Bacterial Gene Regulation

Question 1

Why has bacterial gene regulation evolved so greatly?

Answer 1

Due to their competitive and ever quickly environments bacteria they have to react efficiently to stimuli (must synthesise proteins).

Question 2

What does it mean that the bacterial genome is a) efficient b) flexible

Answer 2

a) Quick protein production

b) Transcription can be altered.

Question 3

What are 4 common structures of the bacterial genome?

Answer 3

1) DNA stored as a single circular chromosome

2) Densely coding (few introns)

3) Operons found commonly

4) Transcriptional units tend to be orientated in the same direction as chromosome replication.

Question 4

What is meant by describing bacterial DNA as densely coding?

Answer 4

There is little space between genes encoding for proteins (short intergenic distances), so are few introns and repeating junk DNA meaning no spliceosome present.

Question 5

What is the average intergenic distance (DNA between genes) in bacteria?

Answer 5

60-70bp

Question 6

What is an operon?

Answer 6

Multiple genes which encode for different RNA molecules are encoded from one structural unit, so these RNA molecules make a protein.

Question 7

What are 3 ways bacterial DNA can be similar to eukaryotic DNA?

Answer 7

1) Can be linear

2) Can have one or more chromosomes

3) Genes can be transcribed as monocistronic transcripts.

Question 8

How are the few introns present in some bacterial DNA removed?

Answer 8

As they are self splicing, so no spliceosomes are still present.

Question 9

How is clashes between the RNA polymerase on each strand of DNA avoided?

Answer 9

> Transcriptional units tend to be orientated in the same direction as chromosome replication (a bit like anti-parallel but just on a circle instead of linear)

> Right branch genes in anticlockwise direction

> Left branch genes in a clockwise direction.

Question 10

What is the average size of an E.coli bacterium and the average length of its genome, why is this an issue if ignored?

Answer 10

> E.coli is a 2um cell
E.coli genome = 4.6 Mb (mega bases) / 1.6mm long

> Not only is the genome so much longer than the cell, as bacteria are usually ready for division they normally have 2-8 copies of their genome present; which is too big for the cell.

Question 11

How is the bacterial genome able to be a) Organised/ compact but still b) flexible/ accessible?

Answer 11

a) DNA is constrained into multiple domains held together by NAPs for organisation.
>These domains are further coiled for compactness.

b) Coiling is independent between loops (domains) so each can be further compressed or relaxed for transcription.

Question 12

What does NAPs stand for?

Answer 12

Nucleoid associated proteins

Question 13

What are NAPs (Nucleoid associated proteins) similar to in eukaryotic cells and why?

Answer 13

Similar to histones as they bind to DNA and help organise it in a compact way.

Question 14

How many NAPs (Nucleoid associated proteins) does E.coli have?

Answer 14

6

Question 15

What are the names of the 6 NAPs (Nucleoid associated proteins) E.coli has and what is their function?

Answer 15

1. H-NS proteins can bridge between adjacent segments of DNA
2. Fis and IHF proteins induce severe bends

3/4. HU (most conserved NAP in bacteria species) can condense DNA into a fibre by wrapping around DNA.

5/6. Dps and CbpA (expressed in stationary phase/ when E.coli are stressed) condense DNA to protect from damage

Question 16

How many base pairs does Unsupercoiled (relaxed) DNA have per turn in the helix?

Answer 16

Approx 10 base pairs per turn.

Question 17

What are the 2 ways DNA can be supercoiled and what occurs in both?

Answer 17

1) Positive supercoil
>Over-twist the helix (twist clockwise, same direction as helix), the DNA gets more coiled and loops form on the left side of the DNA.

2) Negative supercoil
>Under-twist the helix (anti-clockwise/ opposite direction of helix), and the DNA gets unwinds and forms a loop on the right side of the DNA

Question 18

What is the difference and similarity between positive and negative supercoiling?

