Essential Cell Biology-4th Ed Test-Bank – Alberts-
ESSENTIAL CELL BIOLOGY, FOURTH EDITIONESSENTIAL CELL BIOLOGY, FOURTH EDITIONCHAPTER 10: MODERN RECOMBINANT DNA TECHNOLOGY© 2014 GARLAND SCIENCE PUBLISHING
10-1 Recombinant DNA technologies involve techniques that permit the creation of custom-made DNA molecules that can be introduced back into living organisms. These technologies were first developed in the ______.(a) 1930s.(b) 1950s.(c) 1970s.(d) 1990s.
Manipulating and Analyzing DNA Molecules
10-2 Which of the following statements about restriction nucleases is false?(a) A reproducible set of DNA fragments will be produced every time a restriction nuclease digests a known piece of DNA.(b) Restriction nucleases recognize specific sequences on single-stranded DNA.(c) Some bacteria use restriction nucleases as protection from foreign DNA.(d) Some restriction nucleases cut in a staggered fashion, leaving short, single-stranded regions of DNA at the ends of the cut molecule.
10-3 You have purified DNA from your recently deceased goldfish. Which of the following restriction nucleases would you use if you wanted to end up with DNA fragments with an average size of 70 kilobase pairs (kb) after complete digestion of the DNA? The recognition sequence for each enzyme is indicated in the right-hand column.(a) Sau3AI GATC(b) BamHI GGATCC(c) NotI GCGGCCGC(d) XzaI GAAGGATCCTTC
10-4 You have a circular plasmid that can be cut by the restriction nuclease HindIII, as diagrammed in Figure Q10-4. Figure Q10-4
If you were to cut this circular piece of DNA with HindIII, which of the answers below best predicts what you would get?(a) one linear piece of DNA(b) two circular pieces of DNA(c) two semicircular pieces of DNA(d) two linear pieces of DNA
10-5 You have a piece of circular DNA that can be cut by the restriction nucleases XhoI and SmaI, as indicated in Figure Q10-5.
If you were to cut this circular piece of DNA with both XhoI and SmaI, how many fragments of DNA would you end up with?(a) 1(b) 2(c) 3(d) 4
10-6 You have a piece of circular DNA that can be cut by the restriction nucleases EcoRI, HindIII, and NotI, as indicated in Figure Q10-6.
Which of the following statements is false?(a) One piece of DNA will be obtained when this DNA is cut by NotI.(b) A piece of DNA that cannot be cut by EcoRI will be obtained by cutting this DNA with both NotI and HindIII.(c) Two DNA fragments that cannot be cut by HindIII will be obtained when this DNA is cut by EcoRI and NotI.(d) Two DNA fragments of unequal size will be created when this DNA is cut by both HindIII and EcoRI.
10-7 You have a circular plasmid that has a single EcoRI site in it, as diagrammed in Figure Q10-7, which also shows the cleavage site for EcoRI. Choose the answer below that best represents what the end of the DNA molecule will look like once you cut the plasmid with EcoRI. Note that only the very ends of the DNA molecule are shown in the answers.
10-8 You have accidentally torn the labels off two tubes, each containing a different plasmid, and now do not know which plasmid is in which tube. Fortunately, you have restriction maps for both plasmids, shown in Figure Q10-8. You have the opportunity to test just one sample from one of your tubes. You have equipment for agarose-gel electrophoresis, a standard set of DNA size markers, and the necessary restriction enzymes.
A. Outline briefly the experiment you would do to determine which plasmid is in which tube.B. Which restriction enzyme or combination of restriction enzymes would you use in this experiment?
10-9 You have a linear piece of DNA that can be cut by the restriction nucleases HindIII and EcoRI, as diagrammed in Figure Q10-9.
If you were to cut this linear DNA with HindIII, what type of DNA fragments do you predict you will obtain?(a) three linear pieces of DNA(b) two linear pieces of DNA, only one of which can be cut by EcoRI(c) two linear pieces of DNA, both of which can be cut by EcoRI (d) two linear pieces of DNA, only one of which can be cut by HindIII
10-10 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.
During gel __________________, DNA fragments can be loaded into one end of an agarose slab to separate the fragments on the basis of charge. As ____________________ is applied across the agarose slab, the DNA molecules, which have a __________________ charge, will migrate toward the ___________________ electrode. Because _________________ DNA fragments will migrate more quickly, they will be found furthest away from the area of the gel where the DNA fragments were loaded. One method to visualize the DNA on the agarose slab involves staining the DNA with a dye that will __________________ under ultraviolet light.
centrifugation negativedigested neutralelectrophoresis positivefluoresce radioactivegravity sequencingirradiate smallerlarger voltage
10-11 During gel electrophoresis, DNA fragments _______________________.(a) travel through a matrix containing a microscopic network of pores.(b) migrate toward a negatively charged electrode.(c) can be visualized without stains or labels.(d) are separated on the basis of their sequence.
10-12 Assume that defects in a hypothetical gene X have been linked to antisocial behavior. Two copies of a defective gene X predispose a child to bad behavior from childhood, whereas a single copy of the gene seems to produce no symptoms until adulthood. Because early treatment can counteract the effects of the gene, a program of voluntary genetic testing is being performed with delinquent prospective parents. Charles S. and Caril Ann F. have been arrested on charges of robbery and assault, and Caril Ann is pregnant with Charles’s child. You obtain DNA samples from Charles, Caril Ann, and the fetus. You digest these samples with NotI and use these samples to perform two Southern blots, which you probe with two different oligonucleotide probes, A and B, that hybridize to different parts of the normal gene X, as shown in Figure Q10-12A. Your results are shown in Figure Q10-12B.
A. Which of the three individuals have defects in gene X?B. Which individuals have a single defective gene and which have two defective copies of the gene?C. Indicate the nature (single base-pair mutation or deletion) and location of each individual’s defects on gene X.
10-13 Figure Q10-13 shows a restriction map of a piece of DNA containing your favorite gene. The arrow indicates the position and orientation of the gene in the DNA. In part (B) of the figure are enlargements showing the portions of the DNA whose sequences have been used to make oligonucleotide probes A, B, C, and D. Which of the oligonucleotides can be used to detect the gene in each of the following?
