- All of the following are true about the measles virus EXCEPT:
a. | It is encased by an envelope of membrane. |
b. | It fuses with the host cell membrane during infection. |
c. | It has progeny that bud out of the host cell. |
d. | It generates a rash of red spots on the skin of infected patients. |
e. | It is never fatal. |
ANS: E DIF: Easy REF: 6.1 TOP: I.B
MSC: Remembering
- The destruction of cells as a result of viral lysis can be observed as a:
a. | tissue culture | d. | plaque |
b. | lawn | e. | host |
c. | soft agar | |
ANS: D DIF: Easy REF: 6.1 TOP: I.B.iii
MSC: Remembering
- __________ can be counted as representing individual infectious virions from a phage suspension.
a. | Plaques | d. | Proteomes |
b. | Viruses | e. | Burst size |
c. | Genomes | |
ANS: A DIF: Easy REF: 6.1 TOP: I.B.iii
MSC: Remembering
- Which of the following is INCORRECT about the giant mimivirus?
a. | It may cause human pneumonia. |
b. | It is as large as some bacteria. |
c. | It conducts DNA repair. |
d. | It conducts protein folding by chaperones. |
e. | It contains reverse transcriptase. |
ANS: E DIF: Medium REF: 6.1 TOP: I.C
MSC: Remembering
- Infectious agents for which the nucleic acid genome is itself the entire infectious particle are called:
a. | prions | d. | bacteria |
b. | viroids | e. | virions |
c. | viruses | |
ANS: B DIF: Easy REF: 6.1 TOP: I.D.i
MSC: Remembering
- Spontaneous __________ diseases can be inherited genetically.
a. | prion | d. | virion |
b. | viroid | e. | bacterial |
c. | virus | |
ANS: A DIF: Easy REF: 6.1 TOP: I.E
MSC: Remembering
- __________ gained notoriety when they were implicated in brain infections, such as what is popularly known as mad cow disease.
a. | Viroids | d. | Bacteria |
b. | Viruses | e. | Virions |
c. | Prions | |
ANS: C DIF: Easy REF: 6.1 TOP: I.E
MSC: Remembering
- An open reading frame corresponds to a series of codons that are translated to amino acids, producing a(n):
a. | RNA | d. | capsid |
b. | protein | e. | envelope |
c. | DNA | |
ANS: B DIF: Easy REF: 6.1 TOP: I.C
MSC: Understanding
- The RNA genome of the potato spindle tuber viroid requires which type of modified host polymerase for replication?
a. | DNA-dependent DNA polymerase | d. | RNA-dependent RNA polymerase |
b. | DNA-dependent RNA polymerase | e. | reverse transcriptase |
c. | RNA-dependent DNA polymerase | |
ANS: D DIF: Medium REF: 6.1 TOP: I.D
MSC: Understanding
- __________ diseases are unique in that they can be transmitted by an infective protein that propagates conformational change of existing molecules without synthesizing entirely new infective molecules.
a. | Viral | d. | Virion |
b. | Viroid | e. | Prion |
c. | Bacterial | |
ANS: E DIF: Easy REF: 6.1 TOP: I.E
MSC: Understanding
- An icosahedral capsid has how many sides?
ANS: E DIF: Easy REF: 6.2 TOP: II.A.i
MSC: Remembering
- Icosahedral viral capsids take the form of a polyhedron with 20 identical __________ faces.
a. | helical | d. | triangular |
b. | filamentous | e. | asymmetrical |
c. | rectangular | |
ANS: D DIF: Easy REF: 6.2 TOP: II.A.i.a
MSC: Remembering
- Virus particles can be observed by standard transmission electron microscopy (TEM), but details of capsid structure require visualization by:
a. | phase-contrast microscopy | d. | negative stain |
b. | scanning electron microscopy | e. | bright field microscopy |
c. | cryo-electron microscopy | |
ANS: C DIF: Medium REF: 6.2 TOP: II.A.i.b
MSC: Remembering
- Some viral species may derive their __________ from intracellular membranes, such as the nuclear membrane or endoplasmic reticulum.
a. | capsid | d. | neck |
b. | genome | e. | tail fibers |
c. | envelope | |
ANS: C DIF: Medium REF: 6.2 TOP: II.A.i.c
MSC: Remembering
- Tegument proteins may be found:
a. | in the capsid | d. | in the nucleus |
b. | between the capsid and envelope | e. | in the spikes |
c. | in the envelope | |
ANS: B DIF: Easy REF: 6.2 TOP: II.A.i.e
MSC: Remembering
- __________ phages have been used to nucleate the growth of crystalline “nanowires” for electronic devices.
a. | Icosahedral | d. | Filamentous |
b. | Complex | e. | Viroid |
c. | Asymmetrical | |
ANS: D DIF: Easy REF: 6.2 TOP: II.A.ii
MSC: Remembering
- In a filamentous virus, the pattern of capsid monomers forms a __________ tube around the genome.
a. | icosahedral | d. | complex |
b. | filamentous | e. | helical |
c. | asymmetrical | |
ANS: E DIF: Easy REF: 6.2 TOP: II.A.ii.a
MSC: Remembering
- In bacteriophage T4, the DNA genome is contained in the head, and binding to the host is facilitated by attachment of the:
a. | capsid | d. | envelope |
b. | neck | e. | tegument |
c. | tail fibers | |
ANS: C DIF: Easy REF: 6.2 TOP: II.B.i
MSC: Remembering
- Large __________ viruses contain so many enzymes that they appear to have evolved from degenerate cells.
a. | icosahedral | d. | bacteriophage |
b. | filamentous | e. | asymmetrical |
c. | complex | |
ANS: E DIF: Medium REF: 6.2 TOP: II.B.ii
MSC: Remembering
- Genes for the capsid proteins are found in the:
a. | plasmid | d. | ribosome |
b. | bacterial chromosome | e. | viral genome |
c. | envelope | |
ANS: E DIF: Medium REF: 6.2 TOP: II.A.i.a
MSC: Understanding
- If something descended from a common ancestor, it is:
a. | monomeric | d. | monosyllabic |
b. | monophyletic | e. | monolithic |
c. | monogrammed | |
ANS: B DIF: Medium REF: 6.3 TOP: III
MSC: Remembering
- __________ of viruses are designated with the suffix viridae.
a. | Genera | d. | Orders |
b. | Families | e. | Classes |
c. | Species | |
ANS: B DIF: Medium REF: 6.3 TOP: III.A
MSC: Remembering
- Viruses may be classified by the __________ method.
a. | Annapolis | d. | Augusta |
b. | Baltimore | e. | Portland |
c. | Orono | |
ANS: B DIF: Easy REF: 6.3 TOP: III.B
MSC: Remembering
- So far, the known mechanisms of replication and mRNA expression define how many fundamental groups of virus species?