Answer 18

> Positive supercoiling coils the DNA more by a single stranded break by Topoisomerase I,II, or IV, causing the supercoiled loop to form on the left side. While negative supercoiling un-coils the DNA by a double stranded break by DNA Gyrase (Topoisomerase II) causing the supercoiled loop to form on the right side.

> Both cause loops of DNA to form which condenses the DNA molecule length, but as DNA is usually in negatively coiled state then positively supercoiling leads to relaxing DNA/ un-condencing.

Question 19

As well as keeping DNA compact, what other mechanisms is supercoiling useful for?

Answer 19

The energy introduced into DNA from supercoiling can be used to make transcription more efficient as energy can be added or removed making the transition from DNA in a closed to open complex easier.

Question 20

How does supercoiling store and release energy?

Answer 20

The cell expends chemical energy which is held in the knot/loop as it’s under tension. If the loop is loosened energy is released, if it is tightened energy is stored (energy is stored when DNA is negatively supercoiled, then when positive supercoiling relaxes the strand it releases energy).

Question 21

What are the 3 enzymes regulate positive supercoils and what 1 enzyme regulates negative supercoils in E.coli?

Answer 21

1) Topoisomerase I, III, IV induce positive supercoils which overwinds DNA.

2) Topoisomerase II (Gyrase) negatively supercoils DNA (underwinds)

Question 22

Is the majority of bacterial DNA negatively or positively supercoiled?

Answer 22

Most bacterial DNA is negatively supercoiled (more as it makes transcription more efficient)

Question 23

What is DNA Gyrase (Topoisomerases II) composed of?

Answer 23

2 protein subunits, GyrB + GyrA

Question 24

How does negative supercoiling occur in E.coli in 3 steps?

Answer 24

1) GyrB/GyrA complex (Gyrase) – GyrB binds DNA and GyrA makes a double-strand break, remaining covalently bound to each end of the break.

2) GyrA (an ATPase) hydrolyses ATP causing a conformational change that passes the intact double strand of DNA through the break, reducing unwinds the helix causing a supercoiled loop to form on the right

3)GyrB re-ligates the break, leaving the DNA negatively supercoiled (unwinding causes new supercoiled loop to form on the right)

Question 25

How is positive supercoiling occur in E.coli in 3 steps?

Answer 25

1) Topoisomerases I, II or IV introduces a single-strand break

2) It holds both ends of the broken single strand and passes the intact strand through, increasing tension by adding more turns to the helix removing the supercoiled loop to the right (made by negative supercoiling)

3) Then re-ligates the broken single strand. The negative supercoil has now been relaxed/ a supercoil has been removed (as all dna is mainly negatively supercoiled, when is positively supercoiled the tension introduced makes dna less condensed).

Question 26

How many Topoisomerases enzymes does E.coli have and what is their overall function?

Answer 26

> 4
To regulate supercoils.

Question 27

Why does bacterial DNA have to be positively supercoiled sometimes? FIND OUT WHY POSITIVE SUPERCOILING IS GOOD

Answer 27

> The majority of bacterial DNA is negatively supercoiled making it very compact (both compact dna but as it is all negatively supercoiled, when is positively supercoiled it “relaxes DNA” by over-winding the DNA to remove supercoiled loops made by negative supercoiling)

> WHY IS IT NEEDED SOMETIMES

Question 28

What is an example of a Quinolone antibiotic that targets Gyrase (Topoisomerases II)?

Answer 28

Ciprofloxacin

Question 29

How does the Quinolone antibiotic ciprofloxacin work and what is its effect?

Answer 29

> Binds to DNA Gyrase (Topoisomerases II) stabilising the covalent complex formed with the broken ends of the DNA and the GyrB subunit.
This means the double stranded DNA remains broken and negative supercoiling cannot occur.

Question 30

What is the distance between the -35 box and -10 box in an optimum promotor region?

Answer 30

17 base pairs.

