Application and Experimental Questions

CHAPTER 19

Application and Experimental Questions E1. What is the functional significance of sticky ends in a cloning experiment? What types of chemical bonds make the ends sticky? Answer: Sticky ends, which are complementary in their DNA sequence, promote the binding of DNA fragments to each other. This binding is due to hydrogen bonding between the sticky ends. E2. Table 19.3 describes the cleavage sites of six different restriction enzymes. After these restriction enzymes have cleaved the DNA, five of them produce sticky ends that can hydrogen bond with complementary sticky ends, as shown in Figure 19.1. The efficiency of sticky ends binding together depends on the number of hydrogen bonds; more hydrogen bonds makes the ends “stickier” and more likely to stay attached. Rank these five restriction enzymes in Table 19.3 (from best to worst) with regard to the efficiency of their sticky ends binding to each other. Answer: Remember that AT base pairs form two hydrogen bonds, while GC base pairs form three hydrogen bonds. The order (from stickiest to least sticky) would be: BamHI = PstI = SacI > EcoRI > ClaI. E3. Describe the important features of cloning vectors. Explain the purpose of selectable marker genes in cloning experiments. Answer: All vectors have the ability to replicate when introduced into a living cell. This ability is due to a DNA sequence known as an origin of replication, which determines the host cell specificity of a vector. Modern vectors also contain convenient restriction sites where geneticists can insert fragments of DNA. These vectors also contain selectable markers, which are genes that confer some selectable advantage for the host cell that carries them. The most common selectable markers are antibiotic-resistance genes, which confer resistance to antibiotics that would normally inhibit the growth of the host cell. E4. How does gene cloning produce many copies of a gene? Answer: In conventional gene cloning, many copies are made because the vector replicates to a high copy number within the cell, and the cells divide to produce many more cells. In PCR, the replication of the DNA to produce many copies is facilitated by primers, deoxyribonucleoside triphosphates (dNTPs), and Taq polymerase. E5. In your own words, describe the series of steps necessary to clone a gene. Your answer should include the use of a probe to identify a bacterial colony that contains the cloned gene of interest. Answer: First, the chromosomal DNA that contains the source of the gene that you want to clone must be obtained from a cell (tissue) sample. A vector must also be obtained. The vector and chromosomal DNA are digested with a restriction enzyme. They are mixed together to allow the sticky ends of the DNA fragments to bind to each other, and DNA ligase is then added to promote covalent bonds, hopefully creating a recombinant vector. The DNA is introduced into a living cell via transformation or transfection. The vector will replicate and the cells will divide to produce a colony of cells that contain “cloned” DNA pieces. To identify colonies that contain the gene you wish to clone, you must use a probe that will specifically identify a colony containing the correct recombinant vector. A probe may be a DNA probe that is complementary to the gene you want to clone or it could be an antibody that recognizes the protein that is encoded by the gene. E6. What is a recombinant vector? How is a recombinant vector constructed? Explain how X-Gal can be used in a method to identify recombinant vectors that contain segments of chromosomal DNA. Answer: A recombinant vector is a vector that has a piece of “foreign” DNA inserted into it. The foreign DNA came from somewhere else, such as the chromosomal DNA of some organism. To construct a recombinant vector, the vector and source of foreign DNA are digested with the same restriction enzyme. The complementary ends of the fragments are allowed to hydrogen bond to each other (i.e., sticky ends are allowed to bind), and then DNA ligase is added to create covalent phosphoester bonds. In some cases, a piece of the foreign DNA will become ligated to the vector, thereby creating a recombinant vector. In other cases, the two ends of the vector ligate back together, restoring the vector to its original structure. As described in Figure 19.2, the insertion of foreign DNA can be detected using X-Gal. As seen here, the insertion of the foreign DNA causes the inactivation of the lacZ gene. The lacZ gene encodes the enzyme β-galactosidase, which converts the colorless compound X-Gal to a blue compound. If the lacZ gene is inactivated by the insertion of foreign DNA, the enzyme will not be produced, and the bacterial colonies will be white. If the vector has simply recircularized, and the lacZ gene remains intact, the enzyme will be produced and the colonies will be blue. E7. If a researcher began with a sample that contained three copies of double-stranded DNA, how many copies would be present after 27 cycles of PCR, assuming 100% efficiency? Answer: 3  227, which equals 4.0 108, or about 400 million copies. E8. Why is a thermostable form of DNA polymerase (e.g., Taq