Gene Regulation in Bacteria 1

Question 1

What 3 important factors are needed to allow transcription to occur in bacteria?

Answer 1

• DNA binding proteins… Proteins that bind to DNA and dictate whether a gene will be transcribed.
• Sequence of DNA… specific motifs and arrangement of bases for the transcriptional machinery to bind and carry out transcription.
• Nutritional environment… availability of nutrients determines in some cases which genes are transcribed as and when needed.

Question 2

Describe the characteristic of translation and transcription in bacteria which differs to that in eukaryotes.

Answer 2

Translation occurs simultaneously to transcription i.e. as the mRNA is produced, it is translated into the protein.
This is possible due to the lack of nuclear envelope.
In eukaryotes, transcription and translation is separated and takes much longer.

Question 3

Outline how RNA is synthesised.

Answer 3

RNA is synthesised from a DNA template, via RNA polymerase in the 5’ to 3’ direction, antiparallel to the DNA strand.

Question 4

What are the three types of RNA, and what are their functions?

Answer 4

mRNA- directing the synthesis of proteins, and carrying the genetic code.
tRNA- carries the amino acid in protein synthesis.
rRNA- structural component of the ribosome.

Question 5

What is the experimental evidence for the importance for the 3’ OH group in phosphodiester bond formation?

Answer 5

If a mutated version… a 3’ deoxyadenosine is given to cells… no phosphodiester bonds are formed, indicating the hydroxyl on the 3’ end is essential for the formation of this bond.

Question 6

How many species of RNA polymerase are there in bacteria?

Answer 6

Only 1, which synthesises all RNA except the short RNA primers needed for DNA replication.

Question 7

Outline what RNA polymerase is.

Answer 7

It is a multi-subunit protein, that recognises a specific nucleotide sequence (promoter) ,where it initiates transcription, and makes a complementary RNA strand copy of the DNA template strand.

Question 8

Outline the structure of RNA polymerase.

Answer 8

It is a multisubunit protein, containing 5 different subunits, each important in its function… sigma, 2x alpha, beta and beta prime.

Question 9

Briefly describe the mechanism of function of RNA polymerase.

Answer 9

The sigma factor recognises and binds to the promoter region, before transcribing the DNA sequence in question.

Question 10

What is the difference between the core enzyme and the holoenzyme?

Answer 10

Core enzyme:

• Has four subunits, 2x identical alpha subunits, beta and beta prime.
• This enzyme lacks specificity i.e. cannot recognise the promoter region on the DNA template.

Holoenzyme:
-The four subunits, with the additional sigma factor, which enables RNA polymerase to recognise the promoter regions on the DNA template, and required for the correct initiation of transcription.

Question 11

What are the main functions of RNA polymerase with respect to transcription?

Answer 11

Binding:
-Must recognise the beginning of a gene.

Initiation:
-Must insert the correct nucleotides into position as dictated by the DNA template.

Elongation:
-Must catalyse the formation of a phosphodiester bond.

Termination:
-Must recognise the end of a gene.

Question 12

What experimental evidence is there for the existence of multiple polypeptide subunits in RNA polymerase?

Answer 12

Cellulose acetate chromatography. Proteins are loaded onto one end and current is passed through, separating the core subunits of the protein. Shows that RNA polymerase is made from multiple components.

Question 13

What is the promoter region?

Answer 13

The region upstream of the start of transcription, which is where the sigma factor of RNA Polymerase will bind. It is a specific DNA sequence recognised specifically by the relative sigma factor.

Question 14

Outline the process of binding in RNA polymerase, in transcription.

Answer 14

The sigma factor of RNA polymerase binds to the promoter region, allowing the other factors to associate once the sigma has bound. They can not attach on their own as they lack the specificity.

Question 15

What is a consensus sequence?

Answer 15

A sequence of DNA that has a similar structure and function across different organisms.

Question 16

What is the importance of the sigma factor?

Answer 16

To recognise the promoter region and lead to the assembly of the rest of the RNA molecule.

Question 17

Compare the structure and importance of the -35 and the -10 regions of the promoter.

Answer 17

The -10 and -35 box are separated by around 17 base pairs.
-35 box:
> The first control sequence.
>Initial point of contact of the holoenzyme sigma factor.