ANS: C DIF: Medium REF: 6.3 TOP: III.B
MSC: Remembering
- David Baltimore proposed that the primary distinction among classes of viruses was the __________ composition and the route used to express messenger RNA.
a. | genome | d. | tegument |
b. | envelope | e. | neck |
c. | capsid | |
ANS: A DIF: Easy REF: 6.3 TOP: III.B.i
MSC: Remembering
- The genome of __________ ssRNA viruses can serve directly as mRNA.
a. | positive-sense (+) | d. | double-stranded |
b. | negative-sense (–) | e. | prion |
c. | all | |
ANS: A DIF: Easy REF: 6.3 TOP: III.B.i.d
MSC: Remembering
- A key factor in the evolution of killer strains of influenza is that they:
a. | only have eight genes | d. | have a segmented genome |
b. | have an RNA genome | e. | have a circular chromosome |
c. | contain reverse transcriptase | |
ANS: D DIF: Medium REF: 6.3 TOP: III.B.i.e
MSC: Remembering
- Which of the following is NOT true of the pararetroviruses?
a. | They have an RNA genome. |
b. | They do not make a DNA intermediate. |
c. | Some have a viral reverse transcriptase. |
d. | Some use a host reverse transcriptase. |
e. | They consist of human and plant pathogens. |
ANS: A DIF: Medium REF: 6.3 TOP: III.B.i.g
MSC: Remembering
- __________ are genes of common ancestry in two genomes that share the same function.
a. | Homologs | d. | Orthologs |
b. | Open reading frames | e. | Lysogenic |
c. | Proteomes | |
ANS: D DIF: Difficult REF: 6.3 TOP: III.C.i.a
MSC: Remembering
- The proteomic tree of bacteriophages is based on comprehensive analysis of viral:
a. | DNAs | d. | capsids |
b. | RNAs | e. | host range |
c. | proteins | |
ANS: C DIF: Easy REF: 6.3 TOP: III.C.ii
MSC: Remembering
- Proteomic analysis predicts seven major evolutionary categories of phage species with subgroups based on:
a. | type of genetic material | d. | transcriptional strategy |
b. | virion structure | e. | replicase gene sequence |
c. | shared hosts | |
ANS: C DIF: Medium REF: 6.3 TOP: III.C.ii
MSC: Understanding
- After the genome is inserted, the phage capsid remains outside, attached to the cell surface, and is referred to as:
a. | an envelope | d. | a ghost |
b. | a neck | e. | temperate |
c. | tail fibers | |
ANS: D DIF: Easy REF: 6.4 TOP: IV.B
MSC: Remembering
- Which of the following steps is NOT part of the life cycle of a lytic phage?
a. | Phage DNA is injected into the bacterial cell. |
b. | The phage DNA integrates into the bacterial chromosome. |
c. | Many copies of phage DNA are made. |
d. | The phage DNA is transcribed, and the resulting mRNA is translated to make capsid proteins. |
e. | All of the above steps are part of the life cycle of a lytic phage. |
ANS: B DIF: Difficult REF: 6.4 TOP: IV.C.i
MSC: Remembering
- The number of virus particles released at lysis is referred to as:
a. | lytic number | d. | temperate number |
b. | burst size | e. | release number |
c. | lysogenic number | |
ANS: B DIF: Easy REF: 6.4 TOP: IV.C.ii.b
MSC: Remembering
- A(n) __________ phage may integrate its genome into that of the host cell.
a. | lytic | d. | asymmetrical |
b. | temperate | e. | oncogenic |
c. | viroid | |
ANS: B DIF: Easy REF: 6.4 TOP: IV.C.iii.a
MSC: Remembering
- The integrated phage genome is called a(n):
a. | temperate | d. | lytic |
b. | lysogen | e. | prophage |
c. | oncogene | |
ANS: E DIF: Easy REF: 6.4 TOP: IV.C.iii.b
MSC: Remembering
- A spacer is a(n):
a. | bacterial gene | d. | inactive viral RNA |
b. | bacterial enzyme | e. | cascade protein |
c. | piece of phage DNA | |
ANS: C DIF: Easy REF: 6.4 TOP: IV.D.iii
MSC: Remembering
- What happens to the virus protein coat when a bacterial cell is infected?
a. | It enters the host cell with the viral genome. |
b. | It remains on the outside of the host cell. |
c. | It enters the host cell separately from the viral genome. |
d. | It is released to attach to and inject another host cell. |
e. | It becomes part of the host cell membrane. |
ANS: B DIF: Easy REF: 6.4 TOP: IV.B
MSC: Understanding
- Which one of the viral gene products is likely to be expressed early in bacteriophage T4 life cycle to facilitate the replication of viral genome?
a. | a DNase that digests host genome |
b. | an RNase that digests viral genome |
c. | a protease that digests viral capsids |
d. | the capsids required to produce mature viral particles |
e. | the ribosomes required for translation of viral mRNA |
ANS: A DIF: Difficult REF: 6.4 TOP: IV.C.i
MSC: Understanding
- Which of the following statements regarding bacterial defenses against phage infections is FALSE?
a. | Resistance to phages can occur through mutations. |
b. | Bacterial restriction endonucleases can cleave viral RNA genome. |
c. | The clustered, regularly interspaced short palindromic repeats inserted in the bacterial genome facilitate cleavage of specific phage DNA. |
d. | Bacterial restriction endonuclease system cleaves DNA lacking appropriate methylation patterns. |
e. | Resistance to phages can occur through altered host receptor proteins. |
ANS: B DIF: Medium REF: 6.4 TOP: IV.D.i | IV.D.ii | IV.D.iii
MSC: Understanding
- The papillomavirus is an example of:
a. | latent infection | d. | transformation |
b. | lytic infection | e. | acute infection |
c. | persistent infection | |
ANS: A DIF: Difficult REF: 6.4 | 6.5 TOP: IV.C.iii | V.B.i.a
MSC: Understanding
- Within a host, receptor molecules can also determine the __________, or tendency to infect a particular tissue type.
a. | tropism | d. | burst size |
b. | host range | e. | plaque |
c. | virulence | |
ANS: A DIF: Easy REF: 6.5 TOP: V.A.i
MSC: Remembering
- The primary factor determining the life cycle of an animal virus is the physical form of the:
a. | envelope | d. | receptor |
b. | capsid | e. | burst |
c. | genome | |
ANS: C DIF: Easy REF: 6.5 TOP: V.B
MSC: Remembering
- The flu is an example of:
a. | latent infection | d. | transformation |
b. | lytic infection | e. | oncogenic infection |
c. | persistent infection | |
ANS: B DIF: Easy REF: 6.5 TOP: V.B
MSC: Remembering
- Which of the following is the most common sexually transmitted disease in the United States, and one of the most common worldwide?