Question 31

What does it mean to say there is a 17bp difference in the -35 box and -10 box?

Answer 31

-35 is where the promoter starts and the sequence is 8 bases long, so ends at 27, this leaves a gap of 17bp between the end of the -35 TGTTGACA and the start of the -10 TATAAT.

Question 32

Why is there a 17bp difference in the -35 box and -10 box in a optimum promotor region?

Answer 32

> The -35 and -10 boxes are bound by the same protein (RNA Polymerase) and 17bp places the 2 sections of DNA at the right distance and on the right face of the DNA so they can both be bound to the binding sites of the same RNA Polymerase molecule.

Question 33

What is the effect of making the 17bp space between -35 and -10 boxes longer or shorter in a promotor region?

Answer 33

> If made longer or shorter, a single protein (e.g. RNA Polymerase) cannot bind to both sites

> If distance is longer the RNA polymerase cannot reach both as is too short
If distance is shorter the two boxes will be at different curvatures/ faces of the helix so RNA Polymerase can’t bind to both.

Question 34

How can the optimum 17bp distance between -35 and -10 box in a optimum promotor region be used to regulate transcription?

Answer 34

> This 17bp rule can be used to regulate transcription, as the size can be altered to change the strength of a promoter
e.g. 17bp is strongest, 18 or 16 are a bit weaker (e.g. by a mutation or supercoiling)

Question 35

How has evolution of promotor regions allowed for efficient trasncription.

Answer 35

Evolution over millions of year has caused the sequences of -35 and -10 boxes and the distance between them by supercoils to be down or upregulated so each promotor makes the exact amount of transcript needed to make the protein without waste.

Question 36

What are the 6 main landmarks of a bacterial DNA structural unit? What is the order of these structures?

Answer 36

1. Promotor region (-35 to -10 box separated by 17bp gap)
2. Transcription start site (+1)
3. Shane Dalgarno sequence
   >Ribosome binding point (recognised by 16s of ribosome)
4. Start codon
   >ATG sequence which codes for methionine amino acid.
5. Stop codon
   >E.g. TAA
6. Transcription Terminator

Question 37

What subunits make up a Core Polymerase enzyme?

Answer 37

2alpha, 1beta, 1*omega subunits

Question 38

What does Core Polymerase need to form a complex with so it can bind with DNA?

Answer 38

A Sigma factor

Question 39

What is the difference between RNA polymerase core and RNA polymerase holoenzyme?

Answer 39

The core is enzymes lacking the sigma factor, while the holoenzyme is enzymes comprising the sigma factor.

Question 40

What happens when RNA polymerase core binds with a sigma factor?

Answer 40

They form RNA polymerase holoenzyme and now the RNA polymerase can recognise promotor regions.

Question 41

What is the most common Sigma factor in bacteria especially E.coli?

Answer 41

Sigma 70

Question 42

What are the 4 steps to transcription initiation involving Sigma 70?

Answer 42

1. RNA polymerase core binds with 𝛔70 to form RNA polymerase holoenzyme.
2. RNA Polymerase holoenzyme binds non-specifically and scans until 𝛔70 subunit recognises a promoter -> Closed complex
   >Looks for -35 and -10 with a 17bp spacing.
3. Unwinds the DNA -> Open complex (made more energy efficient by negative supercoiling)
4. The 𝛔 factor is kicked out and transcription starts

Question 43

What are the 4 stages of transcription?

Answer 43

Initiation, promoter clearance, elongation, and termination.

Question 44

What are the 2 different mechanisms for transcription termination in bacteria?

Answer 44

1) Rho dependent termination

2) Rho independent termination

Question 45

What is it called when transcription ends randomly before it should?

Answer 45

Premature termination (barely happens in bacteria)

Question 46

Where do Termination signals occur in DNA?

Answer 46

After a gene

Question 47

What are the 4 steps to Rho-dependent termination?