-10 box:
> The Pribnow box.
> Site of initial DNA unwinding/ melting.
> Melting converts the closed complex to an open complex… transcription bubble.
> This is where the core part of RNA polymerase binds.

Question 18

What is the base sequence for the -35 and -10 regions?

Answer 18

• 35: 5’-TTGACAT-3’

- 10: 5’-TATAAT-3’

Question 19

What region does the Pribnow box correspond to?

Answer 19

The -10 box.

Question 20

Compare strong promoters to weak promoters.

Answer 20

Strength of a promoter is based the affinity of RNA polymerase and the sigma factor for the consensus sequence… i.e. how much the consensus sequence matches the RNA polymerase.
STRONG- High affinity for the RNA polymerase to the DNA. Has AT rich regions, which are easier fo the endosome to break apart.
RNA coding genes also have an UP-elements usually as well as other promoters… helps to bind RNA efficiently and for transcription to proceed effectively.

WEAK- Weak affinity between RNA polymerase and the DNA. Some sequence deviation, and no UP -element. Weak promoters give a low transcription and thus a low level of protein product.

Question 21

Give an example of a weak promoter and its significance.

Answer 21

Lac operon… sequence deviation at the -35 and -10 box…
-10: TATGTT… GT in place of AA… more difficult to separate and transcribe the gene, helping with control of the sophisticated nutrient regulation of the operon. Not a hindrance… helps with control of the sophisticated nutrient regulation of the operon.

-35: TTTACAC…

Question 22

What experimental evidence is there for these promoter regions?

Answer 22

DNase 1 footprint experiment… using deoxyribonuclease 1 enzyme which digests the DNA, cleaving the phosphodiester bond. Can see from the footprint… the region where RNA polymerase has bound… as no cleavage occurred in this region.

Question 23

Outline the basic process involved in DNA footprinting.

Answer 23

Two different test tubes… both with piece of cloned DNA and the DNase1 enzyme. but only one with the RNA polymerase enzyme. One strand of the cloned DNA is radioactive allowing it to be visualised later.
When DNase is added, will begin to cleave the DNA fragment, on average every base. However if RNA polymerase is bound, then protects the DNA in this region from cleavage… so get a footprint in this region when products are electrophoresed on an acrylamide gel. Can see RNA pol covers the start site of transcription, promoter region and is facing in the right direction for transcription.

In different lanes, put different things… one lane for each… cleaving G, cleaving purines (A +G), cleving pyrimidines (C+T), and cleaving C. Don’t need one for A and T as can work out from the others.

Question 24

What are the roles of the alpha, beta and beta prime subunits of the RNA polymerase?

Answer 24

alpha- assembly of the core enzyme… initiates RANP assembly by dimerising to form a platform on which the beta subunits can interact.

beta- Template DNA binding.

beta prime- Nucleotide binding and catalytic activity (phosphodiester bond).

Question 25

What experimental evidence is there for which subunit binds the nucleotide?

Answer 25

Add radio-labelled nucleotides to RNA polymerase then electrophorese… can see that only one subunit contains the radio-labelled base- Beta subunit…. the one with the nucleotide binding site.

Question 26

What is the first stage of RNA polymerase binding?

Answer 26

Sigma 70 recognises and binds to a specific DNA sequence, the promoter region at -10 (orientates RNA polymerase), and -35 (recognition site) consensus sequences.

Question 27

What is a closed promoter complex, and when does it occur?

Answer 27

When the proteins have bound to the DNA, and nothing has happened yet.
The two DNA strands are still hydrogen bonded together, and proteins are lined up correctly , ready to move along strand when transcritn starts.

Question 28

What is an open promoter complex, and when does it occur?

Answer 28

It occurs when the two strands start to open up at the -10 region, allowing transcription to occur.
Therefore there is unwinding at the front of the enzyme, and rewinding of the DNA molecule behind the enzyme. This forms a transcription bubble… which is the region where the two strands are separated anr RNA polymerase is transcribing the gene.

Question 29

What occurs in the elongation phase of RNA transcription?