a. | human papillomavirus (HPV) | d. | syphilis |
b. | human immunodeficiency virus (HIV) | e. | chlamydia |
c. | gonorrhea | |
ANS: A DIF: Medium REF: 6.5 TOP: V.B.i
MSC: Remembering
- In contrast to animal viruses and bacteriophages, plant viruses infect cells by mechanisms that do NOT involve specific:
a. | receptors | d. | vectors |
b. | membranes | e. | hosts |
c. | envelopes | |
ANS: A DIF: Medium REF: 6.5 TOP: V.C.i.a
MSC: Remembering
- Which of the following is FALSE concerning HIV?
a. | It is the causative agent of AIDS. |
b. | It makes a DNA copy of its RNA genome. |
c. | Reverse transcriptase is translated from an early gene. |
d. | The virion contains two copies of the HIV genome. |
e. | The viral RNA is copied into double-stranded DNA. |
ANS: C DIF: Medium REF: 6.5 TOP: V.B.iii
MSC: Understanding
- The period of time when virions are virtually undetectable inside the infected cell is the:
a. | eclipse period | d. | burst period |
b. | latent period | e. | lag period |
c. | rise period | |
ANS: A DIF: Easy REF: 6.6 TOP: VI.A.iii
MSC: Remembering
- HIV is classified in a category of slowly replicating retroviruses termed:
a. | picornaviruses | d. | lentiviruses |
b. | papillomaviruses | e. | pararetroviruses |
c. | oncogenic viruses | |
ANS: D DIF: Difficult REF: 6.6 TOP: VI.B
MSC: Remembering
- Hosts may benefit from infection by persistent viruses because:
a. | hosts never come down with illness when infected by persistent viruses |
b. | persistent viruses could encode growth factors beneficial to infected hosts |
c. | persistent viruses shorten hosts’ generation time, resulting in hosts that outcompete their competitors |
d. | persistent viruses could encode toxins that kill off competing bacteria |
e. | persistent viruses could improve hosts’ efficiency in uptake of nutrients |
ANS: D DIF: Difficult REF: 6.7 TOP: VII.B.iii
MSC: Understanding
SHORT ANSWER
- Why are viruses not affected by antibiotics? Why are there so few antiviral drugs?
ANS:
Viruses are not cellular entities. Common antibiotics disrupt cell wall synthesis or interact with ribosomes and disrupt the translation process. Viruses do not have either cell walls or ribosomes, so these antibiotics would be ineffective.
Viruses that infect humans commandeer the host cell to make more viruses. One cannot target the host machinery because then the host would be harmed. There are few viral targets that can be attacked without harming the host cell.
DIF: Difficult REF: 6.1 TOP: I.B.vi MSC: Understanding
- What are prions and how do they cause disease?
ANS:
Prions are aberrant forms of naturally occurring proteins that assume an abnormal conformation. They cause disease by binding to normally folded proteins of the same type and altering their conformation. Harmful aggregates of these proteins cause cell death.
DIF: Medium REF: 6.1 TOP: I.E.i MSC: Understanding
- What is the advantage of symmetry in viral particles?
ANS:
Symmetry provides a way to form a package out of repeating protein units. This requires a small number of genes, since the same protein unit is repeated. Viral genomes are generally very small, so they have few genes. They can also devote more resources to copying their genome for viral replication if it takes few resources to make the capsid proteins.
DIF: Medium REF: 6.2 TOP: II.A MSC: Understanding
- What is the viral envelope composed of? How does it differ from the bacterial cell envelope?
ANS:
The viral envelope is composed of the phospholipid portion of the membrane from the host cell from which it came and proteins encoded by the virus genome. The bacterial cell envelope refers to the layers outside of the cell membrane.
DIF: Medium REF: 6.2 TOP: II.A.i.c MSC: Understanding
- Compare and contrast the Baltimore classification of viruses to the proteomic analysis of bacteriophages. Which more accurately reflects the significant impact of phage evolution?
ANS:
The Baltimore classification divides the viruses into seven groups based on the genome of the virus and the route used to get to mRNA. Proteomic analysis is a comparison of the proteins encoded by the genomes. The proteomic tree also yields seven categories of phages. These categories, however, group the phages by host bacteria. Phages with different types of genomes may be grouped together. The proteomic tree analogy more truly reflects the evolutionary process. Viruses that share a host have the ability to exchange genetic information.
DIF: Difficult REF: 6.3 TOP: III.B.i | III.C.ii
MSC: Understanding
- Mosaic genomes are genomes that are derived from different sources. How does this occur in viruses?
ANS:
When different viruses coinfect a host, they have the opportunity to exchange genetic information through recombination or reassortment of segmented genomes. Viruses may also pick up genes from the host cell, as seen in the oncogenic viruses. The genomic content of a virus is truly influenced by its host range.
DIF: Difficult REF: 6.3 TOP: III.C MSC: Understanding
- Some viruses require an RNA-dependent RNA polymerase. What does that mean? Using the same vernacular, what would you call the host cell RNA polymerase (RNA pol)? What would you call reverse transcriptase (RTase)?
ANS:
An RNA-dependent RNA polymerase means that the polymerase requires an RNA template (RNA-dep) to synthesize RNA (RNA pol). The host cell RNA pol would be called a DNA-dependent RNA polymerase, since it requires a DNA template to synthesize RNA. RTase would be referred to as an RNA-dependent DNA polymerase since it utilizes an RNA template to synthesize DNA.
DIF: Difficult REF: 6.3 TOP: III.B.i MSC: Applying
- Why do many RNA viruses encode their own RNA-dependent RNA polymerase and package them in viral particles? How do we take advantage of these viral-specific polymerases?
ANS:
Animal or bacterial cells lack RNA-dependent RNA polymerase and thus cannot replicate RNA regardless of whether they are single- or double-stranded RNA. The initial replication of viral RNA genomes have to be accomplished by existing RNA-dependent RNA polymerase. Subsequent replication and/or transcription of viral RNA genome and mRNA could be accomplished by viral RNA-dependent RNA polymerase produced in the hosts.
These viral-specific polymerases could be the targets for antiviral therapies with lowered side effects. They could also be utilized for in vitro replication/amplification of RNA molecules.
DIF: Difficult REF: 6.3 TOP: III.B.i MSC: Applying
- What determines whether a temperate phage will go through the lysogenic cycle or switch to the lytic cycle?
ANS:
Proteins bind DNA to repress viral replication genes during lysogeny. The switch to the lytic cycle can occur randomly or can be triggered by environmental stresses that damage the cell’s DNA. Events that threaten a cell’s survival can trigger the lytic cycle. If the host cells are growing well, the virus can continue to survive and be propagated as a prophage.
