Prokaryotic Molecular Biology (1-3)

Question 1

What is the B form of DNA?

Answer 1

The great majority of DNA in living cells occurs as the B form

B form:
→ two polynucleotide chains are in opposite orientation
→ regular right-handed double helix
→ diameter of 2nm and making a complete turn every 3.4nm
→ ~10.5 base pairs per turn of the helix

Flexibilities:
→ the number of base pairs per turn can be altered
→ the helix in the cell is not straight but coiled in 3D space
→ there are certain sequence features where bends occur

Question 2

What is supercoiling?

Answer 2

Occurs when additional turns are introduces or removed into the DNA double helix
→ introduced - positive supercoiling
→ removed - negative supercoiling

The DNA in the circular genome of E. coli is supercoiled

Question 3

How is DNA torsional stress (from twisting force) accomodated?

Answer 3

→ formation of superhelices
→ altering number of base paris per turn of helix

Expressed by linking number (L) → the total number of times that the two strands of the double helix of a closed molecular cross each other when constrained to lie in a plane

Question 4

What are topoisomerases?

Answer 4

Enzymes that catalyse changed in DNA topology
→ type I, type II

Question 5

How do type I topoisomerases work?

Answer 5

Break one strand of the DNA, pass the complete strand through the gap and change the linking number by +/- 1
→ relaxes -ve supercoiled DNA

Question 6

How do type II topoisomerases work?

Answer 6

Break both strands of the DNA, pass another part of the helix through the gap and change the linking number by +/- 2
→ e.g. DNA gyros - essential for E. coli DNA replication

Question 7

How is DNA organised in E. coli?

Answer 7

In E. coli the single circular DNA molecule is organised into a series of supercoiled loops that radiate from a central focus protein core
→ highly organised structure

Question 8

What does the protein component of the E. coli nucleic include?

Answer 8

→ DNA gyrase and DNA topoisomerase I - maintain the supercoiled state of the DNA
→ at least 4 proteins, the most abundant being HU, that are involved in packaging the DNA - forms tetramers covering 1/5 of the genomes

Archaea differ → don’t have proteins related to HU, instead more similar to eukaryotic histones

Question 9

What are the types of prokaryotic genome?

Answer 9

Linear
Multipartite → genomes divided into two or more DNA molecules
Plasmids → small DNA molecule that usually codes for non-essential genes

Question 10

What are genomic islands?

Answer 10

Regions of genome that have a very different genetic signal
→ horizontally acquired
→ often mutates so masking/destroying their transmission and integration modes
→ can confer fitness to occupy a particular ecological niche
→ some linked to pathogenesis

Question 11

What is the basic unit of replication for bacteria/archaea?

Answer 11

Replicon
→ a DNA molecule or sequence with a functional origin of replication
→ each replicon must be replicated at least once per cell division cycle

Question 12

What are the features that link replication to the cell cycle in all organisms?

Answer 12

→ initiation of replication commutes cell to a subsequent division
→ cell division cannot occur until the round of replication associated with a particular initiation has been completed

Question 13

What is the oriC?

Answer 13

The single origin of a bacterial chromosome where replication is initiated
→ bidirectional replication leading to a theta structure

Question 14

How is bacterial replication initiated?

Answer 14

Initiation commences with the binding of ~20 monomers of DnaA to the RH part of oriC
→ forms the closed complex
→ open complex forms when the 3 AT-rich 13bp repeats melt

DnaB helicase is loaded onto the melted DNA with DnaC
→ ATP is hydrolysed and DnaC released

DnaB unwinds the DNA bidirectionally in a process that requires single strand binding protein (SSB) and DNA gyrase

Primase synthesises a primer RNA molecule on both strands and replication commences

Question 15

What is the purpose of single binding protein (SSB)?

Answer 15

Prevents activating damage repair mechanisms when genome is single-stranded
→ single-stranded genomes are a sign of infection/foreign so usually degraded

Question 16

What does DnaA do in initiating bacterial replication?

