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LF206: Molecular Cell Biology > Lecture 1: Bacterial Genomes > Flashcards

Lecture 1: Bacterial Genomes Flashcards

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1
Q

What is the traditional view of a prokaryotic genome?

A

Single, circular DNA molecule localized within the nucleoid

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2
Q

What is the nucleoid?

A

A lightly staining region of the cell where the circular DNA molecule in prokaryotic cells is localized.

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3
Q

What are the characteristics of the B form of DNA?

A
  • 2 polynucleotide chains in opposite orientation
  • Regular right handed double helix
  • 2nm diameter, makes complete turn every 3.4nm
  • ~10.5 bp / turn of helix
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4
Q

What flexibilities are present within the basic B-form of DNA?

A
  • Numbers of base pairs per turn of the helix can be altered
  • Helix in cell is not straight but coiled in 3D space
  • Certain sequence features where bends occur
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5
Q

What are the characteristics of DNA in the B form?

A
  • It consists of two chains in opposite orientation arranged in a right-handed double helix.
  • The diameter is approximately 2 nm, and it makes a complete turn every 3.4 nm.
  • There are approximately 10.5 base pairs per turn of the helix.
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6
Q

What is supercoiling in DNA, and how does it occur?

A
  • Occurs when additional turns are introduced into the DNA double helix (positive supercoiling) or if turns are removed (negative supercoiling)
  • In the circular genome of E. coli, DNA is supercoiled, which helps compact the DNA and regulate gene expression.
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7
Q

How is torsional stress accommodated in DNA molecules?

A
  • Torsional stress in DNA molecules is accommodated in two ways:
    • Formation of superhelices
    • Altering the number of base pairs per turn of the helix
  • These responses are expressed by the linking number (L), which represents the total number of times that the two strands of the double helix of a closed molecule cross each other when constrained to lie in a plan
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8
Q

What are Type I topoisomerases and how do they function?

A
  • Type I topoisomerases break one strand of DNA, pass the other strand through the gap, and then seal the break.
  • This process changes the linking number by ±1.
  • An example is Topoisomerase I of E. coli, which relaxes negatively supercoiled DNA.
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9
Q
A
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10
Q

What is the nucleoid?

A

A lightly staining region of the cell where the circular DNA molecule in prokaryotic cells is localized.

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11
Q

What are the characteristics of the B form of DNA?

A
  • 2 polynucleotide chains in opposite orientation
  • Regular right handed double helix
  • 2nm diameter, makes complete turn every 3.4nm
  • ~10.5 bp / turn of helix
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12
Q

What flexibilities are present within the basic B-form of DNA?

A
  • Numbers of base pairs per turn of the helix can be altered
  • Helix in cell is not straight but coiled in 3D space
  • Certain sequence features where bends occur
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13
Q

What are the characteristics of DNA in the B form?

A
  • It consists of two chains in opposite orientation arranged in a right-handed double helix.
  • The diameter is approximately 2 nm, and it makes a complete turn every 3.4 nm.
  • There are approximately 10.5 base pairs per turn of the helix.
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14
Q

What is supercoiling in DNA, and how does it occur?

A
  • Occurs when additional turns are introduced into the DNA double helix (positive supercoiling) or if turns are removed (negative supercoiling)
  • In the circular genome of E. coli, DNA is supercoiled, which helps compact the DNA and regulate gene expression.
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15
Q

How is torsional stress accommodated in DNA molecules?

A
  • Torsional stress in DNA molecules is accommodated in two ways:
    • Formation of superhelices
    • Altering the number of base pairs per turn of the helix
  • These responses are expressed by the linking number (L), which represents the total number of times that the two strands of the double helix of a closed molecule cross each other when constrained to lie in a plan
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16
Q

What are Type I topoisomerases and how do they function?

A
  • Type I topoisomerases break one strand of DNA, pass the other strand through the gap, and then seal the break.
  • This process changes the linking number by ±1.
  • An example is Topoisomerase I of E. coli, which relaxes negatively supercoiled DNA.
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17
Q

What is the function of Type I topoisomerases in DNA topology?