DIF: Difficult REF: 6.4 TOP: IV.C.iii MSC: Understanding
- How can a virus be used in biotechnology to transfer genes from one bacterial cell to another?
ANS:
In the process of transduction, a lysogenic phage is triggered to begin the lytic cycle. When this occurs, the viral DNA is excised from the host chromosome, and sometimes some of the host chromosome is excised with the viral DNA. A combination of viral DNA and host DNA can be packaged into a viral capsid, which then infects a new host and brings both viral and host DNA to the new host. In other cases, a viral infection causes host DNA to be cut into small pieces, and some of those pieces are accidentally packaged into viral capsids. This host DNA in the viral capsid can be transferred to a new host when the virus infects the new host.
DIF: Difficult REF: 6.4 TOP: IV.C.iv.a MSC: Applying
- What determines the host range and tropism of a virus?
ANS:
Viruses must attach to specific host cell receptors to cause infection. Only host cells with the specific receptors that a particular virus recognizes can be infected with that viral species. Receptors may also only be found on specific tissues, meaning that the receptors also determine tropism, or which tissues can be infected.
DIF: Medium REF: 6.4 | 6.5 TOP: IV.A | V.A.i MSC: Understanding
- How does genome uncoating occur with animal viruses?
ANS:
In extracellular uncoating, the virus attaches to the host cell, then inserts its DNA into the host, similar to bacteriophage. Intracellular uncoating can occur in several different ways. The virus can bind host receptors that cause the viral envelope to fuse with the host cell membrane. The genome is then uncoated and released into the host cell. Some viruses are taken up by endocytosis. The host forms an endocytic vesicle around the virus and brings the virus into the cell. The vesicle fuses with a lysosome whose acidic environment activates entry of the capsid into the cytoplasm. Some viruses enter the host by endocytosis and are then transported to the nucleus, where the viral genome is injected into the nucleus.
DIF: Difficult REF: 6.5 TOP: V.A.ii MSC: Remembering
- Define oncogenes and oncogenic viruses, and explain how they can be used in gene therapy.
ANS:
Oncogenes are genes that, when expressed, cause uncontrolled proliferation of the infected cell and ultimately cancer. These oncogenes may be activated by viral transfer of host genes to abnormal chromosome locations. Alternatively, viruses may carry these oncogenes in their genome. Viruses that lead to oncogene expression are referred to as oncogenic viruses.
In gene therapy, a virus is used as a vehicle to transfer a functional copy of a gene to an individual to “correct” a genetic defect.
DIF: Difficult REF: 6.5 TOP: V.B.iv.a MSC: Understanding
- What is necessary for bacteriophage and animal virus entry into a cell? How does that differ from plant viral entry?
ANS:
There is a specific receptor on the host cell to which a bacteriophage or animal virus must bind to initiate viral entry into the host. There is no receptor involved in plant viral entry. It is instead gained by either contact with damaged tissues, transmission by an animal vector, or transmission through seed.
DIF: Medium REF: 6.5 TOP: V.B | V.C MSC: Understanding
- How is a plaque assay used to count virions in a solution?
ANS:
Dilutions of a solution containing an unknown number of viruses is diluted, then mixed with host cells. The mixture is added to soft agar, then poured onto a plate of solid medium. When the virions infect host cells, the host cells will lyse and release more virions to infect neighboring host cells. This cycle continues until a visible clearing, a plaque, can be seen within the lawn of bacteria. The plaques can be counted and multiplied by the dilution factor to determine plaque-forming units/ml.
DIF: Medium REF: 6.6 TOP: VI.C MSC: Remembering
- During the eclipse period of a viral infection, why are the virions undetectable in the growth medium?
ANS:
The virions attach to cell-surface receptors and inject their DNA. Since all or most of the virions are attached to the host-cell receptors, there are few present in the growth medium.
DIF: Medium REF: 6.6 TOP: VI.A.iii MSC: Understanding
- Why was the advent of tissue culture heralded as the end of the “monkey era”? What types of things are we able to do with it?
ANS:
Monkeys were being used to culture polio virus. This was very expensive and labor-intensive, and involved the use of many animals. Tissue culture affords us the option to avoid animal use. It has become the method of choice to test chemical and antibiotic reagents. It is also used to study animal and plant viruses and to develop vaccines and antiviral chemicals.
DIF: Medium REF: 6.6 TOP: VI.B.iii MSC: Understanding
- To count the number of viruses in a solution, one can perform a plaque assay. Describe how a plaque is generated in the case of a lytic virus and how a virus that does not lyse its host can be detected.
ANS:
A plaque represents the area where one virus initially infected one cell. It looks like a clear zone on a lawn of uninfected cells. In the case of a lytic virus, the plaque appears because the cells lyse when the viruses are released. In the case of a nonlytic virus, the cells are not lysed, but when infected they contain components of the replicating virus. Fluorescent antibodies targeting specific viral proteins can be reacted with the cells.
DIF: Medium REF: 6.6 TOP: VI.C | VI.D MSC: Understanding
- The myxovirus was introduced to Australia to control the rabbit population. How does the virus keep the rabbit population relatively low without killing all of the rabbits?
ANS:
If there is a large population of rabbits, the rate of transmission of the virus increases. Since they are living closer together, it is more likely that they can pass the virus to other rabbits. Once the population decreases and they are more spread out, they are less likely to come in contact with rabbits carrying the virus. Some of the remaining rabbits are resistant to the virus and others don’t come in contact with it. As long as the population density is low, the virus is transmitted at a low rate. If the population density is high, the virus is more likely to be transmitted to others.
DIF: Difficult REF: 6.7 TOP: VII.B.i MSC: Understanding
- Discuss the role that marine viruses play in carbon balance.
ANS:
The carbon cycle in a marine environment is usually described as grazers consuming phytoplankton and then being consumed by carnivores. One must realize that viral infection of each of these leads to lysis and the formation of detritus. This organic material is consumed by bacteria that respire, returning carbon dioxide to the atmosphere.
DIF: Medium REF: 6.7 TOP: VII.B.ii MSC: Understanding
CHAPTER 7: Genomes and Chromosomes
MULTIPLE CHOICE
- The intestinal bacterium __________ was the initial focus of efforts to understand genes and genetics.
a. | Salmonella enterica | d. | Mycoplasma genitalium |
b. | Escherichia coli | e. | Streptococcus pneumoniae |
c. | Staphylococcus aureus | |
ANS: B DIF: Easy REF: Introduction TOP: I
MSC: Remembering
- Intestinal microbes contain __________ times more genes than the human genome?