A. A Southern blot of genomic DNA cut with HindIII.B. A Northern blot.
10-14 A double-stranded DNA molecule can be separated into single strands by heating it to 90°C because _______________________.(a) heat disrupts the hydrogen bonds holding the sugar–phosphate backbone together.(b) DNA is negatively charged.(c) heat disrupts hydrogen-bonding between complementary nucleotides.(d) DNA is positively charged.
10-15 You are interested in a single-stranded DNA molecule that contains the following sequence:
5′- …..GATTGCAT…. -3′
Which molecule can be used as a probe that will hybridize to your sequence of interest?(a) 5′-GATTGCAT-3′(b) 5′-TACGTTAG-3′(c) 5′-CTAACGTA-3′(d) 5′-ATGCAATC-3′
10-16 During DNA renaturation, two DNA strands will ________.(a) break the covalent bonds that hold the nucleotides together while maintaining the hydrogen bonds that hold the two strands together.(b) break the hydrogen bonds that hold the two strands together with no effect on the covalent bonds that hold the nucleotides together.(c) re-form a double helix if the two strands have complementary sequences.(d) re-form a double helix if the two strands are identical in sequence.
10-17 Which of the following statements about gel-transfer hybridization (or Southern blotting) is false?(a) This technique involves the transfer of DNA molecules from gel onto nitrocellulose paper or nylon paper.(b) In this technique, single-stranded DNA is separated by electrophoresis.(c) A labeled DNA probe binds to the DNA by hybridization.(d) The DNA that is separated on a gel is not labeled.
DNA Cloning in Bacteria
10-18 DNA ligase is an enzyme used when making recombinant DNA molecules in the lab. In what normal cellular process is DNA ligase involved?(a) none, it is only found in virally infected cells(b) transcription(c) transformation(d) DNA replication
10-19 Figure Q10-19 shows the cleavage sites of several restriction nucleases. Figure Q10-19
You cut a vector using the PciI restriction nuclease. Which of the following restriction nucleases will generate a fragment that can be ligated into this cut vector with the addition of only ligase and ATP?(a) HindIII(b) NcoI(c) MmeI(d) NspV
10-20 Figure Q10-20 depicts a strategy by which a DNA fragment produced by cutting with the EcoRI restriction nuclease can be joined to a DNA fragment produced by cutting DNA with the HaeIII restriction nuclease.
Note that cutting DNA with EcoRI produces a staggered end, whereas cutting DNA with HaeIII produces a blunt end. Why must polymerase be added in this reaction?(a) Polymerase will fill in the staggered end to create a blunt end.(b) Polymerase is needed to seal nicks in the DNA backbone.(c) Polymerase will add nucleotides to the end produced by the HaeIII restriction nuclease.(d) Without polymerase, there will not be enough energy for the reaction to proceed.
10-21 DNA can be introduced into bacteria by a mechanism called ____________.(a) transcription.(b) ligation.(c) replication.(d) transformation.
10-22 Figure Q10-22 shows the recognition sequences and sites of cleavage for the restriction enzymes SalI, XhoI, PstI, and SmaI, and also a plasmid with the sites of cleavage for these enzymes marked.
A. After which of the five treatments listed below can the plasmid shown in Figure Q10-22 re-form into a circle simply by treating it with DNA ligase? Assume that after treatment any small pieces of DNA are removed, and it is the larger portion of plasmid that will reassemble into a circle.
After digestion with __________.1. SalI alone2. SalI and XhoI3. SalI and PstI4. SalI and SmaI5. SmaI and PstI
B. After which of the treatments can the plasmid re-form into a circle by the addition of DNA ligase after treating the cut DNA with DNA polymerase in a mixture containing the four deoxyribonucleotides? Again, assume that you are trying to get the larger portion of plasmid to reassemble into a circle.
10-23 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.
Two fragments of DNA can be joined together by __________________. Restriction enzymes that cut DNA straight across the double helix produce fragments of DNA with __________________. A fragment of DNA is inserted into a __________________ in order to be cloned in bacteria. A __________________ library contains a collection of DNA clones derived from mRNAs. A __________________ library contains a collection of DNA clones derived from chromosomal DNA.
blunt ends DNA polymerase RNAcDNA genomic staggered endsDNA ligase probe vector
10-24 Name three features that a cloning vector for use in bacteria must contain. Explain your answers.
10-25 A plasmid ______________.(a) can confer antibiotic resistance to a bacterium.(b) is a single-stranded circular DNA molecule that can undergo horizontal transfer among bacteria.(c) is a tool designed in the lab and never found in naturally occurring bacteria.(d) always becomes part of the bacterial chromosome during transformation.
10-26 Which of the following statements about DNA libraries is true?(a) Production of a DNA library involves the direct insertion of short DNA fragments into bacteria through transformation.(b) By placing the library DNA into bacteria, the bacteria can be used to amplify the desired DNA fragments from the DNA library.(c) Individual bacteria that have taken up most of the library DNA are selected for during the construction of a DNA library.(d) The library DNA within the bacteria will only be replicated when it hybridizes to a DNA probe.
10-27 Which of the following statements about genomic DNA libraries is false?(a) The larger the size of the fragments used to make the library, the fewer colonies you will have to examine to find a clone that hybridizes to your probe.(b) The larger the size of the fragments used to make the library, the more difficult it will be to find your gene of interest once you have identified a clone that hybridizes to your probe.(c) The larger the genome of the organism from which a library is derived, the larger the fragments inserted into the vector will tend to be.(d) The smaller the gene you are seeking, the more likely it is that the gene will be found on a single clone.
10-28 A DNA library has been constructed by purifying chromosomal DNA from mice, cutting the DNA with the restriction enzyme NotI, and inserting the fragments into the NotI site of a plasmid vector. What information cannot be retrieved from this library?(a) gene regulatory sequences(b) intron sequences(c) sequences of the telomeres (the ends of the chromosomes)(d) amino acid sequences of proteins
10-29 You want to design a DNA probe used for hybridization to isolate a clone from a cDNA library. Which of the following concerns about DNA probe design is the most legitimate?(a) You must be careful when designing your probe to take into account which DNA strand was transcribed in mRNA and choose a probe complementary to the mRNA.(b) You must be careful not to include any DNA sequences in your probe that are upstream (5′) of the AUG start codon.(c) You must make sure that all the DNA sequences in your probe lie within an exon, and do not span two exons.(d) You must make sure that all the DNA sequences in your probe are not located downstream (3′) of the polyadenylation signal.
