected to detect transformants. Prototrophic cells are used for donor DNA extraction; then this DNA is cloned and the clones are added to an auxotrophic recipient culture. Successful transformants are identified by plating the recipient culture on minimal medium and looking for colonies. What experimental design would you use to make sure that a colony that you hope is a transformant is not, in fact, a. a prototrophic cell that has entered the recipient culture as a contaminant? b. a revertant (mutation back to prototrophy by a second mutation in the originally mutated gene) of the auxotrophic mutation? a. To ensure that a colony is not, in fact, a prototrophic contaminant, the prototrophic line should be sensitive to a drug to which the recipient is resistant. A simple additional marker would also achieve the same end. b. Use a nonrevertible auxotroph as the recipient (such as one containing a deletion.) 27. A cloned fragment of DNA was sequenced by using the dideoxy chaintermination method. A part of the autoradiogram of the sequencing gel is represented here. a. Deduce the nucleotide sequence of the DNA nucleotide chain synthesized from the primer. Label the 5 and 3 ends. b. Deduce the nucleotide sequence of the DNA nucleotide chain used as the template strand. Label the 5 and 3 ends. c. Write out the nucleotide sequence of the DNA double helix (label the 5 and 3 ends). d. How many of the six reading frames are open as far as you can tell?

10 Chapter Ten 309 a. The gel can be read from the bottom to the top in a 5 -to-3 direction. The sequence is 5 T T C G A A A G G T G A C C C C T G G A C C T T T A G A 3 b. By complementarity, the template was 3 A A G C T T T C C A C T G G G G A C C T G G A A A T C T 5 c. The double helix is 5 T T C G A A A G G T G A C C C C T G G A C C T T T A G A 3 3 A A G C T T T C C A C T G G G G A C C T G G A A A T C T 5 d. Open reading frames have no stop codons. There are three frames for each strand, for a total of six possible reading frames. For the strand read from the gel, the transcript would be 5 U C U A A A G G U C C A G G G G U C A C C U U U C G A A 3 And for the template strand 5 U U C G A A A G G U G A C C C C U G G A C C U U U A G A 3 Stop codons are in bold and underlined. The two stop codons in the mrna read from the template strand are both in the same reading frame. There are a total of four open reading frames of the six possible. 28. The cdna clone for the human gene encoding tyrosinase was radioactively labeled and used in a Southern analysis of EcoRI-digested genomic DNA of wild-type mice. Three mouse fragments were found to be radioactive (were bound by the probe). When albino mice were used in this Southern analysis, no genomic fragments bound to the probe. Explain these results in relation to the nature of the wild-type and mutant mouse alleles. The region of DNA that encodes tyrosinase in normal mouse genomic DNA contains two EcoRI sites. Thus, after EcoRI digestion, three different-sized fragments hybridize to the cdna clone. When genomic DNA from certain albino mice is subjected to similar analysis, there are no DNA fragments that contain complementary sequences to the same cdna. This indicates that these mice lack the ability to produce tyrosinase because the DNA that encodes the enzyme must be deleted.

11 310 Chapter Ten 29. Transgenic tobacco plants were obtained in which the vector Ti plasmid was designed to insert the gene of interest plus an adjacent kanamycin-resistance gene. The inheritance of chromosomal insertion was followed by testing progeny for kanamycin resistance. Two plants typified the results obtained generally. When plant 1 was backcrossed with wild-type tobacco, 50 percent of the progeny were kanamycin resistant and 50 percent were sensitive. When plant 2 was backcrossed with the wild type, 75 percent of the progeny were kanamycin resistant and 25 percent were sensitive. What must have been the difference between the two transgenic plants? What would you predict about the situation regarding the gene of interest? Plant 1 shows the typical inheritance for a dominant gene that is heterozygous. Assuming kanamycin resistance is dominant to kanamycin sensitivity, the cross can be outlined as follows: kan R /kan S kan S /kan S 1/2 kan R /kan S 1/2 kan S /kan S This would suggest that the gene of interest would be inserted once into the genome. Plant 2 shows a 3:1 ratio in the progeny of the backcross. This suggests that there have been two unlinked insertions of the kan R gene and presumably the gene of interest as well. kan R1 /kan S1 ; kan R2 /kan S2 kan S1 /kan S1 ; kan S2 /kan S2 1/4 kan R1 /kan S1 ; kan R2 /kan S2 1/4 kan R1 /kan S1 ; kan S2 /kan S2 1/4 kan S1 /kan S1 ; kan R2 /kan S2 1/4 kan S1 /kan S1 ; kan S2 /kan S2 30. A cystic-fibrosis mutation in a certain pedigree is due to a single nucleotide-pair change. This change destroys an EcoRI restriction site normally found in this position. How would you use this information in counseling members of this family about their likelihood of being carriers? State the precise experiments needed. Assume that you find that a woman in this family is a carrier, and it transpires that she is married to an unrelated man who also is a heterozygote for cystic fibrosis, but, in his case, it is a different mutation in the same gene. How would you counsel this couple about the risks of a child s having cystic fibrosis? Assuming that the DNA from this region is cloned, it could be used as a probe to detect this RFLP on Southern blots. DNA from individuals within

