Molecular Biology Of The Cell 6th Edition by Bruce Alberts – Test Bank

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MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION

CHAPTER 5: DNA REPLICATION, REPAIR, AND RECOMBINATION

© Garland Science 2015

 

1.1.         Which of the following is correct regarding the mutation rate of genomic DNA in different organisms?

  1. Human cells have a much higher mutation rate compared to bacteria when the rate is normalized to a single round of replication over the same length of DNA.
  2. Mutation rates limit the number of essential genes in an organism’s genome.
  3. Mutations in the somatic cells cannot be lethal.
  4. Even if the mutation rate was 10 times higher than its current value, germ-cell stability in humans would not have been affected.
  5. All of the above.

 

1.2.         The mutation rate in bacteria is about 3 nucleotide changes per 10 billion nucleotides per cell generation. Under laboratory conditions, bacteria such as Escherichia coli can divide and double in number about every 40 minutes. If a single Escherichia coli cell is allowed to exponentially divide for 10 hours in this manner, how many mutations would you expect to observe on average in the genome (4.5 million nucleotide pairs) of each of the resulting bacteria compared to the original cell? Assume all mutations are neutral; that is, they do not affect the cell-division time.

  1. Less than 0.001
  2. About 0.02
  3. One or two
  4. About 10
  5. About 100

 

1.3.         On average, errors occur in DNA synthesis only once in every ten billion nucleotides incorporated. Which of the following does NOT contribute to this high fidelity of DNA synthesis?

  1. Complementary base-pairing between the nucleotides
  2. “Tightening” of the DNA polymerase enzyme around its active site to ensure correct pairing before monomer incorporation
  3. Exonucleolytic proofreading by the 3′-to-5′ exonuclease activity of the enzyme to correct mispairing even after monomer incorporation
  4. A strand-directed mismatch repair system that detects and resolves mismatches soon after DNA replication
  5. All of the above mechanisms DO contribute to the fidelity.

 

1.4.         The nuclear DNA polymerases in human cells …

  1. polymerize about 1000 nucleotides per second during DNA replication in vivo.
  2. are incapable of 3′-to-5′ exonuclease activity.
  3. are capable of 3′-to-5′ DNA polymerase activity.
  4. have a single active site that is used for both polymerization and editing.
  5. are unable to initiate polymerization de novo (i.e. in the absence of a primer).

 

1.5.         What is the main source of the free energy for the mechanical work performed by DNA helicases during DNA replication in our cells?

  1. The hydrogen-bonding energy in the DNA double helix
  2. Thermal energy in the nucleus
  3. ATP hydrolysis by the helicase
  4. The energy of SSB binding to single-stranded DNA
  5. ATP hydrolysis by DNA topoisomerases

 

1.6.         During DNA replication in the cell, DNA primase makes short primers that are then extended by the replicative DNA polymerases. These primers …

  1. are made up of DNA.
  2. generally have a higher number of mutations compared to their neighboring DNA.
  3. are made more frequently in the leading strand than the lagging strand.
  4. are joined to the neighboring DNA by DNA ligase.
  5. provide a 3′-phosphate group for the DNA polymerases to extend.

 

1.7.         DNA ligases are used in both DNA replication and repair to seal breaks in the DNA. But DNA damage can result in single- or double-strand breaks that are not normal ligase substrates. These need to be processed first before a ligase can act on them. One of the enzymes that is recruited to some of such breaks is called PNK. It has two separate activities on the DNA, both of which can help provide a canonical ligase substrate. Which of the following activities would you expect PNK to have in this context?

  1. 5′ kinase (phosphorylation of a free 5′-OH group) and 3′ kinase
  2. 5′ phosphatase (dephosphorylation to create a free 5′-OH group) and 3′ phosphatase
  3. 3′ kinase and 3′ phosphatase
  4. 5′ phosphatase and 3′ kinase
  5. 5′ kinase and 3′ phosphatase

 

1.8.         Fill in the gap in the following paragraph using what you know about the activities of the proteins involved in DNA replication.

“Mitochondrial DNA replication requires a set of proteins similar to those used for the replication of the nuclear genome. However, mitochondria lack a dedicated DNA … and use the mitochondrial RNA polymerase instead.”

 

1.9.         During DNA replication, the single-strand DNA-binding (SSB) proteins …

  1. are generally found more on the leading strand than the lagging strand.
  2. bind cooperatively to single-stranded DNA and cover the bases to prevent base-pairing.
  3. prevent the folding of the single-stranded DNA.
  4. bind cooperatively to short hairpin helices that readily form in the single-stranded DNA.
  5. All of the above.

 

1.10.       This protein is present at every replication fork and prevents DNA polymerase from dissociating, but does not impede the rapid movement of the enzyme. Which of the following is true regarding this protein?

  1. It self-assembles onto DNA at the replication fork.
  2. It is assembled on DNA as soon as DNA polymerase runs into a double-strand region of DNA.
  3. Its assembly normally follows the synthesis of a new primer by the DNA primase.
  4. It disassembles from DNA as soon as DNA polymerase runs into a double-strand region.
  5. All of the above.

 

1.11.       At the replication fork, the template for the lagging strand is thought to loop around. This looping would allow the lagging-strand polymerase to move along with the rest of the replication fork instead of in the opposite direction. The single-strand part of the loop is bound by the single-strand DNA-binding (SSB) proteins. As each Okazaki fragment is synthesized toward completion, how does the size of the loop change? What about the size of the SSB-bound part of the loop?

  1. Increases; increases.
  2. Increases; decreases.
  3. Decreases; increases.
  4. Decreases; decreases.
  5. Decreases; does not change.

 

1.12.       The Dam methylase is responsible for methylating the adenine base in GATC sequences in Escherichia coli. Imagine two E. coli strains, one without any active Dam methylase, and the other with a hyperactive version of the enzyme that operates faster than the wild-type enzyme. Which of these strains would you expect to show a “mutator” phenotype?

  1. Both of the strains
  2. Neither of them
  3. Only the first strain
  4. Only the second strain

 

1.13.       In the following schematic drawing, two DNA molecules are shown before and after the action of a protein that is also involved in the process of DNA replication. What is this protein called?

 

  1. DNA ligase
  2. DNA helicase
  3. DNA polymerase I
  4. DNA topoisomerase I
  5. DNA topoisomerase II

 

1.14.       What do the enzymes topoisomerase I and topoisomerase II have in common?

  1. They both have nuclease activity.
  2. They both create double-strand DNA breaks.
  3. They both require ATP hydrolysis for their function.
  4. They both can create winding (tension) in an initially relaxed DNA molecule.
  5. All of the above.

 

1.15.       In Escherichia coli, replication of DNA can occur throughout the cell cycle while the cell is also actively transcribing its genes. This means collisions between replication forks and RNA polymerases are inevitable. Depending on the orientation of the genes, collisions can be rear-end (when both machines are traveling in the same direction) or head-on (when they are traveling in opposite directions). In the following paragraph, match each of the letters (A to D) to one appropriate number below. Do not use a number more than once. Your answer would be a four-digit number composed of digits 1 to 5 only, e.g. 1253.

“Typically, in a rear-end collision, the (A) of RNA polymerase collides with the (B) in the replication fork. In contrast, in a head-on collision, the (C) of RNA polymerase hits the (D) in the fork.”

  1. front edge (of RNA polymerase)
  2. rear edge (of RNA polymerase)
  3. DNA helicase
  4. leading-strand DNA polymerase
  5. lagging-strand DNA polymerase

 

1.16.       Which of the following features is common between the replication origins in Escherichia coli and Saccharomyces cerevisiae?

  1. They both normally exist in one copy per genome.
  2. Both are specified by DNA sequences of tens of thousands of nucleotide pairs.
  3. Both contain sequences that attract initiator proteins, as well as stretches of DNA rich in A-T base pairs.
  4. Both contain GATC repeats that are methylated to prevent the inappropriate “firing” of the origin.
  5. All of the above.

 

1.17.       You have found a strain of Escherichia coli that has an unusually short doubling time of only 15 minutes, despite the fact that its complete DNA replication should take almost 35 minutes. You also find that there is only one replication origin on its chromosome from which two forks originate, just like the normal process described for E. coli. However, you discover that the origin of replication in this strain has a significantly shorter “refractory period,” resulting in the reactivation of the origin before the previous round of replication is over. Based on this model, if you examine the chromosomes of this strain (under conditions of fast growth), how many replication forks would you expect to observe per chromosome on average?

  1. Two, just like the wild-type strain
  2. Four
  3. Six
  4. Eight
  5. Ten

 

1.18.       The telomerase enzyme in human cells …

  1. has an RNA component.
  2. extends the telomeres by its RNA polymerase activity.
  3. polymerizes the telomeric DNA sequences without using any template.
  4. removes telomeric DNA from the ends of the chromosomes.
  5. creates the “end-replication” problem.

 

1.19.       If this protein complex does not function normally, the ends of the eukaryotic chromosomes would activate the cell’s DNA damage response, causing chromosomal fusions and other genomic anomalies. What is this protein complex called?

  1. Telomerase
  2. T-loop
  3. ORC
  4. Shelterin
  5. RecA

 

1.20.       Which of the following schematic drawings better depicts the end of mammalian chromosomal DNA?

 

 

 

 

 

 

1.21.       Which of the following spontaneous lesions in DNA occurs most frequently in a mammalian cell?

  1. Depurination
  2. Cytosine deamination
  3. Guanine oxidation
  4. Guanine alkylation
  5. Depyrimidination

 

1.22.       DNA glycosylases constitute an enzyme family found in all three domains of life. They can …

  1. add sugar moieties to DNA.
  2. remove sugar moieties from DNA.
  3. add a purine or pyrimidine base to DNA.
  4. remove a purine or pyrimidine base from DNA.
  5. remove a nucleotide from DNA.