Answer 16

DnaA binding bends DNA
→ causing stress
→ forms open complex
→ AT rich regions melt and separate

Question 17

What controls whether a round of prokaryotic replication is initiated?

Answer 17

Dam methylase methylates the adenine residues GATC motifs in OriC
→ replication will only initiate if all 14 copies of GATC in OriC are methylated

→ newly synthesised strands don’t have methylated GATCs

Question 18

How is prokaryotic replication terminated?

Answer 18

Termination occurs when DNA replication is completed
→ the 2 replication forks began at OriC and move in opposite directions approaching one another
→ they fuse in a region opposite to OriC - the terminus region a replication fork trap pausing DNA at a certain site

Question 19

What are DNA terminator sites (ter)?

Answer 19

Polar in their action
→ arrest a fork approaching from one direction but not the other
→ tus a terminator protein must e bound to the ter site to halt the fork
→ spread out across the genome
→ allows one direction of replication to be slow and one fast

Question 20

How is a fork arrested from only one direction in prokaryotic DNA replication?

Answer 20

The Tus protein binds to ter sites and then interacts with a replication form heading in one direction only halting the fork
→ fork arrest results from inhibition of helicase-mediated unwinding of the DNA duplex at the apex of the fork

Question 21

What is the Meselson-Radding model of how recombination starts?

Answer 21

Cleavage → one strand is cleaved by an endonuclease
Chain displacement → DNA synthesis displaces a chain
Invasion → the ss chain invades a homologous ds DNA molecule - catalysed by RecA
Chain removal → the displaced chain is digested
Ligation → produces a Holliday junction
Branch migration → increases heteroduplex, catalysed by RuvAB
Isomerisation → the strands of the Holiiday 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 (won’t produce exactly the same thing on both strands)

Question 22

What is RecBCD’s role in recombination?

Answer 22

Binds tightly to the end of a dsDNA substrate and unwinds it using its helices activity
→ as it unwinds the helix it degrades both ssDNA strands using its dual 5’ → 3’ and 3’ → 5’ exonuclease activates

→ moves in steps of 23bp long - quantum inchworm, until it meets the Chi site where it produces a ssDNA tail with a 3’ end and RecA binds - strand invasion
→ Chi sites are sites of recombination in E.coli genome
→ when RecBCD encounters chi site its enzymatic activities are dramatically altered - 3’ to 5’ exonuclease inhibited, 5’ to 3’ exonuclease stimulated (helices activity unaffected)
→ the outcome is that RecBCD produces a ssDNA tail with a 3’ end and RecA binds to the ssDNA tail which invades

RecBCD has
→ ssDNA exonuclease (5’ to 3’ and 3’ to 5’)
→ ssDNA endonuclease
→ dsDNA exonuclease
→ DNA-dependent ATPase
→ DNA helices (prefers blunt dsDNA ends)

Question 23

What is the hin region?

Answer 23

H-inversion region - 995bp long bounded by two 14bp inverted repeats
→ encodes for an invertase which catalyses inversion

Structure of H1 and H2 genes:
→ the H1 gene has its own promoter and operator and is physically separated from the hin region
→ the H2 gene is an 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
Phase 1: H1 expressed
Phase 2: H2 expressed, H1 repressed

Question 24

What is the purpose of homologous recombination?

Answer 24

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
→ Holliday proposed a widely accepted model for homologous recombination but has no clear idea of how invasion occurs - Meselson-Radding model can address this problem

Question 25

What is transposition?

Answer 25

Inversion sequences → many DNA sequences which don’t transpose but can alter their orientation with a DNA molecule, inversions sequences control gene expression in some organisms

e.g. phase variation in Salmonella spp.
→ most can produce 2 different types of flagellum phase 1 and 2 H antigens
→ growing phase 1 cells will produce some phase 2 cells and vice versa
→ phase is determined by the orientations of the hin (H-inversion) region

Question 26

Why is DNA repair needed?