Answer 47

1. RNA Polymerase passes termination signal which causes transcription of G-C rich sequence of RNA.
2. Rho binds to this G-C rich RNA and it wraps around the Rho hexamer.
3. Rho ATPase activity drives spooling (winding) of RNA which pulls the Rho closer to RNA polymerase.
4. Rho catches RNA polymerase (still transcribing) and contact between the 2 causes RNA polymerase to stop transcription and dissociate.

Question 48

What are the 4 steps to Rho-independent termination?

Answer 48

1. Termination signal after the gene is a sequence e.g. a run of Gs/Cs.
2. Transcription of RNA with a run of complimentary base pairs e.g. Gs/Cs
3. RNA folds to form a stem loop (double stranded loop) very quickly right behind RNA polymerase.
4. This spontaneous stem loop formation causes the RNA to touch RNA polymerase making it stop transcribing and dissociate from DNA.

Question 49

What is a Sigma Factor?

Answer 49

Component of RNA polymerase holoenzyme which recognises the promotor

Question 50

How could a prokaryotic cell alter which promotor a RNA polymerase holoenzyme recognises?

Answer 50

To recognise a different promotor, change the type of sigma factor

Question 51

What are 4 ways the expression and activity of sigma factors can be regulated and how do they work?

Answer 51

1) Anti-sigma factors
>Bind and inhibit their cognate sigma factor

2) Proteolysis
>Break down of sigma factors (degradation)

3) Anti-anti sigma factors
>Restore sigma factor function after bound to an anti-sigma factor.

4) Secondary structure mRNA
>mRNA can interact with itself to cause activation or degradation of sigma factors in specific environmental conditions.

Question 52

How many Sigma factors does E.coli have?

Answer 52

7 (one housekeeping and six alternative)

Question 53

What are the 7 sigma factors for E.coli and what is their function?

Answer 53

1. Sigma 70 (Sigma A)
   >Housekeeping
2. Sigma 38 (Sigma S)
   >General stress response
3. Sigma 32 (Sigma H)
   >Heat shock/ cytoplasmic stress
4. Sigma 28 (Sigma F)
   >Flagella and motility
5. Sigma 24 (Sigma E)
   >Extracytoplasmic stress (e.g. heat or proteins extracellular)
6. Sigma 19 (Sigma I)
   >Ferric citrate transport (ferric citrate transporter for iron)

7.Sigma 54 (Sigma N)
>Nitrogen-related (controls nitrogen assimilation)

Question 54

How are Sigma factors characterized?

Answer 54

Sigma factors are characterized by their molecular weight e.g. sigma70 has a molecular weight of 70kDa (Dalton)

Question 55

What is a regulon?

Answer 55

A group of genes or operons that are turned on or off in response to the same signal by the same regulatory protein

Question 56

How is Sigma 70 linked to the regulation of the 6 other alternative Sigma factors in E.coli?

Answer 56

Sigma 70 recognises the promotors of most of the 6 alternative sigma factors so the mRNA is constantly produced for them, then due to higher regulation in the correct conditions the needed sigma factor will be produced.

Question 57

What sigma factor is most abundant during a) exponential growth b) stationary phase?

Answer 57

a) Sigma 70

b) Sigma 38

Question 58

What are the two anti-sigma factors for Sigma 70 and how do they work?

Answer 58

1) Rsd
>Rsd binds to region 4 (C terminal of sigma 70) and blocks sigma 70 from binding to RNAP core.

2) HscC
>Less specific and can bind anywhere on Sigma 70, blocking interaction with RNA polymerase core

> These allow the lower affinity Sigma 38 to bind to RNAP instead of Sigma 70.

Question 59

If a E.coli cell is exposed to a general stressful environment how does it react in terms of sigma factors?

