Exam 2: Prokaryotic Transcription Flashcards

1
Q

dNTP

A

Deoxynuceloside triphosphate

precursor to DNA

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2
Q

NTP

A

Nuceloside triphosphate

precursor to RNA

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3
Q

Both transcription and replication use a __________ to determine which _________ to incorporate into a new polynucelotide chain

A

Both transcription and replication use a \_DNA template\_ to determine which \_nucleotide triphosphates (NTP/dNTP)\_ to incorporate into a new polynucelotide chain

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4
Q

Both transcription and replication use enzymes that move in the ________ direction to synthesise polynucleotides in the _________ direction

A

Both transcription and replication use enzymes that move in the _3’ to 5’_ direction to synthesise polynucleotides in the _5’ to 3’_ direction

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5
Q

What must be done to the DNA template so that it is suitable to be replicated or transcribed?

A

dsDNA must be unwound at the site of replication or transcription

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6
Q

How is transcription similar to replication?

A

Both use DNA template to incorporate nucleotide triphosphates into a polynucelotide chain.

Both process use enzymes that synthesize polynucletides in the the 5’ to 3’ direction

Both processes need to unwind DNA

Both processes begin only at specific sites in the DNA

Both processes require the step-wise assembly of a multi-component protein complex for initiation

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7
Q

How does the DNA template in transcription differ from that of replication?

A

Transcription is asymmetric: for any gene, only one stand is transcribed because only one complimentary strand will give you the mRNA product you want.

In transcription, the template strand can be the top strand or the bottom strand, unlike in replication where all DNA is replicated and both strands are templates for new daughter strands.

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8
Q

The mRNA product is identical to the __________ DNA strand (aside from T->U conversion and use of NTPs rather than dNTPs)

A

coding strand

or sense strand

or non-template strand

is the complimentary strand to the non-coding (or template, antisense, anticoding, or transcribed strand) strand from which the nascent transcript is synthesised, and so contains a sequence analagous to the RNA product

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9
Q

What is different about the number of copies of transcription vs replication?

A

Transcription makes varying numbers (upto 1000s) of copies, depending on the gene, whereas replication makes only one copy.

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10
Q

How many strands are copied during transcription? replication?

A

In transcription, only ONE strand is copied (the coding strand). In replication, two strands (the parallel and antiparallel strands of dsDNA.

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11
Q

de novo transcription in RNA polymerase means? Is this term applicable to DNA polymerase?

A

RNA polymerase can initiate WITHOUT A PRIMER, DNA polymerase cannot and requires RNA primers such as Okazaki fragments on the lagging stand.

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12
Q

How much material is transcribed during transcription? Replication?

A

In transcription, only genes, about 2% of the genome. In replication, the whole genome

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13
Q

What cell cycle does transcription take place in? replication?

A

Most transcription takes place in G0, G1 and G2 phases. Replication takes place in S phase only.

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14
Q

Where does transcription begin? replication?

A

Transcription begins at the promoter sequences. Replication begins at the origin of replication.

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15
Q

How does process of transcription differ from replication?

A

Only a small fraction of the genome is transcribed

RNA polymerase can initiate de novo (WITHOUT A PRIMER).

Only a small part of the dsDNA is unwound at any given time

As RNA is synthesised, the new strand DOES NOT remain hydrogen-bonded to the template

Genes are transcribed at different levels depending on expression, in replication all DNA is copied as a unit in the same quantity

Most transcription (aside from histone production occuring primarily in S phase) takes place during G0 (post-miotic) , G1, or G2.

The product (RNA) of transcription is heavily modified.

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16
Q

What is the primary macromolecule and precursors used in translation? replication?

A

RNA polymerase and NTPs (nucleoside triphosphates, sometimes written ribonucleoside triphosphates, rNTPs, however that is redundant) are used in translation. DNA polymerase and dNTPs (deoxynucelotide triphosphates) are used in replication.

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17
Q

Primase and NTPs are needed for:

replication

translation

both

neither

A

replication. translation uses RNA polymerase which DOES NOT REQUIRE A PRIMER, so it does not need primase.

Both require NTPs, for replication to make primers, for translation to make the RNA product.

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18
Q

What proofreading activity does RNA polymerase engage in?

A

FOR THE PURPOSES OF THIS CLASS, RNA POLYMERASE HAS NO PROOF-READING ACTIVITY.

