Gene Regulation in Bacteria

Question 1

What are the three main types of RNA in bacteria and their roles?

Answer 1

mRNA (messenger RNA): is the complementary copy of the base sequence encoded in the genome. It carries the genetic code into the cytoplasm to ribosomes in order to initiate protein synthesis.

rRNA (ribosomal RNA): structural components of the ribosome that assist in translation.

tRNA (transfer RNA): Associates to amino acid residues to ensure that the right one is incorporated to the polypeptide chain during synthesis.

Question 2

What key feature of 3’-deoxyadenosine is able to show that RNA is synthesised from the 5’ to 3’ direction?

Answer 2

3’deoxyadenosine, when added to cells, is phosphorylated and three P are added to it. It then imitates a RNA adenine but without the 3’ OH group. When this is used, synthesis of RNA doesn’t occur. This indicated that the directionality of RNA synthesis was 5’ to 3’ and the 3’ OH group was necessary for this.

Question 3

Why are bacteria able to simultaneously transcribe and translate mRNA, as opposed to eukaryotic cells?

Answer 3

Eukaryotic cells consist of membrane-bound organelles that compartmentalise the transcriptional and translational processes. In prokaryotes, like bacteria, there are no organelles and all enzymes exist within one cytosol. Therefore, translation is able to occur simultaneously on newly transcribed mRNA.

Question 4

Which subunit of RNA Pol binds to the promoter sequence?

Answer 4

The sigma factor binds to the promoter, which is located just upstream of the start of the gene.

Question 5

What are the roles of the -35 and -10 regions for the sigma factor?

Answer 5

The -35 region in the DNA is the recognition site for the promoter, whereas the -10 region ensures that the RNA Pol will be assemble in the right orientation. Once the sigma factor binds, the rest of the subunits associate around it.

Question 6

Briefly describe the basics of the DNase 1 fingerprinting experiment.

Answer 6

DNase 1 fingerprinting is an experiment that shows RNA Pol binds to DNA in the process of transcription. DNase 1 is an enzyme that cleaves DNA phosphodiester bonds.

The experiment consists of two test tubes: TT1 with DNA and DNase acts as a control, whereas TT2 contains DNA, RNAP and DNase 1. The premise is that TT1 only contains DNA and DNase 1 and therefore, the DNase has a clear run to cleave all of the phosphodiester bonds in the fragment. On the other hand, TT2 contains RNAP as well. DNase cannot cleave the DNA if it is obstructed by a RNAP that has bound to the DNA.

The results are analysed by electrophoresis to separate out fragments that are different sizes, even by 1 bp. The gel of TT1 shows all of the nucleotides are present as single bases. But TT2, there is a distinct ‘footprint’, which shows that some nucleotides are not present. This shows that RNAP did in fact bind to DNA and obstruct DNase from cleaving some phosphodiester bonds.

Question 7

How has the DNase 1 fingerprinting expanded our understanding of the protein interactions at the promoter region?

Answer 7

Fingerprinting can identify the specific part of the DNA sequence where the RNAP binds, which is going to be the promoter region. It could also show if there is a consensus of binding location between different strains or species.

Question 8

Describe the roles played by different subunits of RNA Pol.

Answer 8

RNA Pol consists of five subunits: 2 alphas, beta, beta’ and the sigma factor.

The alpha subunits are responsible for assembly of the core enzyme into a holoenzyme, and for promoter recognition.

The beta subunit plays a role in template binding and keeping the DNA attached to the enzyme.

The beta’ subunit is the catalytic domain, responsible for formation of the phosphodiester bond.

When the RNAP is bound to the DNA in its whole, this is called the closed promoter complex. When the DNA becomes unwound, it is called the open promoter complex. Then transcription can begin.

Question 9

What are the key phases of the transcription process?

Answer 9

Transcription starts with the initiation phase, whereby the correct nucleotide is inserted, according to the DNA template.
Then the elongation phase is where the phosphodiester bond is formed. In this phase, the RNAP must bind the correct ribonucleoside triphosphate before forming the phosphodiester bond. Then the RNAP moves along one nucleotide at a time.

