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BIOC2001 Dina > Gene regulation in bacteria > Flashcards

Gene regulation in bacteria Flashcards

1
Q

What are the three important factors for transcription?

A
  1. DNA-binding proteins
  2. specific DNA sequences
  3. environment - nutrient availability
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2
Q

What are the three main types of RNA?

A

mRNA
tRNA
rRNA

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3
Q

What are the functions of the three main types of RNA?

A

mRNA - directs the synthesis of proteins
tRNA - carries the amino acid
rRNA - structural component of the ribosome

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4
Q

What is the function of RNA polymerase?

A

synthesises RNA from a DNA template

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5
Q

What is the experimental evidence that the 3’ hydroxyl group is used to create the phosphodiester link?

A

3’ deoxyadenosine was phosphorylated, but no mRNA was produced

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6
Q

What are the subunits of bacterial RNA polymerases?

A

2α, β, β’, σ

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7
Q

Of which subunits is the core enzyme comprised?

A

α2ββ’

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8
Q

Of which subunits is the holoenzyme comprised?

A

α2ββ’σ

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9
Q

How is separation of the core subunits achieved?

A

by use of cellulose acetate chromatography

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10
Q

What are the role of RNA polymerase?

A
  • binding = recognise the beginning of a gene
  • initiation = insert correct nucleotide into position, as dictated by the DNA template
  • elongation = catalyse the formation of a phosphodiester link
  • termination = recognise the end of a gene
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11
Q

What is the main event involved in binding?

A

the sigma (σ) sybunit of RNA polymerase recognises and binds to the promoter

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12
Q

What are the major sites on the DNA?

A

+1 signifies the start of transcription

-10 and -35 sequences make up the promoter

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13
Q

Describe the sequences that make up the promoter

A
  • 35 region = recognition site

- 10 region = orientates RNA polymerase

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14
Q

What is the consensus sequence of the -35 region?

A

TTGACAT

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15
Q

What is the consensus sequence of the -10 region?

A

TATAAT

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16
Q

Approximately how many base pairs separate the -35 and -10 regions?

A

17 bp

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17
Q

How does the sequence -10 region achieve its function?

A

AT-rich - the two hydrogen bonds enable the strands to separate more easily so that RNA polymerase can catalyse the formation of the complementary RNA strand

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18
Q

What is a DNase footprinting assay?

A
  • detects DNA-protein interaction using the fact that a protein bound to DNA will often protect the DNA from enzymatic cleavage
  • this makes it possible to locate a protein binding site on a particular DNA molecule
  • the method uses the enzyme deoxyribonuclease (DNase) to cut the radioactively end-labelled DNA
  • this is followed by gel electrophoresis to detect the resulting cleavage pattern
  • the cleavage pattern of the DNA in the absence of DNA binding protein (free DNA) is compared to the cleavage pattern of DNA in the presence of DNA binding protein
  • if the protein binds DNA, the binding site is protected from enzymatic cleavage
  • this protection will result in a clear area on the gel, which is referred to as the ‘footprint’
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19
Q

What is the role of the α subunit?

A

assembly of core enzyme and promoter recognition

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20
Q

What is the role of the β subunit?

A

nucleotide binding and catalytic activity (phosphodiester bond)

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21
Q

What is the role of the β’ subunit?

A

template binding

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22
Q

What is a transcription bubble?

A

a molecular structure that forms when a limited portion of the DNA double strand is unwound to allow RNA polymerase to bind to the exposed DNA and begin synthesising a new strand of RNA

23
Q

At what point does the sigma factor dissociate from the DNA?

A

when the RNA chain is eight nucleotides long

24
Q

What are the main stages of elongation?

A
  • binding of correct nucleoside triphosphates
  • phosphodiester bond formation
  • RNA polymerase moves one nucleotide at a time along the DNA
  • melting and movement creates transcription bubble
  • 5’ end of RNA chain is displaced as DNA helix reforms
25
Q

What are the main stages of elongation?

A
  • RNA synthesis must end
  • newly synthesised RNA must be released
  • RNA polymerase must dissociate from the DNA template
26
Q

What is Rho-independent termination?

A
  • recognition of the terminator sequence
  • GC-rich regions are inverted repeats of one another (terminator region)
  • followed by a sequence that is AT-rich
  • complementary sequences have the potential to hydrogen bond to one another to form a hairpin loop of RNA (very strong binding)
  • this is followed by the weak A-U bonds
  • it is more energetically favourable for DNA to rehybridise to release the RNA chain
  • RNA polymerase is released and the DNA helix reforms
27
Q

What is Rho-dependent termination?

A
  • rho is a hexamer protein
  • binds to the transcript after the protein has been translated and the ribosome has left
  • has ATPase activity, but the mechanism is not clear
  • terminators are C-rich and ~40 nucleotides long
28
Q

What is the equivalent of the leader region?

A

UTR

29
Q

What is the equivalent of the protein-coding region?

A

ORF

30
Q

What is the function of the trailer?