However…..:

Proofreading also occurs in mRNA translation for protein synthesis. Proofreading begins with fraying of the misincorporated nucleotide away from the DNA template, which pauses transcription. Subsequent backtracking of RNAP by one position enables nucleolytic cleavage of an RNA dinucleotide that contains the misincorporated nucleotide. Since cleavage occurs at the same active site that is used for polymerization, the RNAP proofreading mechanism differs from that used by DNAPs, which contain a distinct nuclease specific active site. This topic is less well understood compared to DNA proofreading.

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19
Q

What proofreading activity does DNA polymerase engage in?

A

HAS PROOF-READING

In bacteria, all three DNA polymerases (I, II and III) have the ability to proofread. In eukaryotes only the polymerases that deal with the elongation (delta, and epsilon) have proofreading ability.

Both use 3’ -> 5’ exonuclease activity. When an incorrect base pair is recognized, DNA polymerase reverses its direction by one base pair of DNA and excises the mismatched base. Following base excision, the polymerase can re-insert the correct base and replication can continue.

The extent of proofreading in DNA replication determines the mutation rate, and is different in different species. For example, loss of proof-reading due to mutations in the DNA polymerase epsilon gene results in a hyper-mutated genotype with >100 mutations per Mbase of DNA in human colorectal cancers.

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20
Q

Compare the proofreading activity of DNA polymerase and RNA polymerase.

A

DNA polymerase has high proof-reading activity while RNA polymerase has none (kind of).

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21
Q

Describe the difference between post processing in replication and transcription.

A

DNA is not processed after replication. In transcription (in eukaryotes), pre-mRNA with exons (coding regions) and introns (non coding regions) are made that go through splicing where introns are removed, 5’ cap and poly(A) tail are added. The mature mRNA complexes with the ribosome and codons are matched to anticodons on tRNA to add a particular amino acid to the nascent polypeptide.

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22
Q

New NTPs are added to the nascent transcript on what end?

A

Nucleotides are added to the 3’-OH group of the nascent oligonucleotide

The NTP is incorporated as a monophosphate, retaining the α-phosphate group

The newly added nucleotide forms a phosphodiester bond. This bond is therefore described as a 3’ -> 5’ phosphodiester bond to indicate the directionality of the growing oligonucleotide

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23
Q

On the coding strand, is the promoter region (ex -35 sequence) closer to the 5’ or 3’ end than the transcriptional start site?

A

5’ end. In transcription, mRNAs are added to the 3’ end, just like replication. The coding strand, also known as the sense strand, IS NOT read from during transcription, and as it came from dsDNA it is antiparallel. It gets its name from being of the same code (less T-U conversion) as the resulting mRNA product.

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24
Q

On the anti-sense strand, is the promoter region (ex -35 sequence) closer to the 5’ or 3’ end than the transcriptional start site?

A

3’ end. Transcription, like replication, adds new NTPs to the growing mRNA (dNTPs in replication) in the 5’ -> 3’ direction. That means the anti-sense (or template) strand is read from the 3’ to 5’ direction, so regions before the start site are towards the 3’ end.

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25
Q

Regulatory sequence

A

Site for the binding of regulatory proteins; the role of regulatory proteins is to influence the rate of transcription. Regulatory sequences can be found in a variety of locations.

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26
Q

Promoter

A

Site for RNA polymerase binding to dsDNA; signals the beginning of transcription

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27
Q

Terminator

A

Signals the end of DNA transcription

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28
Q

Ribosome binding site

A

Site for ribosome binding; translation begins near this site in the mRNA.

In eukaryotes, the ribosome scans the mRNA for a start codon.

In prokaryotes the 5’ untranslated region (5’ UTR) proceeding the start codon includes the ribosome binding site.

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29
Q

Start codon

A

Specifies the first amino acid in a polypeptide sequence. The start codon is the first codon of a messenger RNA (mRNA) transcript translated by a ribosome. The start codon always codes for methionine (Met) in eukaryotes and a modified formylmethionine (fMet) in prokaryotes. The most common start codon is AUG.

The start codon is often preceded by a 5’ untranslated region (5’ UTR). In prokaryotes this includes the ribosome binding site.

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30
Q

Codon

A

3-nucleotide sequences within the mRNA that specify particular amino acids. The sequence of codons within mRNA determines the sequence of amino acids within a polypeptide.

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31
Q

Stop codon

A

Specifies the end of polypeptide synthesis

in RNA:

UAG (“amber”)

UAA (“ochre”)

UGA (“opal”)

in DNA:

TAG (“amber”)

TAA (“ochre”)

TGA (“opal” or “umber”)

(remember, these are only different because of the T/U change in DNA/RNA. Thymine is just “5-methyluracil”!)