Answer 29

The formation of the polymer… Phosphodiester bond formation between the nucleoside triphosphate, one at a time.
The 5’ end of the RNA chain is displaced as DNA helix reforms. As it is displaced, translation can begin (protein synthesis).

Question 30

What happens in transcription termination, and what are the main two types?

Answer 30

The RNA polymerase dissociates from the DNA, and the double strands join together again as the helix reforms. The newly synthesized RNA must be released.
Rho independent, and Rho dependent termination.

Question 31

What is rho independent termination?

Answer 31

Intrinsic termination… When a sequence intrinsic in the DNA code, dictates termination by forming a structure that physically blocking the RNA polymerase, leading to termination. This is the most common form.

Termination sequence has two inverted repeats… High GC content, so strong binding. When this region is transcribed into ssRNA… these regions will H-bond together to form a stem loop structure… very strong due to GC bonding. This physically blocks the RNA polymerase, causing it to stall, and the mRNA to dislodge from the RNA polymerase active site, releasing the RNA chain and the RNA pol.

Question 32

What is rho dependent transcription?

Answer 32

Termination hat is dependent upon the hexamer protein Rho.

The Rho proteins binds a C-rich rho recognition site… terminator site, near the 5; end of the nascent RNA (40bp lng), and using ATP ase activity, moves along RNA until it reaches the paused RNA polymerase at the termination site. The RNA-dependent ATPase-dependent helicase activity to unwind the DNa/ RNA hybrid., causing release of the mRNA.
Hydrolysis of ATp is used to move the growing 3’ ed of the RNA chain out of the active site of the RNA polymerase…. so no 3’OH group held by RNA polymerase sitting on the template strand… so nothing to add nucleotides to.

Question 33

Compare the mRNA in bacteria to that of eukaryotes.

Answer 33

In bacteria, no introns, just leader and trailer sections, flanking either side of the protein coding region. These are not translated into proteins. No splicing in bacterial cells.

Question 34

What is the significance of the leader and trailer regions in the bacterial mRNA?

Answer 34

Leader region- Ribosome binding for the translation of the protein.

Trialer region-
Where the Rho independent terminator will occur… involved in transcription termination.

Question 35

How does control of bacterial gene expression vary with different proteins, i.e. housekeeping proteins vs metabolic proteins etc.

Answer 35

Some proteins are always present at a constant amount… i.e. housekeeping genes - for DNA, RNA synthesis etc. cell’s basic functions.

Other proteins will be present in varying amounts depending on the demand… i.e. for proteins involved in sugar metabolism, and amino acid synthesis.

Question 36

What is an operon?

Answer 36

A unit made up of linked genes, involved in the same metabolic process, that are generally transcribed together.

Question 37

What is the benefit of operons to bacteria?

Answer 37

Saves time and energy, as proteins not transcribed when they are not needed.

Question 38

How are the proteins in an operon transcribed?

Answer 38

Into one long mRNA… polycistronic mRNA.

Question 39

How is polycistronic mRNA from an operon converted into the individual proteins?

Answer 39

One promoter is used to control the transcription of the genes… saves space on the genome, and can respond very quickly when there is a particular food source available … however in each gene/ protein code, there will be a start codon for each of the proteins to be synthesised, and one stop codon (i.e. AUG). Ribosomal subunits attach to each initiation site, and can begin translating the mRNA into separate peptides.

Question 40

What are the two types of operon?

Answer 40

Inducible:
- Transcription must be turned on when required.

Repressible:
-Transcription can be turned off when they have enough of nutrient, so don’t waste time and energy aking more than is needed.

Question 41

What is the basic concept of control in operons?

Answer 41

All operson involve a small molecule and a regulatory protein.

Question 42

What kind of operon is the lac operon?

Answer 42

An inducible operon ie. turned on when lactose is present and glucose is low.

Question 43

Give an overview of the importance of the lac operon and how and when it functions.

Answer 43

Important to conserve energy and time for the bacteria. To transcribe the genes needed in the metabolism of lactose to be able to use lactose as an energy source when glucose is low and lactose is present.

Question 44

Describe the main components of the lac operon.

Answer 44

Structural genes... LacZ, LacY and LacA.
Promoter Plac.
Operator region. 
LacI... repressor protein. 
cAMP/CAP complex. 
Allolactose inducer.