A
  • Type II topoisomerases break both strands of the DNA, pass another part of the helix through the gap, and change the linking number by ±2.
  • An example is DNA gyrase of E. coli, which creates negative supercoils in DNA.
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18
Q

What is the role of Type II topoisomerases, specifically DNA gyrase of E. coli, in DNA topology?

A
  • Type II topoisomerases, such as DNA gyrase in E. coli, break both strands of the DNA, enabling another portion of the helix to pass through the gap, thereby altering the linking number by ±2.
  • DNA gyrase is a heterotetramer composed of two subunits, A and B.
  • It utilizes ATP to create negative supercoils in DNA.
  • DNA gyrase plays a crucial role in DNA replication by opening up the DNA strands, facilitating processes such as unwinding and supercoiling, which are essential for DNA replication to proceed efficiently.
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19
Q

How is DNA organised in bacteria?

A
  • In E. coli the single circular DNA molecule is organized into a series of supercoiled loops (40-50) that radiate from a central protein core
  • Highly organised structure
20
Q

What are the protein components of the E. coli nucleoid and their functions?

A
  • DNA gyrase and DNA topoisomerase I: maintain the supercoiled state of the DNA.
  • HU (heat unstable) proteins: most abundant proteins involved in packaging the DNA.
    • HU proteins form tetramers around which the DNA is wound, approximately in units of 60 base pairs, covering about 1/5 of the genome.
  • A single E. coli cell contains approximately 13,000 HU proteins.
  • Unlike E. coli, Archaea lack proteins related to HU and instead have proteins related to eukaryotic histones.
21
Q

What was the traditional view of prokaryotic genomes?

A

Single, circular DNA molecule, localized within the nucleoid (a lightly staining region of the cell)

22
Q

What variations exist in bacterial genomes regarding their structure?

A
  • Some bacteria possess linear genomes instead of circular ones.
    • Borrelia burgdorferi (causative agent of Lyme disease)
    • Streptomyces coelicolor (a producer of antibiotics)
    • Agrobacterium tumefaciens (causing plant tumors).
  • Others have multipartite genomes (their genomes are divided into two or more DNA molecules)
  • Plasmids are often small DNA molecules that typically carry non-essential genes, but some large plasmids may also contain essential genes
  • The structure of bacterial genomes represents a spectrum rather than a clear discontinuity
23
Q

What role does horizontal gene transfer (HGT) play in bacterial evolution, and what are some of its mechanisms?

A
  • HGT: transfer of genetic material between different organisms, rather than from parent to offspring
    • Important in acquiring new traits and adapting to changing environments
  • One mechanism of HGT involves prophages (phage genomes integrated into the bacterial chromosome)
  • Many bacterial genomes contain phage-like elements (i.e. cryptic prophages)
    • These prophages can contribute to pathogenesis and other important phenotypes by transferring genes encoding virulence factors, antibiotic resistance, or metabolic capabilities
24
Q

What are the roles of DNA polymerase III and DNA polymerase I in DNA replication, particularly regarding the leading and lagging strands?

A
  • DNA polymerase III (Pol III) is the primary enzyme responsible for synthesizing DNA during replication.
  • It synthesizes the leading strand continuously in the 5’ to 3’ direction, matching the template strand’s 3’ to 5’ direction.
  • On the lagging strand, DNA replication occurs discontinuously in the form of Okazaki fragments, which are short, synthesized DNA pieces.
  • DNA polymerase I (Pol I) primarily fills the gaps left between the Okazaki fragments, ensuring continuous DNA synthesis on the lagging strand.
  • Additionally, Pol I has a crucial role in removing RNA primers from the Okazaki fragments by its 5’ to 3’ exonuclease activity and replacing them with DNA nucleotides.
  • This process of gap filling and primer removal ensures the completion of both leading and lagging strands during DNA replication.
25
Q

What are the fundamental features linking replication to the cell cycle in bacteria and Archaea?

