Chapter 9

Base excision repair is particularly useful for removing bases from DNA that have spontaneously BLANK 2. During base excision repair, a specific glycosylase catalyses Without base excision repair, the presence of unnatural bases may cause mutations or more serious effects. For example, the presence of interferes with DNA synthesis, and is thus lethal to the cell.

1. deaminated 2. formation of AP site 3. 3-methyladenine

Transposable elements differ from plasmids in that only plasmids A. may exist autonomously, not integrated into host DNA. B. consist of double-stranded DNA .C. may code for antibiotic resistance. D. may be transmitted horizontally from one bacterium into another.

. may exist autonomously, not integrated into host DNA.

Association of RecA with an incoming piece of foreign DNA is required for a cell to do which of the following? A. swap the incoming DNA with a homologous region on the chromosome B. transcribe the DNA into mRNA C. metabolize the DNA as a food source D. maintain the DNA as a plasmid E. degrade one strand of a dsDNA molecule

. swap the incoming DNA with a homologous region on the chromosome

A strain of Mycobacterium tuberculosis that is deficient in nonhomologous end joining repair should be more sensitive to damage by

X rays

One sign of horizontal gene transfer is A. uniform codon usage throughout the genome. B. a GC base ratio different from flanking chromosomal DNA. C. pathogenicity via gene loss. D. gene duplication.

a GC base ratio different from flanking chromosomal DNA.

Paralogs arise within a genome by which process? integration of prophages acquisition of pseudogenes divergence of a duplicated gene gain of genomic islands

divergence of a duplicated gene

A paralog is a __________ gene in __________ species, while an ortholog is a duplicated gene in a __________ species.

duplicated; one; two

What type of mutation is most likely to occur as a result of nonhomologous end joining repair? A. frameshift mutations B. large-scale inversions C. transpositions D. single base-pair mismatches E. pyrimidine dimers

frameshift mutations

Genes that share a common ancestry are broadly referred to as . They are further classified as if they exist within the same species but have distinct functions, while usually have similar functions but are found within different species.

homologs, paralogs, orthologs

1.Many Japanese people consume a diet rich in seaweed, including the edible red alga Porphyra, which is used for preparing sushi. Although humans cannot digest the seaweed polysaccharides (porphyran and agarose), certain marine Bacteroidetes do possess the necessary CAZymes. Curiously, Japanese individuals frequently harbor seaweed-digesting Bacteroides plebeius in their gut microbiomes, while individuals from North America do not. Bacteroides plebeius is not a marine bacterium, and its close relatives cannot digest seaweed. What is the mostly likely explanation for how B. plebeius acquired functional porphyranase and agarase genes? 2.Further analyses revealed that a region downstream of the seaweed-digesting genes in Bacteroides plebeius contained sequences homologous to conserved relaxase/mobilization proteins. These data provide additional evidence that the genes were acquired by...

horizontal transfer of genes from a marine bacterium 2. conjugation

Genes can be lost during genomic evolution through A. mutation. B. transposons leaving and taking the genes with them. C. duplication and divergence. D. lack of repair to the gene when it is no longer needed.

lack of repair to the gene when it is no longer needed.

1. DNA mismatch repair is the primary mechanism for correcting misinserted base pairs caused by 2.For the mismatch repair system to work correctly, newly replicated DNA must 3. An E. coli strain with a mutation rate of about 10-10 per base pair replicated would be considered a BLANK strain.

1. DNA POLYMERASE 2. remain unmethylated long enough for mismatches to be repaired. become methylated before the subsequent round of replication. 3. normal

1. How do Phaeobacter compete for limited resources in the coastal salt marshes? A. Phaeobacter degrade plant material, making it unavailable to other microbes. B. Phaeobacter secrete the compound indigoidine that kills other bacteria. C. Phaeobacter secrete the compound penicillin that kills other bacteria. D. Phaeobacter can grow more quickly than other bacteria. 2. ILLUSTRATION Alison Buchan and her team used transposon mutagenesis to create a large pool of mutant Phaeobacter. Three of these mutants are shown in the figure. Why did culture II not have a zone of clearance around it?

