Lecture 46 Flashcards

1
Q

Some of the key differences between bacterial and eukaryotic cells are that bacterial cells do not have _____ (the protein around which eukaryotic DNA is wrapped/condensed), and bacterial cells simultaneously carry out transcription and translation (should make sense since there is no nucleus to isolate transcription process).

A

Histones

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2
Q

Bacterial genes are typically organized into ______, a transcriptional unit with multiple genes. This is not true for eukaryotic genes which are organized into singletons. Also, bacterial genes do not typically contain ______, so the transcript requires little or no processing before translantion –> another reason these processes can occur simultaneously in bacterial cells.

A

Operons

Introns

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3
Q

Most bacterial have a single circular dsDNA molecule and plasmids. However, this is not always the case. For example, ______ ______ (causes Lyme disease) has a single _____ dsDNA chromosome, and _____ _____ (causes cholera) has TWO circular dsDNA chromosomes.

A

Borrelia burgdoferi

Vibrio cholerae

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4
Q

Bacterial DNA (B-DNA) typically has about ____ bp per turn in the relaxed (minimal energy) state. However, the stress of under or over-winding causes supercoiling. Is positively or negatively supercoiled DNA required for transcription?

A

about 10 bp/turn (10.6)

Negatively (underwound) supercoiled is required for transcription.

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5
Q

Topoisomerase ___ (aka gyrase) is the enzyme responsible for negatively supercoiling B-DNA, while Topoisomerase ___ is responsible for positively supercoiling. An indication that a balance between these two is necessary in bacteria is if you introduce a point mutation in one of the enzymes, the other will acquire a compensatory mutation.

A

Topoisomerase II

Topoisomerase I

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6
Q

______ is the class of antibiotics that acts on Topoisomerase II –> positive supercoiling cannot be undone –> B-DNA replication cannot continue

A

Quinolones

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7
Q

The main DNA polymerase in B-DNA replication is DNA Pol ___. The one that fills in the gaps in the lagging strand and is involved in DNA damage repair is DNA Pol __.

A

Pol III

Pol I

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8
Q

What kind of supercoiling does the action of helicase induce, and what enzyme must act upstream to compensate?

A

Positive supercoiling is induced, so Topoisomerase II must act upstream to compensate.

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9
Q

Bacterial DNA can acquire genetic variations through both homologous and non-homologous recombination. E. coli require a minimum _____ bp sequence of homology for homologous recombination.

A

300bp

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10
Q

There are two types of bacterial transposable gene elements (TGEs). the first, an _____ _____, is a single 15-40bp reading frame that codes for transposase and is flanked by inverted repeat sequences (IR-left and IR-right). The second, a ______, is typically larger, about 40bp, and contains basically the same structure but with multiple reading frames that typically include a gene conferring antibiotic resistance.

A

Insertion Sequence

Transposon

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11
Q

Bacterial cells will always have only one type of flagellar filament. Salmonella flagellar phase variation is a good example of how TGEs cause knock out of gene expression, and thus a variation in phenotype. Normally, the flj__ reading frame (Phase 2/type B flagella) is transcribed along with the flj___ reading frame (translational repressor of flj__). However, the ___ segement of the operon contains the promoter for the flj__ and flj__ genes, and this segment can be inverted –> if it is, the flj__ gene does not produce the Phase 2 flagella, and he flj__ gene does not produce the repressor for the flj__ gene, which codes for the Phase 1/Type C flagella. This flagella phase variation provides a mechanism for evading the host immune response that had initially recognized the Phase 2/Type B flagella.

A

fljB

fljA

fljC

H segment (hin gene)

fljB and fljA

fljB

fljA

fljC

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12
Q

Griffith’s classic experiment in 1928, using mice and Pneumococcal smooth (S) and rough (R) strains, showed horizontal gene transfer. How did this experiment work?

A

Mouse was injected with R-strain –> mouse lived

Mouse injected with S-strain –> mouse died

Mouse injected with heat-killed S-strain –> mouse lived

Mouse injected with heat-killed S-strain + live R-strain –> mouse died.

So it must be the case that the R-strain acquired virulence from the heat-killed S-strain.

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