10-30 You want to design a DNA probe used for hybridization to isolate a clone from a cDNA library. Which of the following statements about DNA probes is true?(a) The shorter the DNA probe used to probe the library, the greater the number of colonies to which the probe might hybridize.(b) A DNA probe that contains sequences that span two exons is better suited to the purpose than a DNA probe that only contains sequences from one exon.(c) A DNA probe that contains sequences immediately upstream of the DNA that codes for the first methionine in the open reading frame will usually not hybridize to clones in a cDNA library.(d) Hybridization of a DNA probe to the plasmid of interest will permit the detection of the clone of interest; labeling of the DNA probe is not necessary.
10-31 You have the amino acid sequence of a protein and wish to search for the gene encoding this protein in a DNA library. Using this protein sequence, you deduce a particular DNA sequence that can encode this protein. Why is it unwise to use only this DNA sequence you have deduced as the probe for isolating the gene encoding your protein of interest from the DNA library?
10-32 What is the main reason for using a cDNA library rather than a genomic library to isolate a human gene from which you wish to make large quantities of the human protein in bacteria?
10-33 Some clones from cDNA libraries can have defects because of the way in which a cDNA library is constructed. For each dilemma (A to D) listed below, indicate which of the outcomes (1 to 4) you might encounter, and explain why. Outcomes may be used more than once.
10-34 You have an oligonucleotide probe that hybridizes to part of gene A from a eukaryotic cell. Indicate whether a cDNA library or a genomic DNA library will be more appropriate for use in the following applications.A. You want to study the promoter of gene A.B. Gene A encodes a tRNA and you wish to isolate a piece of DNA containing the full-length sequence of the tRNA.C. You discover that gene A is alternatively spliced and you want to see which predicted alternative splice products the cell actually produces.D. You want to find both gene A and the genes located near gene A on the chromosome.E. You want to express gene A in bacteria to produce lots of protein A.
Note: The following codon table is to be used for Problems 10-35, 10-36, 10-49, 10-60, 10-61, and 10-62.
10-35 Which of the restriction nucleases listed below can potentially cleave a segment of cDNA that encodes the peptide KIGDACF?
Essential Cell Biology-4th Ed Test-Bank – Alberts-
10-36 Your biochemist friend has isolated a protein he thinks is responsible for making you feel sleepy. Since he knows you’re taking Cell Biology, he wants you to help him isolate the gene encoding this protein. Unfortunately, because your friend could only isolate small amounts of protein, he was only able to obtain three short stretches of amino acid sequence from the protein:(a) H-C-W-K-M(b) R-S-L-L-S(c) D-A-Q-W-Y
For each of the three peptides above, you design a set of DNA oligonucleotide probes that can be used to detect the gene in a library. Which of the three sets of oligonucleotide probes would be preferable for screening a library? Explain. (Refer to the codon table above Q10-35.)
10-37 Your friend has isolated a protein present in the cheek cells of all straight-A seniors at your school. She says that this protein helps you remember everything you read and therefore will help cut down on the number of hours needed to study for exams. She sequences the protein, which she calls “geniuszyme,” and designs a probe to isolate the gene that encodes it. She has consulted with a company expert at this process and they have 100% confidence that her probe will work. She also obtains a high-quality liver cDNA library from the company and uses her probe to try to isolate the gene for geniuszyme. Unfortunately, she is unable to isolate any clones.A. What is the likely explanation for her failure?B. Not to be discouraged, your friend has obtained some genomic DNA isolated from the nuclei of liver cells and has made a genomic library from that DNA. Do you expect she will succeed in isolating the geniuszyme gene from this library? Why or why not?
DNA Cloning by PCR
10-38 PCR was invented in _______.(a) the 1800s.(b) the 1950s.(c) the 1980s.(d) 2009.
10-39 Starting with one double-stranded DNA molecule, how many cycles of PCR would you have to perform to produce about 100 double-stranded copies (assuming 100% efficiency per cycle)?(a) 2(b) 7(c) 25(d) 100
10-40 Which of the following statements about PCR is false?(a) PCR uses a DNA polymerase from a thermophilic bacterium.(b) PCR is particularly powerful because after each cycle of replication, there is a linear increase in the amount of DNA available.(c) For PCR, every round of replication is preceded by the denaturation of the double-stranded DNA molecules.(d) The PCR will generate a pool of double-stranded DNA molecules, most of which will have DNA from primers at the 5′ ends.
10-41 PCR involves a heating step, followed by a cooling step, and then DNA synthesis. What is the primary reason for why this cooling step is necessary?(a) Cooling the reaction ensures the integrity of the covalent bonds holding the nucleotides together in the DNA strand.(b) Cooling the reaction gives the DNA polymerase an opportunity to rest from the previous cycle so that it will be ready for the next round of synthesis.(c) Transcription takes place during the cooling step.(d) Cooling the reaction brings the temperature down to a level that is compatible with the short primers forming stable hydrogen bonds with the DNA to be amplified.
10-42 Why is a heat-stable DNA polymerase from a thermophilic bacterium (the Taq polymerase) used in the polymerase chain reaction rather than a DNA polymerase from E. coli or humans?
10-43 Which of the following limits the use of PCR to detect and isolate genes?(a) The sequence at the beginning and end of the DNA to be amplified must be known.(b) It also produces large numbers of copies of sequences beyond the 5′ or 3′ end of the desired sequence.(c) It cannot be used to amplify cDNAs or mRNAs.(d) It will amplify only sequences present in multiple copies in the DNA sample.
10-44 You want to amplify the DNA between the two stretches of sequence shown in Figure Q10-44. Of the listed primers, choose the pair that will allow you to amplify the DNA by PCR. Figure Q10-44
10-45 Your friend works at the Centers for Disease Control and Prevention and has discovered a brand-new virus that has recently been introduced into the human population. She has just developed a new assay that allows her to detect the virus by using PCR products made from the blood of infected patients. The assay uses primers in the PCR assay that hybridize to sequences in the viral genome.
