Viruses, Transposable Elements, and Cancer Practice Test

Bacteriophages Questions

1. Six deletion mutants within the A gene of the rII region of phage T4 were tested in all pairwise combinations for wild-type recombinants. In the following table, + = recombination, 0 = no recombination. Construct a topological map for these deletions. (Hint: see Solved Problem 11.3.)



2. Phage MS2 is a single-stranded RNA virus of E. coli. After infecting a cell, the phage RNA (the "plus" strand) is made into a double-stranded replicative intermediate form ("plus-minus") from which "plus" RNA is synthesized. The "minus" strands when isolated are not infective. Phage X174 is a single-stranded DNA virus of E. coli. When injected into a bacterium, the same events as described for MS2 occur, but the "minus" strands when isolated are infective. Devise a reasonable hypothesis to account for these observations.
3. The DNA of bacteriophage T4 contains approximately 200,000 nucleotide pairs. The rII region of the T4 genome occupies about 1%of its total genetic length. Benzer has found that about 300 sites are separable by recombination within the rII region. Determine the average number of nucleotides in each recon.
4. The molecular weight of DNA in phage T4 is estimated to be 160 × 10⁶. The average molecular weight of each nucleotide is approximately 400. The total genetic map of T4 is calculated to be approximately 2500 recombination units long. With what frequency are r⁺ recombinants expected to be formed when two different r mutants (with mutations at adjacent nucleotides) are crossed?
5. A number of mutations were found in the rII region of phage T4. From the recombination data shown in the table below, determine whether each mutant is a point defect or a deletion (+ = recombination, 0 = no recombination). Two of the four mutants have been known to undergo backmutation; the other two have never been observed to backmutate.



6. Escherichia coli strain B is doubly infected with two rII mutants of phage T4. A 6 × 10⁷ dilution of the lysate is plated on E. coli B. A 2 × 10⁵ dilution is plated on strain K. Twelve plaques appeared on strain K, 16 on strain B. Calculate the amount of recombination between these two mutants.
7. Anonlytic response usually is observed in lysogenic (λ) E. coli cells when conjugated with nonlysogenic Hfr donors or in crosses of Hfr (λ) × F^– (λ). The donated prophage is almost never inherited by the recombinants. Lysis is very anomalous in crosses of Hfr (λ) × F^–. Explain these observations.
8. Temperate phages such as lambda sometimes produce turbid plaques on lambda-sensitive indicator cells; virulent phages that cannot lysogenize always produce clear plaques on cells of their host range.     (a)  Offer an explanation for the turbid plaques.     (b)  Some lambda mutants produce only clear plaques. What genetic locus is most likely mutant in these cases?
9. When bacterial DNA is damaged by a mutagenic agent, excision repair normally operates to repair the lesion. This process is less than 100% efficient, however, so that some residual lesions remain unrepaired. If these lesions delay replication of DNA, an error-prone "SOS repair" system becomes operative, involving activation and increased production of a multifunctional protein called RecA protein (for "recombination"). RecA protein interferes with cell partition, resulting in elongation of cells into filaments. RecA protein also cleaves lambda repressor; this repressor must remain intact for the virus to remain dormant as a prophage. E. coli strain B is lysogenic for lambda; strain A is not lysogenic for lambda. This knowledge led Moreau, Bailone, and Devoret to devise a "prophage induction test" or "inductest" for potential carcinogens. Lysogenic strain B of E. coli is made defective in its excision repair system and genetically modified to make the cell envelopes permeable to a wide variety of test chemicals. This special strain is mixed with indicator strain A and rat liver extract; the mixture is then plated; the medium is covered with a thin layer of indicator bacteria interspersed with a few lysogenic bacteria. The test chemical is applied to a filter paper disk and placed in the center of the plate for a "spot test."     (a)  After incubation, how is DNA damage assayed?     (b)  Why is strain A required as an indicator?     (c)  What advantage does an inductest have over an Ames test?     (d)  Explain the selective advantage of lysogenic induction.     (e)  Genetic engineers have spliced the gene for galacto-kinase into a bacterial chromosome, thereby creating an organism for assaying mutagens by an enzymatic activity test. Where was this gene inserted into the chromosome and how does the system work?
10. The single-stranded phage X174 of E. coli contains 5386 nucleotides coding for 11 proteins with a combined molecular weight of 262,000 bp.     (a)  If an average amino acid has a molecular weight of 110, by how many amino acids is the coding capacity of the phage exceeded?     (b)  How can X174 code for more proteins than it has coding triplets?     (c)  Several animal viruses make more proteins than for which they seem to have coding triplets. Suggest some ways by which they might accomplish this feat if a single reading frame is used.