ANS: D DIF: Medium REF: Introduction TOP: I
MSC: Remembering
- Gene transfer from parent to child is called:
a. | expression | d. | transformation |
b. | horizontal transmission | e. | recombination |
c. | vertical transmission | |
ANS: C DIF: Easy REF: 7.1 TOP: I.A.i
MSC: Remembering
- Bacteria are capable of __________ transmission—the transfer of genetic information from one cell to another.
a. | vertical | d. | transformation |
b. | horizontal | e. | pseudogene |
c. | recombination | |
ANS: B DIF: Easy REF: 7.1 TOP: I.A.ii
MSC: Remembering
- __________ allowed genes to be mapped relative to each other based on the time of transfer.
a. | Transformation | d. | Polycistrons |
b. | Transduction | e. | Polymerase chain reaction |
c. | Conjugation | |
ANS: C DIF: Easy REF: 7.1 TOP: I.B.i
MSC: Remembering
- It takes approximately how many minutes to move the entire E. coli chromosome from one cell to another?
ANS: D DIF: Medium REF: 7.1 TOP: I.B.i
MSC: Remembering
- The smallest cellular genomes identified thus far are those of:
a. | E. coli | d. | Mycoplasma |
b. | Staphylococcus | e. | yeast |
c. | Streptococcus | |
ANS: D DIF: Easy REF: 7.2 TOP: II.A.i
MSC: Remembering
- __________ in the DNA helix provide DNA-binding proteins access to base sequences buried in the center of the molecule, so that proteins can interact with the bases without the strands being separated.
a. | Cohesive ends | d. | Attractions |
b. | Grooves | e. | Repulsions |
c. | Nicks | |
ANS: B DIF: Easy REF: 7.2 TOP: II.C
MSC: Remembering
- __________ are bicyclic bases and __________ are monocyclic.
a. | Pyrimidines; purines | d. | Thymines; uracils |
b. | Adenines; guanines | e. | Purines; pyrimidines |
c. | Thymines; cytosines | |
ANS: E DIF: Difficult REF: 7.2 TOP: II.C.ii.a
MSC: Remembering
- Bacteria and archaea growing at extreme pH or temperature protect their DNA from denaturation through the use of:
a. | pseudogenes | d. | introns |
b. | Okazaki fragments | e. | ligase |
c. | DNA-binding proteins | |
ANS: C DIF: Easy REF: 7.2 TOP: II.C.iii
MSC: Remembering
- In a test tube at chromosome-packing density, DNA is almost a:
a. | solid | d. | liquid |
b. | gel | e. | plasma |
c. | gas | |
ANS: B DIF: Easy REF: 7.2 TOP: II.E
MSC: Remembering
- A nucleoid gently released from E. coli appears as 30–100 tightly wound loops, each defined by anchoring:
a. | histones | d. | gyrase |
b. | histone-like proteins | e. | topoisomerase |
c. | supercoils | |
ANS: B DIF: Easy REF: 7.2 TOP: II.F.i
MSC: Remembering
- The nucleoids of bacteria and most archaea, as well as the nuclear DNA of eukaryotes, are kept __________ supercoiled.
a. | positively | d. | reverse |
b. | neutrally | e. | extra |
c. | negatively | |
ANS: C DIF: Easy REF: 7.2 TOP: II.F.iii
MSC: Remembering
- Supercoiling in bacteria is typically introduced by an enzyme called:
a. | gyrase | d. | polymerase |
b. | helicase | e. | endonuclease |
c. | ligase | |
ANS: A DIF: Easy REF: 7.2 TOP: II.G.ii
MSC: Remembering
- The thermal stability of the helix is predominantly due to the __________ of the hydrophobic base pairs.
a. | attraction | d. | stacking and hydrogen bonding |
b. | binding | e. | transformation |
c. | repulsion and hydrogen bonding | |
ANS: D DIF: Medium REF: 7.2 TOP: II.C
MSC: Understanding
- Which of the following is NOT true about the two strands of DNA in a double helix?
a. | The base pairs of one strand interact with the base pairs of the other. |
b. | They are antiparallel. |
c. | They are complementary. |
d. | They have equal amounts of adenine and guanine. |
e. | They have base-stacking interactions. |
ANS: D DIF: Easy REF: 7.2 TOP: II.C
MSC: Applying
- Hyperthermophilic archaea possess an unusual gyrase called __________, which introduces positive supercoils into the chromosome in order to protect the DNA from __________.
a. | DNA ligase; thermal denaturation |
b. | DNA helicase; thermal denaturation |
c. | reverse DNA gyrase; excessive gene expression |
d. | reverse DNA gyrase; thermal denaturation |
e. | reverse transcriptase; excessive gene expression |
ANS: D DIF: Difficult REF: 7.2 TOP: II.G
MSC: Applying
- The semiconservative mechanism of replication provides for each daughter duplex to be checked for __________, based on its parent strand.
a. | pseudogenes | d. | redundancy |
b. | introns | e. | accuracy |
c. | plasmids | |
ANS: E DIF: Easy REF: 7.3 TOP: III.A.i
MSC: Remembering
- Enzymes that synthesize DNA or RNA can synthesize in a __________ direction.
a. | forward | d. | reverse |
b. | 3′-to-5′ | e. | 5′-to-3′ or 3′-to-5′ |
c. | 5′-to-3′ | |
ANS: C DIF: Easy REF: 7.3 TOP: III.A.i.b
MSC: Remembering
- The lagging strand is synthesized __________ , while the leading strand can be synthesized __________.
a. | continuously; discontinuously | d. | 3′-to-5′; 5′-to-3′ |
b. | discontinuously; continuously | e. | quickly; slowly |
c. | 5′-to-3′; 3′-to-5′ | |
ANS: B DIF: Difficult REF: 7.3 TOP: III.B.i
MSC: Remembering
- Initiation of DNA replication is controlled by DNA __________ and by the binding of a specific initiator protein to the origin sequence.
a. | restriction | d. | methylation |
b. | ligase | e. | gyrase |
c. | helicase | |
ANS: D DIF: Difficult REF: 7.3 TOP: III.C.i
MSC: Remembering
- When the chromosome replicates, how is the newly made strand related to its template strand?
a. | The two strands have identical sequences and are parallel to each other. |
b. | The two strands have complementary sequences and are parallel to each other. |
c. | The two strands have identical sequences and are antiparallel to each other. |
d. | The two strands have complementary sequences and are antiparallel to each other. |
e. | The two strands have identical sequences and are antiparallel to each other, except that U replaces T. |
ANS: D DIF: Difficult REF: 7.3 TOP: III.A
MSC: Understanding
- Which of the following is NOT true of DNA replication in E. coli?
a. | DnaA triggers initiation |
b. | It is bidirectional. |
c. | One new strand is synthesized discontinuously. |
d. | It is conservative. |
e. | Synthesis occurs in the 5′-to-3′ direction. |
ANS: D DIF: Easy REF: 7.3 TOP: III.A.i
MSC: Understanding
- The primer in DNA replication is:
a. | an RNA primer with a free 3′ OH group | d. | a DNA primer with a free 5′ OH group |
b. | an RNA primer with a free 5′ OH group | e. | a telomere with a free 5′ phosphate |
c. | a DNA primer with a free 3′ OH group | |
ANS: A DIF: Medium REF: 7.3 TOP: III.C.iii
MSC: Understanding
- The sliding clamp is used to __________ and is loaded by ___________.