Your friend is distraught because of the result she obtained (see Figure Q10-45) when she looked at PCR products made using the blood of three patients suffering from the viral disease, using her own blood, and using a leaf from her petunia plant.
You advise your friend not to panic, as you believe she is missing an important control. Which one of the choices listed below is the best control for clarifying the results of her assay? Explain your answer.
(a) a PCR assay using blood from a patient who is newly infected but does not yet show symptoms(b) a PCR assay using blood from a dog(c) a PCR assay using blood from an uninfected person(d) repeating the experiments she has already done with a new tube of polymerase
10-46 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.
A nuclease hydrolyzes the __________________ bonds in a nucleic acid. Nucleases that cut DNA only at specific short sequences are known as __________________. DNA composed of sequences from different sources is known as __________________. __________________ can be used to separate DNA fragments of different sizes. Millions of copies of a DNA sequence can be made entirely in vitro by the __________________ technique.
DNA sequencing phosphodiesterendonucleases polymerase chain reactionexonucleases recombinant DNAgel electrophoresis restriction nucleaseshydrogen ribonucleasesnucleic acid hybridization
Exploring and Exploiting Gene Function
10-47 Why is an excess of normal deoxyribonucleoside triphosphate molecules (dNTPs) needed during dideoxy sequencing?(a) DNA polymerase uses the dNTPs to synthesize a DNA molecule complementary to the molecule being sequenced.(b) dNTPs are consumed as energy to fuel the sequencing reactions.(c) When dNTP levels are too low, there will be very few chain-termination events.(d) The dNTPs can hybridize to the fragment to be sequenced and serve as primers for DNA polymerase.
10-48 Why are dideoxyribonucleoside triphosphates used during DNA sequencing?(a) They cannot be incorporated into DNA by DNA polymerase.(b) They are incorporated into DNA particularly well by DNA polymerases from thermophilic bacteria.(c) Incorporation of a dideoxyribonucleoside triphosphate leads to the termination of replication for that strand.(d) Dideoxyribonucleoside triphosphates are more stable than deoxyribonucleoside triphosphates.
10-49 You have sequenced a short piece of DNA and produced the gel shown in Figure Q10-49. Figure Q10-49
A. What is the sequence of the DNA, starting from the 5′ end?B. If you knew that this sequence is from the middle of a protein-coding cDNA clone, what amino acid sequence can be deduced from this sequence? (Refer to the codon table above Q10-35.)
10-50 You have sequenced a fragment of DNA and produced the gel shown in Figure Q10-50. Near the top of the gel, there is a section where there are bands in all four lanes (indicated by the arrow). Which of the following mishaps would account for this phenomenon? Explain your answer. Figure Q10-50
(a) You mistakenly added all four dideoxynucleotides to one of the reactions.(b) You forgot to add deoxynucleotides to the reactions.(c) Your primer hybridizes to more than one area of the fragment of DNA you are sequencing.(d) A restriction nuclease cut a fraction of the DNA you are sequencing.
10-51 With fully automated Sanger sequencing, all four chain-terminating ddNTPs can be added into a single reaction. This is different from the traditional slab gel Sanger sequencing, where a different reaction had to be carried out for each ddNTP. The mixing of all four ddNTPs can be carried out because ______________.(a) the fully automated Sanger sequencing reactions are loaded onto a capillary gel.(b) the fully automated Sanger sequencing reactions utilize ddNTPs each labeled with a different fluorescent tag, which allows all four ddNTPs to be incorporated into a single molecule of DNA. (c) the fully automated Sanger sequencing reactions generate a set of products, each of which carries a single fluorescent tag whose color reveals the identity of the base that is at the end of the product.(d) the fully automated Sanger sequencing reactions do not require DNA polymerase because the bases are read as the DNA is pulled through a tiny pore at the end of the capillary gel.
10-52 Second-generation sequencing differs from Sanger sequencing because _____________.(a) second-generation sequencing does not depend on chain-terminator ddNTPs.(b) second-generation sequencing does not require DNA polymerase.(c) for the cost per base sequenced, second-generation sequencing is much more expensive than Sanger sequencing.(d) second-generation sequencing can sequence tens of millions of pieces of DNA at the same time on a single glass slide.
10-53 Which of the following techniques is not appropriate if you want to examine the transcriptome of a specific tissue?(a) in situ hybridization(b) production of a cDNA library(c) RNA-Seq(d) microarray analysis
10-54 You are interested in understanding how the brain works, and are using the fruit fly Drosophila as a model system to study brain development. You perform a microarray analysis to try to determine genes expressed in the fly brain. For your microarray experiment, you first prepare cDNA from fly brains and label it with a red fluorochrome. Then you isolate cDNA from whole flies and label it with a green fluorochrome. Next you hybridize these cDNA populations to a microarray containing the Drosophila genes. From this you obtain a list of genes that are specifically enriched in the brain (those that show up as a red spot on the microarray).
You are disappointed because your favorite fly gene, tubby, does not appear on this list, even though you have repeated the microarray experiment 10 times and did not encounter any technical difficulties. The reason you thought tubby would appear on this list is that you believe tubby is important for brain development, because flies mutant in tubby have no brains. Not to be discouraged, you perform in situ analysis using the tubby DNA as a probe, and see that it is expressed at high levels in the fly brain of normal flies but not expressed in animals lacking the tubby gene.
Why do you think tubby did not show up as a gene specifically enriched in the brain in your microarray experiment?
10-55 You create a recombinant DNA molecule that fuses the coding sequence of green fluorescent protein to the regulatory DNA sequences that control the expression of your favorite genes. Which of the following pieces of information can you NOT gain by examining the expression of this reporter gene?(a) the tissue where the protein encoded by this gene is expressed(b) the cell in which the protein encoded by this gene is expressed(c) the specific location within the cell of the protein encoded by this gene(d) when, during an organism’s development, this gene is expressed
10-56 Which of the following statements about RNA interference (or RNAi) is false?(a) RNAi is a natural mechanism used to regulate genes.(b) During the process of RNAi, hybridization of a small RNA molecule with the mRNA degrades the mRNA.(c) Because RNAi depends on the introduction of a double-stranded RNA into a cell or an organism, it is not a process that can cause heritable changes in gene expression.(d) In C. elegans, RNAi can be introduced into the animals by feeding them with bacteria that produce the inhibitory RNA molecules.