polymerase) used in PCR? Is it necessary to use a thermostable form of DNA polymerase in the techniques of dideoxy DNA sequencing or site-directed mutagenesis? Answer: A thermostable form of DNA polymerase (e.g., Taq polymerase) is used in PCR because each PCR cycle involves a heating step to denature the DNA. This heating step would inactivate most forms of DNA polymerase. However, Taq polymerase is thermostable and can remain functional after many cycles of heating and cooling. It is not necessary to use a thermostable form of DNA polymerase in the techniques of dideoxy sequencing or site-directed mutagenesis. In these methods, DNA polymerase can be added after the annealing step, and the sample can be incubated at a temperature that does not inactivate most forms of DNA polymerase. E9. Describe how you could produce many copies of the β-globin cDNA from the rat using RT-PCR. Answer: First, you would isolate mRNA from rat red blood cells. The mRNA would be mixed with reverse transcriptase and nucleotides to create a complementary strand of DNA. Reverse transcriptase also needs a primer. This could be a primer that is known to be complementary to the β-globin mRNA. Alternatively, mature mRNAs have a polyA tail, so one could add a primer that consists of many T’s, called a poly-dT primer. After the complementary DNA strand has been made, the sample would then be mixed with primers, Taq polymerase, and nucleotides and subjected to the standard PCR protocol. Note: the PCR reaction would have two kinds of primers. One primer would be complementary to the 5 end of the mRNA and would be unique to the β-globin sequence. The other primer would be complementary to the 3 end. This second primer could be a poly-dT primer or it could be a unique primer that would bind slightly upstream from the polyA-tail region. E10. What type of detector is used for real-time PCR? Explain how the level of fluorescence correlates with the level of PCR product. Answer: The detector is a molecule whose fluorescence increases as the PCR product increases. One example is TaqMan. It has a fluorescent molecule and a quencher, which are connected by a primer that binds to the DNA that is amplified. When PCR is occurring, Taq polymerase digests the primer and thereby separates the fluorescent molecule from the quencher. Therefore, the fluorescence increases as the PCR product increases. Alternatively, other detectors intercalate into the DNA, which causes their fluorescence to increase. In this case, as the PCR product increases, the fluorescence also increases because the detectors are intercalating into the newly made DNA. E11. What phase of PCR (exponential, linear, or plateau) is analyzed to quantitate the amount of DNA or RNA in a sample? Explain why this phase is chosen. Answer: The exponential phase of PCR is chosen because it is during this phase that the amount of PCR product is proportional to the amount of the original DNA in the sample. E12. Let’s suppose that you have recently cloned a gene, which we will call gene X, from corn. You use this cloned gene to probe genomic DNA from corn in a Southern blot experiment under conditions of low and high stringency. The following results were obtained: [Insert Text Art 19.5] What do these results mean? Answer: One interpretation would be that the gene is part of a gene family. In this case, the family would contain four homologous members. At high stringency, the probe binds only to the gene that is its closest match, but at low stringency it recognizes the three other homologous genes. E13. What is a DNA library? Do you think that this is an appropriate name? Answer: A DNA library is a collection of recombinant vectors that contain different pieces of DNA from a source of chromosomal DNA. Because it is a diverse collection of many different DNA pieces, the name library seems appropriate. E14. Some vectors used in cloning experiments contain bacterial promoters that are adjacent to unique cloning sites. This makes it possible to insert a gene sequence next to the bacterial promoter and express the gene in bacterial cells. These are called expression vectors. If you wanted to express a eukaryotic protein in bacterial cells, would you clone genomic DNA or cDNA into the expression vector? Explain your choice. Answer: It would be necessary to use cDNA so the gene would not carry any introns. Bacterial cells do not contain spliceosomes (which are described in Chapter 14). To express a eukaryotic protein in bacteria, a researcher would clone cDNA into bacteria, because the cDNA does not contain introns. E15. Southern and Northern blotting depend on the phenomenon of hybridization. In these two techniques, explain why hybridization occurs. Which member of the hybrid is labeled? Answer: Hybridization occurs due to the hydrogen bonding of complementary sequences. Due to the chemical properties of DNA and RNA strands, they form double-stranded regions when the base sequences are complementary. In a Southern and Northern experiment, the cloned DNA is labeled and used as a probe. E16. In Southern, Northern, and Western blotting, what is the purpose of gel electrophoresis? Answer: The purpose of gel electrophoresis is to separate