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LSM2232 > DLW01 - Prokaryotic Transcription > Flashcards

DLW01 - Prokaryotic Transcription Flashcards

1
Q

Define the coding strand

A

The coding strand is the DNA strand that has the same sequence as the mRNA

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2
Q

Define the promoter

A

The promoter is a region of DNA where RNA polymerase binds to initiate transcription

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3
Q

Define the terminator

A

The terminator is a sequence of DNA that causes RNA polymerase to terminate transcription

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4
Q

Define the transcription unit

A

The sequence between sites of initiation and termination by RNA polymerase.

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5
Q

The transcription unit only includes one gene. TRUE or FALSE

A

FALSE.

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6
Q

The transcription unit may contain multiple genes. TRUE or FALSE

A

TRUE.

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7
Q

Define the start point (+1)

A

The start point is the position on DNA corresponding to the first base incorporated into RNA

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8
Q

RNA is synthesised in what direction?

A

5’ to 3’

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9
Q

The rate of transcription is faster than the rate of DNA replication. TRUE or FALSE

A

FALSE. The rate of transcription is about 20 times slower than replication.

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10
Q

Define and describe nascent RNA

A

Nascent RNA is an RNA chain that is still being synthesised. Its 3’ end is still paired with DNA.

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11
Q

What are the three stages of transcription?

A

Initiation, elongation, and termination.

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12
Q

The formation of a closed complex is reversible. TRUE or FALSE

A

TRUE.

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13
Q

List the subunits of bacterial RNA Polymerase holoenzyme

A

2 alpha subunits, 1 beta, 1 beta prime, and 1 omega

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14
Q

Distinguish a holoenzyme from an apoenzyme.

A

An apoenzyme is the inactive form of an enzyme, lacking association with its cofactor/coenzyme. A holoenzyme, on the other hand, is a complete and catalytically active enzyme.

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15
Q

State the function of the alpha subunits of bacterial RNA Pol holoenyzme.

A

Facilitates the assembly of the holoenzyme and interacts with the promoter and some regulatory factors through its C-terminal domain (CTD).

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16
Q

State the function of the beta subunits of bacterial RNA Pol holoenyzme.

A

Accounts for catalysis.

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17
Q

State the function of the omega subunit of bacterial RNA Pol holoenyzme.

A

Facilitates the assembly of the holoenzyme.

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18
Q

How does the sigma factor affect RNA Polymerase?

A

The sigma factor reduces the affinity of RNA Pol for general DNA and increases its affinity for promoters.

Sigma factor assists RNA polymerase in melting DNA at transcription start site.

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19
Q

Briefly describe the formation of an open complex in transcription.

A

RNA Pol first binds to the promoter as a closed complex, where the DNA is still double stranded. RNA Polymerase then separates the DNA strands, resulting in an open complex (tight binding).

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20
Q

List the function(s) of the sigma factor.

A
  1. It increases the affinity of RNA Pol for promoters instead of general DNA.
  2. It aids in the melting/unwinding of DNA during the formation of the open complex.
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21
Q

Define the ternary complex.

A

The ternary complex is the complex in the initiation of transcription that consists of RNA polymerase, DNA, and nascent RNA.

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22
Q

The sigma factor binds to the promoter region at both the -35 and -10 box. TRUE or FALSE

A

TRUE. Sigma has important conserved regions that interacts with the promoter at both the -10 and -35 regions.

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23
Q

The sigma factor is only released when the general elongation complex is formed. TRUE or FALSE

A

FALSE. The sigma factor is actually released earlier: right before the formation of the initial elongation complex.

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24
Q

Describe the premature termination of transcription.

A

When the nascent RNA formed is too short (8-9bp), the nascent RNA is very unstable. RNA Polymerase then goes through abortive initiation by restarting transcription again.

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25
Q

State the length of nascent RNA that corresponds with the formation of the ternary complex.

A

10 bases, because it means that abortive synthesis has already been bypassed.

26
Q

State the length of nascent RNA that corresponds with the formation of the general elongation complex.

A

15-20 bases

27
Q

Describe the contraction of the transcription initiation complex.

A
  1. The initiation complex contacts at -55 to +20
  2. It contracts to -30 to +20 when the sigma factor dissociates.
  3. It further contracts to -20 to +20 in the formation of the general elongation complex.
28
Q

How long is the promoter clearance time?

A

1-2 seconds

29
Q

Define the promoter clearance time.

A

The promoter clearance time is defined as the duration the current polymerase takes to leave the promoter such that another promoter can be initiated.

30
Q

Which terminus (N- or C-) hides the DNA binding domain of the sigma factor?

A

The N-terminus

31
Q

What induces a conformational change in the sigma factor such that it binds to promoter DNA?

