Microbiology chap 8 Flashcards

The properties of a cell that are determined by its DNA composition are its

A. phenotype.
B. genotype.
C. metabolism.
D. nucleoid.

2. The source of variation among microorganisms that were once identical is
A. antibiotic resistance.
B. virulence factors.
C. sigma factors.
D. mutation.

3. The characteristics displayed by an organism in any given environment is its
A. genotype.
B. archaetype.
C. mutatotype.
D. phenotype.

4. Which change in a gene's DNA sequence would have the least effect on the eventual amino acid sequence produced from it?
A. Deletion of two consecutive nucleotides

B. Addition of one nucleotide

C. Addition/deletion of three consecutive nucleotides

5. The designation his ^- refers to
A. the genotype of a bacterium that lacks a functional gene for histidine synthesis.
B. the genotype of a bacterium that has a functional gene for histidine synthesis.
C. the opposite of a hers gene.
D. bacteria that are auxotrophic for histidine.
E. the genotype of a bacterium that lacks a functional gene for histidine synthesis AND bacteria that are auxotrophic for histidine.

6. Segments of DNA capable of moving from one area in the DNA to another are called
A. base analogs.
B. intercalating agents.
C. transposons.

7. Transposons
A. are informally known as jumping genes.
B. may cause insertion mutations.
C. may cause knockout mutations.
D. were first recognized in plants.
E. All of the choices are correct.

8. Chemical mutagens often act by altering the
A. alkyl groups of the nucleobase.
B. nucleobase sequence.
C. number of binding sites on the nucleobase.
D. hydrogen bonding properties of the nucleobase.

9. The largest group of chemical mutagens consists of
A. radiation.
B. base analogs.
C. nitrous acid.
D. alkylating agents.

10. Chemical mutagens that mimic the naturally occurring bases are called
A. nitrogen mustards.
B. alkylating agents.
C. base analogs.
D. nitrous oxide.

11. Planar molecules used as chemical mutagens are called
A. nitrous oxide.
B. base analogs.
C. alkylating agents.
D. intercalating agents.

12. Intercalating agents
A. act during DNA synthesis.
B. often result in frameshift mutations.

C. only act in dormant cells.

D. alter the hydrogen bonding properties of the bases.
E. act during DNA synthesis AND often result in frameshift mutations.

13. Irradiation of cells with ultraviolet light may cause
A. four nucleotides to covalently bind together.

B. thymine dimers.
C. adenine complementary base pairing with cytosine.
D. the addition of uracil.

14. On which of the following DNA strands would UV radiation have the most effect?
A. AACCGGG
B. TATATACG
C. AUAUCGAU
D. AATTAGTTC
E. TATATACG AND AATTAGTTC

15. Thymine dimers are dealt with by
A. no repair mechanisms.
B. photoreactivation repair.
C. SOS repair.
D. excision repair.
E. photoreactivation repair AND excision repair.

16. The formation of a covalent bond between two adjacent thymines is caused by
A. mustard gas.
B. alkylating agents.
C. microwave radiation.
D. UV radiation.

17. X-rays
A. have no effect on DNA.
B. cause thymine trimers.
C. cause single and double strand breaks in DNA molecules.
D. make the DNA radioactive.

18. DNA repair mechanisms occur
A. only in prokaryotes.
B. only in eukaryotes.
C. in both eukaryotes and prokaryotes.
D. in neither eukaryotes or prokaryotes.

19. Which is not true about mismatch repair?
A. It utilizes an endonuclease.
B. It requires DNA polymerase and DNA ligase.
C. It utilizes the state of methylation of the DNA to differentiate between strands.
D. It removes both strands in the mismatch area.

20. Antibiotics
A. cause mutations to occur.
B. may act as alkylating mutagens.
C. provide an environment in which preexisting mutants survive.

21. Prokaryotic cell mutations can be observed very quickly because the prokaryotic chromosome is
A. diploid.
B. polyploid.
C. haploid.

22. The diploid character of eukaryotic cells may mask the appearance of a mutation since
A. this may be a frameshift.

B. the mutation is often reversible.
C. The mutation may create inverted repeats.

D. the matching chromosome may carry the dominant gene.

23. Direct selection involves inoculating cells onto growth media on which
A. the mutant but not the parental cell type will grow.
B. the mutation will be reversed.
C. the nutrients necessary for mutation to occur are present.
D. the mutagen is present.

24. Among the easiest of the mutations to isolate are those which
A. involve haploid chromosomes.
B. involve antibiotic resistance.
C. allow populations to be measured.
D. use an indirect method for measurement.
E. involve haploid chromosomes AND involve antibiotic resistance.

