Gene Regulation in Bacteria

Question 1

What are the 3 important factors when DNA is transcribed into RNA?

Answer 1

1) DNA-binding proteins
2) Specific DNA sequences that are recognised by these
3) The environment in the sense of the availability of nutrients

Question 2

In bacteria RNA synthesis and protein synthesis take place at the same time. Why, and how can this be visualised?

Answer 2

Bacterial cells only live for 20-30 mins so the process needs to be very efficient, so there is no nuclear membrane (unlike in eukaryotes), so the 2 processes are not separated.
This can be visualised on an electron micrograph, where ‘christmas tree’ structures can be seen.

Question 3

Can deoxyriboadenosine be used as a building block to form an RNA strand?

Answer 3

No, as it does not have have the 3’ hydroxyl group to form a phosphodiester bond with the next nucleotide (by liberation of a pyrophosphate). This experiment showed that ribonucleoside triphosphates are the building blocks for RNA synthesis.

Question 4

What are all the subunits that come together to form the whole RNA polymerase ‘holoenzyme’? How can they be separated?

Answer 4

There are: 2 alpha subunits, a beta subunit, and a beta prime subunit, which come together to form the whole enzyme. There is also a sigma factor. The subunits can be separated by cellulose acetate chromatography.

Question 5

What are the 4 things that RNA polymerase must do?

Answer 5

1) BINDING- recognise the beginning of a gene
2) INITIATION- insert the correct nucleotide into position (as dictated by the DNA template)
3) ELONGATION- catalyse the formation of the phosphodiester bond
4) TERMINATION- must recognise the end of a gene

Question 6

What unit of the RNA polymerase recognises and binds to the promoter at the beginning of the gene to be transcribed?

Answer 6

The sigma unit - it binds between -35 and -10 in the DNA promoter region. Other factors associate once sigma is bound. Sigma factor dissociates when the RNA chain is 8 nucleotides long (so is only needed initially).

Question 7

What are the -35 and -10 regions in the promoter?

Answer 7

The -35 region is the recognition site, and the -10 region orientates the RNA polymerase to allow it to only transcribe forwards in the correct direction.

Question 8

How does the composition of the -10 region allow efficient transcription?

Answer 8

The consensus sequence for the -10 region is TATAAT, and as it contains no G’s or C’s and is very close to the start of transcription, the composition aids the unwinding of the DNA strands due to the fewer hydrogen bonds.

Question 9

Explain the DNase 1 footprint experiment. Why was it done, what are the principles behind it, and how are the results visualised?

Answer 9

The experiment was done to prove that RNA polymerase binds to the promoter region. One test tube contained just DNA (a control), and the second test tube contained DNA with added RNA polymerase. DNase 1 was added to both tubes and is able to cleave phosphodiester bonds in the DNA. If the RNA polymerase is bound however, then it is unable to cleave the bonds (the DNA is protected). You can visualise the results by running the products (with radioactive labels) side by side on a polyacromide gel - allows you to see 1 base pair differences. Any gaps represent where the RNA polymerase has bound.

Question 10

What can increase the strength of a promoter?

Answer 10

UP-elements (upstream promoter elements) - allow RNA polymerase to bind effectively and transcription proceeds effectively.

Question 11

Why is the lac promoter weak?

Answer 11

As the -10 region has a G (is a variant) and there is no UP-element. This ‘weakness’ is actually needed to allow sophisticated ‘nutrient’ regulation.

Question 12

What is the function of the beta subunit of RNA polymerase? How was this confirmed?

Answer 12

Binds the nucleotide and controls the catalytic activity (forms the phosphodiester bond). This subunit binding the nucleotide was confirmed by adding radioactive labelled nucleotides and seeing which subunit is radioactive.

Question 13

What is the function of the beta prime subunit of RNA polymerase?

Answer 13

Involved in template DNA binding.

Question 14

What is the function of the alpha subunit of RNA polymerase?

Answer 14

Assembles the core enzyme and is involved in promoter recognition.

Question 15

What 2 things must happen in order to terminate RNA synthesis?