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32
Q

Polycistronic

A

A bacterial mRNA may be polycistronic, which means it encodes two or more polypeptides

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33
Q

Promoter Region

A

In bacteria, the promoter contains two short sequence elements approximately -10 and -35 nucleotides upstream from the transcription start site.

The sequence at -10 (the -10 element) has the consensus sequence TATAAT.

The sequence at -35 (the -35 element) has the consensus sequence TTGACA.

The above consensus sequences, while conserved on average, are not found intact in most promoters. On average, only 3 to 4 of the 6 base pairs in each consensus sequence are found in any given promoter. Few natural promoters have been identified to date that possess intact consensus sequences at both the -10 and -35; artificial promoters with complete conservation of the -10 and -35 elements have been found to transcribe at lower frequencies than those with a few mismatches with the consensus.

The optimal spacing between the -35 and -10 sequences is 17 bp.

Some promoters contain one or more upstream promoter element (UP element) subsite (ex the -42 region or -52).

It should be noted that the above promoter sequences are recognized only by RNA polymerase holoenzyme containing sigma-70. RNA polymerase holoenzymes containing other sigma factors recognize different core promoter sequences.

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34
Q

Conservation of promoter region consensus sequences

A

The consensus sequences TATAAT (-10 element or Pribnow box) and TTGACA (-35 element), while conserved on average, are not found intact in most promoters. On average, only 3 to 4 of the 6 base pairs in each consensus sequence are found in any given promoter. Few natural promoters have been identified to date that possess intact consensus sequences at both the -10 and -35; artificial promoters with complete conservation of the -10 and -35 elements have been found to transcribe at lower frequencies than those with a few mismatches with the consensus.

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35
Q

What protein motif is most common in DNA binding domains?

A

α-helix-turn-α-helix (or helix-turn-helix for short)

α-helix fit into the major groove of DNA to recognise sequences, the side chains extend out of the helix to contact the edges of base pairs

36
Q

Explain the recognition region of the helix-turn-helix motif

A

One helix (the “recognition helix”, 7-9 amino acid residues in length) sits in the major groove and makes specific interactions with certain base pair sequences. It is held there by the non-specific interactions of the second helix with the backbone.

37
Q

What is the major DNA binding protein motif? What are two other motifs of DNA binding proteins?

A

The major motif is the helix-turn-helix motif. Two other motifs are the zinc finger and leucine zipper motifs:

38
Q

Holoenzyme

A

An enzyme together with the cofactor(s) required for activity is called a holoenzyme. The term holoenzyme can also be applied to enzymes that contain multiple protein subunits, such as the DNA/RNA polymerases; here the holoenzyme is the complete complex containing all the subunits needed for activity.

39
Q

RNA polymerase holoenzyme slides along _______ until its \________ recognises a \________, which makes RNA polymerase holoenzyme forms a \_____ ____\__. An \_____ _____\__ is formed and transcription begins. \__________ is released as elongation commences.

A

RNA polymerase holoenzyme slides along _dsDNA_ until its \_σ-factor (sigma-factor)_ recognises a _promoter_, which makes RNA polymerase holoenzyme forms a _closed complex_. An _open complex_ is formed and transcription begins. \_σ-factor (sigma-factor)_ is released as elongation commences.

40
Q

Which end of mRNA exits RNA polymerase first?

A

5’ end, as new NTPs are added to the 3’ end of the nascent transcription

41
Q

Which strand of dsDNA does RNA polymerase read from, the parallel or antiparallel strand?

A

Either one, not both, reading in a 5’ -> 3’ direction. Either strand of dsDNA can be the template strand for a gene.

42
Q

Stem loop

A

Stem-loop intramolecular base pairing is a pattern that can occur in single-stranded DNA or, more commonly, in RNA. The structure is also known as a hairpin or hairpin loop. It occurs when two regions of the same strand, usually complementary in nucleotide sequence when read in opposite directions, base-pair to form a double helix that ends in an unpaired loop. The resulting structure is a key building block of many RNA secondary structures

43
Q

ρ (rho)-dependent termination

A

ρ (rho)-protein, a helicase, recognises the rut site on a mRNA and binds to it. It makes its way towards RNA polymerase travelling in the 5’ -> 3’ direction of the nascent transcript. Upstream, RNA polymerase translates a sequence that forms a stem loop in the mRNA, causing polymerase to pause. This pause enables ρ-protein to “catch up” to the site of transcription, separating the DNA-RNA hybrid which releases the mRNA.