1.23.       Upon heavy damage to the cell’s DNA, the normal replicative DNA polymerases may stall when encountering damaged DNA, triggering the use of backup translesion polymerases. These backup polymerases …

  1. lack 3′-to-5′ exonucleolytic proofreading activity.
  2. are replaced by the replicative polymerases after adding only a few nucleotides.
  3. can create mutations even on undamaged DNA.
  4. may recognize specific DNA damage and add the appropriate nucleotide to restore the original sequence.
  5. All of the above.

 

1.24.       What are the products of deamination of cytosine and 5-methyl cytosine, respectively?

  1. Thymine and uracil
  2. Thymine in both cases
  3. Uracil and thymine
  4. Uracil in both cases
  5. Xanthine and hypoxanthine

 

1.25.       Which of the following repair pathways can accurately repair a double-strand break?

  1. Base excision repair
  2. Nucleotide excision repair
  3. Direct chemical reversal
  4. Homologous recombination
  5. Nonhomologous end joining

 

1.26.       This protein folds into a doughnut shape that can encircle DNA. It can load on the DNA only when the DNA is broken in both strands, so that the DNA can thread through the hole in the protein. Which of the following proteins do you think matches this description?

  1. PCNA, the sliding clamp for DNA polymerases at the replication forks
  2. Ku, the protein that recognizes DNA ends and can initiate nonhomologous end joining
  3. MCM, the helicase critical for the initiation and elongation of replication
  4. Topoisomerase II, which can create or relax superhelical tension in DNA
  5. RecA/Rad51, which carries out strand invasion in homologous recombination

 

1.27.       In contrast to vertebrates, there is very little DNA methylation in the genomes of invertebrates such as Drosophila melanogaster and Caenorhabditis elegans. Indicate whether you expect each of the following statements to be true (T) or false (F) regarding 5′-CG-3′ dinucleotide sequences in the genome. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT.

(  )  On average, approximately one dinucleotide out of every 16 in the human genome is a CG dinucleotide.

(  )  On average, approximately one dinucleotide out of every 256 in the D. melanogaster genome is a CG dinucleotide.

(  )  The proportion of CG dinucleotides in the human genome is more than that of C. elegans.

(  )  The proportion of CG dinucleotides in the human genome is more than that expected by chance.

 

1.28.       In which phases of the eukaryotic cell cycle does homologous recombination often occur to repair DNA damage?

  1. G1 and S phases
  2. S and G2 phases
  3. G2 and M phases
  4. M and G1 phases
  5. G1 and G2 phases

 

1.29.       Which of the following is NOT correct regarding homologous recombination and its regulation?

  1. Loss of heterozygosity can occur if a broken chromosome is repaired using a sister chromatid instead of its homologous chromosome.
  2. Repair of double-strand breaks by homologous recombination is favored during or soon after DNA replication.
  3. Homologous recombination can rescue broken or stalled replication forks in S phase.
  4. Excessive use of homologous recombination by human cells can lead to cancer.
  5. Low usage of homologous recombination by human cells can lead to cancer.

 

1.30.       In the following schematic drawing of a Holliday junction that undergoes branch migration, cutting at which combination of the sites a to d would generate a crossover?

  1. a and b
  2. a and c
  3. a and d
  4. b and c
  5. b and d

 

1.31.       In meiosis, a crossover in one position is thought to inhibit crossing-over in the neighboring regions. This regulatory mechanism …

  1. results in a very uneven distribution of crossover points along each chromosome.
  2. ensures that even small chromosomes undergo at least one crossover.
  3. controls how the Holliday junctions are resolved.
  4. All of the above.

 

1.32.       What group of mobile genetic elements is largely responsible for the resistance of the modern strains of pathogenic bacteria to common antibiotics?

  1. DNA-only transposons
  2. Retroviral-like retrotransposons
  3. Nonretroviral retrotransposons
  4. Site-specific recombinases

 

1.33.       DNA-only transposons …

  1. can be recognized by the presence of short inverted repeats at each end.
  2. often encode a transposase that mediates the transposition process.
  3. leave double-strand breaks in the donor chromosome.
  4. can move by a cut-and-paste mechanism.
  5. All of the above.

1.34.       Which of the following is true regarding retroviral-like retrotransposons?

  1. They encode both a reverse transcriptase and an RNA polymerase.
  2. They have directly repeated long terminal repeats at their two ends when integrated into chromosomal DNA.
  3. Their genomic RNA can be translated to produce viral coat proteins.
  4. They leave double-strand breaks in the original donor DNA.
  5. The Alu element in our genome is an example of retroviral-like retrotransposons.

 

1.35.       Which of the following is NOT correct regarding long and short interspersed nuclear elements?

  1. Each of them encodes a reverse transcriptase.
  2. They rely on the cellular transcription machinery to produce their RNA transcript.
  3. They use one of the strands in the target DNA as a primer to synthesize their DNA.
  4. Together, they make up about 40% of our genome.
  5. They can move into new regions of genome that do not have any homology with their DNA.

 

1.36.       Phase variation helps protect the bacterium Salmonella typhimurium against the immune system of its host by switching the orientation of a certain promoter. This process …

  1. is carried out through a DNA transposition mechanism.
  2. is irreversible.
  3. can often result in the excision of the promoter from the chromosome altogether.
  4. is mediated by enzymes that form transient covalent bonds with the DNA.

 

1.37.       A replication fork is shown schematically below. The strand labeled A is called the … strand.

 

1.38.       The sliding clamp and the DNA helicase that function at the replication fork both have three-dimensional structures resembling a ring with a central hole through which DNA is threaded. Which of these proteins, the clamp (C) or the helicase (H), do you think has a wider hole in its structure? Write down C or H as your answer.

 

1.39.       Indicate true (T) and false (F) statements below regarding the initiation of replication in human cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF.

(  )  Tens of thousands of replication origins are used each time a cell in our body replicates its DNA.

(  )  Different cells in our body use different sets of replication origins.

(  )  Both replication forks in a replication bubble are normally active in replication.

(  )  Gene expression and chromatin structure can affect the choice of the origins to use as well as the order in which they are activated.

 

1.40.       What combination of the following events normally prevents the origins of replication in the yeast Saccharomyces cerevisiae from “firing” more than once during the cell cycle? Your answer is a two-letter string composed of letters A to E only, e.g. AB. Order the letters in your answer alphabetically.

(A) The helicase-loading proteins Cdc6 and Cdt1 are only active in S phase.

(B) The helicase Mcm1 can be delivered to the origin recognition complex (ORC) only in S phase.

(C) The ORC can only become active (by dephosphorylation) in G1 phase.

(D) The helicase can only become active (by phosphorylation) in S phase.

(E) The ORC can bind to the origin only in S phase.

 

1.41.       Examples of two general types of DNA damage are shown in the following drawing. Which type of damage (1 or 2) is more common in our cells?

 

1.42.       In cells that are exposed to sunlight, ultraviolet (UV) light can result in covalent linkage between two adjacent DNA bases. If not repaired in time, such dimers can lead to inheritable mutations. Consider the sequence 5′-GGTATGATCATTATAA-3′ in the chromosome of a cell that is exposed to intense sunlight. How many possible dimers can form in the region of DNA double helix corresponding to this sequence?

 

1.43.       Indicate whether each of the following DNA lesions is typically repaired via the base excision (B) or nucleotide excision (N) repair pathway? Your answer would be a four-letter string composed of letters B and N only, e.g. BBBB.

  1. Deaminated cytosines
  2. Depurinated residues
  3. Thymine dimers
  4. Bulky guanine adducts

 

1.44.       Indicate true (T) and false (F) descriptions below regarding the nucleotide excision repair pathway. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF.

(  )  It involves recognition of distortions in the DNA double helix rather than specific base changes.

(  )  It involves endonucleolytic cleavage and helicase-mediated strand removal.

(  )  It involves cleavage by the AP endonuclease.

(  )  It is coupled to the DNA transcription machinery of the cell.

 

1.45.       Fill in the blank in the following paragraph. Do NOT use abbreviations.

 

“In human cells, the predominant pathway to repair double-strand breaks is …, in which the broken ends are simply rejoined with the concomitant loss of a few nucleotides. This leaves scars at the breakage sites. This pathway can potentially create chromosome translocations.”

 

3.46.       Sort the following steps in the order that they normally happen during the process of repairing double-strand breaks by homologous recombination. Your answer would be a six-letter string composed of letters A to F only, e.g. DEFABC.

(A) Ligation

(B) DNA synthesis using undamaged DNA as the template

(C) DNA synthesis using original DNA as the template

(D) Release of the invading strand

(E) Strand invasion

(F) Nuclease digestion (resection)

 

1.47.       The RecA/Rad51 protein carries out strand exchange during homologous recombination. Indicate true (T) and false (F) statements about this process below. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF.

(  )  RecA hydrolyzes ATP upon binding to the invading DNA strand.

(  )  RecA forces the invading strand into a conformation that fully mimics the geometry of a long DNA double helix.

(  )  Sampling of the homologous duplex by the invading strand is likely to occur in triplet nucleotide blocks.

(  )  RecA-bound, invading, single-stranded DNA binds and destabilizes the homologous duplex to allow the sampling of its sequence by base-pairing.