Answer 26

DNA in the cell is continuously exposed to damage
→ damage is any change from the normal nucleotide sequence and supercoiled double helical state
→ causes: physical and chemical agents in the environment e.g. UV light, free radicals produced during metabolism, errors in DNA replication

Question 27

What are the 2 classes DNA damage falls into?

Answer 27

Single base changes → produces mutations but have no effect on physical process of transcription or replication, replication errors due to keto-enol type tautomerisation, deamination of cytosine to uracil, incorporation of U rather than T during replication, chemical modification of bases
Structural distortions → may impede transcription and/or replication, single strand breaks, covalent modification of bases e.g. alkylation, removal of a base, inter strand and intrastrand covalent bonds

Question 28

What are some nucleotide base tautomers?

Answer 28

Standard base tautomers (structural isomers - readily interconvert)
→ standard base pairing arrangements - thymine (keto) adenine (amino), cytosine (amino) guanine (keto)
→ anomalous base pairing arrangements - thymine (enol) guanine (keto), cytosine (imino) adenine (amino)

Question 29

What is an example of structural distortion?

Answer 29

Thymine dimer formation caused by UV light
→ two adjacent thymines on the same strand become covalently linked in a cyclobutane structure or a (6-4) photoproduct

Question 30

How is DNA damage dealt with?

Answer 30

Cells have systems which recognise mismatches and structural distortions in DNA and resolve these with a range of repair processes
Direct repair → reversal or simple removal of the damage
Mismatch repair → detection and repair of mismatched bases
Excision repair → recognition of the damage followed by excision of a patch of DNA and its replacement by undamaged DNA
Tolerance systems → allow DNA replication to process through damages regions of DNA
Retrieval systems → recombinational processes to repair damaged DNA

Question 31

Whats an example of direct repair?

Answer 31

Photolyase → photo reactivation is an example of direct repair
→ this process repairs any UV-induced intrastrand pyrimidine dimer (almost always thymine dimers)
→ the enzyme deoxyribopyrimidine photolysis binds specifically to pyrimidine (thymine) dimers in the dark
→ photolyase contains two chromophore that absorb light energy in the range 300-600nm - absorbed energy is used to split cyclobutane structures

Question 32

What is an example of mismatch repair?

Answer 32

Uracil DNA glycosylase (UDG)
→ uracil is occasionally incorporated into DNA instead of thymine
→ UDG removes uracil base form the nucleotide making an AP site (apurinic or apyrimidinic)
→ AP endonuclease makes a break in the phosophodiester backbone adjacent (5’) to the AP site
→ DNA polymerase I binds to the break and lays down new DNA and the gap is sealed by DNA ligase

The mut system → MutS recognises mismatched and short insertion/deletions (indels) on semi-methylated DNA and binds to them
→ MutL binds and stabilises the complex
→ the MutS-MutL complex activates MutH
→ MutH locates a nearby methyl group and nicks the newly synthesised strand opposite the methyl group
→ MutU (helicase II) unwinds the DNA from the nick in the direction of the mismatch
→ DNA PolI degrades and replaces the unwound DNA and DNA ligase seals the single strand break

Question 33

What is an example of excision repair?

Answer 33

In E. coli there are 3 excision repair modes:
→ very short patch (deals with mismatches between bases)
→ short patch ~20 nucleotides
→ long patch 1,500-10,000 bps
Both short and long patch repair utilise the repair endonuclease
→ encoded by the uvrA, uvrB and uvrC genes

Enzyme uvrABC binds to damaged regions makes and incision on both sides of the damage
UvrD (alias MutU DNA helicase II) separates strands
As in mismatch repair DNA polI replaces the DNA ligase fills the gap
Short patch repair accounts for 99% of bulky lesions repair events
Long patch repair is an inducible activity

Question 34

Whats an example of a tolerance system?