A
  • Initiation of replication commits the cell to a subsequent division.
  • Cell division cannot occur until the round of replication associated with a particular initiation has been completed.
  • Bacteria and Archaea have a single origin of replication (oriC), indicating a single replicon.
  • Replication in these organisms is bidirectional, leading to the formation of a theta structure.
  • Each replicon, represented by a DNA molecule or sequence with a functional origin of replication, must be replicated at least once per cell division cycle to ensure proper genetic inheritance and cell division.
26
Q

How has the origin of replication (oriC) in E. coli been characterized?

A
  • Through experimental methods.
  • DNA from E. coli is digested with a restriction enzyme and then ligated into a plasmid lacking an origin of replication.
  • The minimum region identified as functioning as an origin is 245 base pairs long and contains specific sequences:
    • 14 copies of the sequence GATC.
    • 4 or 5 copies of a 9-base-pair sequence located in the right-hand two-thirds of oriC.
    • 3 copies of an AT-rich 13-base-pair sequence found in the left-hand one-third of oriC
27
Q

How does initiation of DNA replication occur in E. coli?

A
  • Begins with the binding of approximately 20 monomers of DnaA to the 4 or 5 copies of a 9-base-pair sequence in the right-hand portion of oriC.
  • This binding forms the closed complex.
  • The open complex is formed when the 3 AT-rich 13-base-pair repeats in the left-hand portion of oriC melt.
  • DnaB helicase is then loaded onto the melted DNA with the assistance of DnaC.
  • ATP is hydrolyzed, and DnaC is released.
  • DnaB helicase unwinds the DNA bidirectionally with the help of Single Strand Binding protein (SSB) and DNA gyrase.
  • Primase synthesizes a primer RNA molecule on both strands, initiating replication.
28
Q

What controls whether a round of replication is initiated in E. coli?

A
  • Methylation of adenine residues within GATC motifs in OriC by the Dam methylase.
  • Replication will only initiate if all 14 copies of GATC in oriC are methylated.
  • After replication, the copies of GATC on the newly synthesized strand are not methylated.
  • Semi-conservative replication results in each new DNA molecule containing one old (methylated) and one new (unmethylated) strand.
  • New DNA molecules are hemi-methylated (”half”-methylated).
  • Hemi-methylated oriC DNA is sequestered at a membrane site and is not available for replication.
  • Re-methylation occurs about 1/3 of the way into the cell cycle.
29
Q

How does replication terminated in E.coli?

A
  • When DNA replication is completed
  • 2 replication forks which began at the origin (oriC), and moved in opposite directions away round the genome, approach one another
  • These forks fuse at the terminus region
30
Q

What is the terminus region?

A
  • A replication fork trap
  • Region opposite oriC w a series of DNA terminator (ter) sites that arrest (pause) fork progression
31
Q

Describe the nature of ter sites

A
  • Polar (will arrest a fork approaching from one direction but not the other)
    • Fork arrest results from inhibition of helicase mediated unwinding of the DNA
      duplex at the apex of the fork
32
Q

What is the role of Tus?

A

A terminator protein, must be bound to the ter site to halt the fork

33
Q

Why is there more than one ter site?

A
  • Acts as a backup to ensure any fork entering the terminus regions don’t exit
  • Ensures replication is terminated
  • The outer terminators are probably used only rarely.
  • It is not known how frequently fork fusion occurs at a specific terminator compared with a fork to the next one
34
Q

How is a fork arrested from only one direction?

A
  • The Tus protein binds to ter sites and then interacts with and halts a replication fork heading in ONE direction only
  • Fork arrest results from inhibition of helicase-mediated unwinding of the DNA duplex at the apex of the fork
35
Q

What is essential for the generation of genetic diversity and DNA repair, and what is the requirement for recombination to occur?

A
  • Homologous recombination is essential for the generation of genetic diversity and DNA repair.
  • For recombination to occur, the two molecules must have homologous regions of the order of 100-500 bp.
36
Q

What are the 2 widely accepted models for homologous recombinaNon and how does one model address the problem of the prior?