 

1.48.       Indicate true (T) and false (F) statements below regarding the use of homologous recombination in meiosis. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF.

(  )  Meiotic recombination starts with a double-strand break caused by errors in DNA replication.

(  )  Meiotic recombination occurs preferentially between DNA from maternal and paternal chromosome pairs.

(  )  Holliday junctions can form during meiotic recombination, sometimes in pairs.

(  )  During meiotic recombination in human cells, the majority of the invading strands are released, leading to no crossover.

 

1.49.       Indicate whether each of the following mobile elements in Drosophila is a DNA-only transposon (D), a retroviral-like retrotransposon (R), or a nonretroviral retrotransposon (N). Your answer would be a four-letter string composed of letters D, R, and N only, e.g. RRRR.

(  )  P elements; these have inverted terminal repeats and move with the help of a transposase enzyme.

(  )  Copia elements; these have directly repeated long terminal repeats and move with the help of a reverse transcriptase and an integrase.

(  )  F elements; these are long interspersed nuclear elements (LINEs).

(  )  Mariner elements; these move via a DNA intermediate.

 

1.50.       Consider three types of mobile genetic elements that are found in our genome: the DNA-only transposons (D), the retroviral-like retrotransposons (R), and the nonretroviral retrotransposons (N). Which type appears to still be active and move in our genome, accounting for a detectable fraction of human mutations? Write down your answer as D, R, or N.

 

1.51.       As shown in the following drawing, a researcher has engineered three pairs of LoxP sites (for conservative site-specific recombination) in a region that contains three reporter genes coding for red, yellow, or cyan fluorescent proteins, respectively. Each type of LoxP sequence (shown as a black, gray, or white arrowhead) is specific, meaning it does not recombine with the other types of LoxP sequences. Upon Cre recombinase activation, depending on which recombination event occurs first (which we assume is random), a number of possible combinations of reporters can remain in the final DNA. For each of the following combinations, indicate whether it can (C) or cannot (N) result from this recombination scheme. Do not consider the re-integration of excised DNA, which happens very rarely. Your answer would be a six-letter string composed of letters C and N only, e.g. CCCCNN.

 

 

(  )  Red and yellow

(  )  Red only

(  )  Yellow only

(  )  Cyan only

(  )  Yellow and cyan

(  )  Red and cyan

 

 

Answers:

  1. Answer: B

Difficulty: 2

Section: The Maintenance of DNA Sequences

Feedback: The mutation rate per nucleotide pair per replication cycle in humans is similar to that of Escherichia coli, and both are extremely low, less than one in a billion. Nevertheless, current mutation rates are thought to limit the number of essential genes in an organism to about 30,000. The stability of the germ-line genome depends on mutation rates, as does somatic-cell stability. Cancers typically arise from somatic mutations.

  1. Answer: B

Difficulty: 3

Section: The Maintenance of DNA Sequences

Feedback: About 15 cell generations pass in 10 hours of growth. Thus, the expected average number of mutations is equal to:

(15 generations) × (3 × 10–10 mutations/generation/nucleotide) × (4.5 × 106 nucleotides/genome)

= ~2 × 10–2 mutations per genome.

This is equivalent to about 98% of the bacterial cells having zero mutations and the remaining 2% having only one mutation each. Please note that the number of mutations in different cells in a given culture will vary widely around the average calculated above.

  1. Answer: E

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: Base-pairing is the basis of all DNA replication and repair. The other three mechanisms each increase accuracy by about 100-fold.

  1. Answer: E

Difficulty: 1

Section: DNA Replication Mechanisms

Feedback: Eukaryotic DNA polymerases are not as fast as their bacterial counterparts, presumably due to the difficulty in replicating through nucleosomes. All known DNA polymerases polymerize DNA from 5′ to 3′ and require a free 3′ end provided by a primer. The self-correcting polymerases have an additional 3′-to-5′ exonuclease activity that takes place at a distinct site in the enzyme.

  1. Answer: C

Difficulty: 1

Section: DNA Replication Mechanisms

Feedback: The helicase at the replication fork binds and hydrolyses ATP to move on the DNA in a stepwise fashion.

  1. Answer: B

Difficulty: 1

Section: DNA Replication Mechanisms

Feedback: The DNA primase enzyme syntheses RNA primers (with free 3′-OH groups) that are extended and later replaced by DNA that is more accurately synthesized. This happens more frequently in the lagging strand.

  1. Answer: E

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: The DNA ligase reaction normally requires a free 3′-OH group and a 5′-phosphate group on the two DNA ends to proceed. If a break has created 3′-phosphate and 5′-OH ends instead, for example, an enzyme such as PNK can dephosphorylate the 3′ end in the upstream molecule, and phosphorylate the 5′ end in the downstream molecule, to create the canonical ends.

  1. Answer: primase

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: Like the RNA polymerases involved in transcription, primases do not require priming and can synthesize RNA molecules that can be used as primers for DNA synthesis by the replicative DNA polymerases. RNA polymerase is used to create RNA primers for mitochondrial DNA replication.

  1. Answer: C

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: The SSB proteins bind cooperatively to the regions of exposed single-stranded DNA (which are routinely found in the lagging strand) and prevent the formation of hairpin helices without blocking the base-pairing potential.

  1. Answer: C

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: The sliding clamp is normally loaded (more frequently on the lagging strand) on the primer–template duplex by the clamp loader, a process that requires ATP hydrolysis by the loader. As soon as the polymerase runs into a double-strand region downstream, the clamp releases the polymerase but is not itself immediately disassembled from DNA.

  1. Answer: A

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: The double-strand region of the loop represents the growing Okazaki fragment and gradually increases in size. Additionally, as each Okazaki fragment is growing, the fork is also traveling, exposing more single-stranded DNA in the loop. The loop size resets to zero upon initiating the synthesis of the next Okazaki fragment.

  1. Answer: A

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: An inactive or hyperactive Dam methylase can interfere with strand-directed mismatch repair, as both make it harder to distinguish between the newly replicated and old DNA strands. Both situations are expected to increase the overall mutation rate, giving rise to a mutator phenotype.

  1. Answer: E

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: This untangling of intertwined DNA is mediated by the enzyme topoisomerase II. This type of reaction helps solve the winding problem during replication.

  1. Answer: A

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: Unlike topoisomerase I, which can only relieve the tension in DNA through introducing a nick (single-strand break) in one of the DNA strands, topoisomerase II can actively introduce or relieve tension by creating double-strand breaks (that remain tightly associated with the enzyme) using energy from ATP hydrolysis.

  1. Answer: 2413

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: The replication fork in E. coli can progress almost 20 times faster than the RNA polymerase, meaning that rear-end collisions are typically between the rear edge of the RNA polymerase and the front edge of the leading-strand DNA polymerase that is traveling (faster) in the same direction on the same template strand. In a head-on collision, however, the front edge of RNA polymerase hits the DNA helicase that is traveling on the same (lagging) strand in the opposite direction. Note that the polymerases translocate with respect to their template in the 3′-to-5′ direction, while the helicase in the fork translocates in the 5′-to-3′ direction.

  1. Answer: C

Difficulty: 1

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: The yeast origins of replication are similar to the bacterial origins in that they are A-T-rich and have particular sequences that attract replication initiator proteins. In higher eukaryotes, in contrast, the determinants of the replication origins are probably less sequence-specific. E. coli and S. cerevisiae both have relatively short origins compared to higher organisms.

  1. Answer: C

Difficulty: 3

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: Even though each round of replication takes 35 minutes, the “firing” interval can be adjusted independently, and even go out of control. In this strain, possibly due to mutations in the methylation pathway, the origins can become activated every 15 minutes, resulting in an average of six forks (three bubbles) per chromosome, the newest of which have started about 30 minutes after the oldest ones.

  1. Answer: A

Difficulty: 1

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: Telomerase uses its RNA component as a template to polymerize DNA sequences at the chromosome ends to solve the “end-replication” problem.

  1. Answer: D

Difficulty: 1

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: With the help of the t-loop structure, shelterin “hides” the telomere ends from the cell’s double-strand break detector and repair systems.

  1. Answer: D

Difficulty: 3

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: The telomeres have 3′ overhangs and are folded in structures called t-loops, which help protect the chromosome ends.

  1. Answer: A

Difficulty: 1

Section: DNA Repair

Feedback: Depurination (that is, the hydrolytic removal of purine bases from DNA) is by far the most common endogenous DNA lesion in our cells. Please refer to Table 5–3 for details.

  1. Answer: D

Difficulty: 1

Section: DNA Repair

Feedback: These enzymes can recognize altered bases in DNA and catalyze their removal by hydrolyzing the glycosidic bond between the base and the deoxyribose sugar.

  1. Answer: E

Difficulty: 1

Section: DNA Repair

Feedback: The translesion polymerases come in different flavors and some of them are very efficient in repairing specific damage, but they generally have a low fidelity as they cannot proofread. The cell thus limits their usage to very few polymerization steps.

  1. Answer: C

Difficulty: 2

Section: DNA Repair

Feedback: The use of thymine instead of uracil by DNA has the advantage that deamination of cytosine (to uracil) can be readily detected as an abnormal base in DNA. However, deamination of 5-methyl cytosine (a modification that is found in vertebrate DNA) can produce thymine and lead to mutations at a higher rate.

  1. Answer: D

Difficulty: 1

Section: DNA Repair

Feedback: Compared to the faithful homologous recombination pathway to repair double-strand breaks, nonhomologous end joining is a “quick and dirty” solution that often results in mutations at the site of repair.