Transposable Elements Questions

11. A given transposable element becomes duplicated at a fairly constant (although usually low) rate. Therefore, over evolutionary time, the descendants of a bacterial cell might be expected to contain thousands of copies of such a transposon. However, the number of copies of bacterial transposons is very low(usually only one or two per cell).     (a)  Offer an explanation for this low copy number.     (b)  Why have most bacterial transposons been isolated from plasmids rather than from the bacterial chromosome?
12. Transposition of a particular transposable element is found to be dependent on reverse transcriptase activity. Propose a mechanism for its transposition.
13. How might a transposition event result in oncogenesis?

Eukaryotic Viruses Questions

14. Give at least two mechanisms whereby RNA viruses produce mRNA.
15. With regard to retroviruses:     (a)  specify their defining characteristic,     (b)  name the enzyme contained in their virions and list three biochemical activities of that enzyme,     (c)  identify the template for synthesis of retroviral mRNA,     (d)  identify the cellular location of their replication,     (e)  specify those attributes suggesting that their DNA-insertion mechanism is related to transposition.
16. The life cycles of eukaryotic viruses and bacteriophages have many similarities, including the establishment of new replication and transcription systems, regulation of gene action (e.g., early vs. late transcription), and synthesis of large quantities of structural proteins. There are certain aspects of viral life cycles, however, that are not (or only rarely) found in the life cycles of phage. Specify some of these unique aspects.

Cancer Questions

17. Many tests of the oncogenic potential of viruses have been made with an established mouse cell line called NIH 3T3 because it had been used for many years to study viral transformation and chemical carcinogens. If cancer is a multistep process, how can the introduction of a single active viral oncogene transform these cells into cancerous cells?
18. Some slow-transforming viruses (such as avian leukemia virus) do not contain oncogenes. Offer an explanation as to how they might transform cells.
19. Cellular protooncogenes usually contain introns; viral oncogenes do not.     (a)  Propose a scenario for the origin of a viral oncogene.     (b)  Why is it more probable that oncogenes originate in cells rather than in viruses?
20. A virus is suspected to be involved in the development of breast cancer in certain strains of mice. The virus is transmitted through the milk to the offspring. In crosses where the female carries the "milk factor" and the male is from a strain that is free of the factor, about 90% of the female progeny develop breast cancer prior to 18 months of age. The virus usually does not initiate cancer development in the infected mouse until she enters the nursing stage, and then only in conjunction with a hormone (estrone) from the ovaries. Males from a virus- infected strain are crossed with females from a virus-free strain.     (a)  Predict the proportion of the offspring from this cross that, if individually isolated from weaning to 18 months of age, will probably exhibit breast cancer.     (b)  Predict the proportion of offspring from this cross that will probably exhibit breast cancer if housed in a group from weaning to 18 months of age.     (c)  Answer part (a) when the reciprocal is made.     (d)  Answer part (b) when the reciprocal cross is made.
21. Another case in which a disease is acquired through the milk (see Problem 11.25) is hemolytic anemia in newborn horses. A mare may produce two or three normal offspring by the same stallion and the next foal may develop severe jaundice within about 96 h after birth and die. Subsequent matings to the same stallion often produces the same effect. Subsequent matings to another stallion could produce normal offspring. It has been found that if nursed for the first few days on a foster mother, the foal will not become ill and develops normally. Evidently, something in the early milk (colostrum) is responsible for this syndrome. If the foal should become ill, and subsequently recovers, the incompatibility is not transmitted to later generations.     (a)  How might this disease be generated?     (b)  How is the acquisition of this disease different from that of breast cancer in mice?