Question 45

What are the structural genes involved in the lac operon?

Answer 45

LacZ- beta-galactosidase… to cleave lactose from its monomers, and for conversion of lactose to its isomer allolactose.

LacY- galactosidease permease… used to transport lactose past the plasma membrane… it is a membrane protein… transporter.

LacA- Transacetylase… to convert toxic thiogalactoside transacetylase metabolite from process… transported out cell by LacY.

Question 46

What region controls the structural genes?

Answer 46

Promoter region, just upstream to the operator.

Question 47

What is the importance of the operator region?

Answer 47

The operator region is the region where the repressor proteins bind, important in the regulation of the lac operon.

Question 48

What is LacI, where is it located and what is its function?

Answer 48

It is a regulatory gene, coding for the repressor protein, which binds to the operator region. Transcribed independently of the operon… has its own promoter… constitutively active. Binds to and blocks the operator region… stops the RNA polymerase from binding.. repressing the operon.

The repressor has two mutually exclusive binding sites, one for allolactose, and one for the operator, but cannot bind both of these at the same time.

Question 49

Describe the characteristics of the repressed operon.

Answer 49

The repressor protein encoded by the LacI gene sits on the operator region, preventing RNA polymerase from binding, so no transcription of the operon. This occurs when there is no lactose present, and high glucose… so low cAMP.

Question 50

Describe the characteristics of the activated operon.

Answer 50

Lactose is present in higher amounts and enters the cell… allolactose is the isomer that acts as the inducer, by binding to the repressor protein, causing a conformational change so that it drops off the operator region, hence allowing RNA polymerase to bind. It can then begin transcribing the genes. Lower glucose levels, mean higher cAMP levels, so can bidn to CAP, binding to CAP site by the RNA polymerase binding site, increasing its affinity, and enhancing transcription.

Question 51

What experimental evidence is there for the repression by the LacI repressor at the operator region

Answer 51

DNase I footprint.
Can be used to see overlapping footprints of the repressor and the RNA polymerase… showing they bind to similar parts of the gene, and so that the repressor covers the binding site for RNA polymerase when it is present.

Question 52

How exactly does the operator region and repressor proteins repress transcription of the operon?

Answer 52

There is a primary/ principal operator region (O1), and an auxiliary operator region (O3) further upstream to the promoter. There are two half sites in each operator, which are inverted repeated motifs, suggesting protein binding occurs in these regions. A repressor protein will bind to each of the half sites, meaning that there are four proteins in total which bind, one to each half site. The auxiliary operator region bends over to form a tetramer with the principal operator region, blocking the promoter region, preventing RNA polymerase from binding.

Question 53

What conformation is the active repressor?

Answer 53

A tetramer.

I.e. two repressor molecules bound to each operator.

Question 54

Outline the mechanism of positive regulation of the lac operon.

Answer 54

The binding of cAMP to CAP, which binds to the CAP site and massively facilitates transcription.

Question 55

Outline the mechanism of negative regulation of the lac operon.

Answer 55

The binding of the lac repressor to the operator… operon is off when the repressor is bound.
Glucose also acts as a catabolite repressor.

Question 56

What is catabolite repression?

Answer 56

When a metabolite of glucose inhibits the transcription of the lac operon…
[glucose] is inversely proportional to [cAMP].

Question 57

Compare the conditions when glucose is present vs when glucose is absent with respect to the level of cAMP.

Answer 57

Glucose present:
- Glucose is immediately phosphorylated to G.6-P when it enters the cell by hexokinase, this traps it in the cell. In this case, adenylyl is inactive.

Glucose not present:
-In glucose absence, the enzyme hexokinase is able to phosphorylate the adenylyl cycle protein, which activates it, allowing it to convert ATP to cAMP… only in glucose absence… as the enzyme is free to phosphorylate the adenylyl cyclase.

Question 58

What is the relevance of cAMP to the lac operon?

Answer 58

Increases the efficiency of the RNA polymerase binding to the promoter, increasing efficiency of transcription.
CAP/cAMP complex can bind to the DNA and bend it >90 degrees around the centre of symmetry to increase the affinity of RNA polymerase to the lac promoter.