  1. Answer: B

Difficulty: 3

Section: DNA Repair

Feedback: The Ku dimer forms a hole that accommodates a DNA double helix. In addition, its structure reflects the fact that it should only bind to free DNA ends (for example, from double-strand breaks) in a fashion similar to threading a needle.

  1. Answer: FFFF

Difficulty: 3

Section: DNA Repair

Feedback: In a random DNA sequence, one out of every 16 dinucleotides is a CG. However, these dinucleotides are found at a much lower frequency in the human genome, which is probably related to the fact that cytosine methylation at these sites (which has a role in transcription regulation) can be followed by deamination to cause a C-to-T substitution. Consistent with this notion, the CG frequency in D. melanogaster and C. elegans is close to the expected value.

  1. Answer: B

Difficulty: 2

Section: Homologous Recombination

Feedback: Homologous recombination is used to repair DNA damage during and shortly after the S phase of the cell cycle, when a sister chromatid is available for faithful repairs.

  1. Answer: A

Difficulty: 2

Section: Homologous Recombination

Feedback: If a homologous chromosome is used (instead of a sister chromatid) to repair a double-strand break, one of the parental alleles of the affected gene can be replaced by the other allele, leading to loss of heterozygosity.

  1. Answer: E

Difficulty: 3

Section: Homologous Recombination

Feedback: Cutting at b and d would result in a crossover. Cutting at a and c would only lead to a heteroduplex (no crossover). The other combinations are not normally chosen.

  1. Answer: B

Difficulty: 2

Section: Homologous Recombination

Feedback: The poorly understood regulatory mechanism of crossover control ensures a roughly even distribution of crossover points along chromosomes. It also ensures that each chromosome undergoes at least one crossover every meiosis. For many organisms, roughly two crossovers per chromosome occur, one on each arm.

  1. Answer: A

Difficulty: 1

Section: Transposition and Conservative Site-Specific Recombination

Feedback: DNA-only transposons that carry genes encoding antibiotic-inactivating enzymes are largely responsible for the spread of antibiotic resistance in bacterial strains.

  1. Answer: E

Difficulty: 1

Section: Transposition and Conservative Site-Specific Recombination

Feedback: DNA-only transposons often encode a transposase to move by a cut-and-paste mechanism that results in a broken donor chromosome. Short inverted repeats of DNA sequence are found at the ends of these transposons.

  1. Answer: B

Difficulty: 1

Section: Transposition and Conservative Site-Specific Recombination

Feedback: Retroviral-like retrotransposons resemble retroviruses in their replication, but lack the ability to produce a protein coat. They have directly repeated long terminal repeats at each end and are transcribed by cellular RNA polymerases. The transcript is then used as a template by a reverse transcriptase (encoded by the element) to make a double-stranded DNA copy that is then integrated into a new site on the chromosome using an integrase (also encoded by the element). This mechanism keeps the original copy unchanged. Alu elements are nonretroviral retrotransposons.

  1. Answer: A

Difficulty: 2

Section: Transposition and Conservative Site-Specific Recombination

Feedback: Most short interspersed nuclear elements (SINEs) do not encode any proteins and spread by pirating enzymes encoded by other transposons, such as long interspersed nuclear elements (LINEs).

  1. Answer: D

Difficulty: 2   

Section: Transposition and Conservative Site-Specific Recombination

Feedback: Phase variation in Salmonella is brought about by a conservative site-specific recombinase encoded by the bacterium. The recombinases that catalyze conservative site-specific recombination form transient, high-energy covalent bonds with the DNA and use this energy to complete the DNA rearrangements, an action reminiscent of topoisomerases.

  1. Answer: lagging

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: Strand A is extended 5′-to-3′ (from right to left) whereas the fork moves in the opposite direction; therefore, the strand should be polymerized discontinuously.

  1. Answer: C

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: The central hole in the clamp can accommodate a double-stranded DNA molecule with extra room for smooth sliding. However, the helicase only allows a single strand of DNA to pass through its central hole.

  1. Answer: TTTT

Difficulty: 2

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: Human cells have probably hundreds of thousands of potential origins, out of which about 40,000 are used each time a cell divides. Depending on gene expression and chromatin state, different sets of origins can be used in different cells of the body. Once an origin is activated, the resulting replication bubble expands on both sides of the origin.

  1. Answer: CD

Difficulty: 3

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: The helicase is delivered to the ORC by active Cdc6 and Cdt1 proteins in G1 phase, but is activated by phosphorylation only in S phase. At the same time, phosphorylation of ORC prevents its reactivation in S phase. ORC can be reactivated again later in G1.

  1. Answer: 2

Difficulty: 1

Section: DNA Repair

Feedback: Depurination (2) and deamination (1) are both common, although depurination is by far the most common endogenous DNA lesion in our cells.

  1. Answer: 5

Difficulty: 3

Section: DNA Repair

Feedback: UV irradiation can cause two adjacent pyrimidine bases (T or C) to form these covalent dimers. Please note that the complementary strand should also be taken into account. The possible sites of damage are TC and TT doublets in the strand shown, and TT, TC, and CC doublets in the complementary strand.

  1. Answer: BBNN

Difficulty: 2

Section: DNA Repair

Feedback: Please note that depurinated residues are repaired through only part of the base excision repair pathway.

  1. Answer: TTFT

Difficulty: 1

Section: DNA Repair

Feedback: In nucleotide excision repair, a multienzyme complex recognizes DNA double-helix distortions caused by damage. This pathway involves cutting the strand containing the damage at two positions flanking the damage. A helicase is also recruited and catalyzes strand removal. This pathway is also linked to the transcription machinery.

  1. Answer: nonhomologous end joining

Difficulty: 1    Section: DNA Repair

Feedback: In nonhomologous end joining, the broken ends are simply brought together and rejoined by DNA ligation, often concomitant with the loss of nucleotides at the site of joining.

  1. Answer: FEBDCA

Difficulty: 2

Section: Homologous Recombination

Feedback: This order of events results in the accurate repair of double-strand breaks.

  1. Answer: FFTT

Difficulty: 2

Section: Homologous Recombination

Feedback: RecA binds cooperatively to the invading strand, forcing blocks of three nucleotides into a conformation mimicking that of three nucleotides in a double helix (although, between adjacent triplets, the DNA backbone is untwisted and stretched out). This complex then binds and destabilizes the target homologous duplex, allowing the sampling to take place by base-pairing interactions. RecA hydrolyzes its bound ATP after the strand exchange is completed.

  1. Answer: FTTT

Difficulty: 2

Section: Homologous Recombination

Feedback: Meiotic recombination requires an initial double-strand break. In contrast to the breaks resulting from DNA damage, this one is provided in a controlled manner by a specialized meiotic protein.

  1. Answer: DRND

Difficulty: 2

Section: Homologous Recombination

Feedback: P elements and mariner elements are DNA-only transposons in different Drosophila species, copia is an abundant retroviral-like retrotransposon, and F elements are nonretroviral retrotransposons.

  1. Answer: N

Difficulty: 1

Section: Transposition and Conservative Site-Specific Recombination

Feedback: Some nonretroviral retrotransposons (such as Alu elements) are still moving in our genome, accounting for perhaps two mutations out of every thousand new human mutations. Although our genome is also littered with relics of retroviral-like retrotransposons, none appear to be active today.

  1. Answer: CNNCNC

Difficulty: 3

Section: Transposition and Conservative Site-Specific Recombination

Feedback: The arrangement of LoxP sites matters. For example, here, it does not allow the final combination (yellow + cyan) because deleting the red reporter requires the use of the white sequence, which also deletes the yellow reporter.

 

 

 

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION

Chapter 17: THE CELL CYCLE

© Garland Science 2015

 

1.1.         Which of the following simplified diagrams better shows the changes in the concentrations of three major cyclin–Cdk complexes (G1/S-Cdk, S-Cdk, and M-Cdk) in the cell in different stages of the cell cycle?

 

 

1.2.         Resveratrol is a natural compound found in red grapes (and red wine) and is thought to have beneficial effects in mammals, such as preventing tumor growth and delaying age-related diseases. In vitro, resveratrol and its derivatives have been shown to cause cell-cycle arrest in S phase and induce apoptosis. You have analyzed the DNA content of cultured cells in the presence and absence of these drugs using fluorescence-activated cell sorting. Compared with control cells (dashed line), which of the following curves do you think better represents the DNA content of cells treated with these compounds?

 

 

Reference: FACS Data Analysis (Questions 3-4)

You have been studying the effect of loss-of-function mutations in the Cdk inhibitor protein (CKI) p21. You add the drug fucoxanthin to cell cultures harboring either wild-type or mutant versions of the p21 gene. Fucoxanthin is known to induce cell-cycle arrest in G1. After a day, you add the thymidine analog BrdU to the culture media, collect the cells after an hour, treat them with anti-BrdU antibody and the fluorescent DNA stain DAPI, and finally subject them to fluorescence-activated cell sorting (FACS). The FACS data can be viewed as a two-dimensional dot plot composed of thousands of dots, in which each cell is represented by one dot at coordinates that correspond to the intensities of the DAPI fluorescence signal (X axis) and BrdU fluorescence signal (Y axis) for that cell. Answer the following question(s) according to the simplified dot plot below, generated from your experiment.

 

 

 

1.3.         Indicate which boxed region (1, 2, or 3) in the FACS plots corresponds better to each of the following phases of the cell cycle. Your answer would be a three-digit number composed of numbers 1 to 3, with each number used once, e.g. 312.