Termination is trigger when the RNAP recognises the end of a gene via a terminator sequence. Newly synthesised mRNA is released and RNAP dissociates from the DNA.

Question 10

Howe do we know that ribonucleoside triphosphate (UTP) bind to the beta subunit of RNA polymerase?

Answer 10

UTP is a radioactively labelled analog of nucleosides that can bind to RNAP. When RNAP is mixed with UTP and then the subunits are separated by cellulose acetate chromatography, it shows the four different subunits in different bands. The band for the beta subunit can be seen to be radioactive, identifying that as the binding site for DNA.

Question 11

Describe the key features of a rho-independent terminator and explain why these features lead to the release of the RNA from the DNA template.

Answer 11

Rho-independent terminator relies on an intrinsic terminator sequence in the RNA. The terminator sequence is GC rich, which creates a loop structure as they are complimentary. The region after this is AT rich, creating a sequence of U’s at the end of the mRNA. The stem loop in the mRNA and the weak A-U bonds weakens the interaction between the RNAP and the DNA. This releases the mRNA strand and the RNAP.

Question 12

Describe the key features of a rho-dependent terminator and explain why these features are thought to lead to the release of RNA from the DNA template.

Answer 12

Rho is a homo-hexamer protein. Rho binds to the ‘rut’ site (Rho utilisation) near the 3’ end of the mRNA. Rho moves along the RNA in the 5’ direction. Then there is ATPase activity where the energy of ATP is released. This leads to the release of mRNA from the RNAP. The details of how it requires ATP and how it terminates transcription is not yet known.

Question 13

Why does an mRNA sequence contain a leader sequence at its 5’ end?

Answer 13

The leader sequence is the first part of the mRNA that is transcribed because it lays just downstream of the promoter and operator sequences - it is involved in ribosome binding and sometimes in regulation of mRNA production.

Question 14

Describe the features of an operon and how the genes are synthesised by the operon.

Answer 14

Operons contain more than one protein-coding gene, but are all under the control of one promoter. Operons are transcribed into a polycistronic mRNA. E.g. a operon with three gene will arranged in a modular way with three start codons and three stop codons for ribosomes to produce three separate proteins. They usually encode all the proteins needed for one metabolic pathway, making it easier for the bacteria to carry these out.

Operons are either inducible or repressible. Inducible operons mean that they can be switched on, whereas repressible operons can be switched off. This is to conserve energy within the bacteria to only use transcribe when needed. These operons are controlled by small molecules and regulatory proteins. Inducible operons are usually for catabolic pathways e.g. sugar breakdowns, whereas repressible operons are used for anabolic reactions.

Question 15

Galactosidase is the first enzyme in the lac operon. Which two reactions are catalysed by beta-galactosidase?

Answer 15

The substrate for beta-galactosidase is lactose. It is responsible for the cleavage of beta-galactiside linkages in lactose, to produce galactose and glucose.

Beta-galactosidase also catalyses the conversion of lactose into allolactose.

Question 16

What is the role of an operator sequence? How do the operator sequences on lac operon allow for a tetramer to form.

Answer 16

The operator is the negative regulatory site that binds the lac repressors. It is an almost palindromic sequence of one half to the other, so it is said to have two ‘half sites’. It allows it to bind two lac repressor on the two repressors. It overlaps the promoter sequence so that when repressors are bound to it, the RNA Pol cannot bind to the promoter - therefore, transcription cannot initiate.

The lax operon has three operators. It has the primary operator, next to the promoter at +11, the auxillary operator centred at -84 and another one at the end of the gene at +412. Mainly, the primary and auxillary operator are used - the two operators bind two lax repressors each. These operators can then bind to form a repressor tetramer and form a loop structure in the DNA>

Question 17

Describe the process of the lac operon’s negative regulation of transcription.

Answer 17

The lax operon is negatively regulated when there is no lactose present in the cell. This means that the repressor is able to freely bind to the operator and block transcription of the lac operon. This is because if lactose is present, the lactose binds to lac repressors and prevent it from being able to bind to operators. Therefore, this would allow RNA Pol to bind and transcription of the lac operon would go ahead.

Question 18

What are two anomalies of the lac operon and how are these overcome?