Answer 34

Inducible error-prone repair
Low-fidelity DNA polymerases (translation synthesis polymerases (TPSs)) can synthesise DNA past damaged bases
→ not efficient at replicating undamaged DNA accurately - most lacking proof-reading ability
Two in E. coli: polymerases IV and V and 5 in human cells
Almost all are members of new DNA pol family and Y-family
In some circumstances make many errors so can generate mutations

Each TSP appears to have a different substrate specificity → human polymerase eta can bypass the major UV photoproduct very efficiently, usually inserting the correct nucleotides - less efficient with most other types of damage
→ defective in the variant form of a highly skin-cancer-prone genetic disorder, xeroderma pigmentosum, - so eta helps prevent UV-induces mutations in cancer

Question 35

Whats an example of a retrieval system?

Answer 35

Daughter strand gap repair
→ does not actually repair damage
→ permits replication to occur successfully
→ relies on other repair processes such as excision repair to repair the damage afterwards

The SOS response in E. coli
→ if it suffers severe DNA damage it activates the expression of a large number of diverse unlinked genes involved in DNA repair, error prone replication etc.
→ all genes and operons under SOS control are subject to repression by the LexA protein - 2 domains: dimerisation and a DNA-binding domain
→ conserved binding site “LexA box’ located within th w-promoter of genes regulated by LexA
→ binding of LexA to the LexA boxes represses expression of SOS operons
RecA proteins is involved in inducing the SOS response
→ responds to DNA damage changes conformation which activates it - inhibition of the 3’-5’ editing in DNA Pol III allowing error-prone DNA replication, interacts with LexA which autoclaves becomes inactive leading to the SOS repose - sulA expression inhibits cell division
→ one DNA damage is repaired RecA* converts back
SOS réponse allows the cell to survive severe DNA damage by allowing DNA replication but at the expense of fidelity - last ditch effort by the cell to replicate with severe DNA damage

Question 36

How is prokaryotic transcription initiation controlled?

Answer 36

Induction - switching on of genes when they are required
Repression - switching off of genes when they are not required

Regulatory proteins interact with DNA
→ repressors - prevent transcription when bound to DNA
→ activators - activate transcription when bound to DNA
converted between inactive and active states by effectors
→ inducers - activate activators or inactivate repressors - switch genes on
→ co-repressors - activate repressors or inactive activators - switch genes off

Question 37

What is a regulon?

Answer 37

Genes associated with a particular physiological function may not be in just one operon
→ these operons may be controlled by a single regulatory protein - together called a regulon
e.g. phosphate or PHO regulon >80 genes in many operons controlled by one regulatory protein

Question 38

How does E. coli use diauxic growth in the lac operon?

Answer 38

E. coli prefers to use glucose as its carbon source
→ if both glucose and an alternative carbon source is available the glucose is used first - diauxic growth
→ has diauxic lag when switching between carbon sources

Question 39

What is the basis of the lac operon?

Answer 39

Glucose represses the synthesis of enzymes that metabolise less preferred carbon sources - catabolite repression
→ if lactose not available and glucose is E. coli will have a small amount of enzymes that metabolise lactose
→ when the cells runs out of glucose there is rapid induction of the enzymes for lactose metabolism

When glucose absent activates adenylate cyclase which turns ATP into cAMP induces CAP binding
→ RNA polymerase binds p
→ lactose present activates allolactose which binds o
→ stimulates expression of lacZ lacY and lacA

On the lac operon lacI → inhibits mRNA production for the proteins encoded by the lac operon - encodes a repressor protein
→ allolactose binds the repressor, stops repression - allows transcription of the lac genes

Question 40

What are trans/cis acting elements?

Answer 40

Trans-acting elements → some genes encoded for diffusible products that regulated gene expression on both DNA molecules
Cis-acting elements → other genes dit not produce a produce but regulated genes on the DNA molecule they were encoded upon

Question 41

What is gene complementation?

Answer 41

In E.coli lac operon
→ both normal and Lac+ inducible operons repressed
→ allolcatose induces expression on both molecules
→ chromosome produces a functional LacZ
→ F’ plasmid produces a functional LacY
→ these complement the two mutated genes so phenotype is normal