Answer 15

The newly synthesised RNA must be released and the RNA polymerase must dissociate from the DNA template.

Question 16

Explain rho independant termination.

Answer 16

The terminator sequence contains inverted repeats that have a high GC content, and once transcribed these 2 inverted repeat sequences can hydrogen bond together in the RNA- a stem loop/hairpin loop structure is formed. There is a string of U’s after this, and as the GC bonds are much stronger than the AU bonds the RNA is pulled away from the polymerase; DNA strands will rather bond together than with the mRNA synthesis. This leaves a string of U’s at the end of the mRNA.

Question 17

Describe rho.

Answer 17

It is a hexamer protein that binds to transcripts only after the rho protein has been translated and the ribosome has left. It has ATPase activity, is C-rich, but its method of termination is unclear.

Question 18

What is an operon?

Answer 18

The clustering together of genes that code for similar proteins involved in the same metabolic pathway (lay side by side on a chromosome). When transcribed produces polycistronic mRNA, with each gene on the mRNA having its own start and stop codon -> allows multiple proteins to be produced from one mRNA, and allows more than one ribosome to bind to the mRNA.

Question 19

What is the benefit of an operon?

Answer 19

All the genes are controlled from one promoter and so less space is taken up, and it ensures that all the genes needed for the pathway are transcribed at the same time together.

Question 20

What 2 things does the control of an operon always involve?

Answer 20

1) A small molecule

2) A regulatory protein

Question 21

Inducible operons are operons in catabolic pathways. What is the small molecule for these operons?

Answer 21

The small molecule is the substrate (or closely related molecule) to the first enzyme in the metabolic pathway.

Question 22

What is the small molecule for the lac operon? What happens when you add more of the small molecule?

Answer 22

Lactose, as the first enzyme in the metabolic pathway is beta-galactosidase. When lactose is added, the amount of the enzyme increases greatly, and if you remove lactose it decreases dramatically.

Question 23

What gene encodes the repressor of the lac operon? When is it needed?

Answer 23

The lac I gene encodes the repressor, and is needed when there is no lactose so do not need transcription of the lac operon.

Question 24

What genes are in the lac operon? What do they encode?

Answer 24

Lac Z - beta galactosidase (breaks down lactose)
Lac Y - lactose permease (transporter)
Lac A - thiogalactoside transacetylase (helps remove toxic metabolites, that are transported out of the cell by the lac Y protein)

Question 25

How does the repressor act to repress transcription of the lac operon?

Answer 25

The repressor binds to the operator region and blocks the promoter binding site for RNA polymerase. This means that it cannot transcribe the DNA.

Question 26

How is the lac operon de-repressed?

Answer 26

When lactose is present, the operon is activated as the lactose binds to the repressor which changes its conformation so it can no longer bind to DNA.

Question 27

Is the lac I gene inducible?

Answer 27

No, it is transcribed from its own promoter and is constitutively expressed (all the time).

Question 28

What experiment could you use to show that the repressor and RNA polymerase have the same binding sites?

Answer 28

The DNase I footprint experiment.

Question 29

What alteration in structure upon repressor binding prevents the RNA polymerase from binding to the promoter?

Answer 29

2 repressor molecules bind to each half sites in both the O1 (principal operator) and O3 (auxillary operator) regions, and this causes a bend in the DNA from the O1 to O3 region. This physically blocks the promoter site from RNA polymerase.

Question 30

Why is glucose metabolism favoured over using lactose?

Answer 30

Glucose metabolism needs no new proteins, so when glucose is high this inhibits lac operon transcription (or when there is both glucose and lactose).

Question 31

How does glucose inhibit lac operon transcription?

Answer 31

By catabolite repression- if glucose is high then cAMP is low (as glucose is phosphorylated to glucose-6-phosphate, meaning that adenylyl cyclase is not phosphorylated and thus cannot produce cAMP from ATP).

Question 32

How do high cAMP levels (when glucose is low) stimulate lac operon transcription?

Answer 32

cAMP binds to CAP and the complex that forms assists RNA polymerase in binding effectively to the promoter by increasing its affinity for the weak lac promoter, by bending the DNA greater than 90 degrees around the centre of symmetry . It does this by interacting with RNA pol’s alpha subunit carboxy-terminal domain.