1. Phaeobacter secrete the compound indigoidine that kills other bacteria. . A transposon in this culture interrupted a Phaeobacter gene required for indigoidine synthesis, thereby allowing the V. fischeri to grow close.

Which of the following is an example of horizontal transmission? A. Cell division B. PCR C. Mitosis D. Transformation

Transformation

An insertion sequence contains a gene for which enzyme?

C. Transposase

1. Correctly order the following steps in bacterial conjugation from start to finish. 2.Upon successful completion of conjugation, the donor cell is and the recipient cell is .3.Why is it incorrect to refer to conjugation as bacterial sex?

1. sex pillus informed that joins an F+ cell to an F- cell sex pilus retracts and a relaxosome forms the F factor is nicked at oriT, and one strand starts to transfer to recipient DNA polymerase 3 synthesizes a replacement strand in the donor transferred DNA strand circularizes and competes replication 2. F+,F+ 3. it does not involve gametes. .The recipient is haploid for the transferred genes. The transferred DNA is not required for normal cellular function.

The phylogenetic tree of Prochlorococcus suggests that early lineages lived deeper in the euphotic zone below the limit of UV light penetration. What DNA repair adaptation currently allows Prochlorococcus cells to live in the upper photic zone? A. nucleotide excisionB. recombinational repairC. methyl mismatch repairD. photolyases How did Prochlorococcus now living in the upper euphotic zone acquire the ability to repair their UV-damaged DNA? A. Prochlorococcus directly acquired the photolyase enzyme from other nearby microbes. B. Prochlorococcus themselves slowly evolved a unique ability to repair DNA over time. C. Prochlorococcus form symbiotic relationships with other microbes that can repair damaged DNA. D. Prochlorococcus likely acquired the photolyase genes through horizontal gene transfer.

1.photolyases 2. Prochlorococcus likely acquired the photolyase genes through horizontal gene transfer.

Pathogenicity islands (PAIs) are often found integrated near A. mRNA genes. B. rRNA genes. C. tRNA genes. D. PAIs are not found integrated near rRNA, mRNA, or tRNA genes.

C. tRNA genes.

What two conditions must be met to produce a heritable mutation?

C.There is a change in the DNA sequence. D.The mutation does not get repaired before the cell divides.

Select the true statement regarding the relationship between genotype and phenotype. A. The phenotype reflects only the portion of the genotype that is expressed. B. Only genes that produce proteins are considered part of an organism's genotype. C. The phenotype represents the entire genome of an organism. D. Some mutations will cause a change in the genotype, while others will not. E. Any change in the genotype will be reflected in the phenotype.

A. The phenotype reflects only the portion of the genotype that is expressed.

In the basic Ames test for mutagenesis, a mutagen is tested to see if it can produce colonies on basic medium that ______ histidine, starting with a _____ strain of bacteria. A. lacks; hisG mutant B. includes; hisG mutant C. lacks; wild-type D. includes; wild-type

A. lacks; hisG mutant

ILLUSTRATION Consider the complex transposon in the above illustration. Suppose you discover a derivative of this element that replicatively inserts into target DNA but then becomes "stuck" as part of a cointegrate. Your graduate student sequences the respective gene products and finds no change in the primary amino acid sequences of the encoded enzymes. In which locus must the defect lie?

A. res

ILLLUSTRATION Based on this gene alignment, predict which of the following descriptions most likely apply to the genes in red. A.They likely encode virulence factors. B.They were likely horizontally transferred. C.They likely encode symbiosis genes. D.They are likely a stable part of the chromosome. E.They likely encode housekeeping genes.

A.They likely encode virulence factors .B.They were likely horizontally transferred.

In bacteria, a partially diploid strain may result from A.specialized transduction. B.acquisition of an F' factor. C.generalized transduction. D.natural transformation (not electroporation). E.acquisition of F factor.

A.specialized transduction. B.acquisition of an F' factor.

ILLUSTRATION Please examine and correctly label this image illustrating the two basic types of transposition.

AReplicative transposition BDuplication caused by previous insertion CTarget sequence DNonreplicative transposition ETransposable element FRecipient DNA

The early stages of genome reduction in bacteria are normally characterized by the proliferation of pseudogenes. These most likely arise as a result of A. transforming DNA. B. frameshift and nonsense mutations. C. transduced DNA.D. conjugative plasmids. E. transposons.