A

ensures that RNA is the correct length and terminates the sequence

31
Q

How many proteins does on the E. coli chromosome code for?

A

~ 4000

32
Q

What is an operon?

A

a functioning unit of genomic DNA containing a cluster of genes under the control of a single promoter

33
Q

Define ‘polycistronic’

A

a single mRNA encoding several different polypeptide chains

34
Q

How many promoters do genes in an operon have?

A

one

35
Q

What type of operon is the lac operon?

A

inducible

36
Q

What is the first enzyme in the lac operon pathway?

A

β-galactosidase

37
Q

What is lactose broken down into?

A

disaccharide; broken down into glucose and galatose

38
Q

What does lacZ encode?

A

β-galactosidase (breaks down lactose)

39
Q

What does lacY encode?

A

lactose permease (transporter)

40
Q

What does lacA encode?

A

thiogalactosidase transacetylase (helps remove toxic metabolites eg. thiogalactosides, which are transported out of the cell by lacY protein)

41
Q

What is lacI?

A
  • repressor protein
  • has its own promoter
  • constitutively expressed
42
Q

How does the repressor work?

A

binds to the operator of lacZ/Y/A to repress the operon

43
Q

What is the role of lactose?

A

lactose binds to the repressor to prevent it from binding (conformational change) and to enable RNA polymerase to bind to the promoter to enable transcription, since the represssor and RNA polymerase binding sites overlap

44
Q

Explain the structure of the active repressor

A
  • a tetramer
  • two repressor molecules bind to each operator
  • two bind to the primary operator and two to the auxiliary operator
  • the repressor molecules associate to form a loop structure and prevent RNA polymerase from binding to the promoter, which is contained within the loop
  • the third operator can also form a loop with the primary operator, although this would prevent the action of the auxiliary operator
45
Q

What is the relationship between [glucose] and [cAMP]?

A

If [glucose] is high then [cAMP] is low

46
Q

Why does this relationship hold?

A
  • when glucose enters the cell, it is converted into glucose-6-phosphate to prevent it from leaving the cell
  • adenyl cyclase (makes cAMP) is inactive
  • in the absence of glucose, adenyl cyclase is phosphorylated and thus activated to produce cAMP
    ATP —> cAMP
47
Q

What is the effect of a high [cAMP]?

A
  • cAMP binds to CAP (catabolite activator protein)
  • cAMP-CAP complex binds to the DNA and interacts with RNA polymerase
  • CAP increases its affinity for the weak lac promoter
  • CAP bends DNA >90 degrees around the centre of symmetry
  • this exposes the DNA so that RNA polymerase can bind effectively
48
Q

Explain the scenario of glucose present, no lactose

A
  • [cAMP] low
  • repressor binds to operator
  • CAP cannot bind to DNA in the absence of cAMP
  • RNA polymerase cannot bind to promoter
  • no lac mRNA produced
49
Q

Explain the scenario of glucose present, lactose present

A
  • [cAMP] low
  • lactose binds to repressor to prevent its binding to the operator
  • CAP cannot bind to DNA in the absence of cAMP
  • RNA polymerase can bind to promoter
  • very little lac mRNA produced
50
Q

Explain the scenario of no glucose present, lactose present

A
  • [cAMP] high
  • lactose binds to repressor to prevent its binding to the operator
  • CAP forms a complex with cAMP and binds to DNA
  • CAP increases the affinity of RNA polymerase for the promoter
  • abundance of mRNA produced
51
Q

What are two ‘anomalies’ of the lac operon?

A
  1. transport of inducer requires permease (lac y)
  2. true inducer is allolactose, not lactose
    lactose —> allolactose, requiring β galactosidase
52
Q

How are these ‘anomalies’ resolved?

A
  • binding of the repressor is never infinitely strong
  • drops off approximately once every cell generation
  • therefore there is always a low level of lac transcription
53
Q

How was genetic analysis of lac operon regulation carried out by Jacob and Monod?

A
  • by use of a chromosome (lac I -) and an F plasmid (lac I +)
  • lac I - = mutation in repressor such that the protein cannot bind to the operator
  • the operator is ‘on’ even in the absence of lactose (constitutive mutation)
  • can lac I + on the F plasmid complement lac I - mutation on the chromosome to restore the regulation of the genes on the chromosome
  • F + plasmid causes the synthesis of wild-type repressor
  • ‘on’ in presence of lactose and ‘off’ in its absence
  • I + is dominant to I - and dictates the action and activation of the operon
  • I + acts in trans - repressor is transported to act on the operon
54
Q

How did Jacob and Monod determine if something tries to bind to the operator or if the operator itself encodes a protein?

A
  • O(c) = constitutive mutation in the operator sequence such that the repressor cannot bind
  • is any information passing between the operon on the chromosome and that on the plasmid?
  • β-galactosidase is non-functional to prevent it from affecting the experiment
  • the action of permease does not affect the experiment
  • mutation on the chromosome operator does not affect the action of the plasmid operator
  • this also proved that the operator was a DNA sequence, since it does not encode any protein

BIOC2001 Dina (6 decks)

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