25. Indirect selection
A. is necessary to isolate auxotrophic mutants.
B. uses media on which the mutant but not the parental cell type will grow.
C. uses media that reverses the mutation.
D. uses media upon which neither the parental cell typeor mutant grows.
E. is necessary to isolate auxotrophic mutants AND uses media upon which neither the parental cell typeor mutant grows.

26. Replica plating
A. is useful for direct selection.
B. is useful for identifying auxotrophs.
C. uses media on which the mutant will not grow and the parental cell type will.
D. is used to store strains of bacteria.
E. is useful for identifying auxotrophs AND uses media on which the mutant will not grow and the parental cell type will.

27. A clever technique that streamlines the identification of auxotrophic mutants is
A. gas chromatography.
B. replica plating.
C. direct selection.
D. reversion.

To increase the proportion of auxotrophic mutants in a population of bacteria, one may use

A. direct selection.
B. replica plating.
C. penicillin enrichment.
D. individual transfer.

29. A quick microbiological test for potential carcinogens was developed by
A. Fleming.
B. Lederberg.
C. Ames.
D. Crick.

To increase the chance of detecting carcinogens in the Ames test, the test substance is treated with

A. penicillin.
B. heat.
C. ground-up rat liver.

D. reverse transcriptase.
E. penicillin AND heat.

31. The Ames test is useful as a rapid screening test to identify those compounds that
A. will respond to chemical agents.
B. are mutagens.
C. respond to the deletion of DNases.
D. will protect an organism from cancer.
E. will respond to chemical agents AND will protect an organism from cancer.

32. Bacteria that have properties of both the donor and recipient cells are the result of
A. UV light.
B. SOS repair.
C. frame shift mutations.
D. genetic recombination.

33. The mechanism by which genes are transferred into bacteria via viruses is called
A. ellipsis.
B. replica plating.
C. transformation.
D. transduction.
E. conjugation.

34. In conjugation the donor cell is recognized by the presence of
A. an F plasmid.
B. a Y chromosome.
C. diploid chromosomes.
D. an SOS response.
E. an F plasmid AND diploid chromosomes.

35. The F plasmid carries the information for
A. antibiotic resistance.
B. recipient cell DNA replication.
C. the Y chromosome.
D. the sex pilus.
E. antibiotic resistance AND the Y chromosome.

36. Competent cells
A. are able to take up naked DNA.
B. are antibiotic resistant.
C. occur naturally.
D. can be created in the laboratory.
E. are able to take up naked DNA, occur naturally, AND can be created in the laboratory.

37. The material responsible for transformation was shown to be DNA by
A. Watson and Crick.
B. Avery, MacLeod, and McCarty.

C. Lederberg.

D. Stanley.

38. In conjugation, transformation, or transduction, the recipient bacteria is most likely to accept donor DNA
A. from any source.
B. from any species of bacteria.
C. from the same species of bacteria.
D. only through plasmids.
E. from any source AND only through plasmids.

39. Gene transfer that requires cell-to-cell contact is
A. transformation.
B. competency.
C. conjugation.
D. functional genomics.

40. Insertion sequences
A. are the simplest type of transposon.
B. code for a transposase enzyme.
C. are characterized by an inverted repeat.
D. can produce pili.
E. are the simplest type of transposon, code for a transposase enzyme, AND are characterized by an inverted repeat.

41. The transfer of vancomycin resistance from Enterococcus faecalis to Staphylococcus aureus is thought to have involved
A. conjugation.
B. transformation.
C. transduction.
D. transposons.
E. conjugation AND transposons.

42. Which is not true about a crown gall tumor?
A. It is a bacterial infection of plants.
B. It requires a plasmid.
C. It produces a large amount of opines that neither the plant nor bacteria synthesizes.
D. It is due to the incorporation of bacterial plasmid DNA into the plant chromosome.
E. All of the choices are true.

43. The study of the crown gall tumor found
A. a bacterial plasmid promoter that was similar to plant promoters.
B. an R plasmid.
C. incorporation of the bacterial chromosome into the plant.
D. incorporation of the plant chromosome into the bacteria.

Is it as effective to take two antibiotics sequentially for an infection as it is to take them simultaneously, so long as the total length of time of the treatment is the same?

A. No. There's always one specific antibiotic that will be the most effective, and that is the only antibiotic that should be used to treat a particular infection.
B.

Yes. So long as the length of time is the same, the two treatments should be essentially the same in terms of effectively eliminating the infection.

C. No. Taken sequentially, the first antibiotic will select for the small portion of the population that will spontaneously mutate towards resistance. Then, the second antibiotic will do the exact same thing-selecting for resistance to the second drug from the few bacterial cells that remained from the first drug treatment.

