Biology, Eighth Edition (Raven)

Chapter 27: Viruses

Post Test

1
Which of the following is/are NOT legitimate structures for a natural virus genome? (p. 526)
A)Double-stranded DNA
B)Single-stranded DNA
C)Double-stranded RNA
D)Single-stranded RNA
E)All of the above are found in examples of natural viruses
2
Under which circumstances might the lysogenic cycle benefit a virus more than the lytic cycle? (p. 529)
A)Because the lytic cycle amplifies the numbers of infections to all susceptible hosts in the area, all hosts will eventually be consumed. The lysogenic cycle puts a brake on this process while still retaining infectious potential.
B)By integrating with the bacterial chromosome, the genetic instructions for the virus become refreshed after one or more replication events during binary fission.
C)Lysogenic infection cycles don't harm their host cells, so they can produce virus particles indefinitely.
D)Formation of a prophage gives the cells new abilities and survivability in an unpredictable environment, causing both the bacterium and the virus to proliferate.
E)Lysogeny causes more mutations to occur in the virus, creating more variants upon which natural selection can operate.
3
Why is a mutant CCR5 receptor often associated with resistance to HIV infection? (p. 530)
A)The mutant receptor binds to and inactivates circulating HIV particles.
B)The immune system in people with the mutated CCR5 receptor is inadvertently activated to recognize HIV particles, conferring resistance to infection.
C)The CCR5 receptor causes a signal transduction effect which inhibits protease activity required for HIV particle assembly.
D)The gene for the mutant CCR5 receptor is on the chromosome next to the gene for an enzyme that digests HIV's gp120 surface protein.
E)HIV interacts with normal CCR5 surface receptors and use them to gain entry to the cell's interior.
4
In 1997, a form of avian influenza capable of infecting humans was named H5N1. The "N" stands for neuraminidase, which helps the virus break free of the host cell upon assembly. What does the H (hemagglutinin) variety do for the virus? (p. 534)
A)It causes fever.
B)Hemagglutinin is involved with viral construction.
C)It is a surface protein which provides entry to the cell interior.
D)The virus needs this enzyme to incorporate into the host cell's genome.
E)It provides for a viral envelope, and acts to bypass the cell's immune defenses.
5
What would be the effect of creating a knockout mutation for the normal PrP gene in a mouse line, then exposing these mice to the infectious PrP prion protein? (p. 536)
A)The mice would have antibodies against the infectious PrP protein.
B)The mice would be completely unable to create more infectious PrP proteins.
C)Because of the knockout mutation, the mice would identify the PrP protein as a foreign invader and launch a dramatic immune response.
D)Without the protection of the normal PrP protein, the mouse would not have defenses against the infectious version, and thus will quickly become ill.
E)All of the above are equally plausible.
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