Answer 59

1. Transcription factors sample metabolic conditions in cell and stop suppressing promotor region of spoS. Sigma 70 can now transcribe mRNA for Sigma 38.
2. mRNA of spoS stops being degraded
3. Ira proteins inhibit RssB taking Sigma 38 to ClpXP. Decreasing proteolysis of Sigma 38.
4. Anti-sigma factors Rsd and HscC expressed blocking Sigma 70 binding to RNAP Core so lower affinity Sigma 30 can bind.

Question 60

What are 3 examples of factors causing general stress to a bacteria cell?

Answer 60

Temperature, antibiotic stress, osmolarity

Question 61

What is a similarity and a difference between Sigma 70 and 38 promotor regions?

Answer 61

> Sigma 70 and 38 promotors have -10 and -35 elements with a 17bp spacing.

> However they have small differences in amino acid sequence

Question 62

What is the abundance of Sigma 38 compared to 70?

Answer 62

Sigma 38 abundance is 1/3 that of Sigma 70.

Question 63

Why are alternative sigma factors under complex regulation?

Answer 63

Bacteria only want to switch from Sigma 70 in the correct environmental conditions.

Question 64

What are 3 ways Sigma 38 is tightly regulated?

Answer 64

1. Transcription is regulated
   >Transcription factors sample metabolic conditions in cell and suppress promotor in nutrient conditions. So Sigma 70 can’t express Sigma 38.
2. RNA is regulated
   >5’UTR region of mRNA is degraded when in nutrient conditions
3. Protein is regulated
   >In stressful, Ira proteins inhibit RssB so Sigma 38 isn’t taken to ClpXP (in nutrient conditions, RssB takes Sigma 38 to be degraded by ClpXP via proteolysis).

Question 65

If an enzyme for example is needed in exponential growth a lot and in stationary phase less, what is the effect on the promotor region of the enzyme?

Answer 65

There will be a strong promotor region for sigma 70 and a weaker promotor region for sigma 38 (promotor has lower affinity for RNA polymerase)

Question 66

What is the gene for Sigma 32 and how is it regulated?

Answer 66

Intrinsic structure of mRNA is used to regulate rpoH, as the secondary structures of the mRNA produced from rpoH are acutely temperature sensitive.

Question 67

What happens to the mRNA for Sigma 32 when temperature is a) low (e.g. 30 degrees/ below) b) high (e.g. 42 degrees if bacteria is swallowed)?

Answer 67

a) mRNA has many intrinsic bonds making it highly structured causing downregulated translation (hard for ribosome to handle). The small amount of rpoH protein produced is sent for degradation by chaperones.

b) High temp melts secondary structure (breaks intrinsic bonds) making it linear, a better substrate for ribosomal translation. mRNA efficiently translated to Sigma 38. High temp causes degradation systems to be busy degrading denatured proteins and avoid Sigma 38 as is stable in heat.

Question 68

What are 3 downstream effects of Sigma 32 in response to heat shock?

Answer 68

> Make cell membrane composition less fluid (more rigid)

> Express different folding chaperones that are more stable in heat.

> Upregulate protein degradation pathways (for denatured proteins)

Question 69

What is a sigma factor cascade?

Answer 69

A complex gene expression program required many sigma factors to be active at different times so the proteins are made in the right sequence.

Question 70

What is the abundance of Sigma 28 compared to Sigma 70 and when does this decrease further?

Answer 70

Is half as abundant as σ70 under most conditions, and decreases upon heat-shock as bacteria don’t need to swim in heat.

Question 71

What is the sigma factor cascade encoding for flagellar assembly in E.coli, what is produced by each operon, and how are the operons expressed in the correct order?

Answer 71

> Genes coding for the proteins are arranged in a series of operons, each class of operons is mediated by a single promotor:

1. Class 2 operons (most important)
   >Controlled by Sigma 70 promotors
   >In normal conditions class 2 proteins can be made; base, hook and motor of flagella.
   >FlgM (anti-sigma factor for 28) is produced, but when hook is complete (final protein produced) it’s secreted.
2. Class 3 operons
   >Controlled by Sigma 28 (activated when FlgM is secreted out).
   >Produces proteins for filament and motility genes so bacteria can sense nutrients and chemical gradients.

