Molecular Biology of the Cell 5th Edition Alberts Johnson Lewis Raff Test Bank

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Molecular Biology of the Cell 5th Edition Alberts Johnson Lewis Raff Test Bank

ISBN-13: 978-0815341055

ISBN-10: 0815341059



Molecular Biology of the Cell 5th Edition Alberts Johnson Lewis Raff Test Bank

ISBN-13: 978-0815341055

ISBN-10: 0815341059




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Manipulating Proteins, DNA, and RNA

Molecular Biology of the Cell, Fifth Edition

Ó 2008 Garland Science Publishing






8-1       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 Q8-1. 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.

  1. Outline briefly the series of steps you would perform to determine which plasmid is in which tube.
  2. Which restriction enzyme or combination of restriction enzymes would you use in this experiment?


Figure Q8-1



8-2       You have sequenced a short piece of DNA and produced the gel shown below:

  1. What is the sequence of the DNA, starting from the 5¢ end?
  2. If you know that this sequence is from the middle of a protein-coding cDNA clone, what amino acid sequence can you deduce from this sequence?



Figure Q8-2


8-3       Figure Q8-3 shows the recognition sequences for the restriction enzymes SalI, XhoI, PstI, and SmaI and a plasmid with the sites of cleavage for these enzymes marked.

  1. Consider all possible digestion reactions with one or two restriction enzymes. After which of the digestions can the plasmid form into a circle again simply by treatment with DNA ligase? Assume that after digestion any small pieces of DNA are removed, and it is only the larger portion of plasmid that you are trying to restore to its circular state.
  2. After which of the possible digestions can the plasmid be restored to a circle by first adding DNA polymerase and the four deoxynucleotides, then treating with DNA ligase? Again assume that you are trying to form the larger portion of plasmid into a circle again.


Figure Q8-3


8-4       You have an oligonucleotide probe that hybridizes to part of gene A from a eucaryotic cell. You can use this probe to isolate a corresponding plasmid from a DNA library harbored by a collection of bacteria. Will a cDNA library or a genomic DNA library be more appropriate for the following applications? Explain.

  1. You want to study the promoter of a gene A.
  2. Gene A encodes a tRNA and you wish to isolate a piece of DNA containing the full-length sequence of the tRNA.
  3. You discover that gene A is alternatively spliced and you want to see which predicted alternative splice products are actually produced in a cell.
  4. You want to find both gene A and the genes located near gene A on the chromosome.
  5. You want to express gene A in bacteria to produce lots of protein A.
  6. You want to perform a phylogenetic comparison to find amino acid sequences important for the function of gene A.


8-5       You want to make an antibody against a C. elegans nematode protein that the DNA shown in Figure Q8-5A encodes. To do this, you will clone the coding sequence into a bacterial expression plasmid, overexpress the protein, purify it, and inject it into a mouse to stimulate the production of antibodies against the protein.

  1. You use polymerase chain reaction (PCR) amplification to amplify the DNA corresponding to the 1800 base pairs of coding sequence; this coding sequence lies between the two flanking sequence shown in Figure Q8-5A. You will then insert the PCR product into the BamHI site of the expression plasmid shown in Figure Q8-5B. Is it best to use genomic DNA or cDNA as template in the PCR reaction? What are the sequences of the two oligonucleotide primers that will allow you to amplify the DNA by PCR and insert it into the BamHI site? (Remember to indicate the 5¢ and 3¢ ends of the primers.)
  2. Briefly describe the series of enzymatic treatments you will use to create the expression plasmid before transferring it into bacterial cells.
  3. Once the correct new expression plasmid is in bacterial cells, you induce high levels of expression of your nematode protein. Briefly describe the series of steps you will perform to go from the cells containing high levels of protein to partly purified protein.
  4. Several weeks after you inject a mouse with your purified protein, you sample its blood serum, which contains antibodies. You know from in situ hybridizations that the mRNA corresponding to your protein is found in gonad cells but not in gut cells. To test whether the mouse has made antibodies against your protein, you isolate protein from gonad cells and from gut cells and perform a Western blot (also known as an immunoblot) with the mouse serum and a fluorescent second antibody that binds mouse antibodies. Your result is shown in Figure Q8-5C. Did the mouse make antibodies against your protein? Does the serum specifically recognize only one nematode protein? (Illustrate your written answers by labeling the blot.) What might you do next if you wanted to use antibodies from this mouse to determine the subcellular location of your protein in fixed cells?