a. | open the helix; clamp loader complex |
b. | open the helix; DNA polIII |
c. | tether the polymerase to the chromosome; clamp loader complex |
d. | tether the polymerase to the chromosome; DNA polI |
e. | tether the polymerase to the chromosome; DNA primase |
ANS: C DIF: Medium REF: 7.3 TOP: III.C.iv
MSC: Understanding
- What type of bonds does DNA polymerase form?
a. | aminoacyl bonds | d. | hydrogen bonds |
b. | peptide bonds | e. | disulfide bonds |
c. | phosphodiester bonds | |
ANS: C DIF: Easy REF: 7.3 TOP: III.D
MSC: Understanding
- Accidental errors during replication are corrected by the DNA proofreading activity intrinsic to:
a. | DNA polI | d. | DNA polIV |
b. | DNA polII | e. | DNA polV |
c. | DNA polIII | |
ANS: C DIF: Easy REF: 7.3 TOP: III.D.i.b
MSC: Understanding
- Okazaki fragments in E. coli are approximately how many bases in length?
a. | 10 | d. | 1,000 |
b. | 50 | e. | 2,000 |
c. | 100 | |
ANS: D DIF: Medium REF: 7.3 TOP: III.D.ii
MSC: Understanding
- Which molecule is responsible for removing torsional stress during DNA replication?
a. | DNA primase | d. | DNA gyrase |
b. | single-stranded binding protein | e. | DNA ligase |
c. | RNase H | |
ANS: D DIF: Easy REF: 7.3 TOP: III.D.iv
MSC: Understanding
- Which of the following is NOT involved in the replication termination process?
a. | ter sites | d. | XerD |
b. | DNA ligase | e. | specific sequences called dif |
c. | XerC | |
ANS: B DIF: Medium REF: 7.3 TOP: III.E
MSC: Understanding
- Two kinds of extragenomic DNA molecules are __________ and the genomes of __________.
a. | bacteriophages; pseudogenes | d. | plasmids; introns |
b. | introns; bacteriophages | e. | introns; exons |
c. | plasmids; bacteriophages | |
ANS: C DIF: Medium REF: 7.4 TOP: IV
MSC: Remembering
- Which of the following is NOT true of plasmids?
a. | They are found in archaea. | d. | They are not supercoiled. |
b. | They are found in bacteria. | e. | They are usually circular. |
c. | They are found in eukaryotes. | |
ANS: D DIF: Difficult REF: 7.4 TOP: IV.A
MSC: Remembering
- __________ is a process that involves cell-to-cell contact to move the plasmid from a donor cell to a recipient cell.
a. | Transformation | d. | Conjugation |
b. | Transduction | e. | Transposon |
c. | Pseudogene | |
ANS: D DIF: Easy REF: 7.4 TOP: IV.C.i
MSC: Remembering
- Plasmids have a high probability of being present in the new generation of cells by all of the following EXCEPT:
a. | being present at a high number so that some copies will end up in each new cell after cell division |
b. | encoding traits such as antibiotic resistance that are required for growth in some environments |
c. | having replicative processes that are not tied to replication of the chromosome |
d. | using actin-like ParM filament to bind to ParR-parC complexes on the plasmid |
e. | carrying host survival genes and self-preservation genes |
ANS: C DIF: Difficult REF: 7.4 TOP: IV.A
MSC: Applying
- Eukaryotic chromosomes are __________, contained within a nucleus, and their separation into the two cells involves the process of __________.
a. | linear; meiosis | d. | circular; meiosis |
b. | linear; mitosis | e. | circular; rolling circle |
c. | circular; mitosis | |
ANS: B DIF: Medium REF: 7.5 TOP: V.A
MSC: Remembering
- Which of the following terms refers to gene duplications that have decayed into nonfunctional entities?
a. | introns | d. | pseudogenes |
b. | contigs | e. | Okazaki fragments |
c. | orthologs | |
ANS: D DIF: Easy REF: 7.5 TOP: V.A.iv
MSC: Remembering
- __________ and __________ are noncoding sequences that make up a large portion of eukaryotic chromosomes.
a. | Exons; introns | d. | Introns; pseudogenes |
b. | Bacteriophages; plasmids | e. | Pseudogenes; bacteriophages |
c. | Plasmids; introns | |
ANS: D DIF: Medium REF: 7.5 TOP: V.A.iv
MSC: Remembering
- Archaea resemble eukaryotes in all of the following EXCEPT:
a. | nuclear membrane | d. | ribosomal components |
b. | DNA-packing proteins | e. | DNA polymerase |
c. | RNA polymerase | |
ANS: A DIF: Medium REF: 7.5 TOP: V.B.ii
MSC: Understanding
- Which of the following is NOT true about histones?
a. | They are rich in arginine and lysine. |
b. | They bind to negatively charged DNA. |
c. | They are positively charged. |
d. | They are used to control expression of bacterial operons. |
e. | They form units with DNA called nucleosomes. |
ANS: D DIF: Difficult REF: 7.5 TOP: V.A
MSC: Applying
- __________ is an enzyme that may have evolved from ancient progenitor cells with an RNA genome. It functions as a __________.
a. | Primase; helicase | d. | Telomerase; reverse transcriptase |
b. | Primase; reverse transcriptase | e. | Telomerase; methylator and acetylator |
c. | Telomerase; helicase | |
ANS: D DIF: Difficult REF: 7.5 TOP: V.A.ii
MSC: Applying
- What role does lysozyme play when isolating DNA?
a. | dissolves the cell membranes |
b. | degrades the peptidoglycan cell wall |
c. | precipitates the proteins in a high-salt solution |
d. | elutes DNA from the DNA-binding column |
e. | performs alcohol precipitation of DNA |
ANS: B DIF: Medium REF: 7.6 TOP: VI.A
MSC: Remembering
- __________ naturally function to cut the DNA of invading plasmids and phages at specific points.
a. | Topoisomerases | d. | Polymerases |
b. | Restriction endonucleases | e. | Gyrases |
c. | Ligases | |
ANS: B DIF: Easy REF: 7.6 TOP: VI.B.i
MSC: Remembering
- Agarose gel electrophoresis separates pieces of linear DNA based on:
a. | size | d. | supercoiling |
b. | charge | e. | methylation |
c. | sequence | |
ANS: A DIF: Easy REF: 7.6 TOP: VI.B.iv
MSC: Remembering
- The ability of cohesive ends of DNA from different organisms to __________ forms the basis of recombinant DNA technology.