10-57 You have been hired to create a cat that will not cause allergic reactions in cat-lovers. Your coworkers have cloned the gene encoding a protein found in cat saliva, expressed the protein in bacteria, and shown that it causes violent allergic reactions in people. But you soon realize that even if you succeed in making a knockout cat lacking this gene, anyone who buys one will easily be able to make more hypoallergenic cats just by breeding them. Which of the following will ensure that people will always have to buy their hypoallergenic cats from you?(a) Inject the modified embryonic stem (ES) cells into embryos that have a genetic defect to prevent the mature adult from reproducing.(b) Implant the injected embryos into a female cat that is sterile as a result of a genetic defect.(c) Sell only the offspring from the first litter of the female cat implanted with the injected embryos.(d) Surgically remove the sexual organs of all the knockouts before you sell them.
10-58 Insulin is a small protein that regulates blood sugar level and is given to patients who suffer from diabetes. Many years ago, diabetics were given insulin that had been purified from pig pancreas. Once recombinant DNA techniques became available, the DNA encoding insulin could be placed into an expression vector and insulin could be produced in bacteria. Which of the following is NOT a reason why purifying insulin from bacteria is a better way to produce insulin for diabetics than using insulin purified from a pig pancreas.(a) Insulin can be easily produced in large quantities from cells carrying the cloned DNA sequence.(b) The creation of transgenic pigs that expressed insulin was very expensive compared to the cost of creating bacteria that expressed insulin.(c) Insulin made from a bacterial culture and then purified will be free of any possible contaminating viruses that pigs (and any other animals) harbor. Since pigs are more closely related to people than bacteria are, their viruses are more likely to be harmful to people than are viruses that might infect bacteria.(d) The pig protein has slight amino acid differences compared to the human protein, so human insulin produced by bacteria will work better in people.
10-59 Which of the following describes a feature found in bacterial expression vectors but not in cloning vectors?(a) origin of replication(b) cleavage sites for restriction nucleases(c) promoter DNA sequences(d) a polyadenylation signal
10-60 You are studying a protein, and a small fragment of its sequence is shown below. You have decided that the glutamine in the protein segment has an important role in its function. You decide to change this glutamine to a lysine using DNA technology with the use of site-directed mutagenesis. You have a plasmid that contains the full-length version of the gene that encodes this protein and wish to create a new DNA molecule that has a change in only one base and that substitutes a lysine for a glutamine. Given the partial mRNA sequence that codes for this stretch of protein below, devise a 21-nucleotide DNA oligonucleotide that can be used to make this mutation. Be sure to label the 5′ and 3′ ends. (Refer to the codon table above Q10-35.)
F D P Q G S H5′-UUCGACCCGCAGGGCAGCCAC–3′
10-61 You are studying a protein that contains the peptide sequence RDWKLVI. The part of the DNA encoding this peptide is included in the sequence shown below.
This sequence does not contain any HindIII restriction enzyme sites; the target sequence for the HindIII restriction nuclease is shown in Figure Q10-61.
Your goal is to create a HindIII site on this plasmid without changing the coding sequence of the protein. Explain how you would do this. (Refer to the codon table above Q10-35.)
10-62 You have created a piece of recombinant DNA by placing a cDNA from a gene you believe is important for the differentiation of liver cells (called LC1) onto an expression plasmid that contains all the sequences necessary for propagation of this DNA in bacteria and for the production of the LC1 protein in bacteria. A picture of this plasmid is shown in Figure 10-62A, with the segment of the DNA containing the LC1 gene depicted as a gray rectangle; the promoter sequence is depicted as a white rectangle. The LC1 protein is phosphorylated on serine 54; the nucleotide sequence of the portion of the DNA that encodes this region is shown below the diagram. All HindIII and SalI restriction sites have also been mapped on the plasmid; the recognition sequences for these restriction nucleases are shown in Figure 10-62B.
A. Given the information above, write out the amino acids 52 to 57, encoded by the nucleotide sequence shown above. Be sure to number the amino acids appropriately. (Hint: Remember, serine is amino acid number 54.) (Refer to the codon table above Q10-35.)B. You want to create a mutant version of the LC1 gene that replaces the serine 54 found on this peptide with a glycine. You want to do this by changing only one nucleotide, and you also want to destroy the HindIII recognition sequence with this change. Write out a 21-nucleotide DNA sequence that can accommodate these changes. Be sure to (i) write out the DNA and label the 5′ and 3′ ends, (ii) underline the mutated HindIII recognition site, and (iii) circle any change made to the original sequence.
How We Know: Sequencing the Human Genome
10-63 You have been asked to consult for a biotech company that is seeking to understand why some fungi can live in very extreme environments, such as the high temperatures inside naturally occurring hot springs. The company has isolated two different fungal species, F. cattoriae and W. gravinius, both of which can grow at temperatures exceeding 95°C. The company has determined the following things about these fungal species:
By sequencing and examining their genomes, the biotech company hopes to understand why these species can live in extreme environments. However, the company only has the resources to sequence one genome, and would like your input as to which species should be sequenced and whether you believe a shotgun strategy will work in this case. (Be sure to explain your answer.)
10-64 Figure Q10-64A depicts the restriction map of one segment of the human genome for four restriction nucleases W, X, Y, and Z. Figure Q10-64B depicts the restriction maps of four individual BAC clones that contain segments of human DNA from the region depicted in Figure Q10-64A.
From this information, how would you order these BAC clones, from left to right?(a) 1, 2, 3, 4(b) 2, 1, 4, 3(c) 3, 4, 2, 1(d) 4, 1, 3, 2
10-2 (b) Restriction nucleases cleave double-stranded DNA.
10-3 Choice (c) is correct. A restriction nuclease that has a 4-base-pair recognition sequence cuts on average once every 44 or 256 base pairs; one that has a 6-base-pair recognition sequence cuts once every 46 or 4096 base pairs; one that has an 8-base-pair recognition sequence cuts once every 48 or 65,536 base pairs; one that has a 12-base-pair recognition sequence cuts once every 412 or 16 million base pairs. Thus, to obtain fragments of about 70 kb in size, you would cut with a nuclease that recognizes an 8-base-pair site.