Question 33

What happens when glucose is present and no lactose?

Answer 33

cAMP is low and so catabolite repression occurs and the lac operon is inhibited. No cAMP-CAP complex forms to increase the affinity of RNA pol for the weak lac promoter.

Question 34

What happens when both glucose and lactose are present?

Answer 34

Catabolite repression still occurs as the presence of glucose means that the lac operon does not need to be transcribed, but some lactose will bind to the repressor which causes the lac operon to be derepressed in this sense, so a little lac mRNA is produced.

Question 35

What happens when there is lactose but no glucose?

Answer 35

There will be high levels of cAMP, which can bind to CAP and allow RNA polymerase to bind effectively to the lac promoter and so transcription of the lac operon is stimulated.

Question 36

What are the two ‘anomalies’ in the idea that lac operon is turned off in the absence of lactose?

Answer 36

1) The transport of the inducer (lactose) requires permease (lac Y)
2) The true inducer is allolactose not lactose, which is made from lactose by beta-galactosidase.

Question 37

Is the lac operon completely turned off in the absence of lactose?

Answer 37

No- there is always a low level of lac transcription as the binding of the repressor is never infinitely strong (will drop off approximately once every cell generation).

Question 38

What did Jacob and Monod do to prove that the repressor was a protein and not DNA?

Answer 38

Created partial diploid cells by introducing an F plasmid into a bacterial cell that contained a lac operon. The operon in the bacterial chromosome was lac I- meaning that there was a mutation in the repressor causing it to be unable to bind to the operator. Wanted to see if the lac I+ on the plasmid could complement lac I- to restore regulation of the lac operon. Found that it can as the lac operon in the chromosome was no longer transcribed in the absence of lactose. This proved that the repressor was a protein as a DNA would not be able to carry this information from plasmid -> chromosome.

Question 39

How did Jacob and Monod show what the repressor protein interacts with? What term did they come up with?

Answer 39

Came up with the term ‘operator’. Created mutation in the operator sequence of the chromosome so that the repressor cannot bind. Wanted to see if the normal operator on the plasmid (O+) could interact with this. Found that information was not passed from the plasmid to the chromosome -> indicated that the operator was a DNA sequence and not a protein.

Question 40

What is IPTG?

Answer 40

A molecular mimic of allolactose, and so triggers transcription of the lac operon.

Question 41

In Jacob and Monod’s experiment investigating the operator region, when ITPG was not present what did they observe?

Answer 41

The lac operon on the chromosome was not repressed. If information passed from the F plasmid to the operator of the chromosome then transcription of the operon would have been inhibited.

Question 42

What is a repressible operon and give an example of one.

Answer 42

A repressible operon is one in which transcription is switched off when there is sufficient product present in the cell as to not waste energy. An example is the Trp operon - as tryptophan concentration increases in the cell, the transcription of the trp gene decreases (opposite to the lac operon).

Question 43

Describe the trp operon.

Answer 43

Has a single promoter, with 5 genes coding for proteins (enzymes) that ultimately produce the amino acid tryptophan. There is a regulation sequence, with the attenuator sequence in the leader DNA encoded by trp a.

Question 44

There are two ways in which the trp operon is regulated. Explain how it is regulated by the aporepressor and corepressor.

Answer 44

The aporepressor is coded for upstream of the operon by the trp A gene, and tryptophan (the co-repressor) binds to the aporepressor to form the whole repressor. This can then bind to the DNA and block the binding site for RNA polymerase. Therefore, the trp operon is under negative control of the active repressor.

Question 45

How is the trp operon regulated by the attenuation mechanism?

Answer 45

This involves incomplete transcription: the attenuator sequence senses if there is enough tryptophan, and if there is then transcription will be stopped and the RNA polymerase will drop off (only 161 bases of the trpL leader sequence is transcribed).

Question 46

What 2 things does attenuation depend on?