(  )  G1 phase

(  )  S phase

(  )  G2 and M phases

 

 

1.4.         Which one of the FACS plots (a or b) would you expect to correspond to the loss-of-function p21 mutants? Write down a or b as your answer.

 

 

1.5.         Sort the following schematic diagrams (A to F) to reflect the order of events in a typical eukaryotic M phase. An interphase cell in G2 phase is drawn on the left for comparison. Your answer would be a six-letter string composed of letters A to F only, e.g. BCEADF.

 

 

 

1.6.         Consider two mammalian cells, one in G1 and the other in G0 (stationary) phase. If they are stimulated to pass the restriction point by the addition of an extracellular proliferation signal, but the signal is then immediately removed, what would you expect to happen?

  1. Both cells will replicate their DNA.
  2. Only the G1 cell will replicate its DNA.
  3. Only the G0 cell will replicate its DNA.
  4. Only the G1 cell will start to replicate its DNA, but will stop halfway through the replication and will not reach G2.
  5. Neither of the cells will replicate their DNA.

 

1.7.         Which of the time points (A to E) in the following schematic drawing of the mammalian cell cycle represents the restriction point?

 

Answer: B                   Difficulty: 1    Section: The Cell-Cycle Control System

Feedback: Under favorable conditions and in the presence of signals to grow and divide, cells in early G1 (or G0) progress through a commitment point near the end of G1 known as Start (in yeasts) or the restriction point (in mammalian cells).

 

1.8.         Indicate whether each of the following descriptions better applies to a Wee1 protein (W) or a Cdc25 protein (C). Your answer would be a four-letter string composed of letters W and C only, e.g. WWCW.

(  )  It is a protein kinase.

(  )  It activates M-Cdk complexes.

(  )  It is activated by M-Cdk complexes.

(  )  Its loss in fission yeast results in small cell size.

 

1.9.         Mammalian Cdk inhibitor proteins (CKIs) can be grouped into two families based on their structural and functional differences. The Cip/Kip family proteins (e.g. p21) have a broad binding specificity. These proteins bind preferentially to already formed cyclin–Cdk complexes and thus enhance complex formation. However, they inhibit the kinase activity of most complexes (e.g. S-Cdks), except in the case of G1-Cdk complexes where no inhibition occurs. Consequently, Cip/Kip family proteins have an overall positive effect on Cdk4/6 activity due to their help in bringing the subunits together. In contrast, the inhibitors of the INK4 family (e.g. p16) bind only to the Cdk subunit of G1-Cdks and prevent binding of both the G1 cyclins and the Cip/Kip family CKIs. Based solely on these findings, would you expect p16 to activate (A) or inactivate (I) the S-Cdks in the presence of limited amounts of p21?

 

1.10.       In the fission yeast Schizosaccharomyces pombe, the proteins Cut1 and Cut2 form a complex that is catalytically inactive. At the onset of anaphase, Cut2 is polyubiquitylated by a large E3 complex containing Cut4, Cut9, Cut23, and other proteins, and is subsequently destroyed. Cut1 then cleaves Rad21, a non-SMC component of a complex that also contains two SMC proteins, thus allowing sister-chromatid separation. Mutations in the cut genes lead to the cut phenotype, in which the cell attempts cytokinesis without chromosome segregation. Indicate which of the following proteins or protein complexes corresponds to or contains the product of the genes cut1 (A), cut2 (B), cut4 (C), and rad21 (D). Your answer would be a four-letter string composed of letters A to D only, e.g. DCAB.

(  )  Securin

(  )  Cohesin

(  )  Anaphase-promoting complex

(  )  Separase

 

1.11.       The following schematic diagram shows the activation of M-Cdk in mitosis. Indicate which proteins below correspond to those indicated as A to D in the diagram. Your answer would be a four-letter string composed of letters A to D only, e.g. BACD.

 

(  )  Wee1

(  )  CAK

(  )  Cdc25

(  )  M-cyclin

 

1.12.       Indicate whether each of the following occurs mainly in G1 phase (G), S phase (S), or G2 phase (H) of the cell cycle. Your answer would be a four-letter string composed of letters G, S, and H only, e.g. HGSS.

(  )  DNA helicase activation

(  )  DNA helicase deposition on DNA at the replication origins

(  )  ORC phosphorylation

(  )  Licensing of replication origins

 

1.13.       Which of the following events occurs in mitotic metaphase?

  1. Nuclear envelope breakdown
  2. Nuclear envelope reassembly
  3. Chromosome attachment to spindle microtubules for the first time
  4. Chromosome alignment at the spindle equator
  5. Mitotic spindle assembly

 

1.14.       Consider two kinesin motor proteins at the mitotic spindle midzone: kinesin-5 is a tetrameric motor that walks toward the plus end of both microtubules to which it is attached via its motor domains; kinesin-14, on the other hand, walks toward the minus end of one microtubule while it is attached to another microtubule via its tail domain. How do these motors affect the length of the spindle?

  1. They both work to shorten the spindle.
  2. Kinesin-5 works to shorten the spindle whereas kinesin-14 works to lengthen it.
  3. Kinesin-5 works to lengthen the spindle whereas kinesin-14 works to shorten it.
  4. They both work to lengthen the spindle.

 

1.15.       How is centrosome duplication similar to DNA replication?

  1. They both use a semiconservative mechanism.
  2. They are initiated at around the same time in the cell cycle, near the G1/S transition.
  3. They are both controlled in such a way that they replicate once and only once per cell cycle.
  4. They are both separated from their sister copies in mitosis.
  5. All of the above.

 

1.16.       Which of the following is more directly driven by active M-Cdk?

  1. Centrosome maturation
  2. Centrosome duplication
  3. Nuclear envelope reassembly
  4. Inactivation of APC/C
  5. Cell cleavage

 

1.17.       Indicate true (T) and false (F) statements below regarding mitosis and the changes that it brings about compared to interphase. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTF.

(  )  Microtubules become greatly stabilized in mitosis compared to interphase.

(  )  The number of γ-tubulin ring complexes in the centrosomes increases greatly during mitosis compared to interphase.

(  )  Chromosomes stabilize mitotic microtubules through activation of Ran monomeric G protein.

(  )  A bipolar spindle can be formed even in the absence of centrosomes.

1.18.       Fill in the blank: “The … is a large structure formed at the centromeric region of each eukaryotic chromosome. It captures spindle microtubules in mitosis, and therefore serves to attach the chromosomes to the spindle poles.”

 

1.19.       Which one of the following chromosomes has formed stable attachment(s) to the spindle microtubules?

 

 

 

1.20.       The cell cycle can be arrested in mitosis when a single sister-chromatid pair is mono-oriented on the mitotic spindle. If at this point a glass microneedle is used to pull the mono-oriented chromosome toward the pole to which it is not attached, the cell proceeds to anaphase. This observation confirms that …

  1. mono-oriented chromosomes are stable.
  2. mechanical tension drives the formation of bi-oriented chromosomes.
  3. bi-oriented chromosomes activate the spindle assembly checkpoint.
  4. lack of mechanical tension at the kinetochore in at least one chromosome prevents entry into anaphase.
  5. mechanical tension at the kinetochore in at least one chromosome is required for entry into anaphase.

 

1.21.       Indicate whether each of the following phosphorylation events typically activates (A) or inactivates (I) the protein that is being phosphorylated. Your answer would be a four-letter string composed of letters A and I only, e.g. IIIA.

(  )  Phosphorylation of Cdc25 by M-Cdk

(  )  Phosphorylation of condensin subunits by M-Cdk

(  )  Phosphorylation of kinesin-5 by Aurora-A

(  )  Phosphorylation of Ndc80 subunits by Aurora-B

 

1.22.       In the following schematic drawing of a vertebrate prometaphase chromosome as seen under a microscope, indicate whether the chromosome is more likely to be closer to pole A or pole B. Write down A or B as your answer.

 

 

Reference: Chromosome Movement in Anaphase (Questions 23-24)

To study chromosome movement during anaphase in mammals, you have injected fluorescently labeled tubulin subunits into cells such that microtubules can be seen under a fluorescence microscope. You then use a relatively strong laser beam to bleach the fluorescent dyes in a limited area of a metaphase cell, as indicated in the schematic diagram below, and follow the progression of mitosis under the microscope by time-lapse imaging. You then use the images to measure the change in distance between various components, and plot the results in the graph below. Answer the following question(s) based on these results.

 

 

 

1.23.       Based on the results from the experiment, which is dominant in this cell: anaphase A or anaphase B? Write down A or B as your answer.

 

1.24.       Based on the results from your experiment, which force is dominant in chromosome movement in this cell: polar ejection force (E), microtubule flux (F), or kinetochore microtubule plus-end depolymerization (P)? Write down E, F, or P as your answer.

 

1.25.       Imagine a prometaphase chromosome pair that is attached to one spindle pole. Which of the following would happen if both arms of the chromosome are severed with a strong laser beam?

  1. All chromosome fragments (the centromere-containing fragment and the arm fragments) would be pushed away from the pole.
  2. Only the two arm fragments would be pulled toward the pole. The kinetochore-containing fragment would remain stationary.
  3. The kinetochore-containing fragment would be pulled toward the pole. The two arm fragments would move away from the pole.
  4. The kinetochore-containing fragment would move away from the pole, but the two arm fragments would be pulled toward the pole.
  5. All chromosome fragments would be pulled toward the pole.

 

1.26.       If cells that have started mitosis are treated with nocodazole, a drug that depolymerizes microtubules, what would you expect to happen? Where would you expect Mad2 protein to be localized?