Figure Q8-5


8-6       Current technologies allow much more facile manipulation and analysis of nucleic acids than proteins. For example, it is straightforward to generate and purify thousands of different DNA molecules for a DNA microarray, but difficult to make an analogous protein microarray. This may reflect fundamental differences between the two kinds of polymers. Colloquially, we might say that DNA double helices are all the same, but each protein has its own personality.

  1. Compare how the shape of a DNA double helix or protein molecule depends on the sequence of monomers.
  2. Compare the means used to detect a specific DNA molecule in a complex mixture of other DNAs with that used to detect a specific protein in a complex protein mixture.
  3. Briefly compare the means used to make many copies of a DNA or protein molecule for biochemical experiments.
  4. Can you unambiguously predict an amino acid sequence from a cDNA sequence? Can you unambiguously predict a cDNA sequence from an amino acid sequence? Explain.


8-7       You are working in a laboratory, trying to identify how the nematode worm C. elegans senses and moves toward specific “attractant” chemicals. You examine an electronic database that compiles mRNA expression profiles of all nematode genes measured by DNA microarray hybridization experiments, and find that an unknown gene is transcribed coordinately with several genes known to be involved in the attractant response. You name this gene Unk1 and set out to learn what it does.

  1. Without doing a single experiment at the lab bench, how can you learn more about what the Unk1 protein might be doing? How might this influence your choice of experiments with Unk1?
  2. You decide to use reverse genetics to learn more about the function of the Unk1 gene. What is the simplest first experiment to do, considering that you are working with nematodes?
  3. You decide to look for proteins that bind to Unk1, which may provide important clues to its function. Briefly describe an approach to find binding partners.
  4. You decide to determine where and when Unk1 is expressed within the animal, to learn more about how it causes the observed phenotypes. Describe an approach to monitor spatiotemporal dynamics of Unk1 expression.


8-8       Human babies use the lactase enzyme to metabolize the milk sugar lactose. Most human adults are lactose-intolerant because this gene is normally turned off in adults. However, many adults of European descent still express lactase. This lactase-persistence trait is tightly linked to a SNP called C/T-13910 located near the lactase gene. Examination of many SNPs flanking the lactase gene provided evidence that lactase persistence has been subject to very strong positive selective pressure and became prevalent relatively recently, about 7500 years ago, which is roughly coincident with the time that dairy farming arose in northern Europe.

  1. Much of the supporting genomic evidence comes from the sizes of haplotype blocks. Are the haplotype blocks surrounding the lactase gene in lactase-persistent individuals bigger or smaller than the average blocks found throughout the genome, or of average size? Are the corresponding haplotype blocks in lactose-intolerant individuals bigger, smaller, or average?
  2. Does the C/T-13910 SNP cause the lactase-persistence trait?
  3. Name an experimental technique that can determine if an unknown adult is lactose-intolerant or lactase-persistent.
  4. Is lactase persistence likely to be a dominant trait or a recessive trait?








































  1. First, digest DNA from one of the tubes with a combination of restriction enzymes that will yield DNA fragments with sizes that are unique to one of the plasmids. Then, load the resulting mixture of DNA fragments on an agarose gel alongside a set of size markers, use electrophoresis to separate DNAs by size, stain the gel to reveal the DNAs, and determine the fragment sizes. By looking at the restriction maps, you should then be able to match your results to one of the plasmids.
  2. 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 sizeRemember, you cannot distinguish them simply by making a single cut (with HindIII or BglII alone) and determining the size of the complete DNA by gel electrophoresis. Nor can you distinguish them by cutting with all four restriction nucleases (or with BamHI and BglII), 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 the location of the BglII site relative to the two BamHI sitesRemember, 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.



  2. Arg-Leu-Thr. You can only use the second reading frame; reading frame 1 contains a stop codon (TAG), as does reading frame 3 (TGA).



  1. Ligase treatment alone can convert the large cleaved plasmid fragments to circular forms after the plasmid is first digested with any of the single restriction enzymes and one of the double enzyme combinations. When SalI and XhoI cut DNA, the staggered ends left behind will match up by base-pairing and thus can be joined by ligase alone. When all other pairs of enzymes cut DNA, the ends will not be complementary; that is, there will not be two sticky ends that can anneal, nor will there be two blunt ends that can be ligated.
  2. In addition to the answers for A, there are two new enzyme pairs that will allow the plasmid to form again into a circle after treatment with DNA polymerase and ligase: SalI + SmaI and XhoI + SmaI. Two blunt ends can be joined by DNA ligase. Digestion with SmaI creates a blunt end. Addition of DNA polymerase and the four deoxynucleotides will fill in the 5¢ overhangs generated by digestion with SalI and XhoI, leaving blunt ends. However, 3¢ overhangs (namely those generated by PstI) will not be filled in, because DNA polymerase can only add nucleotides to a recessed 3¢ end that is paired with a longer template strand (a 5¢ overhang). DNA ligase will not join 3¢ overhangs from PstI digestion to any of the blunt ends created by the other enzymes.