Question 59

How does cAMP influence the lac operon?

Answer 59

cAMP binds to CAP (catabolite activator protein- transcriptional factor)… Used to assist RNA polymerase in effective binding to the weak promoter, binds to CAP site upstream of the promoter region, where it can also bind to the alpha-Carboxy-terminal -domain subunit of the RNA polymerase, locking it into place, bringin the RNA polymerase to site, and keeping it effectively bound… increasing the affinity of the RNA polymerase for the promoter.

Question 60

What level of transcription will occur in the following scenarios:

• LOW lactose, HIGH glucose (low cAMP).
• HIGH lactose, HIGH glucose (low cAMP).
• HIGH lactose, LOW glucose (high cAMP).

Answer 60

• None… as repressor is blocking the operator.
• Low levels… as the repressor is not there, but low cAMP, so the RNA polymerase still has a low affinity for the promoter.
• High levels… RNA polymerase is efficiently bound due to cAMP/CAP complex, and can transcribe to a high rate.

Question 61

What are two anomalies regarding the lac operon?

Answer 61

• Allolactose not lactose acts as the true inducer molecule to remove repressor from the operator region… this requires the enzyme from LacZ beta galactosidase, to convert it to the isomer.
• Also, lactose requires permease to get into the cell in the first place… which is encoded by LacY gene, transcirbed by the lac operon.

So if there wasn’t the lac operon in the first place, you couldn’t get lactose in or operon activated.

Question 62

What is the solution for the two anomalies of the lac operon?

Answer 62

In order for the lac operon to get activated, it requires a low level of transcription of the lac operon… as the proteins encoded by LacZ and LacY are needed at low levels.

Therefore, as the binding of the repressor is never infinitely strong, i.e doesn’t never come off… once every cell generation, the repressor will drop off, so always a low level of lac transcription. Essential to enable low level of protein synthesis otherwise would have no activation of the lac operon.

Therefore, the lac operon is tightly regulated, but there is always a low level of transcription.

Question 63

Who investigated the distinction between the regulatory substances and sites where they act to change gene expression?

Answer 63

Jacob and Monod.

Question 64

What did Jacob and Monod recognise?

Answer 64

The distinction between the regulatory substances and the sites at which they act to change the gene expression.

Question 65

Briefly describe the test used by Jacob and Monod.

Answer 65

Diploid cells created, with bacterial chromosomes, and an additional plasmid, so effectively have two lac operons.
Carried out complementation tests to investigate the functions and characteristics of the repressor and operator, by mutating one and seeing if a functional copy could complement the mutation.

Question 66

Outline the complementation test for the repressor.

Answer 66

One copy of the lac genes carries a mutation in LacI, but the other is a wild type… So one repressor protein will be mutated… so would lead to constitutive transcription in a haploid cell…. however in a diploid cell, where there is a functional repressor gene… the phenotype is normal as it complements the mutation, so the repressor from the plasmid can bind to the operator on the other chromosome, repressing the operon. So wild type is dominant and the repressor is said to act in trans… because can influence other genes not directly next to it… other genomic material… crossing over of genetic material.

Question 67

What is a complementation test?

Answer 67

When two strains of an organism with homozygous recessive mutations crossed… wild type phenotype is established. I.e. the functional gene complements the cmuateted one… as restores wild type function to it.

Question 68

Outline the complementation test for the operator.

Answer 68

When operator site is mutated, no complementation is seen if the plasmid has a functional operator region… shows it acts in cis… LacOc mutation… as operator constitutively active… RNA polymerase can bind and transcribe with no control.
LacZ also mutated in the chromosome but functional on the plasmid, as an extra measure to ensure correct information is obtained, and to be able to distinguish more easily between the Plasmid and the chromosome.

Question 69

Which elements of the lac operon act in cis and which in trans?

Answer 69

Cis- the operator… indicating it is a DNA sequence… cannot move between genetic elements.

Trans- the repressor… indicating it is a protein, can move between genetic elements.

Question 70

What is the purpose of IPTG in the complementation tests.

Answer 70

Is a molecular mimic of lactose/ allolactose inducer… can bind to the operator and cause it to fall off the operator.