  1. The cells would arrest in prometaphase; Mad2 would be localized to the spindle poles.
  2. The cells would arrest in telophase; Mad2 would be localized to the spindle poles.
  3. The cells would immediately enter anaphase, finish mitosis, and enter G1; Mad2 would be localized to the spindle poles.
  4. The cells would arrest in prometaphase; Mad2 would be localized to almost all kinetochores.
  5. The cells would arrest in telophase; Mad2 would be localized to almost all kinetochores.

 

1.27.       Treatment of dividing cells with a low dose of the antifungal drug benomyl, which destabilizes microtubules, slows down correct spindle assembly. But at such doses, the spindle is eventually formed and the cells survive. However, mutations in some genes confer benomyl sensitivity: the mutant cells die because they fail to arrest the cell cycle in the presence of unattached kinetochores and progress through anaphase, with disastrous consequences. Which of the following would you expect to be a benomyl-sensitive mutant?

  1. A loss-of-function mutant in the gene encoding Mad2.
  2. A mutation causing the overexpression of Cdc20.
  3. A loss-of-function mutation in the gene encoding a kinase that inhibits Cdc20–APC/C.
  4. A loss-of-function mutation in the gene encoding a tubulin subunit.
  5. All of the above.

 

1.28.       Which of the following motor proteins are more directly involved in anaphase B?

  1. Kinesin-5 on interpolar microtubules and dynein on kinetochore microtubules
  2. Kinesin-13 on kinetochore microtubules and dynein on astral microtubules
  3. Kinesins 4 and 10 on interpolar and astral microtubules and dynein on kinetochore microtubules
  4. Kinesin-5 on interpolar microtubules and kinesins 4 and 10 on interpolar and astral microtubules
  5. Kinesin-5 on interpolar microtubules and dynein on astral microtubules

 

1.29.       The microtubule-binding protein Patronin binds to the minus ends of microtubules at the spindle pole and protects them from the effect of catastrophe factors. If Patronin activity is inhibited by injecting an anti-Patronin antibody into Drosophila melanogaster embryos prior to cellularization, anaphase B is suppressed and the spindles are significantly shorter. Which of the forces (1 or 2) in the following schematic diagram do you think is mostly responsible for anaphase B in this organism at this stage? Does the effect of Patronin inhibition resemble that of kinesin-5 inhibition (I) or overactivation (O)?

 

 

  1. 1; I
  2. 1; O
  3. 2; I
  4. 2; O

 

1.30.       Mutant Saccharomyces cerevisiae cells that lack the gene encoding securin can divide more or less normally by mitosis, without significant chromosome segregation defects. Cells harboring a nondegradable version of securin, on the other hand, arrest in metaphase as expected, since they cannot activate separase to enter anaphase. Similarly, cells lacking Cdc20 arrest in metaphase, since they cannot activate APC/C. Finally, cells lacking both securin and Cdc20 arrest in anaphase: they manage to separate sister chromatids, but do not progress much further. These results suggest that in wild-type cells, …

  1. degradation of securin is necessary to trigger sister-chromatid separation.
  2. degradation of securin is sufficient to trigger sister-chromatid separation.
  3. Cdc20–APC/C is NOT necessary for sister-chromatid separation.
  4. Cdc20–APC/C is NOT necessary for later events in anaphase.
  5. All of the above.

 

1.31.       Indicate true (T) and false (F) statements below regarding cytokinesis in animal cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF.

(  )  The force for cytokinesis is generated by kinesin motors on microtubule bundles that form the contractile ring.

(  )  As the contractile ring constricts, its thickness increases to keep a constant volume.

(  )  The midbody forms from bundles of actin and myosin II.

(  )  Local activation of Ran GTPase triggers the assembly and contraction of the contractile ring.

 

1.32.       Formin nucleates the growth of parallel actin bundles in the cell. Additionally, myosin motor activity is positively regulated by phosphorylation. The monomeric G protein RhoA is important in cytokinesis, because it directly or indirectly …

  1. activates formins and inactivates myosin light-chain phosphatase.
  2. inactivates formins and activates myosin light-chain kinases.
  3. activates formins as well as myosin light-chain phosphatases.
  4. inactivates formins as well as myosin light-chain kinases.
  5. None of the above.

 

Reference: Contractile-Ring Positioning (Questions 33-35)

Three models for contractile-ring positioning in animal cells are presented in the schematic diagrams below. Answer the following question(s) according to these models.

 

 

 

 

 

1.33.       In classic experiments carried out in the 1960s and schematized below, an egg of the sand dollar Echinarachnius parma was made into a torus shape using a glass bead on the end of a needle, to study the influence of microtubules on cleavage-furrow positioning. The cells were allowed to undergo two rounds of mitosis. After the second mitosis, an extra furrow (indicated by a red arrowhead) was observed between the two asters that did not have a spindle between them. Which of the models (1 to 3) is more consistent with this observation? Write down 1, 2, or 3 as your answer.

 

 

1.34.       In experiments on the flattened eggs of the sea urchin Tripneustes gratilla, cytokinesis was analyzed after placing a small physical barrier (e.g. a glass needle or an oil droplet) between the spindle and various areas of the cell cortex. Blocks that were located over the equatorial plane prevented furrow formation in the proximal membrane, whereas blocks located in other areas did not affect furrowing. Which model (1 to 3) is NOT supported by these results? Write down 1, 2, or 3 as your answer.

 

 

1.35.       In the early embryos of Caenorhabditis elegans, defects in the formation of astral microtubules increase myosin activity throughout the cell cortex. If the spindle is forced to one side of the cell, the cortical myosin activity is observed mostly at the opposite side of the cell. Which model (1, 2, or 3) better predicts these observations? Write down 1, 2, or 3 as your answer.

 

1.36.       Indicate true (T) and false (F) statements below regarding cytokinesis in eukaryotic cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF.

(  )  The preprophase band, composed of microtubules and actin filaments, marks the site of cytokinesis in plant cells.

(  )  In some cell types, the site of contractile-ring formation is determined before mitosis.

(  )  The early cell plate in dividing plant cells appears after phragmoplast formation.

(  )  The membrane required for the newly formed cell plate in plant cells is provided by endocytosis from the equatorial plasma membrane.

 

1.37.       In most mammalian cells, low M-cyclin protein levels are maintained during G1 phase. What is mainly responsible for maintaining these low levels?

  1. Cdc20–APC/C
  2. Cdh1–APC/C
  3. Skp2–SCF
  4. β-trCP–SCF
  5. p27

 

1.38.       In the following schematic diagram of an early Drosophila embryo, which step (A to C) corresponds to cellularization? Write down A, B, or C as your answer.

 

 

 

1.39.       Indicate whether each of the following, when active, tends to activate (A) or inactivate (I) M-Cdks. Your answer would be a four-letter string composed of letters A and I only, e.g. IIAI.

(  )  M-Cdk

(  )  Cdc20–APC/C

(  )  Cdh1–APC/C

(  )  Sic1

 

1.40.       How is Cdc20–APC/C similar to Cdh1–APC/C?

  1. They are both active throughout interphase.
  2. They are both inhibited by M-Cdk.
  3. They both inhibit M-Cdk activity.
  4. They are both activated suddenly at the onset of mitosis.
  5. They are both inactivated soon after anaphase.

 

1.41.       Which division in meiosis is more similar to mitosis? In which division do sister chromatids separate from each other?

  1. Meiosis I; meiosis I
  2. Meiosis I; meiosis II
  3. Meiosis II; meiosis I
  4. Meiosis II; meiosis II

 

1.42.       Which stage is usually the longest in meiosis?

  1. Prophase I
  2. Prometaphase I
  3. Telophase I
  4. Prophase II
  5. Prometaphase II

 

1.43.       Indicate true (T) and false (F) statements below regarding meiosis in eukaryotic cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF.

(  )  Bivalents form before prophase I.

(  )  Crossing-over begins before the synaptonemal complex assembly.

(  )  Chiasmata can first be seen under the microscope after the disassembly of the synaptonemal complexes.

(  )  All recombination events lead to crossovers.

 

1.44.       In the following schematic diagram of a meiotic bivalent in diplotene, indicate whether each of the following chromatid pairs have undergone DNA exchange (E) or not (N). The different color of the final form of chromatid 3 is for clarity only. Your answer would be a four-letter string composed of letters E and N only, e.g. NENE.

 

 

(  )  Chromatids 1 and 3

(  )  Chromatids 1 and 4

(  )  Chromatids 2 and 3

(  )  Chromatids 2 and 4

 

1.45.       Loss of the gene encoding shugoshin in many multicellular organisms leads to sterility, suggesting defects in meiosis. What would you expect to occur in meiotic cells lacking shugoshin?

  1. The homologs fail to separate in anaphase I.
  2. The sister chromatids fail to separate in anaphase II.
  3. All chromatids separate prematurely in anaphase I.
  4. Removal of cohesion between homolog arms fails in prophase I.
  5. Removal of cohesion between sister chromatids fails in prophase II.

 

1.46.       Indicate whether each of the following descriptions better applies to mitogens (M), growth factors (G), or survival factors (S). Your answer would be a four-letter string composed of letters M, G, or S only, e.g. GMSG.

(  )  They unblock cell-cycle progression

(  )  They suppress apoptosis

(  )  They trigger a wave of G1/S-Cdk activity

(  )  They inhibit the degradation of cellular macromolecules

 

1.47.       Which one better supports cell proliferation when added to fibroblast cultures: serum or plasma? This activity is due to the presence of mitogens in this fluid. What is responsible for making these mitogens?