the many DNA fragments, RNA molecules, or proteins that were obtained from the sample you want to analyze. This separation is based on molecular mass and allows you to identify the molecular mass of the DNA fragment, RNA molecule, or protein that is being recognized by the probe. E17. What is the purpose of a Northern blotting experiment? What types of information can it tell you about the transcription of a gene? Answer: The purpose of a Northern blotting experiment is to identify a specific RNA within a mixture of many RNA molecules, using a fragment of cloned DNA as a probe. It can tell you if a gene is transcribed in a particular cell or at a particular stage of development. It can also tell you if a pre-mRNA is alternatively spliced into two or more mRNAs of different sizes. E18. Let’s suppose an X-linked gene in mice exists as two alleles, which we will call B and b. X inactivation, a process in which one X chromosome is turned off, occurs in the somatic cells of female mammals (see Chapter 4). Allele B encodes an mRNA that is 900 nucleotides long, while allele b contains a small deletion that shortens the mRNA to a length of 825 nucleotides. Draw the expected results of a Northern blot using mRNA isolated from somatic tissue of the following mice: Lane 1. mRNA from an XbY male mouse Lane 2. mRNA from an XbXb female mouse Lane 3. mRNA from an XBXb female mouse. Note: The sample taken from the female mouse is not from a clone of cells. Answer: The Northern blot is shown here. The female mouse expresses the same total amount of this mRNA compared to the male. In the heterozygous female, there would be 50% of the 900 bp band and 50% of the 825 bp band. E19. The method of Northern blotting can be used to determine the amount of a particular RNA transcribed in a given cell type and the size of the mRNA. Alternative splicing (discussed in Chapter 17) can produce mRNAs from the same gene that have different lengths. A Northern blot is shown here using a DNA probe that is complementary to the mRNA encoded by a particular gene. The mRNA in Lanes 1–4 was isolated from different cell types. [Insert Text Art 19.6] Explain these results. Answer: It appears that this mRNA is alternatively spliced to create a high molecular mass and a lower molecular mass product. Nerve cells produce a very large amount of the larger mRNA, whereas spleen cells produce a moderate amount of the smaller mRNA. Both types are produced in small amounts by the muscle cells. It appears that kidney cells do not transcribe this gene. E20. Southern blotting can be used to detect the presence of repetitive sequences, such as transposable elements, that are present in multiple copies within the chromosomal DNA of an organism. (Note: Transposable elements are described in Chapter 21.) In the Southern blot shown here, chromosomal DNA was isolated from three different strains of Baker’s yeast, digested with a restriction enzyme, run on a gel, blotted, and then probed with a radioactive DNA probe that is complementary to a transposable element called the Ty element. [Insert Text Art 19.7] Explain, in a general way, why the banding patterns are not the same in lanes 1, 2, and 3. Answer: Restriction enzymes recognize many sequences throughout the chromosomal DNA. If two fragments from different samples have the same molecular mass in a Southern blot, it is likely (though not certain) that the two fragments are found at the same chromosomal site in the genome. In this Southern blot, most of the transposable elements are found at the same sites within the genomes of these different yeast strains. However, a couple of bands are different among the three strains. These results indicate that the Ty element may occasionally transpose to a new location or that chromosomal changes (point mutations, chromosomal rearrangements, deletions, and so on) may have slightly changed the genomes among these three strains of yeast in a way that changes the distances between the restriction sites. E21. In Chapter 8, Figure 8.7 describes the evolution of the globin gene family. All of the genes in this family are homologous to each other, though the degree of sequence similarity varies depending on the time of divergence. Genes that have diverged more recently have sequences that are more similar. For example, the 1 and 2 genes have DNA sequences that are more similar to each other compared to the 1 and ξ genes. In a Southern blotting experiment, the degree of sequence similarity can be discerned by varying the stringency of hybridization. At high temperature (i.e., high stringency), the probe will only recognize genes that are a perfect or very close match. At lower temperature, however, homologous genes with lower degrees of similarities can be detected because slight mismatches are tolerated. If a Southern blot was conducted on a sample of human chromosomal DNA, and a probe was used that was a perfect match to the β-globin gene, rank the following genes (from those that would be detected at high stringency down to those that would only be detected at low stringency) as they would appear in a Southern blot experiment: Mb, 1, β, γA, δ, and ε. Answer: 1. β (detected at the highest stringency) 2. 