Answer 18

Two anomalies of the lac operon come from the fact that you need transcription of the lac operon to induce the transcription of the lac operon.

Transport of the inducer to the operon requires permease, which is encoded by lac y. Also, the true inducer of lac operon is actually allolactose, not lactose. Lactose is converted into allolactose by beta-galactosidase, another enzyme that is transcribed from the lac operon.

This is solved if you consider that the binding of the repressor is not infinitely strong and there are drop off. This means that there is some leakage and lac transcription is always occurring at low levels.

Question 19

Describe the growth curve of E.coli grown on both glucose and lactose.

Answer 19

If glucose is present in the growth media, E.coli will preferentially metabolise glucose. Glucose inhibits the lac operon transcription. So, E. coli displays a diauxic growth when both gluocse and lactose are present. This means the gluocse is used first until it runs out, showing an upward trend in population size. Then there is a lag phase, where the glucose has run out and lac operon is being transcribed. This is shown as a plateau. Then There is another upward trend in population size as the E. coli start to metabolise lactose.

Question 20

Describe the role of adenylyl cyclase in positive control of the lac operon.

Answer 20

When glucose is absent, adenylyl cyclase is activated and [cAMP] increases. The protein that regulates adenylyl cyclase is IIA(glc). When there is no glucose, IIA(glc) interacts to the transporter complex that usually transports glucose. It is then phophorylated and interacts with adenylyl cyclase. This then starts to catalyse the synthesis of cAMP from ATP.

cAMP then goes on to increase the transcription of lac operon.

Question 21

Why is the CAP-cAMP complex essential for efficient transcription of the lac operon?

Answer 21

cAMP binds to CAP (catabolite activator protein) and this interaction enhances the transcription of lac operon. CAP-cAMP binds to the CAP binding site upstream of the lac operon. The CAP binds to this site more efficiently when it is in the complex.

CAP increases the affinity of the RNA Pol for the weak lac promoter by interacting with the C terminal domains of the alpha subunits, therefore enhancing the level of transcription. It is thought that CAP bends the DNA and this change in conformation is responsible for the increased affinity.

Question 22

What is the proposed model for the regulation of adenylyl cyclase activity?

Answer 22

Adenylyl cyclase has two domains: the N terminal catalytic domain and the C terminal regulatory domain. The IIA(glc) protein binds to the regulatory. It is thought that when it has bound, there is low activity but high activity is reached when IIA(glc) is phosphorylated. Then it produces a higher concentration cAMP.

Question 23

Discuss why the presence of glucose inhibits lac operon transcription.

Answer 23

When glucose is present in the cell, it is converted into glucose-6-phosphate when it enters. This means that the IIA(glc) is dephosphorylated and this blocks the activity of adenylyl cyclase. Therefore, [cAMP] remains low and therefore [CAp-cAMP] remains low.

Question 24

What effect would an Oc mutant lac operator have on the transcription of the lac operon? (In the absence of glucose)

Answer 24

Oc (c=constitutive) have a mutant operator that is non-functional. Therefore, it continuously transcribed the lac operon even in the absence of lactose. Oc mutants would not be able to bind the lac repressor to its lac operator and therefore RNA Pol could bind to the promoter and initiate transcription.

Question 25

What effect would a mutant lac repressor (lac i) have on the transcription of the lac operon? (In the absence of glucose)

Answer 25

If the lac repressor was mutated, then it would not be able to bind to the operator region of the lac operon. Therefore, RNA Pol could bind to the promoter and initiate transcription.

Question 26

Describe how the use of merodiploid cells led to the prediction of the existence of the operator and repressor.

Answer 26

Merodiploid cells are those that carry two copies of any chromosomal segment. For example, in this case, cells that carry one wild-type and one mutated lac operon. The two constitutive mutants could be differentiation as either acting cis or trans.

The merodiploid containing the inactive lac i mutant cannot make the repressor protein. However, there is enough repressor in the cell to exert the effect onto both operons. This means that the repressor acts in trans. This means that the protein is able to cross the cell to exert its efect.