Answers

Vocabulary

1. capsid
2. bacteriophage or phage
3. host restriction
4. lytic
5. prophage
6. envelope
7. persistent
8. retroviruses
9. transposons
10. oncogenes

Multiple Choice

1. e (b, c)
2. e
3. b
4. a
5. d
6. c
7. c
8. a
9. e
10. e

Bacteriophages

1.  



2. The DNA "minus" strand can serve as a template and can utilize the bacterial enzyme DNA polymerase for replication. The "minus" strand of RNA does not code for the enzyme that replicates RNA (RNA synthetase), and this enzyme is absent in uninfected bacteria. The "plus" strand of RNA carries the coded instructions for this enzyme and acts first as mRNA for enzyme synthesis. In the presence of the enzyme, single-stranded RNA can form a complementary strand and becomes a double-helical replicative form.
3. Approximately 7 nucleotides per recon.
4. 0.00625% of all progeny are expected to be r⁺ recombinants.
5. m₁ andm₂ are overlapping deletions; m₃ is within m₁; m₄ is within m₂.
6. 0.5% recombination
7. Repressor is already present in the cytoplasm of the F^– (λ) recipient cell. It binds to the operators that prevent prophage induction (vegetative reproduction of the phage) in either the F^– (λ) or the Hfr (λ) donated chromosome segment. Early enzymes are therefore not produced, and recombination (leading to the inheritance of the donated lambda genes) cannot occur. Nonlysogenic F^– cells do not contain repressor. There are so few repressor molecules in a lysogenized cell that it is unlikely that free repressor would be bound to the newly synthesized donor fragment that moves almost immediately through the pilus into the F^– recipient. When the prophage from Hfr (λ) enters the F^– cell, a race occurs between the production of lambda repressor and an early protein of vegetative phage development. The outcome of this race is not predictable; hence, lysis is unpredictable in such crosses.
8. (a)  Turbid plaques are due to secondary growth of lysogenized bacteria derived from the lambda-sensitive indicator strain.     (b)  Mutations of the gene coding for the lambda repressor of lytic activity are likely to produce an inactive (nonfunctional) repressor; hence, these mutants cannot lysogenize the indicator strain.
9. (a)  DNA damage activates RecA protein, which then cleaves lambda repressor and opens up the viral genome for replication (induction). The cell bursts and releases viruses that infect and lyse the indicator strain A, causing plaques (holes) to appear in the bacterial "lawn" surrounding the paper disk. (b) If a cell of strain B is induced to lysis, the viruses cannot multiply in other cells of the same strain because active lambda repressor is present in these cells as a product of their prophages. Therefore, a nonlysogenic strain (A) is required to indicate how many viruses have been induced by the chemical treatment. (c) The inductest can assay a potential carcinogen at doses that would kill the tester bacteria in an Ames test (giving a false-negative reaction). The Ames test only detects the rare backmutations of his^– to his⁺ , whereas DNA damage at any site can initiate lysogenic induction (a mass effect, independent of cell survival by toxic chemicals).     (d)  If DNA of the host cell cannot replicate, the cell is likely to die. Under these conditions it would be advantageous for a prophage to enter the lytic cycle and thereby possibly infect a "healthier" cell (like a "rat leaving a sinking ship").     (e)  The gene for galactokinase was inserted adjacent to (and under the control of) the lambda repressor. When the mutagen damages DNA, RecA protein is activated and cleaves the repressor; this opens the operon to RNA polymerase and allows synthesis of the enzyme galactokinase, the activity of which can be quantitated spectrophotometrically when supplied with its substrate.
10. (a)  (262,000 molecular weight/110 molecular weight per amino acid) – (5386 bases/3 bases per codon) = 586.5 amino acids     (b)  Some phage genes are overlapping; i.e., the same sequence can be transcribed in different reading frames. The only structural feature responsible for gene overlap is the location of each AUG start codon     (c)  By alternative intron cleavage sites from the same primary transcript, two proteins could be produced that have the same N terminus but different C termini. A polyprotein could also be enzymatically cleaved in more than one way to produce different products.