44
Q

ρ (rho)-independent termination

A

RNA polymerase translates a sequence that forms a stem loop in the mRNA followed by a chain of URNA-ADNA base pairs. The bonds between uracil and adenine are very weak. A protein bound to RNA polymerase (nusA) binds to the stem-loop structure tightly enough to cause the polymerase to temporarily stall. This pausing of the polymerase coincides with transcription of the poly-uracil sequence. The weak Adenine-Uracil bonds lower the energy of destabilization for the RNA-DNA duplex, allowing it to unwind and dissociate from the RNA polymerase. This is also known as intrinsic termination.

45
Q

Gene regulation

A

Regulation of gene expression includes a wide range of mechanisms that are used by cells to increase or decrease the production of specific gene products (protein or RNA), and is informally termed gene regulation. Sophisticated programs of gene expression are widely observed in biology, for example to trigger developmental pathways, respond to environmental stimuli, or adapt to new food sources. Virtually any step of gene expression can be modulated, from transcriptional initiation, to RNA processing, and to the post-translational modification of a protein.

Gene regulation is essential for viruses, prokaryotes and eukaryotes as it increases the versatility and adaptability of an organism by allowing the cell to express protein when needed.

46
Q

Why regulate genes?

A

Gene regulation is essential for viruses, prokaryotes and eukaryotes as it increases the versatility and adaptability of an organism by allowing the cell to express protein when needed.

The benefit of regulating genes is that encoded proteins will be produced only when required to avoid wasting valuable energy and resources.

47
Q

Constitutive gene

A

Constitutive genes are unregulated by the organism, wth essentially constant levels of expression.

Usually necessary for the survival of the organism.

48
Q

What is the most common way to regulate gene expression?

A

Transcriptional regulation, the rate of RNA synthesis can be increased or decreased using repressors and activators.

49
Q

Repressor protein

A

Binds to dsDNA and inhibits transcription.

Negative control.

50
Q

Activator protein

A

Binds to dsDNA and increases transcription.

Positive control.

51
Q

Negative control

A

Repressor proteins bind to dsDNA and inhibit transcription

52
Q

Positive control

A

Activator proteins bind to dsDNA and increase transcription

53
Q

Effector molecules

A

Small effector molecules, such as inducers, corepressors, and inhibitors bind to regulatory proteins (not directly to DNA) and affect transcription regulation in inducible/repressible genes

54
Q

How to inducers affect transcription?

A

Inducers increase transcription by inhibiting repressors and/or activating activators in induceible genes:

55
Q

How do corepressors affect transcription?

A

Corepressors activate repressors and reduce transcription in repressible genes

56
Q

How do inhibitors affect transcription?

A

Inhibitors inhibit activators and reduce transcription in repressible genes

57
Q

Operon

A

An operon is a regulatory unit consisting of a few structural genes under the control of one promoter.

It encodes polycistronic mRNA that contains the coding sequence for two or more structural genes.

This allows a bacterium to coordinately regulate a group of genes that encode proteins with a common function (ex lac operon)

An operon contains several different regions: promoter; terminator; structural genes; operator

58
Q

What regions does the operon contain?

A

An operon contains several different regions: promoter; terminator; structural genes; operator

59
Q

What are the structural genes of the lac operon?

A

Structural genes:

lacZ -> Encodes β-galactosidase

lacY -> Encodes lactose permease

lacA -> Encodes transacetylase

60
Q

lacZ

A

lacZ encodes β-galactosidase (LacZ), an intracellular enzyme that cleaves the disaccharide lactose into glucose and galactose

61
Q

lacY

A

lacY encodes lactose permease (LacY), a transmembrane symporter that pumps β-galactosides (including lactose, shown) into the cell using a proton (H+) gradient in the same direction

62
Q

lacA

A

lacA encodes galactoside O-acetyltransferase (LacA), an enzyme that transfers an acetyl group from acetyl-CoA to β-galactosides. Functional significance remains unclear

63
Q

What are the DNA elements of the lac operon?

A

DNA elements

■ Promoter: Binds RNA polymerase

■ Operator: Binds the lac repressor protein

■ CAP site: Binds the Catabolite Activator Protein (CAP)

■ Terminator: Signals end of transcription

64
Q

Describe the roles of the structural genes of the lac operon as a system

A

lacY encodes lactose permiase, a structrural protein that allows lactose and similar moecules to diffuse through the membrane at a faster rate, which are then broken down by β-galactosidase, the structural protein of lacZ for further catabolism. A byproduct of β-galactosidase is allolactase, a small effector molecule involved in pathway regulation.