A
  • Holliday (no idea of how invasion occurs)
  • Meselson-Radding model (addresses problem of invasion)
37
Q

Describe the Meselson-Radding model

A
  • Cleavage: one strand is cleaved by an endonuclease
  • Chain displacement: DNA synthesis displaces a chain
  • Invasion: the single stranded chain invades a homologous double-stranded DNA molecule
    • Catalysed by RecA
  • Chain removal: the displaced chain is digested
  • Ligation – produces a Holliday junction
  • Branch migration - increases heteroduplex, catalyzed by RuvAB
  • Isomerization - the strands of the Holliday junction spontaneously cross and uncross, does not require catalysis
  • Resolution - The crossed strands of the Holliday junction cleaved by RuvC. Products depend on configuration of junction at cleavage
  • Unlike Holliday model outcome is be asymmetric
38
Q

List all the enzymes involved in homologous recombination and their function

A
  • RecA: catalyses strand invasion
  • RuvAB: catalyses branch migration (increases heteroduplex)
  • RuvC: cleaves the crossed strands of the Holliday junction
39
Q

What’s the main difference b/w Holliday model n the Medelson-Redding model?

A

Holliday model results in symmetric heteroduplex whilst Medelson-Radding model results in an asymmetric heteroduplex

40
Q

What is RecBCD and its 5 functions?

A
  • RecBCD is involved in ~99% of all recombination in E. coli. It has:
    • ssDNA exonuclease (5’→3’ and 3’→5’)
    • ssDNA endonuclease
    • dsDNA exonuclease
    • DNA-dependent ATPase
    • DNA helicase **(prefers blunt dsDNA ends)
41
Q

Describe the function of RecBCD in homologous recombination

A
  • RecBCD binds tightly to the end of a dsDNA substrate
    • Unwinds it using its helicase activity
    • Degrades both ssDNA strands using its dual 5’→3’ and 3’ → 5’ exonuclease activities
  • RecBCD moves in steps and each step is 23 bp long in a unique mechanism of action
  • Called the “quantum inchworm” (on handout)
  • RESULT
    • ssDNA tail with a 3’ end and RecA binds to the ssDNA tail, which invades RecA coated ssDNA drives strand invasion
42
Q

What is a chi (c) site and how many are found in E.coli?

A
  • 5’ GCTGGTGG 3’ -
  • Sites of recombination
  • 1009 found in E.coli genome
43
Q

Give an example of how inversion sequences control gene expression

A
  • Phase variation in Salmonella spp
    • Most Salmonella spp. can produce two different types of flagellum
      • Phase 1
      • Phase 2 H antigens
    • Both are expressed n will always hv a mix
    • Phase is determined by the orientation of the hin (H-inversion) region
  • Hin region
    • Is 995 bp long and bounded by two 14 bp inverted repeats
    • Encodes for an invertase which catalyzes inversion
44
Q

What is the structure of H1 n H2 genes?

A
  • The H1 gene: own promoter and operator and is physically separated from the hin region
  • The H2 gene: operon with the rep gene that encodes a repressor for the H1 gene
    • The promoter for the H2-rep operon lies within the hin region
45
Q

Describe the expression patterns and regulatory mechanisms during Phase 1 and Phase 2 of the Hin-mediated inversion system.

A
  • Phase 1:
    • H1 is expressed due to inversion.
    • When the Hin region is inverted, the H2 promoter isn’t in the correct orientation, so Repressor (Rep) is not produced.
    • H1 isn’t repressed in this phase.
  • Phase 2 (H2 expressed, H1 repressed - no inversion):
    • H2 is expressed, and H1 is repressed.
    • The Hin region isn’t inverted, allowing the H2 promoter to be in the correct orientation.
    • Both H2 and Repressor (Rep) are produced.
    • Repressor binds to H1’s operator, leading to the repression of H1.
46
Q

What do phase-variable genes encode for?

A

Surface components (e.g. capsule, fimbria, LPS, flagella)

LF206: Molecular Cell Biology (16 decks)

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