10-8 A. You would first digest your sample with a combination of restriction enzymes selected so that they give a set of fragment sizes that could have come from only one of the plasmids. Then you would run the resulting mixture of DNA fragments on a gel alongside a set of size markers and determine the size of each fragment. By looking at the restriction maps, you should then be able to match your results to one of the plasmids.B. Digestion with any of the following combinations will enable you to distinguish which plasmid you have: HindIII + BglII; EcoRI + BglII; EcoRI + BglII + HindIII. The plasmids are the same size, so you cannot distinguish between them simply by making a single cut (with HindIII) and determining the size of the complete DNA by gel electrophoresis. Nor can you distinguish them by cutting with all four restriction nucleases, because the set of fragment sizes produced from both plasmids will be the same. Cutting with BamHI or EcoRI on their own is not sufficient because you will get bands of the same size from both plasmid A and plasmid B. The only difference between the two plasmids is in the location of the BglII site relative to the two BamHI sites, so if you cut with an enzyme that cuts outside the BamHI fragment and with BglII, you will get different-sized fragments from the two plasmids.
10-10 During gel electrophoresis, DNA fragments can be loaded into one end of an agarose slab to separate the fragments on the basis of charge. As voltage is applied across the agarose slab, the DNA molecules, which have a negative charge, will migrate toward the positive electrode. Because smaller DNA fragments will migrate more quickly, they will be found furthest away from the area of the slab where the DNA fragments were loaded. One method to visualize the DNA on the agarose slab involves staining the DNA with a dye that will fluoresce under ultraviolet light.
10-12 A. All three are affected.B. The two parents have a single defective copy of the gene; the fetus has two defective copies.C. As seen in the two blots in Figure Q10-12B, Caril Ann and Charles each have one full-length copy of gene X (the bands at the top of the gel), which hybridizes with both probe A and probe B. The fetus does not. The blot with probe A shows that Caril Ann and the fetus have a short fragment of gene X that hybridizes with probe A only, indicating that this copy of gene X has a deletion somewhere other than in the region recognized by probe A. The second blot (with probe B) shows that Charles and the fetus have a short fragment that hybridizes with probe B, indicating that this copy of gene X has a deletion somewhere other than in the region recognized by probe B. Because the shortened gene found in Charles does not show up on the probe A blot, this deletion must be in the region of A; similarly, because the shortened gene found in Caril Ann does not show up on the probe B blot, her deletion must be in the region of B. The fetus has inherited two defective copies of gene X, one from each parent.
10-13 A. All four oligonucleotide probes.B. Oligonucleotide probe B and oligonucleotide probe D. Both the upper and lower strands of DNA are present in genomic Southern blots, so all four oligonucleotides will hybridize to either Southern blot. (Oligonucleotides A and B will still be able to hybridize to genomic DNA cut with BglII, because they can still base-pair with the individual fragments that result from the digest.) Northern blots contain only RNA, which has the sequence of the upper strand of the DNA. Hence, only oligonucleotide probe B and oligonucleotide probe D will hybridize to a Northern blot.
10-16 Choice (c) is correct. DNA strands should not renature if their sequences are identical [choice (d)]. The breaking of the hydrogen bonds between two DNA strands is the process of denaturation [choice (b)].
10-17 (b) During Southern blotting, double-stranded DNA is loaded onto the agarose gel. The DNA becomes denatured (and thus single-stranded) as it gets transferred by the alkali solution from the gel to the nitrocellulose or nylon sheet.
10-19 (b) However, you will not be able to excise the fragment after ligation, because you will destroy both the PciI site and the NcoI site, creating a new site with the sequence
10-22 A. 1 and 2. When SalI and XhoI cut DNA, the staggered ends left behind will match up by base-pairing and can therefore be joined by ligase alone.B. 1, 2, and 4. SmaI cuts and leaves a blunt end. Addition of DNA polymerase and the four deoxyribonucleotides will fill in the 5′ overhangs generated by digestion with SalI and XhoI, leaving blunt ends. DNA ligase joins the blunt ends. However, 3′ overhangs (that is, those generated by PstI) will not be filled in, because DNA polymerase moves in a 5′-to-3′ direction. DNA ligase will not join 3′ overhangs to blunt-ended DNA, which are the situations presented in treatments 3 and 5.
10-23 Two fragments of DNA can be joined together by DNA ligase. Restriction enzymes that cut DNA straight across the double helix produce fragments of DNA with blunt ends. A fragment of DNA is inserted into a vector in order to be cloned in bacteria. A cDNA library contains a collection of DNA clones derived from mRNAs. A genomic library contains a collection of DNA clones derived from chromosomal DNA.
10-24 A cloning vector for use in bacteria must contain the following:1. a bacterial replication origin (to allow the plasmid to be replicated)2. at least one unique restriction site (to allow easy insertion of foreign DNA)3. an antibiotic-resistance gene or some other selectable marker gene (to allow selection for bacteria that have taken up the recombinant plasmids)
10-25 Choice (a) is correct. Plasmids that are found in naturally occurring bacteria [choice (c)] can confer antibiotic resistance to a bacterium; plasmids in the lab often carry an antibiotic-resistance gene so that scientists can select for bacteria that are carrying their recombinant DNA molecule. Plasmids are made of double-stranded DNA [choice (b)], and typically have their own replication origin, allowing for plasmids to replicate independently of the bacterial chromosome [choice (d)].
10-26 Choice (b) is correct. To make a DNA library, the fragmented DNA must first be placed into an appropriate plasmid before it can be placed into the bacteria through transformation [choice (a)]. Individual bacteria in a library should only carry a single DNA fragment from the library [choice (c)]. Although DNA probes are used to identify the bacterial colony that contains the DNA library fragment of interest, this hybridization is done on dead bacteria and is not involved in replication of the desired DNA [choice (d)].
10-27 (c) The sizes of the fragments left after a restriction digest do not depend on the total size of the genome; they depend on the sequence of the genome and the frequency with which the restriction enzyme recognition site is found in the genome. Choices (a) and (b) are true: as a limiting case, think of what would happen if a fragment the size of the entire genome were inserted into the bacterial vector. You would have to screen only one colony to find the clone that hybridized to your probe, but it would be very difficult to find out where on the insert your gene of interest lay. Choice (d) is true: the larger the gene you are seeking, the more likely it is that there will be a restriction fragment in the gene (or that the gene will be broken if the DNA was fragmented by random shearing), and hence the less likely it is that the entire gene will be found in one clone.