65
Q

What negatively regulates lac?

A

lac repressor, encoded by lacI, a constitutively expressed regulatory gene.

66
Q

Where is the lacI gene?

A

A constitutive (always-expressed) regulatory gene ahead of the lac operon (not considered to be part of lac) with its own promotor (the i promoter) that negatively regulates lac. It does this by encoding the lac repressor in small quantities (only about 10 proteins per cell!)

67
Q

i promoter

A

Promoter region of the lacI regulatory gene

68
Q

What is the state of the lac repressor in the absence of lactose in the environment?

A

Bound to lacO (lac operator) inhibiting transcripton

69
Q

What is the state of the lac repressor with the presence of lactose in the environment?

A

Some lactose is converted to allolactose (an effector molecule for the lac operon) which binds to lac repressor and inhibits its function. In this way, allolactose is an inducer to lac.

70
Q

What positively regulates lac?

A

CAP (Catabolite Activator Protein)

binds to the CAP site at the beginning of the lac operon, facilitated by the cAMP small effector molecule

71
Q

cAMP

A

E. coli deprived of glucose produce cAMP using adenynyl cyclase, which serves as an internal signal to activate expression of genes for importing and metabolizing other sugars. cAMP exerts this effect by binding the transcription factor CRP (cAMP regulatory protein) , also known as CAP (catabolite activator protein).

72
Q

CAP

A

cAMP receptor protein (CRP; also known as catabolite activator protein, CAP) is a regulatory protein in bacteria. CAP protein binds cAMP, which causes a conformational change that allows CAP to bind tightly to a specific DNA site in the promoters of the genes it controls. In Escherichia coli, CAP can regulate the transcription of more than 100 genes, including lac.

CAP binds to a DNA site located upstream of core promoter elements and activates transcription through protein-protein interactions between “activating region 1” of CRP and the C-terminal domain of RNA polymerase alpha subunit.

73
Q

How does CAP effect lac transcription?

A

Transcription is inhibited by lack of CAP binding. CAP binding enhances promoter binding by polymerase.

74
Q

Under what cellular conditions is does CAP bind to the CAP site?

A

Lack of D-glucose. D-glucose inhibits the enzyme adenylyl cyclase, which converts ATP to cAMP. cAMP promotes CAP binding, which promotes lac transcription.

When the cell doesn’t need to use lactose (because it has glucose), lac is inhibited to “save energy” that would othewise be wasted building lactose-catabolising machinery that isn’t needed in the presence of glucose.

75
Q

Define the state of the lac operon in the presence of:

no lactose, no glucose

A
76
Q

Define the state of the lac operon in the presence of:

lactose and glucose

A
77
Q

Define the state of the lac operon in the presence of:

lactose, no glucose

A
78
Q

Define the state of the lac operon in the presence of:

glucose, no lactose

A
79
Q

In terms of regulation, when is the lac operon expressed?

A

lac operon expressed only when repressor is off and the activator is on.

lactose present, but no glucose

80
Q

What does the cell do when exposed to both glucose and lactose?

A

E. coli uses glucose first.
Only when glucose is depleted does E. coli use lactose (creating a lactose, no glucose scenario)

81
Q

Diauxic growth

A

Diauxic growth is the sequential use of two sugars by bacteria

82
Q

How many lacO sites are there in E. coli?

A

E. coli’s lac operon has three operator sites:

O3, O1, O2

83
Q

How does the lac repressor bind to lacO?

A

lac repressor must bind to O1 plus either O2 or O3 to cause repression

A loop in the DNA brings operator sites closer together

84
Q

How many sites does lac repressor bind to?

A

Two, each dimer of the lac repressor tetramer binds to one of the lac**O sites. lac repressor must bind to O1 plus either O2 or O3 to cause repression:

85
Q

Describe the difference potential knock-out mutations within the lac**O site and its effect on lac transcription

A

_*lac* repressor must bind to O1 plus either O2 or O3 to cause repression_

If no knockouts, supression is strong (1300x supression).

If either O2 or O3 is knocked out leaving O1 intact, then O1 in combination with O2 or O3 still has strong suppression (440-700x supression).

If both O2 and O3 are knocked out leaving O1 intact, O1 is not very strong alone but will still supress (18x supression).

If O1 is knocked out in any combination, supression is weak (1x supression=no supression)