3. γA and ε

4. 1 5. Mb (detected only at the lowest stringency) E22. In the Western blot shown here, polypeptides were isolated from red blood cells and muscle cells from two different individuals. One individual was normal, and the other individual suffered from a disease known as thalassemia, which involves a defect in hemoglobin. In the Western blot, the gel blot was exposed to an antibody that recognizes β-globin, which is one of the polypeptides that constitute hemoglobin. [Insert Text Art 19.8] Explain these results. Answer: Lane 1 shows that β-globin is made in normal red blood cells. In red blood cells from a thalassemia patient (lane 2), however, very little is made. Perhaps this person is homozygous for a down promoter mutation, which diminishes the transcription of the gene. As shown in lanes 3 and 4, β-globin is not made in muscle cells. E23. Let’s suppose a researcher was interested in the effects of mutations on the expression of a structural gene that encodes a polypeptide that is 472 amino acids in length. This polypeptide is expressed in leaf cells of Arabidopsis thaliana. Because the average molecular mass of an amino acid is 120 Daltons, this protein has a molecular mass of approximately 56,640 Daltons. Make a drawing that shows the expected results of a Western blot using polypeptides isolated from the leaf cells that were obtained from the following individuals: Lane 1. A normal plant Lane 2. A plant that is homozygous for a deletion that removes the promoter for this gene Lane 3. A plant that is heterozygous in which one gene is normal and the other gene has a mutation that introduces an early stop codon at codon 112 Lane 4. A plant that is homozygous for a mutation that introduces an early stop codon at codon 112 Lane 5. A plant that is homozygous for a mutation that changes codon 108 from a phenylalanine codon into a leucine codon Answer: The Western blot is shown here. The sample in lane 2 came from a plant that was homozygous for a mutation that prevented the expression of this polypeptide. Therefore, no protein was observed in this lane. The sample in lane 4 came from a plant that is homozygous for a mutation that introduces an early stop codon into the coding sequence; thus, the polypeptide is shorter than normal (13.3 kDa). The sample in lane 3 was from a heterozygote that expresses about 50% of each type of polypeptide. Finally, the sample in lane 5 came from a plant that is homozygous for a mutation that changed one amino acid to another amino acid. This type of mutation, termed a missense mutation, may not be detectable on gel. However, a single amino acid substitution could affect polypeptide function.

E24. If you wanted to know if a protein was made during a particular stage of development, what technique would you choose? Answer: Western blotting. E25. Explain the basis for using an antibody as a probe in a Western blotting experiment. Answer: The products of structural genes are proteins with a particular amino acid sequence. Antibodies can specifically recognize proteins due to their amino acid sequence. Therefore, an antibody can detect whether or not a cell is making a particular type of protein. E26. Starting with pig cells and a probe that is the human β-globin gene, describe how you would clone the β-globin gene from pigs. You may assume that you have available all of the materials needed in a cloning experiment. How would you confirm that a putative clone really contained a β-globin gene? Answer: You would first make a DNA library using chromosomal DNA from pig cells. You would then radiolabel a portion of a strand from the human β-globin gene and use it as a probe in a colony hybridization experiment; each colony would contain a different cloned piece of the pig genome. You would identify “hot” colonies that hybridize to the human β-globin probe. You would then go back to the master plate and pick these colonies and grow them in a test tube. You would then isolate the plasmid DNA from the colonies and subject the DNA to DNA sequencing. By comparing the DNA sequences of the human β-globin gene and the putative clones, you could determine if the putative clones were homologous to the human clone and likely to be the pig homolog of the β- globin gene. E27. A cloned gene fragment contains a regulatory element that is recognized by a regulatory transcription factor. Previous experiments have shown that the presence of a hormone results in transcriptional activation by this transcription factor. To