B. frameshift and nonsense mutations.

Over time, the genome of a species A. rapidly acquires harmful mutations. B. may change due to mutation or gene exchange. C. remains unchanged D. doubles or triples to produce polyploidy genomes

B. may change due to mutation or gene exchange.

ILLUSTRATION Examine and correctly label the following illustration of a bacterial restriction and modification system.

Bacterial chromosome Bacteriophage Phage DNA Methylation sites Restriction enzyme

Which of the following enzymes is used in methyl mismatch repair, nucleotide excision repair, and base excision repair? A. DNA polymerase I B. UvrA C. Photolyase D. MutS

DNA polymerase I

ILLUSTRATION : What type of DNA damage is shown below? A. Pyrimidine dimer B. Depurination C. Inversion D. Deamination of cytosine

Depurination

Which of the following does NOT cause mutations? A. Exposure to antibiotics B. UV irradiation C. Chemical mutagens D. Spontaneous mutations during DNA replication

Exposure to antibiotics

ILLUSTRATION Please examine and correctly label the following model of horizontal gene transfer among the gut microbiota.

Phage released Transformation Transduction Conjugation Exit from lysogeny

If a cytosine deaminates and becomes a uracil, which enzyme will cleave the uracil from the DNA backbone in the first step of a repair process? A. Glycosylase B. MutH C. RecA D. AP endonuclease

Glycosylase

CRISPR-Cas9 may be used as a gene-editing tool to repair mutant genes (such as the gene that causes Huntington's disease in humans). This technique has been used in bacteria, mice, and even human embryos. First, CRISPR-Cas9 is used to target and remove the mutant gene. Second, naturally occurring repair enzymes will add the wild-type gene into the gap left from the site where the mutant gene was removed. Which of these bacterial repair mechanisms would most likely be responsible for repairing the DNA when used to edit bacterial genes? A. UmuDC translesion bypass synthesis in SOS B. methyl mismatch repair C. base excision D. RecA recombination repair E. nucleotide excision

RecA recombination repair

Which of the following DNA repair mechanisms is error prone? A. Methyl mismatch repair B. SOS repair C. Photoreactivation D. Nucleotide excision repair

SOS repair

ILLUSTRATION: Please correctly label the functions of the proteins that participate in the SOS response in E. coli.

SOS repressor Translesion bypass replication Inhibits cell division Coprotease Nucleotide excision repair

ILLUSTRATION Consider the reaction shown in the above illustration. This event is a potential source of mutations and results from exposure to A. X-rays. B. reactive oxygen species. C. base analogs. D. UV light .E. acridine orange.

UV light

The enzyme photolyase repairs DNA damage caused by A. alkylating agents. B. deaminating agents. C. denaturation. D. UV radiation.

UV radiation

Which of the following describes a change in an organism's genotype (rather than its phenotype)? A. loss of polar flagella by Vibrio cholerae during colonization of its host B. lack of red pigmentation by Serratia marcescens colonies on nutrient agar C. increased generation time of Salmonella at 4°C vs. 37°C D. initiation of endospore formation by Bacillus subtilis during nutrient limitation E. acquisition by Corynebacterium of a prophage that encodes diphtheria toxin

acquisition by Corynebacterium of a prophage that encodes diphtheria toxin

The retractable type IV pilus expressed by the marine bacterium Vibrio cholerae allows uptake of what? any single-stranded DNA any free double-stranded DNAspecies-specific DNA only conjugative plasmids membrane vesicles containing DNA

any free double-stranded DNA

CRISPR uses pieces of phage DNA incorporated into the genome A. as a replicative transposon site. B. as the source of an RNA that will guide Cas proteins to cleave the foreign DNA. C. to direct specialized transduction. D. to direct restriction enzyme activity to cleave foreign DNA with homologous sequences.

as the source of an RNA that will guide Cas proteins to cleave the foreign DNA.