Question 72

Do bacteria living in more complex environments have more sigma factors usually?

Answer 72

Yes as bacteria living in more complex environments tend to have a larger genome so contain more sigma factors to regulate the range of responses to the changing environment.

Question 73

What system allows bacteria to sense and respond to their environment via regulating gene expression?

Answer 73

Two-component systems (TCS) cause gene upregulation or suppression after sensing the environmental conditions.

Question 74

What makes up a Two-component system (TCS)?

Answer 74

At their simplest (most are simple), they consist of a sensor (called a Histidine Kinase, HK) and an effector (called a Response Regulator, RR)

Question 75

What is the difference in function of sigma factors and two-component systems?

Answer 75

Sigma factors allow bacteria to change entire regulons, two component systems can control 1 gene or many (tend to control local level though).

Question 76

Do bacteria living in more complex environments have more TCSs?

Answer 76

It is a linear relationship between genome size and TCS presence, so as bacteria in more complex environments usually have a larger genome and need more fine tuning of genes to respond to the environment.

Question 77

In a simple Two-component system (TCS) where is histidine kinase (sensor/ receptor) found?

Answer 77

Is often membrane bound.

Question 78

What are the 3 steps to a simple Two-component system being activated?

Answer 78

1. Autophosphorylation of Histidine Kinase sensor
   >Catalytic ATPase domain receives signal, binds ATP and removes phosphate to phosphorylate a conserved histidine residue (in the same protein so is called autophosphorylation) in the DHp domain (dimerisation with phosphorylated histidine domain)
2. Phosphorylation of RR by Histidine Kinase
   >Phosphorylated histidine kinase transiently interacts with its cognate response regulator and the phosphoryl group is transferred to the receiver domain of the RR from the HK via a phosphorylation cascade.
3. Phosphorylated RR is an active effector protein which alters gene expression.

Question 79

What are 3 additional proteins which can be added to histidine kinase to alter its function?

Answer 79

1. PAS
   >Small ligand binding domain
2. GAF
   >Small ligand binding domain
3. HAMP
   >Domain involved in transmitting a conformational change along proteins used to transmit signal from outside to ATPase catalytic domain on inside.

Question 80

What are 5 different structures found in response regulators and their functions?

Answer 80

1. Some RRs are just a receiver domain
   >Get phosphorylated then function is unknown.
2. Most common form: Receiver domain and a DNA-binding domain
   >Act as transcription factor- bind upstream of specific gene and can up or down regulate expression of these genes.
3. Receiver domain, AAA+ (ATPase) and a DNA binding domain
   >Activity to modulate Sigma 54 dependant genes.
4. A receiver domain fused to GGDEF instead of a DNA binding domain
   >GGDEF: An enzymatic domain which produces c-di-GMP switching sessile to motile.
5. Receiver domain with a Methyltransferase
   >Involved in chemotactic response (move in response to chemical concentrations).

Question 81

What are 4 different roles accessory proteins can have on Two-component system?

Answer 81

1. Can have a separate sensory protein that transmits to HK
   >A HK which responds to multiple signals.
2. A scaffold protein can facilitate transmission to RR
   >Sit between HK and RR and facilitate the transfer of phosphate.
3. A connector protein can swap signals to another TCS – allows different systems to work together in certain circumstances
   >So activation of one TCS can activate another, so can control many sets of genes in one time.
4. Allostery – RR without output that feeds into another TCS
   >HK works with a RR that doesn’t have an effector domain, this feeds into another TCS

Question 82

What is an example of when a Two-component system activates a response regulator?

Answer 82

The RstAB system, where RstB is the hitidine kinase sensor and RstA is the response regulator

Question 83

How does a response regulator find the correct DNA sequence in the RstAB two-component system in 4 steps?