a. | be methylated | d. | repel |
b. | be cut | e. | be sequenced |
c. | base-pair | |
ANS: C DIF: Easy REF: 7.6 TOP: VI.C
MSC: Remembering
- In the early days of cloning, DNA was __________ into E. coli and then could be transferred to another organism using a __________.
a. | conjugated; contig | d. | transformed; contig |
b. | conjugated; shuttle vector | e. | transformed; polymerase chain reaction |
c. | transformed; shuttle vector | |
ANS: C DIF: Medium REF: 7.6 TOP: VI.C.iv
MSC: Remembering
- What is the correct order of the steps in a cycle of PCR?
a. | annealing, denaturation, extension | d. | denaturation, annealing, extension |
b. | denaturation, extension, annealing | e. | annealing, extension, denaturation |
c. | extension, annealing, denaturation | |
ANS: D DIF: Medium REF: 7.6 TOP: VI.D
MSC: Remembering
- Pyrosequencing has aided in the understanding of:
a. | uses for polymerase chain reaction |
b. | the expression of the LacZ genes |
c. | use of plasmids in both the hospital and in the research arena |
d. | colony collapse disorder and genomes of possible pathogens present in collapsed hives that may be a cause of the problem |
e. | development of fluorescent labels for chain termination |
ANS: D DIF: Easy REF: 7.6 TOP: VI
MSC: Understanding
- Which of the following is NOT a palindrome?
a. | toot | d. | radar |
b. | rotor | e. | toto |
c. | racecar | |
ANS: E DIF: Easy REF: 7.6 TOP: VI.B.iii
MSC: Applying
- Pieces of DNA that have been cut with restriction enzymes can then be “grafted” together using the enzyme:
a. | DNA gyrase | d. | RNA polymerase |
b. | DNA ligase | e. | reverse transcriptase |
c. | DNA primase | |
ANS: B DIF: Easy REF: 7.6 TOP: VI.C.i
MSC: Applying
- The Sanger method of DNA sequencing involves __________ that bring about chain termination.
a. | 2’ deoxyribonucleotides | d. | 2’ deoxyadenosine |
b. | 3’ deoxyribonucleotides | e. | 3’ dideoxyadenosine |
c. | 2’, 3’ dideoxyribonucleotides | |
ANS: C DIF: Medium REF: 7.6 TOP: VI.E.i
MSC: Applying
SHORT ANSWER
- DNA contains structural genes and control sequences. Describe each.
ANS:
Structural genes are those that are transcribed to make RNA, which in turn may be translated to produce a protein. Control sequences are regions that do not code for a protein. They are regions involved in regulation of gene expression. Examples of control sequences are an RNA polymerase–binding site, an enhancer-binding site, an inducer-binding site, or a repressor-binding site. In other words, they are nucleotide sequences that interact with proteins to regulate activation or inactivation of a promoter.
DIF: Medium REF: 7.1 TOP: I.A.iv MSC: Understanding
- A genome is referred to by its length. If a genome is said to be 5 kb, what does that mean? If it were 5 kb, what would that tell you?
ANS:
A 5 kb genome would be a piece of DNA that is 5 kilobases long.
DIF: Medium REF: 7.2 TOP: II.A.i MSC: Understanding
- Describe how the name of a gene and its gene product are denoted.
ANS:
Bacterial gene names are written in italics with lowercase letters. The gene is named using an abbreviation of either the enzyme product’s name or a protein’s function. The genes involved in tryptophan biosynthesis, for example, are referred to as trp genes. There is more than one gene, so they are referred to as trpA, trpB, and so on. The gene product is not italicized, and the first letter is capitalized. The product of the trpA gene is, therefore, TrpA.
DIF: Medium REF: 7.2 TOP: II.B MSC: Understanding
- Describe the similarities and differences between DNA and RNA.
ANS:
Similarities: They are both made of nucleotides held together by phosphodiester bonds. They both contain the nitrogen bases adenine, guanine, and cytosine. Differences: DNA is usually double-stranded, RNA is usually single-stranded. The sugar in DNA is deoxyribose; the sugar in RNA is ribose. DNA has the nitrogen base thymine; RNA has the nitrogen base uracil.
DIF: Medium REF: 7.2 TOP: II.D MSC: Understanding
- Explain how understanding topoisomerases has helped healthcare.
ANS:
Gyrase (topoisomerase type II) was discovered as a result of studying mutants with resistance to nalidixic acid and oxolinic acid, two antibiotics. Because of this discovery, gyrases became a target for further antibiotic development. Fluoroquinolones were developed to inhibit gyrase and are now commonly used.
DIF: Medium REF: 7.2 TOP: II.G MSC: Understanding
- When E. coli was lysed and its chromosome released, researchers found that it was 1,500 times as long as the cell. List three questions that emerge from this finding.
ANS:
Answers vary, but possible responses include: How could this all be packaged into such a small space? Knowing that the generation time is 20 minutes in a laboratory setting, how is this done? How does the chromosome avoid getting tangled? How are parts expressed?
DIF: Difficult REF: 7.2 TOP: II.E MSC: Applying
- Differentiate between topoisomerase type I and topoisomerase type II. What is the role they play in bacterial chromosome structure?
ANS:
Topoisomerase I will make a single strand break in the DNA, thus relaxing supercoils. Topoisomerase II (e.g., gyrase) will introduce double-strand breaks, use ATP and introduce negative supercoils. By using both enzymes, DNA is packed into a smaller volume to fit inside the cell and then maintained with slight negative supercoils for access for transcription.
DIF: Difficult REF: 7.2 TOP: II.G MSC: Analyzing
- It takes 40 minutes for E. coli chromosome replication but only 20 minutes for cell division. How is this possible?
ANS:
A partially replicated chromosome can start new rounds of replication at the two daughter chromosome origins of replication before the first round of replication is complete. This overlapping of generations allows the cell to accommodate the 40-minute DNA replication within the 20-minute cell division time.
DIF: Medium REF: 7.3 TOP: III.B MSC: Understanding
- What determines when DNA replication begins?
ANS:
DNA methylation, and thus the actions of SeqA and DnaA, control the timing of replication. Initiation of replication is activated by the DnaA protein and inhibited by the SeqA protein.
SeqA binds hemimethylated (newly replicated) origins and prevents another initiation too soon after replication. Once the new strand is methylated, SeqA decreases binding at the origin, which allows DnaA to bind. The replication initiator protein is DnaA. As the cell grows, the amount of DnaA increases and binds the origin. This increases melting of DNA at the origin, which initiates replication.
DIF: Difficult REF: 7.3 TOP: III.C MSC: Understanding
- Describe the role of the two RNA polymerases that are involved early on in the DNA replication process.