  1. Serum; red blood cells
  2. Serum; platelets
  3. Plasma; red blood cells
  4. Plasma; white blood cells
  5. Plasma; platelets

 

1.48.       Which of the following is an inhibitory extracellular signal for cell proliferation?

  1. EGF
  2. TGFβ
  3. Erythropoietin
  4. IGF
  5. PDGF

 

1.49.       Which of the following cell populations in our body has the highest mitotic index?

  1. Neurons
  2. Hepatocytes
  3. Red blood cells
  4. Fibroblasts
  5. Skeletal myocytes

 

1.50.       Which of the following proteins is the product of an immediate early gene expressed following mitogenic stimulation of cell-cycle entry?

  1. E2F
  2. Rb
  3. Myc
  4. G1-cyclins
  5. All of the above

 

1.51.       Which of the following events occurs at the onset of S phase in a mitogen-stimulated vertebrate cell?

  1. Activation of Cdh1–APC/C
  2. Activation of Rb
  3. Activation of geminin
  4. Activation of Cdc20
  5. All of the above

 

1.52.       Which of the following events contributes to driving the mammalian cell past the restriction point of the cell cycle?

  1. Phosphorylation of Cdh1–APC/C by G1/S-Cdk
  2. Destruction of CKIs that target S-Cdks
  3. Phosphorylation of Rb by G1-Cdk, G1/S-Cdk, and S-Cdk
  4. Activation of E2F gene expression by active E2F protein
  5. All of the above

 

1.53.       In the following schematic diagram of a typical eukaryotic cell cycle, choose two major time points (among A to E) at which the cell-cycle control system normally arrests the cycle if DNA damage is detected. Your answer would be a two-letter string composed of letters A to E only in alphabetical order, e.g. CE.

 

 

 

1.54.       A cell has been subjected to ultraviolet irradiation, causing a significant number of mutations in the genome. Which of the following would you NOT expect to occur as a result?

  1. Activation of the protein kinase ATR
  2. Activation of the protein kinase Chk1
  3. Inactivation of the protein phosphatase Cdc25
  4. Binding of p53 to Mdm2
  5. Stabilization of p53

 

1.55.       In a multicellular organism such as a mammal, loss-of-function mutations in many genes can contribute to the development of cancer. These genes are therefore called tumor suppressors. In their absence, the cell fails to stop progression through the cell cycle under conditions in which normal cells would arrest, paving the way for tumorigenesis. Indicate whether each of the following proteins is (Y) or is not (N) expected to be the product of a tumor suppressor gene based on its function in the cell cycle. Your answer would be a four-letter string composed of letters Y and N only, e.g. YNNN.

(  )  Rb

(  )  Myc

(  )  p53

(  )  p21

 

1.56.       Indicate true (T) and false (F) statements below regarding cellular control of growth and division. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTF.

(  )  Replicative cell senescence usually arises due to the accumulation of mutations in genes encoding S- and M-cyclins.

(  )  Replicative cell senescence is caused by the induction of apoptosis by p53.

(  )  Excessive mitogenic stimulation can result in Mdm2 activation and the induction of apoptosis or cell-cycle arrest.

(  )  The Mdm2 inhibitor Arf induces cell-cycle progression.

 

1.57.       You have synchronized a culture of HeLa cells so that the cells are all at the same stage in the cell cycle. You then treat the cells with serum containing nutrients and growth factors in the presence or absence of the drug rapamycin. The distribution of cell sizes in the two cell populations is shown below. Based on these results, do you think rapamycin is an activator or inhibitor of the TOR complex? Is the S6 kinase expected to be up-regulated or down-regulated in the presence of rapamycin?

 

  1. Activator; up-regulated
  2. Activator; down-regulated
  3. Inhibitor; up-regulated
  4. Inhibitor; down-regulated

 

 

Answers

  1. Answer: A

Difficulty: 2

Section: The Cell-Cycle Control System

Feedback: The G1/S-Cdk activity (solid black curve) peaks in late G1 and falls in S phase. The S-Cdk levels (solid gray curve) remain elevated through S and G2 until early M phase. M-Cdk (dashed gray curve) accumulates before entry into mitosis, but their level falls in mid-mitosis.

  1. Answer: B

Difficulty: 3

Section: The Cell-Cycle Control System

Feedback: S-phase arrest will increase the proportion of cells in this phase; that is, cells with a relative cellular DNA content of between 1 and 2.

  1. Answer: 213

Difficulty: 3

Section: The Cell-Cycle Control System

Feedback: After a short incubation with BrdU, cells that are replicating their DNA (in S phase) would incorporate the compound and would show a high BrdU signal compared to cells that are in G1, G2, or M phases. The DAPI signal intensity represents the total DNA content, which increases from G1 to S to G2 phases.

  1. Answer: a

Difficulty: 3

Section: The Cell-Cycle Control System

Feedback: Given the significant difference between the results from wild-type and p21 mutant cells, fucoxanthin appears to arrest the cell cycle (in G1) by activating a p21-dependent pathway. Loss of p21 (plot a) therefore blocks the cell-cycle arrest induced by the drug.

  1. Answer: CFEDAB

Difficulty: 2

Section: Overview of the Cell Cycle

Feedback: The M phase includes mitotic prophase (C), prometaphase (F), metaphase (E), anaphase (D), telophase (A), and finally cytokinesis (leading to B).

  1. Answer: A

Difficulty: 1

Section: Overview of the Cell Cycle

Feedback: Once a cell passes the restriction point, it is committed to DNA replication (and cell division) even if the stimulating signals are removed.

  1. Answer: B

Difficulty: 1

Section: The Cell-Cycle Control System

Feedback: Under favorable conditions and in the presence of signals to grow and divide, cells in early G1 (or G0) progress through a commitment point near the end of G1 known as Start (in yeasts) or the restriction point (in mammalian cells).

  1. Answer: WCCW

Difficulty: 2

Section: The Cell-Cycle Control System

Feedback: Wee1 kinases phosphorylate and inactivate cyclin–Cdk complexes; Cdc25 phosphatases do the opposite. In a positive feedback loop at the onset of mitosis, M-Cdk inactivates its inactivator, Wee1, and activates its activator, a Cdc25. The small cell size in wee mutants can result from Wee1 depletion.

  1. Answer: I

Difficulty: 3

Section: The Cell-Cycle Control System

Feedback: In the absence of p16, p21 is sequestered by the G1-Cdk complex. Active p16, however, prevents p21 from binding, making it available to inhibit the S-Cdk complexes.

  1. Answer: BDCA

Difficulty: 2

Section: S Phase

Feedback: Rad21 is a Scc1 homolog, which is cleaved by separase in the metaphase–anaphase transition.

  1. Answer: CBDA

Difficulty: 2

Section: Mitosis

Feedback: Activation of M-Cdk is sharpened by positive feedback loops in which M-Cdk activates its activator (Cdc25) and inactivates its inactivator (Wee1) by phosphorylation.

  1. Answer: SGSG

Difficulty: 2

Section: S Phase

Feedback: The loading of inactive DNA helicases onto the replication origins and the formation of the prereplicative complex (preRC) occur in late mitosis and early G1. In S phase, the helicases are activated to initiate DNA replication, while the origin is prevented from refiring by origin recognition complex (ORC) phosphorylation and other mechanisms.

  1. Answer: D

Difficulty: 1

Section: Mitosis

Feedback: The alignment of chromosomes at the cell equator is the hallmark of metaphase.

  1. Answer: C

Difficulty: 2

Section: Mitosis

Feedback: While kinesin-5 pushes the two poles of the spindle apart by walking toward the plus end of antiparallel microtubules, kinesin-14 pulls the two poles toward each other by walking toward the minus end of interpolar microtubules in the midzone.

  1. Answer: E

Difficulty: 1

Section: Mitosis

Feedback: The centrosomes, like the chromosomes, are duplicated only once during each cell cycle. Centrosome duplication is triggered by the activation of G1/S-Cdk. Both centrosome duplication and DNA replication employ a semiconservative mechanism. The two new centrosomes remain together on one side of the cell until entry into mitosis.

  1. Answer: A

Difficulty: 1

Section: Mitosis

Feedback: The M-Cdk is responsible for most events in early mitosis, including spindle assembly and centrosome maturation. It is inactivated later in mitosis, resulting in the events of telophase and cytokinesis.

  1. Answer: FTTT

Difficulty: 2

Section: Mitosis

Feedback:  Mitosis is accompanied by a dramatic reorganization of the microtubule cytoskeleton, with a significant increase in microtubule instability. At the same time, centrosome maturation greatly increases microtubule nucleation. The capture of microtubules by chromosomes stabilizes microtubules in part by activating Ran, a monomeric GTPase. The ability of mitotic chromosomes to stabilize and organize microtubules makes it possible to form bipolar spindles even in the absence of centrosomes.

  1. Answer: kinetochore

Difficulty: 1

Section: Mitosis

Feedback: Kinetochores are giant protein structures that attach sister chromatids to the mitotic spindle.

  1. Answer: C

Difficulty: 1

Section: Mitosis

Feedback: Bi-oriented chromosomes are stabilized due to the unique tension created.

  1. Answer: D

Difficulty: 2

Section: Mitosis

Feedback: This important experiment confirms the importance of mechanical tension within the kinetochores in the spindle assembly checkpoint mechanism.