A

The sigma factor exposes its DNA binding regions when it associates with the core enzyme.

32
Q

The DNA binding domains of the sigma factor is located on the C-terminal. TRUE or FALSE?

A

TRUE.

33
Q

Define the consensus sequence

A

A consensus sequence is an idealised sequence in which each position represents the base most found when many actual sequences are compared.

34
Q

What is the consensus sequence of the hexamer at -10?

A

TATAAT

35
Q

What is the consensus sequence of hexamer at -35?

A

TTGACA

36
Q

What is the consensus sequence at which the sigma factor binds?

A

TTGACA (-35)

TATAAT (-10)

37
Q

What is the ideal spacing between the -10 and -35 regions?

A

17 bases

38
Q

How many sigma factors are there in E.coli?

A

Seven

39
Q

Name the sigma factor that is required for general transcription in E. coli

A

sigma^(70)

40
Q

Briefly describe the function of sigma^(32)

A

It is used for heat-shock regulated gene expression.

41
Q

Briefly describe the function of sigma^(S)

A

It is used for gene expression in the stationary phase

42
Q

Briefly describe the function of sigma^(F)

A

It is used for expression of flagellar operons

43
Q

Briefly describe the function of sigma^(54)

A

It is used for nitrogen regulated gene expression

44
Q

Define the terminator sequence.

A

A terminator is a DNA sequence that causes RNA polymerase to terminate transcription. It locates before the point at which the last base is added to the RNA.

45
Q

Define antitermination

A

Antitermination causes the RNA polymerase to continue transcription past the terminator sequence, an event called readthrough.

46
Q

State the two types of termination

A

Intrinsic termination and rho-dependent termination (example of extrinsic termination).

47
Q

Define intrinsic termination

A

An intrinsic termination occurs when the core enzyme can terminate at certain sites in the absence of any other factors.

48
Q

In termination, a hairpin loop is formed. Describe the hairpin loop.

A

The hairpin loop consists of palindromic GC-rich sites (which bind to each other) and a U-rich region.

49
Q

Describe the process of extrinsic termination, using the Rho-dependent terminator as an example.

A

The Rho factor is a terminator protein that binds to a rut (rho utilisation site) on a nascent RNA.
Rho translocates along the RNA while the RNA polymerase pauses at the hairpin loop.
When Rho catches up to the RNA polymerase, it causes the RNA to dissociate from DNA, releasing all components.

50
Q

mRNA is transcribed, translated AND degraded at the same time in prokaryotes. TRUE or FALSE

A

TRUE

51
Q

Bacterial mRNA is typically more stable than eukaryotic mRNA. TRUE or FALSE.

A

FALSE. Bacterial mRNA has a half-life of a few minutes, while eukaryotic mRNA can persist for several hours.

52
Q

Define monocistronic mRNA

A

“Monocistronic” describes mRNA that encodes only one protein.

53
Q

Define polycistronic

A

“Polycistronic” describes mRNA that encodes more than one protein.

54
Q

The intercistronic distance (distance between two coding regions) is always more than 1. TRUE or FALSE

A

FALSE. The intercistronic distance can actually be negative, in which the coding regions overlap.

55
Q

What is the aim of conducting DNA footprinting?

A

DNA footprinting identifies DNA binding sites for proteins.

56
Q

Briefly describe DNA footprinting.

A
  1. The protein and DNA is first allowed to interact.
  2. Partial attack of DNAse I is carried out
  3. The cleaved DNA is isolated, denatured, and run on gel electrophoresis. Electrophoresis is carried out with two samples: one with DNAse treated protein-DNA, the other with only DNAse treated DNA.
  4. Missing bands identify the binding site of the protein

** Note that the target DNA must be labelled only on ONE end.

57
Q

Summarise the concept behind DNA footprinting

A

Protein binding blocks phosphodiester bonds from attack (from nucleases or chemicals), revealing the protein’s precise recognition site as a protected zone.

58
Q

What is the aim of conducting EMSA?

A

EMSA is able to distinguish the sizes of DNA binding proteins. It is also able to trace the amounts of sequence specific DNA binding protein.

59
Q

Summarise the concept behind EMSA

A

The larger the bound protein, the greater the retardation of the DNA molecule.

60
Q

What is EMSA?

A

Electrophoresis Mobility Shift Assay

61
Q

Suggest how you would determine the sequence of a DNA region that binds to a specific protein X.

A
  1. Protein X should be incubated with a large pool of short dsDNA, each with a randomly generated nucleotide sequence.
  2. Protein-DNA complexes are then isolated (EMSA can be used here)
  3. The protein is removed from the DNA strand and the DNA fragment can be sequenced.

LSM2232 (9 decks)

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