  1. Genomic library. cDNAs are produced from mRNAs. Therefore, the promoters will not be included in a cDNA library.
  2. Genomic library. cDNAs are usually constructed by using an oligo-dT primer, and tRNAs do not have poly-A tails. Therefore, tRNA sequences are not usually found in cDNA libraries. If the cDNA library were made using random primers, it would be unlikely to contain the full-length version of the tRNA.
  3. cDNA library. Because cDNAs are produced from mRNAs, isolating cDNAs would tell you which splice variants are produced in a cell.
  4. Genomic library. Some genomic DNA fragments will probably contain the genes next to your gene of interest; cDNAs will not.
  5. cDNA library. Bacteria do not have the ability to remove introns, which may exist in DNA isolated from a genomic library.
  6. cDNA library. A phylogenetic comparison of proteins requires amino acid sequences or nucleotide coding sequences from orthologous or homologous proteins. It is often difficult to use genomic sequences to find unambiguously the coding sequences, sprinkled among the more abundant intron and regulatory region sequences.



  1. It is best to use cDNA as template, because the bacteria will be unable to splice out introns in the genomic DNA. The appropriate PCR primers are 5¢-GGATCCGACCTGTGGAAGC-3¢ and     5¢-GGATCCTCAATCCCGTATG-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. Both primers have the BamHI recognition site added to the 5’ end to allow cleavage by BamHI, which will generate complementary single-stranded ends that can hybridize to the ends of the BamHI-cut plasmid.
  2. First, use a thermophilic DNA polymerase for PCR amplification. Second, use BamHI to digest both the PCR product and the plasmid. Third, use DNA ligase to join the two digested DNA molecules together.
  3. First, use ultrasonic vibration or another method to break open the cells. Second, use preparative ultracentrifugation to remove the whole cells and cellular debris from the protein extract. Third, use column chromatography to fractionate the proteins in the extract and thereby separate the desired nematode protein from other bacterial proteins. Because the nematode protein is tagged with multiple histidines, it will bind to a metal-affinity chromatography matrix. Alternatively, another matrix can be used, for example an ion-exchange matrix.
  4. Yes, the mouse made antibodies against the desired target protein, which is the band seen only in the gonad lane. No, the serum does not specifically recognize only one nematode protein; it also recognizes two proteins found in both gonad and gut cells (Figure A8-5C). To determine the subcellular location of the target protein you cannot use this serum because it recognizes additional proteins. To make a cleaner, more specific antibody preparation, you can create a hybridoma cell line that makes a unique monoclonal antibody, which is likely to recognize only one protein. Alternatively, you can make an affinity chromatography matrix containing the protein of interest, and use affinity chromatography to purify the desired antibodies out of the mouse serum.



Figure A8-5C



  1. DNA contains only 4 different monomers, and the shape of the double helix is almost uniform, having only a relatively subtle dependence on the nucleotide sequence. Protein contains 20 different monomers and the shapes of proteins differ markedly depending on their amino acid sequence. Some proteins have long thin rodlike structures and others have globular structures. All proteins have unique detailed surface contours with different charge and hydrophobicity properties that allow them to bind to other molecules and perform their specific unique catalytic or structural roles.
  2. Any specific DNA sequence can be uniquely detected simply by hybridization to a complementary DNA or RNA. It is easy, inexpensive, and fast to make a DNA or RNA with any desired sequence (see below). A protein can be uniquely detected either with a mass spectrometer (MS) or with a specific antibody. MS equipment costs hundreds of thousands of dollars and demands highly trained operators. Production of antibodies is expensive and time-consuming, and usually requires the killing of laboratory animals.
  3. To make many copies of a DNA molecule, we can use DNA cloning or PCR amplification. PCR requires only two small chemically synthesized oligonucleotide primers and a small automated reaction that takes a few hours. DNA cloning requires restriction enzymes, ligase, and a host plasmid; once a recombinant plasmid has been created, we can make nearly limitless supplies by growing bacteria harboring the plasmid and performing a simple standard DNA purification, regardless of nucleotide sequence. To make many copies of a protein, we can overexpress a gene from a recombinant expression plasmid, then purify the accumulated protein away from other proteins with the use of a series of chromatographic columns. It takes dedicated trial-and-error and patience to find chromatography conditions that are uniquely suited to purification of the protein of interest, because each protein behaves differently.
  4. Yes, you can unambiguously predict an amino acid sequence from a cDNA sequence, because each codon specifies a unique amino acid. However, you cannot unambiguously predict a cDNA sequence from an amino acid sequence, because most amino acids are specified by more than one nucleotide triplet, or codon.