  1. Answer: AAAI

Difficulty: 2

Section: Mitosis

Feedback: M-Cdk activates the Cdc25 phosphatase in a positive feedback loop. It also phosphorylates condensin subunits, stimulating DNA binding and supercoiling activity. Along with Aurora-A, it also phosphorylates kinesins such as kinesin-5, stimulating microtubule binding and motor activity. Without tension in the kinetochore, Aurora-B phosphorylates and inactivates components of Ndc80, lowering its affinity for microtubule plus ends.

  1. Answer: A

Difficulty: 2

Section: Mitosis

Feedback: The polar wind, mediated by plus-end directed kinesin motors, gets stronger as the chromosome approaches a pole, and pushes the chromosome arms toward the opposite pole.

  1. Answer: A

Difficulty: 2

Feedback:  In this example, the two poles are pushed away from each other only slightly during anaphase.

  1. Answer: F

Difficulty: 3

Feedback: In these cells, the chromosomes seem to approach the nearby pole as fast as the bleached zone approaches the pole, meaning that the microtubule subunit flux is enough to explain the poleward chromosome movement.

  1. Answer: C

Difficulty: 2

Section: Mitosis

Feedback: The result of such an experiment has provided evidence for polar ejection forces acting on chromosome arms.

  1. Answer: D

Difficulty: 2

Section: Mitosis

Feedback: A cell treated with nocodazole fails to assemble a functional spindle, and therefore arrests in prometaphase with the chromosomes left unattached. Mad2 is recruited to unattached kinetochores and activates the spindle assembly checkpoint, preventing the onset of anaphase.

  1. Answer: E

Difficulty: 3

Section: Mitosis

Feedback: All of these mutations make the cells more sensitive to the benzimidazole drug.

  1. Answer: E

Difficulty: 1

Section: Mitosis

Feedback:  Two major forces act in anaphase B. Kinesin-5 pushes the two poles away from each other, and dynein pulls the poles toward the cell periphery.

  1. Answer: A

Difficulty: 3

Section: Mitosis

Feedback: The force generated by kinesin-5 to lengthen the spindle in anaphase B is counteracted by catastrophe factors at the poles, which are inhibited by Patronin. Inhibition of Patronin would therefore result in anaphase B suppression.

  1. Answer: A

Difficulty: 3

Section: Mitosis

Feedback: Both Cdc20 activation and securin destruction are necessary for sister-chromatid separation because cells in which either of these fail to occur are observed to arrest in metaphase. Securin degradation does not appear to be sufficient for sister-chromatid separation, as the cells lacking securin are more or less healthy, meaning other mechanisms also control the separation. Cdc20 appears to be necessary for anaphase beyond sister-chromatid separation, since the double knockouts still arrest in anaphase despite lacking securin.

  1. Answer: FFFF

Difficulty: 1

Section: Cytokinesis

Feedback: The contractile ring is based on actin and myosin II, constricts in a dynamic fashion, and gradually disassembles. RhoA GTPase helps assemble this structure.

  1. Answer: A

Difficulty: 1

Section: Cytokinesis

Feedback: This G protein activates formin as well as multiple protein kinases, including the Rho-activated kinase Rock. Rock phosphorylates regulatory myosin light chains and also inhibits myosin light-chain phosphatase by phosphorylation.

  1. Answer: 1

Difficulty: 2

Section: Cytokinesis

Feedback: The astral stimulation model (1) predicts this outcome. The central spindle stimulation model (2) predicts the formation of two furrows when there are two spindles. However, it is also likely that a midzone-like structure formed between the two spindles in these experiments, and so model 2 could also be correct here. The astral relaxation model (3) predicts the formation of two furrows in the first division.

  1. Answer: 3

Difficulty: 2

Section: Cytokinesis

Feedback: The astral relaxation model (3) cannot explain these results.

  1. Answer: 3

Difficulty: 2

Section: Cytokinesis

Feedback: These observations are in better agreement with the astral relaxation model (3).

  1. Answer: TTTF

Difficulty: 2

Section: Mitosis

Feedback: In plant cells, the preprophase band assembles just before mitosis and marks the site of cytokinesis. The assembly of the cell plate between the two daughter cells is then guided by phragmoplast. The membrane required for the formation of cell plate comes from the Golgi apparatus.

  1. Answer: B

Difficulty: 1

Section: Mitosis

Feedback: Cdh1–APC/C activity increases in late mitosis and keeps the M-Cdk activity low via M-cyclin degradation during G1.

  1. Answer: C

Difficulty: 1

Section: Cytokinesis

Feedback: Cellularization is the coordinated cytokinesis that occurs after 13 mitoses in early fruit-fly development.

  1. Answer: AIII

Difficulty: 2

Section: Mitosis

Feedback:  Both Cdc20–APC/C and Cdh1–APC/C target M-cyclins for degradation. M-Cdk, in turn, activates the former and inactivates the latter. Sic1 in budding yeast is an example of a CKI that also inhibits M-Cdks. M-Cdk activates the expression of M-cyclin genes in a positive feedback loop, although its activity is also regulated by negative feedback, for example through Cdc20–APC/C mentioned above.

  1. Answer: C

Difficulty: 2

Section: Mitosis

Feedback: Although regulated differently, both APC/C complexes target M-cyclins for degradation.

  1. Answer: D

Difficulty: 2

Section: Meiosis

Feedback: Meiosis II resembles mitosis, in which sister chromatids segregate during anaphase.

  1. Answer: A

Difficulty: 1

Section: Meiosis

Feedback: Prophase I is by far the longest stage in meiosis.

  1. Answer: FTTF

Difficulty: 2

Section: Meiosis

Feedback: Bivalents form and recombination begins during leptotene in prophase I. The synaptonemal complex is formed during zygotene and pachytene, and disassembled in diplotene, which is when chiasmata become visible. Only some recombination events lead to crossovers; that is, reciprocal exchange of DNA.

  1. Answer: NNEN

Difficulty: 2

Section: Meiosis

Feedback: There are two chiasmata in this drawing, both corresponding to exchange between chromatids 2 and 3.

  1. Answer: C

Difficulty: 1

Section: Meiosis

Feedback: In the absence of shugoshin, cohesin complexes at the centromeric regions are no longer protected from phosphorylation. The phosphorylated cohesin subunits are then cleaved by separase in anaphase I. Consequently, the sister chromatids segregate randomly in anaphase II.

  1. Answer: MSMG

Difficulty: 1

Section: Control of Cell Division and Cell Growth

Feedback: Mitogens stimulate cell division by promoting cell cycle progression. Survival factors promote cell survival by suppressing apoptosis. Growth factors stimulate increase in cell mass by promoting anabolic metabolism.

  1. Answer: B

Difficulty: 1

Section: Control of Cell Division and Cell Growth

Feedback: When blood clots, the incorporated platelets release the contents of their secretory vesicles, containing platelet-derived growth factor (PDGF) among other proteins.

  1. Answer: B

Difficulty: 1

Section: Control of Cell Division and Cell Growth

Feedback: Transforming growth factor-β (TGFβ) and its relatives are amongst the best-understood inhibitory signal proteins.

  1. Answer: D

Difficulty: 1

Section: Control of Cell Division and Cell Growth

Feedback: Since they spend less time than the other cells in G1, fibroblasts have a high mitotic index. Note that not all cells with a rapid turnover have a high mitotic index.

  1. Answer: C

Difficulty: 1

Section: Control of Cell Division and Cell Growth

Feedback: Once produced, Myc activates the transcription of delayed-response genes such as G1-cyclins. These in turn activate transcription regulators of the E2F family.

  1. Answer: C

Difficulty: 1

Section: Control of Cell Division and Cell Growth

Feedback: Once the activity of Cdh1–APC/C falls in late G1, geminin starts to accumulate.

  1. Answer: E

Difficulty: 1

Section: Control of Cell Division and Cell Growth

Feedback: All of these processes ensure a sharp transition from G1 to S phase, as well as sustained S-Cdk activity afterward.

  1. Answer: BC

Difficulty: 1

Section: Control of Cell Division and Cell Growth

Feedback: These two transitions are the major time points at which DNA damage is detected and cell-cycle arrest occurs, although the cell can sense DNA damage and arrest the cell cycle at other points throughout the cycle, including in S phase.

  1. Answer: D

Difficulty: 2

Section: Control of Cell Division and Cell Growth

Feedback: Activation of the DNA damage response involves the two upstream kinases ATM and ATR, which phosphorylate and activate the Chk1 and Chk2 kinases. Cdc25 phosphatases can be inhibited by phosphorylation in this pathway, whereas p53 phosphorylation stabilizes the protein by reducing its interaction with its inhibitor Mdm2.

  1. Answer: YNYY

Difficulty: 2

Section: Control of Cell Division and Cell Growth

Feedback: Loss-of-function mutations in p53, Rb, and p21 are found in human cancers. Myc, on the other hand, has the opposite effect: gain-of-function mutations in Myc can drive cancer.

  1. Answer: FFFF

Difficulty: 2

Section: Control of Cell Division and Cell Growth

Feedback: Replicative cell senescence is a result of the shortening of telomeres in somatic cells that lack telomerase activity. Eventually, the unprotected chromosome ends trigger cell-cycle arrest or apoptosis through p53 activation. Excessive stimulation by mitogens can also activate p53, since the cell-cycle inhibitor Arf is activated under excessive Myc or Ras activity and inhibits the p53 inhibitor Mdm2.

  1. Answer: D

Difficulty: 3

Section: Control of Cell Division and Cell Growth

Feedback: Rapamycin inhibits the mammalian TOR complex that is activated in response to growth factors in a PI 3-kinase-dependent manner. As a result, cell growth is retarded in a number of ways, including by inhibition of TOR-mediated activation of S6 kinase.

 

 

 

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