12 Chapter Ten 311 this pedigree would be isolated (typically from blood samples containing white blood cells) and restricted with EcoRI, and Southern blots would be performed. Individuals with this mutant CF allele would have one band that would be larger (owing to the missing EcoRI site) when compared with wild type. Individuals that inherited this larger EcoRI fragment would, at minimum, be carriers for cystic fibrosis. In the specific case discussed in this problem, a woman that is heterozygous for this specific allele marries a man that is heterozygous for a different mutated CF allele. Just knowing that both are heterozygous, it is possible to predict that there is a 25 percent chance of their child s having CF. However, because the mother s allele is detectable on a Southern blot, it would be possible to test whether the fetus inherited this allele. DNA from the fetus (through either CVS or amniocentesis) could be isolated and tested for this specific EcoRI fragment. If the fetus did not inherit this allele, there would be a 0 percent chance of its having CF. On the other hand, if the fetus inherited this allele, there would be a 50 percent chance the child will have CF. 31. Bacterial glucuronidase converts a colorless substance called X-Gluc into a bright blue indigo pigment. The gene for glucuronidase also works in plants if given a plant promoter region. How would you use this gene as a reporter gene to find the tissues in which a plant gene that you have just cloned is normally active? (Assume that X-Gluc is easily taken up by the plant tissues.) The promoter and control regions of the plant gene of interest must be cloned and joined in the correct orientation with the glucuronidase gene. This places the reporter gene under the same transcriptional control as the gene of interest. The companion text discusses the methodology used to create transgenic plants. Transform plant cells with the reporter gene construct, and as discussed in the text, grow into transgenic plants. The glucuronidase gene will now be expressed in the same developmental pattern as the gene of interest and its expression can easily be monitored by bathing the plant in an X-Gluc solution and assaying for the blue reaction product. 32. The plant Arabidopsis thaliana was transformed by using the Ti plasmid into which a kanamycin-resistance gene had been inserted in the T-DNA region. Two kanamycin-resistant colonies (A and B) were selected, and plants were regenerated from them. The plants were allowed to self-pollinate, and the results were as follows: Plant A selfed 3/4 progeny resistant to kanamycin 1/4 progeny sensitive to kanamycin Plant B selfed 5/16 progeny resistant to kanamycin 1/16 progeny sensitive to kanamycin a. Draw the relevant plant chromosomes in both plants. b. Explain the two different ratios.

13 312 Chapter Ten a. and b. During Ti plasmid transformation, the kanamycin gene will insert randomly into the plant chromosomes. Colony A, when selfed, has 3/4 kanamycin-resistant progeny, and colony B, when selfed, has 15/16 kanamycin-resistant progeny. This suggests that there was a single insertion into one chromosome in colony A and two independent insertions on separate chromosomes in colony B. This can be schematically represented by showing a single insertion within one of the pair of chromosome A for colony A. Chromosomes A T-DNA kanamycin R T-DNA and two independent insertions into one of each of the pairs of chromosomes B and C for colony B. Chromosomes B Chromosomes C T-DNA T-DNA kanamycin R T-DNA kanamycin R Genetically, this can be represented as T-DNA Colony A Colony B kan RA /kan SA kan RB /kan SB ; kan RC /kan SC When these are selfed kan RA /kan SA kan RA /kan SA 1/4 kan RA /kan RA 1/2 kan RA /kan SA 1/4 kan SA /kan SA kan RB /kan SB ; kan RC /kan SC kan RB /kan SB ; kan RC /kan SC 9/16 kan RB / ; kan RC / 3/16 kan RB / ; kan SC /kan SC 3/16 kan SB /kan SB ; kan RC / 1/16 kan SB /kan SB ; kan SC /kan SC

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