Answer 46

1) The folding pattern of the leader mRNA- if 1 base pairs to 2 then there is no transcription, 2 to 3 it is transcribed, and 3 to 4 then then only the leader mRNA is transcribed.
2) The extent to which the ribosome has translated the leader mRNA- as region 1 of the leader mRNA encodes a leader peptide that has 2 tryptophans.

Question 47

When region 3 of the leader mRNA is bound to region 4, why is there transcription of the leader mRNA only?

Answer 47

As rho-independant termination occurs -> the hairpin/stem loop pulls the mRNA away from DNA as the G-C bonds in the loop are stronger than the following A-U bonds, and so the DNA strands will preferentially bind together rather than with the mRNA strand.

Question 48

Why is the extent to which the ribosome has translated the leader mRNA important? Relate this to when tryptophan is high and low and the position of the ribosome.

Answer 48

Region 1 of the leader mRNA has 2 trp codons in the leader sequence, and so if there is a high concentration of tryptophan tRNA will easily be able to bring 2 tryptophan residues and the leader sequence is transcribed. This will mean that the ribosome is covering region 2 of the leader mRNA, so regions 3 and 4 are forced to hydrogen bond and form a step loop -> RNA polymerase is released and transcription stopped.
Conversely, when tryptophan is low no tRNA will be able to bring tryptophan in order to allow translation of the leader mRNA. This means that the ribosome is stuck in region 1, 2+3 form a loop, and region 4 is not affected so RNA polymerase can still move along the DNA and transcribe.

Question 49

What are the two components in the regulatory systems of bacteria?

Answer 49

1) An activator (sensor) protein

2) A regulator protein

Question 50

Explain how phosphate levels can be regulated in a cell.

Answer 50

When phosphate levels are normal, the PhoR protein that spans the inner membrane is bound to phosphate, but under low phosphate conditions this periplamsic phosphate will be lost. This causes a conformational change in PhoR, resulting in it gaining a phosphate from ATP in the cell (a histidine in the cytoplasmic domain is phosphorylated). This activates the protein, and means it can phosphorylate PhoB, which can bind to genes next to promoter regions and allows RNA pol to bind and transcribe the genes.

Question 51

Give some examples of genes that can have increased expression as a result of PhoB/PhoR under low phosphate conditions.

Answer 51

phoA - alkaline phosphatase
phoS - phosphate binding protein
phoE - a porin, so more phosphate can get into cell
ugpB- phosphate transporter

Question 52

Large numbers/a quorum of what bacterium can cause fish and squid to luminesce? Why is this beneficial?

Answer 52

Vibrio fischeri- beneficial to host as prey comes up thinking it is moonlight, and good environment for the bacteria to live.

Question 53

What signalling molecule causes the luminescence?

Answer 53

VAI

Question 54

What is the first gene of the operon for the bioluminescnece?

Answer 54

lux I - encodes the VAI synthase that can synthesise VAI

Question 55

Why is the signalling molecule (VAI) auto inducing?

Answer 55

As it binds to the lux R protein (regulator), causing its shape to change so it can bind to the promoter -> results in more VAI synthase which then results in more of the small molecule VAI.

Question 56

What genes encode luciferase?

Answer 56

LuxA and luxB

Question 57

What genes encode luciferin (substrate of luciferase)? What reaction allows the production of light?

Answer 57

LuxC, luxD and luxE. Luciferin is oxidised by luciferase in the presence of FADH2 and this produces the light.

Question 58

What demonstrates the symbiotic relationship between the bacteria and squid?

Answer 58

The greater the number of bacteria, the more the epithelial cells of the light organ expand -> can accommodate more bacteria.

Question 59

How is mature tRNA produced?

Answer 59

Processing of the primary transcript (multiple tRNAs) occurs, and the flanking sequences are removed by ribonucleases. Ribonuclease P can make a cleavage at the 5’ end of the tRNA (which separates the tRNA molecules), and then ribonuclease D starts from the 3’ end and cleaves one phosphodiester bond at a time until it reaches CCA sequence.

Question 60

What enzyme cleaves the double stranded region of rRNA to produce the final processed RNA?

Answer 60

Ribonuclease III - shortens the strands