Transposable Elements

11. (a)  Most of the DNA in bacteria, unlike the DNA in eukaryotic cells, is coding information. There is relatively little DNA that is not serving some function. Thus, the movement of most transposons to a new location would inactivate one or more vital genes, causing cell death or weakening the cell so that it cannot compete with normal cells.     (b)  Plasmids rarely are essential to their host cells, and therefore could tolerate the integration of transposable elements without interfering with vital gene functions.
12. The transposable element must first be copied into RNA. This RNA is then copied by reverse transcriptase into cDNA. This double-stranded cDNA is then capable of integrating into the genome at a different site.
13. Transposition may result in several types of insertional and recombinational mutation events: insertion in a coding region of a protooncogene, resulting in loss of function; insertion into the promoter of a protooncogene, resulting in loss of function; insertion near the controlling regions of a gene and acting as a fortuitous promoter, resulting in altered gene regulation; various types of recombination events that would result in translocations or deletions, upon "jumping out" a transposon may take nearby gene sequences with it, creating a deletion.

Eukaryotic Viruses

14. (1)  Minus-strand RNA viruses transport into the cell a replicase enzyme that synthesizes mRNA from the (–) strand template.     (2)  Plus-strand RNA viruses (other than retroviruses) use their infective strand as a template for synthesizing mRNA using host RNA polymerase.     (3)  Retroviruses use their (+) strand as a template for DNA synthesis, which is then transcribed into mRNA.     (4)  Double-stranded RNA viruses bring a replicase into the host cell that copies double-stranded RNA and synthesizes a (+) strand that functions as mRNA.
15. (a)  They replicate from a DNA intermediate.     (b)  RNA-dependent DNA polymerase (reverse transcriptase). Enzyme functions: (1) converting the single-stranded viral RNA to a DNA-RNA hybrid, (2) digesting RNA from a DNA-RNA hybrid, and (3) copying a primed single-stranded DNA to form a double-stranded DNA.     (c)  The double-stranded DNA that is formed by reverse transcription from retroviral RNA.     (d)  The DNA-RNA hybrid is made in the cytoplasm. The hybrid is converted to double-stranded DNA and becomes inserted into a host chromosome. Messenger RNA is made from the proviral DNA in the nucleus by host RNA polymerase.     (e)  Their integrated proviral DNA is terminated at each end by a long terminal repeat and a short inverted repeat (like a composite transposon), which in turn is flanked by a short, direct repeat (like a target sequence).
16. (1) Viral proteins usually enter the infected cell along with the viral genome.     (2)  The RNA of some viruses is converted to DNA.     (3)  The mRNA of viruses is processed just like the cellular mRNA of their eukaryotic hosts.     (4)  Polyproteins are produced by some viruses.

Cancer

17. An established cell line has already experienced one or more early steps (e.g., immortalization) in the induction of neoplastic transformation before exposure to the effects of the oncogene.
18. These viruses might become integrated into the host DNA near a cellular protooncogene and activate it, via a viral enhancer sequence, to become an oncogene.
19. (a)  A retrovirus becomes integrated as a provirus adjacent to a cellular protooncogene. The provirus and the adjacent protooncogene are transcribed into a single transcript. The RNA transcript is processed to remove introns and becomes packaged into a viral capsid. The infective virus is released from the host cell and infects another cell.     (b)  Each of the steps in part (a) involves known genetic mechanisms. There is no known mechanism by which introns from viruses can be inserted into cellular protooncogenes.
20. (a) (b)  None of the progeny is expected to develop breast cancer because non-infected females have nursed them.     (c)  None of the progeny is expected to develop breast cancer because, in isolation, the infected females could never produce a litter and subsequently enter a lactation period, a prerequisite for expression of the milk factor.     (d)  50% females × 90%of females develop breast cancer = approximately 45%.
21. (a)  This disease is similar to the Rh blood group system incompatibility between a human mother and her baby. In this case, antibodies are transferred to the offspring through the milk rather than across the placenta.     (b)  The particular stallion that is used has an immediate effect on the character. This is not true in the acquisition of breast cancer in mice. The incompatibility disease in horses cannot be transmitted to later generations, so there is no evidence of a specifically self-duplicating particle like the infective agent that causes cancer in mice.