  1. You can look in sequence databases compiled from many organisms to try to find other genes that encode proteins with similar amino acid sequences. If you find a related protein that has been studied in another organism, it may suggest a function for Unk1. For example, if Unk1 is very similar in amino acid sequence to a known protein kinase, then it probably functions as one.
  2. The simplest first experiment is to knock out Unk1 gene expression by using RNAi and to evaluate the phenotype of the resulting organisms. Operationally, you can do this by obtaining a bacterial strain harboring a DNA plasmid that directs the expression of double-stranded RNA with the same sequence as Unk1. You can feed these bacteria to nematodes, which will take up the RNA into their cells, where it will direct the degradation of Unk1 mRNA and alteration of chromatin at the Unk1 gene locus.
  3. There are two or more acceptable answers. (1) You can discover the protein–protein interactions by using a two-hybrid technique in yeast cells. Unk1 can be fused to a DNA-binding domain (DBD) and used as bait. Prey molecules can be fished out of a DNA library in which nematode cDNAs are fused to a gene encoding a yeast transcriptional activation domain (TA). One yeast strain among thousands that contains both Unk1-DBD and an Unk1-interacting partner fused to AS can be identified and isolated because it will transcribe the reporter gene. (2) You can identify the protein–protein interactions with mass spectrometry (MS) of an Unk1-containing multiprotein complex. First, Unk1 and its associated proteins can be partly purified from nematodes. This can be done with an antibody raised against Unk1 or by fusing a protein “tag” to the Unk1 protein expressed in nematodes; in both cases, the Unk1 protein can be fished out of a complex protein mixture with antibody-linked beads that bind Unk1 or its tag. Then the partly purified mixture can be subjected to MS analysis. Because the genome of nematodes has been sequenced, computer algorithms can search for matches between (a) the measured mass-to-charge ratios of peptides from co-purifying proteins and (b) the mass-to-charge ratios of peptides predicted to arise from particular proteins encoded by the nematode genome.
  4. There are several acceptable answers. (1) In situ hybridization of the Unk1 mRNA can detect the mRNA in different cells at different times of development. (2) An antibody that specifically binds the Unk1 protein can be used for immunofluorescence to detect the protein in different cells at different times of development. (3) A reporter gene can be put under the control of regulatory sequences from upstream and downstream of the Unk1 coding sequence, and production of the reporter can be monitored in animals. For example, a live transgenic animal harboring a GFP reporter gene controlled by Unk1 regulatory sequences can be monitored with a microscope to watch where and when the Unk1 gene is turned on.



  1. The haplotype blocks surrounding the lactase gene in lactase-persistent individuals are much bigger than average haplotype blocks. The corresponding haplotype blocks in lactose-intolerant individuals are of average size. (Actually, the haplotype block containing the lactase gene in lactase-persistent individuals of European descent is 1000–2000 kilobases, in comparison with an average block of only 2 kilobases! This provides one of the strongest genomic signatures of evolutionary history in humans reported so far. When a genetic variant or allele arises that greatly benefits its carriers and enables them to produce many more viable offspring than others in the population, this allele will become the most prevalent one in a population. If it happens relatively quickly, over tens of generations, then there will few meioses during which crossovers can occur to separate this allele from the flanking SNPs. Over time, the genetic linkage between this allele and flanking SNPs will degrade as a result of recombination and new mutations. Thus, a large and prevalent haplotype block is evidence for the recent rise of a particular allele under strong positive selective pressure. It is fascinating that this lactase-persistence trait seems to have been strongly selected in humans only where and when they domesticated mammals for dairy farming!)
  2. The information provided is insufficient to conclude that the C/T-13910 SNP causes the lactase-persistence trait. A correlation between a SNP and a trait, by itself, never proves a causal relationship, and this is especially true if the SNP falls in a large haplotype block.
  3. Either Northern blotting or RT-PCR can determine whether an unknown adult is lactose-intolerant or lactase-persistent, by measuring whether the lactase mRNA is absent or present, respectively.
  4. Lactase persistence is likely to be a dominant trait, because heterozygous individuals will make some lactase enzyme and will thus be able to digest milk.




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