Test Bank 5th-Ed Molecular Biology of the Cell Study Aid by Robert-Weaver 2008

Test Bank 5th-Ed Molecular Biology of the Cell Study Aid by Robert-Weaver 2008


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Test Bank 5th-Ed Molecular Biology of the Cell Study Aid by Robert-Weaver 2008

Sample Chapter No 5                                                              


DNA, Replication, Repair, and Recombination

Molecular Biology of the Cell, Fifth Edition

Ó 2008 Garland Science Publishing







5-1       The mutation rate is about 1 mutation per 109 nucleotides per cell division in most organisms, from E. coli bacteria through nematode worms up to humans. You aspire to develop an improved superorganism with a reduced mutation rate. Through a clever genetic screen, you isolate a mutant strain of E. coli that suffers only about 1 mutation per 1012 nucleotides per cell division. You imagine that this superorganism will have superior survival and proliferation properties, and might take over the world. A colleague suggests you do the following experiment: Put a gene encoding yellow fluorescent protein (YFP) in the original strain and a gene encoding cyan fluorescent protein (CFP) in the mutant strain. Start with a single cell from each strain and allow those cells to proliferate under optimal conditions for a month. Then mix equal numbers of cells from each strain together so that 50% of the cells are yellow and 50% are cyan, and split the mixture into two parts. Culture “Mix 1” in rich growth medium for four weeks, diluting the cells and adding fresh medium every two days. Culture “Mix 2” in a similar way, except add excess salt during the first week, add copper during the second week, grow at an elevated temperature during the third week, and reduce the nutrients in the medium during the fourth week. Finally, examine the ratio of yellow to cyan cells in the mixes. You find that 55% of the cells in Mix 1 are cyan, but only 40% of the cells in Mix 2 are cyan. Explain these observations. Suggest a reason why the mutation rate is 10-9 in all wild-type organisms tested.


5-2       A pregnant mouse is exposed to high levels of a chemical. Many of the mice in her litter are deformed, but when they are interbred with each other, all their offspring are normal. For each of the statements below, explain why it is a likely or an unlikely explanation of the observations.

  1. The mother mouse’s germ cells were mutated.
  2. In the deformed mice, germ cells but not somatic cells were mutated.
  3. The toxic chemical affects development but is not mutagenic.
  4. In the deformed mice, somatic cells but not germ cells were mutated.


5-3       Imagine that you join a laboratory that has established a technique for examining DNA replication in a cellular extract. To the cellular protein extract, you simply add nucleotides, a small amount of radiolabeled 32P-dGTP to aid visualization of the synthesized DNA, and a 4000-base-pair linear double-stranded DNA molecule that contains an origin of replication in the middle. After allowing the reaction to proceed for 30 minutes at 30°C, you boil the mixture to denature the proteins and the DNA strands, separate the components on an acrylamide gel, and detect the radiolabeled DNA products. This complete reaction is shown in lane 2 on the gel in Figure Q5-3.

  1. You perform the assay by yourself for the first time late on a Friday night, and no one else is in lab. The tube labeled “nucleotide mixture” has only enough for a single reaction, but you find a tube with a label that reads “dATP, dTTP, dGTP, dCTP” and you use this in a second reaction. You find the first reaction looks like lane 2 and the second reaction looks like lane 3. Why was no DNA synthesized in the second reaction?
  2. Your lab partner has recently isolated several mutant strains that can replicate their DNA and proliferate normally at 30°C but exhibit no DNA replication when grown at 40°C. He calls these mutants tsr1 through tsr4, for Temperature Sensitive Replication. The wild-type strain replicates efficiently at both temperatures. You believe that you can use the biochemical assay of cell extracts to identify which genes are defective in the new mutants. First you grow wild-type cells and tsr1 cells at 40°C, make the extracts, incubate the DNA replication reaction at 40°C, and detect the products on a gel. You observe the pattern shown in lanes 4 and 5. You are so excited by the results, you run to your lab partner and tell him that you know what enzyme is defective. He is delighted by your results but says that there are two different enzymes that can account for your results. What are they?
  3. Heartened by your initial success in narrowing down the identity of the defective genes in the new mutants, you test the mutants called tsr2, tsr3, and tsr4. You find they all give the pattern shown in lane 6: they fail to make detectable DNA products. Name two of the several proteins that, when defective, might yield the pattern in lane 6.
  4. To explore these mutants further, you mix the extracts together. You find that tsr2 plus tsr3 gives a wild-type pattern shown in lane 7, but tsr2 plus tsr4 still gives the mutant pattern in lane 8. What is the simplest explanation for these results?

Figure Q5-3


5-4       Strand-directed mismatch repair occurs soon after the passage of a replication fork to repair mismatched base pairs and preferentially restore the original sequence. How does the repair system know which is the original sequence and which is the new, mutated sequence? To identify the newly synthesized strand, procaryotes take advantage of the transient absence of methylation. Eucaryotes cannot use hemimethylation as a clue and instead use single-stranded DNA nicks to identify the newly synthesized strand. Consider a 1 kb stretch of double-stranded DNA before replication, with its “top” strand and “bottom” strand. To convince yourself that replication generates a new “bottom” strand and a new “top” strand that both contain nicks (transiently), draw a replication bubble that started in the middle of the DNA and extends almost to the edge. Show the origin of replication, the replication fork(s), the leading and lagging strands, and the 5’ and 3’ ends of all DNA fragments.