10-28 (c) The very ends of all of the chromosomes are unlikely to be NotI sites, meaning that the fragments containing the ends of the chromosomes will not be able to insert into the bacterial vector (because they have not been cut by NotI at both ends) and will be lost from the library. All sequences present in genomic DNA (which includes regulatory sequences and introns) should be present in a genomic library. The coding sequence of the gene (and hence the amino acid sequence of the encoded protein) is also present in a genomic clone, although it is interrupted by intron sequences and is therefore somewhat difficult (but not impossible) to determine.
10-29 Choice (d) is correct. mRNAs are converted into cDNAs, which are double-stranded, so it is not necessary to worry about which strand the probe will hybridize with [choice (a)]. The transcriptional and translational start sites of an mRNA molecule differ, so including some sequences 5′ of the AUG start codon is not necessarily a problem [choice (b)]. Because cDNAs are made from mRNAs, the exon-splicing events have already occurred so it shouldn’t matter if your probe sequences span two exons or lie within a single exon [choice (c)].
10-30 (a) The shorter the DNA probe, the more likely it is that that particular sequence will show up in the genome at random. cDNA libraries contain sequences represented by exons, so it does not really matter whether or not the probe spans two exons [choice (b)]. mRNAs usually have 5′ untranslated regions that should be represented in a cDNA library, so choice (c) is not true. DNA probes are usually labeled (for example, with radioactivity) for visualization [choice (d)].
10-31 Because most amino acids can be encoded by more than one codon, a given sequence of amino acids could be encoded by several different nucleotide sequences. Probes corresponding to all these possible sequences have to be synthesized, to be sure of including the one that corresponds to the actual nucleotide sequence of the gene and thus will hybridize with it.
10-32 The gene isolated from a genomic library would still contain introns, and bacteria do not contain the biochemical machinery for removing introns by RNA splicing. The same gene isolated from the cDNA library will already have had its introns removed.
10-33 A. Outcome 1 would occur. If the mRNA is degraded from the 5′ end, it will still be reverse-transcribed and will end up in the library as a clone lacking its 5′ end.B. Outcome 4 would occur. If the mRNA is degraded from the 3′ end, it will lack its 3′ poly-A tail. In the construction of a cDNA library, only molecules that still have their poly-A tail will be reverse-transcribed, so mRNAs lacking their 3′ end will not be represented in the library.C. Outcome 1 would occur. If the 5′ end hybridizes to sequences in the middle of the gene, the “hairpin” formed when the single-stranded DNA loops back on itself to form the primer for DNA polymerase will be very large. After this loop has been digested, the remaining double-stranded DNA fragment will lack the 5′ end of the gene.D. Outcome 2 would occur. If the gene has a long stretch of internal As, the poly T primer used in the reverse transcription step can hybridize to the internal poly-A stretch rather than to the poly-A tail, and the resulting cDNA will have lost its 3′ end.
10-34 A. Genomic library. (cDNAs are produced from mRNAs; therefore, the promoters will not be included in a cDNA library.)B. Genomic library. (cDNAs are usually constructed by using an oligo dT primer; tRNAs do not have poly-A tails. If the cDNA library were made using random primers, it would be unlikely to contain the full-length version of the tRNA.)C. cDNA library. (Because cDNAs are produced from mRNAs, isolating cDNAs would tell you which splice variants are produced in a cell.)D. Genomic library. (A genomic DNA fragment can contain the genes next to your gene of interest; a cDNA will not.)E. cDNA library. (Bacteria do not have the ability to remove introns, which may exist in DNA isolated from a genomic library.)
10-35 The nucleotide sequences that can encode the peptide KIGDACF are shown below.
The enzyme NsiI cleaves at ATGCAT.
10-36 (a) H-C-W-K-M. There is the least amount of degeneracy in the nucleotides that could code for this peptide; see below.
10-37 A. Geniuszyme is not expressed in the liver. Because cDNA is made from mRNA, a cDNA library reflects only those genes that are actively transcribed in a particular tissue.B. Yes, she should be able to isolate the gene, because genomic DNA is essentially the same in all tissues.
10-39 (b) After 7 cycles of PCR, you would have 128 molecules of DNA if the reaction were 100% efficient.
10-41 Choice (d) is correct. During PCR, the reaction is heated to a temperature that will break the hydrogen bonds holding the strands together but should not be harmful to the covalent bonds holding the nucleotides together [choice (a)]. The DNA polymerase used in PCR is a thermostable enzyme that can tolerate the high temperatures used in PCR [choice (b)]. Transcription involves the production of RNA from DNA and does not occur during PCR [choice (c)].
10-42 The PCR technique involves heating the reaction at the beginning of each cycle to separate the newly synthesized DNA into single strands so that they can act as templates for the next round of DNA synthesis. Using a heat-stable polymerase avoids having to add it afresh for each round of DNA replication.
10-43 (a) To construct primers that will bracket the desired gene, you have to know the sequence at the beginning and end of the DNA to be copied.
10-44 The appropriate PCR primers are primer 1 (5′-GACCTGTGGAAGC-3′) and primer 8 (5′-TCAATCCCGTATG-3′). The first primer will hybridize to the bottom strand and prime synthesis in the rightward direction. The second primer will hybridize to the top strand and prime synthesis in the leftward direction. (Remember that strands pair antiparallel.)
The middle two primers in each list (primers 2, 3, 6, and 7) would not hybridize to either strand. The remaining pair of primers (4 and 5) would hybridize, but would prime synthesis in the wrong direction—that is, outward, away from the central segment of DNA. Each wrong choice has been made at one time or another in most laboratories that use PCR. In most cases, the confusion arises because the conventions for writing nucleotide sequences have been ignored. By convention, nucleotide sequences are written 5′ to 3′, with the 5′ end on the left. For double-stranded DNA, the 5′ end of the top strand is on the left.