D. It depends. Provided that the majority of the infectious agent is killed off by the first drug, the likelihood that the few that are left would not also be killed by the second drug is low. However, simultaneous treatment should be more effective at eliminating all the microbes in the shortest time possible, and with the least probability of selection for multiple drug resistance mutations.

55. Strong chemical mutagens may be used to treat cancer cells. Is this a good or bad idea?
A. Good-kill those cancer cells as quickly as possible to cure the patient!
B. Bad-these mutagens will also affect the non-cancerous cells, possibly leading to new cancerous states!
C. Good and bad-they're very good at killing cancer cells, but depending on mode of administration, they could also be dangerous to non-cancerous cells.
D. Bad-the cancer cells are already mutated. We don't want to mutate them more and make them more cancerous!

56. Every 24 hours, every genome in every cell of the human body is damaged 10,000 times or more. Given the possible DNA repair mechanisms, which order listed below would be most effective at repairing these as quickly as possible in order to prevent mutations from being carried forward in DNA replication?
A. Proofreading by DNA polymerase, glycosylase enzyme activities, excision repair, SOS repair

B. SOS repair, excision repair, glycosylase enzyme activities, proofreading by DNA polymerase

C. SOS repair, proofreading by DNA polymerase, glycosylase enzyme activities, excision repair

D. Glycosylase enzyme activities, SOS repair, proofreading by DNA polymerase, excision repair

To maximize the number of thymine dimer mutations following UV exposure, should you keep human cells in tissue culture in the dark, in the light, or does it matter at all?

A. The dark-light will activate the photorepair systems that can break thymine dimers induced by UV light.
B. The light-it's important to keep on producing the thymine dimers by keeping the plate exposed to light as much as possible.
C. It's best to alternate light and dark every hour to increase the chances that thymine dimers will form in the human cells, but still keep the photorepair systems from correcting them as they are formed.

D. It doesn't matter-human cells don't possess the enzymes needed for photorepair of thymine dimers.

58. Two bacterial genes are transduced simultaneously. What does this suggest about their proximity to each other within the original host genome?
A. Not a thing-it's highly likely that two separate virus particles were carrying each gene, and that they coinfected the new target cell at the same time, delivering their genetic payloads. This could mean the two original genes might not even be from the same original host cell!

B. It's highly likely that the two genes are located next to each other in the original host cell chromosome. Since transduction relies on either mispackaging of bits of host cell DNA into non-functional virus units, or improper excision of lysogenic phage DNA from a host cell chromosome (carrying parts of the host cell DNA with it), the genes must lie close to each other to be transduced into a new cell simultaneously.

C. They must be within five gene lengths of each other, but not necessarily immediately adjacent. If they were immediately adjacent, the transposons that facilitate the transfer of genetic information between the two cells wouldn't be able to 'jump' into them.

D. It doesn't mean anything. Transduction relies on the ability of a cell to take up foreign DNA. It's possible here that the cell has simply taken up two separate bits of DNA at the same time from the surrounding environment.

59. DNA transfer by conjugation is more efficient in a liquid medium setting, subjected to very mild agitation (stirring), rather than on an agar plate format. Why?
A. Direct cell-to-cell contact is required for this process, and this is more likely to be achieved in the plate format than in the fluid format (especially for relatively non-motile types of bacteria).
B. Direct cell-to-cell contact isn't required for this process, so the ability to secrete the DNA into the surrounding fluid medium makes the process more efficient than the dry surface of an agar plate.
C. Direct cell-to-cell contact is required for this process, and this is more likely to be achieved in the fluid liquid format than on an agar plate (especially for relatively non-motile types of bacteria).
D. Trick question-it can take place with the same degree of efficiency on either format. It doesn't matter!

60. Some bacteria have a higher incidence rate of thymine dimer mutations following exposure to UV light than others. What might be going on here to lead to this outcome?
A. They may simply have a higher proportion of T nucleotides next to each other in their DNA sequences than other bacteria, leading to more possible dimers being formed.
B. They may have a stronger expression of photoreactivation enzymes, leading to more thymine dimers being formed and retained.
C. They may have a weaker expression of photoreactivation enzymes, leading to more thymine dimers being formed and retained.
D. They may simply have a higher proportion of T nucleotides next to each other in their DNA sequences than other bacteria, leading to more possible dimers being formed AND they may have a stronger expression of photoreactivation enzymes, leading to more thymine dimers being formed and retained.
E. They may simply have a higher proportion of T nucleotides next to each other in their DNA sequences than other bacteria, leading to more possible dimers being formed AND they may have a weaker expression of photoreactivation enzymes, leading to more thymine dimers being formed and retained.