A mutation always results in A. a change in phenotype. B. a change in genotype. C. beneficial effects to the organism. D. harmful effects to the organism.

change in genotype

Which type of DNA uptake is dependent on transferable plasmids?

conjugation

Based on gel electrophoresis, a nonfunctional protein is found to have a smaller molecular weight than its wild-type counterpart. A likely explanation for this observation is a A. nonsense mutation in the DNA coding for the protein. B. duplication of the region of the DNA encoding the protein. C. missense mutation in the DNA coding for the protein. D. silent mutation in the DNA encoding for the protein.

nonsense mutation in the DNA coding for the protein.

illustration Consider the data from a hypothetical Ames test below. Then match the following chemicals to the best description of their mutagenicity: mutagenic to bacteria only, mutagenic to both humans and bacteria, or not mutagenic at all. monosodium glutamate (MSG) caffeine tobacco smoke the fatty acid 4-hydroxyhexenal

not mutagenic : monosodium glutamate (MSG) Bacteria only: caffeine humans and bacteria: tobacco smoke the fatty acid 4-hydroxyhexenal

Bacteria may donate DNA to other bacteria of the same A. or of a different species, some eukaryotic cells, and bacteriophages. B. or of a different species. C. or of a different species, and also some eukaryotic cells .D. species only.

or of a different species, some eukaryotic cells, and bacteriophages.

While horizontal gene transfer can be beneficial, there are also risks. Bacteria protect themselves from foreign DNA through A.DNA repair mechanisms protecting the genome. B. recombination. C. restriction endonucleases that cleave unmethylated DNA at specific sites. D. transposable elements disrupting any transferred DNA.

restriction endonucleases that cleave unmethylated DNA at specific sites.

V. cholerae is capable of transformation, which means it can do what? . take up DNA through conjugation with F-factor plasmids B. take up naked DNA C. transfer DNA through nanotubes D. take up DNA through transduction with phages 2. Label the diagram of naked DNA uptake by V. cholerae. dsDNA unwinds; one strand enters cytoplasm; other strand is degraded. BComEA binds dsDNA and pulls it through the pilus. CRecA recruits ssDNA to homologous site for integration. DPilus binds dsDNA and retracts.

take up naked DNA

Which type of DNA uptake is dependent on viruses?

transduction

Consider the descriptions of genetic exchange methods below and match each description to the proper type of exchange. This method would be affected by DNase, an enzyme that can degrade naked DNA that is unprotected. This method of DNA exchange cannot occur using the recipient cells mixed with cell-free extract only. This method can occur with cell-free extract only and is not affected by DNase.

transduction: This method can occur with cell-free extract only and is not affected by DNase. conjugation: This method of DNA exchange cannot occur using the recipient cells mixed with cell-free extract only. transformation : This method would be affected by DNase, an enzyme that can degrade naked DNA that is unprotected.

The process of importing free DNA from the environment into cells is called

transformation

What are the basic parts of an insertion sequence? A.ligase gene B.transposase coding region C.random DNA of 100-200 bp D.two flanking inverted repeats E.target sequence F.coding region for polymerase Sort the following events according to whether they occur in replicative transposition, nonreplicative transposition, or both: Donor DNA retains a copy of insertion sequence. Recombination separates donor and target DNAs. Hairpin structures form on each end of insertion sequence. Insertion sequence "jumps" into recipient DNA. Donor DNA loses insertion sequence. DNA polymerase fills in gaps. Transposase is required. Target sequence gets duplicated. DNA ligase seals nicks. Transposases are multifunctional enzymes. During nonreplicative transposition, the transposase does all of the following except

transposasae coding region and two flanking inverted repeats Both: DNA polymerase fills in gaps. Transposase is required. Target sequence gets duplicated. DNA ligase seals nicks. Nonreplicative : Hairpin structures form on each end of insertion sequence. Insertion sequence "jumps" into recipient DNA. Donor DNA loses insertion sequence. Replicative: Donor DNA retains a copy of insertion sequence. Recombination separates donor and target DNAs. fill in single-stranded gaps that occur in the target region of the recipient DNA.

The above illustration shows an example of a(n) mutation that results in a(n) mutation. GOES TO TGA WHICH IS STOP CODON

transversion nonsense

Gene transfer agents, when viewed using electron microscopy, could easily be mistaken for membrane vesicles. viruses .naked DNA .sex pili. nanotubes.

viruses