Answer 83

1. RstB (HK) phosphorylates itself when activated, transfers phosphate group to monomeric RstA (response regulator) causing dimerization of 2 RstA at N terminal forming a complex of 2 RstA response regulators bound at N terminal with 2 sperate C terminals (DNA binding domains)
2. RstA complex scans DNA for specific sequence, when one N terminal binds to this sequence, conformational change allows the other N terminal to look for the same sequence nearby.
3. If the 2nd DNA binding domain finds the same sequence nearby the RstA complex bind together, dimerizing the DNA so the response regulator (RstA) can regulate the gene.
4. If the 2nd DNA binding domain (C terminal) doesn’t find the same DNA sequence, the RstA complex disassociates and searches again.

Question 84

What are the 2 outcomes of a response regulator binding bound to DNA and interacting with the C-terminal domain of RNAP?

Answer 84

1. Enhance RNA polymerase to enhance gene expression
   >Occurs if the specific DNA sequence the RR is bound to is close to or slightly overlapping with the promotor region that RNAP binds to but not covering it completely so RR can be in contact with RNAPR
2. Inhibit RNA polymerase to decrease gene expression
   >Occurs if specific DNA sequence RR binds to overlaps with the promotor for RNAP for a gene. As the bound RR blocks RNA polymerase binding to the promotor.

Question 85

What 2 contacts can a response regulator bound to DNA make?

Answer 85

RR bound to DNA makes contacts with the C-terminal domain of RNAP ⍺-subunit or with the σ factor to activate transcription.

Question 86

What is an example of a system which uses a complex two component system to be regulated in E.coli?

Answer 86

Acid resistance in E.coli

Question 87

Describe the interplay between two-component systems and RpoS in the acid stress response of E. coli. (Short answer question)

Answer 87

1. TCS system 1 (EvgSa-EvgA)
   >EvgSA is a histidine kinase that senses low pH and activates/ phosphorylates itself.
   >EvgSA phosphorylates EvgA response regulator.
   >EvgA upregulates transcription of 2 genes: SafA and ydeO
2. TCS system 2 (PhoQ-Phop) causes general response to stress.
   >SafA is a membrane-bound connector protein that binds to and activates PhoQ (HK for second TCS).
   >Activated PhoQ phosphorylates itself and then PhoP (RR for second TCS).
   >Activated PhoP upregulates IraM (Ira gene) which sequesters RssB, preventing degradation of Sigma 38 (for general stress response).
3. 1st TCS interacts with sigma 38 (rpoS) to have specific response to acid.
   >ydeO was upregulated by EvgA from the first TCS.
   >ydeO binds with Sigma 38 allowing it to bind to gadE gene for a specific acid response.
   >gadE expression activates a metabolic pathway which reduces H_ in the cell by secreting it out keeping the bacterial cytoplasm relatively neutral (specific acid response)

Question 88

What is the effect of a) Negative b) Positive Supercoiling on the 17bp gap between -35 and -10 boxes?

Answer 88

1) Can result in the tightening of the DNA structure. This can bring the -35 and -10 regions closer together, effectively reducing the spacing between them. As a result, the promoter sequence may become more accessible and favourable for binding of RNA polymerase, enhancing transcription initiation.

2) Overwinding of the DNA helix, can increase the spacing between the -35 and -10 regions. This can lead to decreased accessibility of the promoter and reduced efficiency of transcription initiation.

Question 89

What does the promotor region contain in E.coli?

Answer 89

Promotor region= contains -35 and -10 box, separated by a 17bp gap.

Question 90

Why does negatively supercoiled DNA have more efficient transcription?

Answer 90

The presence of negative supercoiling ahead of the replication or transcription machinery helps to unwind the DNA strands more easily. This is because the underwound DNA structure reduces the torsional strain and facilitates strand separation.