ANS:
The first RNA polymerase is the so-called housekeeping RNA pol, which is responsible for most of the RNA synthesis in a cell. It transcribes a portion of the DNA at oriC to produce RNA that helps separate the two strands of DNA at the origin. The second RNA pol is called DNA primase (DnaG), and it is the RNA pol that synthesizes the short primers required for DNA replication.
DIF: Medium REF: 7.3 TOP: III.C.iii MSC: Understanding
- Why is replication of the lagging DNA strand a problem, and how is this problem overcome?
ANS:
DNA polymerase can only add nucleotides onto a 3′ OH end of a growing DNA chain, so DNA must be made in the 5′-to-3′ direction. The two strands of DNA are antiparallel, meaning the leading strand can easily be replicated toward the replication fork as it opens, but the lagging strand can’t be made in that direction. The solution is the production of Okazaki fragments. Primase places a new primer as the replication fork opens and DNA polymerase can extend the primer. This process continues, and the lagging strand ends up containing many short sections of RNA primer, which must eventually be removed.
DIF: Medium REF: 7.3 TOP: III.D MSC: Understanding
- Describe three functions of DNA polymerase III.
ANS:
It is the main polymerizing enzyme for DNA replication, adding new nucleotides to the growing DNA chain and creating phosphodiester bonds. It has proofreading activity to find mispaired bases. It has exonuclease activity, which allows it to break a phosphodiester bond and remove mispaired bases.
DIF: Medium REF: 7.3 TOP: III.D.i MSC: Understanding
- Describe the role that DNA Pol I plays in replication.
ANS:
The lagging strand has RNA primers left behind after the DNA Pol III replicates the Okazaki fragments. DNA Pol I, with its 5′-to- 3′ exonuclease activity, is able to remove the RNA primer as it adds to the existing 3′ end of the DNA.
DIF: Medium REF: 7.3 TOP: III.D.iii MSC: Understanding
- In addition to primases and polymerases, name three other proteins required for initiation of and elongation during replication. Describe their functions.
ANS:
Any three of the following (except polymerase and primase): DnaA acts as an initiator of replication by recognizing the oriC sequence. DnaB acts as a helicase and is loaded onto the DNA via the action of DnaC, the helicase loader. Once loaded, the helicase recruits primase. Once the DNA is opened and primed, the sliding clamp is loaded by the clamp loader. Two molecules of polymerase are added at this time. The clamp loader loads these proteins and then the sliding clamp keeps the polymerase bound to the DNA. Once replication has been initiated and in order to elongate, single-stranded binding proteins hold the single strands of DNA to prevent them from reannealing to each other. On the lagging strand, DNA ligase creates the phosphodiester bond between the Okazaki fragments.
DIF: Difficult REF: 7.3 TOP: III.C | III.D MSC: Analyzing
- How can plasmids ensure that they are passed on from one generation to the next?
ANS:
Some plasmids are high-copy-number, meaning a cell may have 50–700 copies, and there is a high probability that both daughter cells will receive at least one copy. Low-copy-number plasmids may have partitioning systems to ensure both daughter cells receive copies. Some plasmids ensure their inheritance by carrying genes whose functions give the bacteria an advantage in certain environments, such as antibiotic resistance.
DIF: Medium REF: 7.4 TOP: IV.B MSC: Remembering
- How are archaeal genomes similar to both bacterial and eukaryotic genomes?
ANS:
Similarities to bacteria include the following: both have operons, reproduce by asexual reproduction, lack nuclear membranes, and have single circular chromosomes. Similarities to eukaryotes include greater similarity to eukaryotes with respect to the following: DNA-packing proteins, RNA polymerase, ribosomes, and DNA polymerase.
DIF: Medium REF: 7.5 TOP: V.B MSC: Understanding
- Describe how pyrosequencing led to an understanding of colony collapse disorder in bees. Why is traditional Sanger sequencing not sufficient for this task?
ANS:
Answers will vary. Pyrosequencing was used to study random sequences in collapsed hives and compare them to DNA sequences in normal hives. As result of obtaining this sequence information, DNA for the Israeli acute paralysis virus was found, as was DNA for the microsporidia group of fungi. In order to obtain all this information with traditional sequencing, some information about possible pathogens and their sequence information would be needed to design primers. Also, the traditional sequencing would have taken much longer to sequence each individual piece of DNA and then compile the information. Pyrosequencing gave the information in months rather than years.
DIF: Medium REF: 7.6 TOP: VI MSC: Understanding
- Humans have an immune system that can distinguish self from nonself. How is the bacterial restriction-modification system similar?
ANS:
Bacterial cells have restriction enzymes that recognize specific sequences of DNA and make double-stranded cuts at these recognition sites. If an organism that produced these restriction enzymes did not have some mechanism by which to protect its own DNA, it would destroy itself by chopping its DNA into bits. Hence, the bacterial cell has a matching modification enzyme that will methylate the recognition site. The methylated DNA cannot be cleaved by the restriction enzyme.
In all likelihood, an invading bacterium or bacteriophage will not be modified similar to the host. The host cell does not destroy itself due to methylation, but any foreign DNA entering the cell that contains the specific recognition site and is not methylated will be cut by the restriction enzymes. In this way, a bacterium also can distinguish self from nonself in its attempts to destroy any foreign DNA.
DIF: Medium REF: 7.6 TOP: VI.B MSC: Understanding
- How has the study of metagenomics allowed discovery of so many new species? How is it possible to know about these organisms without being able to grow them in a laboratory?
ANS:
Metagenomics is the use of modern genomic techniques such as rapid DNA sequencing to study microbial communities directly in their natural environments, without the need for laboratory cultivation of the microbes. This is a breakthrough because it is estimated that at least 99% of microbial species can’t currently be cultivated in lab. The genes for 16S rRNA can be sequenced directly from environmental samples. This information can be used to determine how the unknown organisms are related to known species and construct phylogenetic trees that include these organisms. Metagenomics can also be used to determine the metabolic capabilities of the unknown organisms.
DIF: Medium REF: 7.6 TOP: VI.G MSC: Understanding
- Describe two ways that the study of bacteria and their processes has led to the development of molecular biology and revolutionized genetics.
ANS:
Any two of the following:
(1) Scientists studying bacterial genetics discovered restriction endonucleases—enzymes used by bacteria to degrade foreign DNA. They were then able to clone genes by precisely cutting DNA with these enzymes and recombining it.
(2) Scientists also discovered plasmids and were able to engineer them and use their antibiotic resistance genes.
(3) By studying thermophilic bacteria, scientists have discovered a variety of heat-stable enzymes, most notably taq polymerase (and other heat-stable DNA polymerases) that are used in polymerase chain reaction.
DIF: Medium REF: 7.6 TOP: VI.B | VI.C | VI.D
MSC: Applying