In the merodiploid cells with the non-functional operator sequence, this prevents the repressor from binding to it. Even though there are repressors present, it cannot bind so the lac products are produced in the absence of lactose. Therefore, the operator acts in cis. It is cis because it exerts its effect only on its own chromosome.

Jacob and Monod then predicted that the operator was controlled by a protein, the repressor, and this protein acts in trans and can cross the cell. Whereas, the operators act as cis because it only has a local effect directly on the DNA it is a part of, so it is integrated in the DNA.

Question 27

Describe the structure of the lac operon.

Answer 27

The most upstream feature is the CAP site, where the CAP binds to promote transcription of the operon.

Then it is the promoter site, followed by the operator site. The promoter site is where RNA Pol binds to initiate transcription and the operator is where the repressor binds.

Three genes of the operon are lac z, lac y, and lac a. Lac z is the most upstream of the genes and encodes for beta-galactosidase. Lac y encodes lactose permease, which is a transmembrane protein that ‘pumps’ lactose into cells. Lac a encodes a transacetylate, an enzyme that catalyses the transfer of acetyl groups. It is not known what role this plays in lactose metabolism.

Question 28

What is the difference between inducible and repressible operons? When are they used?

Answer 28

Inducible operons are where the presence of the substrate induces/upregulates transcription of the operon. Inducible operons are usually for catabolism.

On the other hand, repressible operons are where transcription is switched off when there is sufficient product, so the presence of the product represses/down regulates transcription. Repressible operons are usually for anabolism.

Question 29

Describe the function and structure of the trp operon.

Answer 29

Trp (tryptophan) operon is controlled by the concentration of trp in the cell - it is a repressible operon.

Trp operon encodes five proteins needed for tryptophan biosynthesis - trpE, trpD, trpC, trpB, and trpA. Thy catalyse the conversion of chorismic acid into tryptophan. At the most upstream end, there are the promoter sequence (for RNA Pol binding) and the operator (repressor binding). Between the P/O sequence and the genes, there are two sequences, termed trpL and trpa, which is the leader sequence.

Upstream of this operon, there is also a trpR sequence that encodes for the repressor.

Question 30

Describe the negative regulation of the trp operon.

Answer 30

trpR encodes aporepressors (apo=not complete) that form dimers and bind a trp each to form the complete repressor. This binds to the operator sequence, which blocks the promoter so that the RNA Pol cannot bind to the DNA. This therefore means there is no transcription of the mRNA and no tryptophan synthesis.

Question 31

What is the process of attenuation in trp operon transcription and what is it for?

Answer 31

Attenuation is another level of regulation on the trp operon. It prevents transcription of the operon when [trp] is high, but instead of blocking the initiation of transcription, it blocks the completion of transcription.

This involves the trpL and trpa of the leader sequence. trpL encodes a short polypeptide of 161 bps and trpa consists of four GC-rich attenuator sequences. These attenuators are complementary and can either form a terminator hairpin loop structure or an anti-terminator loop, depending on whether there is high or low [trp].

Out of the four attenuator sequences, if the 3rd and 4th sequence complementary bind to form a hairpin loop this is called the terminator loop. This forms if the RNA Pol is transcribing at a high speed through the DNA, which would mean that [trp] is high so the transcription is repressed this way. On the other hand, when [trp] is low, transcription by RNA Pol of the attenuator sequence is slower and the anti-terminator loop forms between the 2nd and 3rd sequence, which allows transcription is continue.

Question 32

What role does simultaneous transcription + translation play in the attenuation process of trp operon?

Answer 32

Attenuation is also affected in the translation stage because the leader sequence contains two trp residues. This means that depending on whether the [trp] is high or low, the speed of translation will differ.

When [trp] is low, the ribosome will stall at the two trp sequences in the mRNA, which is very close to the 1st attenuator sequence. This allows regions 2 and 3 to form the stem loop, therefore forming the anti-terminator sequence and allowing transcription of trp to occur.

When [trp] is high, this means that the ribosome is not delayed in trying to find trp residues for its translation so it forms the 1/2 and 3/4 stem loops, leading to termination of transcription.

Question 33

What is a regulon and how is it transcribed?

Answer 33

A regulon is a set of genes that are regulated as a unit. Unlike an operon, genes in a regulon have their own promoters which are all similar in sequences and so will be stimulated by the same signal.