study this effect, you conduct a gel retardation assay and obtain the following results: [Insert Text Art 19.9] Explain the action of the hormone. Answer: In this case, the transcription factor binds to the regulatory element when the hormone is present. Therefore, the hormone promotes the binding of the transcription factor to the DNA and thereby promotes transcriptional activation. E28. Describe the rationale behind a gel retardation assay. Answer: The rationale behind a gel retardation assay is that a segment of DNA with a protein bound to it will migrate more slowly through a gel than will the same DNA without any bound protein. A shift in a DNA band to a higher molecular mass provides a way to identify DNA-binding proteins. E29. A gel retardation assay can be used to study the binding of proteins to a segment of DNA. In the experiment shown here, a gel retardation assay was used to examine the requirements for the binding of RNA polymerase II (from eukaryotic cells) to the promoter of a structural gene. The assembly of RNA polymerase II at the core promoter is described in Chapter 14 (Figure 14.14). In this experiment, the segment of DNA containing a promoter sequence was 1,100 bp in length. The fragment was mixed with various combinations of proteins and then subjected to a gel retardation assay. [Insert Text Art 19.10] Explain which proteins (TFIID, TFIIB, and/or RNA polymerase II) are able to bind to this DNA fragment by themselves. Which transcription factors (i.e., TFIID and/or TFIIB) are needed for the binding of RNA polymerase II? Answer: TFIID can bind to this DNA fragment by itself, as seen in lane 2. However, TFIIB and RNA polymerase II cannot bind to the DNA by themselves (lanes 3 and 4). As seen in lane 5, TFIIB can bind, if TFIID is also present, because the mobility shift is higher than TFIID alone (compare lanes 2 and 5). In contrast, RNA polymerase II cannot bind to the DNA when only TFIID is present. The mobility shift in lane 6 is the same as that found in lane 2, indicating that only TFIID is bound. Finally, in lane 7, when all three components are present, the mobility shift is higher than when both TFIIB and TFIID are present (compare lanes 5 and 7). These results mean that all three proteins are bound to the DNA. Taken together, the results indicate that TFIID can bind by itself, TFIIB needs TFIID to bind, and RNA polymerase II needs both proteins to bind to the DNA. E30. As described in Chapter 17 (Figure 17.7), certain regulatory transcription factors bind to the DNA and transactivate RNA polymerase II. When glucocorticoid binds to the glucocorticoid receptor (a regulatory transcription factor), this changes the conformation of the receptor and allows it ultimately to bind to the DNA. The glucocorticoid receptor binds to a DNA sequence called a glucocorticoid response element (GRE). In contrast, other regulatory transcription factors, such as the CREB protein, do not require hormone binding in order to bind to DNA. The CREB protein can bind to the DNA in the absence of any hormone, but it will not transactivate RNA polymerase II unless the CREB protein is phosphorylated. (Phosphorylation is stimulated by certain hormones.) The CREB protein binds to a DNA sequence called a cAMP response element (CRE). With these ideas in mind, draw the expected results of a gel retardation assay conducted on the following samples: Lane 1. A 600 bp fragment containing a GRE, plus the glucocorticoid receptor Lane 2. A 600 bp fragment containing a GRE, plus the glucocorticoid receptor, plus glucocorticoid hormone Lane 3. A 600 bp fragment containing a GRE, plus the CREB protein Lane 4. A 700 bp fragment containing a CRE, plus the CREB protein Lane 5. A 700 bp fragment containing a CRE, plus the CREB protein, plus a hormone (such as epinephrine) that causes the phosphorylation of the CREB protein Lane 6. A 700 bp fragment containing a CRE, plus the glucocorticoid receptor, plus glucocorticoid hormone Answer: The glucocorticoid receptor will bind to GREs if glucocorticoid hormone is also present (lane 2). The glucocorticoid receptor does not bind without hormone (lane 1) and it does not bind to CREs (lane 6). The CREB protein will bind to CREs with or without hormone (lanes 4 and 5), but it will not bind to GREs (lane 3). The expected results are shown here. In this drawing, the binding of CREB protein to the 700 bp fragment results in a complex with a higher mass compared to the glucocorticoid receptor binding to the 600 bp fragment.