5-5       Yeast cells that are heterozygous for the ADE2 gene are useful for following the distribution of alleles after meiosis. The meiotic products in yeast are called spores, and are analogous to sperm and eggs. All four spores derived from meiosis in a single cell are packaged together as a tetrad, and they can be separated under the microscope and deposited at specific locations on an agar pad. When the agar pad is placed on appropriate medium in a Petri dish, a colony of cells grows from each spore. Colonies that have wild-type ADE2 are white and those that have mutant ade2 are red. Most tetrads give two white and two red colonies, as would be expected from simple Mendelian genetics.

  1. Sometimes a tetrad yields three white and one red spore colony, or vice versa. What is the name for the phenomenon that causes this pattern, and how does it arise?
  2. When certain yeast mutants are sent through meiosis, a few of the spore colonies exhibit sectoring. A sectored colony has about half red cells and half white cells. Consider all of the enzymatic steps in homologous recombination. What part of homologous recombination is defective in these “PMS” (for Post-Meiotic Segregation) mutants?
  3. Mutations in human homologs of the yeast PMS genes contribute to the development of some cancers, as a result of their effects on mitotic DNA replication. Why do you think this is?


5-6       Chromosomal DNA sequences are maintained and replicated with such high fidelity that the sequence of the human genome is changed by only about three nucleotides each time a cell divides. You mention this astonishing fact to your grandfather, who has just read an article in the newspaper about human genetics. He exclaims that this seems to be a very high mutation rate because changing the function of three proteins every time a cell divides should quickly lead to a defective individual. Explain the mistake in your grandfather’s reasoning.


5-7       The bacterium E. coli replicates its genome in about 20 minutes under ideal growth conditions. Surprisingly, the fruit fly Drosophila can replicate its genome in only 3 minutes. How can this be? For each statement below, indicate whether it is true and whether it can explain the difference in the duration of replication.

  1. The Drosophila genome is smaller than the E. coli genome.
  2. Eucaryotic DNA polymerase synthesizes DNA at a much faster rate than procaryotic DNA polymerase.
  3. The nuclear membrane keeps the Drosophila DNA concentrated in one place in the cell.
  4. Drosophila DNA contains more origins of replication than E. coli DNA.
  5. Eucaryotes have more than one kind of DNA polymerase.


5-8       Some retrotransposons and retroviruses integrate preferentially into regions of the chromosome that are (a) packaged in euchromatin and (b) located outside the regions of genes that encode the amino acid sequences of proteins. Why might these mobile genetic elements have evolved this strategy?














5-1       The experiment is a microcosm of natural selection and evolution. You found that under optimal conditions, to which the laboratory strain had adapted over the course of decades, the strain with a lower mutation rate (your mutant) was able to proliferate more than the strain with the higher mutation rate (the original strain). This suggests that in a stable and optimal environment, mutations are detrimental to proliferation and fitness of a population. Yet under suboptimal conditions, when grown with a series of environmental challenges and changes, the strain with the higher mutation rate outgrew the strain with the lower mutation rate. This suggests that in a changing and suboptimal environment, a higher mutation rate provides a diversity of genotypes in the population, such that a subpopulation is more likely to be able to thrive when confronted with a novel environmental challenge. The shared mutation rate of 10–9 in all organisms seems to be a compromise between selective pressures: high-fidelity maintenance of nucleotide sequences is necessary to avoid accumulating mutations, most of which will be detrimental to the function of the encoded gene product, but mutations generate the diversity upon which natural selection acts and yield genetic variants that may be better suited for novel circumstances.


5-2       Statements C and D are likely explanations, whereas A and B cannot explain the observations. Recall that germ cells (eggs and sperm) contain the only genetic material that is passed on to the next generation.

  1. If the mother mouse’s germ cells are mutated when she is pregnant, it will have no effect on the fetuses that she was already carrying.
  2. If the fetal germ cells but not somatic cells were mutated in utero, the baby mice would be unlikely to be deformed but might yield deformed offspring, which is the opposite of what was observed.
  3. If the toxic chemical affected development but is not mutagenic, then abnormal fetal development could cause the babies to be deformed but would have no effect on the genetic material that they pass on to their offspring.
  4. If the fetal somatic cells but not germ cells were mutated in utero, mutations in the somatic cells may cause the deformations, yet the unmutated DNA in the germ cells would be passed on to their offspring, which will thus not be deformed.