Question 34

What is the pho regulon and what kind of proteins are encoded in the regulon in E. coli?

Answer 34

The pho regulon is the set of genes that regulates the [phosphate] in cell. Therefore, it includes proteins for sensing the [phosphate], proteins that act as transcription factors for the regulon, and ~30 genes that encodes proteins that increase the [phosphate].

In E.coli, there is a two-part regulation system of [phosphate] - it requires sensors (Pho R) and regulators (Pho B). Pho R is a histidine kinase that is responsible for sensing phosphate and for initiating Pho B when the [phosphate] is low. Pho B is then able to bind to DNA and assist RNA Pol in initiating and enhancing transcription of the genes required to increase [phosphate].

Question 35

Describe how the PstSCAB complex is able to act under sufficient phosphate concentration.

Answer 35

PstSCAB is a transmembrane protein that sits in the periplasmic membrane. It consists of PstS, PstC, PstA and PstB that make up the PstSCAB, as well as Pho U.
Pst = Phosphate-specific transporter

PstS is the phosphate binding protein that is free in the periplasmic space. When PstS has bound a phosphate, it moves through the periplasm to the membrane and binds to PstC and PstA (which are the part of the complex that sit inside the periplasmic membrane). This in tern interacts with PstB, which is a dimer that sits on the intracellular side of the membrane. The whole complex then acts as a channel for the phosphate to pass through into the cell. When there is sufficient [phosphate], PhoR is not needed so PhoU inactivates it while this happens.

Question 36

Describe how the PstSCAB complex is able to signal to PhoR when there is insufficient phosphate concentration.

Answer 36

When there is little to no phosphate, PstS does not interact with the rest of the complex and so the phosphate transport channel closes. This change in conformation in turn changes the conformation of PhoU. This signals to PhoR, which is next to PhoU, that there is a change. PhoR auto-phosphorylates itself on a histidine residue and thereby activates. Active PhoR is then able to phosphorylate PhoB on an arginine residue (requires ATP). The now activated PhoB can go on to bind to the DNA to initiate transcription.

Question 37

What role of the activator PhoB play in promoting transcription of the pho operon?

Answer 37

When the PhoB is activated by phosphorylation, it is able to form a dimer, which can then go on to bind to the Pho box. The Pho box is upstream of the promoter and allows the PhoB dimer is recruit RNA Pol by interacting with the sigma factor and assisting its assembly.

Question 38

Describe the symbiotic relationship between bacteria and Vibrio fischeri.

Answer 38

The bacteria colonise the squid’s light organ and the luminescence produced by the bacteria provides a camouflage effect for the squid when hunting its pray. In turn, the squid provides the bacteria with a nutrient rich environment within its organ. In small numbers, the bacteria do not fluoresce as it requires a high density of bacteria (the quorum) in order to lead to luminescence.

Question 39

Describe the autoinduction of the lux operon transcription.

Answer 39

Each bacterium produces a signalling molecule, called the auto-inducer (3-oxo-C6). 3-oxoC6 increases the transcription of the lux operon. Therefore, the more bacteria there are, the more 3-oxo-C6 there is and the more transcription of the lux operon there is. The luxI gene encodes 3-oxo-C6 synthetase, which catalyses the synthesis of 3-oxo-C6. It is called auto-induction of it increases transcription from the products of its own operon.

It is able to up-regulate transcription because it binds to the DNA binding protein luxR. LuxR forms a dimer, with each bound to 3-oxo-C6, and binds to the lux box of the operon to increase transcription.

Question 40

Which protein plays a role in light production in V. fischeri and what is the function of each of these proteins in the production of luminescence?

Answer 40

luxA and luxB encode for the two subunits of luciferase, the enzyme termed a mono-oxygenase because it converts aldehydes into carboxylic acids - a reaction that produces light. luxC, luxD and luxE all encode a part of the fatty acid reductase complex, which will catalyse the reaction to produce long chain aldehyde substrates for luciferase. luxG encodes a flaxin reductase, which is responsible for converting FMN to FMNH2 (which is also required for the luciferase reaction).