10-45 (c) Primers can sometimes hybridize to unintended sequences and produce unintended products. The appropriate control for your friend’s experiment would be DNA from an uninfected person; in that way, she would be able to determine whether the bands present in the PCR from her blood truly correspond to product generated from viral DNA rather than cross-hybridization to DNA sequences in the human genome, because the bands would be absent from a person uninfected by the virus in the former case only. Doing PCR from an infected but asymptomatic person would not be useful [choice (a)], because it would not allow your friend to distinguish whether she is infected. Although doing PCR from dog blood [choice (b)] should not give any viral bands, any nonspecific products from a dog would probably be different from those in your friend. The absence of PCR fragments in the petunia lane suggests that there is no viral contaminant in any of your friend’s reagents, so using a new tube of polymerase is not the solution [choice (d)].
10-46 A nuclease hydrolyzes the phosphodiester bonds in a nucleic acid. Nucleases that cut DNA only at specific short sequences are known as restriction nucleases. DNA composed of sequences from different sources is known as recombinant DNA. Gel electrophoresis can be used to separate DNA fragments of different sizes. Millions of copies of a DNA sequence can be made entirely in vitro by the polymerase chain reaction technique.
10-49 A. 5′-TAGACTGACCTG-3′ B. Arg-Leu-Thr [only the second reading frame can be used, because reading frame 1 contains a stop codon (TAG), as does reading frame 3 (TGA)].
10-50 (d) If some of the DNA templates you are sequencing are cut at one specific site (as would be the case if a restriction enzyme cut the DNA), the polymerase will stop when it comes to the end of the DNA, giving rise to at least some product of one particular size in all the reaction mixtures. If this were the case, all four lanes will have a band of this particular size. In addition, you would get a normal sequence from the full-length templates, and a normal sequence from those templates in which the polymerase incorporated a dideoxynucleotide before encountering the end. The other options are incorrect: if you added all four dideoxynucleotides to one of the reactions [choice (a)], that lane would have a band at every position because the polymerase would stop at As, Cs, Gs, and Ts instead of at only one type of nucleotide. If you forgot to add deoxynucleotides to the reactions [choice (b)], you would not get any polymerization, and all of your lanes would be blank. If your primer hybridized to more than one part of the fragment of DNA you were sequencing [choice (c)], your gel would look as though two different sequences had been superimposed on each other.
10-51 Choice (c) is correct. Although it is true that fully automated Sanger sequencing requires a capillary gel, this is not why all four ddNTPs can be mixed in a single reaction [choice (a)]. Ideally, a single ddNTP is incorporated into each product; all four ddNTPs should not be incorporated into a single product [choice (b)]. Fully automated Sanger sequencing still requires DNA polymerase, and reads the fluorescent ddNTP at the end of each product [choice (d)].
10-52 (d) Second-generation sequencing is faster and cheaper because millions of reactions can be carried out in parallel on a solid support.
10-53 (a) Although in situ hybridization can tell you a great deal about the expression patterns of a particular gene, this technique is not easily adapted to simultaneously monitor the thousands of genes expressed in a particular tissue.
10-54 The tubby gene is expressed in all tissues, including the brain. A red spot on a microarray is indicative of a difference in gene expression between the two RNA samples being compared. You may expect in this experiment that the tubby gene would be a yellow spot (a gene expressed at equal levels in both samples). An in situ experiment looks at the RNA level directly in the tissue of interest, which is why in this case you see ample levels of tubby RNA.
10-55 (c) The information for localizing proteins within a cell is found on the protein product and not in the regulatory DNA sequences. If you were to fuse your reporter gene to the DNA sequences that encode the protein to produce a GFP fusion protein, then you might determine the specific location of the protein within the cell.
10-57 (d) Neutering all the knockout animals that you sell is the only option of the four listed that will prevent happy pet owners from becoming happy pet breeders. The situation described in choice (a) will not allow you to make any knockout cats because the first litter (which will at best have a few mosaics in which one copy of the gene has been knocked out in the germ cells) will be sterile and you will not be able to mate them. The genotype of the female cat in which you implant the embryos has no effect on the genotype of the embryos, which is why choice (b) is incorrect. Choice (c) is incorrect because the first litter will yield mosaic cats that still have one copy of the allergen-producing gene in their cells and are therefore not hypoallergenic.
10-58 (b) Transgenic pigs were not created for the purification of insulin. Instead, scientists were purifying the endogenous insulin present in the pig pancreas that is made from the normal pig insulin gene.
10-59 Choice (c) is correct. Origins of replication [choice (a)] and cleavage sites for restriction nucleases [choice (b)] are found in both cloning vectors and expression vectors. Bacterial mRNAs do not undergo polyadenylation [choice (d)].
10-60 Either 5′-TTCGACCCGAAGGGCAGCCAC–3′ or 5′-GTGGCTGCCCTTCGGGTCGAA-3′ would work. The changed nucleotide is indicated in gray in the sequences above. The codon for glutamine (CAG) has been changed to AAG, which codes for lysine.
10-61 The new DNA sequence is shown in Figure A10-61.
The change in the sequence is indicated by the rectangle; the HindIII recognition sequence has been underlined. The peptide encoded by this piece of DNA is indicated above the DNA sequence. Note that this DNA will still encode leucine, despite the change from a CUA codon to a CUU codon.
10-62 A. 52 alanine, 53 valine, 54 serine, 55 leucine, 56 tryptophan, 57 cysteine. See Figure 10-63A for how this answer was derived.B. Either of the two nucleotide sequences shown in Figure A10-63B is correct. The peptide encoded is shown above the top sequence, for reference. The bottom nucleotide sequence is the reverse complement of the top sequence.
10-63 Even though the genome of F. cattoriae is smaller, the W. gravinius genome is more attractive to sequence because it contains less repetitive DNA. Repetitive DNA makes the assembly of sequenced fragments difficult. Shotgun sequencing would not be an unrealistic approach for W. gravinius, because the genome of W. gravinius contains little repetitive DNA and is relatively small. The genome of H. influenzae is 1.83 megabases long and was successfully sequenced with the shotgun approach. (For comparison, the genome of S. cerevisiae is 14 megabases long.)
10-64 (b) The order of the BAC clones relative to the segment of human DNA is shown in Figure A10-64.
Essential Cell Biology-4th Ed Test-Bank – Alberts- 2014