  1. No DNA is synthesized in the second reaction because you included only deoxyribonucleotides for DNA polymerase and not ribonucleotides for RNA synthesis by primase. No DNA replication occurred because the extract was unable to synthesize RNA primers.
  2. Ligase and RNase. The pattern on the gel includes fragments of 2000 bp, which is half the length of the DNA template. These fragments are the expected length of completed leading strand products. Other observed fragments are about 500 bp long, which is about the expected length of Okazaki fragments from the lagging strand (different organisms have different average lengths, ranging from about 150 to 1000 bp). The newly synthesized fragments were not joined together either because enzymes failed to remove the RNA primers and re-synthesize DNA at those locations or because ligase failed to stitch the all-DNA fragments together.
  3. Lane 6 indicates that no DNA replication has occurred, which might mean that one of the following proteins was defective: initiator proteins, helicase, primase, SSB, DNA polymerase.
  4. The simplest explanation is that tsr2 and tsr3 contain mutations in different genes, so that the mixture has a functional copy of every relevant gene and thus contains all of the proteins it needs for normal DNA synthesis. This is called complementation. Along these lines, a simple explanation of the tsr2 plus tsr4 mixture is that both strains contain mutations in the same gene, so that neither strain is able to provide a functional version of the corresponding protein and the mixed extract is defective.


  • A bidirectional replication bubble has two forks, such that each strand of the original DNA duplex has its newly synthesized complementary strand produced by a combination of leading strand and lagging strand synthesis. Thus, all new strands will contain nicks transiently before RNase degrades the primers, DNA polymerase fills the gaps, and ligase joins the DNA fragments.


Figure A5-4



  1. The phenomenon is called gene conversion. It arises from the formation of heteroduplex DNA during homologous recombination, which occurs much more frequently in meiotic cells than in cells growing mitotically. Heteroduplex DNA contains one strand from one chromosome and another strand from the other chromosome. If the two strands have slightly different sequences, then the heteroduplex can contain a mismatch. Depending on which strand is used as the template for the mismatch repair machinery, the duplex can be restored to give 2:2 segregation or converted to give 3:1 segregation.
  2. Mismatch repair. The DNA heteroduplex that arises during homologous recombination often contains mismatched base pairs, especially for a heterozygous gene. Usually, mismatch repair enzymes act to restore perfectly base paired strands, which sometimes leads to gene conversion (see A above). If DNA mismatches are not repaired during meiosis, as a result of defects in the repair machinery, then spores will sometimes contain two different versions of the ADE2 gene—one sequence on the “top” strand of the duplex and a different sequence on the “bottom” strand. Because DNA replication is semi-conservative, the two daughter cells produced by the original spore will be different—one with ADE2 and one with ade2. As each cell proliferates, it will form more cells identical to itself, with one side of the colony filled with white cells and the other filled with red cells.
  3. PMS genes in yeast and humans are needed for mismatch repair. When mismatch repair is defective, DNA replication has reduced fidelity and thus cells are more likely to accumulate mutations. The origin and development of cancer cells depends on the accumulation of many mutations, so circumstances that increase the mutation rate are likely to increase the likelihood of cancer.


5-6       Grandfather assumed that all nucleotide changes lead to changes in amino acid sequences of proteins. The vast majority of DNA in the human genome does not encode the amino acid sequences of proteins, but instead serves some regulatory role (telling when and where to turn a gene on or off) or structural role (helping to package or segregate chromosomes), or is simply “junk” or ancestral remnant DNA. Even when a nucleotide change occurs in the protein-coding region of a gene, the mutation might not change which amino acid is incorporated; this is because most amino acids are specified by more than one nucleotide triplet. Even when an amino acid is changed, the function of the protein might not be altered; this is because some amino acids are not critical for protein function or their role is flexible, for example any small hydrophobic amino acid might suffice at a particular location in a given protein.



  1. False. The Drosophila genome is larger than the E. coli genome, so the size difference cannot explain faster replication in Drosophila.
  2. False. Eucaryotic DNA polymerase synthesizes DNA at a much slower rate than procaryotic DNA polymerase, probably because of the challenges of negotiating through nucleosomes, so the enzymatic rates of individual polymerase enzymes cannot explain faster replication in Drosophila.
  3. True. The nuclear membrane keeps the Drosophila DNA concentrated in one place in the cell, but this does not explain faster replication in Drosophila.
  4. True. Drosophila DNA contains many origins of replication to allow many chromosomal regions to be replicated simultaneously, whereas E. coli DNA has only one origin of replication. This fact alone can explain the difference in overall time required to duplicate the entire genomes of the two species.
  5. True. Eucaryotes have more than one kind of DNA polymerase, but this does not explain faster replication in Drosophila.


5-8       The most evolutionarily successful mobile genetic elements are those that are best at reproducing themselves. To increase the number of copies of a particular element, the element must meet two criteria: it must not kill its host, and it must maximize its ability to continue reproducing. If an element inserts into the coding region of a gene, it might disable the gene and thereby confer a selective disadvantage in the reproduction or survival of its host. Thus, elements better able to avoid insertion into coding regions were probably better able to increase their copy number throughout the human population. If an element inserts into a heterochromatic region of a chromosome, its genes may not be expressed and therefore it may become immobile. Elements that gained the ability to direct insertion into euchromatin would be more likely to maintain mobility and thereby increase their copy number over time.



Test Bank 5th-Ed Molecular Biology of the Cell Study Aid by Robert-Weaver 2008

Test Bank 5th-Ed Molecular Biology of the Cell Study Aid by Robert-Weaver 2008