E31. In the technique of DNase I footprinting, the binding of a protein to a region of DNA will protect that region from digestion by DNase I by blocking the ability of DNase I to gain access to the phosphodiester linkages in the DNA. In the DNase I footprinting experiment shown here, a researcher began with a sample of cloned DNA that was 400 bp in length. This DNA contained a eukaryotic promoter for RNA polymerase II. The assembly of RNA polymerase II at the core promoter is described in Chapter 14. For the sample loaded in lane 1, no proteins were added. For the sample loaded in lane 2, the 400 bp fragment was mixed with RNA polymerase II plus TFIID and TFIIB. [Insert Text Art 19.11] Which region of this 400 bp fragment of DNA is bound by RNA polymerase II and TFIID and TFIIB? Answer: The region of the gel from about 350 bp to 175 bp does not contain any bands. This is the region being covered up; the “footprint” is about 175 bp long. E32. Explain the rationale behind a DNase I footprinting experiment. Answer: The rationale behind a footprinting experiment has to do with accessibility. If a protein is bound to the DNA, it will cover up the part of the DNA where it is bound. This region of the DNA will be inaccessible to the actions of chemicals or enzymes that cleave the DNA, such as DNase I. E33. DNA sequencing can help us to identify mutations within genes. The following data are derived from an experiment in which a normal gene and a mutant gene have been sequenced: [Insert Text Art 19.12] Locate and describe the mutation. Answer: The A closest to the bottom of the gel is changed to a G. E34. A sample of DNA was subjected to automated DNA sequencing as shown here. [Insert Text Art 19.13] What is the sequence of this DNA segment? Answer: AGGTCGGTTGCCATCGCAATAATTTCTGCCTGAACCCAATA E35. A portion of the coding sequence of a cloned gene is shown here: 5’–GCCCCCGATCTACATCATTACGGCGAT–3’ 3’–CGGGGGCTAGATGTAGTAATGCCGCTA–5’ This portion of the gene encodes a polypeptide with the amino acid sequence alanine-proline–aspartic acid–leucine–histidine–histidine–tyrosine–glycine–aspartic acid. Using the method of site-directed mutagenesis, a researcher wants to change the leucine codon into an arginine codon, with an oligonucleotide that is 19 nucleotides long. What is the sequence of the oligonucleotide that should be made? You should designate the 5’ and 3’ ends of the oligonucleotide in your answer. Note: The mismatch should be in the middle of the oligonucleotide, and a one-base mismatch is preferable over a two- or three-base mismatch. Use the bottom strand as the template strand for this site-directed mutagenesis experiment. Answer: 5–CCCCCGATCGACATCATTA–3. The mutagenic base is underlined. E36. Let’s suppose you want to use site-directed mutagenesis to investigate a DNA sequence that functions as a regulatory element for hormone binding. From previous work, you have narrowed down the response element to a sequence of DNA that is 20 bp in length with the following sequence: 5’–GGACTGACTTATCCATCGGT–3’ 3’–CCTGACTGAATAGGTAGCCA–5’ As a strategy to pinpoint the actual regulatory element sequence, you decide to make 10 different site-directed mutants and then analyze their effects by a gel retardation assay. What mutations would you make? What results would you expect to obtain? Answer: There are lots of different strategies one could follow. For example, you could mutate every other base and see what happens. It would be best to make very nonconservative mutations such as a purine for a pyrimidine or a pyrimidine for a purine. If the mutation prevents protein binding in a gel retardation assay, then the mutation is probably within the regulatory element. If the mutation has no effect on protein binding, it probably is outside the regulatory element. E37. Site-directed mutagenesis can also be used to explore the structure and function of proteins. For example, changes can be made to the coding sequence of a gene to determine how alterations in the amino acid sequence affect the function of a protein. Let’s suppose that you are interested in the functional importance of one glutamic acid residue within a protein you are studying. By site-directed mutagenesis, you make mutant proteins in which this glutamic acid codon has been changed to other codons. You then test the encoded mutant proteins for functionality. The results are as follows: Functionality Normal protein 100% Mutant proteins containing: Tyrosine 5% Phenylalanine 3% Aspartic acid 94% Glycine 4% From these results, what would you conclude about the functional significance of the glutamic acid residue within the protein? Answer: You would conclude that it might be important. The only amino acid substitution that gave a substantial amount of functional activity was an aspartate. Glutamatic acid and aspartic acid have very similar amino acid side chains (see Chapter 15); they both contain a carboxyl (COOH) group. Based on these results, you would suspect that a carboxyl group at this location in the protein might be important for its function.